Gxjxjx

1. ⃗
r′(t) = (x′(t), y′(t)), t ∈ [a, b]
⃗
r′(t) = (x′(t), y′(t), z′(t)), t ∈ [a, b]
Speed in the time t (a scalar; shows how fast the movement is)
v = ∥ ⃗
r′(t)∥ = (x′(t))2
+ (y′(t))2
, t ∈ [a, b]
v = ∥ ⃗
r′(t)∥ = (x′(t))2
+ (y′(t))2
+ (z′(t))2
t ∈ [a, b]
Velocity in the time t (a vector tangent to the curve)

2. ⃗
r : [a, b] → ℝ3
⃗
r(a)
⃗
r(b)
⃗
r(t) = (x(t), y(t), z(t)), s . t . x, y, z : [a, b] → ℝ
⃗
r
a = t0 b = tn
l1
l ≈ l1
n = 1
Approximation by pieces of straight lines
SMOOTH

3. ⃗
r : [a, b] → ℝ3
⃗
r(a)
⃗
r(b)
⃗
r(t) = (x(t), y(t), z(t)), s . t . x, y, z : [a, b] → ℝ
⃗
r
a = t0 b = tn
l1
n = 2
t1
l2
l ≈ l1 + l2
⃗
r(t1)
SMOOTH

4. ⃗
r : [a, b] → ℝ3
⃗
r(a)
⃗
r(b)
⃗
r(t) = (x(t), y(t), z(t)), s . t . x, y, z : [a, b] → ℝ
⃗
r
a = t0 b = tn
l1
l ≈ l1 + l2 + l3
n = 3
t1 t2
l2 l3
⃗
r(t1)
⃗
r(t2)
SMOOTH

5. ⃗
r : [a, b] → ℝ3
⃗
r(a)
⃗
r(b)
⃗
r(t) = (x(t), y(t), z(t)), s . t . x, y, z : [a, b] → ℝ
⃗
r
a = t0 b = tn
l1
l ≈ l1 + l2 + l3 + l4
n = 4
⃗
r(t1)
⃗
r(t2)
⃗
r(t3)
l2
l3
t1 t2 t3
l4
SMOOTH

6. ⃗
r : [a, b] → ℝ3
⃗
r(a)
⃗
r(b)
⃗
r(t) = (x(t), y(t), z(t)), s . t . x, y, z : [a, b] → ℝ
⃗
r
a = t0 b = tn l1
l ≈ l1 + l2 + l3 + l4 + l5 + l6
n = 6
⃗
r(t3)
t1 t2 t3 t4 t5
⃗
r(t1)
⃗
r(t2)
⃗
r(t4) ⃗
r(t5)
l2 l3
l4 l5
l6
SMOOTH

7. ⃗
r : [a, b] → ℝ3
⃗
r(a)
⃗
r(b)
⃗
r(t) = (x(t), y(t), z(t)), s . t . x, y, z : [a, b] → ℝ
⃗
r
a = t0 b = tn
l ≈
n
∑
i=0
li
l1
n → ∞
t1 t2
…
l2
… ln
SMOOTH

8. ⃗
r : [a, b] → ℝ3
⃗
r(a)
⃗
r(b)
⃗
r(t) = (x(t), y(t), z(t)), s . t . x, y, z : [a, b] → ℝ
⃗
r
a = t0 b = tn
l1
n → ∞
t1 t2
…
l2
… ln
SMOOTH
l ≈
n
∑
i=0
li =
n
∑
i=1
| ⃗
r(ti) − ⃗
r(ti−1)|
the origin
⃗
r(ti−1) ⃗
r(ti)
position vectors

9. ⃗
r : [a, b] → ℝ3
⃗
r(a)
⃗
r(b)
⃗
r(t) = (x(t), y(t), z(t)), s . t . x, y, z : [a, b] → ℝ
⃗
r
a = t0 b = tn
l ≈
n
∑
i=0
li =
n
∑
i=1
| ⃗
r(ti) − ⃗
r(ti−1)| =
n
∑
i=1
⃗
r(ti) − ⃗
r(ti−1)
Δti
Δti
l1
n → ∞
t1 t2
…
l2
… ln
SMOOTH
→
b
∫
a
| ⃗
r′(t)|dt
n → ∞

10. in R3:
generally:
(ti) − !
r(ti−1)
∆ti
"
"
"
" ∆ti
M FQMp2`;2`` KQi BMi2;`H2M
b
ˆ
a
d
dt
!
r(t)
dtX
(t)2 + y!(t)2 + z!(t)2,
2b p
d =
distance
generalised sum speed s
time
Corresponds to the well-known with constant speed s
d = st
d =
n
∑
i=1
siΔti
If the speed is constant on really small time intervals:
i?2 `+ H2M;i? Q7 i?2 +m`p2 +QKTmi2b b
l =
b
ˆ
a
#
x!(t)2 + y!(t)2 + z!(t)2dt.
RN
Distance Speed Time Formula

11. Line integrals
15.3 and 15.4

12. γ1
γ2
γ3
γ4
γ5
piece-wise smooth curve
⃗
r(t) = (x(t), y(t))
⃗
r(t) = (x(t), y(t), z(t))

13. In Sections 11 and 12
Line integrals / Curve integrals
of functions of vector fields
KbbM 7ƺ` F`QTT2M K QK 7mMFiBQM
K,b pQHvK QK f(x, y, z) = 1 B ?2H
Em`pBMi2;`H2` p 7mMFiBQM2`
2+FMb
´
γ
fdsc /2 #2` FM` Fm`pM
PK n = 2- HHib´ Fm`pM ` THM
b´ Fm`pM ` THM Q+? f ` 2M i
bmiQK `2M UK2/ i2+F2MV p /2M
x, y)X
` p p2FiQ`7 Hi
´
γ
F · d
r U+B`F
i
F : Rn
→ Rn
#2bF`Bp2` `#2i2i
arc-length element
vector differential
area, mass, arc length work
h?2Q`2K k9X G´i F p` 2ii FQMb2`piBpi p2FiQ`7 Hi
γ ∈ D K2HHM
a Q+?
b ; HH2`,
ˆ
γ
F · d
r = Φ(
b) −
LQi2` HBF?2i2M K2/ MHvb2Mb mpm/bibX SQi2MiB
7mMFiBQM2MX aib2M BKTHB+2`` ii Fm`pBMi2;`H2M K2H
Fm`pX 6ƺ` bHmiM Fm`pQ` γ FHHb Fm`pBMi2;`H2M p
F
Q+? #2i2+FMb
˛
γ
F · d
r.
Circulations:
only (piece wise) smooth curves, and continuous functions

14. C : [a, b] ! R2
, ~
r(t) = (x(t), y(t)), f : R2
! R
tk)) · lk ⇡ f(~
r(tk)) ·
p
(x(tk) x(tk 1))2 + (y(tk) y(tk 1))2
= f(~
r(tk)) ·
s✓
x(tk) x(tk 1)
tk tk 1
◆2
+
✓
y(tk) y(tk 1)
tk tk 1
◆2
· (tk tk 1).
e total area is a Riemann sum
n
X
k=1
f(~
r(tk)) ·
s✓
xk
tk
◆2
+
✓
yk
tk
◆2
· tk
ch converges to the integral when we let n tend to infinity
Z
C
f ds =
b
Z
a
f(~
r(t)) · |~
r 0
(t)|dt =
b
Z
a
f(~
r(t)) ·
p
(x0(t))2 + (y0(t))2dt
di↵erential ds = |~
r 0
(t)|dt is calles arc length element.
k=1
k k
which converges to the integral when we let n tend to infinity
Z
C
f ds =
b
Z
a
f(~
r(t)) · |~
r 0
(t)|dt =
b
Z
a
f(~
r(t)) ·
p
(x0(t))2 + (y0(t))2dt
The di↵erential ds = |~
r 0
(t)|dt is calles arc length element.
Example. Consider the vector field
~
F(x, y) = (2x + ey
+ cos(x + y2
), xey
+ 2y cos(x + y2
) + 1).
Show that the field is conservative by determining its potential.
1
C : [a, b] ! R3
, ~
r(t) = (x(t), y(t), z(t)), f : R3
! R
Z
C
f ds =
b
Z
a
f(~
r(t)) · |~
r 0
(t)|dt =
b
Z
a
f(~
r(t)) ·
p
(x0(t))2 + (y0(t))2 + (y0(t))2dt
notation
formula for computations
C : [a, b] ! R3
, ~
r(t) = (x(t), y(t), z(t)), f : R3
! R
Z
C
f ds =
b
Z
a
f(~
r(t)) · |~
r 0
(t)|dt =
b
Z
a
f(~
r(t)) ·
p
(x0(t))2 + (y0(t))2 + (z0(t))2dt
notation
formula for computations
n
X
k=1
f(~
r(tk)) ·
s✓
xk
tk
◆2
+
✓
yk
tk
◆2
· tk
which converges to the integral when we let n tend to infinity
Z
C
f ds =
b
Z
a
f(~
r(t)) · |~
r 0
(t)|dt =
b
Z
a
f(~
r(t)) ·
p
(x0(t))2 + (y0(t))2dt
The di↵erential ds = |~
r 0
(t)|dt is calles arc length element.
Example. Consider the vector field
~
F(x, y) = (2x + ey
+ cos(x + y2
), xey
+ 2y cos(x + y2
) + 1).
Show that the field is conservative by determining its potential.
Path independence: If ~
F is a smooth vector fie
nected domain D and its line integrals are pa
conservative.
Assume that the line integrals between each two po
Fix some point P0 = (x0, y0, z0) and define for eac
(x, y, z) =
Z
~
F · d~
r

15. Properties of line integrals of functions
Line integrals are independent of the parametrisation of the curve (Chain rule)
Line integrals are independent of the orientation of the curve
∫
γ
f ds =
∫
γ1
f ds +
∫
γ2
f ds +
∫
γ3
f ds +
∫
γ4
f ds +
∫
γ5
f ds
γ1
γ2
γ3
γ4
γ5
piece-wise smooth curve
They describe the mass of the curve if f describes the density in each point
They describe the length of the curve if f is constant equal to 1
l(γ) =
∫
γ
ds
mass(γ) =
∫
γ
ρ(x, y, z) ds
Additivity
Applications
and surface area of the vertical screen
for plane curves and functions of two variables

16. x
y
z
plane z = y
1
1
-1
(0, 1, 1)
⃗
r(t) = (cos t, sin t), t ∈ [0,π]
⃗
v (t) = (cos t, sin t, sin t), t ∈ [0,π]
Compute the line integral where is the half circle
∫γ
y ds γ
γ = {(x, y) ; x2
+ y2
= 1, y ⩾ 0 }
Compute the area of
the vertical screen
between the half circle
in the upper half-plane
and the plane z = y
Compute the total mass
of the half circle
in the upper half-plane
if the density in each
point is expressed
by .
(x, y)
ρ(x, y) = y

17. ProblemLm
fyds
f any x7y 1
y
1 Parameterize the curve
p
Xlt cost A
yet Sint often
2 Express all the elements in the
formula by t
f Pitt f xlthy.CH yCe
sintOxlftl
Sint
Y'Ct cost
pythagorean
IN'Itlkftsinth cost
L

18. 3 compute the int
yds Sint dt f cost
I
th f z
C answer
mmr
Let's take another paramehisation
I
of 8
cµ coset
yet sincat
Otte
Iii
0
Fx ty 2t t
2
yds
sinczttgdt fk.tk If

19. Determine the value of
where is the intersection between
the cylinder
(for some positive a) and the plane
starting at and ending at .
∫
γ
xy ds
γ
x2
+ y2
= a2
z = x
(0, a, 0) (a, 0, a)

20. Problem2m
fxyds
y Etykal
from Co a o to
2 L
a O a
um
Paramehisation of R
x t acoste range
yltkasinte CO a
079
F
zcepacost act't O t
asiata p
11
a O a
9
I
L The values of
acoff a line int of
asint O p functions do
U not depend on
orientation of the curve
so I Can choose to
integrate from 0 to
E
The arc
length element
x e asint y'Ctkacost Ect asint

21. I ri t l f It tasinT
pythagorean id
fa211 sr.net
aJ1tsinT
The integrand
f x
y Z e x
y
f Fct x t y Ct
a'cost sint
mum
x
yds a'cost sint a EFI dt
meat
at tri y
answer aI 2r
hT

22. The primitive f to www.T
tnnt1thnt
change variables
t.SE e
Eat.dt f
t nit
H
1zfrudu z.uzI C
ru t C
1zfEt3

23. Curve C is the intersection between surfaces and .
Determine the total mass of the curve if the density in the point is
expressed by .
x2
+ z2
= 1 y = x2
(x, y, z)
ρ(x, y, z) = 1 + 4x2
z2
x2
+ z2
= 1
y = x2
z
y
x
y
C : [a, b] ! R3
, ~
r(t) = (x(t), y(t), z(t)), f : R3
! R
Z
C
f ds =
b
Z
a
f(~
r(t)) · |~
r 0
(t)|dt =
b
Z
a
f(~
r(t)) ·
p
(x0(t))2 + (y0(t))2 + (z0(t))2dt
C : [a, b] ! R2
, ~
r(t) = (x(t), y(t)), f : R2
! R
f(~
r(tk)) · lk ⇡ f(~
r(tk)) ·
p
(x(tk) x(tk 1))2 + (y(tk) y(tk 1))2
= f(~
r(tk)) ·
s✓
x(tk) x(tk 1)
tk tk 1
◆2
+
✓
y(tk) y(tk 1)
tk tk 1
◆2
· (tk tk 1).
The total area is a Riemann sum
n
X
k=1
f(~
r(tk)) ·
s✓
xk
tk
◆2
+
✓
yk
tk
◆2
· tk
which converges to the integral when we let n tend to infinity

24. Problem3m play z ds
planet r
y E
The parametrization of V
fG
Often 21T
The arc
length element
x t Sint
y Ctf 2 cost sent
2 Kt cost
IT'CtH f t4cos4simtt
tr
the integrand
Pythagorean id
qcxlthy.lt zCt1 fTt4xltD4zCtD

25. f1t4ws2tsint
mm
play Hds TEE't flthoftsidt
Fut taunt
dt
q
sin2d
2sindcost
S It in2tY1dt
YEi.i'i
m
me
sin2
1
qs2L
F
t
t
at

26. EE at

27. z = x2
z = 2 − x2
− 2y2
x y
Curve C is the part of the intersection between surfaces
and which is situated in the first octant ( ). Determine the
total mass of the curve if the density in the point is expressed by
.
z = 2 − x2
− 2y2
z = x2
x, y, z ⩾ 0
(x, y, z)
ρ(x, y, z) = xy
(0,1,0)
(1,0,1)
C : [a, b] ! R3
, ~
r(t) = (x(t), y(t), z(t)), f : R3
! R
Z
C
f ds =
b
Z
a
f(~
r(t)) · |~
r 0
(t)|dt =
b
Z
a
f(~
r(t)) ·
p
(x0(t))2 + (y0(t))2 + (z0(t))2dt
C : [a, b] ! R2
, ~
r(t) = (x(t), y(t)), f : R2
! R
f(~
r(tk)) · lk ⇡ f(~
r(tk)) ·
p
(x(tk) x(tk 1))2 + (y(tk) y(tk 1))2
= f(~
r(tk)) ·
s✓
x(tk) x(tk 1)
tk tk 1
◆2
+
✓
y(tk) y(tk 1)
tk tk 1
◆2
· (tk tk 1).
The total area is a Riemann sum
n
X
k=1
f(~
r(tk)) ·
s✓
xk
tk
◆2
+
✓
yk
tk
◆2
· tk

28. probl.cm
party 2 Ids
C
Z 2 2y2
o
PGH xy
k
i I
acts 1
me
are
length element ft za
t'T't
x'Ct I f 2T
ylCtI I 7
z t It
It'aH r2 Ea
T
ffi

29. Fitt
y t
p
x et y t 2 t x t
y Ct
t FA
me
mass c play Hds
tray
dt
t f2 G 4t44 dt
G 2 tf I 4 t't 4th
trz uyadt ffffu.IS
e o
se I
t I s I

30. u
raids 4k
the.EE
EiztEI tHtEI
answer
m E
J TE ds is computed by
invers sub it
I
J fr cos u.fr cos u du
2 fueseu du 2J
cos2 u th
l f
du

31. cos2t cosk sink
mn
2cos'd
1
www.zz
du 111dm
sinu_

32. In Sections 11 and 12
Line integrals / Curve integrals
of functions of vector fields
KbbM 7ƺ` F`QTT2M K QK 7mMFiBQM
K,b pQHvK QK f(x, y, z) = 1 B ?2H
Em`pBMi2;`H2` p 7mMFiBQM2`
2+FMb
´
γ
fdsc /2 #2` FM` Fm`pM
PK n = 2- HHib´ Fm`pM ` THM
b´ Fm`pM ` THM Q+? f ` 2M i
bmiQK `2M UK2/ i2+F2MV p /2M
x, y)X
` p p2FiQ`7 Hi
´
γ
F · d
r U+B`F
i
F : Rn
→ Rn
#2bF`Bp2` `#2i2i
arc-length element
vector differential
area, mass, arc length work
h?2Q`2K k9X G´i F p` 2ii FQMb2`piBpi p2FiQ`7 Hi
γ ∈ D K2HHM
a Q+?
b ; HH2`,
ˆ
γ
F · d
r = Φ(
b) −
LQi2` HBF?2i2M K2/ MHvb2Mb mpm/bibX SQi2MiB
7mMFiBQM2MX aib2M BKTHB+2`` ii Fm`pBMi2;`H2M K2H
Fm`pX 6ƺ` bHmiM Fm`pQ` γ FHHb Fm`pBMi2;`H2M p
F
Q+? #2i2+FMb
˛
γ
F · d
r.
Circulations:
only (piece wise) smooth curves, smooth vector fields, and continuous functions

33. What is the work done
by the forces of the vector field
by moving a particle
along the curve?

34. ⃗
r(t) = (x(t), y(t))
⃗
r′(t) = (x′(t), y′(t))
⃗
F(x, y) = (P(x, y), Q(x, y))
Hi
´
γ
F · d
r U+B`FmHiBQM2`
¸
γ
F · d
r QK Fm`p
#2bF`Bp2` `#2i2i bQK 7 Hi2ib F`7i2` mi` ii` 7ƺ
γX
˜
Y
f(x, y, z)dS 7ƺ` 2M #2;` Mb/ i`2p`B#2H7mM
KTFi vi B `mKK2iX a´/M BMi2;`H2` FM #2b
p`D2 TmMFi p Y - 2HH2` viMb Kbb QK f #2bF
What is the work done by the forces of the vector field by moving a particle along the curve?

35. ⃗
r(t) = (x(t), y(t))
⃗
r′(t) = (x′(t), y′(t))
⃗
F( ⃗
r(t))
α
W = Fd
dx + (x2
+ y2
) dy
len av di↵erentialformen xy dx+(x2
+y2
)dy
magnitude of the force
distance
(negative if in the opposite direction)
W = Fd
F = |~
F| cos ↵ = |~
F| ·
~
F · ~
r 0
(t)
|~
F| · |~
r 0(t)|
= ~
F ·
~
r 0
(t)
|~
r 0(t)|
d = ds = |~
r 0
(t)| dt
s =
b
Z
a
~
F(~
r(t))·
~
r 0
(t)
|~
r 0(t)|
·|~
r 0
(t)| dt =
b
Z
a
~
F(~
r(t))·~
r 0
(t) dt =
Z
C
~
F · d~
r
Z
W = Fd
s ↵ = |~
F| ·
~
F · ~
r 0
(t)
|~
F| · |~
r 0(t)|
= ~
F ·
~
r 0
(t)
|~
r 0(t)|
d = ds = |~
r 0
(t)| dt
·
~
r 0
(t)
|~
r 0(t)|
·|~
r 0
(t)| dt =
b
Z
a
~
F(~
r(t))·~
r 0
(t) dt =
Z
C
~
F · d~
r
Z
xy dx + (x2
+ y2
) dy
t0 = a tn = b
F = |~
F| cos ↵ = |~
F| ·
~
F · ~
r 0
(t)
|~
F| · |~
r 0(t)|
= ~
F ·
~
r 0
(t)
|~
r 0(t)|
d = ds = |~
r 0
(t)| dt
W =
Z
C
F ds =
b
Z
a
~
F(~
r(t))·
~
r 0
(t)
|~
r 0(t)|
·|~
r 0
(t)| dt =
b
Z
a
~
F(~
r(t))·~
r 0
(t) dt =
Z
C
~
F · d~
r
Z
formula for computations notation
⃗
r

36. Z
C
~
F · d~
r =
b
Z
a
~
F(~
r(t)) · ~
r 0
(t)dt
W = Fd
formula for computations
notation
vector differential
Line integral of vector field over the line C with parametrisation
⃗
F : ℝn
→ ℝn
⃗
r : [a, b] → ℝn
W = Fd
d~
r = ~
r 0
(t) dt
F = |~
F| cos ↵ = |~
F| ·
~
F · ~
r 0
(t)
|~
F| · |~
r 0(t)|
= ~
F ·
~
r 0
(t)
|~
r 0(t)|
d = ds = |~
r 0
(t)| dt
Computes the work done by the forces of the vector field by moving a particle along the curve.

37. Properties of line integrals of vector fields
Line integrals are independent of the parametrisation of the curve (Chain rule)
γ1
γ2
γ3
γ4
γ5
piece-wise smooth curve
Additivity
∫
γ
⃗
F ⋅ d ⃗
r =
∫
γ1
⃗
F ⋅ d ⃗
r +
∫
γ2
⃗
F ⋅ d ⃗
r +
∫
γ3
⃗
F ⋅ d ⃗
r +
∫
γ4
⃗
F ⋅ d ⃗
r +
∫
γ5
⃗
F ⋅ d ⃗
r
Line integrals are dependent of the orientation of the curve
∫
−γ
⃗
F ⋅ d ⃗
r = −
∫
γ
⃗
F ⋅ d ⃗
r
Orientation:
counterclockwise: positive
clockwise: negative

38. Five methods for computing line integrals of vector fields
From the definition, with help of parametrisation of the curve
Differential form (if we integrate over horizontal and/or vertical line pieces)
Fundamental Theorem for conservative v.f.
Green’s Theorem for 2D v.f. (Section 17)
Stokes’ Theorem for 3D v.f. (Section 19)
a
!
F(!
r(t)) · !
r !
(t)dt
M ` pB H´i2` H M;/2M p /2HBMi2`pHH2M B T`iBiBQM2M ;´ KQi MQHHX
6ƺ` 2ii BM}MBi2bBKHi iB/bBMi2`pHH dt- b´ #HB` !
r !(t)dt 2M BM}MBi2bBKH bi` +
pB bFmHH2 FmMM #2i2+FM d!
rX JM Mp M/2` / `7ƺ` #2i2+FMBM;2M
ˆ
γ
!
F · d!
r =
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt 7ƺ` 2M T`K
17i2`bQK
!
r(t) = (x(t), y(t), z(t)), d!
r = (dx, dy, dz)
!
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
b´ FM Fm`pBMi2;`H2M p 2ii p2FiQ`7 Hi Q+Fb´ bF`Bpb
ˆ
γ
!
F · d!
r
#$ %
.Bz2`2MiBH7Q`K
=
b
ˆ
a
(P(x(t), y(t), z(t))x!
(t) + Q(x(t), y(t), z(t))y!
(t) + R(x(
pBHF2i KQiBp2`` 2M MMM #2i2+FMBM; 7ƺ` Fm`pBMi2;`H2M,
ˆ
γ
!
F · d!
r =
ˆ
γ
Pdx + Qdy + Rdz
!
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
b´ FM Fm`pBMi2;`H2M p 2ii p2FiQ`7 Hi Q+Fb´ bF`Bpb
ˆ
γ
!
F · d!
r
#$ %
.Bz2`2MiBH7Q`K
=
b
ˆ
a
(P(x(t), y(t), z(t))x!
(t) + Q(x(t), y(t), z(t))y!
(t) + R(x(t
pBHF2i KQiBp2`` 2M MMM #2i2+FMBM; 7ƺ` Fm`pBMi2;`H2M,
ˆ
γ
!
F · d!
r =
ˆ
γ
Pdx + Qdy + Rdz
a2Mbi2 7Q`K2HM UB k.@p`BMiV Mp M/ K2/ 7ƺ`/2H iBHH mTT;B7i2` bQK /K
1t2KT2H NjX U/Kb R8X9 mTT;B7i 3V 1pHm2`
˛
γ
x2
y2
dx + x3
ydy
/ ` γ ` Fp/`i2M K2/ ?ƺ`M B (0, 0), (1, 0), (1, 1), (0, 1) Q`B2Mi2`/2 KQim
oB /2H` mTT Fm`pM γ B 7v` Fm`pQ`, γ = γ1 ∪γ2 ∪γ3 ∪γ4 bQK B #BH/2MX UhX2t
dy = 0 27i2`bQK Fm`pM ` ?Q`BbQMi2HH- pBHF2i #2iv/2` BM;2M 7ƺ` M/`BM; B y@H
UV 1D bHmi2M U#V aHmi2M- 2D 2MF2H U+V aHmi2M- 2MF2H
oB bF Mm pBb ii 7ƺ` FQMb2`piBp p2FiQ`7 Hi b´ ` Fm`pBMi2;`H2M K2HHM ip´ TmMFi2` Q#2`Q2M@
/2 p pBHF2M Fm`p K2HHM /2bb #´/ TmMFi2` KM p HD2` Q+? ii #2` FMBM; p Fm`pBMi2;`H2`
2MF2Hi FM ;ƺ`b K2/ ?D HT p p2FiQ`7 Hi2ib TQi2MiBHX
h?2Q`2K k9X G´i
F p` 2ii FQMb2`piBpi p2FiQ`7 Hi B D K2/ TQi2MiBH ΦX 6ƺ` p`D2 Fm`p
γ ∈ D K2HHM
a Q+?
b ; HH2`,
ˆ
γ
F · d
r = Φ(
b) − Φ(
a).
LQi2` HBF?2i2M K2/ MHvb2Mb mpm/bibX SQi2MiBH2M ` KQibp`B;?2i2M iBHH /2M T`BKBiBp
7mMFiBQM2MX aib2M BKTHB+2`` ii Fm`pBMi2;`H2M K2HHM ip´ TmMFi2` ` Q#2`Q2M/2 p pH p
Fm`pX 6ƺ` bHmiM Fm`pQ` γ FHHb Fm`pBMi2;`H2M p
F H M;b γ 7ƺ` +B`FmHiBQM2M p
F H M;b γ
Q+? #2i2+FMb
˛
γ
F · d
r.

39. Let and let the curve C be given by its parametrisation:
⃗
F(x, y) = (x, xy)
x(t) = t, y(t) = t2
, 0 ⩽ t ⩽ 1.
Compute the line integral of the vector field over the curve.
Z
C
~
F · d~
r =
b
Z
a
~
F(~
r(t)) · ~
r 0
(t)dt
W = Fd

40. Let and let the curve C be given by its parametrisation:
⃗
F(x, y) = (x, xy)
x(t) = t, y(t) = t2
, 0 ⩽ t ⩽ 1.
Compute the line integral of the vector field over the curve.

41. problemh
fcf.DE
Early e x
Fy
Cc
x
ty Ete
F di
ft
dt
0 def of Ft
ECTCt E Ht yet
t
i
Htt 2t4ldt t
f
t
4
fo

42. Let and let C be the quarter
of the unit circle from the point (1,0) to the point (0,1).
Compute the line integral of the vector field over the curve C.
⃗
F(x, y) = (xy, x2
+ y2
)
Z
C
~
F · d~
r =
b
Z
a
~
F(~
r(t)) · ~
r 0
(t)dt
W = Fd
0 0

43. Let and let C be the quarter
of the unit circle from the point (1,0) to the point (0,1).
Compute the line integral of the vector field over the curve C.
⃗
F(x, y) = (xy, x2
+ y2
)
C
iH `#2i2i #HB` _B2KMMbmKKM
n
!
k=1
!
F(!
r(tk)) ·
d!
r
dt
(tk)∆tk
QMp2`;2`` KQi BMi2;`H2M
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt

44. Problem2m Fahy
pg ftp4
C a
quarter of the unit circle
from 1,07 to 0,1
Parametrisation of C
E E E ate't
cthykth tsiytaf.co
gtfQ
E Fct E Ht y A
xCtIylt7gfxCtDtCyCtD2J
ux
Cwst
tz1FlCt1
E PCH cost
sent
t
It
cost

45. E d F sin'tcost cost dt
sight a t cost dt
sh du
f singe sent
Iz t I answer

46. Problem 3
Early 2k43g 2xty
F'Ct 2T teE
fr
di
IHrHD.rIeeatW
23t def of E
ECTCt F xlt yet
I def of
261143gCt 2xltlty.lt
21211431
3
2.2ft t
8tt3t 4t t
T
ftp.r
YHM ft tf
ft't Gt3 112T't3t5 16t4l8t43t5
mm

47. F did 6th 18 t't 3 t
5
dt
fo t9 tt I1f
l6zt9z 1z 32gt30 6g2
3

48. Z
C
~
F · d~
r =
b
Z
a
~
F(~
r(t)) · ~
r 0
(t)dt
W = Fd
F = |~
F| cos ↵ = |~
F| ·
~
F · ~
r 0
(t)
|~
F| · |~
r 0(t)|
= ~
F ·
~
r 0
(t)
|~
r 0(t)|
d = ds = |~
r 0
(t)| dt
I
C
~
F · d~
r
~
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
~
r(t) = (x(t), y(t), z(t)), d~
r = (dx, dy, dz)
dx = x0
(t)dt, dy = y0
(t)dt, dz = z0
(t)dt
Z
C
~
F·d~
r =
b
Z
a
(P(x(t), y(t), z(t))x0
(t)+Q(x(t), y(t), z(t))y0
(t)+R(x(t), y(t), z(t))z0
(t))dt
Z
~
F · d~
r =
Z
Pdx + Qdy + Rdz
C
~
F · d~
r
~
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
~
r(t) = (x(t), y(t), z(t)), d~
r = (dx, dy, dz)
dx = x0
(t)dt, dy = y0
(t)dt, dz = z0
(t)dt
Z
C
~
F·d~
r =
b
Z
a
(P(x(t), y(t), z(t))x0
(t)+Q(x(t), y(t), z(t))y0
(t)+R(x(t), y(t), z(t))z0
(t))dt
Z
C
~
F · d~
r =
Z
C
Pdx + Qdy + Rdz
~
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
~
r(t) = (x(t), y(t), z(t)), d~
r = (dx, dy, dz)
dx = x0
(t)dt, dy = y0
(t)dt, dz = z0
(t)dt
Z
C
~
F·d~
r =
b
Z
a
(P(x(t), y(t), z(t))x0
(t)+Q(x(t), y(t), z(t))y0
(t)+R(x(t), y(t), z(
Z
C
~
F · d~
r =
Z
C
Pdx + Qdy + Rdz
~
r(t) = (x(t), y(t), z(t)), d~
r = (dx, dy, dz)
dx = x0
(t)dt, dy = y0
(t)dt, dz = z0
(t)dt
Z
C
~
F·d~
r =
b
Z
a
(P(x(t), y(t), z(t))x0
(t)+Q(x(t), y(t), z(t))y0
(t)+R(x(t), y(t), z(t))z0
(t))dt
Z
C
~
F · d~
r =
Z
C
Pdx + Qdy + Rdz
~
r(t) = (x(t), y(t), z(t)), d~
r = (dx, dy, dz)
dx = x0
(t)dt, dy = y0
(t)dt, dz = z0
(t)dt
Z
C
~
F·d~
r =
b
Z
a
(P(x(t), y(t), z(t))x0
(t)+Q(x(t), y(t), z(t))y0
(t)+R(x(t), y(t), z(t))z0
(t))dt
Z
C
~
F · d~
r =
Z
C
Pdx + Qdy + Rdz

49. Z
C
~
F · d~
r =
b
Z
a
~
F(~
r(t)) · ~
r 0
(t)dt
W = Fd
F = |~
F| cos ↵ = |~
F| ·
~
F · ~
r 0
(t)
|~
F| · |~
r 0(t)|
= ~
F ·
~
r 0
(t)
|~
r 0(t)|
d = ds = |~
r 0
(t)| dt
I
C
~
F · d~
r
~
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
~
r(t) = (x(t), y(t), z(t)), d~
r = (dx, dy, dz)
dx = x0
(t)dt, dy = y0
(t)dt, dz = z0
(t)dt
Z
C
~
F·d~
r =
b
Z
a
(P(x(t), y(t), z(t))x0
(t)+Q(x(t), y(t), z(t))y0
(t)+R(x(t), y(t), z(t))z0
(t))dt
Z
~
F · d~
r =
Z
Pdx + Qdy + Rdz
C
~
F · d~
r
~
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
~
r(t) = (x(t), y(t), z(t)), d~
r = (dx, dy, dz)
dx = x0
(t)dt, dy = y0
(t)dt, dz = z0
(t)dt
Z
C
~
F·d~
r =
b
Z
a
(P(x(t), y(t), z(t))x0
(t)+Q(x(t), y(t), z(t))y0
(t)+R(x(t), y(t), z(t))z0
(t))dt
Z
C
~
F · d~
r =
Z
C
Pdx + Qdy + Rdz
~
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
~
r(t) = (x(t), y(t), z(t)), d~
r = (dx, dy, dz)
dx = x0
(t)dt, dy = y0
(t)dt, dz = z0
(t)dt
Z
C
~
F·d~
r =
b
Z
a
(P(x(t), y(t), z(t))x0
(t)+Q(x(t), y(t), z(t))y0
(t)+R(x(t), y(t), z(
Z
C
~
F · d~
r =
Z
C
Pdx + Qdy + Rdz
~
r(t) = (x(t), y(t), z(t)), d~
r = (dx, dy, dz)
dx = x0
(t)dt, dy = y0
(t)dt, dz = z0
(t)dt
Z
C
~
F·d~
r =
b
Z
a
(P(x(t), y(t), z(t))x0
(t)+Q(x(t), y(t), z(t))y0
(t)+R(x(t), y(t), z(t))z0
(t))dt
Z
C
~
F · d~
r =
Z
C
Pdx + Qdy + Rdz
~
r(t) = (x(t), y(t), z(t)), d~
r = (dx, dy, dz)
dx = x0
(t)dt, dy = y0
(t)dt, dz = z0
(t)dt
Z
C
~
F·d~
r =
b
Z
a
(P(x(t), y(t), z(t))x0
(t)+Q(x(t), y(t), z(t))y0
(t)+R(x(t), y(t), z(t))z0
(t))dt
Z
C
~
F · d~
r =
Z
C
Pdx + Qdy + Rdz
The differential form

50. Example. Compute
I
x2
y2
dx + x3
ydy
where is a square with vertices in (0, 0), (1, 0), (1, 1), (0, 1) oriented coun-
terclockwise.
= 1 [ 2 [ 3 [ 4
1 : y = 0, dy = 0, 0 6 x 6 1
Z
1
x2
y2
dx + x3
ydy =
1
Z
0
x2
· 0dx = 0
2 : x = 1, dx = 0, 0 6 y 6 1
Z
2
x2
y2
dx + x3
ydy =
1
Z
0
ydy =

y2
2
1
0
=
1
2

51. γ1
γ2
γ3
γ4
x
y
1
1
Example. Compute
I
x2
y2
dx + x3
ydy
where is a square with vertices in (0, 0), (1, 0), (1, 1), (0, 1) oriented coun-
terclockwise.
= 1 [ 2 [ 3 [ 4
1 : y = 0, dy = 0, 0 6 x 6 1
Z
1
x2
y2
dx + x3
ydy =
1
Z
0
x2
· 0dx = 0
2 : x = 1, dx = 0, 0 6 y 6 1
Z 1
Z  2 1
= 1 [ 2 [
1 : y = 0, dy = 0, 0 6 x 6 1
Z
1
x2
y2
dx + x3
ydy
2 : x = 1, dx = 0, 0 6 y 6 1
Z
2
x2
y2
dx + x3
ydy =
Z
0
3 : y = 1, dy = 0, x varies from 1
Z
3
x2
y2
dx + x3
ydy =
0
Z
1
: x = 0, dx = 0, y varies from 1
2 0
2 0 2
3 : y = 1, dy = 0, x varies from 1 to 0
Z
3
x2
y2
dx + x3
ydy =
0
Z
1
x2
dx =

x3
3
0
1
=
1
3
4 : x = 0, dx = 0, y varies from 1 to 0
Z
4
x2
y2
dx + x3
ydy =
0
Z
1
0 · ydy = 0
I
x2
y2
dx + x3
ydy = 0 +
1
2
1
3
+ 0 =
1
6
.
1
2
x y dx + x ydy =
0
ydy =
2 0
=
2
3 : y = 1, dy = 0, x varies from 1 to 0
Z
3
x2
y2
dx + x3
ydy =
0
Z
1
x2
dx =

x3
3
0
1
=
1
3
4 : x = 0, dx = 0, y varies from 1 to 0
Z
4
x2
y2
dx + x3
ydy =
0
Z
1
0 · ydy = 0
I
x2
y2
dx + x3
ydy = 0 +
1
2
1
3
+ 0 =
1
6
.
1
Example. Compute
I
x2
y2
dx + x3
ydy
where is a square with vertices in (0, 0), (1, 0), (1, 1), (0, 1) oriented coun-
terclockwise.
= 1 [ 2 [ 3 [ 4
1 : y = 0, dy = 0, 0 6 x 6 1
Z
1
x2
y2
dx + x3
ydy =
1
Z
0
x2
· 0dx = 0
2 : x = 1, dx = 0, 0 6 y 6 1
Z
2 2 3
1
Z 
y2 1
1
terclockwise.
= 1 [ 2 [ 3 [ 4
1 : y = 0, dy = 0, 0 6 x 6 1
Z
1
x2
y2
dx + x3
ydy =
1
Z
0
x2
· 0dx = 0
2 : x = 1, dx = 0, 0 6 y 6 1
Z
2
x2
y2
dx + x3
ydy =
1
Z
0
ydy =

y2
2
1
0
=
1
2
3 : y = 1, dy = 0, x varies from 1 to 0
Z
3
x2
y2
dx + x3
ydy =
0
Z
1
x2
dx =

x3
3
0
1
=
1
3
4 : x = 0, dx = 0, y varies from 1 to 0
Z 0
Z
2 0
2 0 2
3 : y = 1, dy = 0, x varies from 1 to 0
Z
3
x2
y2
dx + x3
ydy =
0
Z
1
x2
dx =

x3
3
0
1
=
1
3
4 : x = 0, dx = 0, y varies from 1 to 0
Z
4
x2
y2
dx + x3
ydy =
0
Z
1
0 · ydy = 0
I
x2
y2
dx + x3
ydy = 0 +
1
2
1
3
+ 0 =
1
6
.
1
2
x2
y2
dx + x3
ydy =
0
ydy =
2 0
=
2
dy = 0, x varies from 1 to 0
Z
3
x2
y2
dx + x3
ydy =
0
Z
1
x2
dx =

x3
3
0
1
=
1
3
dx = 0, y varies from 1 to 0
Z
4
x2
y2
dx + x3
ydy =
0
Z
1
0 · ydy = 0
I
x2
y2
dx + x3
ydy = 0 +
1
2
1
3
+ 0 =
1
6
.
1
= 1 [ 2 [ 3 [ 4
1 : y = 0, dy = 0, 0 6 x 6 1
Z
1
x2
y2
dx + x3
ydy =
1
Z
0
x2
·
2 : x = 1, dx = 0, 0 6 y 6 1
Z
2
x2
y2
dx + x3
ydy =
1
Z
0
ydy =
3 : y = 1, dy = 0, x varies from 1 to 0
Z
2 2 3
0
Z
2
I
x2
y2
dx + x3
ydy
vertices in (0, 0), (1, 0), (1, 1), (0, 1) oriented coun-
= 1 [ 2 [ 3 [ 4
0 6 x 6 1
x2
y2
dx + x3
ydy =
1
Z
0
x2
· 0dx = 0
0 6 y 6 1
1
Z  1
3 : y = 1, dy = 0, x varies from 1 to 0
Z
3
x2
y2
dx + x3
ydy =
0
Z
1
x2
dx =

x3
3
0
1
=
4 : x = 0, dx = 0, y varies from 1 to 0
Z
4
x2
y2
dx + x3
ydy =
0
Z
1
0 · ydy = 0
I
x2
y2
dx + x3
ydy = 0 +
1
2
1
3
+ 0 =
1
6
.

52. Fundamental Theorem of Calculus
b
∫
a
f′(x)dx = f(b) − f(a)
x
y
y = f′(x)
a b
The boundary of the domain: two points a and b
The integral of the derivative
over a domain—the sum of
local changes in function f in
the entire domain—depends
only on the values of f on the
boundary of the domain.
f′
f(x) =
x
∫
a
f′(t)dt

53. ∫
γ
∇Φ ⋅ d ⃗
r = Φ( ⃗
b ) − Φ( ⃗
a )
⃗
a = (a1, a2)
⃗
b = (b1, b2)
γ
[α, β] p` 2M T`K2i`Bb2`BM; p Fm`pM γ 7`´M !
a = !
r(α) iBHH !
b = !
r(β)X
bKKMb iiMBM;2M p Φ = Φ(x, y, z) K2/ !
r(t) = (x(t), y(t), z(t)) ; HH2`
d
dt
Φ(!
r(t)) =
∂Φ
∂x
(!
r(t))x!
(t) +
∂Φ
∂y
(!
r(t))y!
(t) +
∂Φ
∂z
(!
r(t))z!
(t) = ∇
AMb iiMBM; p /2ii bK#M/ B Fm`pBMi2;`H2M p !
F ;2` /´ U2MHB;i MH
ˆ
γ
!
F · d!
r =
β
ˆ
α
!
F(!
r(t)) · !
r !
(t) dt =
β
ˆ
α
∇Φ(!
r(t)) · !
r !
(t) dt =
β
ˆ
α
= Φ(!
r(β)) − Φ(!
r(α)) = Φ(!
b) − Φ(!
a)
.2i ` #` 7ƺ` FQMb2`piBp p2FiQ`7 Hi bQK Fm`pBMi2;`H2` ` p ;Q#2
p2FiQ`7 Hi /2}MB2`i T´ 2ii #´;pBb bKKM? M;M/2 QK`´/2 DX
A
B
6B;m` eN, 1M K M;/ U/2M ;`´ /2H2M B #BH/2Mc /2 pBi /2H`M BHHmbi`2`
K M;/2MV FHHb #´;pBb bKKM? M;M/2 QK /2i 7ƺ` p`D2 T` TmMFi
K M;/2M }MMb /2i 2M ;Hii UC1V Fm`p UHB;;M/2 7mHHbi M/B;i B K M;
#BH/2MV bQK 7ƺ`2M` TmMFi2`MX
Mi; ii Fm`pBMi2;`H2M K2HHM ip´ TmMFi2` ` p ;Q#2`Q2M/2X 6Bt2` T
Let ~
F be a conservative vektor field in an (arc-) connected set D, with po-
tential . For each smooth curve 2 D between ~
a and ~
b we have:
Z
~
F · d~
r = (~
b) (~
a).
I
~
F · d~
r = 0
Proof (Here for a 3D-field, but the proof is identical for other dimensions.)
Let ~
r(t), t 2 [↵, ] be a parametrisation of the curve from ~
a = ~
r(↵) to
~
b = ~
r( ). According to the Chain Rule for the composition of = (x, y, z)
with ~
r(t) = (x(t), y(t), z(t)) we have
d
dt
(~
r(t)) =
@
@x
(~
r(t))x0
(t) +
@
@y
(~
r(t))y0
(t) +
@
@z
(~
r(t))z0
(t) = r (~
r(t)) · ~
r 0
(t).
We replace the field in the integral by the gradient of the potential and we
Fundamental Theorem for Conservative Fields

54. Electric field
~
E(x, y) =
1
x2 + y2
(x, y) (x, y) 6= (0, 0)
is conservative and its potential functions are:
(x, y) =
1
2
ln(x2
+ y2
) + C, C 2 R.
r = (P, Q) )
✓
@
@x
=
x
x2 + y2
and
@
@y
=
y
x2 + y2
◆
.
The first condition gives (variable substitution: the numerator is almost
derivative of the denominator):
(x, y) =
Z
x
x2 + y2
dx =
1
2
ln(x2
+ y2
) + (y).
Putting in this expression in the second condition gives:
y
x2 + y2
=
@
@y
✓
1
2
ln(x2
+ y2
) + (y)
◆
=
y
x2 + y2
+ 0
(y)
which gives (y) = 0 ) (y) = C so all the potential functions to ~
E are:
(x, y) =
1
2
ln(x2
+ y2
) + C, C 2 R.
ctric field
~
E(x, y) =
1
x2 + y2
(x, y) (x, y) 6= (0, 0)
nservative and its potential functions are:
(x, y) =
1
2
ln(x2
+ y2
) + C, C 2 R.
r = (P, Q) )
✓
@
@x
=
x
x2 + y2
and
@
@y
=
y
x2 + y2
◆
.
first condition gives (variable substitution: the numerator is almost
vative of the denominator):
(x, y) =
Z
x
x2 + y2
dx =
1
2
ln(x2
+ y2
) + (y).
ing in this expression in the second condition gives:
y
x2 + y2
=
@
@y
✓
1
2
ln(x2
+ y2
) + (y)
◆
=
y
x2 + y2
+ 0
(y)
tric field
~
E(x, y) =
1
x2 + y2
(x, y) (x, y) 6= (0, 0)
nservative and its potential functions are:
(x, y) =
1
2
ln(x2
+ y2
) + C, C 2 R.
r = (P, Q) )
✓
@
@x
=
x
x2 + y2
and
@
@y
=
y
x2 + y2
◆
.
first condition gives (variable substitution: the numerator is almost
ative of the denominator):
(x, y) =
Z
x
x2 + y2
dx =
1
2
ln(x2
+ y2
) + (y).
ng in this expression in the second condition gives:
y
x2 + y2
=
@
@y
✓
1
2
ln(x2
+ y2
) + (y)
◆
=
y
x2 + y2
+ 0
(y)
Electric field
~
E(x, y) =
1
x2 + y2
(x, y) (x, y) 6= (0, 0)
is conservative and its potential functions are:
(x, y) =
1
2
ln(x2
+ y2
) + C, C 2 R.
r = (P, Q) )
✓
@
@x
=
x
x2 + y2
and
@
@y
=
y
x2 + y2
◆
.
The first condition gives (variable substitution: the numerator is almost
derivative of the denominator):
(x, y) =
Z
x
x2 + y2
dx =
1
2
ln(x2
+ y2
) + (y).
Putting in this expression in the second condition gives:
y
x2 + y2
=
@
@y
✓
1
2
ln(x2
+ y2
) + (y)
◆
=
y
x2 + y2
+ 0
(y)
Electrostatic field
⃗
a = (a1, a2)
⃗
b = (b1, b2)
γ
pBHF2i ;2` ψ(y) = 0 ⇒ ψ(y) = C Q+? HH TQi2MiBH7mMFiB
Φ(x, y) =
1
2
HM(x2
+ y2
) + C,
.2ii #2iv/2` B bvMM2`?2i ii 2FpBTQi2MiBHFm`pQ` 7ƺ` /2i
Q`B;QX .2bbmiQK- QK γ ` 2M Fm`p K2/ bi`iTmMFi2M B
a = (
;2` mpm/bib2M ii,
ˆ
γ
E · d
r = Φ(b1, b2) − Φ(a1, a2) =
1
2
HM(b2
1 +
aKKM7iiMBM;bpBb, :`22Mb bib FM iBHH KTb #` 7ƺ`
aib2M QKpM/H` Fm`pBMi2;`H2M p 2ii p2FiQ`7 Hi ƺp2` 2M
7 Hi2ib Uk.V@`QiiBQM ƺp2` U/2i THiiV QK`´/2i D BMQK Fm
pBHF2i ;2` ψ(y) = 0 ⇒ ψ(y) = C Q+? HH TQi2MiBH7mMFiBQM2` iBHH
E ?` 7Q`K2M,
Φ(x, y) =
1
2
HM(x2
+ y2
) + C, C ∈ R.
.2ii #2iv/2` B bvMM2`?2i ii 2FpBTQi2MiBHFm`pQ` 7ƺ` /2ii 7 Hi ` +B`FH` K2/ K
Q`B;QX .2bbmiQK- QK γ ` 2M Fm`p K2/ bi`iTmMFi2M B
a = (a1, a2) Q+? bHmiTmMFi2M B
;2` mpm/bib2M ii,
ˆ
γ
E · d
r = Φ(b1, b2) − Φ(a1, a2) =
1
2
HM(b2
1 + b2
2) −
1
2
HM(a2
1 + a2
2).
aKKM7iiMBM;bpBb, :`22Mb bib FM iBHH KTb #` 7ƺ` k.@7 Hi- HHib´
F(x, y) =
aib2M QKpM/H` Fm`pBMi2;`H2M p 2ii p2FiQ`7 Hi ƺp2` 2M bHmi2M Fm`p iBHH /m##2HB
2MF2Hi FM ;ƺ`b K2/ ?D HT p p2FiQ`7 Hi2ib TQi2MiBHX
h?2Q`2K k9X G´i
F p` 2ii FQMb2`piBpi p2FiQ`7 Hi B D K2/ TQi
γ ∈ D K2HHM
a Q+?
b ; HH2`,
ˆ
γ
F · d
r = Φ(
b) − Φ(
a).
LQi2` HBF?2i2M K2/ MHvb2Mb mpm/bibX SQi2MiBH2M ` KQibp
7mMFiBQM2MX aib2M BKTHB+2`` ii Fm`pBMi2;`H2M K2HHM ip´ TmMFi
Fm`pX 6ƺ` bHmiM Fm`pQ` γ FHHb Fm`pBMi2;`H2M p
F H M;b γ 7ƺ` +B
Q+? #2i2+FMb
˛
γ
F · d
r.
PK 7 Hi2i ` FQMb2`piBpi ; HH2` HHib´ ii QpMbi´2M/2 BMi
U/´ M/TmMFi2`M bKKM7HH2` K2/ p`M/`VX

55. the field is conservative
(it has a potential)
line integrals of the field
are path independent
circulations of the field
are equal to zero
⇐
⇐
⇐
⇔
⃗
a
⃗
b
γ1
γ2
∮
γ
⃗
F ⋅ d ⃗
r =
∫
γ1−γ2
⃗
F ⋅ d ⃗
r =
∫
γ1
⃗
F ⋅ d ⃗
r −
∫
γ2
⃗
F ⋅ d ⃗
r = 0
Fundamental Theorem (V121)
In the next slide
2MF2Hi FM ;ƺ`b K2/ ?D HT p p2FiQ`7 Hi2ib TQi2MiBHX
h?2Q`2K k9X G´i
F p` 2ii FQMb2`piBpi p2FiQ`7 Hi B D K2/ TQi2MiBH Φ
γ ∈ D K2HHM
a Q+?
b ; HH2`,
ˆ
γ
F · d
r = Φ(
b) − Φ(
a).
LQi2` HBF?2i2M K2/ MHvb2Mb mpm/bibX SQi2MiBH2M ` KQibp`B;?2i2
7mMFiBQM2MX aib2M BKTHB+2`` ii Fm`pBMi2;`H2M K2HHM ip´ TmMFi2` ` Q#
Fm`pX 6ƺ` bHmiM Fm`pQ` γ FHHb Fm`pBMi2;`H2M p
F H M;b γ 7ƺ` +B`FmHiB
Q+? #2i2+FMb
˛
γ
F · d
r.
PK 7 Hi2i ` FQMb2`piBpi ; HH2` HHib´ ii QpMbi´2M/2 BMi2;`H
U/´ M/TmMFi2`M bKKM7HH2` K2/ p`M/`VX
2pBbX U6Q`KmH2`b ? ` 7ƺ` j.@7 Hi- K2M #2pBb2i ` B/2MiBbFi 7ƺ` HH /BK
[α, β] p` 2M T`K2i`Bb2`BM; p Fm`pM γ 7`´M !
a = !
r(α) iBHH !
b = !
r(β)X
bKKMb iiMBM;2M p Φ = Φ(x, y, z) K2/ !
r(t) = (x(t), y(t), z(t)) ; HH2`
d
dt
Φ(!
r(t)) =
∂Φ
∂x
(!
r(t))x!
(t) +
∂Φ
∂y
(!
r(t))y!
(t) +
∂Φ
∂z
(!
r(t))z!
(t) = ∇
AMb iiMBM; p /2ii bK#M/ B Fm`pBMi2;`H2M p !
F ;2` /´ U2MHB;i M
ˆ
γ
!
F · d!
r =
β
ˆ
α
!
F(!
r(t)) · !
r !
(t) dt =
β
ˆ
α
∇Φ(!
r(t)) · !
r !
(t) dt =
β
ˆ
α
= Φ(!
r(β)) − Φ(!
r(α)) = Φ(!
b) − Φ(!
a)
.2i ` #` 7ƺ` FQMb2`piBp p2FiQ`7 Hi bQK Fm`pBMi2;`H2` ` p ;Q#2
p2FiQ`7 Hi /2}MB2`i T´ 2ii #´;pBb bKKM? M;M/2 QK`´/2 DX
A
B
(arc-) connected set
Three equivalent conditions for fields in (arc-) connected domains

56. Example. Compute
R
~
F · d~
r if is a half circle ~
r(t) = (2 + cos t, 1 + sin t),
t 2 [0, ⇡] and ~
F(x, y) = (y + 2x, x).
Because
@P
@y
= 1 =
@Q
@x
, we suspect that the field is conservative and we try
to find the potential .
8
:
@
@x
= y + 2x
@
@y
= x
The second equation gives (x, y) = xy+f(x),
plugged in in the first equation gives
0
x = y + f0
(x) = y + 2x )
f0
(x) = 2x ) f(x) = x2
+ C.
We get (x, y) = xy + x2
+ C. The starting point of the curve is (3, 1) and
the ending point is (1, 1) so, because ~
F is conservative on R2
, we can use the
Fundamental Theorem:
Z
~
F · d~
r = (1, 1) (3, 1) = 1 · 1 + 12
(3 · 1 + 32
) = 10.

57. (1,1) (3,1)
aib k8X
ˆ
γ
!
F · d!
r ` p ;Q#2`Q2M/2 T´ D ⇐⇒ !
F ` FQMb2`piBpi T´ DX
1t2KT2H N8X 2` FM
´
γ
!
F · d!
r QK γ ` ?Hp+B`F2HM !
r(t) = (2 + +Qb t, 1 + bBM t), t ∈ [0, π] Q+?
!
F(x, y) = (y + 2x, x)X
17i2`bQK
∂P
∂y
= 1 =
∂Q
∂x
- b´ KBbbi MF2` pB ii 7 Hi2i ` FQMb2`piBpi Q+? pB pBHH / `7ƺ` ?Bii 2M
TQi2MiBH ΦX







∂Φ
∂x
= y + 2x
∂Φ
∂y
= x
M/` 2FpiBQM2M ;2` Φ(x, y) = xy + f(x)- pBHF2i 27i2`
BMb iiMBM; B 7ƺ`bi 2FpiBQM2M ;2`
Φ!
x = y + f!(x) = y + 2x ⇒ f!(x) = 2x ⇒
f(x) = x2 + CX
oB ?` HHib´ Φ(x, y) = xy + x2 + CX Em`pMb bi`iTmMFi ` (3, 1) Q+? /2bb bHmiTmMFi ` (1, 1)-
Q+? 27i2`bQK !
F ` FQMb2`piBpi T´ R2- b´ ; HH2`,
ˆ
γ
!
F · d!
r = Φ(1, 1) − Φ(3, 1) = 1 · 1 + 12
− (3 · 1 + 32
) = −10.
aKKM7iiMBM;, A/; ?` pB H `i Qbb i`2 K2iQ/2` 7ƺ` ii #2` FM Fm`pBMi2;`H2` p p2FiQ`7 Hi-
Q+? /2bb K2iQ/2` iBHH KTb B QHBF bBimiBQM2` U7bi B#HM/ FM ~2` K2iQ/2` Tbb bKK
mTT;B7i5V,
Ç .B`2Fi 7`´M /2}MBiBQM2M- KX?XX T`K2i`Bb2`BM; p Fm`pM Q+? bFH `T`Q/mFi- bQK B 1t@
2KT2H N9X J2iQ/2M 7mM;2`` i2Q`2iBbFi HHiB/ pB/ BMi2;`2`BM; ƺp2` Fm`pQ` K2/ F M/ T`@
K2i`Bb2`BM;- K2M B T`FiBF2M K´bi2 7 Hi2ib FQKTQM2Mib7mMFiBQM2` p` M´;Q`HmM/ 2MFH-
MM`b 7´` pB 2M `BFiB;i ?2KbF 2MF2H BMi2;`H ii Hƺb- pBHF2i FM p` T`Q#H2KiBbFiX 6 Hi2i
γ
1t2KT2H N8X 2` FM
´
γ
!
F · d!
r QK γ ` ?Hp+B`F2HM !
r(t) = (2 + +Qb t, 1 + bBM
!
F(x, y) = (y + 2x, x)X
17i2`bQK
∂P
∂y
= 1 =
∂Q
∂x
- b´ KBbbi MF2` pB ii 7 Hi2i ` FQMb2`piBpi Q+? pB p
TQi2MiBH ΦX







∂Φ
∂x
= y + 2x
∂Φ
∂y
= x
M/` 2FpiBQM2M ;2` Φ(x, y) = xy +
BMb iiMBM; B 7ƺ`bi 2FpiBQM2M ;2`
Φ!
x = y + f!(x) = y + 2x ⇒
f(x) = x2 + CX
oB ?` HHib´ Φ(x, y) = xy + x2 + CX Em`pMb bi`iTmMFi ` (3, 1) Q+? /2bb b
Q+? 27i2`bQK !
F ` FQMb2`piBpi T´ R2- b´ ; HH2`,
ˆ
γ
!
F · d!
r = Φ(1, 1) − Φ(3, 1) = 1 · 1 + 12
− (3 · 1 + 32
) = −10
aKKM7iiMBM;, A/; ?` pB H `i Qbb i`2 K2iQ/2` 7ƺ` ii #2` FM Fm`pBMi2;`
Q+? /2bb K2iQ/2` iBHH KTb B QHBF bBimiBQM2` U7bi B#HM/ FM ~2` K2iQ
mTT;B7i5V,
Ç .B`2Fi 7`´M /2}MBiBQM2M- KX?XX T`K2i`Bb2`BM; p Fm`pM Q+? bFH `T
ˆ
γ
!
F · d!
r = Φ(1, 1) − Φ(3, 1) = 1 · 1 + 12
− (3 · 1 + 32
) = −10.
KKM7iiMBM;, A/; ?` pB H `i Qbb i`2 K2iQ/2` 7ƺ` ii #2` FM Fm`pBMi2;`H2` p p2Fi
? /2bb K2iQ/2` iBHH KTb B QHBF bBimiBQM2` U7bi B#HM/ FM ~2` K2iQ/2` Tbb b
T;B7i5V,
Ç .B`2Fi 7`´M /2}MBiBQM2M- KX?XX T`K2i`Bb2`BM; p Fm`pM Q+? bFH `T`Q/mFi- bQK
2KT2H N9X J2iQ/2M 7mM;2`` i2Q`2iBbFi HHiB/ pB/ BMi2;`2`BM; ƺp2` Fm`pQ` K2/ F M/
K2i`Bb2`BM;- K2M B T`FiBF2M K´bi2 7 Hi2ib FQKTQM2Mib7mMFiBQM2` p` M´;Q`HmM/
MM`b 7´` pB 2M `BFiB;i ?2KbF 2MF2H BMi2;`H ii Hƺb- pBHF2i FM p` T`Q#H2KiBbFiX
FM p` k. 2HH2` j.- #2?ƺp2` BMi2 p` FQMb2`piBpiX
Ç 1MHB;i 7Q`K2HM
´
γ
Pdx+Qdy, 7mM;2`` # bi QK Fm`pM #2bi´` p ~2` p2`iBFH Q+? ?
i2HH #Bi` bQK B 1t2KT2H Nj- / ` pB FM `2/m+2` T`Q#H2K2i iBHH ~2` 2Mp`B#2H@T`
6mM;2`` 7ƺ` HH bQ`i2`b k.@p2FiQ`7 HiX
· 1 + 12
− (3 · 1 + 32
) = −10.
2` 7ƺ` ii #2` FM Fm`pBMi2;`H2` p p2FiQ`7 Hi-
U7bi B#HM/ FM ~2` K2iQ/2` Tbb bKK
2`BM; p Fm`pM Q+? bFH `T`Q/mFi- bQK B 1t@
/ pB/ BMi2;`2`BM; ƺp2` Fm`pQ` K2/ F M/ T`@
FQKTQM2Mib7mMFiBQM2` p` M´;Q`HmM/ 2MFH-
ii Hƺb- pBHF2i FM p` T`Q#H2KiBbFiX 6 Hi2i
Mb2`piBpiX
K Fm`pM #2bi´` p ~2` p2`iBFH Q+? ?Q`BbQM@
/m+2` T`Q#H2K2i iBHH ~2` 2Mp`B#2H@T`Q#H2KX

58. Compute and if C is
the unit circle
oriented counterclockwise and
∮
C
⃗
F ⋅ d ⃗
r
∮
C
⃗
G ⋅ d ⃗
r
x2
+ y2
= 1
⃗
F(x, y) = (x2
ex3
+y3
, y2
ex3
+y3
)
⃗
G(x, y) = (x2
ex3
+y3
+ y, y2
ex3
+y3
− x)

59. Appendix to Problem 5
mm
r
a no
no
x varies from 3
to 1
REALUsageofyathindependenceffF.drffr.DE
p yt2x
I
L
Pdx 1
f 1t2x dx
3
x 1 931 2 3 9

60. Prothom Ci x'tight counterclockwise
EGiyI x ye
x
y Rex't's'ty y2ex4y3 x
1 F is conservative 8Iy 8
op
Fyi 3g exky
3 2y2ex4y
inner
der
E IS conservative because it
satisfies the condition and
it is defined and smooth
on entire 1122
E.dr
4 Icing

61. fly 3x2y'ex T
11C No
3x2y2ex4y3 Fang
GTxiy Early t y x
I.dF t
ydx xdy
HO mm
tTCxiy7 y x
21T
ftp.drtq
tiCrctDoorYt7dt O
0
Ht cost
yCtl
sintsOEtE2ITdef.oftttT
t tTfcost.sint E sint cost
tsinnt.co

62. tf Mtl T Ct sin't cos't
I
Idt

63. Five methods for computing line integrals of vector fields
From the definition, with help of parametrisation of the curve
Differential form (if we integrate over horizontal and/or vertical line pieces)
Fundamental Theorem for conservative v.f.
Green’s Theorem for 2D v.f. (Section 17)
Stokes’ Theorem for 3D v.f. (Section 19)
a
!
F(!
r(t)) · !
r !
(t)dt
M ` pB H´i2` H M;/2M p /2HBMi2`pHH2M B T`iBiBQM2M ;´ KQi MQHHX
6ƺ` 2ii BM}MBi2bBKHi iB/bBMi2`pHH dt- b´ #HB` !
r !(t)dt 2M BM}MBi2bBKH bi` +
pB bFmHH2 FmMM #2i2+FM d!
rX JM Mp M/2` / `7ƺ` #2i2+FMBM;2M
ˆ
γ
!
F · d!
r =
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt 7ƺ` 2M T`K
17i2`bQK
!
r(t) = (x(t), y(t), z(t)), d!
r = (dx, dy, dz)
!
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
b´ FM Fm`pBMi2;`H2M p 2ii p2FiQ`7 Hi Q+Fb´ bF`Bpb
ˆ
γ
!
F · d!
r
#$ %
.Bz2`2MiBH7Q`K
=
b
ˆ
a
(P(x(t), y(t), z(t))x!
(t) + Q(x(t), y(t), z(t))y!
(t) + R(x(
pBHF2i KQiBp2`` 2M MMM #2i2+FMBM; 7ƺ` Fm`pBMi2;`H2M,
ˆ
γ
!
F · d!
r =
ˆ
γ
Pdx + Qdy + Rdz
!
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
b´ FM Fm`pBMi2;`H2M p 2ii p2FiQ`7 Hi Q+Fb´ bF`Bpb
ˆ
γ
!
F · d!
r
#$ %
.Bz2`2MiBH7Q`K
=
b
ˆ
a
(P(x(t), y(t), z(t))x!
(t) + Q(x(t), y(t), z(t))y!
(t) + R(x(t
pBHF2i KQiBp2`` 2M MMM #2i2+FMBM; 7ƺ` Fm`pBMi2;`H2M,
ˆ
γ
!
F · d!
r =
ˆ
γ
Pdx + Qdy + Rdz
a2Mbi2 7Q`K2HM UB k.@p`BMiV Mp M/ K2/ 7ƺ`/2H iBHH mTT;B7i2` bQK /K
1t2KT2H NjX U/Kb R8X9 mTT;B7i 3V 1pHm2`
˛
γ
x2
y2
dx + x3
ydy
/ ` γ ` Fp/`i2M K2/ ?ƺ`M B (0, 0), (1, 0), (1, 1), (0, 1) Q`B2Mi2`/2 KQim
oB /2H` mTT Fm`pM γ B 7v` Fm`pQ`, γ = γ1 ∪γ2 ∪γ3 ∪γ4 bQK B #BH/2MX UhX2t
dy = 0 27i2`bQK Fm`pM ` ?Q`BbQMi2HH- pBHF2i #2iv/2` BM;2M 7ƺ` M/`BM; B y@H
UV 1D bHmi2M U#V aHmi2M- 2D 2MF2H U+V aHmi2M- 2MF2H
oB bF Mm pBb ii 7ƺ` FQMb2`piBp p2FiQ`7 Hi b´ ` Fm`pBMi2;`H2M K2HHM ip´ TmMFi2` Q#2`Q2M@
/2 p pBHF2M Fm`p K2HHM /2bb #´/ TmMFi2` KM p HD2` Q+? ii #2` FMBM; p Fm`pBMi2;`H2`
2MF2Hi FM ;ƺ`b K2/ ?D HT p p2FiQ`7 Hi2ib TQi2MiBHX
h?2Q`2K k9X G´i
F p` 2ii FQMb2`piBpi p2FiQ`7 Hi B D K2/ TQi2MiBH ΦX 6ƺ` p`D2 Fm`p
γ ∈ D K2HHM
a Q+?
b ; HH2`,
ˆ
γ
F · d
r = Φ(
b) − Φ(
a).
LQi2` HBF?2i2M K2/ MHvb2Mb mpm/bibX SQi2MiBH2M ` KQibp`B;?2i2M iBHH /2M T`BKBiBp
7mMFiBQM2MX aib2M BKTHB+2`` ii Fm`pBMi2;`H2M K2HHM ip´ TmMFi2` ` Q#2`Q2M/2 p pH p
Fm`pX 6ƺ` bHmiM Fm`pQ` γ FHHb Fm`pBMi2;`H2M p
F H M;b γ 7ƺ` +B`FmHiBQM2M p
F H M;b γ
Q+? #2i2+FMb
˛
γ
F · d
r.

64. ˆ
γ
!
F · d!
r =
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt 7ƺ` 2M
17i2`bQK
!
r(t) = (x(t), y(t), z(t)), d!
r = (dx, dy,
!
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z
b´ FM Fm`pBMi2;`H2M p 2ii p2FiQ`7 Hi Q+Fb´ bF`Bpb
ˆ
γ
!
F · d!
r
#$ %
.Bz2`2MiBH7Q`K
=
b
ˆ
a
(P(x(t), y(t), z(t))x!
(t) + Q(x(t), y(t), z(t))y!
(t) +
pBHF2i KQiBp2`` 2M MMM #2i2+FMBM; 7ƺ` Fm`pBMi2;`H2M,
ˆ
!
F · d!
r =
ˆ
Pdx + Qdy + Rdz
= ⃗
0 ≠ ⃗
0
the field is probably conservative
(for sure if it is defined and C1 in the whole space)
the field is definitely not conservative
Is the curve closed?
YES NO
answer 0
UV 1D bHmi2M U#V aHmi2M- 2D 2MF2H U+V aHmi2M- 2MF2H
F Mm pBb ii 7ƺ` FQMb2`piBp p2FiQ`7 Hi b´ ` Fm`pBMi2;`H2M K2HHM ip´ TmMFi2` Q#2`Q2M@
p pBHF2M Fm`p K2HHM /2bb #´/ TmMFi2` KM p HD2` Q+? ii #2` FMBM; p Fm`pBMi2;`H2`
2Hi FM ;ƺ`b K2/ ?D HT p p2FiQ`7 Hi2ib TQi2MiBHX
b k9X G´i
F p` 2ii FQMb2`piBpi p2FiQ`7 Hi B D K2/ TQi2MiBH ΦX 6ƺ` p`D2 Fm`p γ ∈ D
HM
a Q+?
b ; HH2`,
ˆ
γ
F · d
r = Φ(
b) − Φ(
a).
2` HBF?2i2M K2/ MHvb2Mb mpm/bibX SQi2MiBH2M ` KQibp`B;?2i2M iBHH /2M T`BKBiBp
FiBQM2MX aib2M BKTHB+2`` ii Fm`pBMi2;`H2M K2HHM ip´ TmMFi2` ` Q#2`Q2M/2 p pH p
pX 6ƺ` bHmiM Fm`pQ` γ FHHb Fm`pBMi2;`H2M p
F H M;b γ 7ƺ` +B`FmHiBQM2M p
F H M;b γ
compute a scalar potential;
answer:
or:
find a simpler path from
the start to the finish
and compute the line
integral “manually”
(see the column to the right)
Is the curve closed?
YES NO
test GREEN (in 2D)
or STOKES (in 3D)
(sometimes, in 2D,
even if the curve is
not closed, it is OK
to “close it”
and apply
Green anyway)
parametrise the curve
and compute “manually”:
.2i iQiH `#2i2i #HB` _B2KMMbmKKM
n
!
k=1
!
F(!
r(tk)) ·
d!
r
dt
(tk)∆tk
bQK FQMp2`;2`` KQi BMi2;`H2M
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt
M ` pB H´i2` H M;/2M p /2HBMi2`pHH2M B T`iBiBQM2M ;´ KQi MQHHX
6ƺ` 2ii BM}MBi2bBKHi iB/bBMi2`pHH dt- b´ #HB` !
r !(t)dt 2M BM}MBi2bB
pB bFmHH2 FmMM #2i2+FM d!
rX JM Mp M/2` / `7ƺ` #2i2+FMBM;2M
ˆ
γ
!
F · d!
r =
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt 7ƺ`
17i2`bQK
!
r(t) = (x(t), y(t), z(t)), d!
r = (dx, d
!
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y
or:
a
M ` pB H´i2` H M;/2M p /2HBMi2`pHH2M B T`iBiBQM2M ;´ KQi MQHHX
6ƺ` 2ii BM}MBi2bBKHi iB/bBMi2`pHH dt- b´ #HB` !
r !(t)dt 2M BM}MBi2bBKH bi` +F H M;b
pB bFmHH2 FmMM #2i2+FM d!
rX JM Mp M/2` / `7ƺ` #2i2+FMBM;2M
ˆ
γ
!
F · d!
r =
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt 7ƺ` 2M T`K2i`Bb2`B
17i2`bQK
!
r(t) = (x(t), y(t), z(t)), d!
r = (dx, dy, dz)
!
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
b´ FM Fm`pBMi2;`H2M p 2ii p2FiQ`7 Hi Q+Fb´ bF`Bpb
ˆ
γ
!
F · d!
r
#$ %
.Bz2`2MiBH7Q`K
=
b
ˆ
a
(P(x(t), y(t), z(t))x!
(t) + Q(x(t), y(t), z(t))y!
(t) + R(x(t), y(t), z
pBHF2i KQiBp2`` 2M MMM #2i2+FMBM; 7ƺ` Fm`pBMi2;`H2M,
ˆ
γ
!
F · d!
r =
ˆ
γ
Pdx + Qdy + Rdz
Line integrals of vector fields
Compute the curl
p 7ƺ`bi Q`/MBM;2MX oB bF Mm bim/2` ip´ bT2+B2HH FQK#BMiBQM2` p /2bb T`iB2HH /2`BpiQ`
HBi2 K2` MQ;;`Mi- 27i2`bQK /2- HBFbQK ;`/B2Mi2M- pBb` bB; ? K´M; Mp M/MBM;bQK`´/2MX
.2}MBiBQM jjX 6ƺ` 2ii p2FiQ`7 Hi !
F(x, y, z) b´ /2}MB2`b /Bp2`;2Mb2M Q+? `QiiBQM2M p
/Bp !
F =
∂P
∂x
+
∂Q
∂y
+
∂R
∂z
`2bTX `Qi !
F =
!
∂R
∂y
−
∂Q
∂z
,
∂P
∂z
−
∂R
∂x
,
∂Q
∂x
−
∂P
∂y
.
LQi2` ii bBbi FQKTQM2Mi2M B `QiiBQMbp2FiQ`M ` bKK bQK BMi2;`M/2M B :`22Mb 7Q`K2HX oB
´i2`FQKK2` iBHH /2ii b2M`2X :`/B2Mi2M- /Bp2`;2Mb2M Q+? `QiiBQM2M FM HH mii`v+Fb 2MF2Hi
K2/ ?D HT p bvK#QH2M M#H
∇ =
!
∂
∂x
,
∂
∂y
,
∂
∂z
bQK ` 2M p2FiQ`p `/ /Bz2`2MiBHQT2`iQ`X 1M QT2`iQ` ` 2M 7mMFiBQM p`b /2}MBiBQMb@ Q+?
p `/2K M;/2` ` K M;/2` p 7mMFiBQM2`X 1M /Bz2`2MiBHQT2`iQ` /2`Bp2`` 2H2K2Mi2M B /2}MB@
iBQMbK M;/2M T´ M´;Qi H KTHB;i b ii UB 7HH2i K2/ M#H T`iB2HHiVX
∇Φ =
!
∂
∂x
,
∂
∂y
,
∂
∂z
Φ =
!
∂Φ
∂x
,
∂Φ
∂y
,
∂Φ
∂z
= ;`/ Φ
∇ · !
F =
!
∂
∂x
,
∂
∂y
,
∂
∂z
· (P, Q, R)T
=
∂P
∂x
+
∂Q
∂y
+
∂R
∂z
= /Bp !
F
∇ × !
F =
#
#
#
#
#
#
#
#
!
e1 !
e2 !
e3
∂
∂x
∂
∂y
∂
∂z
P Q R
#
#
#
#
#
#
#
#
=
!
∂R
∂y
−
∂Q
∂z
,
∂P
∂z
−
∂R
∂x
,
∂Q
∂x
−
∂P
∂y
= `Qi !
F.
curl

65. ˆ
γ
!
F · d!
r =
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt 7ƺ` 2M
17i2`bQK
!
r(t) = (x(t), y(t), z(t)), d!
r = (dx, dy,
!
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z
b´ FM Fm`pBMi2;`H2M p 2ii p2FiQ`7 Hi Q+Fb´ bF`Bpb
ˆ
γ
!
F · d!
r
#$ %
.Bz2`2MiBH7Q`K
=
b
ˆ
a
(P(x(t), y(t), z(t))x!
(t) + Q(x(t), y(t), z(t))y!
(t) +
pBHF2i KQiBp2`` 2M MMM #2i2+FMBM; 7ƺ` Fm`pBMi2;`H2M,
ˆ
!
F · d!
r =
ˆ
Pdx + Qdy + Rdz
= ⃗
0 ≠ ⃗
0
the field is probably conservative
(for sure if it is defined and C1 in the whole space)
the field is definitely not conservative
Is the curve closed?
YES NO
answer 0
UV 1D bHmi2M U#V aHmi2M- 2D 2MF2H U+V aHmi2M- 2MF2H
F Mm pBb ii 7ƺ` FQMb2`piBp p2FiQ`7 Hi b´ ` Fm`pBMi2;`H2M K2HHM ip´ TmMFi2` Q#2`Q2M@
p pBHF2M Fm`p K2HHM /2bb #´/ TmMFi2` KM p HD2` Q+? ii #2` FMBM; p Fm`pBMi2;`H2`
2Hi FM ;ƺ`b K2/ ?D HT p p2FiQ`7 Hi2ib TQi2MiBHX
b k9X G´i
F p` 2ii FQMb2`piBpi p2FiQ`7 Hi B D K2/ TQi2MiBH ΦX 6ƺ` p`D2 Fm`p γ ∈ D
HM
a Q+?
b ; HH2`,
ˆ
γ
F · d
r = Φ(
b) − Φ(
a).
2` HBF?2i2M K2/ MHvb2Mb mpm/bibX SQi2MiBH2M ` KQibp`B;?2i2M iBHH /2M T`BKBiBp
FiBQM2MX aib2M BKTHB+2`` ii Fm`pBMi2;`H2M K2HHM ip´ TmMFi2` ` Q#2`Q2M/2 p pH p
pX 6ƺ` bHmiM Fm`pQ` γ FHHb Fm`pBMi2;`H2M p
F H M;b γ 7ƺ` +B`FmHiBQM2M p
F H M;b γ
compute a scalar potential;
answer:
or:
find a simpler path from
the start to the finish
and compute the line
integral “manually”
(see the column to the right)
Is the curve closed?
YES NO
test GREEN (in 2D)
or STOKES (in 3D)
(sometimes, in 2D,
even if the curve is
not closed, it is OK
to “close it”
and apply
Green anyway)
parametrise the curve
and compute “manually”:
.2i iQiH `#2i2i #HB` _B2KMMbmKKM
n
!
k=1
!
F(!
r(tk)) ·
d!
r
dt
(tk)∆tk
bQK FQMp2`;2`` KQi BMi2;`H2M
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt
M ` pB H´i2` H M;/2M p /2HBMi2`pHH2M B T`iBiBQM2M ;´ KQi MQHHX
6ƺ` 2ii BM}MBi2bBKHi iB/bBMi2`pHH dt- b´ #HB` !
r !(t)dt 2M BM}MBi2bB
pB bFmHH2 FmMM #2i2+FM d!
rX JM Mp M/2` / `7ƺ` #2i2+FMBM;2M
ˆ
γ
!
F · d!
r =
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt 7ƺ`
17i2`bQK
!
r(t) = (x(t), y(t), z(t)), d!
r = (dx, d
!
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y
or:
a
M ` pB H´i2` H M;/2M p /2HBMi2`pHH2M B T`iBiBQM2M ;´ KQi MQHHX
6ƺ` 2ii BM}MBi2bBKHi iB/bBMi2`pHH dt- b´ #HB` !
r !(t)dt 2M BM}MBi2bBKH bi` +F H M;b
pB bFmHH2 FmMM #2i2+FM d!
rX JM Mp M/2` / `7ƺ` #2i2+FMBM;2M
ˆ
γ
!
F · d!
r =
b
ˆ
a
!
F(!
r(t)) · !
r !
(t)dt 7ƺ` 2M T`K2i`Bb2`B
17i2`bQK
!
r(t) = (x(t), y(t), z(t)), d!
r = (dx, dy, dz)
!
F(x, y, z) = (P(x, y, z), Q(x, y, z), R(x, y, z))
b´ FM Fm`pBMi2;`H2M p 2ii p2FiQ`7 Hi Q+Fb´ bF`Bpb
ˆ
γ
!
F · d!
r
#$ %
.Bz2`2MiBH7Q`K
=
b
ˆ
a
(P(x(t), y(t), z(t))x!
(t) + Q(x(t), y(t), z(t))y!
(t) + R(x(t), y(t), z
pBHF2i KQiBp2`` 2M MMM #2i2+FMBM; 7ƺ` Fm`pBMi2;`H2M,
ˆ
γ
!
F · d!
r =
ˆ
γ
Pdx + Qdy + Rdz
Line integrals of vector fields
Compute the curl
p 7ƺ`bi Q`/MBM;2MX oB bF Mm bim/2` ip´ bT2+B2HH FQK#BMiBQM2` p /2bb T`iB2HH /2`BpiQ`
HBi2 K2` MQ;;`Mi- 27i2`bQK /2- HBFbQK ;`/B2Mi2M- pBb` bB; ? K´M; Mp M/MBM;bQK`´/2MX
.2}MBiBQM jjX 6ƺ` 2ii p2FiQ`7 Hi !
F(x, y, z) b´ /2}MB2`b /Bp2`;2Mb2M Q+? `QiiBQM2M p
/Bp !
F =
∂P
∂x
+
∂Q
∂y
+
∂R
∂z
`2bTX `Qi !
F =
!
∂R
∂y
−
∂Q
∂z
,
∂P
∂z
−
∂R
∂x
,
∂Q
∂x
−
∂P
∂y
.
LQi2` ii bBbi FQKTQM2Mi2M B `QiiBQMbp2FiQ`M ` bKK bQK BMi2;`M/2M B :`22Mb 7Q`K2HX oB
´i2`FQKK2` iBHH /2ii b2M`2X :`/B2Mi2M- /Bp2`;2Mb2M Q+? `QiiBQM2M FM HH mii`v+Fb 2MF2Hi
K2/ ?D HT p bvK#QH2M M#H
∇ =
!
∂
∂x
,
∂
∂y
,
∂
∂z
bQK ` 2M p2FiQ`p `/ /Bz2`2MiBHQT2`iQ`X 1M QT2`iQ` ` 2M 7mMFiBQM p`b /2}MBiBQMb@ Q+?
p `/2K M;/2` ` K M;/2` p 7mMFiBQM2`X 1M /Bz2`2MiBHQT2`iQ` /2`Bp2`` 2H2K2Mi2M B /2}MB@
iBQMbK M;/2M T´ M´;Qi H KTHB;i b ii UB 7HH2i K2/ M#H T`iB2HHiVX
∇Φ =
!
∂
∂x
,
∂
∂y
,
∂
∂z
Φ =
!
∂Φ
∂x
,
∂Φ
∂y
,
∂Φ
∂z
= ;`/ Φ
∇ · !
F =
!
∂
∂x
,
∂
∂y
,
∂
∂z
· (P, Q, R)T
=
∂P
∂x
+
∂Q
∂y
+
∂R
∂z
= /Bp !
F
∇ × !
F =
#
#
#
#
#
#
#
#
!
e1 !
e2 !
e3
∂
∂x
∂
∂y
∂
∂z
P Q R
#
#
#
#
#
#
#
#
=
!
∂R
∂y
−
∂Q
∂z
,
∂P
∂z
−
∂R
∂x
,
∂Q
∂x
−
∂P
∂y
= `Qi !
F.
curl

66. Flux integrals, Notation, definition, computations and applications
15.6

67. https://en.wikipedia.org/wiki/Flux
⃗
F
⃗
N
α
ΔSi
F
{
ZZ
Y
~
F · N̂dS =
ZZ
D
~
F(~
r(s, t)) ·
✓
@
@
Flux =
n
X
i=1
F Si
F = |~
F| cos ↵ = |~
F|
~
F · ~
N
|~
F|| ~
N|
= ~
F · N̂
n
X
i=1
F Si !
ZZ
Y
~
F · N̂dS
Flux =
n
X
i=1
F Si
F = |~
F| cos ↵ = |~
F|
~
F · ~
N
|~
F|| ~
N|
= ~
F · N̂
n
X
i=1
F Si !
ZZ
Y
~
F · N̂dS
Flux =
n
X
i=1
F Si
F = |~
F| cos ↵ = |~
F|
~
F · ~
N
|~
F|| ~
N|
= ~
F · N̂
n
X
i=1
F Si !
ZZ
Y
~
F · N̂dS
F = |~
F| cos ↵ = |~
F|
~
F · ~
N
|~
F|| ~
N|
= ~
F · N̂
n
X
i=1
F Si !
ZZ
Y
~
F · N̂dS
ZZ
Y
~
F · d~
S
d~
S = N̂dS
Flux =
n
X
i=1
F Si
F = |~
F| cos ↵ = |~
F|
~
F · ~
N
|~
F|| ~
N|
= ~
F · N̂
n
X
i=1
F Si !
ZZ
Y
~
F · N̂dS
ZZ
Y
~
F · d~
S

68. ̂
v =
1
∥ ⃗
v ∥
⃗
v
To normalize a vector (i.e., to transform it into a vector with the same direction
but with length equal to 1), just scale the vector by the inverse of the its length
2 T`K2i`Bb2`b bQK
r(s, t) Q+? ii /2 T`iB2HH /2`BpiQ`M
∂
r
∂s
×
∂
r
∂t
HHib´ FM pB b ii
N̂ =
∂
r
∂s
×
∂
r
∂t
!
!
!
!
∂
r
∂s
×
∂
r
∂t
!
!
!
!
.
pBb bB; T2F ´i 72H ?´HH b´ FM ?2Hi 2MF2Hi #vi THib T´
b´
¨
Y
!
F · N̂ dS = ±
¨
D
!
F(x, y, f(x, y)) ·
(f!
1(x, y), f!
2(x, y), −1)
!
1 + (f!
1(x, y))2 + (f!
2(x, y))2
= ±
¨
D
!
F(x, y, f(x, y)) · (f!
1(x, y), f!
2(x, y), −1) dxdy.
`` FTBiH2i ii 2M vi FmM/2 T`K2i`Bb2`b bQK
r(s, t) Q+? ii /2
2HH K2/ viMX o2FiQ`M
∂
r
∂s
×
∂
r
∂t
2M MQ`KHp2FiQ` iBHH viMX HHib´ FM pB b ii
N̂ =
∂
r
∂s
×
∂
r
∂t
!
!
!
!
∂
r
∂s
×
∂
r
∂t
!
!
!
!
.
2/ /2MM /2}MBiBQM bFmHH2 pBb bB; T2F ´i 72H ?´HH b´ FM ?2Hi

69. b´
¨
Y
!
F · N̂ dS = ±
¨
Y
!
F ·
(f!
1, f!
2, −1)
!
1 + (f!
1)2 + (f!
2)2
1 + (f!
1)2 + (f!
2)2 dxdy
= ±
¨
Y
!
F · (f!
1, f!
2, −1) dxdy.
A /2i M/` 2t2KTH2i #2?ƺp/2 pB BMi2 ?2HH2` Mp M/ M´;QM T`K2i`Bb2`BM; i+F p`2
K2i`B2` bQK T`Q#H2K2i ?/2X PK viM ;2b p f(x, y, z) = 0 FM 7ƺHDM/2 7Q`K2H Mp M/
¨
Y
!
F · N̂ dS =
¨
Y
!
F ·
∇f
|f!
3|
dxdy.
f : ℝ2
→ ℝ
f : ℝ3
→ ℝ
D
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y))
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2dxdy
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y)) ·
| ~
N|
|N3|
dxdy,
mass(Y ) =
ZZ
Y
⇢(x, y, z) dS
• Normal to the graph surface z = f(x, y):
~
N1 = (f0
x, f0
y, 1), ~
N2 = ( f0
x, f0
y, 1)
ZZ
Y
~
F · N̂ dS = ±
ZZ
D
~
F ·
(f0
x, f0
y, 1)
q
1 + (f0
x)2 + (f0
y)2
q
1 + (f0
x)2 + (f0
y)2 dxdy
= ±
ZZ
D
~
F · (f0
x, f0
y, 1) dxdy
Flux =
n
X
i=1
F Si

70. ¨
!
F · N̂dS pBHF2i B#HM/ #2i2+FMb
¨
Y
!
F · d!
S b´ d!
S = N̂dSX
i2;`H2` ` ~ƺ/2bBMi2;`H2`X PK viM Y ` bHmi2M Mp M/b Q7i #2i2+FMBM;`@
‹
Y
!
F · N̂dS Q+?
‹
Y
!
F · d!
S UD K7ƺ` K2/ Fm`pBMi2;`H2`V
ƺ` N̂ QpM Q+? 7Q`K2HM 7ƺ` dS 7`´M 7ƺ`` FTBiH2i 7´` pB,
¨
Y
!
F · N̂dS =
¨
D
!
F(!
r(s, t)) ·




∂!
r
∂s
×
∂!
r
∂t
$
$
$
$
∂!
r
∂s
×
∂!
r
∂t
$
$
$
$




$
$
$
$
∂!
r
∂s
×
∂!
r
∂t
$
$
$
$ dsdt
Ƙ =
¨
D
!
F(!
r(s, t)) ·
(
∂!
r
∂s
×
∂!
r
∂t
)
dsdt.
FiBQM2M p viM Y T´ xy@THM2iX
Direct from the definition, with the (unit) normal to the surface
If the surface is parameterised by ⃗
r : D → ℝ3
, D ⊂ ℝ2
b´
¨
Y
!
F · N̂ dS = ±
¨
D
!
F(x, y, f(x, y)) ·
(f!
1(x, y), f
!
1 + (f!
1(x, y)
= ±
¨
D
!
F(x, y, f(x, y)) · (f!
1(x, y), f!
2(x,
A /2i M/` 2t2KTH2i #2?ƺp/2 pB BMi2 ?2HH2` Mp M/
K2i`B2` bQK T`Q#H2K2i ?/2X PK viM ;2b p f(x, y,
¨
Y
!
F · N̂ dS =
¨
D
!
F
‹
Y
!
F · N̂dS Q+?
‹
Y
!
F · d!
S UD K7ƺ` K2/ Fm`pBMi2;`H2`
´M 7Q`K2HM 7ƺ` N̂ QpM Q+? 7Q`K2HM 7ƺ` dS 7`´M 7ƺ`` FTBiH2i 7´` pB,
¨
Y
!
F · N̂dS =
¨
D
!
F(!
r(s, t)) ·




∂!
r
∂s
×
∂!
r
∂t
$
$
$
$
∂!
r
∂s
×
∂!
r
∂t
$
$
$
$




$
$
$
$
∂!
r
∂s
×
∂!
r
∂t
$
$
$
$ dsdt
Ƙ =
¨
D
!
F(!
r(s, t)) ·
(
∂!
r
∂s
×
∂!
r
∂t
)
dsdt.
` D ` T`QD2FiBQM2M p viM Y T´ xy@THM2iX
;M` `2bi2M p FTBiH2i iBHH ii ` FM M´;` 2t2KT2H T´ ~ƺ/2bBMi2;`H2`X
tKTH2 N3X 2` FM 7ƺH/2i mi m` QK`´/2i K = {(x, y, z) ∈ R3 : x2 + y2 ⩽ z ⩽ 4} 7ƺ` 7 Hi2

71. ¨
!
F · N̂dS pBHF2i B#HM/ #2i2+FMb
¨
Y
!
F · d!
S b´ d!
S = N̂dSX
i2;`H2` ` ~ƺ/2bBMi2;`H2`X PK viM Y ` bHmi2M Mp M/b Q7i #2i2+FMBM;`@
‹
Y
!
F · N̂dS Q+?
‹
Y
!
F · d!
S UD K7ƺ` K2/ Fm`pBMi2;`H2`V
ƺ` N̂ QpM Q+? 7Q`K2HM 7ƺ` dS 7`´M 7ƺ`` FTBiH2i 7´` pB,
¨
Y
!
F · N̂dS =
¨
D
!
F(!
r(s, t)) ·




∂!
r
∂s
×
∂!
r
∂t
$
$
$
$
∂!
r
∂s
×
∂!
r
∂t
$
$
$
$




$
$
$
$
∂!
r
∂s
×
∂!
r
∂t
$
$
$
$ dsdt
Ƙ =
¨
D
!
F(!
r(s, t)) ·
(
∂!
r
∂s
×
∂!
r
∂t
)
dsdt.
FiBQM2M p viM Y T´ xy@THM2iX
Direct from the definition, with the (unit) normal to the surface
If the surface is parameterised by ⃗
r : D → ℝ3
, D ⊂ ℝ2
If the surface is a graph to f : D → ℝ, D ⊂ ℝ2
b´
¨
Y
!
F · N̂ dS = ±
¨
D
!
F(x, y, f(x, y)) ·
(f!
1(x, y), f
!
1 + (f!
1(x, y)
= ±
¨
D
!
F(x, y, f(x, y)) · (f!
1(x, y), f!
2(x,
A /2i M/` 2t2KTH2i #2?ƺp/2 pB BMi2 ?2HH2` Mp M/
K2i`B2` bQK T`Q#H2K2i ?/2X PK viM ;2b p f(x, y,
¨
Y
!
F · N̂ dS =
¨
D
!
F
b´
¨
Y
!
F · N̂ dS = ±
¨
D
!
F(x, y, f(x, y)) ·
(f!
1(x, y), f!
2(x, y), −1)
!
1 + (f!
1(x, y))2 + (f!
2(x, y))2
1 + (f!
1(x, y))2 +
= ±
¨
D
!
F(x, y, f(x, y)) · (f!
1(x, y), f!
2(x, y), −1) dxdy.
‹
Y
!
F · N̂dS Q+?
‹
Y
!
F · d!
S UD K7ƺ` K2/ Fm`pBMi2;`H2`
´M 7Q`K2HM 7ƺ` N̂ QpM Q+? 7Q`K2HM 7ƺ` dS 7`´M 7ƺ`` FTBiH2i 7´` pB,
¨
Y
!
F · N̂dS =
¨
D
!
F(!
r(s, t)) ·




∂!
r
∂s
×
∂!
r
∂t
$
$
$
$
∂!
r
∂s
×
∂!
r
∂t
$
$
$
$




$
$
$
$
∂!
r
∂s
×
∂!
r
∂t
$
$
$
$ dsdt
Ƙ =
¨
D
!
F(!
r(s, t)) ·
(
∂!
r
∂s
×
∂!
r
∂t
)
dsdt.
` D ` T`QD2FiBQM2M p viM Y T´ xy@THM2iX
;M` `2bi2M p FTBiH2i iBHH ii ` FM M´;` 2t2KT2H T´ ~ƺ/2bBMi2;`H2`X
tKTH2 N3X 2` FM 7ƺH/2i mi m` QK`´/2i K = {(x, y, z) ∈ R3 : x2 + y2 ⩽ z ⩽ 4} 7ƺ` 7 Hi2
Needs to be normalized
We’re fine
We’re fine

72. Properties of surface integrals of vector fields (flux integrals)
Surface integrals are independent of the parametrisation of the surface (Chain rule)
They describe the flux of the vector field through the surface Y
Surface integrals are dependent of the orientation of the surface
∬
−Y
⃗
F ⋅ d ⃗
S = −
∬
Y
⃗
F ⋅ d ⃗
S
Y1 Y2
Y4
Y3
piece-wise smooth surface
Additivity
∬
Y
⃗
F ⋅ d ⃗
S =
∬
Y1
⃗
F ⋅ d ⃗
S +
∬
Y2
⃗
F ⋅ d ⃗
S +
∬
Y3
⃗
F ⋅ d ⃗
S +
∬
Y4
⃗
F ⋅ d ⃗
S

73. Compute the flux of the vector field
up through the surface
.
⃗
F(x, y, z) = (x, 3y, x + 3y)
z = 1 − x + y, 0 ⩽ x2
+ y2
⩽ 4, x, y ⩾ 0
D
f(x, y) = 1 − x + y
ZZ
Y
~
F · N̂ dS =
ZZ
D
~
F(x, y, f(x, y)) · ( f0
1(x, y), f0
2(x, y), 1) dxdy
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y))
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2dxdy
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y)) ·
| ~
N|
|N3|
dxdy
ZZ

74. problemL FCxiy.it x 3
xt3y
feast h
HY 2
4 2
f
E D8 If iI dxdy
x'tyk4
x
yz0
ff x t3yXdxdy 2ffxdxdy
D O
polar word
Fiesta J.IE
resordroeo
dxdy
rdrdO0ErE2
OE0EIz2 fEcosodO dr
2 fsino
JF.IE o
2 Ez answer
in

75. Problem 1. Compute the flux of the vector field ~
F = (x, y, 3) out of the domain K = {(x, y, z) ∈ R3
: x2
+ y2
⩽ z ⩽ 4}.
Solution
/2 BMi2;`H2` ` ~ƺ/2bBMi2;`H2`X PK viM Y ` bHmi2M Mp M/b Q7i #2i2+FMBM;`@
‹
Y
!
F · N̂dS Q+?
‹
Y
!
F · d!
S UD K7ƺ` K2/ Fm`pBMi2;`H2`V
2HM 7ƺ` N̂ QpM Q+? 7Q`K2HM 7ƺ` dS 7`´M 7ƺ`` FTBiH2i 7´` pB,
¨
Y
!
F · N̂dS =
¨
D
!
F(!
r(s, t)) ·




∂!
r
∂s
×
∂!
r
∂t
$
$
$
$
∂!
r
∂s
×
∂!
r
∂t
$
$
$
$




$
$
$
$
∂!
r
∂s
×
∂!
r
∂t
$
$
$
$ dsdt
Ƙ =
¨
D
!
F(!
r(s, t)) ·
(
∂!
r
∂s
×
∂!
r
∂t
)
dsdt.
T`QD2FiBQM2M p viM Y T´ xy@THM2iX
2bi2M p FTBiH2i iBHH ii ` FM M´;` 2t2KT2H T´ ~ƺ/2bBMi2;`H2`X
N3X 2` FM 7ƺH/2i mi m` QK`´/2i K = {(x, y, z) ∈ R3 : x2 + y2 ⩽ z ⩽ 4} 7ƺ` 7 Hi2i
3)X
#2;` Mbb p ip´ viQ`X .2Hb THM2i z = 4- Q+?
#QHQB/2M z = x2+y2X hQiH ~ƺ/2i ` bmKKM
i2;`H2`M ƺp2` /2bb #´/ viQ`X 6ƺ` THM2i `
Fi/2 MQ`KH2M N̂ = (0, 0, 1) Q+? ~ƺ/2bBMi2;`@
N̂ dS =
¨
Y1
(x, y, 3) · (0, 0, 1) dS =
¨
Y1
3 dS
= 3
¨
Y1
dS = 3π22
= 12π
1 ` 2M +B`F2HbFBp K2/ `/B2 2X
Y2
Y1
x
y
f
(x,
y)
/viM Y2 ?` T`K2i`Bb2`BM;2M !
r(s, t) = (s, t, s2 + t2) / ` T`K2i``Mb /2}MBiBQMb@
D = {(s, t) ∈ R2 : s2 + t2 ⩽ 4}X Lm ; HH2`
1, 0, 2s),
∂!
r
∂t
= (0, 1, 2t), Q+?
∂!
r
∂s
×
∂!
r
∂t
=
$
$
$
$
$
$
!
e1 !
e2 !
e3
1 0 2s
0 1 2t
$
$
$
$
$
$
= (−2s, −2t, 1).
R8j
Figur 1: Picture to Problem 1.
The boundary of the domain consists of two surfaces: the plane z = 4 and the paraboloid z = x2
+ y2
. The total flux is
the sum of fluxes through both surfaces. For the plane we use the normal pointing upwards N̂ = (0, 0, 1) and the flux is
ZZ
Y1
~
F · N̂ dS =
ZZ
Y1
(x, y, 3) · (0, 0, 1) dS =
ZZ
Y1
3 dS = 3
ZZ
Y1
dS = 3π22
= 12π
because Y1 is a disk with radius 2.
The surface of the paraboloid Y2 has parametrisation ~
r(s, t) = (s, t, s2
+t2
) with the domain D = {(s, t) ∈ R2
: s2
+t2
⩽ 4}.
We have
∂~
r
∂s
= (1, 0, 2s),
∂~
r
∂t
= (0, 1, 2t), and
∂~
r
∂s
×
∂~
r
∂t
=
~
e1 ~
e2 ~
e3
1 0 2s
0 1 2t
= (−2s, −2t, 1).
Because the z-component is positive, this vector is directed inside the domain K. We have to use the other normal
(2s, 2t, −1). The flux through Y2 is:
ZZ
Y2
~
F · N̂ dS = −
ZZ
D
~
F(~
r(s, t)) ·
∂~
r
∂s
×
∂~
r
∂t
dsdt =
ZZ
D
(s, t, 3) · (2s, 2t, −1) dsdt =
=
ZZ
D
(2s2
+ 2t2
− 3) dsdt =
2π
Z
0
2
Z
0
(2r2
− 3)r drdθ = 2π
2r4
4
−
3r2
2
2
0
= 2π
24
2
−
3
2
· 22
= 4π.

76. The total flux is 12π + 4π = 16π.
Answer: The flux is 16π.
Problem 2. Compute the flux of the vector field ~
F(x, y, z) = (2x, y, 0) down through the surface Y with the following
parametric definition:
~
r(s, t) = (3s2
, −3t2
, 2s + t) for 0 ⩽ s ⩽ 1, 0 ⩽ t ⩽ 1.
Solution We compute the flux of the vector field ~
F through the surface Y using the formula:
ZZ
Y
~
F · N̂dS =
ZZ
D
~
F(~
r(s, t)) ·
∂~
r
∂s
×
∂~
r
∂t
dsdt.
The parametric description of Y is ~
r(s, t) = (3s2
, −3t2
, 2s + t) with the range of the parameters
D = {(s, t) ∈ R2
: 0 ⩽ s ⩽ 1, 0 ⩽ t ⩽ 1}. We get thus
∂~
r
∂s
= (6s, 0, 2),
∂~
r
∂t
= (0, −6t, 1), and
∂~
r
∂s
×
∂~
r
∂t
=
~
e1 ~
e2 ~
e3
6s 0 2
0 −6t 1
= (12t, −6s, −36st).
Because the z-component is negative (s, t ⩾ 0 gives −36st ⩽ 0) the normal vector computed above points downwards,
exactly as it should do. The value of the field on the surface expressed in terms of the parameters is:
~
F(~
r(s, t)) = (2x(s, t), y(s, t), 0) = (6s2
, −3t2
, 0).
The flux through Y is thus (apply Fubini’s theorem for rectangles; separate the variables):
ZZ
Y
~
F · N̂ dS =
ZZ
D
~
F(~
r(s, t)) ·
∂~
r
∂s
×
∂~
r
∂t
dsdt =
ZZ
D
(6s2
, −3t2
, 0) · (12t, −6s, −36st) dsdt
=
1
Z
0
1
Z
0
(72s2
t + 18st2
) dsdt = 72
1
Z
0
s2
ds ·
1
Z
0
tdt + 18
1
Z
0
sds ·
1
Z
0
t2
dt =
= 72
s3
3
1
0
·
t2
2
1
0
+ 18
s2
2
1
0
·
t3
3
1
0
=
72
6
+
18
6
= 12 + 3 = 15.
Answer: The flux is equal to 15.
Problem 3. Compute the flux of the vector field ~
F = (x+y, z, 0) out of the sphere S with radius R and centre in the origin.
Solution The outer unit normal in point ~
r = (x, y, z) on the sphere can be written as N̂ = ~
r/R. (Why? Because
|~
r| =
p
x2 + y2 + z2 = R on the sphere.) The flux is thus:
ZZ
S
~
F · N̂ dS =
ZZ
S
(x + y, z, 0) · (x, y, z)
1
R
dS =
ZZ
S
(x2
+ xy + yz)
1
R
dS.
Because of the symmetry of the sphere along the origin, and because the integrands are odd w.r.t y:
ZZ
S
xy dS =
ZZ
S
yz dS = 0,

77. so just integration of x2
is left. We could parameterise the surface with spherical coordinates but we can use a trick instead.
Because of symmetry there is:
ZZ
S
x2
dS =
ZZ
S
y2
dS =
ZZ
S
z2
dS =
1
3
ZZ
S
(x2
+ y2
+ z2
) dS
⇒
ZZ
S
~
F · N̂ dS =
1
3
ZZ
S
(x2
+ y2
+ z2
)
| {z }
R2
1
R
dS =
R
3
ZZ
S
dS
| {z }
Area of S
=
R
3
· 4πR2
=
4πR3
3
.
Answer: The flux is
4πR3
3
.

78. Problem3 ECxiyizt CE.fi
TyYix2
y2 2 2EZE2
fyfe
D8
Methodtm By paramehisation
Tt H
F sit x s t yes t 2 Cs t
Di0EsE2lTg
2EtE2T
FCfCsitDfp.F
recess resins f
fI
fE
wss rsn
GIs
f resins Fraess O

79. off O o 1
I o
N
O O I
F
tEs H
Reis rends e I Kooism
21kg to 1
pythagorean id
F de II
I d t d s Area D
Lpf 2IT 4
q

80. Methods from the definition
ff
ds
fHcxi E ds
m
IS
y7ds Eff
qq.git.ir
8lTO
are

81. 12.9
Taylor’s formula, intro

82. Tangent line through the point (a, f (a))
y = f(a) + f′(a)(x − a)
Tangent plane through the point (a, b, f (a, b))
z = f(a, b) + ∇f(a, b)[
x − a
y − b]
z = f(a, b) + f′
x(a, b)(x − a) + f′
y(a, b)(y − b)
z = f(a, b) + [f′
x(a, b), f′
y(a, b)][
x − a
y − b]
y = mx + b
Ax + By + Cz + D = 0
1 × 2
2 × 1

83. f(a + h) = f(a) + f′(a)h +
f′′(a)
2!
h2
+
f′′′(a)
3!
h3
+ R3(a, h)
f(x) = f(a) + f′(a)(x − a) +
f′′(a)
2!
(x − a)2
+
f′′′(a)
3!
(x − a)3
+ R3(a, x)
f(a + h, b + k) = f(a, b) + ∇f(a, b) ⋅ (h, k) + ?

84. Hessian matrix: the several-variable counterpart of the second derivative
∂2
f
∂x2
∂2
f
∂y∂x
∂2
f
∂x∂y
∂2
f
∂2y
=
ℋ( )
f
[
f′′
11 f′′
12
f′′
21 f′′
22]

85. ∂2
f
∂x2
∂2
f
∂y∂x
∂2
f
∂z∂x
∂2
f
∂x∂y
∂2
f
∂2y
∂2
f
∂z∂y
∂2
f
∂x∂z
∂2
f
∂y∂z
∂2
f
∂2z
=
ℋ( )
f
Hessian matrix: the several-variable counterpart of the second derivative
If then the Hessian matrix is a matrix.
f : ℝn
→ ℝ n × n

86. f′′(a)
2!
h2
1
2
(h, k)T
ℋf(a, b)(h, k)
1
2
(h, k, l)T
ℋf(a, b, c)(h, k, l)
f : ℝ → ℝ
f : ℝ2
→ ℝ
f : ℝ3
→ ℝ

87. Quadratic forms: defined by square matrices
n = 2
n = 3
Q(h, k, l) = f!!
11h2
+ f!!
22k2
+ f!!
33l2
+ 2f!!
12hk + 2f!!
13hl + 2f!!
23kl
o`7ƺ` ` FM` KM b´ K2/ Fp/`iBbF 7Q`K2` .2i ` ip´ ;´M;2` 27i2` p`M/` Ki`BbKmHiB@
THBFiBQM bQK M2/M,
A =
!
a11 a12
a21 a22
, Q(h, k) = (h, k)T
A(h, k)
Q(h, k) = (h, k)
!
a11 a12
a21 a22
!
h
k
= (h, k)
!
ha11 + ka12
ha21 + ka22
= h2
a11 + hka12 + kha21 + k2
a22
A =


a11 a12 a13
a21 a22 a23
a31 a32 a33

 , Q(h, k, l) = (h, k, l)T
A(h, k, l)
Q(h, k, l) = (h, k, l)


a11 a12 a13
a21 a22 a23
a31 a32 a33




h
k
l

 = (h, k, l)


ha11 + ka12 + la13
ha21 + ka22 + la23
ha31 + ka32 + la33

 =
h2
a11 + hka12 + hla13 + kha21 + k2
a22 + kla23 + lha31 + lka32 + l2
a33.
PK Ki`Bb2M A /2bbmiQK ` bvKK2i`BbF U2MHB;i ?mpm//B;QMH2M- HHib´ aij = aji 7ƺ` p`D2 T`
BM/2t2` (i, j)V- FM KM b ii B?QT i2`K2`M bQK BMM2?´HH2` aij K2/ i2`K2`M bQK BMM2?´HH2`
aji U/2 ` KmHiBTHB+2`/2 K2/ bKK FQKTQM2Mi2` p iBHHp tip2FiQ`MVX
Q(h, k) = f!!
11h2
+ f!!
22k2
+ 2f!!
12hk
Q(h, k, l) = f!!
11h2
+ f!!
22k2
+ f!!
33l2
+ 2f!!
12hk + 2f!!
13hl + 2f!!
23kl
o`7ƺ` ` FM` KM b´ K2/ Fp/`iBbF 7Q`K2` .2i ` ip´ ;´M;2` 27i2` p`M/` Ki`BbKmHiB@
THBFiBQM bQK M2/M,
A =
!
a11 a12
a21 a22
, Q(h, k) = (h, k)T
A(h, k)
Q(h, k) = (h, k)
!
a11 a12
a21 a22
!
h
k
= (h, k)
!
ha11 + ka12
ha21 + ka22
= h2
a11 + hka12 + kha21 + k2
a22
A =


a11 a12 a13
a21 a22 a23
a31 a32 a33

 , Q(h, k, l) = (h, k, l)T
A(h, k, l)
Q(h, k, l) = (h, k, l)


a11 a12 a13
a21 a22 a23
a31 a32 a33




h
k
l

 = (h, k, l)


ha11 + ka12 + la13
ha21 + ka22 + la23
ha31 + ka32 + la33

 =
h2
a11 + hka12 + hla13 + kha21 + k2
a22 + kla23 + lha31 + lka32 + l2
a33.
PK Ki`Bb2M A /2bbmiQK ` bvKK2i`BbF U2MHB;i ?mpm//B;QMH2M- HHib´ aij = aji 7ƺ` p`D2 T`
BM/2t2` (i, j)V- FM KM b ii B?QT i2`K2`M bQK BMM2?´HH2` aij K2/ i2`K2`M bQK BMM2?´HH2`
aji U/2 ` KmHiBTHB+2`/2 K2/ bKK FQKTQM2Mi2` p iBHHp tip2FiQ`MVX
o`7ƺ` ` FM` KM b´ K2/ Fp/`iBbF 7Q`K2` .2i ` ip´ ;´M;2` 27i2` p`M/` Ki`BbKmHiB@
THBFiBQM bQK M2/M,
A =
!
a11 a12
a21 a22
, Q(h, k) = (h, k)T
A(h, k)
Q(h, k) = (h, k)
!
a11 a12
a21 a22
!
h
k
= (h, k)
!
a11h + a12k
a21h + a22k
= a11h2
+ a12hk + a21kh + a22k2
A =


a11 a12 a13
a21 a22 a23
a31 a32 a33

 , Q(h, k, l) = (h, k, l)T
A(h, k, l)
Q(h, k, l) = (h, k, l)


a11 a12 a13
a21 a22 a23
a31 a32 a33




h
k
l

 = (h, k, l)


a11h + a12k + a13l
a21h + a22k + a23l
a31h + a32k + a33l

 =
a11h2
+ a12hk + a13hl + a21kh + a22k2
+ a23kl + a31lh + a32lk + a33l2
.
A =
!
a11 a12
a21 a22
, Q(h, k) = (h, k)T
A(h, k)
Q(h, k) = (h, k)
!
a11 a12
a21 a22
!
h
k
= (h, k)
!
a11h + a12k
a21h + a22k
= a11h2
+ a12hk + a21kh + a22k2
A =


a11 a12 a13
a21 a22 a23
a31 a32 a33

 , Q(h, k, l) = (h, k, l)T
A(h, k, l)
Q(h, k, l) = (h, k, l)


a11 a12 a13
a21 a22 a23
a31 a32 a33




h
k
l

 = (h, k, l)


a11h + a12k + a13l
a21h + a22k + a23l
a31h + a32k + a33l

 =
a11h2
+ a12hk + a13hl + a21kh + a22k2
+ a23kl + a31lh + a32lk + a33l2
.

88. 003 h
h b h QCh K
5h46 hkt 4 E
Q h k
00210
th k G f ke QChik d
000
3hIt5kt8
dgneey
4h Kt 2h et 14kt

89. Taylor’s formula, degree 2

90. f( ⃗
a + ⃗
h ) = f( ⃗
a ) + ∇f( ⃗
a ) ⋅ ⃗
h + 1
2
⃗
h T
ℋf( ⃗
a ) ⃗
h + 𝒪(| ⃗
h |3
)
f(a + h, b + k) = f(a, b) + ∇f(a, b) ⋅ (h, k)+
1
2
(h, k)T
ℋf(a, b)(h, k) + 𝒪( h2
+ k2
3
) .
f(a + h) = f(a) + f′(a)h+
1
2
f′′(a)h2
+ 𝒪(h3
)
⇒ F!
(0) = hf!
1(a, b) + kf!
2(a, b)
F!!
(0) = h2
f!!
11(a, b) + 2hkf!!
12(a, b) + k2
f!!
22(a, b)
J2M- 27i2`bQK 6 ` 2M/BK2MbBQM2HH- ;2` hvHQ`b 7Q`K2H bKiB/B;i
f(a + h, b + k) = F(1) = F(0) + F!
(0) · 1 +
F!!(0)
2!
· 12
+ . . .
= f(a, b)
! # $
f(!
a)
+ f!
1(a, b)h + f!
2(a, b)k
! # $
∇f(!
a)·!
h
+
1
2
(f!!
11(a, b)h2
+ 2f!!
12(a, b)hk + f!!
22(a, b)k2
)
! # $
!
h T Hf (!
a)!
h
+ . . .
/ ` Hf (!
a) ` 2bbBMKi`Bb2M
%
f!!
11(a, b) f!!
12(a, b)
f!!
21(a, b) f!!
22(a, b)
UMQi2` ii f!!
12 = f!!
21VX oB ?` ? `H2ii
hvHQ`TQHvMQK2i p Q`/MBM; ip´X
LQi2` ii `2bii2`K2M ?` biQ`H2FbQ`/MBM;2M O((h2 + k2)3/2)- 27i2`bQK h3, h2k, hk2 Q+? k3 HH
` mTT´i #2;` Mb/2 p (h2 + k2)3/2X ` bF`Bp2` D; hvHQ`7Q`K2HM 7ƺ` ip´@ Q+? i`2p`B#2H@
7mMFiBQM2` T´ Ki`Bb7Q`KX 6ƺ` ii 7´ BM ?2H 7Q`KH2` T´ 2M `/- mi2H KM` D; `;mK2Mi2M 7ƺ`
/2`BpiQ`M- K2M /2i bF bi´ f!
1(a, b) QXbXpX B /2i 7ƺ`bi 7HH2i Q+? f!
1(a, b, c) QXbXpX B /2i M/`
7HH2iX
f(a + h, b + k) = f(a, b) + (f!
1, f!
2)
%
h
k
+
1
2
(h, k)
%
f!!
11 f!!
12
f!!
21 f!!
22
%
h
k
+ O((h2
+ k2
)3/2
)
f(a+h, b+k, c+l) = f(a, b, c)+(f!
1, f!
2, f!
3)


h
k
l

+
1
2
(h, k, l)


f!!
11 f!!
12 f!!
13
f!!
21 f!!
22 f!!
23
f!!
31 f!!
32 f!!
33




h
k
l

+O((h2
+k2
+l2
)3/2
)
r?2`2 Hf (!
a) Bb i?2 2bbBM Ki`Bt
!
f!!
11(a, b) f!!
12(a, b)
f!!
21(a, b) f!!
22(a, b)
U a+?r
f!!
12 = f!!
21VX
q2 ?p2 /2`Bp2/ i?2 hvHQ`Ƕb TQHvMQKBH Q7 /2;`22 irQX h?2 `2bi i2`K Bb
h3, h2k, hk2 M/ k3 `2 H2bb i?M Q` 2[mH iQ (h2 + k2)3/2X
` bF`Bp2` D; hvHQ`7Q`K2HM 7ƺ` ip´@ Q+? i`2p`B#2H7mMFiBQM2` T´ K
?2H 7Q`KH2` T´ 2M `/- mi2H KM` D; `;mK2Mi2M 7ƺ` /2`BpiQ`M- K2M
B /2i 7ƺ`bi 7HH2i Q+? f!
1(a, b, c) QXbXpX B /2i M/` 7HH2iX
f(a + h, b + k) = f(a, b) + (f!
1, f!
2)
!
h
k
+
1
2
(h, k)
!
f!!
11 f!!
12
f!!
21 f!!
22
!
h
k
f(a + h, b + k, c + l) = f(a, b, c) + ∇f(a, b, c) ⋅ (h, k, l)+
1
2
(h, k, l)T
ℋf(a, b, c)(h, k, l) + 𝒪( h2
+ k2
+ l2
3
) .

91. 6ƺ` 7mMFiBQM2` bQK ` iBHH` +FHB;i K´M; ;´M;2` /2`Bp2`#` ` 2bbBMKi`Bb2`M bvKK2i`Bb@
F Ua+?r`x bib5V- HHib´ f!!
ij = f!!
ji 7ƺ` HH T` BM/2t2` i, jX . `7ƺ` ` /2M Fp/`iBbF /2H2M
UFp/`iBbF 7Q`K2MV B mip2+FHBM;2M 7ƺHDM/2 7ƺ` ip´@ Q+? i`2p`B#2H7mMFiBQM2`,
Q(h, k) = f!!
11h2
+ f!!
22k2
+ 2f!!
12hk
Q(h, k, l) = f!!
11h2
+ f!!
22k2
+ f!!
33l2
+ 2f!!
12hk + 2f!!
13hl + 2f!!
23kl.
1tKTH2 8RX .2i2`KBM2 hvHQ`Ƕb TQHvMQKBH Q7 b2+QM/ /2;`22 7Q`
f(x, y) = exy
+ x2
+ 2xy3
+ 3y
`QmM/ i?2 TQBMi (2, 0)X q2 +QKTmi2 i?2 T`iBH /2`BpiBp2b,
f!
x = yexy + 2x + 2y3
f!
y = xexy + 6xy2 + 3
f!!
xx = y2exy + 2
f!!
xy = (xy + 1)exy + 6y2
f!!
yy = x2exy + 12xy
⇒
f(2, 0) = 5
f!
x(2, 0) = 4
f!
y(2, 0) = 5
f!!
xx(2, 0) = 2
f!!
xy(2, 0) = 1
f!!
yy(2, 0) = 4
hvHQ`Ƕb TQHvMQKBH Q7 b2+QM/ /2;`22 Bb i?mb,
f(2 + h, k) = 5 + 4h + 5k +
1
2
(2h2
+ 2hk + 4k2
).
Q(h, k) = (h, k)
!
a11 a12
a21 a22
!
h
k
= (h, k)
!
a11h + a12k
a21h + a22k
= a11h2
+ a12hk + a21kh + a22k2
A =


a11 a12 a13
a21 a22 a23
a31 a32 a33

 , Q(h, k, l) = (h, k, l)T
A(h, k, l)
Q(h, k, l) = (h, k, l)


a11 a12 a13
a21 a22 a23
a31 a32 a33




h
k
l

 = (h, k, l)


a11h + a12k + a13l
a21h + a22k + a23l
a31h + a32k + a33l

 =
a11h2
+ a12hk + a13hl + a21kh + a22k2
+ a23kl + a31lh + a32lk + a33l2
.
PK Ki`Bb2M A /2bbmiQK ` bvKK2i`BbF U2MHB;i ?mpm//B;QMH2M- HHib´ aij = aji 7ƺ` p`D2 T`
BM/2t2` (i, j)V- FM KM b ii B?QT i2`K2`M bQK BMM2?´HH2` aij K2/ i2`K2`M bQK BMM2?´HH2`
aji U/2 ` KmHiBTHB+2`/2 K2/ bKK FQKTQM2Mi2` p iBHHp tip2FiQ`MVX
3y
2a12hk
f(a + h, b + k) ≈ f(a, b) + f′
1(a, b)h + f′
2(a, b)k
∇f(a,b)⋅(h,k)
+
1
2
(f′′
11(a, b)h2
+ 2f′′
12(a, b)hk + f′′
22(a, b)k2
)
Q(h,k)=(h,k) Tℋf(a,b)(h,k)
double!

92. Vector fields, some examples
15.1

93. gradient to a scalar field
wind velocity
fluid flow
electric field
magnetic field
gravitational force
Magnitude (length, intensity) and direction

94. 13.1-9
f(x, y) = x2
ye−(x2
+y2
)
x
y
https://www.geogebra.org/m/cXgNb58T
Vector fields in 2D

95. Wind velocity on a weather map

96. https://commons.wikimedia.org/wiki/User:Cronholm144

97. Electric field
https://commons.wikimedia.org/wiki/File:VFPt_cylindrical_magnet_thumb.svg
made by Geek3

98. Magnetic field
https://commons.wikimedia.org/wiki/File:VFPt_cylindrical_magnet_thumb.svg
made by Geek3

99. Functions (or, for plane fields, just P and Q) are
called the component functions of the vector field.
Vector field is continuous / / , … , smooth (has derivatives of all orders)
if its component functions have the same property.
All our fields will be differentiable as many times as needed.
P, Q, R : ℝ3
→ ℝ
C1
C2

100. Functions (or, for plane fields, just P and Q) are
called the component functions of the vector field.
Vector field is continuous / / , … , smooth (has derivatives of all orders)
if its component functions have the same property.
All our fields will be differentiable as many times as needed.
P, Q, R : ℝ3
→ ℝ
C1
C2
Draw 2 examples (1 chaotic, one smooth)

101. Juan Carlos Ponce Campuzano: https://www.geogebra.org/u/jcponce
⃗
F : ℝ2
→ ℝ2
Vector fields
https://www.geogebra.org/m/cXgNb58T
⃗
F : ℝ2
→ ℝ2
seen as points seen as vectors

102. Juan Carlos Ponce Campuzano: https://www.geogebra.org/u/jcponce
⃗
F : ℝ3
→ ℝ3
Vector fields
⃗
F : ℝ3
→ ℝ3
https://www.geogebra.org/m/KKB2Ndez

103. Functions (or, for plane fields, just P and Q) are
called the component functions of the vector field.
Vector field is continuous / / , … , smooth (has derivatives of all orders)
if its component functions have the same property.
All our fields will be differentiable as many times as needed.
P, Q, R : ℝ3
→ ℝ
C1
C2
The domain of a vector field is equal to the least set where all its
component functions are defined (i.e., the intersection of the domains
of all its component functions.)

104. 1 Chaotic
p
p
g
e
r
is
a r
d
H did
I QGiyk2
d
2 Constant
e.g E'Giy 1,2
p
ry n r 7 Party I
n n
y
i i
i i i a
a e n
l I I n I l X
i e n e L
i i i
i i c

105. 3 Fay x
y y
Playtex
run
n Quay y
i n i i n
i
f i
Th T
a e n
I k l I 7 I X
i n e I
i b l I i V
i i c
V

106. 12.2
Functions of several variables, limit and continuity

107. f(x) =
sin x
x
, Df = ℝ∖{0}
lim
x→0
sin x
x
= 1 indeterminate form

108. x
y
a
f(a)
L1
L2
lim
x→a−
f(x) = L1
lim
x→a+
f(x) = L2
lim
x→a
f(x) does not exist (left limit is different from the right limit)
f : ℝ → ℝ
one-sided limits (left and right limit)

109. x
y
a
f(a)
L
lim
x→a−
f(x) = L = lim
x→a+
f(x)
lim
x→a
f(x) = L
f : ℝ → ℝ

110. x
y
a
f(a)
lim
x→a−
f(x) = f(a) = lim
x→a+
f(x)
lim
x→a
f(x) = f(a)
f : ℝ → ℝ
Function f is continuous in a

111. x
y
a
f(a)
L1
L2
lim
x→b−
f(x) = L1
lim
x→c+
f(x) = L2
lim
x→a
f(x) cannot be considered: a is not a limit point of Df
f : ℝ → ℝ
b c

112. x
y
z
1
1
1
(a, b)
z = f(x, y) (a, b, f(a, b))
Df
f : ℝ2
→ ℝ
Limit and continuity in (a, b)

113. lim
(x,y)→(a,b)
f(x, y) = L
The limit of f , as (x, y) approaches (a, b), is L if
1. The point (a, b) is a limit point of the domain of f
* In every neighbourhood of (a, b) you can find a point of the domain of f
different from (a, b); not an isolated point of the domain ;
* It may—but does not have to—belong to the domain of f.
Df
2. The values f (x, y) can get arbitrarily close to L if only
the arguments (x, y) are close enough the point (a, b).

114. x
y
z
1
1
1
(a, b)
z = f(x, y)
Df
(a, b, f(a, b))
L
lim
(x,y)→(a,b)
f(x, y) = L
Limit

115. x
y
z
1
1
1
(a, b)
z = f(x, y) (a, b, f(a, b))
Df
L
Continuity
lim
(x,y)→(a,b)
f(x, y) = f(a, b)

116. (a, b, c), MQi Dmbi 7`QK /Bz2`2Mi /B`2+iBQMb QM i?2 THM2 `QmM/ i?2 TQBMi- #mi HbQ 7`QK Qi?2`
/B`2+iBQMb BM i?2 bm``QmM/BM; bT+2X USB+im`2, MB- rBi? E2vLQi2XV
:` Mbp `/2 Q+? FQMiBMmBi2i 7ƺ` 7mMFiBQM2` 7`´M Rn
iBHH R
JM FM 2MF2Hi #2pBb 7`´M /2}MBiBQM2M ii HH HBMD ` 7mMFiBQM2` f(x, y) = ax + by + c ` FQM@
iBMm2`HB;X .2bbmiQK mTT7vHH2` ;` Mbp `/2M p 7mMFiBQM2` p ~2` p`B#H2` bKK ` FM2`2;H2`
bQK 7ƺ` 7mMFiBQM2` p 2M p`B#2H- pBHF2i H2/2` iBHH K´M; Mv FQMiBMm2`HB; 7mMFiBQM2`Xj
A7 HBK
!
x→!
a
f(
x) = L M/ HBK
!
x→!
a
g(
x) = M, i?2M,
HBK
!
x→!
a
(f(
x) + g(
x)) = L + M, HBK
!
x→!
a
f(
x)g(
x) = LM, HBK
!
x→!
a
f(
x)
g(
x)
=
L
M
(B7 M != 0)
A7- KQ`2Qp2`- F : R → R Bb 7mM+iBQM Q7 QM2 p`B#H2 M/ F Bb +QMiBMmQmb BM t = L i?2M,
HBK
!
x→!
a
F(f(
x)) = F(L).
.2M bBbi `2;2HM b2` FMbF2 KBM/`2 bD HpFH`i mi M /2 M/`- K2M D; i`Q` ii /2 ~2bi bFmHH2
Mp M/ /2M BMimBiBpiX ` FQKK2` 2t2KT2H T´ Mp M/MBM; p `2;H2`M QpM- K2/ f = f(x, y)X
JM FM 7ƺ`bi´b iBHH KT /2bb `2;H2` #` QK mii`v+F2i ` #2bi Ki- 7ƺ` #` B b´/M 7HH FM
KM Fƺ` BMb iiMBM;,
j
JX?XX `2;H2`M RĜ9 pBb` KM 2MF2Hi ii #HXX TQHvMQK7mMFiBQM2`- `iBQM2HH 7mMFiBQM2` Q+? 2M ?2H /2H M/`
` FQMiBMm2`HB;c KM bF #` Fi bB; 7ƺ` Q#2bi K/ mii`v+F- K2M pB H ` Qbb bi`t ?m` KM FM ?Mi2` /2bbX
Plug-in cases

117. Continuous functions of two variables
Constant functions f (x, y) = c and linear functions f (x, y) = ax + by + c (proof from the definition)
The theorem gives more continuous functions: polynomials and rational functions
p(x, y) = 3x + 6y − 7
p(x, y) = 3x2
+ 6y2
− 7xy − x + 6y + 14
p(x, y) = x3
+ y3
+ x2
y + xy2
− 3x2
+ 6y2
− 7xy − 7x − y + 14
p(x, y) = 3x4
+ x2
y2
− x2
y − x2
+ xy − 2x + 5y − 9
degree 1
degree 2
degree 3
degree 4
Same for polynomials in more variables:
p(x, y, z) = x5
+ 7y5
− 5x3
yz − x2
yz + 2x2
z2
− xyz + z2
− x + 1

118. Plug-in cases
RX
HBK
(x,y)→(2,3)
(4x + 2y + x2
) = 4 · 2 + 2 · 3 + 22
= 8 + 6 + 4 = 18.
kX
HBK
(x,y)→(5,2)
x2
y3
= 52
· 23
= 25 · 8 = 200.
jX
HBK
(x,y)→(3,2)
x + 2y
x2 − y
=
3 + 2 · 2
32 − 2
=
7
7
= 1.
9X
HBK
(x,y)→(π,6)
bBM
x
y
= bBM
π
6
=
1
2
.
2`2 r2 ?p2 F(t) = bBM t M/ f(x, y) = x
y X h?2 7mM+iBQM F Bb /2}M2/ 7Q` HH `2H MmK#2`b
M/ i?2 7mM+iBQM f Bb /2}M2/ BM i?2 r?QH2 THM2 2t+2Ti QM i?2 x@tBb #2+mb2 i?2`2 y = 0X
q2 ;2i L = HBK(x,y)→(π,6)
x
y = π
6 X h?2 TB+im`2 BHHmbi`i2b ?Qr i?2 +QKTQbBiBQM rQ`FbX
R2 R R
f F
F ∘ f
(x, y) t
f(x, y) =
x
y
F(t) = sin t
F ∘ f(x, y) = F( f(x, y)) = F
(
x
y)
= sin
x
y
6B;m` Rj, M 2tKTH2 7Q` i?2 `mH2 HBK
x→
a F(f(
x)) = F(L)X USB+im`2, MB- rBi? E2vLQi2XV
RX
HBK
(x,y)→(2,3)
(4x + 2y + x2
) = 4 · 2 + 2 · 3 + 22
= 8 + 6 + 4 = 18.
kX
HBK
(x,y)→(5,2)
x2
y3
= 52
· 23
= 25 · 8 = 200.
jX
HBK
(x,y)→(3,2)
x + 2y
x2 − y
=
3 + 2 · 2
32 − 2
=
7
7
= 1.
9X
HBK
(x,y)→(π,6)
bBM
x
y
= bBM
π
6
=
1
2
.
2`2 r2 ?p2 F(t) = bBM t M/ f(x, y) = x
y X h?2 7mM+iBQM F Bb /2}M2/ 7Q` HH `
M/ i?2 7mM+iBQM f Bb /2}M2/ BM i?2 r?QH2 THM2 2t+2Ti QM i?2 x@tBb #2+mb2
q2 ;2i L = HBK(x,y)→(π,6)
x
y = π
6 X h?2 TB+im`2 BHHmbi`i2b ?Qr i?2 +QKTQbBiBQM
R2 R R
f F
F ∘ f
(x, y) t
f(x, y) =
x
y
F(t) = sin t
F ∘ f(x, y) = F( f(x, y)) = F
(
x
y )
= sin
x
y
6B;m` Rj, M 2tKTH2 7Q` i?2 `mH2 HBK
x→
a F(f(
x)) = F(L)X USB+im`2, MB- rBi?

119. f : ℝ → ℝ f : ℝ2
→ ℝ
Limit in a point exists if the left
limit and the right limit exist and
they are the same. Just two things
to control: nice!
Limit in a point exists if the values
along ALL the curves towards the
point approach the same number:
COMPLICATED!
(a, b)

120. Two different approaches
To show that limit in the point does not exist: it is enough to
point two directions (curves) leading to different results.
To show that the limit in the point does exist: change to
polar coordinates around this point and show that limit with
r tending to zero exists and that it does not depend on
𝜽
,
meaning we get the same value along all directions.

121. Example 456,0
does not exist
my Df Rh Cosa
approach 0,0
along the
y axis
cotton co y
77 o
7 X
T ferry 0
along y axis
approach 0,0
along the x axis
x o
f Gio 0
artifice.gr ye fimo
I 1

122. x z
0 1
-1
values of the
function are
shown
on the z-axis
6B;m` R9, GBKBi HBK(x,y)→(0,0)
2xy
x2+y2 /Q2b MQi 2tBbiX USB+im`2, MB- rBi? E2vLQi2XV
1tKTH2 kyX a?Qr i?i i?2 HBKBi HBK
(x,y)→(0,0)
x4y2
(x4 + y2)2
/Q2b MQi 2tBbiX
q2 i2bi r?i ?TT2Mb HQM; i?2 7QHHQrBM; bi`B;?i HBM2b, (x, y) = (t, kt) M/ (x, y) = (0, t)X h?Bb
;Bp2b,
HBK
t→0
t4(kt)2
(t4 + (kt)2)2
= HBK
t→0
k2t6
(t4 + k2t2)2
= HBK
t→0
k2t2
(t2 + k2)2
= 0, k != 0
6Q` (x, y) = (0, t) M/ (x, y) = (t, 0) i?2 7mM+iBQM ?b pHm2 x2`Q- bQ i?2 HBKBi Bb 2[mH iQ x2`Q
HQM; i?2b2 HBM2bX Qr2p2`- r?2M r2 2tKBM2 r?i ?TT2Mb HQM; i?2 T`#QH (x, y) = (t, t2)
r2 ;2i /Bz2`2Mi `2bmHi,
HBK
t→0
t4(t2)2
(t4 + (t2)2)2
= HBK
t→0
t8
(2t4)2
=
1
4
, bQ i?2 HBKBi /Q2b MQi 2tBbiX
A 7HH2M /´ KM 7´` bKK ;` Mbp `/2 H M;b QHBF HBMD2` FM KM KBbbi MF ii ;` Mbp `/2i
2tBbi2``X 6ƺ` ii #2pBb ii /2i ` 7HH2i- Mp M/2` KM TQH `i FQQ`/BMi#vi2 Q+? KM pBb` ii
QK pbi´M/2i K2HHM Q`B;Q U2HH2` 2M MMM TmMFi / ` KM mM/2`bƺF2` ;` Mbp `/2i- K2M Q7ibi
` /2i Q`B;QV Q+? TmMFi2` bQK M `K` bB; Q`B;Q ;´` KQi MQHH- /´ ;´` p2M pbi´M/2i K2HHM
p `/2M B /2bb TmMFi2` Q+? /2i iBHHi MFi ;` Mbp `/2i KQi MQHH- bQK B 7ƺHDM/2 2t2KT2HX
xy

123. probbemta.ua
t5y I
Problem 4
Ey
Lim
Gay co o X4ty2
Steps check the coordinate axes
along x axis t O
thing
oe

124. along the y axis Ost
Fino't o
StepI check all the lines
y
kx through the origin
t key's
k 0
the
mo tE hm p
K2
steps check any Ct t2
Eso't ee
L
an iaa edn

125. Cx y Ct t
mo't
t I

126. 1tKTH2 kRX a?Qr i?i HBK
(x,y)→(0,0)
xy
!
x2 + 3y2
= 0X
kd
q2 i2bi HH i?2 bi`B;?i HBM2b (x, y) = (t, kt) M/ (x, y) = (0, t)- M/ (x, y) = (t, 0)- ;2iiBM; `2bmHi
x2`Q 7Q` HH Q7 i?2KX h?Bb Bb- ?Qr2p2`- MQi T`QQ7 #2+mb2 i?2`2 `2 Qi?2` Ti?b iQ i?2 Q`B;BM-
MQi Dmbi i?2 bi`B;?i QM2bX
hQ b?Qr i?i i?2 HBKBi 2tBbib M/ Bb 2[mH iQ x2`Q- r2 +?M;2 iQ TQH` +QQ`/BMi2b,
x = r +Qb θ, y = r bBM θ M/ b?Qr U++Q`/BM; iQ i?2 i?2Q`v QM TX k8V i?i
HBK
r→0
r +Qb θ · r bBM θ
!
(r +Qb θ)2 + 3(r bBM θ)2
= 0 MQ Kii2` pHm2 Q7 θ.
r +Qb θ · r bBM θ
!
(r +Qb θ)2 + 3(r bBM θ)2
=
r2 +Qb θ · bBM θ
r
!
(+Qb θ)2 + 3(bBM θ)2
=
r2 +Qb θ · bBM θ
r
!
1 + 2(bBM θ)2
=
r +Qb θ · bBM θ
!
1 + 2(bBM θ)2
.
AM bi2T Hbi #v QM2 r2 mb2/ i?2 Svi?;Q`2M A/2MiBiv BM i?2 /2MQKBMiQ`X q2 `2 MQr `2/v iQ
b?Qr i?i i?2 2tT`2bbBQM i2M/b iQ x2`Q r?2M r i2M/b iQ x2`Q- M/ i?Bb MQ Kii2` r?i i?2 pHm2
Q7 θX q2 +M mb2 7Q` BMbiM+2 i?2 b[m22xBM; i?2Q`2KX
X
ii pBb ii 2ii ;` Mbp `/2 2tBbi2`` ` Q7i ;MbF bp´`iX A#HM/ FM pB /Q+F `2/m+2` T`Q#H2K2i

127. Problem 0
a Yo o
O
x r cos Q
y
r sine
oI IsinFF
Pythagorean
identity
tying fg.gr
a
we use
O
o
gykG
and less
of a
1 cos f f E I

128. squeezing
theorem
0 E r cos ol f r
r
of T two
O O

129. Problemt Adams
12.2pr.IQ
How can the function
fGig txEyEI CqyH
be defined in the
origin so that
rum
it becomes continuous at all
points of the xy plane
Catch fCxI
hh
Of _RYO
Icx1
x O
X O
the
Lim
X't y 3y3
Gyncao
exists

130. I
g
L
W
T T
polar coordinates
noose
code sins y
rsinolim
limr4cos
r 0 r 0
I bounded
0 between
D sand
GyfisTao fGiy L
s
fcx.ge
xEIx CxiyHco
g1
Cxiy 6 o
Problem 2 Show that the
following
limit does not exist
Sinay
him Adams 12.2 9
Gaynor

131. iE
tsin
E
e
ng
z
z o y x t
E
gcx.sk
IyT
a
x
t.tt
tIEEy
x

132. By. Dr Hania Uscka-Wehlou
The plane ℝ2 and the 3-space ℝ3 : points and vectors
ch 10

133. Cartesian coordinate system
Cartesian coordinates = rectangular coordinates
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
origin
quadrant I
quadrant II
quadrant III quadrant IV
x-axis
y-axis
René Descartes
1596—1650

134. Cartesian coordinate system
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
A = (1, 1)
B = (5, 1)

135. x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
A = (1, 1)
B = (5, 1)
!
AB
x = 4
y = 2
!
AB = (5, 1) (1, 1) = (5 1, 1 ( 1)) = (4, 2)
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
A = (1, 1)
B = (5, 1)
!
AB
x = 4
y = 2
!
AB = (5, 1) (1, 1) = (5 1, 1 ( 1)) = (4, 2)
Change in x and change in y
coordinates
components

136. x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
A = (1, 1)
B = (5, 1)
!
BA
E = ( 2, 1)
F = ( 2, 3)
!
EF
G = (4, 3)
H = ( 2, 3)
!
GH
!
BA = ( 4, 2) =
!
AB
6 5 4 3 2 1 0 1 2 3 4 5
4
3
2
1
0
1
2
A = (1, 1)
B
!
BA
E = ( 2, 1)
F = ( 2, 3)
!
EF
!
BA = ( 4, 2) =
!
AB
x
4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
A = (1, 1)
!
BA
= ( 2, 1)
= ( 2, 3)
!
EF
!
BA = ( 4, 2) =
!
AB
!
EF = (0, 2)
!
GH = ( 6, 0)

137. x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
A = (1, 1)
B = (5, 1)
!
AB
C = ( 3, 1)
!
CD
D = (1, 3)
x = 4
y = 2
!
CD = (1, 3) ( 3, 1) = (1 ( 3), 3 1) = (4, 2),
x
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
A = (1, 1)
B = (5, 1)
!
AB
C = ( 3, 1)
!
CD
D = (1, 3)
x = 4
y = 2
!
CD = (1, 3) ( 3, 1) = (1 ( 3), 3 1) = (4, 2),
Identification of vectors with the same coordinates

138. ⃗
v = (v1, v2) v = (v1, v2)
Notation

139. Vector addition and Vector scaling (scalar multiplication)
Example. If u = (3, 1) and v = (2, 4) are two vectors.
x
y
2 1 0 1 2 3 4 5 6
2
1
0
1
2
3
4
5
6
v
u

140. Example. If u = (3, 1) and v = (2, 4) are two vectors then
u + v = (5, 5).
x
y
2 1 0 1 2 3 4 5 6
2
1
0
1
2
3
4
5
6
u
u
v
u + v Vector addition

141. Vector scaling
(scalar multiplication)
Example. If u = (3, 1) and v = (2, 4) are two vectors then
1
2
u = (
3
2
,
1
2
),
1
2
v = ( 1, 2),
3
2
v = (3, 6).
x
y
2 1 0 1 2 3 4 5 6
2
1
0
1
2
3
4
5
6
u
v

142. Example. If u = (3, 1) and v = (2, 4) are two vectors then
u v = u + ( v) = (1, 3).
x
y
2 1 0 1 2 3 4 5 6
2
1
0
1
2
3
4
5
6
u
v
u v
Vector subtraction

143. Identification of points and position vectors
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
!
OB
B = (5, 1)

144. Idag kommer vi att arbeta enbart i det tredimensionella rummet. Det
betecknas R3
. Alla punkter i rummet har tre koordinater, som i bilden
nedan: P(3, 0, 5) och Q( 5, 5, 7). Origo har koordinater (0, 0, 0). Generellt
heter koordinaterna (x, y, z) och axlarna ritas som i bilden nedan (z-axeln
vertikalt; x och y-axlarna kan ritas som i bilderna nedan: i båda fall ligger
x, y och z- axlarna enligt högerhandsregeln). Vektorer i R3
har också tre
koordinater: x, y och z som beskriver förflyttning i x, y och z leden.
10 5
5 10
10
5
5 10
1
5
P(3, 0, 5)
Q( 5, 5, 7)
(0, 0, 0)
Origin
x
y
z
z
2
4
2
0
3
5
2
4
x
y
z
3
5 =
2
4
0
2
3
3
5
10 5
5 10
10
5
5 10
1
(0, 0, 0)
Origin
x
y
z
2
4
2
2
0
3
5
2
4
2
0
1
3
5
2
4
x
y
z
3
5 =
2
4
0
2
3
3
5
Cartesian coordinate system

145. ℝn
⃗
x = (x1, x2, …, xn)
x = (x1, x2, …, xn)

146. ℝn
⃗
x = (x1, x2, …, xn)
x = (x1, x2, …, xn)
⃗
y = (y1, y2, …, yn)
⃗
x + ⃗
y = (x1 + y1, x2 + y2, …, xn + yn)
λ ⃗
x = (λx1, λx2, … , λxn)

147. x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
P = (x1, y1)
Q = (x2, y2)
|x2 x1|
|y2 y1|
d(P, Q) =
p
(x1 x2)2 + (y1 y2)2
=
p
v2
1 + v2
2
Distance between points = length of a vector
Pythagorean theorem

148. x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
P = (x1, y1)
Q = (x2, y2)
|x2 x1|
|y2 y1|
d(P, Q) =
p
(x1 x2)2 + (y1 y2)2
=
p
v2
1 + v2
2
Distance between points = length of a vector
d(P, Q) = ∥ ⃗
PQ ∥ = ∥ ⃗
OQ − ⃗
OP ∥

149. Distance between points = length of a vector
P = (x1, y1, z1), Q = (x2, y2, z2)
d(P, Q) = | ⃗
PQ | = (x2 − x1)2
+ (y2 − y1)2
+ (z2 − z1)2

150. | ⃗
v | |v|
∥ ⃗
v ∥ ∥v∥
Length / Norm

151. ℝn
x = (x1, x2, …, xn)
y = (y1, y2, …, yn)
d(x, y) = (y1 − x1)2
+ (y2 − x2)2
+ … + (yn − xn)2

152. Scaling: a number “times” a vector gives a vector
Dot product: a vector “times” a vector gives a number
Cross product: a vector “times” a vector gives a vector
Scalar triple product: a vector “times” a vector “times” a vector gives a number
αv
v ⋅ u
v × u
w ⋅ (v × u)
only in ℝ3
only in ℝ3

153. u
v
α
Dot product in R2
and R3
u · v = kuk · kvk cos(↵), ↵ 2 [0, ⇡]
Not for computations, but for geometrical interpre-
tation in R2
or in R3
. Vectors u and v:
• are orthogonal i↵ u · v = 0,
• form an acute angle i↵ u · v 0,
• form an obtuse angle i↵ u · v 0.
Dot product gives us a test for orthogonality.
Moreover: u · u = kuk2
.
Cosine theorem gives
u · v =
kuk2
+ kvk2
kv uk2
2
.
If u = (u1, u2, u3) and v = (v1, v2, v3) then
Dot product in R2
and R3
u · v = kuk · kvk cos(↵), ↵ 2 [0, ⇡]
Not for computations, but for geometrical interpre-
tation in R2
or in R3
. Vectors u and v:
• are orthogonal i↵ u · v = 0,
• form an acute angle i↵ u · v 0,
• form an obtuse angle i↵ u · v 0.
Dot product gives us a test for orthogonality.
Moreover: u · u = kuk2
.
Cosine theorem gives
3.2.2 För alla u, v, w 2 Rn
och k 2 R gäller att
= v · u
+ v) = w · u + w · v
) = (ku) · v
= u · u.
sats är viktig:
För u, v 2 Rn
gäller att
|u · v|  kuk · kvk.
et, den så kallade Cauchy–Schwarz olikhet, låter oss definiera
an två vektorer u och v som följande tal mellan 0 och ⇡:
✓ = arccos
✓
u · v
kuk · kvk
◆
.
u · v = u1v1 + u2v2 + . . . + unvn.
Theorem. 3.2.2 För alla u, v, w 2 Rn
och k 2 R gäller att
• u · v = v · u
• w · (u + v) = w · u + w · v
• k(u · v) = (ku) · v
• kuk2
= u · u.
Följande sats är viktig:
Theorem. För u, v 2 Rn
gäller att
|u · v|  kuk · kvk.
Denna olikhet, den så kallade Cauchy–Schwarz olikhet, låter oss de
vinkeln mellan två vektorer u och v som följande tal mellan 0 och ⇡:
✓
u · v
◆

154. u
v
α
Dot product in R2
and R3
u · v = kuk · kvk cos(↵), ↵ 2 [0, ⇡]
Not for computations, but for geometrical interpre-
tation in R2
or in R3
. Vectors u and v:
• are orthogonal i↵ u · v = 0,
• form an acute angle i↵ u · v 0,
• form an obtuse angle i↵ u · v 0.
Dot product gives us a test for orthogonality.
Moreover: u · u = kuk2
.
Cosine theorem gives
u · v =
kuk2
+ kvk2
kv uk2
2
.
If u = (u1, u2, u3) and v = (v1, v2, v3) then
Dot product in R2
and R3
u · v = kuk · kvk cos(↵), ↵ 2 [0, ⇡]
Not for computations, but for geometrical interpre-
tation in R2
or in R3
. Vectors u and v:
• are i↵ u · v = 0,
• form an acute angle i↵ u · v 0,
• form an obtuse angle i↵ u · v 0.
Dot product gives us a test for orthogonality.
Moreover: u · u = kuk2
.
Cosine theorem gives
Dot product in R2
and R3
u · v = kuk · kvk cos(↵), ↵ 2 [0, ⇡]
geometrical interpretation, vectors u and v:
• are orthogonal i↵ u · v = 0,
• form an acute angle i↵ u · v 0,
• form an obtuse angle i↵ u · v 0.
Dot product gives us a test for orthogonality.
Moreover: u · u = kuk2
.
Cosine theorem gives
u
v
α

155. u
v
α
Dot product in R2
and R3
u · v = kuk · kvk cos(↵), ↵ 2 [0, ⇡]
Not for computations, but for geometrical interpre-
tation in R2
or in R3
. Vectors u and v:
• are orthogonal i↵ u · v = 0,
• form an acute angle i↵ u · v 0,
• form an obtuse angle i↵ u · v 0.
Dot product gives us a test for orthogonality.
Moreover: u · u = kuk2
.
Cosine theorem gives
u · v =
kuk2
+ kvk2
kv uk2
2
.
If u = (u1, u2, u3) and v = (v1, v2, v3) then
Dot product in R2
and R3
u · v = kuk · kvk cos(↵), ↵ 2 [0, ⇡]
Not for computations, but for geometrical interpre-
tation in R2
or in R3
. Vectors u and v:
• are orthogonal i↵ u · v = 0,
• form an acute angle i↵ u · v 0,
• form an obtuse angle i↵ u · v 0.
Dot product gives us a test for orthogonality.
Moreover: u · u = kuk2
.
Cosine theorem gives
Not for computations, but for geometrical interpre-
tation in R2
or in R3
. Vectors u and v:
• are orthogonal i↵ u · v = 0,
• form an acute angle i↵ u · v 0,
• form an obtuse angle i↵ u · v 0.
Dot product gives us a test for orthogonality.
Moreover: u · u = kuk2
.
u · v =
kuk2
+ kvk2
kv uk2
2
.
If u = (u1, u2, u3) and v = (v1, v2, v3) then
u · v = u1v1 + u2v2 + u3v3.
Perfect for computations.
tation in R or in R . Vectors u and v:
• are orthogonal i↵ u · v = 0,
• form an acute angle i↵ u · v 0,
• form an obtuse angle i↵ u · v 0.
Dot product gives us a test for orthogonality.
Moreover: u · u = kuk2
.
Cosine theorem gives
u · v =
kuk2
+ kvk2
kv uk2
2
.
If u = (u1, u2, u3) and v = (v1, v2, v3) then
u · v = u1v1 + u2v2 + u3v3.
Perfect for computations.

156. Dot product: properties
1 2 3 1 2 3
u · v = u1v1 + u2v2 + u3v3.
Perfect for computations.
Theorem. For all u, v, w 2 Rn
and k 2 R holds
• u · v = v · u
• w · (u + v) = w · u + w · v
• k(u · v) = (ku) · v
• kuk2
= u · u.

157. Example. Show that the vectors u = (4, 1) and v = ( 1, 4) are orthogonal:
u · v = u1v1 + u2v2 = 4 · ( 1) + 1 · 4 = 4 + 4 = 0.
Two straight lines with slopes m1 and m2 are perpendicular i↵ m1 ·m2 = 1.
Vector u = (4, 1) has the same direction as the line y = 1
4
x 5
4
(they both
have the slope m1 = 1
4
) and vector v = ( 1, 4) has the same direction as
theline y = 4x + 3 (they both have the slope m2 = 4). We huve just
shown that the vectors are orthogonal (and it is also clear from the picture).
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
2
1
0
1
2
3
4
1 5
y = 4x + 3
v = ( 1, 4)
u = (4, 1)
4
theline y = 4x + 3 (they both have the slope m2 = 4). We huve just
shown that the vectors are orthogonal (and it is also clear from the picture).
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
y = 1
4
x 5
4
y = 4x + 3
v = ( 1, 4)
u = (4, 1)

158. Vectors in the plane orthogonal to (a, b)
(a, b)
(b, − a)
(−b, a)
These two and all their scalings

159. Vectors in the 3-space orthogonal to (a, b, c)
(−b, a, 0)
(b, − a, 0)
(0, c, − b)
(0, − c, b)
(c, 0, − a)
(−c, 0, a)
and many, many more

160. ⃗
x = (x1, x2, …, xn)
⃗
y = (y1, y2, …, yn)
⃗
x ⋅ ⃗
y = x1y1 + x2y2 + … + xnyn

161. ADAMS ESSEX: Calculus: a Complete Course, 9th Edition. Chapter 10 – page 582 October 15, 2016
582 CHAPTER 10 Vectors and Coordinate Geometry in 3-Space
PROOF Refer to Figure 10.20 and apply the Cosine Law to the triangle with the ar-
rows u, v, and u � v as sides:
juj2
C jvj2
� 2juj jvj cos D ju � vj2
D .u � v/ .u � v/
D u .u � v/ � v .u � v/
D u u � u v � v u C v v
D juj2
C jvj2
� 2u v
Hence, jujjvj cos D u v, as claimed.
u � v
u
v
v
u � v
u
Figure 10.20 Applying the Cosine Law to
a triangle reveals the relationship between
dot the product and angle between vectors
EXAMPLE 4 Find the angle between the vectors u D 2i C j � 2k and v D
3i � 2j � k.
Solution Solving the formula u v D jujjvj cos for , we obtain
D cos�1 u v
jujjvj
D cos�1
.2/.3/ C .1/.�2/ C .�2/.�1/
3
p
14
D cos�1 2
p
14
57:69ı
:
It is sometimes useful to project one vector along another. We define both scalar and
vector projections of u in the direction of v:
DEFINITION
4
Scalar and vector projections
The scalar projection s of any vector u in the direction of a nonzero vector v
is the dot product of u with a unit vector in the direction of v. Thus, it is the
number
s D
u v
jvj
D juj cos ;
where is the angle between u and v.
The vector projection, uv, of u in the direction of v (see Figure 10.21)
is the scalar multiple of a unit vector O
v in the direction of v, by the scalar
projection of u in the direction of v; that is,
vector projection of u along v D uv D
u v
jvj
O
v D
u v
jvj2
v:
Note that jsj is the length of the line segment along the line of v obtained by dropping
perpendiculars to that line from the tail and head of u. (See Figure 10.21.) Also, s is
negative if 90ı
.
It is often necessary to express a vector as a sum of two other vectors parallel and
perpendicular to a given direction.
uv
v
u
s
Figure 10.21 The scalar projection s and
the vector projection uv of vector u along
vector v
EXAMPLE 5 Express the vector 3i C j as a sum of vectors u C v, where u is
parallel to the vector i C j and v is perpendicular to u.
Solution
METHOD I (Using vector projection) Note that u must be the vector projection of
3i C j in the direction of i C j. Thus,
u D
.3i C j/ .i C j/
ji C jj2
.i C j/ D
4
2
.i C j/ D 2i C 2j
v D 3i C j � u D i � j:
ADAM
9780134154367_Calculus 602 05/12/16 3:48 pm

162. ADAMS ESSEX: Calculus: a Complete Course, 9th Edition. Chapter 10 – page 583 October 15, 2016
SECTION 10.2: Vectors 583
METHOD II (From basic principles) Since u is parallel to iCj and v is perpendicular
to u, we have
u D t.i C j/ and v .i C j/ D 0;
for some scalar t. We want u C v D 3i C j. Take the dot product of this equation with
i C j:
u .i C j/ C v .i C j/ D .3i C j/ .i C j/
t.i C j/ .i C j/ C 0 D 4:
Thus 2t D 4, so t D 2. Therefore,
u D 2i C 2j and v D 3i C j � u D i � j:
Vectors in n-Space
All the above ideas make sense for vectors in spaces of any dimension. Vectors in Rn
can be expressed as linear combinations of the n unit vectors
e1 from the origin to the point .1; 0; 0; : : : ; 0/
e2 from the origin to the point .0; 1; 0; : : : ; 0/
:
:
:
en from the origin to the point .0; 0; 0; : : : ; 1/:
These vectors constitute a standard basis in Rn
. The n-vector x with components
x1; x2; : : : ; xn is expressed in the form
x D x1e1 C x2e2 C C xnen:
The length of x is jxj D
p
x1
2 C x2
2 C C xn
2. The angle between two vectors x
and y is
D cos1 x y
jxjjyj
;
where
x y D x1y1 C x2y2 C C xnyn:
We will not make much use of n-vectors for n 3, but you should be aware that
everything said up until now for 2-vectors or 3-vectors extends to n-vectors.
9780134154367_Calculus 603 05/12/16 3:48 pm

163. Cross product
If u = (u1, u2, u3) and v = (v1, v2, v3) are two elements of R3
we define their
cross product as
u ⇥ v = (u2v3 u3v2, u3v1 u1v3, u1v2 u2v1).
Här är det alltså: en vektor gånger en vektor ger en annan vektor. I skalärprodukt
var det en vektor gånger en vektor ger en skalär, alltså ett tal. Definitionen
för vektorprodukt kan vara lite svår att komma ihåg men det finns minnes-
regler. En av dem är:
u ⇥ v =
e1 e2 e3
u1 u2 u3
v1 v2 v3
.
Här är e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Determinanten är inte
egentligen definierad för vektorer, men om vi räknar med Sarrus regel blir
det rätt svar.
Example. Om u = (1, 1, 0) och v = (1, 1, 1) är
u ⇥ v = (u2v3 u3v2, u3v1 u1v3, u1v2 u2v1).
är det alltså: en vektor gånger en vektor ger en annan vektor. I skalä
det en vektor gånger en vektor ger en skalär, alltså ett tal. Definit
vektorprodukt kan vara lite svår att komma ihåg men det finns m
er. En av dem är:
u ⇥ v =
e1 e2 e3
u1 u2 u3
v1 v2 v3
.
är e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Determinanten ä
ntligen definierad för vektorer, men om vi räknar med Sarrus reg
rätt svar.
If u = (u1, u2, u3) and v = (v1, v2, v3) are two elements of R3
we define their
cross product as
u ⇥ v = (u2v3 u3v2, u3v1 u1v3, u1v2 u2v1).
Här är det alltså: en vektor gånger en vektor ger en annan vektor. I skalärprodukt
var det en vektor gånger en vektor ger en skalär, alltså ett tal. Definitionen
för vektorprodukt kan vara lite svår att komma ihåg men det finns minnes-
regler. En av dem är:
u ⇥ v =
e1 e2 e3
u1 u2 u3
v1 v2 v3
.
Här är e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Determinanten är inte
egentligen definierad för vektorer, men om vi räknar med Sarrus regel blir
det rätt svar.

164. Cross product
If u = (u1, u2, u3) and v = (v1, v2, v3) are two elements of R3
we define their
cross product as
u ⇥ v = (u2v3 u3v2, u3v1 u1v3, u1v2 u2v1).
Här är det alltså: en vektor gånger en vektor ger en annan vektor. I skalärprodukt
var det en vektor gånger en vektor ger en skalär, alltså ett tal. Definitionen
för vektorprodukt kan vara lite svår att komma ihåg men det finns minnes-
regler. En av dem är:
u ⇥ v =
e1 e2 e3
u1 u2 u3
v1 v2 v3
.
Här är e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Determinanten är inte
egentligen definierad för vektorer, men om vi räknar med Sarrus regel blir
det rätt svar.
Example. Om u = (1, 1, 0) och v = (1, 1, 1) är
u ⇥ v = (u2v3 u3v2, u3v1 u1v3, u1v2 u2v1).
är det alltså: en vektor gånger en vektor ger en annan vektor. I skalä
det en vektor gånger en vektor ger en skalär, alltså ett tal. Definit
vektorprodukt kan vara lite svår att komma ihåg men det finns m
er. En av dem är:
u ⇥ v =
e1 e2 e3
u1 u2 u3
v1 v2 v3
.
är e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Determinanten ä
ntligen definierad för vektorer, men om vi räknar med Sarrus reg
rätt svar.
If u = (u1, u2, u3) and v = (v1, v2, v3) are two elements of R3
we define their
cross product as
u ⇥ v = (u2v3 u3v2, u3v1 u1v3, u1v2 u2v1).
Här är det alltså: en vektor gånger en vektor ger en annan vektor. I skalärprodukt
var det en vektor gånger en vektor ger en skalär, alltså ett tal. Definitionen
för vektorprodukt kan vara lite svår att komma ihåg men det finns minnes-
regler. En av dem är:
u ⇥ v =
e1 e2 e3
u1 u2 u3
v1 v2 v3
.
Här är e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Determinanten är inte
egentligen definierad för vektorer, men om vi räknar med Sarrus regel blir
det rätt svar.
x y
z
⃗
i ⃗
j
⃗
k
⃗
i × ⃗
j = ⃗
k

165. u ⇥ v = u1 u2 u3
v1 v2 v3
.
Här är e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1). Determinanten är inte
egentligen definierad för vektorer, men om vi räknar med Sarrus regel blir
det rätt svar.
Example. If u = (1, 1, 0) and v = (1, 1, 1) then
u ⇥ v =
e1 e2 e3
1 1 0
1 1 1
=
✓
1 0
1 1
,
1 0
1 1
,
1 1
1 1
◆
= (1, 1, 2).
We also get
v ⇥ u = ( 1, 1, 2).
Theorem. If u, v, w 2 R3
then
• u ⇥ v = v ⇥ u (anti-commutativity).
• u · (u ⇥ v) = 0.

166. We also get
v ⇥ u = ( 1, 1, 2).
Theorem. If u, v, w 2 R3
then
• u ⇥ v = v ⇥ u (anti-commutativity).
• u · (u ⇥ v) = 0.
• ku ⇥ vk2
= kuk2
kvk2
(u · v)2
.
• u ⇥ (v + w) = u ⇥ v + u ⇥ w (left distributive law)
• (v + w) ⇥ u = v ⇥ u + w ⇥ u (right distributive law)
• k(u ⇥ v) = (ku) ⇥ v = u ⇥ (kv)
• u ⇥ 0 = 0, u ⇥ u = 0.
Cross product: properties
¬√
¬√

167. u
v
α
Geometrical interpretation of cross product
The lengths of the cross product of u, v 2 R3
being ku⇥vk computes in the
following way:
ku ⇥ vk = kuk · kvk sin(↵)
Cross product u ⇥ v is a vector which:
• is orthogonal to both u and v,
• has length equal to the area of the parallelogram formed by the
vectors u and v,
• is oriented so that the vectors u, v and u ⇥ v are arranged
according to the right-hand rule.
Skalär trippelprodukt
Om u, v, w 2 R3
så är
u · (v ⇥ w) = u1 ·
v2 v3
w2 w3
u2 ·
v1 v3
w1 w3
+ u3 ·
v1 v2
w1 w2
=
u1 u2 u3
v1 v2 v3
w1 w2 w3
.
v w
x y
z
⃗
i ⃗
j
⃗
k
⃗
i × ⃗
j = ⃗
k
spanned

168. A till den gula parallellogrammen som spänns upp på vektorerna beräknas
m.h.a. areor av rektanglar och rätvinkliga trianglar i bilden:
A = (a + b)(c + d) (2 · bc + 2 ·
bd
2
+ 2 ·
ac
2
) =
= ac + ad + bc + bd 2bc bd ac = ad bc =
a b
c d
Arean är lika med determinanten till matrisen med vektorerna som kolonner!
a b
b a
c
d
d
c
A
bc
bc
bd
2
bd
2
ac/2
ac/2
A = (a + b)(c + d) (2 · bc + 2 ·
bd
2
+ 2 ·
ac
2
) =
= ac + ad + bc + bd 2bc bd ac = ad bc =
a b
c d
Arean är lika med determinanten till matrisen med vektorerna som kolonner!
b a
c
d
A
bc
bd
2
bd
2
ac/2

169. Scalar triple product: volume of a parallelepiped
If u, v, w 2 R3
then
u · (v ⇥ w) = u1 ·
v2 v3
w2 w3
u2 ·
v1 v3
w1 w3
+ u3 ·
v1 v2
w1 w2
=
u1 u2 u3
v1 v2 v3
w1 w2 w3
.
v
w
u
v w
u cos
u, v, w 2 R3
lie in the same plane i↵ u · (v ⇥ w) = 0.
Scalar triple product: (signed) volume of a parallelepiped
If u, v, w 2 R3
then
u · (v ⇥ w) = u1 ·
v2 v3
w2 w3
u2 ·
v1 v3
w1 w3
+ u3 ·
v1 v2
w1 w2
=
u1 u2 u3
v1 v2 v3
w1 w2 w3
.
v
w
u
v w
u cos
u, v, w 2 R3
lie in the same plane i↵ u · (v ⇥ w) = 0.
the parallelepiped spanned by vectors u, v and w

170. kapitel 2 och 3 (till föreläsningar F5 och F7).
Straight lines in R2
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
l
y = 1
4
x 5
4
m = y
x
= 1
4
, b = 5
4
x
5
+ y
5/4
= 1
x + 4y + 5 = 0
n = ( 1, 4)
v = (4, 1)
x0 = (1, 1)
l :
(
x = 1 + 4t
y = 1 + t.
gen). Bara det sista sättet går att generaliseras till linjer i rumm
• (m, b)-equation: y = mx + b where m = y
x
is the slope of
b is the intercept with the y-axis. Such equation is uniq
that each non-vertical line has exactly one (m, b)-equation
• Intercept form: x
b1
+ y
b2
= 1 where b1 is the intercept betw
and the x-axis and b2 is the intercept with the y-axis. Such
also unique for each line.
• Normal equation for line l through x0 = (x0, y0) and ortho
vector n = (A, B): Ax + By + C = 0 where C = Ax0
If the point x = (x, y) belongs to l then the vectors n =
x x0 = (x x0, y y0) must be orthogonala, which mea
dot product must be zero:
(A, B)·(x x0, y y0) = 0 , Ax Ax0+By By0 = 0
Such equation is called normal equation because orthogo
dicular) vectors are also called normal vectors. Ekvatio
att generaliseras för plan i R3
, men den har ingen motsvar
jer i R3
. Sådan ekvation är inte entydig, eftersom det
många vektorer som är ortogonala mot linjen l (alla skal

171. kapitel 2 och 3 (till föreläsningar F5 och F7).
Straight lines in R2
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
l
y = 1
4
x 5
4
m = y
x
= 1
4
, b = 5
4
x
5
+ y
5/4
= 1
x + 4y + 5 = 0
n = ( 1, 4)
v = (4, 1)
x0 = (1, 1)
l :
(
x = 1 + 4t
y = 1 + t.
also unique for each line.
• Normal equation for line l through x0 = (x0, y0) and orthogonal to the
vector n = (A, B): Ax + By + C = 0 where C = Ax0 By0. Why?
If the point x = (x, y) belongs to l then the vectors n = (A, B) and
x x0 = (x x0, y y0) must be orthogonala, which means that their
dot product must be zero:
(A, B)·(x x0, y y0) = 0 , Ax Ax0+By By0 = 0 , Ax+By+
Such equation is called normal equation because orthogonal (perpen-
dicular) vectors are also called normal vectors. Ekvationen kommer
att generaliseras för plan i R3
, men den har ingen motsvarighet för lin-
jer i R3
. Sådan ekvation är inte entydig, eftersom det finns väldigt
många vektorer som är ortogonala mot linjen l (alla skalningar av n,
alltså vektorer som är parallella med n, men har en annan längd och
möjligen annan riktning).
• Equation in point-vector form: x = x0 + tv, t 2 R where x0 = (x0, y0)
is a point on the line and v = (v1, v2) is the direction vector to the line.
Sådan ekvation är långt ifrån entydig, eftersom man kan både välja
en annan punkt på linjen och en längre eller kortare vektor parallell
mot linjen i.s.f. v. Sådan ekvation skrivs ofta på följande form:
– punkt-vektorform där man sätter in konkreta koordinater:
(x, y) = (x0, y0) + t(v1, v2), t 2 R

172. kapitel 2 och 3 (till föreläsningar F5 och F7).
Straight lines in R2
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
l
y = 1
4
x 5
4
m = y
x
= 1
4
, b = 5
4
x
5
+ y
5/4
= 1
x + 4y + 5 = 0
n = ( 1, 4)
v = (4, 1)
x0 = (1, 1)
l :
(
x = 1 + 4t
y = 1 + t.
that each non-vertical line has exactly
• Intercept form: x
b1
+ y
b2
= 1 where b1 i
and the x-axis and b2 is the intercept w
also unique for each line.
• Normal equation for line l through x0
vector n = (A, B): Ax + By + C = 0
If the point x = (x, y) belongs to l th
x x0 = (x x0, y y0) must be ortho
dot product must be zero:
(A, B)·(x x0, y y0) = 0 , Ax A
Such equation is called normal equati
dicular) vectors are also called norma
att generaliseras för plan i R3
, men den
jer i R3
. Sådan ekvation är inte enty
många vektorer som är ortogonala mo
alltså vektorer som är parallella med n
möjligen annan riktning).
• Equation in point-vector form: x = x0
is a point on the line and v = (v , v ) is

173. kapitel 2 och 3 (till föreläsningar F5 och F7).
Straight lines in R2
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
l
y = 1
4
x 5
4
m = y
x
= 1
4
, b = 5
4
x
5
+ y
5/4
= 1
x + 4y + 5 = 0
n = ( 1, 4)
v = (4, 1)
x0 = (1, 1)
l :
(
x = 1 + 4t
y = 1 + t.
möjligen annan riktning).
• Equation in point-vector form: x = x0 + tv, t 2 R where x0 = (x0, y0)
is a point on the line and v = (v1, v2) is the direction vector to the line.
Sådan ekvation är långt ifrån entydig, eftersom man kan både välja
en annan punkt på linjen och en längre eller kortare vektor parallell
mot linjen i.s.f. v. Sådan ekvation skrivs ofta på följande form:
– punkt-vektorform där man sätter in konkreta koordinater:
(x, y) = (x0, y0) + t(v1, v2), t 2 R
– parameterform där man tydligt separerar koordinaterna:
l :
(
x = x0 + v1t
y = y0 + v2t
, t 2 R.
61

174. Planes in R3
• Normal equation of the plane ⇡ through x0 = (x0, y0, z0) and ort
to the vector n = (a, b, c):
ax + by + cz + d = 0 with d = ax0 by0 cz0.
Why? If a point x = (x, y, z) belongs to ⇡ then the vectors n =
and x x0 = (x x0, y y0, z z0) must be orthogonal, which
that their dot product must be zero:
(a, b, c)·(x x0, y y0, z z0) = 0 , ax ax0+by by0+cz
, ax + by + cz + d = 0.
n = (a, b, c)
x0 = (x0, y0, z0) x = (x, y, z)
Planes in R3
• Normal equation of the plane ⇡ through x0 = (x0, y0, z0) and orthogonal
to the vector n = (a, b, c):
ax + by + cz + d = 0 with d = ax0 by0 cz0.
Why? If a point x = (x, y, z) belongs to ⇡ then the vectors n = (a, b, c)
and x x0 = (x x0, y y0, z z0) must be orthogonal, which means
that their dot product must be zero:
(a, b, c)·(x x0, y y0, z z0) = 0 , ax ax0+by by0+cz cz0 = 0
, ax + by + cz + d = 0.
• Normal equation of the plane ⇡ through x0
to the vector n = (a, b, c):
ax + by + cz + d = 0 with d =
Why? If a point x = (x, y, z) belongs to ⇡
and x x0 = (x x0, y y0, z z0) must
that their dot product must be zero:
(a, b, c)·(x x0, y y0, z z0) = 0 , a
, ax + by + cz +
normal equation

175. Planes in R3
• Normal equation of the plane ⇡ through x0 = (x0, y0, z0) and ort
to the vector n = (a, b, c):
ax + by + cz + d = 0 with d = ax0 by0 cz0.
Why? If a point x = (x, y, z) belongs to ⇡ then the vectors n =
and x x0 = (x x0, y y0, z z0) must be orthogonal, which
that their dot product must be zero:
(a, b, c)·(x x0, y y0, z z0) = 0 , ax ax0+by by0+cz
, ax + by + cz + d = 0.
n = (a, b, c)
x0 = (x0, y0, z0) x = (x, y, z)
Planes in R3
• Normal equation of the plane ⇡ through x0 = (x0, y0, z0) and orthogonal
to the vector n = (a, b, c):
ax + by + cz + d = 0 with d = ax0 by0 cz0.
Why? If a point x = (x, y, z) belongs to ⇡ then the vectors n = (a, b, c)
and x x0 = (x x0, y y0, z z0) must be orthogonal, which means
that their dot product must be zero:
(a, b, c)·(x x0, y y0, z z0) = 0 , ax ax0+by by0+cz cz0 = 0
, ax + by + cz + d = 0.
• Normal equation of the plane ⇡ through x0
to the vector n = (a, b, c):
ax + by + cz + d = 0 with d =
Why? If a point x = (x, y, z) belongs to ⇡
and x x0 = (x x0, y y0, z z0) must
that their dot product must be zero:
(a, b, c)·(x x0, y y0, z z0) = 0 , a
, ax + by + cz +
x0 = (x0, y0, z0) x = (x, y, z)
Planes in R3
• Normal equation of the plane ⇡ through x0 = (x0, y0, z0) and orthogonal
to the vector n = (a, b, c):
ax + by + cz + d = 0 with d = ax0 by0 cz0.
Why? If a point x = (x, y, z) belongs to ⇡ then the vectors n = (a, b, c)
and x x0 = (x x0, y y0, z z0) must be orthogonal, which means
that their dot product must be zero:
(a, b, c)·(x x0, y y0, z z0) = 0 , ax ax0+by by0+cz cz0 = 0
, ax + by + cz + d = 0.
normal equation

176. Planes in R3
• Normal equation of the plane ⇡ through x0 = (x0, y0, z0) and ort
to the vector n = (a, b, c):
ax + by + cz + d = 0 with d = ax0 by0 cz0.
Why? If a point x = (x, y, z) belongs to ⇡ then the vectors n =
and x x0 = (x x0, y y0, z z0) must be orthogonal, which
that their dot product must be zero:
(a, b, c)·(x x0, y y0, z z0) = 0 , ax ax0+by by0+cz
, ax + by + cz + d = 0.
x0 = (x0, y0, z0)
v
u
t = 0
t = 1
t = 1
t = 2
s = 1
s = 0
s = 1
s = 2
s = 3
x = x0 + v + 3u
• Ekvation på punkt-vektorform: x = x0 + tv + su, t, s 2 R där x0 =
(x0, y0, z0) är en punkt i planet och v = (v1, v2, v3) samt u = (u1, u2, u3)
är två vektorer som genererar planet (alltså är parallella med planet
men inte parallella med varandra). Sådan ekvation är långt ifrån
entydig, eftersom man kan både välja en annan punkt på planet och
två andra vektorer i planet som inte är parallella. Sådan ekvation skrivs
ofta på följande form:
– punkt-vektorform där man sätter in konkreta koordinater:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3) + s(u1, u2, u3), t, s 2 R
– parameterform där man tydligt separerar koordinaterna:
⇡ :
8
:
x = x0 + v1t + u1s
y = y0 + v2t + u2s , t, s 2 R.
parametric equation
x0 = (x0, y0, z0)
v
u
t = 0
t = 1
t = 1
t = 2
s = 1
s = 0
s = 1
s = 2
s = 3
x = x0 + v + 3u
• Point-vector equation:
x = x0 + tv + su, t, s 2 R
where x0 = (x0, y0, z0) is a point in the plane, and v = (v1, v2, v3) and
u = (u1, u2, u3) are two vectors generating the plane.
– The same equation coordinate-wise:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3) + s(u1, u2, u3), t, s 2 R

177. Planes in R3
• Normal equation of the plane ⇡ through x0 = (x0, y0, z0) and ort
to the vector n = (a, b, c):
ax + by + cz + d = 0 with d = ax0 by0 cz0.
Why? If a point x = (x, y, z) belongs to ⇡ then the vectors n =
and x x0 = (x x0, y y0, z z0) must be orthogonal, which
that their dot product must be zero:
(a, b, c)·(x x0, y y0, z z0) = 0 , ax ax0+by by0+cz
, ax + by + cz + d = 0.
x0 = (x0, y0, z0)
v
u
t = 0
t = 1
t = 1
t = 2
s = 1
s = 0
s = 1
s = 2
s = 3
x = x0 + v + 3u
• Ekvation på punkt-vektorform: x = x0 + tv + su, t, s 2 R där x0 =
(x0, y0, z0) är en punkt i planet och v = (v1, v2, v3) samt u = (u1, u2, u3)
är två vektorer som genererar planet (alltså är parallella med planet
men inte parallella med varandra). Sådan ekvation är långt ifrån
entydig, eftersom man kan både välja en annan punkt på planet och
två andra vektorer i planet som inte är parallella. Sådan ekvation skrivs
ofta på följande form:
– punkt-vektorform där man sätter in konkreta koordinater:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3) + s(u1, u2, u3), t, s 2 R
– parameterform där man tydligt separerar koordinaterna:
⇡ :
8
:
x = x0 + v1t + u1s
y = y0 + v2t + u2s , t, s 2 R.
parametric equation
x0 = (x0, y0, z0)
v
u
t = 0
t = 1
t = 1
t = 2
s = 1
s = 0
s = 1
s = 2
s = 3
• Point-vector equation:
x = x0 + tv + su, t, s 2 R
where x0 = (x0, y0, z0) is a point in the plane, and v = (v1, v2, v3) and
u = (u1, u2, u3) are two vectors generating the plane.
– The same equation coordinate-wise:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3) + s(u1, u2, u3), t, s 2 R
– Coordinate-wise, in three raws, one for each coordinate:
⇡ :
8
:
x = x0 + v1t + u1s
y = y0 + v2t + u2s
z = z0 + v3t + u3s
, t, s 2 R.

178. Planes in R3
• Normal equation of the plane ⇡ through x0 = (x0, y0, z0) and ort
to the vector n = (a, b, c):
ax + by + cz + d = 0 with d = ax0 by0 cz0.
Why? If a point x = (x, y, z) belongs to ⇡ then the vectors n =
and x x0 = (x x0, y y0, z z0) must be orthogonal, which
that their dot product must be zero:
(a, b, c)·(x x0, y y0, z z0) = 0 , ax ax0+by by0+cz
, ax + by + cz + d = 0.
x0 = (x0, y0, z0)
v
u
t = 0
t = 1
t = 1
t = 2
s = 1
s = 0
s = 1
s = 2
s = 3
x = x0 + v + 3u
• Ekvation på punkt-vektorform: x = x0 + tv + su, t, s 2 R där x0 =
(x0, y0, z0) är en punkt i planet och v = (v1, v2, v3) samt u = (u1, u2, u3)
är två vektorer som genererar planet (alltså är parallella med planet
men inte parallella med varandra). Sådan ekvation är långt ifrån
entydig, eftersom man kan både välja en annan punkt på planet och
två andra vektorer i planet som inte är parallella. Sådan ekvation skrivs
ofta på följande form:
– punkt-vektorform där man sätter in konkreta koordinater:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3) + s(u1, u2, u3), t, s 2 R
– parameterform där man tydligt separerar koordinaterna:
⇡ :
8
:
x = x0 + v1t + u1s
y = y0 + v2t + u2s , t, s 2 R.
parametric equation
x0 = (x0, y0, z0)
v
u
t = 0
t = 1
t = 1
t = 2
s = 1
s = 0
s = 1
s = 2
s = 3
• Point-vector equation:
x = x0 + tv + su, t, s 2 R
where x0 = (x0, y0, z0) is a point in the plane, and v = (v1, v2, v3) and
u = (u1, u2, u3) are two vectors generating the plane.
– The same equation coordinate-wise:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3) + s(u1, u2, u3), t, s 2 R
– Coordinate-wise, in three raws, one for each coordinate:
⇡ :
8
:
x = x0 + v1t + u1s
y = y0 + v2t + u2s
z = z0 + v3t + u3s
, t, s 2 R.
x0 = (x0, y0, z0)
v
u
t = 0
t = 1
t = 1
t = 2
s = 1
s = 0
s = 1
s = 2
s = 3
• Point-vector equation:
x = x0 + tv + su, t, s 2 R
where x0 = (x0, y0, z0) is a point in the plane, and v = (v1, v2, v3) and
u = (u1, u2, u3) are two vectors generating the plane.
– The same equation coordinate-wise:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3) + s(u1, u2, u3), t, s 2 R
– Coordinate-wise, in three raws, one for each coordinate:
⇡ :
8
:
x = x0 + v1t + u1s
y = y0 + v2t + u2s
z = z0 + v3t + u3s
, t, s 2 R.

179. y
z
x
(x0, y0, z0)
⃗
v = [v1, v2, v3]
Straight lines in R3
• Point-vector equation:
x = x0 + tv, t 2 R
where x0 = (x0, y0, z0) is a point on the line and v = (v1, v2, v3) is the
direction vector of the line.
– The same equation coordinate-wise:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3), t 2 R
– Coordinate-wise, in three raws, one for each coordinate:
l :
8
:
x = x0 + v1t
y = y0 + v2t
z = z0 + v3t
, t 2 R.
• Standardekvation för linjen genom punkten x0 = (x0, y0, z0) och med
riktningsvektorn v = (v1, v2, v3) där alla vektorns koordinater är noll-
skilda. Då kan vi lösa ut parameter t från alla tre rader i ekvationssys-
temet ovan och jämföra dem med varandra. Ekvationen blir då:
Straight lines in R3
• Point-vector equation:
x = x0 + tv, t 2 R
where x0 = (x0, y0, z0) is a point on the line and v = (v1, v2, v3) is the
direction vector of the line.
– The same equation coordinate-wise:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3), t 2 R
– Coordinate-wise, in three raws, one for each coordinate:
l :
8
:
x = x0 + v1t
y = y0 + v2t
z = z0 + v3t
, t 2 R.
• Standardekvation för linjen genom punkten x = (x , y , z ) och med

180. Straight lines in R3
• Point-vector equation:
x = x0 + tv, t 2 R
where x0 = (x0, y0, z0) is a point on the line and v = (v1, v2, v3) is the
direction vector of the line.
– The same equation coordinate-wise:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3), t 2 R
– Coordinate-wise, in three raws, one for each coordinate:
l :
8
:
x = x0 + v1t
y = y0 + v2t
z = z0 + v3t
, t 2 R.
• Standardekvation för linjen genom punkten x0 = (x0, y0, z0) och med
riktningsvektorn v = (v1, v2, v3) där alla vektorns koordinater är noll-
skilda. Då kan vi lösa ut parameter t från alla tre rader i ekvationssys-
temet ovan och jämföra dem med varandra. Ekvationen blir då:
y
z
x
(x0, y0, z0)
⃗
v = [v1, v2, v3]
direction vector of the line.
– The same equation coordinate-wise:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3), t 2 R
– Coordinate-wise, in three raws, one for each coordinate:
l :
8
:
x = x0 + v1t
y = y0 + v2t
z = z0 + v3t
, t 2 R.
• Standard equation for the line through the point x0 = (x0, y0, z0) and
with direction vector v = (v1, v2, v3) with all non-zero coordinates. We
eliminate the parameter t and get:
x x0
v1
=
y y0
v2
=
z z0
v3
.
If for example v2 = 0 (and the two others are non-zero) we get the
following equations:
x x0
v1
=
z z0
v3
, y = y0.
The y-coordinate is constant for all the points on the line, which means
that the line lies in the plane y = y0 which is parallel to the xz-plane.
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3), t 2 R
– Coordinate-wise, in three raws, one for each coordinate:
l :
8
:
x = x0 + v1t
y = y0 + v2t
z = z0 + v3t
, t 2 R.
Standard equation for the line through the point x0 = (x0, y0, z0) and
with direction vector v = (v1, v2, v3) with all non-zero coordinates. We
eliminate the parameter t and get:
x x0
v1
=
y y0
v2
=
z z0
v3
.
If for example v2 = 0 (and the two others are non-zero) we get the
following equations:

181. x x0
v1
=
y y0
v2
=
z z0
v3
.
If for example v2 = 0 (and the two others are non-zero) we get the
following equations:
x x0
v1
=
z z0
v3
, y = y0.
The y-coordinate is constant for all the points on the line, which means
that the line lies in the plane y = y0 which is parallel to the xz-plane.
1
Straight lines in R3
• Point-vector equation:
x = x0 + tv, t 2 R
where x0 = (x0, y0, z0) is a point on the line and v = (v1, v2, v3) is the
direction vector of the line.
– The same equation coordinate-wise:
(x, y, z) = (x0, y0, z0) + t(v1, v2, v3), t 2 R
– Coordinate-wise, in three raws, one for each coordinate:
l :
8
:
x = x0 + v1t
y = y0 + v2t
z = z0 + v3t
, t 2 R.
• Standardekvation för linjen genom punkten x0 = (x0, y0, z0) och med
riktningsvektorn v = (v1, v2, v3) där alla vektorns koordinater är noll-
skilda. Då kan vi lösa ut parameter t från alla tre rader i ekvationssys-
temet ovan och jämföra dem med varandra. Ekvationen blir då:
y
z
x
(x0, y0, z0)
⃗
v = [v1, 0, v3]
(0, y0, 0)
y = y0

182. Robert A. Adams, Christopher Essex: Calculus, a complete course. 9th edition
13.1
Extreme values of functions of several variables

183. x
y
max
min
CP, no extremum
global min
global max
a singular point
(no derivative there)
CP — critical point (the derivative equal to zero)
y = f(x)

184. Possibilities:
in a CP in an interior point of the domain
on some endpoint of the domain
in a singular point

185. Local min in (a, b) Local max in (a, b)
(a, b)
(a + h, b + k)
(a, b, f(a, b))
(a + h, b + k, f(a + h, b + k))
δ
(a, b)
(a + h, b + k)
(a, b, f(a, b))
(a + h, b + k, f(a + h, b + k))
δ
f(a + h, b + k) f(a, b)
(h, k) ≠ (0,0)
for all
If
such that h2
+ k2
δ
f(a + h, b + k) f(a, b)
(h, k) ≠ (0,0)
h2
+ k2
δ
for all
If
such that

186. 6mM+iBQM f : R2
→ R ?b HQ+H KtBKmK BM (a, b)
B7 i?2`2 2tBbi δ 0 bm+? i?i
f(x, y) ⩽ f(a, b)
7Q` HH (x, y) bm+? i?i
!
(x − a)2 + (y − b)2 δ.
6mM+iBQM f : R3
→ R ?b HQ+H KtBKmK BM (a, b, c)
B7 i?2`2 2tBbi δ 0 bm+? i?i
f(a + h, b + k, c + l) ⩽ f(a, b, c)
7Q` HH (h, k, l) bm+? i?i
!
h2 + k2 + l2 δ. (a, b)
(x, y)
(a, b, f(a, b))
(x, y, f(x, y))
δ

187. 6mM+iBQM f : R3
→ R ?b HQ+H KtBKmK BM (a, b, c)
B7 i?2`2 2tBbi δ 0 bm+? i?i
f(a + h, b + k, c + l) ⩽ f(a, b, c)
7Q` HH (h, k, l) bm+? i?i
!
h2 + k2 + l2 δ.
6mM+iBQM f : Rn
→ R ?b HQ+H KtBKmK BM
a
B7 i?2`2 2tBbi δ 0 bm+? i?i
f(
a +
h) ⩽ f(
a)
7Q` HH
h bm+? i?i
#
h# δ.

188. 6mM+iBQM f : Rn
→ R ?b HQ+H KtBKmK BM
a
B7 i?2`2 2tBbi δ 0 bm+? i?i
f(
a +
h) ⩽ f(
a)
7Q` HH
h bm+? i?i
#
h# δ.
m` p` /2i B 1Mp`B#2HFHFvH
A 2Mp`B#2HMHvb ?` pB #´/2 7ƺ`bi@ Q+? M/` /2`Bpii2bi2M Ub2 #BH/2M T´ M bi bB@
/VX .2i 7ƺ`bi ` # bi U;2` HHiB/ bp`2iV- K2/M /2i M/` BMi2 ` p;ƺ`M/2 B 7HH2M /´
f!!(a) = 0 B 2M biiBQM ` TmMFiX h2bi2i ` B b´/M 7HH BM/2+BbBp2X 6ƺ` ~2`p`B#2H7mMFiBQ@
M2` }MMb /2i BM;2i 7ƺ`bi /2`Bpii2biX oB ?` 2M#`i 2ii M/` /2`Bpii2bi- bQK
?` 7ƺ``2bi2M bKK #2;` MbMBM; bQK /2M 2M/BK@KQibp`B;?2i2MX
L´;` 2t2KT2H 7ƺ` 7mMFiBQM2` p ip´ p`B#H2` bQK KM FH`` miM ii M@

189. f : ℝ2
→ ℝ

190. f : ℝ2
→ ℝ
∇f(a, b) = (0,0)
z = f(a, b) + ∇f(a, b)[
x − a
y − b]

191. f : ℝ2
→ ℝ
∇f(a, b) = (0,0)
z = f(a, b) + ∇f(a, b)[
x − a
y − b]
Critical points (CP): the point where the gradient is zero.
For functions of two variables: the tangent plane in these points is horizontal.

192. Possibilities:
in a CP in an interior point of the domain
in a boundary point of the domain
in a singular point

193. f(x, y) = 1 − x2
− y2
z = 1 − x2
− y2
z2
= 1 − x2
− y2
x2
+ y2
+ z2
= 1
x
y
1
1
max{f(x, y); (x, y) ∈ Df} = f(0,0) = 1
min{f(x, y); (x, y) ∈ Df} = f(1,0) = 0

194. f(x, y) =
1
x2 + y2 + 1
max{f(x, y); (x, y) ∈ Df} = f(0,0) = 1
inf{f(x, y); (x, y) ∈ Df} = lim
x2
+y2
→∞
f(x, y) = 0 Minimum does not exist.

195. x
y
x
y
(3, − 2)
x2
+ y2
− 6x + 4y − z + 10 = 0
f(x, y) = (x − 3)2
+ (y + 2)2
− 3
max{f(x, y); (x, y) ∈ Df}
min{f(x, y); (x, y) ∈ Df} = f(3, − 2) = − 3
does not exist

196. Minimum in a singular point
BM; T`iB2HH /2`BpiQ` / `X .2ii ` 2M j.@KQibp`B;?
p`B#2HMHvb UFQMiBMm2`HB;- K2M 2D /2`Bp2`#` B MQHHVX
Cross section between the g
is the graph of th
y = 0 ⇒ z =
Because the functio
f(x, y) = g(x
f(x, y)
f(x, y) = x2
+ y2
x
y
z
min{f(x, y); (x, y) ∈ Df} = f(0,0) = 0

197. 6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
f(x, y) = x + y
x + y − z = 0
Zerodamage, Jacobmelgaard

198. s
Robert A. Adams, Christopher Essex: Calculus, a complete course. 9th edition
13.1
Critical points and their classification

199. f : ℝ2
→ ℝ
∇f(a, b) = (0,0)
z = f(a, b) + ∇f(a, b)[
x − a
y − b]
Critical points (CP): the point where the gradient is zero.
For functions of two variables: the tangent plane in these points is horizontal.

200. Necessary condition, but not sufficient
function f
derivative f’
second derivative f’’
decreases decreases decreases decreases
increases increases increases increases
increases decreases
— 0 + + 0 — + 0 + — 0 —
has a minimum
has a maximum
POSITIVE NEGATIVE ZERO ZERO
CP, min CP, max
CP, no extremum CP, no extremum

201. If f is differentiable and it has a local extremum in an interior point of its domain,
then this point must be a CP.

202. If f is differentiable and it has a local extremum in an interior point of its domain,
then this point must be a CP.
But inverse is not true.
17i2`bQK HH HQFH 2ti`2KTmMFi2` !
a iBHH f B pBHF f ` T`iB2HHi /2`Bp2`#` Q+Fb´ ` F`BiBbF
TmMFi2`- b´ FM HQFH 2ti`2KTmMFi2` #` BMi` z B F`BiBbF TmMFi2`- bBM;mH ` TmMFi2` U∇f
2tBbi2`` 2DV Q+? `M/TmMFi2` iBHH Df X C K7ƺ` bBimiBQM2M B 2Mp`B#2HMHvb2MX
.2i ` BMi2 b F2`i ii 2M F`BiBbF TmMFi ` 2M HQFH 2ti`2KTmMFiX HHK Mi FM 2M 7mMFiBQM ? i`2
ivT2` p mib22M/2 F`BM; 2M F`BiBbF TmMFiX
x
y
f(x, y)
UV GQFHi KBMBKmK
x
y
f(x, y)
U#V GQFHi KtBKmK
x
y
f(x, y)
U+V a/2HTmMFi
J2/ ?D HT p hvHQ`TT`QtBKiBQM2M p 2M 7mMFiBQM bF pB i 7`K 2M K2iQ/ 7ƺ` ii F`Fi2`Bb2`
F`BiBbF TmMFi2`X
1
3
2
6B;m` j3, AHHmbi`iBQM iBHH 7ƺ`bi i`2 2t2KT2H, 2M U+B`FmH `V T`#QHQB/ z = x2+y2+1- 2M U+B`FmH `V
T`#QHQB/ z = 3 − x2 − y2 Q+? 2M ?vT2`#QHBbF T`#QHQB/ z = x2 − y2 + 2 US`BM;H2fa/2HVX
local minimum local maximum saddle point

203. If f is differentiable and it has a local extremum in an interior point of its domain,
then this point must be a CP.
But inverse is not true.
17i2`bQK HH HQFH 2ti`2KTmMFi2` !
a iBHH f B pBHF f ` T`iB2HHi /2`Bp2`#` Q+Fb´ ` F`BiBbF
TmMFi2`- b´ FM HQFH 2ti`2KTmMFi2` #` BMi` z B F`BiBbF TmMFi2`- bBM;mH ` TmMFi2` U∇f
2tBbi2`` 2DV Q+? `M/TmMFi2` iBHH Df X C K7ƺ` bBimiBQM2M B 2Mp`B#2HMHvb2MX
.2i ` BMi2 b F2`i ii 2M F`BiBbF TmMFi ` 2M HQFH 2ti`2KTmMFiX HHK Mi FM 2M 7mMFiBQM ? i`2
ivT2` p mib22M/2 F`BM; 2M F`BiBbF TmMFiX
x
y
f(x, y)
UV GQFHi KBMBKmK
x
y
f(x, y)
U#V GQFHi KtBKmK
x
y
f(x, y)
U+V a/2HTmMFi
J2/ ?D HT p hvHQ`TT`QtBKiBQM2M p 2M 7mMFiBQM bF pB i 7`K 2M K2iQ/ 7ƺ` ii F`Fi2`Bb2`
F`BiBbF TmMFi2`X
1
3
2
6B;m` j3, AHHmbi`iBQM iBHH 7ƺ`bi i`2 2t2KT2H, 2M U+B`FmH `V T`#QHQB/ z = x2+y2+1- 2M U+B`FmH `V
T`#QHQB/ z = 3 − x2 − y2 Q+? 2M ?vT2`#QHBbF T`#QHQB/ z = x2 − y2 + 2 US`BM;H2fa/2HVX
local minimum local maximum saddle point
b´ FM HQFH 2ti`2KTmMFi2` #` BMi` z B F`BiBbF TmMFi2`- bBM;mH ` TmMFi2` U∇f
2DV Q+? `M/TmMFi2` iBHH Df X C K7ƺ` bBimiBQM2M B 2Mp`B#2HMHvb2MX
Mi2 b F2`i ii 2M F`BiBbF TmMFi ` 2M HQFH 2ti`2KTmMFiX HHK Mi FM 2M 7mMFiBQM ? i`2
mib22M/2 F`BM; 2M F`BiBbF TmMFiX
x
y
f(x, y)
V GQFHi KBMBKmK
x
y
f(x, y)
U#V GQFHi KtBKmK
x
y
f(x, y)
U+V a/2HTmMFi
T p hvHQ`TT`QtBKiBQM2M p 2M 7mMFiBQM bF pB i 7`K 2M K2iQ/ 7ƺ` ii F`Fi2`Bb2`
TmMFi2`X
1
3
2
, V +B`+mH` T`#QHQB/ z = x2 + y2 + 1- #V +B`+mH` T`#QHQB/ z = 3 − x2 − y2-
2`#QHB+ T`#QHQB/ z = x2 − y2 + 2X
QFH 2ti`2KTmMFi2` !
a iBHH f B pBHF f ` T`iB2HHi /2`Bp2`#` Q+Fb´ ` F`BiBbF
M HQFH 2ti`2KTmMFi2` #` BMi` z B F`BiBbF TmMFi2`- bBM;mH ` TmMFi2` U∇f
? `M/TmMFi2` iBHH Df X C K7ƺ` bBimiBQM2M B 2Mp`B#2HMHvb2MX
2`i ii 2M F`BiBbF TmMFi ` 2M HQFH 2ti`2KTmMFiX HHK Mi FM 2M 7mMFiBQM ? i`2
M/2 F`BM; 2M F`BiBbF TmMFiX
x
y
f(x, y)
Hi KBMBKmK
x
y
f(x, y)
U#V GQFHi KtBKmK
x
y
f(x, y)
U+V a/2HTmMFi
vHQ`TT`QtBKiBQM2M p 2M 7mMFiBQM bF pB i 7`K 2M K2iQ/ 7ƺ` ii F`Fi2`Bb2`
3
2
+mH` T`#QHQB/ z = x2 + y2 + 1- #V +B`+mH` T`#QHQB/ z = 3 − x2 − y2-
2 2
2tBbi2`` 2DV Q+? `M/TmMFi2` iBHH Df X
.2i ` BMi2 b F2`i ii 2M F`BiBbF TmMFi `
ivT2` p mib22M/2 F`BM; 2M F`BiBbF TmMF
x
y
f(x, y)
UV GQFHi KBMBKmK U#
J2/ ?D HT p hvHQ`TT`QtBKiBQM2M p
F`BiBbF TmMFi2`X
1
6B;m` j3, V +B`+mH` T`#QHQB/ z = x2 + y
+V ?vT2`#QHB+ T`#QHQB/ z = x2 − y2 + 2X

204. Robert A. Adams, Christopher Essex: Calculus, a complete course. 9th edition
13.1
Second derivative test for C3 functions of several variables

205. x
y
y = f(x)
f ∈ C3

206. x
y
y = f(x)
a
y = f(a)
f ∈ C3

207. x
y
y = f(x)
a
y = f(a)
f′(a) = 0
y = f(a) + f′(a)(x − a)
f ∈ C3

208. x
y
y = f(x)
a
y = f(a)
f′(a) = 0
y = f(a) + f′(a)(x − a)
f ∈ C3

209. x
y
y = f(x)
a
y = f(a)
f′(a) = 0
y = f(a) + f′(a)(x − a)
f(x) = f(a) + f′(a)(x − a) +
f′′(a)
2!
(x − a)2
+ +R3(a, x)
f ∈ C3

210. x
y
y = f(x)
a
y = f(a)
f′(a) = 0
y = f(a) + f′(a)(x − a)
f(x) = f(a) + f′(a)(x − a) +
f′′(a)
2!
(x − a)2
+ R3(a, x)
f ∈ C3

211. x
y
y = f(x)
a
y = f(a)
f′(a) = 0
y = f(a) + f′(a)(x − a)
f(x) = f(a) + f′(a)(x − a) +
f′′(a)
2!
(x − a)2
+ R3(a, x)
f(x) − f(a) ≈
1
2
f′′(a)(x − a)2
f ∈ C3

212. f(x) − f(a) ≈
1
2
f′′(a)(x − a)2
Second derivative test
If a is a CP (critical point) for a C3-function f then:
f′′(a) 0
f′′(a) 0
f′′(a) = 0
f has local maximum in a
f has local minimum in a
the test is indecisive
⇒
⇒
⇒

213. f : ℝ2
→ ℝ

214. f : ℝ2
→ ℝ
∇f(a, b) = (0,0)
z = f(a, b) + ∇f(a, b)[
x − a
y − b]

215. f : ℝ2
→ ℝ
In the CPs we have horizontal tangent planes, so
f(a + h, b + k) = f(a, b) + ∇f(a, b) ⋅ (h, k)+
1
2
(h, k)T
ℋf(a, b)(h, k) + 𝒪( h2
+ k2
3
) .
f(a + h, b + k) − f(a, b) ≈
1
2
(h, k)T
ℋf(a, b)(h, k)
f ∈ C3

216. Second derivative test
If (a, b) is a CP (critical point) for a C3-function f then:
ℋf(a, b)
ℋf(a, b)
ℋf(a, b)
f has local maximum in (a, b)
f has local minimum in (a, b)
the test is indecisive
⇒
⇒
⇒
f(a + h, b + k) − f(a, b) ≈
1
2
(h, k)T
ℋf(a, b)(h, k)
ℋf(a, b) ⇒ f has a saddle in (a, b)
is negative definite
is positive definite
is indefinite
is none of the above

217. f : ℝn
→ ℝ, C3
tBKmK B !
a ⇐ f(!
a + !
h) − f(!
a) ⩽ 0 7ƺ` HH !
h b´ ii |!
h| δ
Fi B !
a ⇐ f(!
a + !
h) − f(!
a) Mi` #´/2 TQbBiBp Q+? M2;iBp
pB p HD2` !
hX
; p f F`BM; !
a ;2`
f(!
a + !
h) = f(!
a) + ∇f(!
a)
! # $
=!
0, B7 f ?b *S BM !
a
·!
h + 1
2
!
hT
Hf (!
a)!
h + O(|!
h|3
).
Q`B;Q ` O(|!
h|3) Q#2iv/HB; D K7ƺ`i K2/ !
hT Hf (!
a)!
h- bQK ?`
22M/2i ?Qb f(!
a + !
h) − f(!
a) #2bi Kb p #2i22M/2i ?Qb !
hT Hf (!
a
QF2MV HHib´- 27i2`bQK `2bii2`K2M ` Q#2iv/HB; HBi2M- ; HH2`,

218. i2i pB p HD2` !
hX
FHBM; p f F`BM; !
a ;2`
f(!
a + !
h) = f(!
a) + ∇f(!
a)
! # $
=!
o, QK f ?` 2M F`BiBbF TmMFi B !
a
·!
h + 1
2
!
hT
Hf (!
a)!
h + O(|!
h|3
).
M ` Q`B;Q ` O(|!
h|3) Q#2iv/HB; D K7ƺ`i K2/ !
hT Hf (!
a)!
h- bQK ?` biQ
#2i22M/2i ?Qb f(!
a + !
h) − f(!
a) #2bi Kb p #2i22M/2i ?Qb !
hT Hf (!
a)!
h
`b#QF2MV HHib´- 27i2`bQK `2bii2`K2M ` Q#2iv/HB; HBi2M- ; HH2`,
f(!
a + !
h) − f(!
a) ≈ 1
2
!
hT
Hf (!
a)!
h.
f : ℝn
→ ℝ, C3
tBKmK B !
a ⇐ f(!
a + !
h) − f(!
a) ⩽ 0 7ƺ` HH !
h b´ ii |!
h| δ
Fi B !
a ⇐ f(!
a + !
h) − f(!
a) Mi` #´/2 TQbBiBp Q+? M2;iBp
pB p HD2` !
hX
; p f F`BM; !
a ;2`
f(!
a + !
h) = f(!
a) + ∇f(!
a)
! # $
=!
0, B7 f ?b *S BM !
a
·!
h + 1
2
!
hT
Hf (!
a)!
h + O(|!
h|3
).
Q`B;Q ` O(|!
h|3) Q#2iv/HB; D K7ƺ`i K2/ !
hT Hf (!
a)!
h- bQK ?`
22M/2i ?Qb f(!
a + !
h) − f(!
a) #2bi Kb p #2i22M/2i ?Qb !
hT Hf (!
a
QF2MV HHib´- 27i2`bQK `2bii2`K2M ` Q#2iv/HB; HBi2M- ; HH2`,

219. Second derivative test
If is a CP (critical point) for a C3-function f then:
⃗
a
ℋf( ⃗
a )
ℋf( ⃗
a )
ℋf( ⃗
a )
f has local maximum in ⃗
a
f has local minimum in ⃗
a
the test is indecisive
⇒
⇒
⇒
f( ⃗
a + ⃗
h ) − f( ⃗
a ) ≈
1
2
⃗
h T
ℋf( ⃗
a ) ⃗
h
ℋf( ⃗
a ) ⇒ f has a saddle in ⃗
a
is negative definite
is positive definite
is indefinite
is none of the above

220. Robert A. Adams, Christopher Essex: Calculus, a complete course. 9th edition
13.1
Second derivative test for C3 functions of two variables

221. Second derivative test
If (a, b) is a CP (critical point) for a C3-function f then:
ℋf(a, b)
ℋf(a, b)
ℋf(a, b)
f has local maximum in (a, b)
f has local minimum in (a, b)
the test is indecisive
⇒
⇒
⇒
f(a + h, b + k) − f(a, b) ≈
1
2
(h, k)T
ℋf(a, b)(h, k)
ℋf(a, b) ⇒ f has a saddle in (a, b)
is negative definite
is positive definite
is indefinite
is none of the above

222. h?2Q`2K ReX Uh?2Q`2K 3- /Kb RyXdV G2i A #2 bvKK2i`B+ n × n@Ki`Bt rBi? `2H
2H2K2Mib- bQ aij ∈ R 7Q` HH i M/ jX G2i Di 7Q` i = 1, 2, . . . , n #2 i?2 /2i2`KBMMib b BM i?2
TB+im`2X
a11 a12 a13 a1n
a21 a22 a23 a2n
a31 a32 a33 a3n
an1 an2 an3 ann
…
…
…
…
………………………………….
D1 D2 D3 Dn
D1 = |a11 |
a11 a12
a21 a22
D2 =
…
Dn = det(A)
6B;m` jd, AHHmbi`iBQM iBHH aib 3 7`´M /Kb RyXdX UBH/, MB- K2/ E2vLQi2XV
Ji`Bt A Bb,
Ç TQbBiBp2 /2}MBi2 B7 Di 0 7Q` HH i
Ç M2;iBp2 /2}MBi2 B7 D1 0, D2 0, D3 0, D4 0, . . . - bQ TQbBiBp2 7Q` 2p2M MmK#2`b M/
M2;iBp2 7Q` Q// MmK#2`b-
Ç BM/2}MBi2 B7 |A| = Dn #= 0- #mi M2Bi?2` Q7 i?2 #Qp2 +QM/BiBQMb ?QH/X
A7 |A| = Dn = 0- i?2M i?2 i?2Q`2K Bb BM+QM+HmbBp2- K2MBM; i?i Bi /Q2b MQi ;Bp2 mb Mv BM7Q`@
KiBQM #Qmi AX

223. h?2Q`2K RdX Ua2+QM/ /2`BpiBp2 i2biVX G2i f : R2 → R #2 C3@ 7mM+iBQM M/ (a, b) #2
*S BM i?2 BMi2`BQ` Q7 Df - BM Qi?2` rQ`/b,
∇f(a, b) = (f!
1(a, b), f!
2(a, b)) = (0, 0).
G2i KQ`2Qp2` D2 #2 i?2 /2i2`KBMMi Q7 i?2 2bbBM Ki`Bt 7Q` f BM i?2 TQBMi (a, b)- BX2X,
D2 =
!
!
!
!
f!!
11(a, b) f!!
12(a, b)
f!!
21(a, b) f!!
22(a, b)
!
!
!
! .
h?2M r2 ?p2 i?2 7QHHQrBM;,
Ç A7 D2 0- i?2M f ?b HQ+H 2ti`2KmK BM (a, b),
Ĝ A7 D1 = f!!
11(a, b) 0 UbQ i?2 2bbBM Ki`Bt Bb TQbBiBp2 /2}MBi2V- i?2M f ?b
HQ+H KBMBKmK BM (a, b)
Ĝ A7 D1 = f!!
11(a, b) 0 UbQ i?2 2bbBM Ki`Bt Bb M2;iBp2 /2}MBi2V- i?2M f ?b
HQ+H KtBKmK BM (a, b)
Ĝ ULQi2, A7 D1 = f!!
11(a, b) = 0- i?2M D2 = 0 − (f!!
12(a, b))2 ⩽ 0- bQ D2 0 Bb
BKTQbbB#H2XV
Ç A7 D2 0 UbQ i?2 2bbBM Ki`Bt Bb BM/2}MBi2V- f ?b b//H2 TQBMi BM (a, b)X
Ç A7 D2 = 0- i?2 i2bi Bb BM/2+BbBp2X

224. Robert A. Adams, Christopher Essex: Calculus, a complete course. 9th edition
13.1
Critical points and their classification: some simple examples

225. Second derivative test
If (a, b) is a CP (critical point) for a C3-function f then:
ℋf(a, b)
ℋf(a, b)
ℋf(a, b)
f has local maximum in (a, b)
f has local minimum in (a, b)
the test is indecisive
⇒
⇒
⇒
f(a + h, b + k) − f(a, b) ≈
1
2
(h, k)T
ℋf(a, b)(h, k)
ℋf(a, b) ⇒ f has a saddle in (a, b)
is negative definite
is positive definite
is indefinite
is none of the above

226. If f is differentiable and it has a local extremum in an interior point of its domain,
then this point must be a CP.
But inverse is not true.
17i2`bQK HH HQFH 2ti`2KTmMFi2` !
a iBHH f B pBHF f ` T`iB2HHi /2`Bp2`#` Q+Fb´ ` F`BiBbF
TmMFi2`- b´ FM HQFH 2ti`2KTmMFi2` #` BMi` z B F`BiBbF TmMFi2`- bBM;mH ` TmMFi2` U∇f
2tBbi2`` 2DV Q+? `M/TmMFi2` iBHH Df X C K7ƺ` bBimiBQM2M B 2Mp`B#2HMHvb2MX
.2i ` BMi2 b F2`i ii 2M F`BiBbF TmMFi ` 2M HQFH 2ti`2KTmMFiX HHK Mi FM 2M 7mMFiBQM ? i`2
ivT2` p mib22M/2 F`BM; 2M F`BiBbF TmMFiX
x
y
f(x, y)
UV GQFHi KBMBKmK
x
y
f(x, y)
U#V GQFHi KtBKmK
x
y
f(x, y)
U+V a/2HTmMFi
J2/ ?D HT p hvHQ`TT`QtBKiBQM2M p 2M 7mMFiBQM bF pB i 7`K 2M K2iQ/ 7ƺ` ii F`Fi2`Bb2`
F`BiBbF TmMFi2`X
1
3
2
6B;m` j3, AHHmbi`iBQM iBHH 7ƺ`bi i`2 2t2KT2H, 2M U+B`FmH `V T`#QHQB/ z = x2+y2+1- 2M U+B`FmH `V
T`#QHQB/ z = 3 − x2 − y2 Q+? 2M ?vT2`#QHBbF T`#QHQB/ z = x2 − y2 + 2 US`BM;H2fa/2HVX
local minimum local maximum saddle point
b´ FM HQFH 2ti`2KTmMFi2` #` BMi` z B F`BiBbF TmMFi2`- bBM;mH ` TmMFi2` U∇f
2DV Q+? `M/TmMFi2` iBHH Df X C K7ƺ` bBimiBQM2M B 2Mp`B#2HMHvb2MX
Mi2 b F2`i ii 2M F`BiBbF TmMFi ` 2M HQFH 2ti`2KTmMFiX HHK Mi FM 2M 7mMFiBQM ? i`2
mib22M/2 F`BM; 2M F`BiBbF TmMFiX
x
y
f(x, y)
V GQFHi KBMBKmK
x
y
f(x, y)
U#V GQFHi KtBKmK
x
y
f(x, y)
U+V a/2HTmMFi
T p hvHQ`TT`QtBKiBQM2M p 2M 7mMFiBQM bF pB i 7`K 2M K2iQ/ 7ƺ` ii F`Fi2`Bb2`
TmMFi2`X
1
3
2
, V +B`+mH` T`#QHQB/ z = x2 + y2 + 1- #V +B`+mH` T`#QHQB/ z = 3 − x2 − y2-
2`#QHB+ T`#QHQB/ z = x2 − y2 + 2X
QFH 2ti`2KTmMFi2` !
a iBHH f B pBHF f ` T`iB2HHi /2`Bp2`#` Q+Fb´ ` F`BiBbF
M HQFH 2ti`2KTmMFi2` #` BMi` z B F`BiBbF TmMFi2`- bBM;mH ` TmMFi2` U∇f
? `M/TmMFi2` iBHH Df X C K7ƺ` bBimiBQM2M B 2Mp`B#2HMHvb2MX
2`i ii 2M F`BiBbF TmMFi ` 2M HQFH 2ti`2KTmMFiX HHK Mi FM 2M 7mMFiBQM ? i`2
M/2 F`BM; 2M F`BiBbF TmMFiX
x
y
f(x, y)
Hi KBMBKmK
x
y
f(x, y)
U#V GQFHi KtBKmK
x
y
f(x, y)
U+V a/2HTmMFi
vHQ`TT`QtBKiBQM2M p 2M 7mMFiBQM bF pB i 7`K 2M K2iQ/ 7ƺ` ii F`Fi2`Bb2`
3
2
+mH` T`#QHQB/ z = x2 + y2 + 1- #V +B`+mH` T`#QHQB/ z = 3 − x2 − y2-
2 2
2tBbi2`` 2DV Q+? `M/TmMFi2` iBHH Df X
.2i ` BMi2 b F2`i ii 2M F`BiBbF TmMFi `
ivT2` p mib22M/2 F`BM; 2M F`BiBbF TmMF
x
y
f(x, y)
UV GQFHi KBMBKmK U#
J2/ ?D HT p hvHQ`TT`QtBKiBQM2M p
F`BiBbF TmMFi2`X
1
6B;m` j3, V +B`+mH` T`#QHQB/ z = x2 + y
+V ?vT2`#QHB+ T`#QHQB/ z = x2 − y2 + 2X

227. UV GQFHi KBMBKmK U#V GQFHi KtBKmK U+V a/2HTmMFi
J2/ ?D HT p hvHQ`TT`QtBKiBQM2M p 2M 7mMFiBQM bF pB i 7`K 2M K2iQ/ 7ƺ` ii F`Fi2`Bb2`
F`BiBbF TmMFi2`X
G´i f p` 2M 7mMFiBQM p n p`B#H2` K2/ 2M F`BiBbF TmMFi B !
aX
89
6B;m` j3, V +B`+mH` T`#QHQB/ z = x2 + y2 + 1- #V +B`+mH` T`#QHQB/ z = 3 − x2 − y2-
+V ?vT2`#QHB+ T`#QHQB/ z = x2 − y2 + 2X
1tKTH2 8k U:`T? VX *HbbB7v i?2 *S (0, 0) iQ f(x, y) = x2 + y2 + 1X
S`iBH /Bz2`2MiBiBQM ;Bp2b,
f!
1(x, y) = 2x
f!
2(x, y) = 2y
UM/ r2 b22 i?i (0, 0) BM/22/ Bb i?2 QMHv *S 7Q` fV
f!!
11 = 2
f!!
12 = 0
f!!
22 = 2
aQ i?2 2bbBM Ki`Bt Bb,
Hf (0, 0) =
!
2 0
0 2
⇒ !
hT
Hf (0, 0)!
h = 2h2
+ 2k2
.
h?2 7Q`K +M QMHv ?p2 TQbBiBp2 pHm2b- r?B+? K2Mb i?i f ?b HQ+H KBMBKmK BM i?2 Q`B;BM-
2t+iHv b r2 2tT2+i2/X
1tKTH2 8j U:`T? #VX *HbbB7v i?2 *S (0, 0) iQ f(x, y) = 3 − x2 − y2X
S`iBH /Bz2`2MiBiBQM ;Bp2b,
f!
1(x, y) = −2x
ivT2` p mib22M/2 F`BM; 2M F`
x
y
f(x, y)
UV GQFHi KBMBKmK
J2/ ?D HT p hvHQ`TT`QtBK
F`BiBbF TmMFi2`X
G´i f p` 2M 7mMFiBQM p n p
1
6B;m` j3, V +B`+mH` T`#QHQB/
+V ?vT2`#QHB+ T`#QHQB/ z = x2
1tKTH2 8k U:`T? VX *Hbb
S`iBH /Bz2`2MiBiBQM ;Bp2b,
f!
1(x, y) = 2x
f!
2(x, y) = 2y
f!!
11 = 2
f!!
12 = 0
f!! = 2

228. ivT2` p mib22M/2 F`BM; 2M F`BiBbF TmMFiX
x
y
f(x, y)
UV GQFHi KBMBKmK
x
y
f(x, y)
U#V GQFHi KtBKmK
J2/ ?D HT p hvHQ`TT`QtBKiBQM2M p 2M 7mMFiBQM bF pB i 7
F`BiBbF TmMFi2`X
G´i f p` 2M 7mMFiBQM p n p`B#H2` K2/ 2M F`BiBbF TmMFi B
89
1
3
6B;m` j3, V +B`+mH` T`#QHQB/ z = x2 + y2 + 1- #V +B`+mH` T`
+V ?vT2`#QHB+ T`#QHQB/ z = x2 − y2 + 2X
1tKTH2 8k U:`T? VX *HbbB7v i?2 *S (0, 0) iQ f(x, y) = x2 +
S`iBH /Bz2`2MiBiBQM ;Bp2b,
f!
1(x, y) = 2x
f!
2(x, y) = 2y
UM/ r2 b22 i?i (0, 0
f!!
11 = 2
f!!
12 = 0
f!!
22 = 2
Hf (0, 0) =
2 0
0 2
⇒ !
hT
Hf (0, 0)!
h = 2h2
+ 2k2
.
h?2 7Q`K +M QMHv ?p2 TQbBiBp2 pHm2b- r?B+? K2Mb i?i f ?b HQ+H KBMBKmK BM i?2 Q`B;BM-
2t+iHv b r2 2tT2+i2/X
1tKTH2 8j U:`T? #VX *HbbB7v i?2 *S (0, 0) iQ f(x, y) = 3 − x2 − y2X
S`iBH /Bz2`2MiBiBQM ;Bp2b,
f!
1(x, y) = −2x
f!
2(x, y) = −2y
UM/ r2 b22 i?i (0, 0) BM/22/ Bb i?2 QMHv *S 7Q` fV
f!!
11 = −2
f!!
12 = 0
f!!
22 = −2
aQ i?2 2bbBM Ki`Bt Bb,
Hf (0, 0) =
!
−2 0
0 −2
⇒ !
hT
Hf (0, 0)!
h = −2h2
− 2k2
.
h?2 7Q`K +M QMHv ?p2 M2;iBp2 pHm2b- r?B+? K2Mb i?i f ?b HQ+H KtBKmK BM i?2 Q`B;BM-
2t+iHv b r2 2tT2+i2/X
3d

229. ivT2` p mib22M/2 F`BM; 2M F`BiBbF TmMFiX
x
y
f(x, y)
UV GQFHi KBMBKmK
x
y
f(x, y)
U#V GQFHi KtBKmK
x
y
f(x, y)
U+V a/2HTmMFi
J2/ ?D HT p hvHQ`TT`QtBKiBQM2M p 2M 7mMFiBQM bF pB i 7`K 2M K2iQ/ 7ƺ` ii F`Fi2`Bb
F`BiBbF TmMFi2`X
G´i f p` 2M 7mMFiBQM p n p`B#H2` K2/ 2M F`BiBbF TmMFi B !
aX
89
1
3
2
6B;m` j3, V +B`+mH` T`#QHQB/ z = x2 + y2 + 1- #V +B`+mH` T`#QHQB/ z = 3 − x2 − y2-
+V ?vT2`#QHB+ T`#QHQB/ z = x2 − y2 + 2X
1tKTH2 8k U:`T? VX *HbbB7v i?2 *S (0, 0) iQ f(x, y) = x2 + y2 + 1X
S`iBH /Bz2`2MiBiBQM ;Bp2b,
f!
1(x, y) = 2x
f!
2(x, y) = 2y
UM/ r2 b22 i?i (0, 0) BM/22/ Bb i?2 QMHv *S 7Q` fV
f!!
11 = 2
f!!
12 = 0
f!!
22 = 2
aQ i?2 2bbBM Ki`Bt Bb,
1tKTH2 89 U:`T? +VX *HbbB7v i?2 *S (0, 0) iQ f(x, y) = x2 − y2 + 2X
S`iBH /Bz2`2MiBiBQM ;Bp2b,
f!
1(x, y) = 2x
f!
2(x, y) = −2y
UM/ r2 b22 i?i (0, 0) BM/22/ Bb i?2 QMHv *S 7Q` fV
f!!
11 = 2
f!!
12 = 0
f!!
22 = −2
aQ i?2 2bbBM Ki`Bt Bb,
Hf (0, 0) =
!
2 0
0 −2
⇒ !
hT
Hf (0, 0)!
h = 2h2
− 2k2
h?2 7Q`K +M ?p2 #Qi? TQbBiBp2 M/ M2;iBp2 pHm2b 7Q` (h, k) BM 2p2`v M2B;?#Q`?QQ/ Q7 i?2
Q`B;BM UTQbBiBp2 pHm2b ;2ib 7Q` 2tKTH2 7Q` (h, 0) M/ M2;iBp2 7Q` (0, k) rBi? Mv MQM@x2`Q h
M/ kV- r?B+? K2Mb i?i f ?b b//H2 TQBMi- 2t+iHv b r2 2tT2+i2/X
Lm 7ƺHD2` M´;` bp´``2 2t2KT2H- / ` pB BMi2 FM b ; B 7ƺ`p ; pBHF2M bQ`i biiBQM ` TmMFi pB 7´`X
#
1 1
$

230. 13.1
Critical points and their classification: more examples 1

231. Second derivative test
If (a, b) is a CP (critical point) for a C3-function f then:
ℋf(a, b)
ℋf(a, b)
ℋf(a, b)
f has local maximum in (a, b)
f has local minimum in (a, b)
the test is indecisive
⇒
⇒
⇒
f(a + h, b + k) − f(a, b) ≈
1
2
(h, k)T
ℋf(a, b)(h, k)
ℋf(a, b) ⇒ f has a saddle in (a, b)
is negative definite
is positive definite
is indefinite
is none of the above

232. Lm 7ƺHD2` M´;` bp´``2 2t2KT2H- / ` pB BMi2 FM b ; B 7ƺ`p ; pBHF2M bQ`i biiBQM ` TmMFi pB 7´`X
1tKTH2 88X *HbbB7v i?2 *Sb (0, 0) M/
#
1
12 , −1
6
$
iQ f(x, y) = 3x2 + 3xy + y2 + y3X
S`iBH /Bz2`2MiBiBQM ;Bp2b,
f!
1(x, y) = 6x + 3y
f!
2(x, y) = 3x + 2y + 3y2 UM/ r2 b22 i?i #Qi? TQBMib BM/22/ `2 *SbXV
f!!
11 = 6
f!!
12 = 3
f!!
22 = 2 + 6y
aQ i?2 2bbBM Ki`B+2b BM i?2 *Sb `2,
Hf (0, 0) =
!
6 3
3 2
⇒ !
hT
Hf (0, 0)!
h = 6h2
+ 6hk + 2k2
= 6
#
h + 1
2 k
$2
+ 1
2 k2
.
h?2 [m/`iB+ 7Q`K +M QMHv ?p2 TQbBiBp2 pHm2b- bQ f ?` HQ+H KBMBKmK BM i?Bb TQBMiX
Hf ( 1
12, −1
6 ) =
!
6 3
3 1
⇒ !
hT
Hf ( 1
12 , −1
6 )!
h = 6h2
+ 6hk + k2
= 6
#
h + 1
2 k
$2
− 1
2 k2
.
h?2 [m/`iB+ 7Q`K +M ?p2 #Qi? TQbBiBp2 M/ M2;iBp2 pHm2b- bQ r2 ;2i b//H2 TQBMi i?2`2X
1

233. 1tKTH2 8dX 6BM/ M/ +HbbB7v HH i?2 *Sb iQ f(x, y) = 3xy − x2 − 3y2 + x − 12X
h?2 *Sb Kmbi biBb7v i?2 7QHHQrBM; bvbi2K Q7 2[miBQMb,
!
f!
1(x, y) = 3y − 2x + 1 = 0
f!
2(x, y) = 3x − 6y = 0
⇒
!
3y − 2x + 1 = 0
x = 2y
⇒
!
3y − 4y + 1 = 0
x = 2y
⇒ (x, y) = (2, 1).
h?2 b2+QM/ Q`/2` T`iBH /2`BpiBp2b `2 f!!
11(x, y) = −2- f!!
12(x, y) = 3- M/ f!!
22(x, y) = −6 bQ
i?2 2bbBM Ki`Bt UrBi? +Q``2bTQM/BM; [m/`iB+ 7Q`K Q(h, k)V Bb
−2 3
3 −6
#
; Q(h, k) = −2h2
+ 6hk − 6k2
= −2(h2
− 3hk) − 6k2
= −2
$%
h − 3
2k
2
− 9
4 k2
'
− 6k2
= −2
%
h − 3
2 k
2
+ 9
2 k2
− 6k2
= −2
%
h − 3
2 k
2
− 3
2 k2
0 ∀(h, k) $= (0, 0).
h?2 2bbBM Ki`Bt Bb M2;iBp2 /2}MBi2- bQ i?2 *S Bb HQ+H KtBKmKX
PK /2M Fp/`iBbF 7Q`K2M ` b2KB/2}MBi 7ƺ` 2M F`BiBbF TmMFi b´ FM /2M F`BiBbF TmMFi2M
2

234. 1tKTH2 8eX 6BM/ M/ +HbbB7v HH i?2 *S iQ f(x, y) = xy − x2y − y2X
q2 +QKTmi2 i?2 T`iBH /2`BpiBp2b BM Q`/2` iQ }M/ i?2 *SbX
f!
1(x, y) = y − 2xy = 0
f!
2(x, y) = x − x2 − 2y = 0
⇒
!
y(1 − 2x) = 0
x − x2 − 2y = 0
h?2 }`bi 2[miBQM ;Bp2b y = 0 Q` 1 − 2x = 0X q2 2tKBM2 i?2 b2+QM/ 2[miBQM BM #Qi? +b2b,
y = 0 ⇒ x − x2
= 0 ⇒
!
x = 0
x = 1
⇒ irQ *Sb `2 (0, 0) M/ (1, 0)
1 − 2x = 0 ⇒ x = 1
2 ⇒ 1
2 −
1
2
#2
− 2y = 0 ⇒ y = 1
8 ⇒ *S BM (1
2 , 1
8 )
q2 /Bz2`2MiBi2 QM2 KQ`2 iBK2 iQ ;2i i?2 2bbBM Ki`BtX
f!!
11(x, y) = −2y
f!!
12(x, y) = 1 − 2x
f!!
22(x, y) = −2
⇒ Hf (x, y) =
$
−2y 1 − 2x
1 − 2x −2
%
q2 ;2i i?2 7QHHQrBM; 2bbBM Ki`B+2b BM i?2 *Sb,
Hf (0, 0) =
$
0 1
1 −2
%
, Hf (1, 0) =
$
0 −1
−1 −2
%
, Hf (1
2 , 1
8 ) =
$
−1
4 0
0 −2
%
q2 2tKBM2 +Q``2bTQM/BM; [m/`iB+ 7Q`Kb,
$
0 1
1 −2
%
;Bp2b Q(h, k) = 2hk − 2k2
= −2
k − h
2
#2
+ h2
2 UBM/2}MBi2- bQ b//H2 TQBMiV
$
0 −1
−1 −2
%
;Bp2b Q(h, k) = −2hk − 2k2
= −2
k + h
2
#2
+ h2
2 UBM/2}MBi2- b//H2V
$
−1
4 0
0 −2
%
;Bp2b Q(h, k) = −h2
4 − k2
UM2;iBp2 /2}MBi2- bQ HQ+H KtBKmKXV
3

235. Problem4e
fCxiy xwe cIt
fx O
product rule chain r
fy
_O
fi zxy.EEIH xy.EE fzeI
2xy.e
HtY4
1 x2
2xyCt x1Gtx1e
Ety4
cfylex2e
WtY4tx2ye
WtY4
f 2g
m m
x'e city I 2y2
x4try F e
Wt
e
Htg
fit
v
u
fi oc vy F

236. whole
y
axis
µ y
c the
y O
fcxiy.org CPfl0 y
0yco
7fCxiyK0 the values
Fathers
too E
eq
are
E ZERO
on
Vin t'adadley
1 a point
B fCx yK0
Is he
fax
dir
PE AT

237. fI up
2x
y
2
3
y
e yY
f 2
y
6
2
y
e
C
Y4tC2xy 2x3y
m
e
City4
f 2x
e City 2y 6xI4xfyt4xyf
2y.EC Hfz
gIt9YI
Uh
f for all CP
1 5 2 2200
t

238. fi
AB
w
2 2
3
e 54 2xy 2x3y
x2 1 y2 z
X X ttt O
in all my
CP
fyy
fyl
e
City
TK

239. fyy
4Ey e Ht E 2Ey e H
Eijie
city4
f 2 iffy
co
i
y o
f
y o I
H fo
Q o 9
e

240. All four Cps are external
points for f
y O D CO neg def MAX
g so D 0 pros def MIN

241. 13.1-9
f(x, y) = x2
ye−(x2
+y2
)
x
y
https://www.geogebra.org/m/cXgNb58T
Vector fields in 2D

242. Multivariable Calculus
Towards and through the vector fields
Robert A. Adams, Christopher Essex: Calculus, a complete course. 8th or 9th edition
Hania Uscka-Wehlou, Ph.D. (2009, Uppsala University: Mathematics)
University teacher in mathematics (Associate Professor / Senior Lecturer) at Mälardalen University, Sweden
13.2
Extreme values for continuous functions on compact domains

243. Compact set = closed and bounded.

244. Compact set = closed and bounded.
Continuous functions on compact sets attain both maximum and minimum.

245. Compact set = closed and bounded.
Continuous functions on compact sets attain both maximum and minimum.
How to find them?
- localize all the critical and singular interior points , compute in these points
- find max / min on the boundary
- compare all the values found above, pick the largest as max, the smallest as min.
⃗
a f( ⃗
a )

246. Compact set = closed and bounded.
Continuous functions on compact sets attain both maximum and minimum.
How to find them?
- localize all the critical and singular interior points , compute in these points
- find max / min on the boundary
- compare all the values found above, pick the largest as max, the smallest as min.
⃗
a f( ⃗
a )
elimination of one variable, parametrisation, Lagrange multipliers

247. R1 R2 R3
neighbour-
hood
an interval
without boundary
(2 end points)
a disk
without boundary
(circle)
a ball
without boundary
(sphere)
open set
(all the points
included with
certain
neighbourhood)
neither open nor closed.
No points are included
with a neighbourhood.
Two boundary points missing.
The set consists only of
boundary points in R2.
Does not exist
Does not exist
open set
(all the points are
included with some
neighbourhood)
open set (all the points are
included with some neighbourhood)
neither open nor closed.
No points are included with a neighbourhood.
Two boundary points missing. The set consists
only of boundary points in R3.
neither open nor closed.
No points are included with neighbourhood.
Many boundary points missing. The set
consists only of boundary points in R3.
Each open ball with centre in a point on
the disk contains both points from the
disk and outside the disk (both on the
“back side” and the “front side”), so it is
by definition a boundary point!
Does not exist

248. 1tKTH2 ekX 6BM/ Kt M/ KBM 7Q` f(x, y) = xy2 − x − y QM i?2 b[m`2
D, 0 ⩽ x ⩽ 2- 0 ⩽ y ⩽ 2X
x
y
2
2
1
1
(2,2)
(0,2)
(0,0)
(
1
2
,1)
(2,
1
4
)
(2,0)
g1
g2
g3
g4
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
x
y
2
2
1
1
(2,2)
(0,2)
(0,0)
(
1
2
,1)
(2,
1
4
)
(2,0)
g1
g2
g3
g4
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2

249. 1tKTH2 ekX 6BM/ Kt M/ KBM 7Q` f(x, y) = xy2 − x − y QM i?2 b[m`2
D, 0 ⩽ x ⩽ 2- 0 ⩽ y ⩽ 2X
x
y
2
2
1
1
(2,2)
(0,2)
(0,0)
(
1
2
,1)
(2,
1
4
)
(2,0)
g1
g2
g3
g4
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
2
1
g1
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
g3(y) = f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
bQ r2 ?p2 Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
jX o2`iB+2bX h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
q2 +QKTmi2 i?2 pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
iiBM2/, f(0, 0) = 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
q2 ;2i i?2 KtBKmK 4 M/ i?2 KBMBKmK −17
8 .
PK 2M 7mMFiBQM f p ~2` p`B#H2` ` /2}MB2`/ T´ 2ii B+F2@FQKTFi QK`´/2 ` /2i BMi2 b F2`i
ii /2M ?` 2ii biƺ`bi 2HH2` KBMbi p `/2- miM /2ii K´bi2 KQiBp2`b T´ M´;Qi b iiX
4x − 3
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
QBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
`bi TQBMi #2HQM;b iQ Qm` b[m`2X
/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
= f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
= f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
= f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
= f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
= 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
KmK 4 M/ i?2 KBMBKmK −17
8 .
x
y
2
2
1
1
(2,2)
(0,2)
(0,0)
(
1
2
,1)
(2,
1
4
)
(2,0)
g1
g2
g3
g4
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2

250. 1tKTH2 ekX 6BM/ Kt M/ KBM 7Q` f(x, y) = xy2 − x − y QM i?2 b[m`2
D, 0 ⩽ x ⩽ 2- 0 ⩽ y ⩽ 2X
x
y
2
2
1
1
(2,2)
(0,2)
(0,0)
(
1
2
,1)
(2,
1
4
)
(2,0)
g1
g2
g3
g4
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
2
1
g1
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
g3(y) = f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
bQ r2 ?p2 Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
jX o2`iB+2bX h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
q2 +QKTmi2 i?2 pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
iiBM2/, f(0, 0) = 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
q2 ;2i i?2 KtBKmK 4 M/ i?2 KBMBKmK −17
8 .
PK 2M 7mMFiBQM f p ~2` p`B#H2` ` /2}MB2`/ T´ 2ii B+F2@FQKTFi QK`´/2 ` /2i BMi2 b F2`i
ii /2M ?` 2ii biƺ`bi 2HH2` KBMbi p `/2- miM /2ii K´bi2 KQiBp2`b T´ M´;Qi b iiX
4x − 3
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
QBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
`bi TQBMi #2HQM;b iQ Qm` b[m`2X
/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
= f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
= f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
= f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
= f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
= 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
KmK 4 M/ i?2 KBMBKmK −17
8 .
x
y
2
2
1
1
(2,2)
(0,2)
(0,0)
(
1
2
,1)
(2,
1
4
)
(2,0)
g1
g2
g3
g4
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
g3(y) = f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
bQ r2 ?p2 Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
jX o2`iB+2bX h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
q2 +QKTmi2 i?2 pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
iiBM2/, f(0, 0) = 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
q2 ;2i i?2 KtBKmK 4 M/ i?2 KBMBKmK −17
8 .
PK 2M 7mMFiBQM f p ~2` p`B#H2` ` /2}MB2`/ T´ 2ii B+F2@FQKTFi QK`´/2 ` /2i BMi2 b F2`i
ii /2M ?` 2ii biƺ`bi 2HH2` KBMbi p `/2- miM /2ii K´bi2 KQiBp2`b T´ M´;Qi b iiX
1tKTH2 ejX p;ƺ` QK 7mMFiBQM2M f(x, y) =
4x − 3
1 + x2 + y2
?` 2ii KtBKmK Q+? 2ii KBMBKmK
4

251. 1tKTH2 ekX 6BM/ Kt M/ KBM 7Q` f(x, y) = xy2 − x − y QM i?2 b[m`2
D, 0 ⩽ x ⩽ 2- 0 ⩽ y ⩽ 2X
x
y
2
2
1
1
(2,2)
(0,2)
(0,0)
(
1
2
,1)
(2,
1
4
)
(2,0)
g1
g2
g3
g4
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
2
1
g1
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
g3(y) = f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
bQ r2 ?p2 Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
jX o2`iB+2bX h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
q2 +QKTmi2 i?2 pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
iiBM2/, f(0, 0) = 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
q2 ;2i i?2 KtBKmK 4 M/ i?2 KBMBKmK −17
8 .
PK 2M 7mMFiBQM f p ~2` p`B#H2` ` /2}MB2`/ T´ 2ii B+F2@FQKTFi QK`´/2 ` /2i BMi2 b F2`i
ii /2M ?` 2ii biƺ`bi 2HH2` KBMbi p `/2- miM /2ii K´bi2 KQiBp2`b T´ M´;Qi b iiX
4x − 3
x
y
2
2
1
1
(2,2)
(0,2)
(0,0)
(
1
2
,1)
(2,
1
4
)
(2,0)
g1
g2
g3
g4
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
g3(y) = f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
bQ r2 ?p2 Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
jX o2`iB+2bX h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
q2 +QKTmi2 i?2 pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
iiBM2/, f(0, 0) = 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
q2 ;2i i?2 KtBKmK 4 M/ i?2 KBMBKmK −17
8 .
PK 2M 7mMFiBQM f p ~2` p`B#H2` ` /2}MB2`/ T´ 2ii B+F2@FQKTFi QK`´/2 ` /2i BMi2 b F2`i
ii /2M ?` 2ii biƺ`bi 2HH2` KBMbi p `/2- miM /2ii K´bi2 KQiBp2`b T´ M´;Qi b iiX
1tKTH2 ejX p;ƺ` QK 7mMFiBQM2M f(x, y) =
4x − 3
1 + x2 + y2
?` 2ii KtBKmK Q+? 2ii KBMBKmK
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
g3(y) = f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
bQ r2 ?p2 Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
jX o2`iB+2bX h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
q2 +QKTmi2 i?2 pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
iiBM2/, f(0, 0) = 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
q2 ;2i i?2 KtBKmK 4 M/ i?2 KBMBKmK −17
8 .
PK 2M 7mMFiBQM f p ~2` p`B#H2` ` /2}MB2`/ T´ 2ii B+F2@FQKTFi QK`´/2 ` /2i BMi2 b F2`i
ii /2M ?` 2ii biƺ`bi 2HH2` KBMbi p `/2- miM /2ii K´bi2 KQiBp2`b T´ M´;Qi b iiX
4x − 3
4

252. 1tKTH2 ekX 6BM/ Kt M/ KBM 7Q` f(x, y) = xy2 − x − y QM i?2 b[m`2
D, 0 ⩽ x ⩽ 2- 0 ⩽ y ⩽ 2X
x
y
2
2
1
1
(2,2)
(0,2)
(0,0)
(
1
2
,1)
(2,
1
4
)
(2,0)
g1
g2
g3
g4
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
2
1
g1
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
g3(y) = f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
bQ r2 ?p2 Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
jX o2`iB+2bX h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
q2 +QKTmi2 i?2 pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
iiBM2/, f(0, 0) = 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
q2 ;2i i?2 KtBKmK 4 M/ i?2 KBMBKmK −17
8 .
PK 2M 7mMFiBQM f p ~2` p`B#H2` ` /2}MB2`/ T´ 2ii B+F2@FQKTFi QK`´/2 ` /2i BMi2 b F2`i
ii /2M ?` 2ii biƺ`bi 2HH2` KBMbi p `/2- miM /2ii K´bi2 KQiBp2`b T´ M´;Qi b iiX
4x − 3
x
y
2
2
1
1
(2,2)
(0,2)
(0,0)
(
1
2
,1)
(2,
1
4
)
(2,0)
g1
g2
g3
g4
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2
6B;m` 9y, AHHmbi`iBQM iBHH mTT;B7i2M B 1t2KT2H ekX
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
g3(y) = f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
bQ r2 ?p2 Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
jX o2`iB+2bX h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
q2 +QKTmi2 i?2 pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
iiBM2/, f(0, 0) = 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
q2 ;2i i?2 KtBKmK 4 M/ i?2 KBMBKmK −17
8 .
PK 2M 7mMFiBQM f p ~2` p`B#H2` ` /2}MB2`/ T´ 2ii B+F2@FQKTFi QK`´/2 ` /2i BMi2 b F2`i
ii /2M ?` 2ii biƺ`bi 2HH2` KBMbi p `/2- miM /2ii K´bi2 KQiBp2`b T´ M´;Qi b iiX
1tKTH2 ejX p;ƺ` QK 7mMFiBQM2M f(x, y) =
4x − 3
1 + x2 + y2
?` 2ii KtBKmK Q+? 2ii KBMBKmK
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D◦,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
g3(y) = f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
bQ r2 ?p2 Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
jX o2`iB+2bX h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
q2 +QKTmi2 i?2 pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
iiBM2/, f(0, 0) = 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
q2 ;2i i?2 KtBKmK 4 M/ i?2 KBMBKmK −17
8 .
PK 2M 7mMFiBQM f p ~2` p`B#H2` ` /2}MB2`/ T´ 2ii B+F2@FQKTFi QK`´/2 ` /2i BMi2 b F2`i
ii /2M ?` 2ii biƺ`bi 2HH2` KBMbi p `/2- miM /2ii K´bi2 KQiBp2`b T´ M´;Qi b iiX
4x − 3
RX AMi2`BQ` TQBMibX q2 +?2+F B7 i?2 7mM+iBQM ?b *Sb BM D ,
f
1(x, y) = y2 − 1 = 0
f
2(x, y) = 2xy − 1 = 0
⇒
y = ±1
x = 1
2y
U*Sb (1
2 , 1) M/ (−1
2 , −1)XV
PMHv i?2 }`bi TQBMi #2HQM;b iQ Qm` b[m`2X
kX h?2 #QmM/`vX q2 T`K2i2`Bb2 i?2 7Qm` bB/2b M/ +?2+F B7 f ?b *Sb i?2`2X
g1(x) = f(x, 0), ;Bp2b g1(x) = −x 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g2(x) = f(x, 2), ;Bp2b g2(x) = 3x − 2 7Q` 0 ⩽ x ⩽ 2 UMQ *SbV
g3(y) = f(0, y), ;Bp2b g3(y) = −y 7Q` 0 ⩽ y ⩽ 2 UMQ *SbV
g3(y) = f(2, y), ;Bp2b g4(y) = 2y2
− 2 − y 7Q` 0 ⩽ y ⩽ 2 U*S BM (2, 1
4 )V
bQ r2 ?p2 Dmbi QM2 *S QM i?2 #QmM/`v, (2, 1
4 )X
jX o2`iB+2bX h?2 b2i ?b 7Qm` p2`iB+2b, (0, 0), (2, 0), (0, 2), (2, 2)X
q2 +QKTmi2 i?2 pHm2b Q7 Qm` 7mM+iBQM BM i?2 bBt TQBMib r?2`2 Bi Bb TQbbB#H2 i?i Kt Q` KBM Bb
iiBM2/, f(0, 0) = 0, f(0, 2) = −2, f(2, 0) = −2, f(2, 2) = 4, f(1
2, 1) = −1, f(2, 1
4 ) = −17
8 X
q2 ;2i i?2 KtBKmK 4 M/ i?2 KBMBKmK −17
8 .
4

253. 6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
T`Q#H2K 7`´M 2M/BKX .2`Bp2`BM; ;2` g!(φ) = − bBM φ + +Qb φ Q+?,
= 0 ⇔ bBM φ = +Qb φ ⇔ iM φ = 1 ⇔ φ =
π
4
2HH2`
5π
4
.
2 p `/2M 7ƺ` x(φ) = +Qb φ Q+? y(φ) = bBM φ,
x = y =
√
2
2 2HH2` x = y = −
√
2
2 .
b Kt T´ D ` f(
√
2
2 ,
√
2
2 ) =
√
2 Q+? KBM f(−
√
2
2 , −
√
2
2 ) = −
√
2X
2bi K biƺ`bi Q+? KBMbi p `/2 p f(x, y) = x + y + x2 + y2 /´ x2 + y2 ⩽ 1X
1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
No CPs! The gradient is constant, (1,1).

254. 6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
T`Q#H2K 7`´M 2M/BKX .2`Bp2`BM; ;2` g!(φ) = − bBM φ + +Qb φ Q+?,
= 0 ⇔ bBM φ = +Qb φ ⇔ iM φ = 1 ⇔ φ =
π
4
2HH2`
5π
4
.
2 p `/2M 7ƺ` x(φ) = +Qb φ Q+? y(φ) = bBM φ,
x = y =
√
2
2 2HH2` x = y = −
√
2
2 .
b Kt T´ D ` f(
√
2
2 ,
√
2
2 ) =
√
2 Q+? KBM f(−
√
2
2 , −
√
2
2 ) = −
√
2X
2bi K biƺ`bi Q+? KBMbi p `/2 p f(x, y) = x + y + x2 + y2 /´ x2 + y2 ⩽ 1X
1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
17i2`bQK 7mMFiBQM2Mb ;`7 ` 2ii THM B `mKK2i UK2/ 2FpiBQM x + y −
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB`
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
No CPs! The gradient is constant, (1,1).

255. 6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
T`Q#H2K 7`´M 2M/BKX .2`Bp2`BM; ;2` g!(φ) = − bBM φ + +Qb φ Q+?,
= 0 ⇔ bBM φ = +Qb φ ⇔ iM φ = 1 ⇔ φ =
π
4
2HH2`
5π
4
.
2 p `/2M 7ƺ` x(φ) = +Qb φ Q+? y(φ) = bBM φ,
x = y =
√
2
2 2HH2` x = y = −
√
2
2 .
b Kt T´ D ` f(
√
2
2 ,
√
2
2 ) =
√
2 Q+? KBM f(−
√
2
2 , −
√
2
2 ) = −
√
2X
2bi K biƺ`bi Q+? KBMbi p `/2 p f(x, y) = x + y + x2 + y2 /´ x2 + y2 ⩽ 1X
1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
17i2`bQK 7mMFiBQM2Mb ;`7 ` 2ii THM B `mKK2i UK2/ 2FpiBQM x + y −
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
No CPs! The gradient is constant, (1,1).
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB`
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
x y
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
√
2
√
2
√ √
2
√
2
√

256. 6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
T`Q#H2K 7`´M 2M/BKX .2`Bp2`BM; ;2` g!(φ) = − bBM φ + +Qb φ Q+?,
= 0 ⇔ bBM φ = +Qb φ ⇔ iM φ = 1 ⇔ φ =
π
4
2HH2`
5π
4
.
2 p `/2M 7ƺ` x(φ) = +Qb φ Q+? y(φ) = bBM φ,
x = y =
√
2
2 2HH2` x = y = −
√
2
2 .
b Kt T´ D ` f(
√
2
2 ,
√
2
2 ) =
√
2 Q+? KBM f(−
√
2
2 , −
√
2
2 ) = −
√
2X
2bi K biƺ`bi Q+? KBMbi p `/2 p f(x, y) = x + y + x2 + y2 /´ x2 + y2 ⩽ 1X
1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
17i2`bQK 7mMFiBQM2Mb ;`7 ` 2ii THM B `mKK2i UK2/ 2FpiBQM x + y −
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
No CPs! The gradient is constant, (1,1).
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB`
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
x y
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
√
2
√
2
√ √
2
√
2
√
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;`/B2M
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQ
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −

257. 6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
T`Q#H2K 7`´M 2M/BKX .2`Bp2`BM; ;2` g!(φ) = − bBM φ + +Qb φ Q+?,
= 0 ⇔ bBM φ = +Qb φ ⇔ iM φ = 1 ⇔ φ =
π
4
2HH2`
5π
4
.
2 p `/2M 7ƺ` x(φ) = +Qb φ Q+? y(φ) = bBM φ,
x = y =
√
2
2 2HH2` x = y = −
√
2
2 .
b Kt T´ D ` f(
√
2
2 ,
√
2
2 ) =
√
2 Q+? KBM f(−
√
2
2 , −
√
2
2 ) = −
√
2X
2bi K biƺ`bi Q+? KBMbi p `/2 p f(x, y) = x + y + x2 + y2 /´ x2 + y2 ⩽ 1X
1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
17i2`bQK 7mMFiBQM2Mb ;`7 ` 2ii THM B `mKK2i UK2/ 2FpiBQM x + y −
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
No CPs! The gradient is constant, (1,1).
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB`
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
x y
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
√
2
√
2
√ √
2
√
2
√
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;`/B2M
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQ
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQHHp2FiQ`MX
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb p2M B #BH/
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X

258. 6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
T`Q#H2K 7`´M 2M/BKX .2`Bp2`BM; ;2` g!(φ) = − bBM φ + +Qb φ Q+?,
= 0 ⇔ bBM φ = +Qb φ ⇔ iM φ = 1 ⇔ φ =
π
4
2HH2`
5π
4
.
2 p `/2M 7ƺ` x(φ) = +Qb φ Q+? y(φ) = bBM φ,
x = y =
√
2
2 2HH2` x = y = −
√
2
2 .
b Kt T´ D ` f(
√
2
2 ,
√
2
2 ) =
√
2 Q+? KBM f(−
√
2
2 , −
√
2
2 ) = −
√
2X
2bi K biƺ`bi Q+? KBMbi p `/2 p f(x, y) = x + y + x2 + y2 /´ x2 + y2 ⩽ 1X
1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
17i2`bQK 7mMFiBQM2Mb ;`7 ` 2ii THM B `mKK2i UK2/ 2FpiBQM x + y −
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
No CPs! The gradient is constant, (1,1).
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB`
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
x y
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
√
2
√
2
√ √
2
√
2
√
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;`/B2M
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQ
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQHHp2FiQ`MX
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb p2M B #BH/
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X

259. 6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
T`Q#H2K 7`´M 2M/BKX .2`Bp2`BM; ;2` g!(φ) = − bBM φ + +Qb φ Q+?,
= 0 ⇔ bBM φ = +Qb φ ⇔ iM φ = 1 ⇔ φ =
π
4
2HH2`
5π
4
.
2 p `/2M 7ƺ` x(φ) = +Qb φ Q+? y(φ) = bBM φ,
x = y =
√
2
2 2HH2` x = y = −
√
2
2 .
b Kt T´ D ` f(
√
2
2 ,
√
2
2 ) =
√
2 Q+? KBM f(−
√
2
2 , −
√
2
2 ) = −
√
2X
2bi K biƺ`bi Q+? KBMbi p `/2 p f(x, y) = x + y + x2 + y2 /´ x2 + y2 ⩽ 1X
1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
17i2`bQK 7mMFiBQM2Mb ;`7 ` 2ii THM B `mKK2i UK2/ 2FpiBQM x + y −
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
No CPs! The gradient is constant, (1,1).
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB`
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
x y
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
√
2
√
2
√ √
2
√
2
√
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;`/B2M
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQ
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQHHp2FiQ`MX
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb p2M B #BH/
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;`/B2Mi2M, ∇f(x, y) =
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQHHp2FiQ`MX .2ii
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb p2M B #BH/2M T´
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X

260. 6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
T`Q#H2K 7`´M 2M/BKX .2`Bp2`BM; ;2` g!(φ) = − bBM φ + +Qb φ Q+?,
= 0 ⇔ bBM φ = +Qb φ ⇔ iM φ = 1 ⇔ φ =
π
4
2HH2`
5π
4
.
2 p `/2M 7ƺ` x(φ) = +Qb φ Q+? y(φ) = bBM φ,
x = y =
√
2
2 2HH2` x = y = −
√
2
2 .
b Kt T´ D ` f(
√
2
2 ,
√
2
2 ) =
√
2 Q+? KBM f(−
√
2
2 , −
√
2
2 ) = −
√
2X
2bi K biƺ`bi Q+? KBMbi p `/2 p f(x, y) = x + y + x2 + y2 /´ x2 + y2 ⩽ 1X
1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
17i2`bQK 7mMFiBQM2Mb ;`7 ` 2ii THM B `mKK2i UK2/ 2FpiBQM x + y −
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
No CPs! The gradient is constant, (1,1).
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB`
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
x y
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
√
2
√
2
√ √
2
√
2
√
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;`/B2M
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQ
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQHHp2FiQ`MX
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb p2M B #BH/
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;`/B2Mi2M, ∇f(x, y) =
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQHHp2FiQ`MX .2ii
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb p2M B #BH/2M T´
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X
(1/ 2, 1/ 2)
(−1/ 2, − 1/ 2)

261. 6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
T`Q#H2K 7`´M 2M/BKX .2`Bp2`BM; ;2` g!(φ) = − bBM φ + +Qb φ Q+?,
= 0 ⇔ bBM φ = +Qb φ ⇔ iM φ = 1 ⇔ φ =
π
4
2HH2`
5π
4
.
2 p `/2M 7ƺ` x(φ) = +Qb φ Q+? y(φ) = bBM φ,
x = y =
√
2
2 2HH2` x = y = −
√
2
2 .
b Kt T´ D ` f(
√
2
2 ,
√
2
2 ) =
√
2 Q+? KBM f(−
√
2
2 , −
√
2
2 ) = −
√
2X
2bi K biƺ`bi Q+? KBMbi p `/2 p f(x, y) = x + y + x2 + y2 /´ x2 + y2 ⩽ 1X
1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
17i2`bQK 7mMFiBQM2Mb ;`7 ` 2ii THM B `mKK2i UK2/ 2FpiBQM x + y −
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
No CPs! The gradient is constant, (1,1).
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB`
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
x y
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5
4
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
√
2
√
2
√ √
2
√
2
√
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;`/B2M
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQ
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQHHp2FiQ`MX
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb p2M B #BH/
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;`/B2Mi2M, ∇f(x, y) =
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQHHp2FiQ`MX .2ii
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb p2M B #BH/2M T´
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X
bi´b BM; F`BiBbF TmMFi2` BMmiB +B`F2HbFBpMX .2i bvMb 7ƺ`bi´b Q+Fb´ T´ ;`/B2Mi2M, ∇f(x, y) =
(f!
x(x, y), f!
y(x, y)) = (1, 1)X :`/B2Mi2M ` FQMbiMi Q+? /2M / `7ƺ` #HB` H/`B; MQHHp2FiQ`MX .2ii
#2iv/2` ii Kt Q+? KBM iBHH 7mMFiBQM2M K´bi2 mTTM´b T´ +B`F2HMX .2ii bvMb p2M B #BH/2M T´
M bi bB/X
lb2 i?2 T`K2i`BbiBQM x(θ) = +Qb θ, y(θ) = bBM θ 7Q` 0 ⩽ θ ⩽ 2πX
h?Bb ?2HTb mb `2bi`B+i i?2 MmK#2` Q7 p`B#H2b iQ QM2,
g(θ) = f(+Qb θ, bBM θ) = +Qb θ + bBM θ.
Ai Bb *H+R@T`Q#H2KX .Bz2`2MiBiBM; ;Bp2b g!(θ) = − bBM θ + +Qb θ M/,
g!
(θ) = 0 ⇔ bBM θ = +Qb θ ⇔ iM θ = 1 ⇔ θ =
π
4
Q`
5π
4
.
h?Bb ;Bp2b i?2 7QHHQrBM; pHm2b 7Q` x(θ) = +Qb θ M/ y(θ) = bBM θ,
x = y =
√
2
2 Q` x = y = −
√
2
2 .
Mbr2`, h?2 Kt QM D Bb f(
√
2
2 ,
√
2
2 ) =
√
2 M/ i?2 KBM Bb f(−
√
2
2 , −
√
2
2 ) = −
√
2X

262. 13.3
Lagrange multipliers 1

263. Another way of determining extreme values for functions
on boundaries of sets

264. 1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
Zerodamage, Jacobmelgaard

265. 1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
The boundary of
the disk (the circle)
is the level curve
on level 0 for
g(x, y) = x2
+ y2
− 1
Zerodamage, Jacobmelgaard

266. Lagrange “case 1”
bQK (x(t), y(t)) K2/ (x(t0), y(t0)) = (a, b) U7ƺHD2` p BKTHB+Bi 7mMFiBQMbbib2MVX 6mMFiBQM2
ϕ(t) = f(x(t), y(t)) ?` /´ KtBKmK B t = t0- b´
(x′(t0), y′(t0))
(a, b)
6B;m` 9R, .2M `ƺ/ Fm`pM pBb` #BpBHHFQ`2i g(x, y) = cX .2 #H´ Fm`pQ`M ` MBp´Fm`pQ` 7ƺ
f(x, y)X SmMFi2M / ` /2i `ƺ/ #BpBHHFQ`2i iM;2`` 2M #H´ MBp´Fm`p ` /2i KtBKBTmMFi2M 7ƺ
JtBKBx2 f KBMBKBx2 f(x, y) mM/2` i?2 +QMbi`BMi g(x, y) = C.
FQ`2i g(x, y) = C /2}MB2`` 2M Fm`p B THM2iX oB #ƺ`D` K2/ 2ii 2t2KT2H / ` /2i- 7ƺ` F
y) = C- ` H ii ii Hƺb mi y bQK 2M 7mMFiBQM p xX S`Q#H2K2i #HB` /´ ;2Mbi 2M/BK2MbBQ
KTH2 e8X 2bi K /2i FQ`ibi2 Mbi´M/2i 7`´M Q`B;Q iBHH Fm`pM xy = 1, x, y 0X
HH KBMBK2` h(x, y) =
!
x2 + y2- 2HH2` 2FpBpH2Mi f(x, y) = x2 +y2 mM/2` #BpBHHFQ`2i xy
Q`2i FM bF`Bpb y = 1
x- b´ pB pBHH HHib´ KBMBK2`
g(x) = f(x, 1
x) = x2
+
1
x2
U7ƺ` x
−3 1


x = −1

267. Lagrange “case 1”
bQK (x(t), y(t)) K2/ (x(t0), y(t0)) = (a, b) U7ƺHD2` p BKTHB+Bi 7mMFiBQMbbib2MVX 6mMFiBQM2
ϕ(t) = f(x(t), y(t)) ?` /´ KtBKmK B t = t0- b´
(x′(t0), y′(t0))
(a, b)
6B;m` 9R, .2M `ƺ/ Fm`pM pBb` #BpBHHFQ`2i g(x, y) = cX .2 #H´ Fm`pQ`M ` MBp´Fm`pQ` 7ƺ
f(x, y)X SmMFi2M / ` /2i `ƺ/ #BpBHHFQ`2i iM;2`` 2M #H´ MBp´Fm`p ` /2i KtBKBTmMFi2M 7ƺ
JtBKBx2 f KBMBKBx2 f(x, y) mM/2` i?2 +QMbi`BMi g(x, y) = C.
FQ`2i g(x, y) = C /2}MB2`` 2M Fm`p B THM2iX oB #ƺ`D` K2/ 2ii 2t2KT2H / ` /2i- 7ƺ` F
y) = C- ` H ii ii Hƺb mi y bQK 2M 7mMFiBQM p xX S`Q#H2K2i #HB` /´ ;2Mbi 2M/BK2MbBQ
KTH2 e8X 2bi K /2i FQ`ibi2 Mbi´M/2i 7`´M Q`B;Q iBHH Fm`pM xy = 1, x, y 0X
HH KBMBK2` h(x, y) =
!
x2 + y2- 2HH2` 2FpBpH2Mi f(x, y) = x2 +y2 mM/2` #BpBHHFQ`2i xy
Q`2i FM bF`Bpb y = 1
x- b´ pB pBHH HHib´ KBMBK2`
g(x) = f(x, 1
x) = x2
+
1
x2
U7ƺ` x
−3 1


x = −1
⃗
r(t) = (x(t), y(t))

268. Lagrange “case 1”
bQK (x(t), y(t)) K2/ (x(t0), y(t0)) = (a, b) U7ƺHD2` p BKTHB+Bi 7mMFiBQMbbib2MVX 6mMFiBQM2
ϕ(t) = f(x(t), y(t)) ?` /´ KtBKmK B t = t0- b´
(x′(t0), y′(t0))
(a, b)
6B;m` 9R, .2M `ƺ/ Fm`pM pBb` #BpBHHFQ`2i g(x, y) = cX .2 #H´ Fm`pQ`M ` MBp´Fm`pQ` 7ƺ
f(x, y)X SmMFi2M / ` /2i `ƺ/ #BpBHHFQ`2i iM;2`` 2M #H´ MBp´Fm`p ` /2i KtBKBTmMFi2M 7ƺ
JtBKBx2 f KBMBKBx2 f(x, y) mM/2` i?2 +QMbi`BMi g(x, y) = C.
FQ`2i g(x, y) = C /2}MB2`` 2M Fm`p B THM2iX oB #ƺ`D` K2/ 2ii 2t2KT2H / ` /2i- 7ƺ` F
y) = C- ` H ii ii Hƺb mi y bQK 2M 7mMFiBQM p xX S`Q#H2K2i #HB` /´ ;2Mbi 2M/BK2MbBQ
KTH2 e8X 2bi K /2i FQ`ibi2 Mbi´M/2i 7`´M Q`B;Q iBHH Fm`pM xy = 1, x, y 0X
HH KBMBK2` h(x, y) =
!
x2 + y2- 2HH2` 2FpBpH2Mi f(x, y) = x2 +y2 mM/2` #BpBHHFQ`2i xy
Q`2i FM bF`Bpb y = 1
x- b´ pB pBHH HHib´ KBMBK2`
g(x) = f(x, 1
x) = x2
+
1
x2
U7ƺ` x
−3 1


x = −1
φ(t) = f( ⃗
r(t))
⃗
r(t) = (x(t), y(t))

269. Lagrange “case 1”
bQK (x(t), y(t)) K2/ (x(t0), y(t0)) = (a, b) U7ƺHD2` p BKTHB+Bi 7mMFiBQMbbib2MVX 6mMFiBQM2
ϕ(t) = f(x(t), y(t)) ?` /´ KtBKmK B t = t0- b´
(x′(t0), y′(t0))
(a, b)
6B;m` 9R, .2M `ƺ/ Fm`pM pBb` #BpBHHFQ`2i g(x, y) = cX .2 #H´ Fm`pQ`M ` MBp´Fm`pQ` 7ƺ
f(x, y)X SmMFi2M / ` /2i `ƺ/ #BpBHHFQ`2i iM;2`` 2M #H´ MBp´Fm`p ` /2i KtBKBTmMFi2M 7ƺ
(x′(t0), y′(t0))
(a, b)
6B;m` 9R, .2M `ƺ/ Fm`pM pBb` #BpBHHFQ`2i g(x, y) = cX .2 #H´ Fm`pQ`M ` MBp´Fm`pQ`
f(x, y)X SmMFi2M / ` /2i `ƺ/ #BpBHHFQ`2i iM;2`` 2M #H´ MBp´Fm`p ` /2i KtBKBTmMFi2M
f(x, y) H M;b #BpBHHFQ`2i- 27i2`bQK d1 d2X UBH/, qBFBT2/BXV
0 = ϕ!
(t0) = f!
1(x(t0), y(t0))x!
(t0) + f!
2(x(t0), y(t0))y!
(t0) = ∇f(a, b) · (x!
(t0), y!
(t0)).
J2M (x!(t0), y!(t0)) ` iM;2Mip2FiQ`M iBHH Fm`pM g(x, y) = 0 B (a, b)X .2MM p2FiQ` ` pBMF2H`
KQi ∇g(a, b)- b´ 2MHB;i 2FpiBQM2M QpM b´ `
JtBKBx2 f KBMBKBx2 f(x, y) mM/2` i?2 +QMbi`BMi g(x, y) = C.
FQ`2i g(x, y) = C /2}MB2`` 2M Fm`p B THM2iX oB #ƺ`D` K2/ 2ii 2t2KT2H / ` /2i- 7ƺ` F
y) = C- ` H ii ii Hƺb mi y bQK 2M 7mMFiBQM p xX S`Q#H2K2i #HB` /´ ;2Mbi 2M/BK2MbBQ
KTH2 e8X 2bi K /2i FQ`ibi2 Mbi´M/2i 7`´M Q`B;Q iBHH Fm`pM xy = 1, x, y 0X
HH KBMBK2` h(x, y) =
!
x2 + y2- 2HH2` 2FpBpH2Mi f(x, y) = x2 +y2 mM/2` #BpBHHFQ`2i xy
Q`2i FM bF`Bpb y = 1
x- b´ pB pBHH HHib´ KBMBK2`
g(x) = f(x, 1
x) = x2
+
1
x2
U7ƺ` x
−3 1


x = −1
φ(t) = f( ⃗
r(t))
⃗
r(t) = (x(t), y(t))

270. Lagrange “case 1”
bQK (x(t), y(t)) K2/ (x(t0), y(t0)) = (a, b) U7ƺHD2` p BKTHB+Bi 7mMFiBQMbbib2MVX 6mMFiBQM2
ϕ(t) = f(x(t), y(t)) ?` /´ KtBKmK B t = t0- b´
(x′(t0), y′(t0))
(a, b)
6B;m` 9R, .2M `ƺ/ Fm`pM pBb` #BpBHHFQ`2i g(x, y) = cX .2 #H´ Fm`pQ`M ` MBp´Fm`pQ` 7ƺ
f(x, y)X SmMFi2M / ` /2i `ƺ/ #BpBHHFQ`2i iM;2`` 2M #H´ MBp´Fm`p ` /2i KtBKBTmMFi2M 7ƺ
(x′(t0), y′(t0))
(a, b)
6B;m` 9R, .2M `ƺ/ Fm`pM pBb` #BpBHHFQ`2i g(x, y) = cX .2 #H´ Fm`pQ`M ` MBp´Fm`pQ`
f(x, y)X SmMFi2M / ` /2i `ƺ/ #BpBHHFQ`2i iM;2`` 2M #H´ MBp´Fm`p ` /2i KtBKBTmMFi2M
f(x, y) H M;b #BpBHHFQ`2i- 27i2`bQK d1 d2X UBH/, qBFBT2/BXV
0 = ϕ!
(t0) = f!
1(x(t0), y(t0))x!
(t0) + f!
2(x(t0), y(t0))y!
(t0) = ∇f(a, b) · (x!
(t0), y!
(t0)).
J2M (x!(t0), y!(t0)) ` iM;2Mip2FiQ`M iBHH Fm`pM g(x, y) = 0 B (a, b)X .2MM p2FiQ` ` pBMF2H`
KQi ∇g(a, b)- b´ 2MHB;i 2FpiBQM2M QpM b´ `
JtBKBx2 f KBMBKBx2 f(x, y) mM/2` i?2 +QMbi`BMi g(x, y) = C.
FQ`2i g(x, y) = C /2}MB2`` 2M Fm`p B THM2iX oB #ƺ`D` K2/ 2ii 2t2KT2H / ` /2i- 7ƺ` F
y) = C- ` H ii ii Hƺb mi y bQK 2M 7mMFiBQM p xX S`Q#H2K2i #HB` /´ ;2Mbi 2M/BK2MbBQ
KTH2 e8X 2bi K /2i FQ`ibi2 Mbi´M/2i 7`´M Q`B;Q iBHH Fm`pM xy = 1, x, y 0X
HH KBMBK2` h(x, y) =
!
x2 + y2- 2HH2` 2FpBpH2Mi f(x, y) = x2 +y2 mM/2` #BpBHHFQ`2i xy
Q`2i FM bF`Bpb y = 1
x- b´ pB pBHH HHib´ KBMBK2`
g(x) = f(x, 1
x) = x2
+
1
x2
U7ƺ` x
−3 1


x = −1
φ(t) = f( ⃗
r(t))
⃗
r(t) = (x(t), y(t))
orthogonal to ∇g(a, b)

271. Lagrange “case 1”
bQK (x(t), y(t)) K2/ (x(t0), y(t0)) = (a, b) U7ƺHD2` p BKTHB+Bi 7mMFiBQMbbib2MVX 6mMFiBQM2
ϕ(t) = f(x(t), y(t)) ?` /´ KtBKmK B t = t0- b´
(x′(t0), y′(t0))
(a, b)
6B;m` 9R, .2M `ƺ/ Fm`pM pBb` #BpBHHFQ`2i g(x, y) = cX .2 #H´ Fm`pQ`M ` MBp´Fm`pQ` 7ƺ
f(x, y)X SmMFi2M / ` /2i `ƺ/ #BpBHHFQ`2i iM;2`` 2M #H´ MBp´Fm`p ` /2i KtBKBTmMFi2M 7ƺ
(x′(t0), y′(t0))
(a, b)
6B;m` 9R, .2M `ƺ/ Fm`pM pBb` #BpBHHFQ`2i g(x, y) = cX .2 #H´ Fm`pQ`M ` MBp´Fm`pQ`
f(x, y)X SmMFi2M / ` /2i `ƺ/ #BpBHHFQ`2i iM;2`` 2M #H´ MBp´Fm`p ` /2i KtBKBTmMFi2M
f(x, y) H M;b #BpBHHFQ`2i- 27i2`bQK d1 d2X UBH/, qBFBT2/BXV
0 = ϕ!
(t0) = f!
1(x(t0), y(t0))x!
(t0) + f!
2(x(t0), y(t0))y!
(t0) = ∇f(a, b) · (x!
(t0), y!
(t0)).
J2M (x!(t0), y!(t0)) ` iM;2Mip2FiQ`M iBHH Fm`pM g(x, y) = 0 B (a, b)X .2MM p2FiQ` ` pBMF2H`
KQi ∇g(a, b)- b´ 2MHB;i 2FpiBQM2M QpM b´ `
JtBKBx2 f KBMBKBx2 f(x, y) mM/2` i?2 +QMbi`BMi g(x, y) = C.
FQ`2i g(x, y) = C /2}MB2`` 2M Fm`p B THM2iX oB #ƺ`D` K2/ 2ii 2t2KT2H / ` /2i- 7ƺ` F
y) = C- ` H ii ii Hƺb mi y bQK 2M 7mMFiBQM p xX S`Q#H2K2i #HB` /´ ;2Mbi 2M/BK2MbBQ
KTH2 e8X 2bi K /2i FQ`ibi2 Mbi´M/2i 7`´M Q`B;Q iBHH Fm`pM xy = 1, x, y 0X
HH KBMBK2` h(x, y) =
!
x2 + y2- 2HH2` 2FpBpH2Mi f(x, y) = x2 +y2 mM/2` #BpBHHFQ`2i xy
Q`2i FM bF`Bpb y = 1
x- b´ pB pBHH HHib´ KBMBK2`
g(x) = f(x, 1
x) = x2
+
1
x2
U7ƺ` x
−3 1


x = −1
φ(t) = f( ⃗
r(t))
⃗
r(t) = (x(t), y(t))
orthogonal to ∇g(a, b)
The gradients and must be parallel!
∇f(a, b) ∇g(a, b)

272. How to find max and min for f on g(x, y) = 0?
RX 1M/@TQBMib Q7 g(x, y) = 0
kX SQBMib r?2`2 ∇g(x, y) = !
0 UQ` bBM;mH`V
jX SQBMib biBb7vBM; i?2 bvbi2K Q7 2[miBQMb
UBM /KbV





f!
1(x, y) = λg!
1(x, y)
f!
2(x, y) = λg!
2(x, y)
g(x, y) = 0.
P`- 2[mBpH2MiHv,





%
%
%
%
%
f!
1(x, y) f!
2(x, y)
g!
1(x, y) g!
2(x, y)
%
%
%
%
%
= 0.
g(x, y) = 0.

273. ∇f∥∇g
∇f∥∇g
How to find max and min for f on g(x, y) = 0?
The point belongs to the level curve
The point belongs to the level curve
RX 1M/@TQBMib Q7 g(x, y) = 0
kX SQBMib r?2`2 ∇g(x, y) = !
0 UQ` bBM;mH`V
jX SQBMib biBb7vBM; i?2 bvbi2K Q7 2[miBQMb
UBM /KbV





f!
1(x, y) = λg!
1(x, y)
f!
2(x, y) = λg!
2(x, y)
g(x, y) = 0.
P`- 2[mBpH2MiHv,





%
%
%
%
%
f!
1(x, y) f!
2(x, y)
g!
1(x, y) g!
2(x, y)
%
%
%
%
%
= 0.
g(x, y) = 0.

274. 1tKTH2 eyX 6BM/ Kt M/ KBM Q7 f(x, y) = x + y QM i?2 /BbF D : x2 + y2 ⩽ 1X
6B;m` jN, 1t2KT2H eyX UBH/, qBFBT2/BXV
The boundary of
the disk (the circle)
is the level curve
on level 0 for
g(x, y) = x2
+ y2
− 1
Zerodamage, Jacobmelgaard

275. 2HH2` ~2` #BpBHHFQ` U+QMbi`BMibVX 1ii ivT2t2KT2H `,
JtBKBx2 f KBMBKBx2 f(x, y) mM/2` i?2 +QMbi`BMi g(x, y) = C.
BpBHHFQ`2i g(x, y) = C /2}MB2`` 2M Fm`p B THM2iX oB #ƺ`D` K2/ 2ii 2t2KT2H / ` /2i- 7ƺ` Fm`pM
g(x, y) = C- ` H ii ii Hƺb mi y bQK 2M 7mMFiBQM p xX S`Q#H2K2i #HB` /´ ;2Mbi 2M/BK2MbBQM2HHiX
1tKTH2 e8X 2bi K /2i FQ`ibi2 Mbi´M/2i 7`´M Q`B;Q iBHH Fm`pM xy = 1, x, y 0X
oB pBHH KBMBK2` h(x, y) =
!
x2 + y2- 2HH2` 2FpBpH2Mi f(x, y) = x2 +y2 mM/2` #BpBHHFQ`2i xy = 1X
oBHHFQ`2i FM bF`Bpb y = 1
x- b´ pB pBHH HHib´ KBMBK2`
g(x) = f(x, 1
x) = x2
+
1
x2
U7ƺ` x 0V
g
(x) = 2x − 2x−3
= 0 ⇒ 2x(1 −
1
x4
) = 0 ⇒





x = −1
x = 0
x = 1
MQi2` ii /2 ip´ 7ƺ`bi KƺDHB; p `/2M T´ x 2D mTT7vHH2` BMBiBHpBHHFQ`2i x, y 0X A x = 1 7´` pB
pbi´M/2i
12 + (1
1 )2 =
√
2 Q+? /2i ` /2i FQ`ibi2 pbi´M/2i 27i2`bQK g(1) = 2 + 6
14 = 8 0X
1ii Hi2`MiBp bQK FM p` Mp M/#`i QK /2i BMi2 ;´` ii Hƺb mi y bQK 2M 7mMFiBQM p x B
g(x, y) = C ` ii Mp M/ 2M T`K2i`Bb2`BM; p g(x, y) = CX PK pB 2t2KT2HpBb pBHH KtBK2`
f(x, y) = x + y, g(x, y) = x2
+ y2
− 1
RX 1M/@TQBMib Q7 g(x, y) = 0
kX SQBMib r?2`2 ∇g(x, y) = !
0 UQ` bBM;mH`V
jX SQBMib biBb7vBM; i?2 bvbi2K Q7 2[miBQMb
UBM /KbV





f!
1(x, y) = λg!
1(x, y)
f!
2(x, y) = λg!
2(x, y)
g(x, y) = 0.
P`- 2[mBpH2MiHv,





%
%
%
%
%
f!
1(x, y) f!
2(x, y)
g!
1(x, y) g!
2(x, y)
%
%
%
%
%
= 0.
g(x, y) = 0.
1FpiBQMbbvbi2K2i FM Q+Fb´ bF`Bpb FQ`i7iii bQK ∇L(x, y, λ) = !
0 / ` L(x, y, λ) = f
λg(x, y) FHHb 7ƺ` GM;`M;27mMFiBQM2MX A };m`2M QpM b2` pB T`#QHQB/2M bQK
f(x, y) = 1
2(x2 + y2) + 9 K2/ iBHH?ƺ`M/2 MBp´Fm`pQ` T`QDB+2`/2 T´ xy−THM2iX oB b

276. Max min of fCx y xty
on x
y L GGy x7y I
f L fy 1
gI 2x
gy 2y
I
1
2
1
2
1
2
Ex 1 Cfhxth O
i

277. Find max and min for
on the circle .
f(x, y) = x2
y
x2
+ y2
= 3
2 2 2 2
KTH2 edX JtBK2` Q+? KBMBK2` f(x, y) = x2y T´ +B`F2HM x2+y2 = 3- HHib´ T´ MBp´Fm`
y) = 0 iBHH g(x, y) = x2 + y2 − 3X
Zerodamage, Jacobmelgaard

278. f(x, y) = x2
y, g(x, y) = x2
+ y2
− 3
RX 1M/@TQBMib Q7 g(x, y) = 0
kX SQBMib r?2`2 ∇g(x, y) = !
0 UQ` bBM;mH`V
jX SQBMib biBb7vBM; i?2 bvbi2K Q7 2[miBQMb
UBM /KbV





f!
1(x, y) = λg!
1(x, y)
f!
2(x, y) = λg!
2(x, y)
g(x, y) = 0.
P`- 2[mBpH2MiHv,





%
%
%
%
%
f!
1(x, y) f!
2(x, y)
g!
1(x, y) g!
2(x, y)
%
%
%
%
%
= 0.
g(x, y) = 0.
1FpiBQMbbvbi2K2i FM Q+Fb´ bF`Bpb FQ`i7iii bQK ∇L(x, y, λ) = !
0 / ` L(x, y, λ) = f
λg(x, y) FHHb 7ƺ` GM;`M;27mMFiBQM2MX A };m`2M QpM b2` pB T`#QHQB/2M bQK
f(x, y) = 1
2(x2 + y2) + 9 K2/ iBHH?ƺ`M/2 MBp´Fm`pQ` T`QDB+2`/2 T´ xy−THM2iX oB b
Find max and min for
on the circle .
f(x, y) = x2
y
x2
+ y2
= 3

279. fCxiyFEy gCxy x2ty2 3
fI 2xy fy
_x2
gI 2x gj 2y
q
In
n
2 25 5 0
Ivx2 2y
y 3
i i
11 A

280. II A
x x x x
f ran 2 far 11 2
f try 2 f tribe
2
Max is 2
Min is 2

281. Lagrange “case 2”
M;2`` T´ bKK b ii
M f(x, y, z) T´ viM
ii pB KtBK2`` f KB@
MBp´viM iBHH 2M MMM
BHHFQ`2i ii ;`/B2Mi2`@
HH2HH B 2ti`2KTmMFi2`
p`M/`VX oB 7´` /´
2FpiBQM2` Q+? 9 p`B@











f!
1(x, y, z) = λg!
1(x, y, z)
f!
2(x, y, z) = λg!
2(x, y, z)
f!
3(x, y, z) = λg!
3(x, y, z)
g(x, y, z) = 0
bi /2i }MMb BM;2M KƺDHB;?2i ii bF`Bp mTT 2M /2i2`KBMMi
Mi2`- K2/M p`D2 ;`/B2Mi ?` i`2 FQKTQM2Mi2`, ?2H?2i2M `
HH mTT 2M T`QTQ`iBQM bQK mii`v+F2` 7FimK ii ;`/B2Mi2`M
∇f∥∇g
Maximize / minimize f (x, y, z) under the constraint g (x, y, z) = 0.
The point belongs to the level surface

282. Lagrange “case 2”
M;2`` T´ bKK b ii
M f(x, y, z) T´ viM
ii pB KtBK2`` f KB@
MBp´viM iBHH 2M MMM
BHHFQ`2i ii ;`/B2Mi2`@
HH2HH B 2ti`2KTmMFi2`
p`M/`VX oB 7´` /´
2FpiBQM2` Q+? 9 p`B@











f!
1(x, y, z) = λg!
1(x, y, z)
f!
2(x, y, z) = λg!
2(x, y, z)
f!
3(x, y, z) = λg!
3(x, y, z)
g(x, y, z) = 0
bi /2i }MMb BM;2M KƺDHB;?2i ii bF`Bp mTT 2M /2i2`KBMMi
Mi2`- K2/M p`D2 ;`/B2Mi ?` i`2 FQKTQM2Mi2`, ?2H?2i2M `
HH mTT 2M T`QTQ`iBQM bQK mii`v+F2` 7FimK ii ;`/B2Mi2`M
∇f∥∇g
Maximize / minimize f (x, y, z) under the constraint g (x, y, z) = 0.
The point belongs to the level surface
Warning: never put a variable expression in denominators without explanations!

283. iBpi Q+? #2`Q` T´ QK pB MHvb2`` 2M K M;/ B R- R 2HH2` R X Ua2 7ƺ`FH`BM;2M B #BH/2M T´ bX RyRV
1tKTH2 eNX .2i2`KBM2 Kt KBM 7Q` f(x, y, z) = x+2y+3z rBi? +QMbi`BMi x2+y2+z2 = 14,
bQ QM i?2 bT?2`2 rBi? +2Mi`2 BM i?2 Q`B;BM M/ `/Bmb
√
14X
lb2 G;`M;2Ƕb K2i?Q/ +b2 kX h?2 2ti`2K2 pHm2b +M Q++m` r?2`2 i?2 ;`/B2Mib Q7 f M/ g
`2 T`HH2HX
∇f = (1, 2, 3), ∇g = (2x, 2y, 2z).
hrQ p2+iQ`b `2 T`HH2H B7 i?2B` +QQ`/BMi2b biBb7v i?2 7QHHQrBM; T`QTQ`iBQM UMQ T`Q#H2Kb ?2`2,
MQ p`B#H2b BM /2MQKBMiQ`bV,
∇f||∇g ⇔
2x
1
=
2y
2
=
2z
3
oHm2b x = y = z = 0 biBb7v i?2 T`QTQ`iBQM- #mi (0, 0, 0) /Q2b MQi #2HQM; iQ i?2 bT?2`2- bQ Bi
+MMQi #2 Qm` TQBMiX A7 ∇g $=
0- i?2 T`QTQ`iBQM BKTHB2b i?i
2x = y =
2
3
z ⇔ (y = 2x M/ z =
3
2
y =
3
2
· 2x = 3x).
SHm;;BM; BM i?2b2 pHm2b BM i?2 +QMbi`BMi g(x, y, z) = 0 ;Bp2b
x2
+ (2x)2
+ (3x)2
= 14 ⇔ 14x2
= 14 ⇔ (x = 1 Q` x = −1).
oHm2 x = 1 ;Bp2b i?2 TQBMi (1, 2, 3) rBi? f(1, 2, 3) = 1 + 2 · 2 + 3 · 3 = 14- r?BH2 x = −1 ;Bp2b
i?2 TQBMi (−1, −2, −3) rBi? f(−1, −2, −3) = −14X
Mbr2`, h?2 Kt pHm2 7Q` f QM i?2 bT?2`2 Bb R9 M/ i?2 KBM Bb −14X

284. Find max and min for
in the ball .
f(x, y, z) = xyz
x2
+ y2
+ z2
⩽ 12
Inside the ball: CPs; on the sphere: Lagrange 2

285. Lagrange “case 2”
M;2`` T´ bKK b ii
M f(x, y, z) T´ viM
ii pB KtBK2`` f KB@
MBp´viM iBHH 2M MMM
BHHFQ`2i ii ;`/B2Mi2`@
HH2HH B 2ti`2KTmMFi2`
p`M/`VX oB 7´` /´
2FpiBQM2` Q+? 9 p`B@











f!
1(x, y, z) = λg!
1(x, y, z)
f!
2(x, y, z) = λg!
2(x, y, z)
f!
3(x, y, z) = λg!
3(x, y, z)
g(x, y, z) = 0
bi /2i }MMb BM;2M KƺDHB;?2i ii bF`Bp mTT 2M /2i2`KBMMi
Mi2`- K2/M p`D2 ;`/B2Mi ?` i`2 FQKTQM2Mi2`, ?2H?2i2M `
HH mTT 2M T`QTQ`iBQM bQK mii`v+F2` 7FimK ii ;`/B2Mi2`M
∇f∥∇g
Maximize / minimize f (x, y, z)
under the constraint g (x, y, z) = 0.
The point belongs to the level surface
Find max and min for
in the ball .
f(x, y, z) = xyz
x2
+ y2
+ z2
⩽ 12

286. non
th mum
t.EE
EEEQgCx
yiz xty2tz2 12
1 CP inside the ball
f o E
E
I
some of x
y
and Z must
be equal to zero
fCx y z in CP
Our function attains both
e and
Feu Tippu
in the bell
in the first octant side of the
first ochend through the
origin

287. The CP are not interesting
2 Lagrange case 2 on the sphere
Of yz
XZ
xy
Og
2x 2y Zz
fHrg x4yi
7 Observe that max and min
are obtained in the points
with X
y 2 0
I can assume that the values
I am looking for are
In
Trimm L m
expressions u
L E

288. Mutt by z
yt
2
y2 z2
x4y2tz 42
3
2
12
2 x 2
I n l
y 2
y 2
y
2
y 2
2 2 z z 2 2 2 2 2
We get 8 points
f 2,2 2 8 and 3 more
f f 2 2 2 8 and 3 more
Answer Max is 8 minus 8

289. Lagrange “case 3”
Maximize / minimize f (x, y, z) under the constraints g (x, y, z) = 0 and h (x, y, z) = 0.

290. g(x, y, z) = 0, h(x, y, z) = 0
∇g(x, y, z)
∇h(x, y, z)
tangent
∇f(x, y, z)

291. Lagrange “case 3”
Maximize / minimize f (x, y, z) under the constraints g (x, y, z) = 0 and h (x, y, z) = 0.
h(x, y, z) = 0X .2 #´/ pBHHFQ`2M /2}MB2`` 2M Fm`p B `mKK2i UbF `MBM;bFm`pM iBHH MBp´viQ` iBHH
ip´ i`2p`B#2H7mMFiBQM2`VX A p`D2 TmMFi (x, y, z) T´ Fm`pM ` #´/2 ∇g(x, y, z) Q+? ∇h(x, y, z)
pBMF2H` i KQi Fm`pM- b´ p2FiQ`M T(x, y, z) = ∇g(x, y, z) × ∇h(x, y, z) ` T`HH2HH K2/ Fm`pMb
iM;2Mi B (x, y, z)X 1MHB;i `2bQM2KM;2i M ` pB BM7ƺ`/2 KmHiTHBFiQ`K2iQ/2M- b´ K´bi2 /2i B
2ti`2KTmMFi2M (a, b, c) ; HH ii ∇f(a, b, c) · T(a, b, c) = 0- /XpXbX
∇f(a, b, c), ∇g(a, b, c), ∇h(a, b, c) HB2 QM i?2 bK2 THM2X
.2 i`2 ;`/B2Mi2`M ` HBMD `i #2`Q2M/2- pBHF2i ` 2FpBpH2Mi K2/ ii /2i2`KBMMi2M p 2M Ki`Bb
K2/ ∇f- ∇g- ∇h T´ p` bBM `/ K´bi2 p` MQHH B 2ii KtBKmKfKBMBKmK,















%
%
%
%
%
%
%
f!
1(x, y, z) f!
2(x, y, z) f!
3(x, y, z)
g!
1(x, y, z) g!
2(x, y, z) g!
3(x, y, z)
h!
1(x, y, z) h!
2(x, y, z) h!
3(x, y, z)
%
%
%
%
%
%
%
= 0
g(x, y, z) = 0
h(x, y, z) = 0.
1tKTH2 dRX 6BM/ Kt M/ KBM 7Q` f(x, y, z) = y mM/2` +QMbi`BMib
2 2 2

292. Scalar triple product: volume of a parallelepiped
If u, v, w 2 R3
then
u · (v ⇥ w) = u1 ·
v2 v3
w2 w3
u2 ·
v1 v3
w1 w3
+ u3 ·
v1 v2
w1 w2
=
u1 u2 u3
v1 v2 v3
w1 w2 w3
.
v
w
u
v w
u cos
u, v, w 2 R3
lie in the same plane i↵ u · (v ⇥ w) = 0.
Scalar triple product: (signed) volume of a parallelepiped
If u, v, w 2 R3
then
u · (v ⇥ w) = u1 ·
v2 v3
w2 w3
u2 ·
v1 v3
w1 w3
+ u3 ·
v1 v2
w1 w2
=
u1 u2 u3
v1 v2 v3
w1 w2 w3
.
v
w
u
v w
u cos
u, v, w 2 R3
lie in the same plane i↵ u · (v ⇥ w) = 0.
the parallelepiped spanned by vectors u, v and w

293. BMF2H` i KQi Fm`pM- b´ p2FiQ`M T(x, y, z) = ∇g(x, y, z) × ∇h(x, y, z) ` T`HH2HH K2/ Fm`p
M;2Mi B (x, y, z)X 1MHB;i `2bQM2KM;2i M ` pB BM7ƺ`/2 KmHiTHBFiQ`K2iQ/2M- b´ K´bi2 /
ti`2KTmMFi2M (a, b, c) ; HH ii ∇f(a, b, c) · T(a, b, c) = 0- /XpXbX
∇f(a, b, c), ∇g(a, b, c), ∇h(a, b, c) HB;;2` B bKK THMX
.2 i`2 ;`/B2Mi2`M ` HBMD `i #2`Q2M/2- pBHF2i ` 2FpBpH2Mi K2/ ii /2i2`KBMMi2M p 2M K
K2/ ∇f- ∇g- ∇h T´ p` bBM `/ K´bi2 p` MQHH B 2ii KtBKmKfKBMBKmK,















%
%
%
%
%
%
%
f!
1(x, y, z) f!
2(x, y, z) f!
3(x, y, z)
g!
1(x, y, z) g!
2(x, y, z) g!
3(x, y, z)
h!
1(x, y, z) h!
2(x, y, z) h!
3(x, y, z)
%
%
%
%
%
%
%
= 0
g(x, y, z) = 0
h(x, y, z) = 0.
1t2KT2H dRX JtBK2` Q+? KBMBK2` f(x, y, z) = y mM/2` #BpBHHFQ`2M
g(x, y, z) = x + y + z − 1 = 0 Q+? h(x, y, z) = x2
+ y2
+ z2
− 1 = 0.
Lagrange “case 3”
Maximize / minimize f (x, y, z) under the constraints g (x, y, z) = 0 and h (x, y, z) = 0.
h(x, y, z) = 0X .2 #´/ pBHHFQ`2M /2}MB2`` 2M Fm`p B `mKK2i UbF `MBM;bFm`pM iBHH MBp´viQ` iBHH
ip´ i`2p`B#2H7mMFiBQM2`VX A p`D2 TmMFi (x, y, z) T´ Fm`pM ` #´/2 ∇g(x, y, z) Q+? ∇h(x, y, z)
pBMF2H` i KQi Fm`pM- b´ p2FiQ`M T(x, y, z) = ∇g(x, y, z) × ∇h(x, y, z) ` T`HH2HH K2/ Fm`pMb
iM;2Mi B (x, y, z)X 1MHB;i `2bQM2KM;2i M ` pB BM7ƺ`/2 KmHiTHBFiQ`K2iQ/2M- b´ K´bi2 /2i B
2ti`2KTmMFi2M (a, b, c) ; HH ii ∇f(a, b, c) · T(a, b, c) = 0- /XpXbX
∇f(a, b, c), ∇g(a, b, c), ∇h(a, b, c) HB2 QM i?2 bK2 THM2X
.2 i`2 ;`/B2Mi2`M ` HBMD `i #2`Q2M/2- pBHF2i ` 2FpBpH2Mi K2/ ii /2i2`KBMMi2M p 2M Ki`Bb
K2/ ∇f- ∇g- ∇h T´ p` bBM `/ K´bi2 p` MQHH B 2ii KtBKmKfKBMBKmK,















%
%
%
%
%
%
%
f!
1(x, y, z) f!
2(x, y, z) f!
3(x, y, z)
g!
1(x, y, z) g!
2(x, y, z) g!
3(x, y, z)
h!
1(x, y, z) h!
2(x, y, z) h!
3(x, y, z)
%
%
%
%
%
%
%
= 0
g(x, y, z) = 0
h(x, y, z) = 0.
1tKTH2 dRX 6BM/ Kt M/ KBM 7Q` f(x, y, z) = y mM/2` +QMbi`BMib
2 2 2

294. BMF2H` i KQi Fm`pM- b´ p2FiQ`M T(x, y, z) = ∇g(x, y, z) × ∇h(x, y, z) ` T`HH2HH K2/ Fm`p
M;2Mi B (x, y, z)X 1MHB;i `2bQM2KM;2i M ` pB BM7ƺ`/2 KmHiTHBFiQ`K2iQ/2M- b´ K´bi2 /
ti`2KTmMFi2M (a, b, c) ; HH ii ∇f(a, b, c) · T(a, b, c) = 0- /XpXbX
∇f(a, b, c), ∇g(a, b, c), ∇h(a, b, c) HB;;2` B bKK THMX
.2 i`2 ;`/B2Mi2`M ` HBMD `i #2`Q2M/2- pBHF2i ` 2FpBpH2Mi K2/ ii /2i2`KBMMi2M p 2M K
K2/ ∇f- ∇g- ∇h T´ p` bBM `/ K´bi2 p` MQHH B 2ii KtBKmKfKBMBKmK,















%
%
%
%
%
%
%
f!
1(x, y, z) f!
2(x, y, z) f!
3(x, y, z)
g!
1(x, y, z) g!
2(x, y, z) g!
3(x, y, z)
h!
1(x, y, z) h!
2(x, y, z) h!
3(x, y, z)
%
%
%
%
%
%
%
= 0
g(x, y, z) = 0
h(x, y, z) = 0.
1t2KT2H dRX JtBK2` Q+? KBMBK2` f(x, y, z) = y mM/2` #BpBHHFQ`2M
g(x, y, z) = x + y + z − 1 = 0 Q+? h(x, y, z) = x2
+ y2
+ z2
− 1 = 0.
Lagrange “case 3”
Maximize / minimize f (x, y, z) under the constraints g (x, y, z) = 0 and h (x, y, z) = 0.
h(x, y, z) = 0X .2 #´/ pBHHFQ`2M /2}MB2`` 2M Fm`p B `mKK2i UbF `MBM;bFm`pM iBHH MBp´viQ` iBHH
ip´ i`2p`B#2H7mMFiBQM2`VX A p`D2 TmMFi (x, y, z) T´ Fm`pM ` #´/2 ∇g(x, y, z) Q+? ∇h(x, y, z)
pBMF2H` i KQi Fm`pM- b´ p2FiQ`M T(x, y, z) = ∇g(x, y, z) × ∇h(x, y, z) ` T`HH2HH K2/ Fm`pMb
iM;2Mi B (x, y, z)X 1MHB;i `2bQM2KM;2i M ` pB BM7ƺ`/2 KmHiTHBFiQ`K2iQ/2M- b´ K´bi2 /2i B
2ti`2KTmMFi2M (a, b, c) ; HH ii ∇f(a, b, c) · T(a, b, c) = 0- /XpXbX
∇f(a, b, c), ∇g(a, b, c), ∇h(a, b, c) HB2 QM i?2 bK2 THM2X
.2 i`2 ;`/B2Mi2`M ` HBMD `i #2`Q2M/2- pBHF2i ` 2FpBpH2Mi K2/ ii /2i2`KBMMi2M p 2M Ki`Bb
K2/ ∇f- ∇g- ∇h T´ p` bBM `/ K´bi2 p` MQHH B 2ii KtBKmKfKBMBKmK,















%
%
%
%
%
%
%
f!
1(x, y, z) f!
2(x, y, z) f!
3(x, y, z)
g!
1(x, y, z) g!
2(x, y, z) g!
3(x, y, z)
h!
1(x, y, z) h!
2(x, y, z) h!
3(x, y, z)
%
%
%
%
%
%
%
= 0
g(x, y, z) = 0
h(x, y, z) = 0.
1tKTH2 dRX 6BM/ Kt M/ KBM 7Q` f(x, y, z) = y mM/2` +QMbi`BMib
2 2 2
the gradients lie
on the same plane
the point belongs
to both level surfaces

295. BMF2H` i KQi Fm`pM- b´ p2FiQ`M T(x, y, z) = ∇g(x, y, z) × ∇h(x, y, z) ` T`HH2HH K2/ Fm`p
M;2Mi B (x, y, z)X 1MHB;i `2bQM2KM;2i M ` pB BM7ƺ`/2 KmHiTHBFiQ`K2iQ/2M- b´ K´bi2 /
ti`2KTmMFi2M (a, b, c) ; HH ii ∇f(a, b, c) · T(a, b, c) = 0- /XpXbX
∇f(a, b, c), ∇g(a, b, c), ∇h(a, b, c) HB;;2` B bKK THMX
.2 i`2 ;`/B2Mi2`M ` HBMD `i #2`Q2M/2- pBHF2i ` 2FpBpH2Mi K2/ ii /2i2`KBMMi2M p 2M K
K2/ ∇f- ∇g- ∇h T´ p` bBM `/ K´bi2 p` MQHH B 2ii KtBKmKfKBMBKmK,















%
%
%
%
%
%
%
f!
1(x, y, z) f!
2(x, y, z) f!
3(x, y, z)
g!
1(x, y, z) g!
2(x, y, z) g!
3(x, y, z)
h!
1(x, y, z) h!
2(x, y, z) h!
3(x, y, z)
%
%
%
%
%
%
%
= 0
g(x, y, z) = 0
h(x, y, z) = 0.
1t2KT2H dRX JtBK2` Q+? KBMBK2` f(x, y, z) = y mM/2` #BpBHHFQ`2M
g(x, y, z) = x + y + z − 1 = 0 Q+? h(x, y, z) = x2
+ y2
+ z2
− 1 = 0.
Lagrange “case 3”
Maximize / minimize f (x, y, z) under the constraints g (x, y, z) = 0 and h (x, y, z) = 0.
h(x, y, z) = 0X .2 #´/ pBHHFQ`2M /2}MB2`` 2M Fm`p B `mKK2i UbF `MBM;bFm`pM iBHH MBp´viQ` iBHH
ip´ i`2p`B#2H7mMFiBQM2`VX A p`D2 TmMFi (x, y, z) T´ Fm`pM ` #´/2 ∇g(x, y, z) Q+? ∇h(x, y, z)
pBMF2H` i KQi Fm`pM- b´ p2FiQ`M T(x, y, z) = ∇g(x, y, z) × ∇h(x, y, z) ` T`HH2HH K2/ Fm`pMb
iM;2Mi B (x, y, z)X 1MHB;i `2bQM2KM;2i M ` pB BM7ƺ`/2 KmHiTHBFiQ`K2iQ/2M- b´ K´bi2 /2i B
2ti`2KTmMFi2M (a, b, c) ; HH ii ∇f(a, b, c) · T(a, b, c) = 0- /XpXbX
∇f(a, b, c), ∇g(a, b, c), ∇h(a, b, c) HB2 QM i?2 bK2 THM2X
.2 i`2 ;`/B2Mi2`M ` HBMD `i #2`Q2M/2- pBHF2i ` 2FpBpH2Mi K2/ ii /2i2`KBMMi2M p 2M Ki`Bb
K2/ ∇f- ∇g- ∇h T´ p` bBM `/ K´bi2 p` MQHH B 2ii KtBKmKfKBMBKmK,















%
%
%
%
%
%
%
f!
1(x, y, z) f!
2(x, y, z) f!
3(x, y, z)
g!
1(x, y, z) g!
2(x, y, z) g!
3(x, y, z)
h!
1(x, y, z) h!
2(x, y, z) h!
3(x, y, z)
%
%
%
%
%
%
%
= 0
g(x, y, z) = 0
h(x, y, z) = 0.
1tKTH2 dRX 6BM/ Kt M/ KBM 7Q` f(x, y, z) = y mM/2` +QMbi`BMib
2 2 2
the gradients lie
on the same plane
the point belongs
to both level surfaces

296. 1tKTH2 dRX 6BM/ Kt M/ KBM 7Q` f(x, y, z) = y mM/2` +QMbi`BMib
g(x, y, z) = x + y + z − 1 = 0 M/ h(x, y, z) = x2
+ y2
+ z2
− 1 = 0.
h?2 2[miBQM g(x, y, z) = 0 /2}M2b THM2 M/ h(x, y, z) = 0 bT?2`2- bQ i?2 BMi2`b2+iBQM
+m`p2 Bb +B`+H2- bQ +QKT+i b2iX h?Bb K2Mb i?i KtBKmK M/ KBMBKmK 2tBbi- #2+mb2 f Bb
+QMiBMmQmbX 2+mb2 ∇f(x, y, z) = (0, 1, 0)- ∇g(x, y, z) = (1, 1, 1) M/ ∇h(x, y, z) = (2x, 2y, 2z)
Kmbi HB2 BM i?2 bK2 THM2- r2 ?p2,
0 =
%
%
%
%
%
%
0 1 0
1 1 1
2x 2y 2z
%
%
%
%
%
%
= 2x − 2z.
q2 ;2i x = zc THm;;2/ BM BM g M/
h ;Bp2b i?2 bvbi2K Q7 2[miBQMb
2x + y = 1
2x2 + y2 = 1
q2 Tmi y = 1 − 2x BM 2x2 + y2 = 1 M/ r2 ;2i,
2x2
+ (1 − 2x)2
= 1 ⇒ 2x2
+ 1 − 4x + 4x2
= 1 ⇒ x(6x − 4) = 0 ⇒
x = 0
x = 2
3
x = 0 ;Bp2b z = 0 M/ y = 1 − 2 · 0 = 1 h?2 pHm2 f(0, 1, 0) = 1 UKtBKmKV
x =
2
3
;Bp2b z =
2
3
M/ y = 1 − 2 ·
2
3
= −
1
3
h?2 pHm2 f(2
3 , −1
3 , 2
3 ) = −
1
3
. UKBMBKmKV
Mbr2`, h?2 KBM 7Q` f QM i?2 +m`p2 Bb −1
3 X h?2 Kt Bb 1X

298. Robert A. Adams, Christopher Essex: Calculus, a complete course. 8th or 9th edition
13.1–3
Summary: optimization

299. Second derivative test
If is a CP (critical point) for a C3-function f then:
⃗
a
ℋf( ⃗
a )
ℋf( ⃗
a )
ℋf( ⃗
a )
f has local maximum in ⃗
a
f has local minimum in ⃗
a
the test is indecisive
⇒
⇒
⇒
f( ⃗
a + ⃗
h ) − f( ⃗
a ) ≈
1
2
⃗
h T
ℋf( ⃗
a ) ⃗
h
ℋf( ⃗
a ) ⇒ f has a saddle in ⃗
a
is negative definite
is positive definite
is indefinite
is none of the above

300. Compact set = closed and bounded.
Continuous functions on compact sets attain both maximum and minimum.
How to find them?
- localize all the critical and singular interior points , compute in these points
- find max / min on the boundary
- compare all the values found above, pick the largest as max, the smallest as min.
⃗
a f( ⃗
a )
elimination of one variable, parametrisation, Lagrange multipliers

301. Lagrange 1
Given and .
Find max and min for f on a level curve for g
f(x, y) g(x, y)

302. ∇f∥∇g
∇f∥∇g
How to find max and min for f (x, y) on g(x, y) = 0?
The point belongs to the level curve
The point belongs to the level curve
RX 1M/@TQBMib Q7 g(x, y) = 0
kX SQBMib r?2`2 ∇g(x, y) = !
0 UQ` bBM;mH`V
jX SQBMib biBb7vBM; i?2 bvbi2K Q7 2[miBQMb
UBM /KbV





f!
1(x, y) = λg!
1(x, y)
f!
2(x, y) = λg!
2(x, y)
g(x, y) = 0.
P`- 2[mBpH2MiHv,





%
%
%
%
%
f!
1(x, y) f!
2(x, y)
g!
1(x, y) g!
2(x, y)
%
%
%
%
%
= 0.
g(x, y) = 0.
Lagrange “case 1”

303. Lagrange 2
Given and .
Find max and min for f on a level SURFACE for g
f(x, y, z) g(x, y, z)

304. Lagrange “case 2”
M;2`` T´ bKK b ii
M f(x, y, z) T´ viM
ii pB KtBK2`` f KB@
MBp´viM iBHH 2M MMM
BHHFQ`2i ii ;`/B2Mi2`@
HH2HH B 2ti`2KTmMFi2`
p`M/`VX oB 7´` /´
2FpiBQM2` Q+? 9 p`B@











f!
1(x, y, z) = λg!
1(x, y, z)
f!
2(x, y, z) = λg!
2(x, y, z)
f!
3(x, y, z) = λg!
3(x, y, z)
g(x, y, z) = 0
bi /2i }MMb BM;2M KƺDHB;?2i ii bF`Bp mTT 2M /2i2`KBMMi
Mi2`- K2/M p`D2 ;`/B2Mi ?` i`2 FQKTQM2Mi2`, ?2H?2i2M `
HH mTT 2M T`QTQ`iBQM bQK mii`v+F2` 7FimK ii ;`/B2Mi2`M
∇f∥∇g
Maximize / minimize f (x, y, z) under the constraint g (x, y, z) = 0.
The point belongs to the level surface
Warning: never put a variable expression in denominators without explanations!

305. Lagrange 3
Given and and .
Find max and min for f on the intersection curve of
the level SURFACES for g and for h.
f(x, y, z) g(x, y, z) h(x, y, z)

306. BMF2H` i KQi Fm`pM- b´ p2FiQ`M T(x, y, z) = ∇g(x, y, z) × ∇h(x, y, z) ` T`HH2HH K2/ Fm`p
M;2Mi B (x, y, z)X 1MHB;i `2bQM2KM;2i M ` pB BM7ƺ`/2 KmHiTHBFiQ`K2iQ/2M- b´ K´bi2 /
ti`2KTmMFi2M (a, b, c) ; HH ii ∇f(a, b, c) · T(a, b, c) = 0- /XpXbX
∇f(a, b, c), ∇g(a, b, c), ∇h(a, b, c) HB;;2` B bKK THMX
.2 i`2 ;`/B2Mi2`M ` HBMD `i #2`Q2M/2- pBHF2i ` 2FpBpH2Mi K2/ ii /2i2`KBMMi2M p 2M K
K2/ ∇f- ∇g- ∇h T´ p` bBM `/ K´bi2 p` MQHH B 2ii KtBKmKfKBMBKmK,















%
%
%
%
%
%
%
f!
1(x, y, z) f!
2(x, y, z) f!
3(x, y, z)
g!
1(x, y, z) g!
2(x, y, z) g!
3(x, y, z)
h!
1(x, y, z) h!
2(x, y, z) h!
3(x, y, z)
%
%
%
%
%
%
%
= 0
g(x, y, z) = 0
h(x, y, z) = 0.
1t2KT2H dRX JtBK2` Q+? KBMBK2` f(x, y, z) = y mM/2` #BpBHHFQ`2M
g(x, y, z) = x + y + z − 1 = 0 Q+? h(x, y, z) = x2
+ y2
+ z2
− 1 = 0.
Lagrange “case 3”
Maximize / minimize f (x, y, z) under the constraints g (x, y, z) = 0 and h (x, y, z) = 0.
h(x, y, z) = 0X .2 #´/ pBHHFQ`2M /2}MB2`` 2M Fm`p B `mKK2i UbF `MBM;bFm`pM iBHH MBp´viQ` iBHH
ip´ i`2p`B#2H7mMFiBQM2`VX A p`D2 TmMFi (x, y, z) T´ Fm`pM ` #´/2 ∇g(x, y, z) Q+? ∇h(x, y, z)
pBMF2H` i KQi Fm`pM- b´ p2FiQ`M T(x, y, z) = ∇g(x, y, z) × ∇h(x, y, z) ` T`HH2HH K2/ Fm`pMb
iM;2Mi B (x, y, z)X 1MHB;i `2bQM2KM;2i M ` pB BM7ƺ`/2 KmHiTHBFiQ`K2iQ/2M- b´ K´bi2 /2i B
2ti`2KTmMFi2M (a, b, c) ; HH ii ∇f(a, b, c) · T(a, b, c) = 0- /XpXbX
∇f(a, b, c), ∇g(a, b, c), ∇h(a, b, c) HB2 QM i?2 bK2 THM2X
.2 i`2 ;`/B2Mi2`M ` HBMD `i #2`Q2M/2- pBHF2i ` 2FpBpH2Mi K2/ ii /2i2`KBMMi2M p 2M Ki`Bb
K2/ ∇f- ∇g- ∇h T´ p` bBM `/ K´bi2 p` MQHH B 2ii KtBKmKfKBMBKmK,















%
%
%
%
%
%
%
f!
1(x, y, z) f!
2(x, y, z) f!
3(x, y, z)
g!
1(x, y, z) g!
2(x, y, z) g!
3(x, y, z)
h!
1(x, y, z) h!
2(x, y, z) h!
3(x, y, z)
%
%
%
%
%
%
%
= 0
g(x, y, z) = 0
h(x, y, z) = 0.
1tKTH2 dRX 6BM/ Kt M/ KBM 7Q` f(x, y, z) = y mM/2` +QMbi`BMib
2 2 2
the gradients lie
on the same plane
the point belongs
to both level surfaces

307. Problem 1. Let f(x, y) = 3x3
+ 3x2
y y3
15x.
a) Find and classify the critical points of f. Use any method taught during the course (the second-derivative test or
completing the square).
b) One of the critical points is (a, b) = (1, 1). Write down the second-degree Taylor approximation of f about this
point and motivate, both with computations and with words, how one can see from this approximation what kind
of critical point (1, 1) is. Use completing the square.
Solution
Part a
The critical points must satisfy the system of equations:
f0
1(x, y) = 9x2
+ 6xy 15 = 0
f0
2(x, y) = 3x2
3y2
= 0
)
(
3x2
+ 2xy 5 = 0
(x y)(x + y) = 0
The second equation gives x = y or x = y. We check what happens with the first equation in both cases:
y = x ) 5x2
5 = 0 )
(
x = 1
x = 1
) CP (1, 1) and ( 1, 1)
y = x ) x2
= 5 )
(
x =
p
5
x =
p
5
) CP (
p
5,
p
5) and (
p
5,
p
5)
We compute the second partial derivatives to obtain the Hessian matrix in each CP.
f00
11(x, y) = 18x + 6y
f00
12(x, y) = 6x
f00
22(x, y) = 6y
) Hf (x, y) =
✓
18x + 6y 6x
6x 6y
◆
This gives the following Hessian matrices in the CP:
Hf (1, 1) =
✓
24 6
6 6
◆
, Hf ( 1, 1) =
✓
24 6
6 6
◆
Hf (
p
5,
p
5) =
✓
12
p
5 6
p
5
6
p
5 6
p
5
◆
, Hf (
p
5,
p
5) =
✓
12
p
5 6
p
5
6
p
5 6
p
5
◆
Now we examine the corresponding quadratic forms (divided by the coefficients 6 in the first two cases and by 6
p
5 in the
remaining two; this doesn’t change the answer, but gives easier computations):

308. • (1, 1) corresponds to Q(h, k) = 4h2
+ 2hk k2
= 4
⇣
h + k
4
⌘2
5k2
4 indefinite: saddle point.
• ( 1, 1) corresponds to Q(h, k) = 4h2
2hk + k2
= 4
⇣
h + k
4
⌘2
+ 5k2
4 indefinite: saddle point.
• (
p
5,
p
5) corresponds to Q(h, k) = 2h2
+ 2hk + k2
= 2
⇣
h + k
2
⌘2
+ k2
2 positive definite: local minimum.
• (
p
5,
p
5) corresponds to Q(h, k) = 2h2
2hk k2
= 2
⇣
h + k
2
⌘2
k2
2 negative definite: local maximum.
Answer: The funktion has four CP: saddle points in (1, 1) and in ( 1, 1), a local maximum in (
p
5,
p
5) and a local
minimum in (
p
5,
p
5).
Part b
The second-degree Taylor approximation of f about (a, b) gives by:
f(a + h, b + k) = f(a, b) + (f0
1, f0
2)
✓
h
k
◆
+ 1
2 (h, k)
✓
f00
11 f00
12
f00
21 f00
22
◆ ✓
h
k
◆
+ O((h2
+ k2
)3/2
).
If (a, b) = (1, 1), we get (all the partial derivatives are computed in part a):
f(1 + h, 1 + k) = f(1, 1) + (0, 0)
✓
h
k
◆
+ 1
2 (h, k)
✓
24 6
6 6
◆ ✓
h
k
◆
+ O((h2
+ k2
)3/2
)
giving
f(1 + h, 1 + k) = f(1, 1) + 1
2 (24h2
+ 12hk 6k2
) + O((h2
+ k2
)3/2
).
Because the remainder is small compared to the second-degree terms, we get
f(1 + h, 1 + k) f(1, 1) ⇡ 3(4h2
+ 2hk k2
)
or, after completing the square (see part a):
f(1 + h, 1 + k) f(1, 1) ⇡ 12
⇣
h + k
4
⌘2
15k2
4 .
Because the right side can be both positive and negative in each neighbourhood of the origin (we get negative values for
any k 6= 0 and h = k/4; positive for k = 0 and any h 6= 0), so the difference to the left can be both positive and negative
in any neighbourhood of (1, 1), which means that f has neither a local minimum or maximum in (1, 1), but a saddle point.
Problem 2. Let f(x, y) = 3x2
+ 3xy + y2
+ y3
. Find and classify all critical points of f. Use any method taught during
the course (the second-derivative test or completing the square).
Solution
We compute the partial derivatives of the function in order to find the CP’s.
f0
1(x, y) = 6x + 3y
f0
2(x, y) = 3x + 2y + 3y2 ) we get the CP’s from
(
6x + 3y = 0
3x + 2y + 3y2
= 0.
The first equation gives 3x = 3
2 y. When putting it in the second equation we get
3
2
y + 2y + 3y2
= 0 ,
1
2
y + 3y2
= 0 , y(
1
2
+ 3y) = 0 , y = 0 or y =
1
6
.

309. We check what value of x we get in each case (the first equation gives x = 1
2 y):
y = 0 ) x = 0 ) CP in (0, 0).
y =
1
6
) x =
1
12
) CP in
✓
1
12
,
1
6
◆
.
We compute the second-order partial derivatives in order to obtain the Hessian matrix.
f00
11(x, y) = 6
f00
12(x, y) = 3
f00
22(x, y) = 2 + 6y
) Hf (x, y) =
✓
6 3
3 2 + 6y
◆
We examine the Hessian and the corresponding quadratic forms in the CP’s:
Hf (0, 0) =
✓
6 3
3 2
◆
gives 6h2
+ 6hk + 2k2
= 6
⇣
h +
1
2
k
⌘2
+
1
2
k2
which only can get positive values, which means that we get a local minimum.
Hf ( 1
12 , 1
6 ) =
✓
6 3
3 1
◆
gives 6h2
+ 6hk + k2
= 6
⇣
h +
1
2
k
⌘2 1
2
k2
which can get both positive and negative values, so the point is a saddle point.
Answer: The function has two critical points: a saddle point in ( 1
12 , 1
6 ) and a min-point in the origin.
Problem 3. Let
f(x, y) = 2e2y
4ex
ey
+ e4x
.
a) Find and classify the critical points of f. Use any method taught during the course (the second-derivative test or
completing the square).
b) One of the critical points is (a, b) = (0, 0). Write down the second-degree Taylor approximation of f about this
point and motivate, both with computations and with words, how one can see from this approximation what kind
of critical point (0, 0) is. Use completing the square.
Solution
Part a
First we look for critical points of f:
(
f0
x(x, y) = 4ex
ey
+ 4e4x
= 0
f0
y(x, y) = 4e2y
4ex
ey
= 0
)
(
ex+y
= e4x
e2y
= ex+y
)
(
x + y = 4x
x + y = 2x
The only solution of the system of equations is (0, 0), so there is just one CP.
Partial derivatives of the second order: f00
xx = 4ex
ey
+ 16e4x
, f00
yy = 8e2y
4ex
ey
, f00
xy = f00
yx = 4ex
ey
. This gives the
Hessian matrix in (0, 0):
✓
12 4
4 4
◆
. The corresponding quadratic form
Q(h, k) = 12h2
8hk + 4k2
= 4(h2
2hk + k2
) + 8h2
= 4(h k)2
+ 8h2
is positive definite, which means that there is a local minimum in the origin. (Or use the test: D1 = 12 0, D2 =
48 16 0.)

310. Answer: Local minimum in (0, 0).
Part b
The second-degree Taylor approximation of f about (a, b) gives by:
f(a + h, b + k) = f(a, b) + (f0
1, f0
2)
✓
h
k
◆
+ 1
2 (h, k)
✓
f00
11 f00
12
f00
21 f00
22
◆ ✓
h
k
◆
+ O((h2
+ k2
)3/2
).
If (a, b) = (0, 0), we get (all the partial derivatives are computed in part a):
f(h, k) = f(0, 0) + (0, 0)
✓
h
k
◆
+ 1
2 (h, k)
✓
12 4
4 4
◆ ✓
h
k
◆
+ O((h2
+ k2
)3/2
)
giving
f(h, k) = f(0, 0) + 1
2 (12h2
8hk + 4k2
) + O((h2
+ k2
)3/2
).
Because the remainder is small compared to the second-degree terms, we get
f(h, k) f(0, 0) ⇡ 6h2
4hk + 2k2
or, after completing the square (see part a):
f(h, k) f(0, 0) ⇡ 2(h k)2
+ 4h2
.
Because the right side can only have positive values for all (h, k) 6= (0, 0), so the difference to the left must also always be
positive, which means that f has a local minimum in (0, 0).
Problem 4. Determine and classify all the critical points for
f(x, y, z) = x3
+ 3x2
+ 4y2
+ 6z2
6xy 6xz + 8yz + 4z.
Solution
All the critical points (CP) must satisfy
8
:
f0
1(x, y, z) = 3x2
+ 6x 6y 6z = 0
f0
2(x, y, z) = 8y 6x + 8z = 0
f0
3(x, y, z) = 12z 6x + 8y + 4 = 0
We begin with z by comparing the expressions 6x 8y which appear in both equations: 8z = 12z + 4 gives 4z = 4, so
z = 1. The next step is plugging in z = 1 in the first two equations. This gives us a 2 ⇥ 2 system of equations with two
unknown: x and y:
(
3x2
+ 6x 6y = 6
8y 6x = 8
We get y = 3
4 x + 1 from the second equation, which plugged in to equation 1 reduces number of variables to one:
3x2
+ 6x 6(3
4 x + 1) = 6 ) 3x2
+ 6x 9
2 x 6 = 6 ) 2x2
+ x = 0 ) x(2x + 1) = 0.
This gives x = 0 or x = 1
2 , so y = 1 resp. y = 5
8 . And, of course, z = 1 in both cases. We get two CPs: (0, 1, 1) and
( 1
2 , 5
8 , 1).
Time for the Hessian matrix:

311. f00
11 = 6x + 6,
f00
21 = 6,
f00
31 = 6,
f00
12 = 6,
f00
22 = 8,
f00
32 = 8,
f00
13 = 6
f00
23 = 8
f00
33 = 12
which gives the Hessian
Hf (x, y, z) =
0
@
6x + 6 6 6
6 8 8
6 8 12
1
A
• In the point (0, 1, 1) the Hessian is
Hf (x, y, z) =
0
@
6 6 6
6 8 8
6 8 12
1
A
and the quadratic form
Q(h, k, l) = 6h2
+ 8k2
+ 12l2
12hk 12hl + 16kl
= 6(h2
2hk 2hl) + 8k2
+ 12l2
+ 16kl
= 6(h2
2h(k + l)) + 8k2
+ 12l2
+ 16kl
= 6(h (k + l))2
6(k + l)2
+ 8k2
+ 12l2
+ 16kl
= 6(h k l)2
6k2
12kl 6l2
+ 8k2
+ 12l2
+ 16kl
= 6(h k l)2
+ 2k2
+ 6l2
+ 4kl
= 6(h k l)2
+ 2(k2
+ 2kl + l2
) + 4l2
= 6(h k l)2
+ 2(k + l)2
+ 4l2
is positive definite, which means that the function has a local minimum there.
• In the point ( 1
2 , 5
8 , 1) the Hessian is
Hf (x, y, z) =
0
@
3 6 6
6 8 8
6 8 12
1
A
and the quadratic form
Q(h, k, l) = 3h2
+ 8k2
+ 12l2
12hk 12hl + 16kl
is indefinite (check it by completing the square or by the method with determinants), which means that the function
has a saddle there.
Answer: Local minimum in (0, 1, 1) and saddle in ( 1
2 , 5
8 , 1).
Optimization with constraints
Problem 1. Find the size of a rectangular box with no top (i.e., one of the six faces is missing) having the least possible
total surface area, knowing that the volume of the box is 32.
Solution
This is an example of an optimisation with constraint. The problem can be solved with at least two methods. You are of
course expected to give just one solution.

312. x
y
z
Figur 1: Illustration to problem about the box.
• Method 1: Find the least possible value for f(x, y, z) = 2yz + 2xz + xy (meaning of the variables as in the picture)
for positive x, y and z such that xyz = 32. The constraint lets us eliminate one variable, say z = 32
xy and then
formulate the problem as an optimisation problem in two variables. Find the minimum of
S(x, y) =
64
x
+
64
y
+ xy
for positive x and y. We must find the CP of S. First we compute the partial derivatives:
@S
@x
(x, y) =
64
x2
+ y,
@S
@y
(x, y) =
64
y2
+ x.
The CP must satisfy the following system of equations:
64 x2
y = 0 och 64 y2
x = 0,
thus xy(x y) = 0. Our variables are positive, so it must be x = y which gives (because 64 x2
y = 0) x = 4 = y.
So there is just one CP: (x, y) = (4, 4).
To determine whether it is a max- or min-point, we examine the Hessian matrix in this point. We must compute the
second order partial derivatives:
@2
S
@x2
(x, y) =
128
x3
,
@2
S
@y2
(x, y) =
128
y3
,
@2
S
@x@y
(x, y) = 1 =
@2
S
@y@x
(x, y).
In the CP:
@2
S
@x2
(4, 4) = 2,
@2
S
@y2
(4, 4) = 2,
@2
S
@x@y
(4, 4) = 1 =
@2
S
@y@x
(4, 4)
so the Hessian matrix in the CP is
HS(4, 4) =
✓
2 1
1 2
◆
and the corresponding quadratic form (here we use the short notation ~
h = (h, k)):
Q(h, k) = ~
hT
HS(4, 4)~
h = 2h2
+ 2hk + 2k2
= 2(h2
+ hk + k2
) = 2(h + 1
2 k)2
+ 3
2 k2
which can be only positive for each (h, k) 6= (0, 0), which means that there is a local minimum in the CP. If x = 4 = y
then z = 32
xy = 2.
Answer: The box must have a square base 4 ⇥ 4 and it must be 2 units high.
• Method 2: We can formulate the problem as:
Minimise f(x, y, z) = 2yz + 2xz + xy under the constraint g(x, y, z) = xyz 32 = 0.

313. We can then apply Lagrange’s method version 2. The funktionen g (a polynomial) has no singular points and f can
obtain any possibly large values when one of the variables is near zero. The minimum for f on the level surface for
g can thus only be obtained in the points where the level surfaces for f and g are tangent to each other, i.e. where
the gradients are parallel. The minimum we look for must thus be obtained in a point where
x 0, y 0, z 0, g(x, y, z) = xyz 32 = 0, rf||rg.
We have
rf = (2z + y, 2z + x, 2y + 2x), rg = (yz, xz, xy).
Because all the variables must be positive, we can write a proportion with a variable expression in denominators
(because we know they cannot be zero). The condition rf||rg implies that:
2z + y
yz
=
2z + x
xz
=
2y + 2x
xy
,
2
y
+
1
z
=
2
x
+
1
z
=
2
x
+
2
y
which immediately gives x = y and z = 1
2 y. When put in the constraint
g(x, y, z) = xyz 32 = 1
2 y3
32 = 0
it gives the answer y = 4, x = 4, z = 2.
Answer: : The box must have a square base 4 ⇥ 4 and it must be 2 units high.
Problem 2. Let
D = {(x, y) | x 0, y 0, x2
+ 4y2
 1 }.
a) Sketch D. Explain briefly how we can see that D is closed and bounded.
b) Find the largest and the smallest values of f(x, y) = x2
+ y on D.
Solution
Part a
Set D is a subset of the first quadrant and it is bounded by the ellipse x2
+4y2
= 1. See the picture which is more accurate
than is demanded from you. The equation of the ellipse can be rewritten so that it becomes clear that the semi-axes are
a = 1 and b = 1
2 , which also can be seen in the picture:
x2
12
+
y2
1
2
2 = 1.
1 0.5 0.5 1 1.5 2
0.2
0.2
0.4
0.6
0.8
1
x
y
Domain D is coloured in green

314. We can see that the domain is contained in a circle around the origin, which means that it is bounded. (You can for
example take circle x2
+ y2
= 2, or any larger circle.)
Because we allow the equalities in the definition of D, all the boundary points belong to D, which means that D is closed.
Part b
The smallest value is, of course, zero because (0, 0) 2 D, f(0, 0) = 0, and f does not have any negative values in the first
quadrant. We just have to find the largest value in D.
We begin by looking for critical points. We get f0
x(x, y) = 2x and f0
y(x, y) = 1. The partial derivative with respect to y
can never be 0 so there are no critical points.
The boundary consists of three pieces and we examine all of them separately:
• On the straight-line piece x = 0, 0  y  1
2 it holds f(0, y) = y. The largest value f obtains on this piece of the
boundary is 1
2 .
• On the straight-line piece y = 0, 0  x  1 it holds f(x, y) = x2
with the largest value 1.
• On the elliptical arc we have x2
= 1 4y2
. This means that it holds
f(x, y) = x2
+ y = (1 4y2
) + y.
Let g(y) = 1 4y2
+ y. Wi shall now maximise g(y) for y in the interval [0, 1
2 ]. We get g0
(y) = 8y + 1. The only
critical point for g is 1
8 . We compute the value of g in the critical point and on the end points of the interval.
g(0) = 1
g
✓
1
8
◆
= 1 4 ·
1
64
+
1
8
= 1 +
1
16
g
✓
1
2
◆
= 1 4 ·
1
4
+
1
2
=
1
2
.
Answer: The largest value on D is 1 + 1
16 . The smallest value on D is 0.
Problem 3. Maximize and minimize f(x, y) = (2x + 3y + 1)2
on the circle x2
+ y2
= 1.
Solution
We must find the largest and the smallest values of f(x, y) = (2x + 3y + 1)2
with constraint g(x, y) = x2
+ y2
1 = 0.
We use Lagrange’s multipliers. Because the circle is a closed curve with non-zero gradient, all the extremums satisfy the
system of equations
8
:
f0
1(x, y) = g0
1(x, y)
f0
2(x, y) = g0
2(x, y)
g(x, y) = 0
)
8
:
2(2x + 3y + 1)2 = 2x
2(2x + 3y + 1)3 = 2y
x2
+ y2
1 = 0
If 6= 0, we divide the second equation with the first one and we get
3
2
=
y
x
) y =
3
2
x ) x2
+
⇣3
2
x
⌘2
1 = 0 )
13
4
x2
= 1
) x = ±
2
p
13
) (x, y) =
⇣
2
p
13
, 3
p
13
⌘
eller
⇣
2
p
13
, 3
p
13
⌘
.

315. The system of equations has also solutions = 0, 2x + 3y + 1 = 0. For these solutions there is obviously f(x, y) = 0. (The
line y = 2
3 x 1
3 intersects the circle in two points, so we know that there are points on the circle where f has the value
0.) The other values are f( 2
p
13
, 3
p
13
) = (
p
13 + 1)2
and f( 2
p
13
, 3
p
13
) = (
p
13 + 1)2
, so the maximum is (
p
13 + 1)2
and the minimum is 0.
Answer: The largest value on the circle is (
p
13 + 1)2
. The smallest value on the circle is 0.

316. Surface integrals, notation
15.5 and 15.6

317. Double integrals
f : D → ℝ, D ⊂ ℝ2
Triple integrals
f : B → ℝ, B ⊂ ℝ3
Sn =
n
∑
i=1
n
∑
j=1
n
∑
k=1
f(x*
ijk
, y*
ijk
, z*
ijk
)ΔxiΔyjΔzk →
∭
B
f(x, y, z)dxdydz
volume
element
dV
Single integrals
f : [a, b] → ℝ
area element dA
Sn =
n
∑
i=1
n
∑
j=1
f(x*
ij
, y*
ij
)ΔxiΔyj →
∬
D
f(x, y)dxdy
Sn =
n
∑
i=1
f(x*
i
)Δxi →
b
∫
a
f(x)dx
length element dx
Riemann sums Notation
Multiple integrals
n → ∞
n → ∞
n → ∞

318. γ1
γ2
γ3
γ4
γ5
piece-wise smooth curve
⃗
r(t) = (x(t), y(t))
⃗
r(t) = (x(t), y(t), z(t))

319. In Sections 11 and 12
Line integrals / Curve integrals
of functions of vector fields
KbbM 7ƺ` F`QTT2M K QK 7mMFiBQM
K,b pQHvK QK f(x, y, z) = 1 B ?2H
Em`pBMi2;`H2` p 7mMFiBQM2`
2+FMb
´
γ
fdsc /2 #2` FM` Fm`pM
PK n = 2- HHib´ Fm`pM ` THM
b´ Fm`pM ` THM Q+? f ` 2M i
bmiQK `2M UK2/ i2+F2MV p /2M
x, y)X
` p p2FiQ`7 Hi
´
γ
F · d
r U+B`F
i
F : Rn
→ Rn
#2bF`Bp2` `#2i2i
arc-length element
vector differential
area, mass, arc length work
h?2Q`2K k9X G´i F p` 2ii FQMb2`piBpi p2FiQ`7 Hi
γ ∈ D K2HHM
a Q+?
b ; HH2`,
ˆ
γ
F · d
r = Φ(
b) −
LQi2` HBF?2i2M K2/ MHvb2Mb mpm/bibX SQi2MiB
7mMFiBQM2MX aib2M BKTHB+2`` ii Fm`pBMi2;`H2M K2H
Fm`pX 6ƺ` bHmiM Fm`pQ` γ FHHb Fm`pBMi2;`H2M p
F
Q+? #2i2+FMb
˛
γ
F · d
r.
Circulations:
only (piece wise) smooth curves, and continuous functions

320. In sections 14 and 15
Surface integrals
of functions of vector fields; flux integrals
surface element differential surface vector ̂
NdS
area, mass flux (e.g. of fluid) across the surface
Q+? +2Mi`mK B Q`B;Q Ub7 `BbF FQQ`/BMi2`VX Em`p
/ Q+? Fm`pQ`M !
r(s0, t) ` Fm`pQ` K2/ FQMbiMi #
i Q+? f 2M 7mMFiBQM bQK ` /2}MB2`/ B HH TmMF
+FMb
¨
Y
f(x, y, z)dS.
/2MM .2H BM S B n biv+F2M /2HviQ` K2/ `2Q
vi2H2K2Mi dS T´ viMX lM/2` iB/bBMi2`pHH2i ∆t b´ bi`ƺKK` ;2MQK
bQK `vKb B 2M +vHBM/2` K2/ #b`2 dS Q+? ?ƺD/
|!
F(X)|∆t +Qb α = |!
F(X)|∆t
!
F(X) · !
N
|!
F(X)|| !
N|
= !
F(X
*vHBM/2`M BMM2?´HH2` KbbM !
F(X) · N̂∆t dS- b´ ~ƺ/2i T2` iB/b2M?
!
F(X) · N̂ dSX 6Hƺ/2i p !
F ;2MQK ?2H viM Y ;2b p viBMi2;`H2M
¨
Y
!
F · N̂dS pBHF2i B#HM/ #2i2+FMb
¨
Y
!
F · d!
S
PpMbi´2M/2 BMi2;`H2` ` ~ƺ/2bBMi2;`H2`X PK viM Y ` bHmi2M
M
‹
Y
!
F · N̂dS Q+?
‹
Y
!
F · d!
S
ˆ
t +Qb α = |!
F(X)|∆t
!
F(X) · !
N
|!
F(X)|| !
N|
= !
F(X) · N̂∆t.
!
F(X) · N̂∆t dS- b´ ~ƺ/2i T2` iB/b2M?2i ;2MQK v
MQK ?2H viM Y ;2b p viBMi2;`H2M
pBHF2i B#HM/ #2i2+FMb
¨
Y
!
F · d!
S
/2bBMi2;`H2`X PK viM Y ` bHmi2M Mp M/b Q
/2` iB/bBMi2`pHH2i ∆t b´ bi`ƺKK` ;2MQK vi2H2K2Mi2i dS /2M Ki2`B
2/ #b`2 dS Q+? ?ƺD/
)|∆t +Qb α = |!
F(X)|∆t
!
F(X) · !
N
|!
F(X)|| !
N|
= !
F(X) · N̂∆t.
bM !
F(X) · N̂∆t dS- b´ ~ƺ/2i T2` iB/b2M?2i ;2MQK vi2H2K2Mi2i dS `
!
F ;2MQK ?2H viM Y ;2b p viBMi2;`H2M
pBHF2i B#HM/ #2i2+FMb
¨
Y
!
F · d!
S b´ d!
S = N̂dSX
` ~ƺ/2bBMi2;`H2`X PK viM Y ` bHmi2M Mp M/b Q7i #2i2+FMBM;`@
‹
only (piece wise) smooth surfaces, and continuous functions
ℝ3

321. f : ℝ3
→ ℝ
z = f(x, y)
(x, y) ∈ D
x
y
z
a
b
c d
Δxi
Δyj
f(x*
ij
, y*
ij
)
(x*
ij
, y*
ij
)
ΔS
ZZ
D
f(x, y) dA
ZZ
Y
f dS
~
F is a smooth vector field defined on an arc con-
s line integrals are path-independent, then ~
F is
als between each two points are path independent.
0, z0) and define for each (x, y, z) 2 D:
(x, y, z) =
Z
~
F · d~
r
Sn =
n
∑
i=1
n
∑
j=1
f(x*
ij
, y*
ij
)ΔxiΔyj
n→∞
→
∬
D
f(x, y)dxdy
Riemann sum:
ZZ
Y
f dS
Path independence: If ~
F is a smooth vector field defined on an arc con-
nected domain D and its line integrals are path-independent, then ~
F is
conservative.
Assume that the line integrals between each two points are path independent.
Fix some point P0 = (x0, y0, z0) and define for each (x, y, z) 2 D:
(x, y, z) =
Z
~
F · d~
r
1
ΔS1, ΔS2, ΔS3, …
(xi, yi, zi) ∈ ΔSi
1tKTH2 NeX S`K2i`Bb2`BM;2M
!
r(s, t) = (R bBM s +Qb t, R bBM s bBM t, R +Qb s), 0 ⩽ s ⩽ π,
KQibp`` 2M b7 ` K2/ `/B2 R Q+? +2Mi`mK B Q`B;Q Ub7 `BbF FQQ`/BM
Fm`pQ` K2/ FQMbiMi H M;/;`/ Q+? Fm`pQ`M !
r(s0, t) ` Fm`pQ` K2/ F
G´i Mm Y p` 2M vi B `mKK2i Q+? f 2M 7mMFiBQM bQK ` /2}MB2`/
1M viBMi2;`H p f ƺp2` Y #2i2+FMb
¨
Y
f(x, y, z)dS.
m` bF pB ;ƺ` 7ƺ` ii #2` FM /2MM .2H BM S B n biv+F2M /2HviQ`
QXbXpX Q+? H´i (xi, yi, zi) p` 2M TmMFi T´ viM YiX uiBMi2;`H2M ?` /´
n
!
i=1
f(xi, yi, zi)∆Si →
¨
Y
f(x, y, z)dS.
Riemann sum:

322. f : ℝ3
→ ℝ
ΔS
ZZ
Y
f dS
Path independence: If ~
F is a smooth vector field defined on an arc con-
nected domain D and its line integrals are path-independent, then ~
F is
conservative.
Assume that the line integrals between each two points are path independent.
Fix some point P0 = (x0, y0, z0) and define for each (x, y, z) 2 D:
(x, y, z) =
Z
~
F · d~
r
1
ΔS1, ΔS2, ΔS3, …
(xi, yi, zi) ∈ ΔSi
1tKTH2 NeX S`K2i`Bb2`BM;2M
!
r(s, t) = (R bBM s +Qb t, R bBM s bBM t, R +Qb s), 0 ⩽ s ⩽ π, 0 ⩽ t ⩽ 2π
KQibp`` 2M b7 ` K2/ `/B2 R Q+? +2Mi`mK B Q`B;Q Ub7 `BbF FQQ`/BMi2`VX Em`pQ`M !
r(s, t0) `
Fm`pQ` K2/ FQMbiMi H M;/;`/ Q+? Fm`pQ`M !
r(s0, t) ` Fm`pQ` K2/ FQMbiMi #`2//;`/X
G´i Mm Y p` 2M vi B `mKK2i Q+? f 2M 7mMFiBQM bQK ` /2}MB2`/ B HH TmMFi2` T´ viM Y X
1M viBMi2;`H p f ƺp2` Y #2i2+FMb
¨
Y
f(x, y, z)dS.
m` bF pB ;ƺ` 7ƺ` ii #2` FM /2MM .2H BM S B n biv+F2M /2HviQ` K2/ `2Q`M ∆S1- ∆S2-
QXbXpX Q+? H´i (xi, yi, zi) p` 2M TmMFi T´ viM YiX uiBMi2;`H2M ?` /´ _B2KMMbmKKM
n
!
i=1
f(xi, yi, zi)∆Si →
¨
Y
f(x, y, z)dS.
Riemann sum:
STRAN
GE

323. f : ℝ3
→ ℝ
ΔS
ZZ
Y
f dS
~
F is a smooth vector field defined on an arc con-
s line integrals are path-independent, then ~
F is
rals between each two points are path independent.
y0, z0) and define for each (x, y, z) 2 D:
(x, y, z) =
Z
~
F · d~
r
1
ΔS1, ΔS2, ΔS3, …
(xi, yi, zi) ∈ ΔSi
2i`Bb2`BM;2M
(R bBM s +Qb t, R bBM s bBM t, R +Qb s), 0 ⩽ s ⩽ π, 0 ⩽ t ⩽ 2π
/ `/B2 R Q+? +2Mi`mK B Q`B;Q Ub7 `BbF FQQ`/BMi2`VX Em`pQ`M !
r(s, t0) `
H M;/;`/ Q+? Fm`pQ`M !
r(s0, t) ` Fm`pQ` K2/ FQMbiMi #`2//;`/X
B `mKK2i Q+? f 2M 7mMFiBQM bQK ` /2}MB2`/ B HH TmMFi2` T´ viM Y X
2` Y #2i2+FMb
¨
Y
f(x, y, z)dS.
ii #2` FM /2MM .2H BM S B n biv+F2M /2HviQ` K2/ `2Q`M ∆S1- ∆S2-
) p` 2M TmMFi T´ viM YiX uiBMi2;`H2M ?` /´ _B2KMMbmKKM
n
!
i=1
f(xi, yi, zi)∆Si →
¨
Y
f(x, y, z)dS.
Riemann sum:
Two applications:
area of the surface
mass of the surface
.2ii ; HH2` QK Y ` 2M ;`7 p 2M 7mMFiBQM z = z(x, y)
iBHH viM Y X A p`D2 TmMFi (x, y, z) T´ viM Y ?` pB
N
LQi2` Q+Fb´ ii `2M p 2M T`K
A =
¨
Y
dS =
¨
D
!
!
!
!
∂
r
∂s
×
R8y
mass(Y ) =
ZZ
Y
⇢(x, y, z) dS

324. f : ℝ3
→ ℝ
ΔS
ZZ
Y
f dS
~
F is a smooth vector field defined on an arc con-
s line integrals are path-independent, then ~
F is
rals between each two points are path independent.
y0, z0) and define for each (x, y, z) 2 D:
(x, y, z) =
Z
~
F · d~
r
1
ΔS1, ΔS2, ΔS3, …
(xi, yi, zi) ∈ ΔSi
2i`Bb2`BM;2M
(R bBM s +Qb t, R bBM s bBM t, R +Qb s), 0 ⩽ s ⩽ π, 0 ⩽ t ⩽ 2π
/ `/B2 R Q+? +2Mi`mK B Q`B;Q Ub7 `BbF FQQ`/BMi2`VX Em`pQ`M !
r(s, t0) `
H M;/;`/ Q+? Fm`pQ`M !
r(s0, t) ` Fm`pQ` K2/ FQMbiMi #`2//;`/X
B `mKK2i Q+? f 2M 7mMFiBQM bQK ` /2}MB2`/ B HH TmMFi2` T´ viM Y X
2` Y #2i2+FMb
¨
Y
f(x, y, z)dS.
ii #2` FM /2MM .2H BM S B n biv+F2M /2HviQ` K2/ `2Q`M ∆S1- ∆S2-
) p` 2M TmMFi T´ viM YiX uiBMi2;`H2M ?` /´ _B2KMMbmKKM
n
!
i=1
f(xi, yi, zi)∆Si →
¨
Y
f(x, y, z)dS.
Riemann sum:
Two applications:
area of the surface
mass of the surface
.2ii ; HH2` QK Y ` 2M ;`7 p 2M 7mMFiBQM z = z(x, y)
iBHH viM Y X A p`D2 TmMFi (x, y, z) T´ viM Y ?` pB
N
LQi2` Q+Fb´ ii `2M p 2M T`K
A =
¨
Y
dS =
¨
D
!
!
!
!
∂
r
∂s
×
R8y
mass(Y ) =
ZZ
Y
⇢(x, y, z) dS
g
m2 m2

325. If the surface Y is defined as graph
of a real-valued functions of two variables
Y = {(x, y, z); (x, y) ∈ D, z = g(x, y)}

326. z = g(x, y)
(x, y) ∈ D
x y
z
D
z
D is closed and bounded (compact)
g is bounded on D
Surface element dS, approximates by
a piece of the tangent plane (a parallelogram)
dA = dxdy
dx
dy
What is the relationship
between dS and dA?
dA = dS cos ↵
Area =
ZZ
D
dS
~
n = ( f0
1(x, y), f0
2(x, y)
~
e3 = (0, 0, 1)
dA = dS cos ↵ = dS
~
n · ~
e3
|~
n||~
e3|
dS =
p
1 + (f0
1(x, y))2 + (f0
2(x,
~
e3 = (0, 0, 1)
dA = dS cos ↵ = dS
~
n · ~
e3
|~
n||~
e3|
=
dS
|~
n|
dS =
p
1 + (f0
1(x, y))2 + (f0
2(x, y))2 dxdy
The total area of the surface z = f(x, y) over D is
S =
ZZ
D
p
1 + (f0
1(x, y))2 + (f0
2(x, y))2 dxdy,
L =
b
Z
a
p
1 + (f0(x))2dx.
1
dA = dS cos ↵
Area =
ZZ
D
dS
~
n = ( f0
1(x, y), f0
2(x, y), 1)
~
e3 = (0, 0, 1)
dA = dS cos ↵ = dS
~
n · ~
e3
|~
n||~
e |
=
dS
|~
n|
dA
dS
= cos ↵ ) dA = dS cos ↵
Area =
ZZ
D
dS
~
n = ( f0
1(x, y), f0
2(x, y), 1)
~
e3 = (0, 0, 1)
dS
Area =
ZZ
D
dS
~
n = ( f0
1(x, y), f0
2(x, y), 1)
~
e3 = (0, 0, 1)
dA = dS cos ↵ = dS
~
n · ~
e3
|~
n||~
e3|
=
dS
|~
n|
dS =
p
1 + (f0
1(x, y))2 + (f0
2(x, y))2 dxdy
The total area of the surface z = f(x, y) over D is
S =
ZZ p
1 + (f0
1(x, y))2 + (f0
2(x, y))2 dxdy,
Video 81
~
n = ( g0
1(x, y), g0
2(x, y), 1)
~
e3 = (0, 0, 1)
dA = dS cos ↵ = dS
~
n · ~
e3
|~
n||~
e3|
=
dS
|~
n|
dS =
p
1 + (f0
1(x, y))2 + (f0
2(x, y))2 dxdy
The total area of the surface z = g(x, y) over D is
S =
ZZ
D
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2 dxdy
~
n = ( g0
1(x, y), g0
2(x, y), 1)
~
e3 = (0, 0, 1)
dA = dS cos ↵ = dS
~
n · ~
e3
|~
n||~
e3|
=
dS
|~
n|
dS =
p
1 + (f0
1(x, y))2 + (f0
2(x, y))2 dxdy
The total area of the surface z = g(x, y) over D is
S =
ZZ p
1 + (g0
1(x, y))2 + (g0
2(x, y))2 dxdy
~
n = ( g0
1(x, y), g0
2(x, y), 1)
~
e3 = (0, 0, 1)
dA = dS cos ↵ = dS
~
n · ~
e3
|~
n||~
e3|
=
dS
|~
n|
dS =
p
1 + (f0
1(x, y))2 + (f0
2(x, y))2 dxdy
The total area of the surface z = g(x, y) over D is
S =
ZZ
D
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2 dxdy
~
n = ( g0
1(x, y), g0
2(x, y), 1)
~
e3 = (0, 0, 1)
dA = dS cos ↵ = dS
~
n · ~
e3
|~
n||~
e3|
=
dS
|~
n|
dS =
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2 dxdy
The total area of the surface z = g(x, y) over D is
S =
ZZ p
1 + (g0 (x, y))2 + (g0 (x, y))2 dxdy

327. n
!
i=1
f(xi, yi, zi)∆Si →
¨
Y
f(x, y, z)dS.
oB FM mii`v+F (xi, yi, zi) K2/ ?D HT p 2M T`K2i`Bb2`BM; p viM Q+? bFmHH2 pBHD FmMM
mii`v+F p2M ∆Si K2/ T`K2i`Bb2`BM;2MX 1MHB;i bKK `2bQM2KM; bQK pB Mp M/2 M ` pB
pBb/2 ii C+Q#BM2M FmM/2 iQHFb bQK `27ƺ`biQ`BM;2M 7ƺ` 2M p#BH/MBM;- b´ ; HH2`
∆Si =
∂!
r
∂s
∆s ×
∂!
r
∂t
∆t
=
∂!
r
∂s
×
∂!
r
∂t
∆s∆t.
HHib´ FM viBMi2;`H2M mii`v+Fb K2/ ?D HT p T`K2i`Bb2`BM;2M p Y bQK
¨
Y
f(x, y, z)dS =
¨
D
f(x(s, t), y(s, t), z(s, t))
∂!
r
∂s
×
∂!
r
∂t
dsdt
#
dS =
∂!
r
∂s
×
∂!
r
∂t
dsdt
$
R9N
Y = {(x, y, z); (x, y) ∈ D, z = g(x, y)}
~
e3 = (0, 0, 1)
dA = dS cos ↵ = dS
~
n · ~
e3
|~
n||~
e3|
=
dS
|~
n|
dS =
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2 dxdy
The total area of the surface z = g(x, y) over D is
S =
ZZ
D
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2 dxdy
L =
b
Z
a
p
1 + (f0(x))2dx
6Q`K2HM 7ƺ` `2M p ;`72M iBHH 2M 7mMFiBQM p ip´ p`B#H2` bQK pB iQ; 7`K bQK 2M iBHH KTMBM;
p /m##2HBMi2;`H2` ` 2ii bT2+BH7HH p 7Q`K2HM QpMX :`72M iBHH 2M 7mMFiBQM ?` T`K2i`Bb2@
BM;2M
r(s, t) = (s, t, g(s, t)) b´
∂
r
∂s
= (1, 0, g!
1(s, t)),
∂
r
∂t
= (0, 1, g!
2(s, t))
+?
dS =
!
!
!
!
∂
r
∂s
×
∂
r
∂t
!
!
!
! dsdt =
e1
e2
e3
1 0 g!
1(s, t)
0 1 g!
2(s, t)
dsdt = |(−g!
1, −g!
2, 1)| dsdt
=
,
(g!
1(s, t))2 + (g!
2(s, t))2 + 12 dsdt
T`2+Bb bQK 7ƺ`p MiiX PK
N = (N1, N2, N3) ` 2M MMM MQ`KHp2FiQ` iBHH viM Y - /´
(−g!
1, −g!
2, 1) =
n = λ
N = (λN1, λN2, λN3)
BHF2i ;2` λN3 = 1- HHib´ λ = 1
N3
X .´,
|
n| = |λ
N| = |λ| · |
N| =
|
N|
|N3|
+? ¨
f(x, y, z)dS =
¨
f(x, y, z(x, y)) ·
|
N|
|N3|
dxdy,
6Q`K2HM 7ƺ` `2M p ;`72M iBHH 2M 7mMFiBQM p ip´ p`B#H2` bQK pB iQ; 7`K bQK 2M iBHH KTMBM;
p /m##2HBMi2;`H2` ` 2ii bT2+BH7HH p 7Q`K2HM QpMX :`72M iBHH 2M 7mMFiBQM ?` T`K2i`Bb2@
`BM;2M
r(s, t) = (s, t, g(s, t)) b´
∂
r
∂s
= (1, 0, g!
1(s, t)),
∂
r
∂t
= (0, 1, g!
2(s, t))
Q+?
dS =
!
!
!
!
∂
r
∂s
×
∂
r
∂t
!
!
!
! dsdt =
e1
e2
e3
1 0 g!
1(s, t)
0 1 g!
2(s, t)
dsdt = |(−g!
1, −g!
2, 1)| dsdt
=
,
(g!
1(s, t))2 + (g!
2(s, t))2 + 12 dsdt
T`2+Bb bQK 7ƺ`p MiiX PK
N = (N1, N2, N3) ` 2M MMM MQ`KHp2FiQ` iBHH viM Y - /´
(−g!
1, −g!
2, 1) =
n = λ
N = (λN1, λN2, λN3)
pBHF2i ;2` λN3 = 1- HHib´ λ = 1
N3
X .´,
|
n| = |λ
N| = |λ| · |
N| =
|
N|
|N3|
Q+? ¨
Y
f(x, y, z)dS =
¨
D
f(x, y, z(x, y)) ·
|
N|
|N3|
dxdy,
6Q`K2HM 7ƺ` `2M p ;`72M iBHH 2M 7mMFiBQM p ip´ p`B#H2` bQK pB iQ; 7`K bQK 2M iBHH KTMBM;
p /m##2HBMi2;`H2` ` 2ii bT2+BH7HH p 7Q`K2HM QpMX :`72M iBHH 2M 7mMFiBQM ?` T`K2i`Bb2@
BM;2M
r(s, t) = (s, t, g(s, t)) b´
∂
r
∂s
= (1, 0, g!
1(s, t)),
∂
r
∂t
= (0, 1, g!
2(s, t))
+?
dS =
!
!
!
!
∂
r
∂s
×
∂
r
∂t
!
!
!
! dsdt =
e1
e2
e3
1 0 g!
1(s, t)
0 1 g!
2(s, t)
dsdt = |(−g!
1, −g!
2, 1)| dsdt
=
,
(g!
1(s, t))2 + (g!
2(s, t))2 + 12 dsdt
`2+Bb bQK 7ƺ`p MiiX PK
N = (N1, N2, N3) ` 2M MMM MQ`KHp2FiQ` iBHH viM Y - /´
(−g!
1, −g!
2, 1) =
n = λ
N = (λN1, λN2, λN3)
BHF2i ;2` λN3 = 1- HHib´ λ = 1
N3
X .´,
|
n| = |λ
N| = |λ| · |
N| =
|
N|
|N3|
+? ¨
f(x, y, z)dS =
¨
f(x, y, z(x, y)) ·
|
N|
|N3|
dxdy,
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y))
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2dxdy
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y)) ·
| ~
N|
|N3|
dxdy,
mass(Y ) =
ZZ
Y
⇢(x, y, z) dS
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y))
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2dxdy
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y)) ·
| ~
N|
|N3|
dxdy
om v2 = 0 (och de två andra är nollskilda) då blir linjens ekvation
följande:
x x0
v1
=
z z0
v3
, y = y0.
y-koordinaten för alla punkten på linjen är konstanta, vilket betyder
att linjen ligget i planet y = y0 som är parallell med xz-planet.
Plan i R3
Vi kommer att lära oss beskriva plan på två sätt: med och utan parame-
trar. Jag kopierar här två sista beskrivningar till linjer i 2D och anpassar
dem till plan och 3D:
n = (a, b, c)
x0 = (x0, y0, z0) x = (x, y, z)
• Normalekvation till planet ⇡ genom x0 = (x0, y0, z0) och ortogonal mot
vektorn n = (a, b, c): ax + by + cz + d = 0 där d = ax0 by0 cz0.
Varför? Om punkten x = (x, y, z) tillhör ⇡ då måste vektorerna n =
(a, b, c) och x x0 = (x x0, y y0, z z0) vara ortogonala, alltså måste
deras skalärprodukt vara lika med noll:
(a, b, c)·(x x0, y y0, z z0) = 0 , ax ax0+by by0+cz cz0 = 0
, ax + by + cz + d = 0.
Sådan ekvation kallas normalekvation eftersom ortogonala vektorer kallas
också normala vektorer. Sådan ekvation är inte entydig, eftersom det
finns väldigt många vektorer som är ortogonala mot planet ⇡ (alla
skalningar av n, alltså vektorer som är parallella med n, men har en
ax + by + cz + d = 0

328. Properties of surface integrals of functions
Surface integrals are independent of the parametrisation of the surface (Chain rule)
Surface integrals are independent of the orientation of the surface
They describe the area of the surface if f is constant equal to 1
Area(Y) =
∬
Y
dS
They describe the mass of the surface if f describes the density in each point
mass(Y) =
∬
Y
ρ(x, y, z) dS
Applications
Additivity
∬
Y
f dS =
∬
Y1
f dS +
∬
Y2
f dS +
∬
Y3
f dS +
∬
Y4
f dS
piece-wise smooth surface
Y1 Y2
Y4 Y3

329. Compute the surface integral
∬
Y
x2
+ y2
+ 1 dS
where Y is the helicoid defined by
x = ρ cos θ, y = ρ sin θ, z = θ
0 ⩽ θ ⩽ 2π, 0 ⩽ ρ ⩽ 1
Y
f(x, y, z)dS.
m` bF pB ;ƺ` 7ƺ` ii #2` FM /2MM .2H BM S B n biv+F2M /2HviQ` K2/ `2Q`M ∆S1- ∆S2-
QXbXpX Q+? H´i (xi, yi, zi) p` 2M TmMFi T´ viM YiX uiBMi2;`H2M ?` /´ _B2KMMbmKKM
n
!
i=1
f(xi, yi, zi)∆Si →
¨
Y
f(x, y, z)dS.
oB FM mii`v+F (xi, yi, zi) K2/ ?D HT p 2M T`K2i`Bb2`BM; p viM Q+? bFmHH2 pBHD FmMM
mii`v+F p2M ∆Si K2/ T`K2i`Bb2`BM;2MX 1MHB;i bKK `2bQM2KM; bQK pB Mp M/2 M ` pB
pBb/2 ii C+Q#BM2M FmM/2 iQHFb bQK `27ƺ`biQ`BM;2M 7ƺ` 2M p#BH/MBM;- b´ ; HH2`
∆Si =
∂!
r
∂s
∆s ×
∂!
r
∂t
∆t
=
∂!
r
∂s
×
∂!
r
∂t
∆s∆t.
HHib´ FM viBMi2;`H2M mii`v+Fb K2/ ?D HT p T`K2i`Bb2`BM;2M p Y bQK
¨
Y
f(x, y, z)dS =
¨
D
f(x(s, t), y(s, t), z(s, t))
∂!
r
∂s
×
∂!
r
∂t
dsdt
#
dS =
∂!
r
∂s
×
∂!
r
∂t
dsdt
$
R9N
⃗
r′
t =
∂ ⃗
r
∂t
(s, t) =
(
∂x
∂t
(s, t),
∂y
∂t
(s, t),
∂z
∂t
(s, t)
)
⃗
r′
s =
∂ ⃗
r
∂s
(s, t) =
(
∂x
∂s
(s, t),
∂y
∂s
(s, t),
∂z
∂s
(s, t)
)

330. ⃗
r(t) = (cos t, sin t, t), t ∈ [0, 4π]
Helix Helicoid
⃗
r(ρ, θ) = (ρ cos t, ρ sin t, t), t ∈ [−π, π], ρ ∈ [−1, 1]

331. pwblem1
fcxiy.tt t
xcsitkscostiylsitf ssu.int
ZCs tI
tDiOEtE2JT
OEsEXT
compute F
5,90
EE5x
Stephen the integrand
fGly 2 f TEH
fE tIsn
TtT
fs qM
team id
Steps the surface element
IYI.fiIEEiYlsitt
snse
f

332. If cost sint o
fret s sont s cost 1
s t
sint cost s
dSIfsrt
ts2TTtdsdtifFsTdsdtmfyffds
fofTstf.f
its dsdt

333. dsdtfm.eu
FIbifippon
ndef
Ifat sets ds
21T I t s
21T ft 1
T
answer

334. x
y
Compute where Y is the graph surface
to for on rectangle .
∬
Y
x dS
g(x, y) = x2
+ y (x, y) [0,1] × [−1,1]
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y))
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2dxdy
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y)) ·
| ~
N|
|N3|
dxdy,
mass(Y ) =
ZZ
Y
⇢(x, y, z) dS
• Normal to the graph surface z = f(x, y):

335. Problem2_ fyfxds
Y Kiya Chyle 0,11 1 1,1
z
x4y
Y gCx yI x
2
µxds f t4x dxdy
does not depend
on
y
f rEdt
t
I fore art
f 68 we r6

336. x
y
x2
+ y2
= 1
z =
x2
2
D
Compute over the part of the parabolic cylinder
which lies inside the cylinder in the first octant.
∬
Y
x dS z =
x2
2
x2
+ y2
= 1
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y))
p
1 + (g0
1(x, y))2 + (g0
2(x, y))2dxdy
ZZ
Y
f(x, y, z)dS =
ZZ
D
f(x, y, g(x, y)) ·
| ~
N|
|N3|
dxdy,
mass(Y ) =
ZZ
Y
⇢(x, y, z) dS

337. Problem3m fyfxds
Y def by gtx g inside the
cylinder Ety L in the first
octant
Surface element gtxyK
F oof Ey 1
exI
fyfxots ffxrfxdxdy
DT.it
j
Fubini for y
ym.ph

338. I FE dy dx
O
2 y fE
yxrFE dx
O ya
t dx
a b Catb a 62
f FHh
HI dt
1
Iz f Ft dt 12 41Area dish i
get f
g t

339. Different coordinate systems
10.6

340. Cartesian coordinate system
Cartesian coordinates = rectangular coordinates
x
y
6 5 4 3 2 1 0 1 2 3 4 5 6
4
3
2
1
0
1
2
3
4
origin
quadrant I
quadrant II
quadrant III quadrant IV
x-axis
y-axis

341. Idag kommer vi att arbeta enbart i det tredimensionella rummet. Det
betecknas R3
. Alla punkter i rummet har tre koordinater, som i bilden
nedan: P(3, 0, 5) och Q( 5, 5, 7). Origo har koordinater (0, 0, 0). Generellt
heter koordinaterna (x, y, z) och axlarna ritas som i bilden nedan (z-axeln
vertikalt; x och y-axlarna kan ritas som i bilderna nedan: i båda fall ligger
x, y och z- axlarna enligt högerhandsregeln). Vektorer i R3
har också tre
koordinater: x, y och z som beskriver förflyttning i x, y och z leden.
10 5
5 10
10
5
5 10
1
5
P(3, 0, 5)
Q( 5, 5, 7)
(0, 0, 0)
Origin
x
y
z
z
2
4
2
0
3
5
2
4
x
y
z
3
5 =
2
4
0
2
3
3
5
10 5
5 10
10
5
5 10
1
(0, 0, 0)
Origin
x
y
z
2
4
2
2
0
3
5
2
4
2
0
1
3
5
2
4
x
y
z
3
5 =
2
4
0
2
3
3
5
Cartesian coordinate system

342. Two ways to address points in R2
Three ways to address points in R3
Cartesian (rectangular) coordinates
Cartesian (rectangular) coordinates
(x, y)
(x, y, z)

343. x
y
(x, y)
x
y
θ
r
Polar coordinates
argument

344. x
y
(x, y)
x
y
θ
r
Polar to Cartesian
x = r cos θ
y = r sin θ
r = x2
+ y2
tan θ =
y
x
If x = 0: y 0 ⇒ θ =
π
2
, y 0 ⇒ θ =
3π
2
Polar coordinates
Cartesian to polar
r ⩾ 0
θ ∈ [0, 2π)
(0, 0) = [0, 0]

345. Example
x
y
θ
4
r
[4, π/4] =

346. Example
x
y
θ
4
r
[4, π/4] =
sin
π
4
= cos
π
4
=
2
2
x = r cos θ
y = r sin θ

347. Example
x
y
θ
4
r
[4, π/4] = (2 2, 2 2)

348. Example
Equation describing a circle with radius 2 in Cartesian coordinates
transforms to the following equation in polar coordinates
x2
+ y2
= 4
r = 2
Generally: equation for any positive constant c describes a circle with centre in the origin.
r = c

349. Example
The following equation in polar coordinates
θ =
π
4
describes the half-line in the plane
y = x, x 0
x
y
Generally: equation for any describes a half-line without the origin.
θ = θ0 θ0 ∈ [0, 2π)
θ

350. Disk D with centre in the origin and radius 2 is described in polar
coordinates as a rectangle E with sides parallel to the coordinate
axes (axis-parallel rectangles; abbreviated to APR):
{(x, y) ∈ ℝ2
; x2
+ y2
≤ 4} = {(r, θ) ∈ ℝ2
; 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}
x
y
D
2
2
r
E
θ
2
2π

351. Disk D with centre in the origin and radius 2 is described in polar
coordinates as a rectangle E with sides parallel to the coordinate
axes (axis-parallel rectangles; abbreviated to APR):
{(x, y) ∈ ℝ2
; x2
+ y2
≤ 4} = {(r, θ) ∈ ℝ2
; 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}
x
y
D
2
2
r
E
θ
2
2π

352. Disk D with centre in the origin and radius 2 is described in polar
coordinates as a rectangle E with sides parallel to the coordinate
axes (axis-parallel rectangles; abbreviated to APR):
{(x, y) ∈ ℝ2
; x2
+ y2
≤ 4} = {(r, θ) ∈ ℝ2
; 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}
x
y
D
2
2
r
E
θ
2
2π

353. Disk D with centre in the origin and radius 2 is described in polar
coordinates as a rectangle E with sides parallel to the coordinate
axes (axis-parallel rectangles; abbreviated to APR):
{(x, y) ∈ ℝ2
; x2
+ y2
≤ 4} = {(r, θ) ∈ ℝ2
; 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}
x
y
D
2
2
r
E
θ
2
2π

354. Disk D with centre in the origin and radius 2 is described in polar
coordinates as a rectangle E with sides parallel to the coordinate
axes (axis-parallel rectangles; abbreviated to APR):
{(x, y) ∈ ℝ2
; x2
+ y2
≤ 4} = {(r, θ) ∈ ℝ2
; 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}
x
y
D
2
2
r
E
θ
2
2π

355. Problems to solve
Identify the curve defined by the equation r =
1
1 − cos θ
Convert the Cartesian coordinates to polar coordinates
(2, − 2)
D = { (x, y); 1 ⩽ x2
+ y2
⩽ 4, 0 ⩽ x ⩽ y ⩽ 3x }
Describe the set D in polar coordinates:

356. x =
1
2
y2
−
1
2
r =
1
1 − cos θ

357. x
y
y = x
y = 3x
D
1
1
D = { (x, y); 1 ⩽ x2
+ y2
⩽ 4, 0 ⩽ x ⩽ y ⩽ 3x }

358. 71 Convert the Cartesian
coordinates 2 2 to polar
coordinates X
M
0 741 f E
T
252 2
ii
am
2
d2 4 4 8
d f8 2fL
r fItfyT f4t4 252
arctan 22 arctanC D
I
4

359. Problem 2 Identify the
curve defined by the equation
r
f
r r
stF yCmutt.b
1 aesop
FH x 1
FF It x 1hr20
I
0
2x y 1
x nzy.LI
fc
panebohe

360. ProblemI
1sx4y KID
k 44
OEXEyEr3X
III NEED
run
Mr
1Etano EB
tannyEetanoetan'S
If
EO

361. On no 1ErE2 OEF
F
p
r i
l l
l l
l l
Il i I
3g I a
y
l l
l l r
1 2

362. In the 3-space

363. x
y
(x, y)
x
y
θ
r
Polar to Cartesian
Cartesian to polar
x = r cos θ
y = r sin θ
r = x2
+ y2
tan θ =
y
x
(x, y, z) = [r, θ, z]
Cylindrical coordinates

364. x y
z = z0
x = r cos θ, y = r sin θ, z = z
z
r0
(r0, θ0, z0)
θ0
r = r0
Equation of the lateral surface of the cylinder:

365. Example
Equation describing a cylinder with radius 2 in Cartesian coordinates
transforms to the following equation in cylindrical coordinates
x2
+ y2
= 4
r = 2

366. Problems to solve
Convert the Cartesian coordinates to cylindrical coordinates
(2, − 2, 1)
Convert the equation written in cylindrical coordinates into
an equation in Cartesian coordinates.
zr = 2 − r2
Identify the surface defined by the equation z = 3 − r2
Identify the surface defined by the equation r2
− 4r cos θ = 5

367. 1 Convert the Cartesian
coordinates 2,02,1 to
cylindrical coordinates
7I
4
r 252
I I I 1 X
r
o
2 2,1 252,7 I
Problemly Convert zr 2 r2
written in
cylindrical coordinates
into an
equation in Cartesian c
2
Eye 2
2
y2

368. Problein z 3
r2 x4y
z
3µ
paraboloid
y
i
I
3
f
c Sy

369. Problem4_
Identify the surface defined by
the equation
r2 4rcosO
5f
x2ty 4x 5
w
x 4 14 4 5 5
x 2
2
5 9
1 212
5
321
circular cylinder with centre
in 2,0 and radius 3
along
the Z axis

370. Two ways to address points in R2
Three ways to address points in R3
Cartesian (rectangular) coordinates
Cartesian (rectangular) coordinates
(x, y)
(x, y, z)
[r, θ], r ⩾ 0, θ ∈ [0,2π)
[r, θ, z], r ⩾ 0, θ ∈ [0,2π)
Polar coordinates
Cylindrical coordinates

371. x
y
z
(x, y, z)
R
Spherical coordinates
R ⩾ 0

372. y
z
(x, y, z)
r
𝜽
|z|
r
Spherical coordinates
x
R
R ⩾ 0
θ ∈ [0,2π)
(x, y,0)

373. y
z
r
𝜽
|z|
r
Spherical coordinates
x
R
R ⩾ 0
θ ∈ [0,2π)
(x, y, z)
(x, y,0)

374. y
z
r
𝜽
|z|
r
Spherical coordinates
x
R
R ⩾ 0
θ ∈ [0,2π)
(x, y, z)
(x, y,0)

375. y
z
r
𝜽
𝝓
|z|
r
Spherical coordinates
x
R
R ⩾ 0
ϕ ∈ [0,π]
θ ∈ [0,2π)
(x, y, z)
(x, y,0)

376. Spherical coordinates
θ
ϕ
longitude
(co)latitude
R ⩾ 0
ϕ ∈ [0,π]
θ ∈ [0,2π)
y
z
r
𝜽
𝝓
|z|
r
(x, y, z) = [R, ϕ, θ]
x
R
𝝓
(x, y, z)
(x, y,0)

377. (x, y)
(x, y, z)
Polar coordinates
(Adams 8.5, p.488)
[r, θ], r ⩾ 0, θ ∈ [0,2π)
Cylindrical coordinates
(Adams 10.6, p.604)
[r, θ, z], r ⩾ 0, θ ∈ [0,2π)
Spherical coordinates
(Adams 10.6, p.606)
[R, ϕ, θ], R ⩾ 0, ϕ ∈ [0,π], θ ∈ [0,2π)
Two ways to address points in R2
Three ways to address points in R3
Cartesian (rectangular) coordinates
Cartesian (rectangular) coordinates

378. y
z
(x, y, z)
r
𝜽
𝝓
|z|
r
(x, y, z) = [R, ϕ, θ]
Spherical coordinates
x
R
𝝓
θ
ϕ
longitude
(co)latitude
R ⩾ 0
ϕ ∈ [0,π]
θ ∈ [0,2π)
(x, y,0)

379. Some important examples

380. x
y
z
(x, y, z)
R
r
𝜽
𝝓
𝝓
|z|
R = R0 a sphere
(x, y,0)

381. x
y
z
(x, y, z)
R
r
𝜽
𝝓
𝝓
|z|
R = R0
θ = θ0
a sphere
a vertical half-plane with edge along the z-axis
(without this edge)

382. x
y
z
(x, y, z)
R
r
𝜽
𝝓
𝝓
|z|
R = R0
θ = θ0
ϕ = ϕ0
a sphere
a vertical half-plane with edge along the z-axis
a cone if ϕ ≠ 0,
π
2
, π
the positive z-axis if ϕ = 0

383. x
y
z
(x, y, z)
R
r
𝜽
𝝓
𝝓
|z|
R = R0
θ = θ0
ϕ = ϕ0
a sphere
a vertical half-plane with edge along the z-axis
the xy-plane if ϕ =
π
2
a cone if ϕ ≠ 0,
π
2
, π
the positive z-axis if ϕ = 0
the negative z-axis if ϕ = π

384. Disk D with centre in the origin and radius 2 is described in polar
coordinates as a rectangle E with sides parallel to the coordinate
axes (axis-parallel rectangles; abbreviated to APR):
{(x, y) ∈ ℝ2
; x2
+ y2
≤ 4} = {(r, θ) ∈ ℝ2
; 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π}
x
y
D
2
2
r
E
θ
2
2π

385. {(x, y, z) ∈ R3
; x2
+ y2
+ z2
≤ 4}
= {(R, ϕ, θ) ∈ R3
; 0 ≤ R ≤ 2, 0 ≤ ϕ ≤ π, 0 ≤ θ 2π}
x
y
z
(x, y, z)
R
r
𝜽
𝝓
𝝓
|z|
Attention!
Ball B with centre in the origin and radius 2 is described in spherical
coordinates as a rectangular box E with edges parallel to the coordinate
axes (axis-parallel rectangular box; abbreviated to APRB):

386. y
z
(x, y, z)
r
𝜽
𝝓
𝝓
|z|
r
From the yellow triangle:
r = R sin ϕ z = R cos ϕ
This gives (plug in r):
Polar coordinates in xy-plane:
x = r cos θ y = r sin θ
x = R sin ϕ cos θ
y = R sin ϕ sin θ
z = R cos ϕ
Conversion in the other direction: Cartesian to spherical:
Spherical to Cartesian:
r = x2
+ y2
= R sin ϕ
R = x2
+ y2
+ z2
tan ϕ =
r
z
=
x2
+ y2
z
tan θ =
y
x
(x, y, z) = [R, ϕ, θ]
Spherical coordinates
If x = 0: y 0 ⇒ θ =
π
2
, y 0 ⇒ θ =
3π
2
z = 0 ⇒ ϕ =
π
2
x
R

387. Problems to solve
Convert the Cartesian coordinates to spherical coordinates
(2, − 2, 1)
Convert the cylindrical coordinates to Cartesian and to spherical coordinates
[2, π/6, − 2]
Convert the spherical coordinates to Cartesian and to cylindrical coordinates
[4, π/3, 2π/3]
(Adams 10.6, problems 1, 2, 3.)
Describe the “ice cream cone” in spherical coordinates
¨
D



q(x,y)
ˆ
p(x,y)
f(x, y, z) dz


 dxdy =
d
ˆ
c



b(y)
ˆ
a(y)



q(x,y)
ˆ
p(x,y)
f(x, y, z) dz


 dx


 dy
jX PK QK`´/2i D ` r@2MF2Hi U`/B2HHiV, #2` FM /m##2HBMi2;`H2M dxdy
KX?XX TQH ` FQQ`/BMi#vi2X
PK pB T´ 2ii 2MF2Hi b ii FM #2bi KK ip `bMBii2M p F`QTT2M pBMF2H` i KQi M´;QM p FQQ`@
/BMitH`M- b´ FM i`BTT2HBMi2;`H2M #2` FMb
˚
K
f(x, y, z) dxdydz =
q
ˆ
p


¨
Dz
f(x, y, z) dxdy

 dz
/ ` Dz ` ip `bMBii2i p F`QTT2M pBMF2H` i KQi z−t2HMX
1tKTH2 3RX 2` FM i`BTT2HBMi2;`H2M
˚
K
z dxdydz
/ ` F`QTT2M K /2}MB2`b p /2 i`2 QHBF?2i2`M
x2 + y2 ⩽ z2, x2 + y2 + z2 ⩽ 1, z ⩾ 0X
.2M 7ƺ`bi QHBF?2i2M KQibp`` BMMM/ƺK2i p 2M +B`FmH `
FQM K2/ bT2ib2M B Q`B;Q Q+? /2 #´/ M/` QHBF?2i2`M
ƺp`2 ?HpM p 2M?2ibb7 `2MX x
y
f
(x,
y)
K
f(x, y, z) dxdydz =
p

Dz
f(x, y, z) dxdy dz
/ ` Dz ` ip `bMBii2i p F`QTT2M pBMF2H` i KQi z−t2HMX
1tKTH2 3RX 2` FM i`BTT2HBMi2;`H2M
˚
K
z dxdydz
/ ` F`QTT2M K /2}MB2`b p /2 i`2 QHBF?2i2`M
x2 + y2 ⩽ z2, x2 + y2 + z2 ⩽ 1, z ⩾ 0X
.2M 7ƺ`bi QHBF?2i2M KQibp`` BMMM/ƺK2i p 2M +B`FmH `
FQM K2/ bT2ib2M B Q`B;Q Q+? /2 #´/ M/` QHBF?2i2`M
ƺp`2 ?HpM p 2M?2ibb7 `2MX x
f
(x,
y)

388. Dz
2M pBMF2H` i KQi z−t2HMX
BMi2;`H2M
dydz
2 i`2 QHBF?2i2`M
1, z ⩾ 0X
BMMM/ƺK2i p 2M +B`FmH `
2 #´/ M/` QHBF?2i2`M
x
y
f
(x,
y)
;Hbbi`miǶRjX oB #2` FM` i`BTT2HBMi2;`H2M K2/ /2M i`2/D2 K2iQ@
QM2M p QK`´/2i B xy@THM2iX S`QD2FiBQM2M p bF `MBM;bFm`pM
/ ` F`QTT2M K /2}MB2`b p /2 i`2 QHBF?2i2`M
x2 + y2 ⩽ z2, x2 + y2 + z2 ⩽ 1, z ⩾ 0X
.2M 7ƺ`bi QHBF?2i2M KQibp`` BMMM/ƺK2i p 2M +B`FmH `
FQM K2/ bT2ib2M B Q`B;Q Q+? /2 #´/ M/` QHBF?2i2`M
ƺp`2 ?HpM p 2M?2ibb7 `2MX
E`QTT2M K ?` 7Q`K2M p 2M Ƕ;Hbbi`miǶRjX oB #2` FM` i`BT
/2M QpM Q+? i` 7`K T`QD2FiBQM2M p QK`´/2i B xy@THM2
?` 2FpiBQM2M
x2
+ y2
= 1 − x2
− y2
⇒ x2
+
b´ D = {(x, y) ∈ R3 : x2 + y2 = 1
2 }X
oB/`2 ` p(x, y) =
'
x2 + y2 Q+? q(x, y) =
'
1 − x2 − y2- b

389. Problemt
Convert the Cartesian coordinates
X y Z
2 2,1 to spherical coordinates
runway
r 2rz
ii i x
at E
2 2
R d 3 arccosts 7
3 HI
4
R fEtf2HF 54 4
7 3
252
tank 1
252
arctan 2E

390. r 252
j
2 p
To
1112 3
Cos of
L
3
arccos
Problem2e Convert the cylindrical
coordinates 2 T 2 to
Cartesian and to spherical words
r 0
2 T 23
1 Cartesian
x
y
z rcoso rsin 0 z
w
o f2E 2 E 2
B L 2

391. r 252
j
2 p
To
1112 3
Cos of
L
3
arccos
Problem2e Convert the cylindrical
coordinates 2 T 2 to
Cartesian and to spherical words
r 0
2 T 23
1 Cartesian
x
y
z rcoso rsin 0 z
w
o f2E 2 E 2
B L 2

392. h
EE
e
2
cos h
sin I
2 Spherical R fo O
242 I
6
R
2
53 11
45 252
tan 4 32 1
If
tantn
0 I

393. y
34T
4
9 n
I
L If
X
Rio O 252,3 E
Problemse Convert the spherical
coordinates 4 E GI to
Cartesian and to
cylindrical
coordinates
a
R 4,03 14 E E
w w
Rsindcoso 4 EE f E B
n
4 Iz Iz 3
niEE ers

394. y
34T
4
9 n
I
L If
X
Rio O 252,3 E
Problemse Convert the spherical
coordinates 4 E GI to
Cartesian and to
cylindrical
coordinates
a
R 4,03 14 E E
w w
Rsindcoso 4 EE f E B
n
4 Iz Iz 3
niEE ers

395. t
It ol
a 4
case
cos 0 12 sin If
2
Cylindrical coord
8 r O z 2B YI 2
r ftrYt3T 53 9
1 2 B
r 4 sin of 4 Ez 2B

396. Problemlt Describe the ice cream
cone in spherical coordinates
2 ZO
X2ty2tz2s1
X'tyee zz
042
Of 0 E 21T
The lateral surface of the cone
has E
OEOTEE
Rectangular box in the
spherical
coordinates
R ol O
0,17 0,141 10,25
mum

546. 12.3
Partial derivatives, introduction 1: the definition and notation

547. S`iB2HH /2`BpiQ`
G´i f : R2 → R Q+? (a, b) p` 2M BM`2 TmMFi iBHH /2}MBiBQMbK M;/2MX oB FQKK2` ii 7ƺ`bƺF
?Bii 2M #` KQibp`B;?2i iBHH /2`BpiM 7`´M 2Mp`B#2HFHFvHX A/; ;ƺ` pB 2ii 7ƺ`bi 7ƺ`bƺF bQK
?2i2` T`iB2HH /2`BpiX oB /2}MB2`` ip´ T`iB2HH /2`BpiQ` B TmMFi2M (a, b), /2M 7ƺ`bi ?2i2`
/2M T`iB2HH /2`BpiM KXXTX9 x Q+? /2M M/` /2M T`iB2HH /2`BpiM KXXTX y Q+? /2
/2}MB2`b K2/ ?D HT p ;` Mbp `/2#2;`2TT2i 7`´M 2Mp`B#2HFHFvHX A `miM M2/M M;2b #´/2
/2}MBiBQM2M Q+? QHBF #2i2+FMBM;` 7ƺ` /2 T`iB2HH /2`BpiQ`M,
∂f
∂x
(a, b) = f!
x(a, b) = f!
1(a, b) = HBK
h→0
f(a + h, b) − f(a, b)
h
= g!
(a)
∂f
∂y
(a, b) = f!
y(a, b) = f!
2(a, b) = HBK
k→0
f(a, b + k) − f(a, b)
k
= h!
(b)
oB/ T`iB2HH /2`Bp2`BM; #2i`Fib HH p`B#H2`- miQK /2M KM /2`Bp2`` K2/ pb22M/2 T´- bQK
FQMbiMi2`X S`iB2HH /2`Bp2`BM; ` / `K2/ 2;2MiHB;2M /2`Bp2`BM; p 2M 7mMFiBQM p 2M p`B#2H
U? ` g(x) = f(x, b) Q+? h(y) = f(a, y)V- b´ bKK /2`Bp2`BM;b`2;H2` ; HH2` bQK 7ƺ` 7mMFiBQM2` p
2M p`B#2H,
1t2KT2H k9X 2bi K f!
x(x, y) Q+? f!
y(x, y) 7ƺ` f(x, y) =
x
x + y
X
6ƺ` ii #2bi KK f! (x, y) #2i`Fi` pB y bQK 2M FQMbiMi Q+? 7´`,
Notation and definition
w.r.t. reads “with respect to”
First-order partial derivatives w.r.t. x and w.r.t. y
∂ reads “del”
g(x) = f(x, b)
h(y) = f(a, y)
f : ℝ2
→ ℝ

548. f : ℝ3
→ ℝ
∂f
∂x
(a, b, c) = f′
x(a, b, c) = f′
1(a, b, c) = lim
h→0
f(a + h, b, c) − f(a, b, c)
h
= g′(a)
∂f
∂y
(a, b, c) = f′
y(a, b, c) = f′
2(a, b, c) = lim
k→0
f(a, b + k, c) − f(a, b, c)
k
= h′(b)
g(x) = f(x, b, c)
h(y) = f(a, y, c)
∂f
∂z
(a, b, c) = f′
z(a, b, c) = f′
3(a, b, c) = lim
l→0
f(a, b, c + l) − f(a, b, c)
l
= φ′(c)
φ(z) = f(a, b, z)
n variables, n partial derivatives of the first order

549. Example 1
f(x, y, z) = x3
y4
z5
Find all the first partial derivatives of the function

550. f G is i
y4z5
8 Gigi
Cx y
z x34
y z
5
4 3
5
EE
x
y
z
p5x3y4z

551. Example 2
Find all the first partial derivatives of the function f(x, y) = ex+y
sin(xy)
∂
∂x
f(x, y) = ex+y
sin(xy) + ex+y
y cos(xy)
Product rule and chain rule
derivative of the inner function
∂
∂y
f(x, y) = ex+y
sin(xy) + ex+y
x cos(xy)
Product rule and chain rule
derivative of the inner function

552. Example 3
Find all the first partial derivatives of the function
U? ` g(x) = f(x, b) Q+? h(y) = f(a, y)V- b´ bKK /2`Bp2`BM;b`2;H2` ; HH2` bQK 7ƺ` 7mMFiBQ
2M p`B#2H,
1tKTH2 k9X *QKTmi2 f!
x(x, y) M/ f!
y(x, y) 7Q` f(x, y) =
x
x + y
X
hQ +QKTmi2 f!
x(x, y) r2 i`2i y b +QMbiMi M/ r2 ;2i UmbBM; i?2 [mQiB2Mi `mH2V,
f!
x(x, y) =
1 · (x + y) − 1 · x
(x + y)2
=
y
(x + y)2
.
hQ +QKTmi2 f!
y(x, y) r2 i`2i x b +QMbiMiX _27Q`KmHi2 f(x, y) = x(x + y)−1,
f!
y(x, y) = x(−1)(x + y)−2
(1) = −
x
(x + y)2
.
A Fm`b2M 2Mp`B#2HMHvb pBb/2 pB ii 2M /2`Bp2`#` 7mMFiBQM ` FQMiBMm2`HB;
T`iB2HHi /2`Bp2`#` 7mMFiBQM p ~2` p`B#H2` #2?ƺp2` /Q+F BMi2 p` FQMiBMm2`HB;- p
pBbb B M bi 2t2KT2HX
1tKTH2 k8X 6mMFiBQM2M f ` /2}MB2`/ bQK f(x, y) =
xy
x2 + y2
7ƺ` (x, y) = (0, 0) Q
U? ` g(x) = f(x, b) Q+? h(y) = f(a, y)V- b´ bKK /2`Bp2`BM;b`2;H2` ; HH2` bQK 7ƺ` 7mMFiBQM2` p
2M p`B#2H,
1tKTH2 k9X *QKTmi2 f!
x(x, y) M/ f!
y(x, y) 7Q` f(x, y) =
x
x + y
X
hQ +QKTmi2 f!
x(x, y) r2 i`2i y b +QMbiMi M/ r2 ;2i UmbBM; i?2 [mQiB2Mi `mH2V,
f!
x(x, y) =
1 · (x + y) − 1 · x
(x + y)2
=
y
(x + y)2
.
hQ +QKTmi2 f!
y(x, y) r2 i`2i x b +QMbiMiX _27Q`KmHi2 f(x, y) = x(x + y)−1,
f!
y(x, y) = x(−1)(x + y)−2
(1) = −
x
(x + y)2
.
A Fm`b2M 2Mp`B#2HMHvb pBb/2 pB ii 2M /2`Bp2`#` 7mMFiBQM ` FQMiBMm2`HB;X 1M
T`iB2HHi /2`Bp2`#` 7mMFiBQM p ~2` p`B#H2` #2?ƺp2` /Q+F BMi2 p` FQMiBMm2`HB;- pBHF2i
pBbb B M bi 2t2KT2HX
1tKTH2 k8X 6mMFiBQM2M f ` /2}MB2`/ bQK f(x, y) =
xy
2 2
7ƺ` (x, y) = (0, 0) Q+? bQK
U? ` g(x) = f(x, b) Q+? h(y) = f(a, y)V- b´ bKK /2`Bp2`BM;b`2;H2` ; HH2` bQK 7ƺ` 7mMFiBQM2` p
2M p`B#2H,
1tKTH2 k9X *QKTmi2 f!
x(x, y) M/ f!
y(x, y) 7Q` f(x, y) =
x
x + y
X
hQ +QKTmi2 f!
x(x, y) r2 i`2i y b +QMbiMi M/ r2 ;2i UmbBM; i?2 [mQiB2Mi `mH2V,
f!
x(x, y) =
1 · (x + y) − 1 · x
(x + y)2
=
y
(x + y)2
.
hQ +QKTmi2 f!
y(x, y) r2 i`2i x b +QMbiMiX _27Q`KmHi2 f(x, y) = x(x + y)−1,
f!
y(x, y) = x(−1)(x + y)−2
(1) = −
x
(x + y)2
.
A Fm`b2M 2Mp`B#2HMHvb pBb/2 pB ii 2M /2`Bp2`#` 7mMFiBQM ` FQMiBMm2`HB;X 1M
T`iB2HHi /2`Bp2`#` 7mMFiBQM p ~2` p`B#H2` #2?ƺp2` /Q+F BMi2 p` FQMiBMm2`HB;- pBHF2i

553. Adamsl2.3 9
wcx.ly z xYlnZ a
gag
tax a.fi Oalnam
on exponential
variable BASE constant BASE
constant EXPONENT variable EXP
8 yenz.x
o
4g
2 e
2 e
t
2e 2e
Fy fyfggke
T Exu.tn enx
nhTFfCe
2 e e2 lne line
Ty Tz zingy
e

554. 0
EzKEEF xLYm
TEC e 2 e
bye
2e
I Ty

555. Three kinds of consequences of the definition:
For geometry: help define the tangent plane to a graph surface
and compute its equation.
For computations: you compute the partial derivative w.r.t. a
variable according to the “normal” rules from Calculus 1; just
pretend for a while that all the other variables are constant. In this
way to compute partial derivatives is the same as computing
regular derivatives of functions of one variable.
For Calculus: makes it possible to discuss differentiability of
functions of several variables and gives a tool for optimisation of
such functions (a process of finding max and min of functions).

556. Tangent plane to the surface z = f (x, y) in the point (a, b, f (a, b))
A plane through the point (a, b, f (a, b))
which contains the tangent line to z = f (a, y) = h (y) in the plane x = a,
through (a, b, f (a, b)), thus the line with the slope h’(b)
which contains the tangent line to z = f (x, b) = g (x) in the plane y = b,
through (a, b, f (a, b)), thus the line with the slope g’(a)

557. a
b
(a, b, f(a, b))
the plane x = a
intersects the graph of the function
along the curve z = f (a , y) = h (y)
tangent to the curve z = f (a , y);
direction vector (0, 1 , fy’(a,b))
tangent to the curve z = f (x , b); direction vector (1, 0 , fx’(a,b))
z = f(x, y)
x
y
z
(a, b)
(1, 0, g′(a))
(0, 1, h′(b))
the plane y = b
intersects the graph to the function
along the curve z = f (x , b) = g (x)
graph: surface
∂f
∂y
(a, b) = h′(b) = lim
k→0
f(a, b + k) − f(a, b)
k
∂f
∂x
(a, b) = g′(a) = lim
h→0
f(a + h, b) − f(a, b)
h

558. Normal vector to the plane: cross product of two direction vectors
2FpiBQM T´ biM/`/7Q`K- HHib´ Ax + By + Cz + D = 0- 7´b ;2MQK 2
A, B, C) iBHH THM2i Q+? 2M TmMFi B THM2i UbQK ;2` p `/2i 7ƺ` DVX A p´`i
Hp2FiQ` iBHH THM2i Π bQK F`vbbT`Q/mFi2M p `BFiMBM;bp2FiQ`2`M8 (1, 0, fx
a, b)),
!
!
!
!
!
!
!
i !
j !
k
1 0 f!
x(a, b)
0 1 f!
y(a, b)
!
!
!
!
!
!
= (−f!
x(a, b), −f!
y(a, b), 1),
p2FiQ`2`M
−
→
N1 = (−f!
x(a, b), −f!
y(a, b), 1) Q+?
−
→
N2 = (f!
x(a, b), f!
y(a, b), −
Q`2` iBHH viM z = f(x, y) B TmMFi2M P = (a, b, f(a, b)), /2M 7ƺ`bi T2F` m
0V Q+? /2M M/` M2/´i Uiv ∆z = −1 0VX
−
→

559. ⃗
N1 = (−f′
x(a, b), − f′
y(a, b), 1)
(a, b, f(a, b)) (x, y, z)
Π
⃗
N2 = (f′
x(a, b), f′
y(a, b), − 1)
(x, y, z) ∈ Π ⇔ (x − a, y − b, z − f(a, b)) ⋅ (f′
x(a, b), f′
y(a, b), − 1) = 0
⇔ (x − a)f′
x(a, b) + (y − b)f′
y(a, b) − z + f(a, b) = 0

560. !
u = (u1, u2, u3) ` !
v · !
u = v1u1 + v2u2 + v3u3- Q+? / `7ƺ` ?`
⇔ (x − a, y − b, z − f(a, b)) · (f!
x(a, b), f!
y(a, b), −1) = 0
z+f(a, b) = 0 ⇔ z = f(a, b) +
∂f
∂x
(a, b)(x − a) +
∂f
∂y
(a, b)(y − b),
BQM 7ƺ` iM;2MiTHM2i Π iBHH ;`7viM z = f(x, y) B TmMFi2M
/`/2FpiBQM 7ƺ` iM;2MiTHM2i iBHH ;`7viM iBHH f(x, y) = −x2−y2
THM2ib 2FpiBQM `
(a, b) +
∂f
∂x
(a, b)(x − a) +
∂f
∂y
(a, b)(y − b).
y = f(a) + f′(a)(x − a)
Tangent plane to the surface z = f (x, y) in the point (a, b, f (a, b))
Tangent line through the point (a, f (a))
z = f(x, y)
y = f(x)

561. Normal line to the surface z = f (x, y) in the point (a, b, f (a, b))
Line through the point (a, b, f (a, b)) and with direction vector (f′
x(a, b), f′
y(a, b), − 1)
x − a
f′
x(a, b)
=
y − b
f′
y(a, b)
=
z − f(a, b)
−1
(x, y, z) = (a, b, f(a, b)) + t(f′
1(a, b), f′
2(a, b), − 1), t ∈ R

562. Y
Find the normal line to the paraboloid z = x2
+ y2
in the point (1,2,5)
∂z
∂x
= 2x,
∂z
∂y
= 2y
⃗
n 1 = (
∂z
∂x
(1,2),
∂z
∂y
(1,2), − 1) = (2 ⋅ 1, 2 ⋅ 2, − 1) = (2, 4, − 1)
a normal vector:
or the opposite vector (as in the picture):
⃗
n 2 = − ⃗
n 1 = (−2, − 4, 1)
Normal line to the surface in point (1,2,5) is the line through this point and with the normal vector as a direction vector:
(x, y, z) = (1, 2, 5) + t(2, 4, − 1), t ∈ ℝ

563. Problem solving
Find equations of the tangent plane and normal line to the graph of the
given function at the point with specified values of a and b:
f(x, y) = x2
− y2
, (a, b) = (−2, 1)
f(x, y) = cos
x
y
, (a, b) = (π, 4)
f(x, y) = ln(x2
+ y2
), (a, b) = (1, − 2)
f(x, y) =
x
x2 + y2
, (a, b) = (1, 2)
f(x, y) = ln(x2
y) + (2x − y)4
− 2y, (a, b) = (1, 1)

564. Problemt fCay E
y
a 6 2 a
E ffa.ee
fO x a7 ffcab7Cy
TmhztLmnmAxt
By1 Cz 10 0
T T T T
constants
x y Z variables
fCa b
f f 2 1 f 212 12 4 1 3
2x 2y
42
1 2 f 2 4
Fy C 2,1 2.1 2
2 3 4 x C 2 2 y 1
z 3 4 xt2
2y 12
z 3 4 8 2y 12 2 4 2y 3

565. 4x 2y z 3
TT
2,1 3
check 4 C 2 121 3 3
8
t2g6n
0
Nonmallin
T
Z
I
cxiyizf f 2 bz t
4i2iItElR
mm

566. Problem2_ fang u
a b JT 4
n
i
xiy 1 EI
coTyy
8 er 4 sin
I II
E
OUTER
NEER
offy
f sin
F YE Eye sing
xy
IT 4 sin
II I
II
32

567. Normal eg of the
tangent plane
Z
z EK IltIECy
TXE.eu
Normal line
t.EEEFfIPwblem3
fGiy7 y
ftp.I
f 1 2
2 51 C
8 kid
4,27
t
Cx g dog

568. Normal eg of the
tangent plane
Z
z EK IltIECy
TXE.eu
Normal line
t.EEEFfIPwblem3
fGiy7 y
ftp.I
f 1 2
2 51 C
8 kid
4,27
t
Cx g dog

569. x f Cx'ty4
2
2
y
W
infer der
C x't
yya
4,4 21 Es c
Normal eat of the tangent plane
z t
z
x 1 y 2
25 z 5 13 x 1 4 y
2
25
02 51 3 3
45182
3x 4y 25z l0
ITI
Normal line

570. ii
Iif
x
y z 1 2 F t t 3 4 25J
t E IR
Problemlt iD
fall 2
f 1 2 bn 14 C 25 en 5
5 a 27
I
Etty off G D
E
lnQECx I t.ly
Tss c

571. ii
Iif
x
y z 1 2 F t t 3 4 25J
t E IR
Problemlt iD
fall 2
f 1 2 bn 14 C 25 en 5
5 a 27
I
Etty off G D
E
lnQECx I t.ly
Tss c

572. it E 1
his
f 2 4 5
EI YY
Zjln5_
x
y Z 1 2 ln 5 t t l 2,4 5
te R
problem5
fcxiyt lncxyltcf.ir 2yc
a 6 1 1
f G 1
bring
1 I 2
2
G y t 4 2x y 2 to
t 8 2x y
3

573. it E 1
his
f 2 4 5
EI YY
Zjln5_
x
y Z 1 2 ln 5 t t l 2,4 5
te R
problem5
fcxiyt lncxyltcf.ir 2yc
a 6 1 1
f G 1
bring
1 I 2
2
G y t 4 2x y 2 to
t 8 2x y
3

574. 1 1
F t 8 2 t 1
2 18 I I 0
Kid
I
j t 4 2
1
C D 2
g
4 2x y j 2
If a 1
f 4 13 2 5
Tangent plane
2 I t lo x 1 5 y L
I t l 0 x lo 5y
t 5
10x 5
y 6
10 59 2 6
41
T P
N L
fT Y Z
fl

575. Gradient
C+Q#BM2M Q+? ;`/B2Mi2M, 7ƺ`bi BMi`Q/mFiBQM
/2i ` T`FiBbFi ii ? /2K `2/M B/;X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+iBQM Q7 n p`B#H2bX
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ /2MQi2/ b ∇f(!
x) Q` ∇f(t)V
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22FbX
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v = 7(t)V
` iDmpFBF` pB HBi2 T´ ip´ /2}MBiBQM2` bQK 2;2MiHB;2M bF FQKK b2M`2 B Fm`b2M U68Ĝ
/
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+iBQM Q7 n p`B#H2bX
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ /2MQi2/ b ∇f(!
x) Q`
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22FbX
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v = 7(t)V
#2 p2+iQ`@pHm2/ 7mM+iBQM rBi? +QKTQM2Mi 7mM+iBQMb f1, . . . fm- BX2X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+iBQM Q7 n p
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ /2MQi2/
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22FbX
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v = 7(t)V
C+Q#BM2M Q+? ;`/B2Mi2M, 7ƺ`bi BMi`Q/mFiBQM
` iDmpFBF` pB HBi2 T´ ip´ /2}MBiBQM2` bQK 2;2MiHB;2M bF FQKK b2M`2 B Fm`b2M U68ĜdVX J2M
/2i ` T`FiBbFi ii ? /2K `2/M B/;X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+iBQM Q7 n p`B#H2bX
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ /2MQi2/ b ∇f(!
x) Q` ∇f(t)V
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22FbX
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v = 7(t)V
` iDmpFBF` pB HBi2 T´ ip´ /2}MBiBQM2` bQK 2;2MiHB;2M bF FQKK
/2i ` T`FiBbFi ii ? /2K `2/M B/;X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+i
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v
C+Q#BM2M Q+? ;`/B2Mi2M, 7ƺ`bi BMi`Q/mFiBQM
` iDmpFBF` pB HBi2 T´ ip´ /2}MBiBQM2` bQK 2;2MiHB;2M bF FQKK b2M`2 B Fm`b2
/2i ` T`FiBbFi ii ? /2K `2/M B/;X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+iBQM Q7 n p`B#
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ /2MQi2/ b ∇f
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22FX
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v = 7(t)V

576. In the 3-space: the graph of f In the plane: the gradient (vector field)
f(x, y) = x2
+ y2
∂f
∂x
(x, y) = 2x,
∂f
∂y
(x, y) = 2y
⃗
F(x, y) = ∇f(x, y) = (2x,2y)

577. Tangent line through the point (a, f (a))
y = f(a) + f′(a)(x − a)
Tangent plane through the point (a, b, f (a, b))
z = f(a, b) + ∇f(a, b)[
x − a
y − b]
z = f(a, b) + f′
x(a, b)(x − a) + f′
y(a, b)(y − b)
z = f(a, b) + [f′
x(a, b), f′
y(a, b)][
x − a
y − b]
y = mx + b
Ax + By + Cz + D = 0
1 × 2
2 × 1

578. ∂f
∂x
= f′
x = fx = f′
1 = f1
z = f(x, y)
∂3
f
∂x3
=
∂
∂x (
∂2
f
∂x2 )
= f′′′
xxx = fxxx = f′′′
111 = f111
∂f
∂y
= f′
y = fy = f′
2 = f2
∂
∂x (
∂f
∂x )
=
∂2
f
∂x2
= f′′
xx = fxx = f′′
11 = f11
∂
∂y (
∂f
∂y)
=
∂2
f
∂y2
= f′′
yy = fyy = f′′
22 = f22
∂
∂y (
∂f
∂x)
=
∂2
f
∂y∂x
= f′′
xy = fxy = f′′
12 = f12
∂
∂x (
∂f
∂y)
=
∂2
f
∂x∂y
= f′′
yx = fyx = f′′
21 = f21
first order (2)
second order (4)
third order (8)
∂3
f
∂y3
=
∂
∂y (
∂2
f
∂y2 )
= f′′′
yyy = fyyy = f′′′
222 = f222
definition
∂3
f
∂y∂x2
= fxxy
∂3
f
∂x∂y∂x
= fxyx
∂3
f
∂x2∂y
= fyxx
∂3
f
∂x∂y2
= fyyx
∂3
f
∂y∂x∂y
= fyxy
∂3
f
∂y2∂x
= fxyy
notation Higher-order partial derivatives

579. fCx y x'y4z5
C 2nd order
I E t.IE EIy'z5f'fjI5x2yhzh
Of
2nd order
12x'y E
fy 42
3
y
z
f 2O 3y3z4
DI 11 3 11 34 3
of
t5xy4z fyz 2Ox3yz4 f 20xyz
mum
I
2nd order
3rd order der 3 9 27

580. fhay z
Xyz
fit
8
x'z
dy
off 2
x
at
Oxozoy
2x

581. ∂3
∂y∂z∂x
f(x, y, z) =
∂
∂y (
∂
∂z (
∂
∂x
(x2
yz)
))
= 2x
f(x, y, z) = x2
yz
∂
∂x
(x2
yz) = 2xyz
∂
∂z
(2xyz) = 2xy
∂
∂y
(2xy) = 2x
∂3
∂x∂y∂z
f(x, y, z) =
∂
∂x (
∂
∂y (
∂
∂z
(x2
yz)
))
= 2x
∂
∂z
(x2
yz) = x2
y
∂
∂y
(x2
y) = x2
∂
∂x
(x2
) = 2x
f132 = f′′′
132 = fxzy = f′′′
xzy f321 = f′′′
321 = fzyx = f′′′
zyx

582. Theorem (Schwarz; Equality of mixed partials)
Let
Suppose that two mixed nth-order partial derivatives of f
involve the same differentiations but in different orders.
If those partials are continuous at a point P,
and if f and all partials of f of order less than n are continuous in
a neighbourhood of P,
then the two mixed partials are equal at the point P.
f : ℝm
→ ℝ .

583. Example
f(x, y, z) = 3x2
y3
z2
+ y5
z
f′′
31 = f′′
13

584. 12.5
The Chain Rule: a general introduction

585. The chain rule: a rule for computations of
the derivatives of composite functions
x ⟼ x2
⟼ sin x2
⟼ ln(sin x2
)
d
dx
(ln(sin x2
)) =
1
sin x2
⋅ cos x2
⋅ 2x

586. E2/D2`2;2HM #2?ƺpb 7ƺ` ii /2`Bp2` bKKMbii 7mMFiBQM2`X A 2Mp`B#2HMHvb }MMb /2i
2M 2M/ p2`bBQM p F2/D2`2;2H- K2M /2i }MMb ?m` K´M; bQK ?2Hbi p2`bBQM2` p F2/D2`2;H2
~2`p`B#2H7mMFiBQM2`X oB FQKK2` ii ;´ B;2MQK M´;` pBFiB;bi2 p2`bBQM2` Q+? H ` Qbb Mp
/2K KX?XX C+Q#BM2` Q+? KX?XX /B;`K bQK ?Qb /KbX
o`BMi y h?2 QM2@/BK2MbBQMH +?BM `mH2 7Q`
f ◦ g : R
g
−→ R
f
−→ R
bvb i?i
d
dt
f(g(t)) = f!(g(t))g!(t)X qBi? MQiiBQM u = f(v)- v = g(t)- r2 +M r`Bi2
du
dt
=
du
dv
dv
dt
, u : R → R
o`BMi : h?2 KQbi ;2M2`H p2`bBQM Q7 +?BM `mH2, rBi? ?2HT Q7 i?2 C+Q#BMX h?2 +?BM
7Q` +QKTQbBi2 7mM+iBQMb +M #2 2tT`2bb2/ rBi? ?2HT Q7 Ki`Bt KmHiBTHB+iBQMX
G2i !
x = !
g(!
t) : Rq → Rn- M/ !
y = !
f(!
x) : Rn → Rm- i?2M
!
f ◦ !
g : Rq !
g
−→ Rn
!
f
−→ Rm

587. Notation for the four variants of the Chain Rule
Ki`BtVX
A 2t2KTH2M M2/M FQKK2` pB ii Mp M/ bKK #2i2+FMBM;` bQK
QpM B /2 ;2M2`2HH 7Q`KH2`M- K2M 7ƺHDM/2 #2i2+FMBM;` 7ƺ` ii BHHm@
bi`2` p/ 7mMFiBQM2`M bi´` 7ƺ`X .´ Mp M/2` pB,
Ç s, `2H@pHm2/ 7mM+iBQM Q7 QM2 `2H p`B#H2-
Ç !
v, p2+iQ`@pHm2/ 7mM+iBQM Q7 QM2 `2H p`B#H2-
Ç f, `2H@pHm2/ 7mM+iBQM Q7 b2p2`H `2H p`B#H2b-
Ç Φ, +?M;2 Q7 p`B#H2b BM R2 Q` BM R3X
Mp M/b 7ƺ` /2`BpiQ` iBHH 2Mp`B#2H7mMFiBQM2` Q+?
∂
∂x
7ƺ` ~2`p`B#2H7mMFiBQM2`X
Ub2 6k, /2H 8 B aib j T´ bX RdV K2/ q = 1 Q+? n = 1 Q+? m = nX
!
v ◦ s : R
s
−→ R
!
v
−→ Rn
+iHv i?2 `2bmHi BM i?2 `Qr i M/ i?2 +QHmKM j BM i?2 Ki`Bt KmHiBTHB+iBQM #Qp2
m M/ j = 1, . . . , q- #2+mb2 i?2 T`Q/m+i Q7 M m × n@Ki`Bt M/ M n × q@Ki`Bt Bb
Ki`BtVX
A 2t2KTH2M M2/M FQKK2` pB ii Mp M/ bKK #2i2+FMBM;` bQK
QpM B /2 ;2M2`2HH 7Q`KH2`M- K2M 7ƺHDM/2 #2i2+FMBM;` 7ƺ` ii BHHm@
bi`2` p/ 7mMFiBQM2`M bi´` 7ƺ`X .´ Mp M/2` pB,
Ç s, `2H@pHm2/ 7mM+iBQM Q7 QM2 `2H p`B#H2-
Ç !
v, p2+iQ`@pHm2/ 7mM+iBQM Q7 QM2 `2H p`B#H2-
Ç f, `2H@pHm2/ 7mM+iBQM Q7 b2p2`H `2H p`B#H2b-
Ç Φ, +?M;2 Q7 p`B#H2b BM R2 Q` BM R3X
Mp M/b 7ƺ` /2`BpiQ` iBHH 2Mp`B#2H7mMFiBQM2` Q+?
∂
∂x
7ƺ` ~2`p`B#2H7mMFiBQM2`X
Ki`BtVX
A 2t2KTH2M M2/M FQKK2` pB ii Mp M/ bKK #2i2+FMBM;` bQK
QpM B /2 ;2M2`2HH 7Q`KH2`M- K2M 7ƺHDM/2 #2i2+FMBM;` 7ƺ` ii BHHm@
bi`2` p/ 7mMFiBQM2`M bi´` 7ƺ`X .´ Mp M/2` pB,
Ç s, `2H@pHm2/ 7mM+iBQM Q7 QM2 `2H p`B#H2-
Ç !
v, p2+iQ`@pHm2/ 7mM+iBQM Q7 QM2 `2H p`B#H2-
Ç f, `2H@pHm2/ 7mM+iBQM Q7 b2p2`H `2H p`B#H2b-
Ç Φ, +?M;2 Q7 p`B#H2b BM R2 Q` BM R3X
Mp M/b 7ƺ` /2`BpiQ` iBHH 2Mp`B#2H7mMFiBQM2` Q+?
∂
∂x
7ƺ` ~2`p`B#2H7mMFiBQM2`X
, Ub2 6k, /2H 8 B aib j T´ bX RdV K2/ q = 1 Q+? n = 1 Q+? m = nX
!
v ◦ s : R
s
−→ R
!
v
−→ Rn
+iHv i?2 `2bmHi BM i?2 `Qr i M/ i?2 +QHmKM j BM i?2 Ki`Bt KmHiBTHB+iBQM #Qp2
m M/ j = 1, . . . , q- #2+mb2 i?2 T`Q/m+i Q7 M m × n@Ki`Bt M/ M n × q@Ki`Bt Bb
Ki`BtVX
A 2t2KTH2M M2/M FQKK2` pB ii Mp M/ bKK #2i2+FMBM;` bQK
QpM B /2 ;2M2`2HH 7Q`KH2`M- K2M 7ƺHDM/2 #2i2+FMBM;` 7ƺ` ii BHHm@
bi`2` p/ 7mMFiBQM2`M bi´` 7ƺ`X .´ Mp M/2` pB,
Ç s, `2H@pHm2/ 7mM+iBQM Q7 QM2 `2H p`B#H2-
Ç !
v, p2+iQ`@pHm2/ 7mM+iBQM Q7 QM2 `2H p`B#H2-
Ç f, `2H@pHm2/ 7mM+iBQM Q7 b2p2`H `2H p`B#H2b-
Ç Φ, +?M;2 Q7 p`B#H2b BM R2 Q` BM R3X
Mp M/b 7ƺ` /2`BpiQ` iBHH 2Mp`B#2H7mMFiBQM2` Q+?
∂
∂x
7ƺ` ~2`p`B#2H7mMFiBQM2`X
⃗
v ′(t) = (x′(t), y′(t), z′(t))
s′(t)
∇f = (f′
x, f′
y, f′
z)
gradient
jacobian matrix
d
dt
∂
∂x

588. Ub2 6k, /2H 8 B aib j T´ bX RdV K2/ q = 1 Q+? n = 1 Q+? m = nX
!
v ◦ s : R
s
−→ R
!
v
−→ Rn
d
dt
(!
v(s(t))) = s!
(t) · !
v !
(s(t)).
2i b´ 17i2`bQK /2`Bp2`BM; p p2FiQ`p `/ 7mMFiBQM2` bF2` FQKTQM2Mibp
iBQM2`M x1, . . . , xn ` pMHB; 2Mp`B#2H7mMFiBQM2` Q+? 7ƺ` /2K ; HH2`
F2/D2`2;2HMX HHib´- QK !
v(t) = (x1(t), . . . , xn(t))- /´ ` !
v !(t) = (x!
1(t), . . .
B,
,
d d
-
! ! ! !
Variant 1
!
v ◦ s : R
s
−→ R
!
v
−→ Rn
d
dt
(!
v(s(t))) = s!
(t) · !
v !
(s(t)).
2`Bp2`BM; p p2FiQ`p `/ 7mMFiBQM2` bF2` FQKTQM2MibpBb Q+? FQK@
n ` pMHB; 2Mp`B#2H7mMFiBQM2` Q+? 7ƺ` /2K ; HH2` /2M pMHB;
- QK !
v(t) = (x1(t), . . . , xn(t))- /´ ` !
v !(t) = (x!
1(t), . . . , x!
n(t)) Q+?
. . ,
d
dt
(xn(s(t)))
-
= (x!
1(s(t))·s!
(t), . . . , x!
n(s(t))·s!
(t)) = s!
(t)·!
v !
(s(t)).
!
v ◦ s : R
s
−→ R
!
v
−→ Rn
d
dt
(!
v(s(t))) = s!
(t) · !
v !
(s(t)).
K /2`Bp2`BM; p p2FiQ`p `/ 7mMFiBQM2` bF2` FQKTQM2MibpBb Q+? FQK@
. , xn ` pMHB; 2Mp`B#2H7mMFiBQM2` Q+? 7ƺ` /2K ; HH2` /2M pMHB;
HHib´- QK !
v(t) = (x1(t), . . . , xn(t))- /´ ` !
v !(t) = (x!
1(t), . . . , x!
n(t)) Q+?
))), . . . ,
d
dt
(xn(s(t)))
-
= (x!
1(s(t))·s!
(t), . . . , x!
n(s(t))·s!
(t)) = s!
(t)·!
v !
(s(
⇒
o`BMi R, Ub2 6k, /2H 8 B aib j T´ bX RdV K2/ q = 1 Q+? n = 1 Q+? m = nX
!
v ◦ s : R
s
−→ R
!
v
−→ Rn
d
dt
(!
v(s(t))) = s!
(t) · !
v !
(s(t)).
o`7ƺ` ` /2i b´ 17i2`bQK /2`Bp2`BM; p p2FiQ`p `/ 7mMFiBQM2` bF2` FQKTQM2MibpBb Q+
TQM2Mi7mMFiBQM2`M x1, . . . , xn ` pMHB; 2Mp`B#2H7mMFiBQM2` Q+? 7ƺ` /2K ; HH2` /2M
2Mp`B#2H@F2/D2`2;2HMX HHib´- QK !
v(t) = (x1(t), . . . , xn(t))- /´ ` !
v !(t) = (x!
1(t), . . . , x!
n(
/ `7ƺ` 7´` pB,
d
dt
(!
v(s(t))) =
,
d
dt
(x1(s(t))), . . . ,
d
dt
(xn(s(t)))
-
= (x!
1(s(t))·s!
(t), . . . , x!
n(s(t))·s!
(t)) = s!
(
A /2i bBbi bi2;2i ?` pB #`miBi mi bFH `2M s!(t) 7`K7ƺ` p2FiQ`M bKi bii BM !
v !(s(t)) BXb
iQ`M K2/ /2`BpiQ`M iBHH FQKTQM2Mi7mMFiBQM2`M- Q+? B /2i M bi bBbi, iBHH KTi /2M 2Mp
∂x
/2H 8 B aib j T´ bX RdV K2/ q = 1 Q+? n = 1 Q+? m = nX
!
v ◦ s : R
s
−→ R
!
v
−→ Rn
d
dt
(!
v(s(t))) = s!
(t) · !
v !
(s(t)).
17i2`bQK /2`Bp2`BM; p p2FiQ`p `/ 7mMFiBQM2` bF2` FQKTQM2MibpBb Q+? FQK@
x1, . . . , xn ` pMHB; 2Mp`B#2H7mMFiBQM2` Q+? 7ƺ` /2K ; HH2` /2M pMHB;
;2HMX HHib´- QK !
v(t) = (x1(t), . . . , xn(t))- /´ ` !
v !(t) = (x!
1(t), . . . , x!
n(t)) Q+?
(x1(s(t))), . . . ,
d
dt
(xn(s(t)))
-
= (x!
1(s(t))·s!
(t), . . . , x!
n(s(t))·s!
(t)) = s!
(t)·!
v !
(s(t)).

589. Variant 2
7ƺHDM/2,
z = s ◦ f : R2 f
−→ R
s
−→ R,
y
x
Q+? s(t) = `+iM tX
`iB2HH /2`BpiQ`M p bKKMbi HHMBM;2M z = s ◦ f 2MHB;i 7Q`KH2
∂z
∂x
=
ds
dt
·
∂f
∂x
,
∂z
∂y
=
ds
dt
·
∂f
∂y
,
1 ! y 1 ! y x2 ! y
z = s ◦ f : R2 f
−→ R
s
−→ R,
f(x, y) =
y
x
Q+? s(t) = `+iM tX
#2` FM` T`iB2HH /2`BpiQ`M p bKKMbi HHMBM;2M z = s ◦ f 2MHB;i 7Q`KH2`M,
∂z
∂x
=
ds
dt
·
∂f
∂x
,
∂z
∂y
=
ds
dt
·
∂f
∂y
,
K ;2`
∂z
∂x
=
1
1 + (y
x)2
·
!
−
y
x2
=
1
x2+y2
x2
·
!
−
y
x2
=
x2
x2 + y2
·
!
−
y
x2
= −
y
x2 + y2
∂z
∂y
=
1
1 + (y
x)2
·
1
x
=
1
x2+y2
x2
·
1
x
=
x2
x2 + y2
·
1
x
=
x
x2 + y2
.
i ` BMi2 bp´`i- K2M KM K´bi2 H ` bB; F MM B;2M bBimiBQM2` / ` KM ?` bKKM
j
1tKTH2 jRX lTT;B7i 8 7`´M /Kb RkXj, Lm T´
´i2`bT2;H` F`Fi `2M ?Qb /2 BM#HM//2 7mMFi
aBimiBQM2M ` 7ƺHDM/2,
z = s ◦ f : R2
/ ` f(x, y) =
y
x
Q+? s(t) = `+iM tX
oB #2` FM` T`iB2HH /2`BpiQ`M p bKKMbi
∂z
∂x
=
ds
dt
·
∂f
∂x
,
bQK ;2`
! !
1tKTH2 jRX lTT;B7i 8 7`´M /Kb RkXj
´i2`bT2;H` F`Fi `2M ?Qb /2 BM#HM//
aBimiBQM2M ` 7ƺHDM/2,
z = s ◦
/ ` f(x, y) =
y
x
Q+? s(t) = `+iM tX
oB #2` FM` T`iB2HH /2`BpiQ`M p bK
∂z
∂x
=
ds
dt
bQK ;2`
Example: z = arctan
y
x
´i2`bT2;H` F`Fi `2M ?Qb /2 BM#HM//2 7mMFiBQM2`M, #2` FM
∂z
∂x
Q+?
∂z
∂x
/ ` z = `+iM
y
x
X
aBimiBQM2M ` 7ƺHDM/2,
z = s ◦ f : R2 f
−→ R
s
−→ R,
/ ` f(x, y) =
y
x
Q+? s(t) = `+iM tX
oB #2` FM` T`iB2HH /2`BpiQ`M p bKKMbi HHMBM;2M z = s ◦ f 2MHB;i 7Q`KH2`M,
∂z
∂x
=
ds
dt
·
∂f
∂x
,
∂z
∂y
=
ds
dt
·
∂f
∂y
,
bQK ;2`
∂z
∂x
=
1
1 + (y
x)2
·
!
−
y
x2
=
1
x2+y2
x2
·
!
−
y
x2
=
x2
x2 + y2
·
!
−
y
x2
= −
y
x2 + y2
∂z
∂y
=
1
1 + (y
x)2
·
1
x
=
1
x2+y2
x2
·
1
x
=
x2
x2 + y2
·
1
x
=
x
x2 + y2
.

590. Variant 3
HHi Kv+F2i HBFi bBimiBQM2M 7`´M 2Mp`B#2HMHvb Q+? 7Q`KH2`M p` ` ii b´ BMimBiBpX 6`´M Q+?
K2/ Mm #HB` /2i `BFiB;i MviiB;i K2/ #´/2 /B;`K bQK ?Qb /Kb Q+? K2/ Ki`BbKmHiBTHBFiBQMX
1tKTH2 jjX
f ◦ !
v : R
!
v
−→ R2 f
−→ R.
q2 ?p2 z(t) = f(x(t), y(t))X h?2 +QKTQbBiBQM Bb MQ`KH U*H+RV 7mM+iBQM Q7 QM2 p`B#H2 M/
Bib /2`BpiBp2 +QKTmi2b K2/ *H+R@K2i?Q/bX q2 #2;BM rBi? `2TH+BM; z(t) rBi? f(x(t), y(t)),
z
(t) = HBK
h→0
z(t + h) − z(t)
h
= HBK
h→0
f(x(t + h), y(t + h)) − f(x(t), y(t))
h
= · · ·
M/ MQr r2 TTHv ivTB+H i`B+F rBi? bm#i`+iBQM M/ //BiBQM Q7 i?2 bK2 i2`K BM Q`/2` iQ
;2i i?2 T`iBH /2`BpiBp2b,
· · · = HBK
h→0
f(x(t + h), y(t + h)) − f(x(t), y(t + h))
h
+ HBK
h→0
f(x(t), y(t + h)) − f(x(t), y(t))
h
= · · ·
M/ MQr r2 +M TTHv i?2 #bB+ +?BM `mH2 7`QK *H+R,
· · · = f
1(x(t), y(t)) · x
(t) + f
2(x(t), y(t)) · y
(t) =
∂f
∂x
,
∂f
∂y
'
· (x
(t), y
(t)) = ∇f(!
v(t)) · !
v
(t)
BM i?Bb rv r2 b22 r?v r2 ;2i i?2 THmb bB;M M/ ?Qr i?2 +?BM `mH2
The most frequent in our future applications
d
dt
[f( ⃗
v (t)] = ∇f( ⃗
v (t)) ⋅ ⃗
v ′(t)
f
x y
t t
f
x y z
t t t
∂f
∂x
dx
dt
+
∂f
∂y
dy
dt
∂f
∂x
dx
dt
+
∂f
∂y
dy
dt
+
∂f
∂z
dz
dt

591. M/ MQr r2 +M TTHv i?2 #bB+ +?BM `mH2 7`QK *H+R,
· · · = f
1(x(t), y(t)) · x
(t) + f
2(x(t), y(t)) · y
(t) =
∂f
∂x
,
∂f
∂y
'
· (x
(t), y
(t)) = ∇f(!
v(t)) · !
v
(t)
BM i?Bb rv r2 b22 r?v r2 ;2i i?2 THmb bB;M M/ ?Qr i?2 +?BM `mH2
+M #2 r`Bii2M /QrM rBi? ?2HT Q7 i?2 Ki`Bt KmHiBTHB+iBQMX
1tKTH2 j9X .2i2`KBM2
d
dt
f(!
v(t)) 7Q` f(x, y, z) = xz + +Qb y M/ !
v(t) = (bBM t, t2, HM(t2 + 1))X
q2 TTHv i?2 +?BM `mH2 p`BMi j,
d
dt
f(!
v(t)) =f
1(!
v(t))x
(t) + f
2(!
v(t))y
(t) + f
3(!
v(t))z
(t)
[f
1(x, y, z) = z, f
2(x, y, z) = − bBM y, f
3(x, y, z) = x]
= HM(t2
+ 1) +Qb t + (− bBM t2
)2t + bBM t ·
2t
1 + t2
.
9R
f
x y z
t t t
MQi?2` rv Q7 bQHpBM; i?2 bK2 T`Q#H2K rQmH/ #2 iQ +QKTmi2 i?2 +QKTQbBiBQM f(!
v(t)) M/
/Bz2`2MiBi2 Bi b 7mM+iBQM Q7 QM2 p`B#H2,
f(!
v(t)) = bBM t HM(t2
+ 1) + +Qb t2
⇒
d
dt
f(!
v(t)) = +Qb t HM(t2
+ 1) + bBM t ·
2t
t2 + 1
− 2t bBM t2
.
x(t) z(t)
y(t)

592. Variant 4
o2`bBQM 9 K2/ q = 2 = n UQ+? q = 3 = n B #BH/2M T´ M bi bB
f ◦ Φ : R2 Φ
−→ R2 f
−→ R,
;2 Q7 p`B#H2b BM R2- Bb /2}M2/ b, x = x(s, t), y = y(s, t).
z
x y
f (x, y) = f (x(s,t), y(s,t))
f
𝛷
BMp2+FHi bQK B M bi 2t2KT2HX .´ ` /2i T`FiBbFi ii Mp M/ /B;`K bQK B Fm`bHB
1tKTH2 jeX o2`bBQM 9 K2/ q = 2 = n UQ+? q = 3 = n B #BH/2M T´ M bi bB/V
HHib´
f ◦ Φ : R2 Φ
−→ R2 f
−→ R,
r?2`2 Φ- +?M;2 Q7 p`B#H2b BM R2- Bb /2}M2/ b, x = x(s, t), y = y(s, t).
z
x y
s t s t
f (x, y) = f (x(s,t), y(s,t))
f
𝛷
6B;m` Rd, E2/D2`2;2HM- o2`bBQM 9 K2/ q = 2 = n Q+? m = 1X Sbb` T2`72Fi pB/ TQ
M 9 K2/ q = 2 = n Q+? m = 1X Sbb` T2`72Fi pB/ TQH `i FQQ`/B@
2vLQi2XV
(s, t)) #2` FMb T´ 7ƺHDM/2 b ii- pBHF2i FM pH bb MiBM;2M 7`´M
b ii bQK KM pH b2` bMMQHBF?2i2`M 7`´M 2ii HBF`ii mib22M/2
KmHiBTHBFiBQM2M,
·
∂x
∂s
+
∂z
∂y
·
∂y
∂s
,
∂z
∂t
=
∂z
∂x
·
∂x
∂t
+
∂z
∂y
·
∂y
∂t
.
h?2 ;`/B2Mi Q7 f Bb ∇f =
!
∂f
∂x
,
∂f
∂y
M/ i?2 C+Q#BM2MKi`Bt
DΦ =





∂x
∂s
∂x
∂t
∂y
∂s
∂y
∂t





BiBQM Bb 2[mH iQ i?2 Ki`Bt T`Q/m+i Q7 i?2b2 irQ,


∂x ∂x

 !
/B;`K5V 2HH2` KX?XX Ki`BbKmHiBTHBFiBQM2M,
∂z
∂s
=
∂z
∂x
·
∂x
∂s
+
∂z
∂y
·
∂y
∂s
,
∂z
∂t
=
∂z
∂x
·
∂x
∂t
+
∂z
∂y
·
∂y
∂t
6Q`ib ii2` K2/ 1t2KT2H je, h?2 ;`/B2Mi Q7 f Bb ∇f =
!
∂f
∂x
,
∂f
∂y
M/ i?
iQ Φ,
DΦ =





∂x
∂s
∂x
∂t
∂y
∂s
∂y
∂t





h?2 /2`BpiBp2 Q7 i?2 +QKTQbBiBQM Bb 2[mH iQ i?2 Ki`Bt T`Q/m+i Q7 i?2b2
!
∂z
∂s
,
∂z
∂t
= ∇f·DΦ =
!
∂f
∂x
,
∂f
∂y
·





∂x
∂s
∂x
∂t
∂y ∂y





=
!
∂z
∂x
·
∂x
∂s
+
∂z
∂y
·
∂y
∂s
,
/B;`KK2i UT´ T`2+Bb bKK b ii bQK KM pH b2` bMMQHBF?2i2`M 7`´M 2ii HBF`ii mib22M/2
/B;`K5V 2HH2` KX?XX Ki`BbKmHiBTHBFiBQM2M,
∂z
∂s
=
∂z
∂x
·
∂x
∂s
+
∂z
∂y
·
∂y
∂s
,
∂z
∂t
=
∂z
∂x
·
∂x
∂t
+
∂z
∂y
·
∂y
∂t
.
6Q`ib ii2` K2/ 1t2KT2H je, h?2 ;`/B2Mi Q7 f Bb ∇f =
!
∂f
∂x
,
∂f
∂y
M/ i?2 C+Q#BM2MKi`Bt
iQ Φ,
DΦ =





∂x
∂s
∂x
∂t
∂y
∂s
∂y
∂t





h?2 /2`BpiBp2 Q7 i?2 +QKTQbBiBQM Bb 2[mH iQ i?2 Ki`Bt T`Q/m+i Q7 i?2b2 irQ,
!
∂z
∂s
,
∂z
∂t
= ∇f·DΦ =
!
∂f
∂x
,
∂f
∂y
·





∂x
∂s
∂x
∂t
∂y
∂s
∂y
∂t





=
!
∂z
∂x
·
∂x
∂s
+
∂z
∂y
·
∂y
∂s
,
∂z
∂x
·
∂x
∂t
+
∂z
∂y
·
∂y
∂t
.
1tKTH2 jdX G´i f(x, y, v)- / ` x = g(u, v) Q+? y = h(u, v)- b´ ii f ?` 2ii /B`2Fi #2`Q2M/2
p p`B#2HM v Q+? ip´ BM/B`2Fi #2`Q2M/2M p p`B#2HM u UpB 7mMFiBQM2`M g Q+? hVX 2bi K
∂
f(x, y, v)X
∂z
∂s
=
∂z
∂x
·
∂x
∂s
+
∂z
∂y
·
∂y
∂s
,
∂z
∂t
=
∂z
∂x
·
∂x
∂t
+
∂z
∂y
·
∂y
∂t
.
`ib ii2` K2/ 1t2KT2H je, h?2 ;`/B2Mi Q7 f Bb ∇f =
!
∂f
∂x
,
∂f
∂y
M/ i?2 C+Q#BM2MKi`Bt
Φ,
DΦ =





∂x
∂s
∂x
∂t
∂y
∂s
∂y
∂t





2 /2`BpiBp2 Q7 i?2 +QKTQbBiBQM Bb 2[mH iQ i?2 Ki`Bt T`Q/m+i Q7 i?2b2 irQ,
z
∂s
,
∂z
∂t
= ∇f·DΦ =
!
∂f
∂x
,
∂f
∂y
·





∂x
∂s
∂x
∂t
∂y
∂s
∂y
∂t





=
!
∂z
∂x
·
∂x
∂s
+
∂z
∂y
·
∂y
∂s
,
∂z
∂x
·
∂x
∂t
+
∂z
∂y
·
∂y
∂t
.
KTH2 jdX G´i f(x, y, v)- / ` x = g(u, v) Q+? y = h(u, v)- b´ ii f ?` 2ii /B`2Fi #2`Q2M/2
p`B#2HM v Q+? ip´ BM/B`2Fi #2`Q2M/2M p p`B#2HM u UpB 7mMFiBQM2`M g Q+? hVX 2bi K
BMp2+FHi bQK B M bi 2t2KT2HX .´ ` /2i T`FiBbFi ii Mp M/ /B;`K bQK B Fm`bHBii2`im`2MeX
1tKTH2 jeX o2`bBQM 9 K2/ q = 2 = n UQ+? q = 3 = n B #BH/2M T´ M bi bB/V Q+? m = 1-
HHib´
f ◦ Φ : R2 Φ
−→ R2 f
−→ R,
r?2`2 Φ- +?M;2 Q7 p`B#H2b BM R2- Bb /2}M2/ b, x = x(s, t), y = y(s, t).
z
x y
s t s t
f (x, y) = f (x(s,t), y(s,t))
f
𝛷

593. f
x y z
a c a c
b b a c
b
f(x, y, z), x = x(a, b, c), y = y(a, b, c), z = z(a, b, c)
∂f
∂a
=
∂f
∂x
∂x
∂a
+
∂f
∂y
∂y
∂a
+
∂f
∂z
∂z
∂a
∂f
∂b
=
∂f
∂x
∂x
∂b
+
∂f
∂y
∂y
∂b
+
∂f
∂z
∂z
∂b
∂f
∂c
=
∂f
∂x
∂x
∂c
+
∂f
∂y
∂y
∂c
+
∂f
∂z
∂z
∂c
R3
→ R3
→ R
f
(a,b,c) (x,y,z)
f ∘ Φ
Φ
Φ(a, b, c) = (x, y, z)
∇( f ∘ Φ) = ∇f ⋅ DΦ =
∂f
∂x
,
∂f
∂y
,
∂f
∂z
⋅
∂x
∂a
∂x
∂b
∂x
∂c
∂y
∂a
∂y
∂b
∂y
∂c
∂z
∂a
∂z
∂b
∂z
∂c
q = 3 = n

594. T = T(x, y, z) = T(x(u, v), y(t), z(w, t)) = T(x(u(s, t), v), y(t), z(w, t))
y
T
x z
u v t w t
s t
∂T
∂t
=
∂T
∂x
⋅
∂x
∂u
⋅
∂u
∂t
+
∂T
∂y
⋅
dy
dt
+
∂T
∂z
⋅
∂z
∂t
The Chain Rule: an example with a diagram

595. 1tKTH2 jNX G2i f(x, y) #2 +QMiBMmQmb 7mM+iBQM rBi? +QMiBMmQmb T`iBH /2`BpiBp2b Q7 }`bi
Q`/2`X amTTQb2 i?i 7Q` HH `2H MmK#2`b t 0 ?QH/b f(tx, ty) = t3f(x, y)X a?Qr i?i Bi BKTHB2b
x
∂f
∂x
+ y
∂f
∂y
= 3f(x, y).
f
u v
t x t y
f (u,v) = f (tx,ty)
6B;m` ky, E2/D2`2;2HM- 1t2KT2H jNX .2i }MMb ip´ p ;` B /B;`KK2i bQK H2/2` iBHH p`B#2HM tX
UBH/, MB- K2/ E2vLQi2XV
q2 mb2 i?2 +?BM `mH2 U++Q`/BM; iQ i?2 /B;`KV iQ t3f(x, y) = f(tx, ty) M/ r2 ;2i,
1tKTH2 jNX G2i f(x, y) #2 +QMiBMmQmb 7mM+iBQM rBi? +QMiBMmQmb T`iBH /2`BpiBp2b Q7 }`bi
Q`/2`X amTTQb2 i?i 7Q` HH `2H MmK#2`b t 0 ?QH/b f(tx, ty) = t3f(x, y)X a?Qr i?i Bi BKTHB2b
x
∂f
∂x
+ y
∂f
∂y
= 3f(x, y).
f
u v
t x t y
f (u,v) = f (tx,ty)
6B;m` ky, E2/D2`2;2HM- 1t2KT2H jNX .2i }MMb ip´ p ;` B /B;`KK2i bQK H2/2` iBHH p`B#2HM tX
UBH/, MB- K2/ E2vLQi2XV
q2 mb2 i?2 +?BM `mH2 U++Q`/BM; iQ i?2 /B;`KV iQ t3f(x, y) = f(tx, ty) M/ r2 ;2i,
3t2
f(x, y) =
d
dt
f(tx, ty) = f!
1(tx, ty)x + f!
2(tx, ty)y
q2 ?p2 HbQ
∂
∂x
f(tx, ty) = f!
1(tx, ty)t M/
∂
∂y
f(tx, ty) = f!
2(tx, ty)t bQ
∂ ∂
x
∂f
∂x
+ y
∂f
∂y
= 3f(x, y).
f
u v
t x t y
f (u,v) = f (tx,ty)
6B;m` ky, E2/D2`2;2HM- 1t2KT2H jNX .2i }MMb ip´ p ;` B /B;`KK2i bQK H2/2` iBHH p`B#2HM tX
UBH/, MB- K2/ E2vLQi2XV
q2 mb2 i?2 +?BM `mH2 U++Q`/BM; iQ i?2 /B;`KV iQ t3f(x, y) = f(tx, ty) M/ r2 ;2i,
3t2
f(x, y) =
d
dt
f(tx, ty) = f!
1(tx, ty)x + f!
2(tx, ty)y
q2 ?p2 HbQ
∂
∂x
f(tx, ty) = f!
1(tx, ty)t M/
∂
∂y
f(tx, ty) = f!
2(tx, ty)t bQ
x
∂
∂x
f(tx, ty) + y
∂
∂y
f(tx, ty) = xf!
1(tx, ty)t + yf!
2(tx, ty)t = t · 3t2
f(x, y) = 3f(tx, ty).
Q`/2`X amTTQb2 i?i 7Q` HH `2H MmK#2`b t 0 ?QH/b f(tx, ty) = t f(x, y)X a?Qr i?i Bi BKTHB2b
x
∂f
∂x
+ y
∂f
∂y
= 3f(x, y).
f
u v
t x t y
f (u,v) = f (tx,ty)
6B;m` ky, E2/D2`2;2HM- 1t2KT2H jNX .2i }MMb ip´ p ;` B /B;`KK2i bQK H2/2` iBHH p`B#2HM tX
UBH/, MB- K2/ E2vLQi2XV
q2 mb2 i?2 +?BM `mH2 U++Q`/BM; iQ i?2 /B;`KV iQ t3f(x, y) = f(tx, ty) M/ r2 ;2i,
3t2
f(x, y) =
d
dt
f(tx, ty) = f!
1(tx, ty)x + f!
2(tx, ty)y
q2 ?p2 HbQ
∂
∂x
f(tx, ty) = f!
1(tx, ty)t M/
∂
∂y
f(tx, ty) = f!
2(tx, ty)t bQ
x
∂
∂x
f(tx, ty) + y
∂
∂y
f(tx, ty) = xf!
1(tx, ty)t + yf!
2(tx, ty)t = t · 3t2
f(x, y) = 3f(tx, ty).
f
u v
t x t y
f (u,v) = f (tx,ty)
6B;m` ky, E2/D2`2;2HM- 1t2KT2H jNX .2i }MMb ip´ p ;` B /B;`KK2i bQK H2/2` iBHH p`B#2HM tX
UBH/, MB- K2/ E2vLQi2XV
q2 mb2 i?2 +?BM `mH2 U++Q`/BM; iQ i?2 /B;`KV iQ t3f(x, y) = f(tx, ty) M/ r2 ;2i,
3t2
f(x, y) =
d
dt
f(tx, ty) = f!
1(tx, ty)x + f!
2(tx, ty)y
q2 ?p2 HbQ
∂
∂x
f(tx, ty) = f!
1(tx, ty)t M/
∂
∂y
f(tx, ty) = f!
2(tx, ty)t bQ
x
∂
∂x
f(tx, ty) + y
∂
∂y
f(tx, ty) = xf!
1(tx, ty)t + yf!
2(tx, ty)t = t · 3t2
f(x, y) = 3f(tx, ty).
u(x, t) = tx, v(y, t) = ty

596. 1. Compute where f is a partially differentiable function.
∂
∂x
(f(xy2
, x3
))
2. Function is a two variable function differentiable in the whole plane.
Let Show using the chain rule for the
composite function h that
f(u, v)
h(x, y, z) = f
(
x
y
,
y
z )
, y 0, z 0.
x
∂h
∂x
+ y
∂h
∂y
+ z
∂h
∂z
= 0.
3. If determine
z = f(x, y), x = 2s + 3t, y = 3s − 2t,
∂2
z
∂s2
,
∂2
z
∂t2
,
∂2
z
∂s∂t
.
4. Solve the PDE with condition by using the
following change of variables: .
∂f
∂x
− 3
∂f
∂y
= x f(0,y) = ey
u = 3x + y, v = x
5. Solve the PDE by using the
following change of variables: .
∂2
f
∂x2
− 4x
∂2
f
∂x∂y
+ 4x2 ∂2
f
∂y2
− 2
∂f
∂y
= 0
u = x2
+ y, v = x

597. 7. Compute where . (All the involved
functions are continuously differentiable.)
∂z
∂u
z = g(x, y), y = f(x), x = h(u, v)
6. Solve the PDE (x 0, y 0) by using the following
change of variables: .
y
∂f
∂y
− x
∂f
∂x
= 0
u = xy, v = x
8. Compute (in two different ways) knowing that .
∂z
∂x
z = arctan
u
v
, u = 2x + y, v = 3x − y

598. Problemt ffCxy3x3
xyia
Hy
u v
E's E
fxlfcxyix.lt u 8EtffddE
fi.y2tfI 3x2

599. Probkff.LY
fCEf HzO
y 0,2 20
Show thatmmm
Ulery My zkY
I
u V
v
fly yitz FE
yt
w
fly Fifty 18
yE t1zo

600. EE ft Ez if
x
I II t
y f Tiff t
E
z
f Yz
0
T.EE E tzi8ff
tz 0

601. ProblemI
E far y x 2st3t
y 3s 2T
assume f has continuous
Compute pot der of order two
al 85,98 4 t
T
z
x 2st3t
x'T E
y c3s 2T
f't f t
3
O O
x
t
y
3
IEEm
df tt8y 38 285
a
8
Eff fsf28Et3
T T

602. e
Eani
g
CELIE ffsf
2 2 IT8 t 3 Ey 6
3 2Ex Eg 3dog
I DE
4 6828
t9oI Text3
1 412x
y
t
fy
2
48 tl28 y
98y
ed 8
o

603. eEtftf I 38 28
IIa
3 2o
313 18 28g
213 85 Zody D
98 6 7 6
0
48 F98 t2
yt4
Schwan 3 412 9
2
I2xyt4y

604. 8 Es Ft off38 2
ii
nfEL t Esl
I
3 2 Ex Ex t 3Ey Ext
2 2 Ey 13 dog D
68 Ixt 68
Schw
6 5 x
6
fI
x 2y
2 3y
6 49 x
y 4x
y Gy
6
2
t 5x y G
y

605. Problem 4
Solve the POE
x3
T T
with condition
fco g
eY by
using the following change of
variables u
3xty
r x
f 3 1
of
z u 8f
Fu Ey _f

606. 38 38 x
flu v
4 u
Thede
fEiEED v
Go back to the old variables
fGiy tYC3x
b
f ay Itf 3 Oty e't
GH te

607. Check if the solution is correct
8 3 fCo y
x ze
ft
qIy e
Y
off 343
9
3e 7
3

608. problem5y
fu
Solve the PDE
fI 4xE txo 2
T yyy
I
Y fxx 4xifyx 4ifyy
2fy.IT
V
f
fu
t.fr y
2x'fq
EumiMfO
tyg fun
Et2x'funter
Effutffx linearity

609. 2x.ffu7xt2 fu t2x.f rut fur
11product rule
2xf2xfuutfw
t2 2x fruitfur
4x'faut 2xf t
2fut 2xfront f w
Schwarz
4xefuut4xfurtfw 2fi
4x'fuut4xfuvtfwt2fu
4 x 2x fun t fun t 4
2
fun 2fu O
t
E u o
f

610. K
p u
constant
Inquisition
fCu v fufu
T
y y diff
fCx y f E y x L xty
Y y diff functions
of 1 var

611. problem6
y.gg qfz o xiy70wM
by using the following change of
variables
5
i
8 Eu off
y tf
off off x
FKEEH EGIEX.io O

612. x off 0
Id
x 0
00
f u v
4 u
Y differentiable
fCxy fCxy answer
I
I
y y Gy x x y Gy y O

613. Problemt Compute
where E
gang y fCx x hCu v
t T
CAH involved functions are
continuously
differentiable
TX
x
y
1h0
Ito
u v
x
Yo
Fu fEFu
E.ykt.EU
yo
gx.h'utgj.f'Cxl
hvF
gihitgifathi

614. Problemse Compute
where E arctan
3 3 3
Two different methods
Methods
of a
tan
Rt R
mt f
3
arctan 112 11
t o t u to
O
f air
it EE

615. 2 3u
2 3x y 3 2xt
3x y
t 2x y
y
5g
13
2
2xyt2y2
Method
qz
2
Gykarytan yt

616. i
i
I

617. Is each vector field a gradient to some function?
Answer by computations.
15.2

618. f(x, y) = x2
+ y2
∂f
∂x
(x, y) = 2x,
∂f
∂y
(x, y) = 2y
⃗
F(x, y) = ∇f(x, y) = (2x,2y)
Level curves are circles. They are orthogonal to the gradient.
Not a coincidence!
In the 3-space: the graph of f In the plane: the gradient (vector field)

619. ⃗
F(x, y) = (2x, 2y)
⃗
F(x, y) = (y, x)
⃗
F(x, y) = (y, − x)
⃗
F(x, y) = (−x, y)
Φ(x, y) = x2
+ y2
Φ(x, y) = −
1
2
x2
+
1
2
y2
Φ(x, y) = xy

620. ⃗
F(x, y) = (y, − x)

621. Is Fdny gradient for
some EL pi R
Let's say that i 1122 R
sit 0oI
I
oI
y
8
I
II a
afoIGyKy ty
constant w r t X
THE
avg.aka
tmcq.gg
constant want
y
Y C function of 1
There is no such so Yh
NOT a gradient

622. A vector field is called conservative if it is a gradient
to some -scalar field (multivariable function) .
This function is then called a (scalar) potential to the vector field.
⃗
F : ℝn
→ ℝn
C2
Φ : ℝn
→ ℝ
⃗
F = (P, Q), ⃗
F = ∇Φ ⇔ P =
∂Φ
∂x
, Q =
∂Φ
∂y
⃗
F = (P, Q, R), ⃗
F = ∇Φ ⇔ P =
∂Φ
∂x
, Q =
∂Φ
∂y
, R =
∂Φ
∂z
If is a potential to then is also a potential to for any constant C.
Φ ⃗
F Φ + C ⃗
F

623. ⃗
F(x, y) = (2x, 2y)
⃗
F(x, y) = (y, x)
⃗
F(x, y) = (y, − x)
⃗
F(x, y) = (−x, y)
conservative conservative
conservative
not conservative
Potential: Potential:
Potential:
Φ(x, y) = x2
+ y2 Φ(x, y) = −
1
2
x2
+
1
2
y2
Φ(x, y) = xy

624. Φ : ℝ2
→ ℝ, (x, y) ∈ DΦ
Φ(x, y) = C
∇Φ(x, y)
(x, y)
⃗
r′(t) = (x′(t), y′(t)) ⃗
r(t) = (x(t), y(t))
The gradient in each point is orthogonal
to the level curve through this point
φ(t) = Φ(x(t), y(t))
is constant on the level curve
φ′(t) = 0
0 = φ′(t) =
∂Φ
∂x
x′(t) +
∂Φ
∂y
y′(t) =
(
∂Φ
∂x
,
∂Φ
∂y )
⋅ (x′(t), y′(t)) = ∇Φ ⋅ ⃗
r′(t)
If vector field is conservative and is its potential
then the level curves to are called equipotential lines for .
⃗
F : ℝ2
→ ℝ2
Φ : ℝ2
→ ℝ
Φ ⃗
F

625. Gradient in each point in the domain of the function : R2
! R is orthogonal
to the level curve through this point.
A: ~
F(x, y) = (x, y), II: (x, y) = x2
2
+ y2
2
;
B: ~
F(x, y) = (y, x), III: (x, y) = xy;
C: ~
F(x, y) = (x, y), I: (x, y) = x2
2
y2
2
.
Conclusion:
Equipotential lines
are orthogonal to
the field lines
for any smooth field
only for conservative fields

626. Curves

627. A curve is an endlessly thin line in the plane or in the space

628. Some curves are graphs to functions ℝ → ℝ
x
y
f : [a, b] → ℝ
y = f(x)
a b
A curve is an endlessly thin line in the plane or in the space

629. not a graph to a function
f : ℝ → ℝ

630. 0
1
time t position ⃗
r(t)
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
A
B
(xB, yB, zB)

631. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
A
B
(xB, yB, zB)
POSITION VECTOR
time t position ⃗
r(t)

632. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(xB, yB, zB)
time t position ⃗
r(t)

633. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(xB, yB, zB)
time t position ⃗
r(t)

634. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(x3, y3, z3)
(xB, yB, zB)
time t position ⃗
r(t)

635. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(xB, yB, zB)
time t position ⃗
r(t)

636. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(xB, yB, zB)
time t position ⃗
r(t)

637. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(xB, yB, zB)
time t position ⃗
r(t)

638. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(x7, y7, z7)
(xB, yB, zB)
time t position ⃗
r(t)

639. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(x7, y7, z7)
(x8, y8, z8)
(xB, yB, zB)
time t position ⃗
r(t)

640. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(x7, y7, z7)
(x8, y8, z8)
(x9, y9, z9)
(xB, yB, zB)
time t position ⃗
r(t)

641. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(x7, y7, z7)
(x8, y8, z8)
(x9, y9, z9)
(x10, y10, z10)
(xB, yB, zB)
time t position ⃗
r(t)

642. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(x7, y7, z7)
(x8, y8, z8)
(x9, y9, z9)
(x10, y10, z10)
(x11, y11, z11)
(xB, yB, zB)
time t position ⃗
r(t)

643. 0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
x
y
z
(xA, yA, zA)
(x1, y1, z1)
A
B
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(x7, y7, z7)
(x8, y8, z8)
(x9, y9, z9)
(x10, y10, z10)
(x11, y11, z11)
(xB, yB, zB)
time t position ⃗
r(t)

644. x
y
z
A
B
0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
(xA, yA, zA)
(x1, y1, z1)
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(x7, y7, z7)
(x8, y8, z8)
(x9, y9, z9)
(x10, y10, z10)
(x11, y11, z11)
(xB, yB, zB)
time t position ⃗
r(t)

645. x
y
z
A
B
0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
(xA, yA, zA)
(x1, y1, z1)
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(x7, y7, z7)
(x8, y8, z8)
(x9, y9, z9)
(x10, y10, z10)
(x11, y11, z11)
(xB, yB, zB)
time t position ⃗
r(t)

646. x
y
z
A
B
⃗
r : [0,1] → ℝ3
⃗
r(0) = (xA, yA, zA)
⃗
r(1) = (xB, yB, zB)
0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
(xA, yA, zA)
(x1, y1, z1)
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(x7, y7, z7)
(x8, y8, z8)
(x9, y9, z9)
(x10, y10, z10)
(x11, y11, z11)
(xB, yB, zB)
time t position ⃗
r(t)

647. x
y
z
A
B
⃗
r : [0,1] → ℝ3
⃗
r(0) = (xA, yA, zA)
⃗
r(1) = (xB, yB, zB)
orientation
⃗
r(t) = (x(t), y(t), z(t)), where x, y, z : [0,1] → ℝ
VECTOR VALUED FUNCTION (x, y, z: component functions)
0
1
t1
t2
t3
t4
t5
t6
t7
t8
t9
t10
t11
(xA, yA, zA)
(x1, y1, z1)
(x2, y2, z2)
(x3, y3, z3)
(x4, y4, z4)
(x5, y5, z5)
(x6, y6, z6)
(x7, y7, z7)
(x8, y8, z8)
(x9, y9, z9)
(x10, y10, z10)
(x11, y11, z11)
(xB, yB, zB)
time t position ⃗
r(t)
PARAMETRIC CURVE (t is a parameter)

648. 7mMFiBQM2`X
x
y
x
y
x
y
y
z
f : [a, b] → R g : [c, d] → R
(x(t), y(t)) = (t, f(t)), t ∈ [a, b] (x(t), y(t)) = (g(t), t), t ∈ [c, d]
r
r
(x(t), y(t)) = (r cos t, r sin t), t ∈ [0,2π)
(x(t), y(t)) = (r cos 2t, r sin 2t), t ∈ [0,π)
(x(t), y(t)) = (r cos
t
r
, r sin
t
r
), t ∈ [0,2πr)
x
(x0, y0, z0)
⃗
v = [v1, v2, v3]
(x(t), y(t), z(t)) = (x0 + tv1, y0 + tv2, z0 + tv3)
t ∈ R
y = f(x) x = g(y)
a b
c
d

649. Plane curves
Curves in the 3-space
⃗
r(t) = (x(t), y(t)), t ∈ [a, b]
⃗
r(t) = (x(t), y(t), z(t)), t ∈ [a, b]
x, y and z are called component functions
the parameter t is often interpreted as time
is often interpreted as the position of a particle in time t
⃗
r(t)

650. Velocity in the time t (a vector tangent to the curve)
⃗
r′(t) = (x′(t), y′(t)), t ∈ [a, b]
⃗
r′(t) = (x′(t), y′(t), z′(t)), t ∈ [a, b]

651. ⃗
r′(t) =
d
dt
⃗
r(t) = lim
h→0
⃗
r(t + h) − ⃗
r(t)
h
= (x′(t), y′(t))
oB FHH` !
r !(t) iM;2Mip2FiQ`M iBHH Fm`pM B TmMFi2M !
r(t)X PK !
r(t) iQHFb bQK T`iBF2HMb H ;2
b´ FM !
r !(t) iQHFb bQK T`iBF2HMb ?biB;?2i Q+? |!
r !(t)| bQK /2bb 7`i pB/ iB/2M tX oB/`2 FM
!
r !!(t) = (x!!(t), y!!(t), z!!(t)) iQHFb bQK T`iBF2HMb ++2H2`iBQM pB/ iB/2M tX a2 K2` 7ƺ`FH`BM; B
#BH/2`M bQK 7ƺHD2`X
origo
⃗
r(t0)
⃗
r(t0 + h)
⃗
r(t0 + h) − ⃗
r(t0)
(x(t0 + h), y(t0 + h))
(x(t0), y(t0))
⃗
r(t) = (x(t), y(t))
d
dt
⃗
r(t0) = lim
h→0
⃗
r(t0 + h) − ⃗
r(t0)
h
= lim
h→0
(x(t0 + h), y(t0 + h)) − (x(t0), y(t0))
h
= lim
h→0
(x(t0 + h) − x(t0), y(t0 + h) − y(t0))
h
= lim
h→0
(
x(t0 + h) − x(t0)
h
,
y(t0 + h) − y(t0)
h
) = ( lim
h→0
x(t0 + h) − x(t0)
h
, lim
h→0
y(t0 + h) − y(t0)
h
)
= (x′(t0), y′(t0))
1
2 3
4 5
6

652. Functions of several variables, introduction
12.1

653. f : ℝn
→ ℝ
Function of several (real) variables
Multivariable function
Scalar field
f(x1, x2, …, xn) ∈ ℝ
f(x1, x2, …, xn) = x1x2 ⋅ … ⋅ xn

654. f(x, y) = x2
+ y2
− xy
f(x, y, z) = xyz + sin(x + y) − z2
f(x, y) = 2
f(x, y) = 6y
f(x, y) =
1
x2 + y2 − xy
f(x, y, z, t) = sin(txy) − t2
+ z2
Examples

655. The parabola y = x2 (a curve) is a graph to the (single-variable) function f (x) = x2
The paraboloid z = x2 + y2 + 1 (a surface) is a graph to the (two-variable) function f (x, y) = x2 + y2 + 1
x
y
TmMFi2`- b´ FM HQFH 2ti`2KTmMFi2` #
2tBbi2`` 2DV Q+? `M/TmMFi2` iBHH Df X
.2i ` BMi2 b F2`i ii 2M F`BiBbF TmMFi
ivT2` p mib22M/2 F`BM; 2M F`BiBbF TmMF
x
y
f(x, y)
1

656. f : ℝ → ℝ f : ℝ2
→ ℝ
Domain: presented on the x-axis;
an interval, a union of intervals; argument: x
Values: presented on the y-axis; real numbers
Graph: (often) a curve in the xy-plane
Γ = {(x, y); x ∈ Df, y = f(x)}
= {(x, f(x)); x ∈ Df}
(x, f(x))
x
f(x)
Domain: presented on the xy-plane;
a plane region; argument: (x, y)
Values: presented on the z-axis; real numbers
Graph: (often) a surface in the 3-space
Γ = {(x, y, z); (x, y) ∈ Df, z = f(x, y)}
= {(x, y, f(x, y)); (x, y) ∈ Df}
vertical-line test
y = f(x) z = f(x, y)

657. f(x, y) = x2
ye−(x2
+y2
)

658. An observation
Graph to a function of one variable is a set in the plane (in 2D).
We can draw it using Calculus 1 methods.
Graph to a function of two variables is a set in the space (in 3D). It
is sometimes possible to draw it.
Graph to a function of three variables is a set in 4D!
We are unable to draw it…
Γ = {(x, y, z, w); (x, y, z) ∈ Df, w = f(x, y, z)} = {(x, y, z, f(x, y, z)); (x, y, z) ∈ Df}

659. Non-degenerate real quadric surfaces
Ellipsoid
Elliptic paraboloid
Hyperbolic paraboloid
Elliptic hyperboloid of one sheet
Elliptic hyperboloid of two sheets
if one of the variables is
not squared
If there is one MINUS
If there is a MINUS, but all three variables squared
One sheet: just one MINUS
Two sheets: two times MINUS (if +1 on the RHS)
(only PLUS)
https://en.wikipedia.org/wiki/Quadric
z =
x2
a2
+
y2
b2
z =
x2
a2
−
y2
b2

660. Non-degenerate real quadric surfaces
Ellipsoid
Elliptic paraboloid
Hyperbolic paraboloid
Elliptic hyperboloid of one sheet
Elliptic hyperboloid of two sheets
if one of the variables is
not squared
If there is one MINUS
If there is a MINUS, but all three variables squared
One sheet: just one MINUS
Two sheets: two times MINUS (if +1 on the RHS)
https://en.wikipedia.org/wiki/Quadric
z =
x2
a2
+
y2
b2
z =
x2
a2
−
y2
b2
(only PLUS)

661. x
y
r
r
not a graph to a function
f : ℝ → ℝ
x2
+ y2
= r2
But we can express y as a function of x anyway.
On the upper arc:
On the lower arc:
f(x) = r2
− x2
, x ∈ [−r, r]
f(x) = − r2
− x2
, x ∈ [−r, r]
y = f(x)
y2
= r2
− x2
x
x
restrictions for the domain: no negative
numbers under the square root

662. not a graph to a function
f : ℝ2
→ ℝ
x2
+ y2
+ z2
= r2
z = f(x, y)
x
y
z
1
1
1
x
x
z2
= r2
− x2
− y2
z = r2
− x2
− y2
z = − r2
− x2
− y2
the upper half sphere
the lower half sphere
restrictions for the domain: no negative
numbers under the square root

663. f(x, y) = r2
− x2
− y2
x
y
r
r
Df = { (x, y) ∈ ℝ2
; r2
− x2
− y2
⩾ 0 }
Df = { (x, y) ∈ ℝ2
; x2
+ y2
⩽ r2
}

664. Do establish the domain, follow the usual rules
No negative numbers under the square root
No zero’s in denominators
Logarithms are not defined for zero or negative numbers
Same as in Calculus 1, but now the domains will be subsets
of the plane, not intervals or unions of intervals.

665. Problem solving
Specify the domains of the functions:
f(x, y) =
x + y
x − y
f(x, y) = xy
f(x, y) =
xy
x2 − y2
f(x, y) =
x
x2 + y2
f(x, y) = 4x2
+ 9y2
− 36
f(x, y) =
1
4x2 + 9y2 − 36
f(x, y) = ln(1 + xy)
f(x, y) = ln
x + y
x − y

666. PYYi.IT
oyaiwfProfbem1
fCxiyI
i
problem2 fcx.gl_TXT
o

667. Problem's fairy y
o.IE oY
ii
z
i
e
i
B

668. Problem 445
Tty 9 4 495
30620
gkiyt yffy.ge 4x49y 3
04x49y2
36 0
describes the
BOUNDARY
y ty
of them
44 192547 1
q G y
t
tE I
t.tl
omEo
0 0 36 0
I Tx
Alta

669. Problem 7
mitxy
O
f cnyt
lncttxyl.gg
i
of Eid
xy I
e if y E
2 if y E
gGt

670. K2/ ?ƺD/Fm`pQ` T´ 2M F`iX
PK BM;2i MMi M;2b ` /2}MBiBQMbK M;/2M D iBHH 2M 7mMFiBQM f p n p`B#H2` /2i biƺ`bi
K M;/ B Rn 7ƺ` pBHF2M f(x1, x2, . . . , xn) ` p H/2}MB2`i 7ƺ` HH (x1, x2, . . . , xn) ∈ DX
1tKTH2 RdX aT2+B7v i?2 /QKBM Q7 i?2 7mM+iBQM f(x, y) = HM
x + y
x − y
.
f Bb /2}M2/ i?2`2 r?2`2
x + y
x − y
0. q2 2tKBM2
irQ +b2b- QM2 7Q` x−y 0 M/ QM2 7Q` x−y 0,
x − y 0 :
x + y
x − y
0
⇒ x + y 0 ⇒ −x y x
x − y 0 :
x + y
x − y
0
⇒ x + y 0 ⇒ x y −x
Df
Df
x
Ƙy
h?2 BM2[mHBiv BM i?2 }`bi +b2 Bb QMHv TQbbB#H2 7Q` x 0 M/ i?2 QM2 BM i?2 b2+QM/ +b2 7Q`
x 0X h?2 /QKBM Df Bb +QHQm`2/ BM i?2 TB+im`2X
kj
y = x
y = − x
−x
x

671. f(x, y) = ln
x + y
x − y
f(x,0) = ln
x + 0
x − 0
= ln 1 = 0
x
y

672. Cone Paraboloid
f(x, y) = x2
+ y2 g(x, y) = x2
+ y2

673. z = 1 − x2
− y2
z2
= 1 − x2
− y2
x2
+ y2
+ z2
= 1
f(x, y) = 1 − x2
− y2

674. PK 7mMFiBQM2M ?` 2M bvKK2i`B p M´;Qi bH; #HB` /2i 2MFH`2 ii `Bi ;`72M iBHH /2MMX 1ii
2t2KT2H ` `/B2HH bvKK2i`B,
1tKTH2 R8X aF2i+? i?2 ;`T? Q7 i?2 7mM+iBQM
f(x, y) = HM(x2
+ y2
), 0 x2
+ y2
⩽ 4.
AM i?2 xy@THM2 i?2 /BbiM+2 r iQ i?2 Q`B;BM Bb
;Bp2M #v r2 = x2 + y2- bQ i?2 7mM+iBQM +M #2
r`Bii2M b
f(x, y) = HM(x2
+ y2
)
= HM(r2
)
= 2 HM(r), r ∈ (0, 2].
q2 i?mb ;2i i?2 ;`T? Q7 f #v `QiiBM; i?2 ;`T?
Q7 z = 2 HM(r) `QmM/ i?2 z@tBbX
2 4 6 8
−2
2
4 z = 2 HM(r)
r
z
q2 ;2i i?2 7QHHQrBM; bm`7+2,
6B;m` RR, h?2 ;`T? Q7 f(x, y) = HM(x2 + y2) 7`QK 1tX R8

675. Problem solving
Sketch the graphs of the functions:
f(x, y) = − x2
− y2
+ 1
f(x, y) = x2
+ y2
+ 5
f(x, y) = x2
+ y2
− 4x + 4y + 10
f(x, y) = x
f(x, y) = 5
f(x, y) =
1
1 + x2 + y2
f(x, y) = y2

676. fCx y f Fity
2
Eye
z2 ty
Z
oUuppe
part
g x
y x2ty2
z
x2tyc paraboloid
mum
f x y e
Tty
SEE'm
Tryin
m ane
z Txt txt
E txt
y
x

677. g
Cx y x4y2
in XZ plane fy 07

678. Graphs of finances of two
X
L
J
y
fkiy
A
z
L
y

679. f
n Z
z
ypk.gg
x
x c f
Hxiy t5
yµ
z ytf
Xc y

680. fcxiyf x
ytl
nzz px
ytf.la
EH
fkyt ffy
4xtytiao
zca.EE
T
y
y
L

681. Level curves
Let where and The set
f : D → ℝ D ⊂ ℝ2
c ∈ ℝ .
M #2 /B{+mHi iQ /`r +2`iBM bm`7+2b BM j. BM biBb7vBM; rvX M Hi2`MiBp2 +M #
r i?2 bQ +HH2/ H2p2H +m`p2b BMbi2/ UBM i?2 xy@THM2- 7Q` /Bz2`2Mi pHm2b Q7 cV,
Lc = {(x, y) ∈ D : f(x, y) = c}
M `2bmHi2`M/2 };m`2M HBFM` 2M iQTQ;`}bF F`i U?ƺD/Fm`pQ`V ƺp2` 7mMFiBQM2MX C K7ƺ`
2`F`iQ`Mb BbQi2`K2` Q+? BbQ#`2`X
KTH2 ReX aF2i+? i?2 H2p2H +m`p2b 7Q` i?2 7mM+iBQM f(x, y) = HM(x2+y2) 7Q` c = {−2, −1,
y
is called the level curve of f on level c.

682. Nivåkurvor (eng. ”level curves”)
9
OBS: nivåkurvorna “lever” i f:s definitionsmängd. De är alltså projektioner av
skärningskurvorna mellan grafytan z = f (x, y) och planet z = c på xy-planet!
z = c
Lc ⊂ ℝ2
Each level curve “lives” in the domain of f.
It is the projection on the xy-plane
of the intersection curve between
the graph surface z = f (x, y) and the plane z = c

683. g(x, y) = x2
+ y2
f(x, y) = x2
+ y2
Nc = {(x, y); x2
+ y2
= c2
} Nc = {(x, y); x2
+ y2
= c}
Circle with radius c Circle with radius c
x
y
x
y
Cone Paraboloid
L L

684. f(x, y) = 1 − x2
− y2
Df = { (x, y) ∈ ℝ2
; 1 − x2
− y2
⩾ 0 }
Level curve on level c: , circle with radius
x2
+ y2
= 1 − c2
1 − c2

685. PK 7mMFiBQM2M ?` 2M bvKK2i`B p M´;Qi bH; #HB` /2i 2MFH`2 ii `Bi ;`72M iBHH /2MMX 1ii
2t2KT2H ` `/B2HH bvKK2i`B,
1tKTH2 R8X aF2i+? i?2 ;`T? Q7 i?2 7mM+iBQM
f(x, y) = HM(x2
+ y2
), 0 x2
+ y2
⩽ 4.
AM i?2 xy@THM2 i?2 /BbiM+2 r iQ i?2 Q`B;BM Bb
;Bp2M #v r2 = x2 + y2- bQ i?2 7mM+iBQM +M #2
r`Bii2M b
f(x, y) = HM(x2
+ y2
)
= HM(r2
)
= 2 HM(r), r ∈ (0, 2].
q2 i?mb ;2i i?2 ;`T? Q7 f #v `QiiBM; i?2 ;`T?
Q7 z = 2 HM(r) `QmM/ i?2 z@tBbX
2 4 6 8
−2
2
4 z = 2 HM(r)
r
z
q2 ;2i i?2 7QHHQrBM; bm`7+2,
6B;m` RR, h?2 ;`T? Q7 f(x, y) = HM(x2 + y2) 7`QK 1tX R8

686. Ai +M #2 /B{+mHi iQ /`r +2`iBM bm`7+2b BM j. BM biBb7vBM; rvX M Hi2`MiBp2 +M #2 iQ
/`r i?2 bQ +HH2/ H2p2H +m`p2b BMbi2/ UBM i?2 xy@THM2- 7Q` /Bz2`2Mi pHm2b Q7 cV,
Nc = {(x, y) ∈ D : f(x, y) = c}
.2M `2bmHi2`M/2 };m`2M HBFM` 2M iQTQ;`}bF F`i U?ƺD/Fm`pQ`V ƺp2` 7mMFiBQM2MX C K7ƺ` p2M
p /2`F`iQ`Mb BbQi2`K2` Q+? BbQ#`2`X
1tKTH2 ReX aF2i+? i?2 H2p2H +m`p2b 7Q` i?2 7mM+iBQM f(x, y) = HM(x2+y2) 7Q` c = {−2, −1, 0, 1}.
h?2 `/BH bvKK2i`v /2KM/b i?i i?2 H2p2H
+m`p2b Kmbi #2 +QM+2Mi`B+H +B`+H2bX 6Q` c = 0
Bi Bb i?2 mMBi +B`+H2X c = 1 ;Bp2b
1 = HM(x2
+ y2
) ⇒ x2
+ y2
= e
⇒ +B`+H2 rBi? `/Bmb
√
e ≈ 1.65
c = −1 M/ c = −2 ;Bp2
x2
+ y2
= e−1
`2bT2+iBp2Hv x2
+ y2
= e−2
1
1
1
1
1
0
0
0
−1
−
1
−
2
x
y
PK pbi´M/2i K2HHM c@p `/2M ` FQMbiMi- bQK B /2ii 2t2KT2H- b´ #2iv/2` ;H2b MBp´Fm`pQ`
ii 7mMFiBQM2M 7ƺ` M/`b bFi Q+? i i MBp´Fm`pQ` ii 7mMFiBQM2M 7ƺ` M/`b bM##i- B MHQ;B
K2/ ?ƺD/Fm`pQ` T´ 2M F`iX
PK BM;2i MMi M;2b ` /2}MBiBQMbK M;/2M D iBHH 2M 7mMFiBQM f p n p`B#H2` /2i biƺ`bi
K M;/ B Rn 7ƺ` pBHF2M f(x1, x2, . . . , xn) ` p H/2}MB2`i 7ƺ` HH (x1, x2, . . . , xn) ∈ DX

687. Problem solving
Describe some level curves of the functions:
f(x, y) =
x − y
x + y
f(x, y) = xy
f(x, y) =
y
x2 + y2
f(x, y) = x2
+ 4y2
f(x, y) = x − y
mHi iQ /`r +2`iBM bm`7+2b BM j. BM biBb7vBM; rvX M Hi2`MiBp2 +M #2 iQ
H2/ H2p2H +m`p2b BMbi2/ UBM i?2 xy@THM2- 7Q` /Bz2`2Mi pHm2b Q7 cV,
Lc = {(x, y) ∈ D : f(x, y) = c}

688. Problemt fCxiy7 x
yI
E x
g
AY e 4
I o_0
z O 2
F 1 1 1
I l l l l l l l l X
Cx y
x
y
c Cx y y x c
All the level curves are the
parallel straight dines
c O
y
_x
c 2
y
X 2
c 4 y x 14

689. ProblemI far y x44y2
my z x44y
T T
X't y L c
4 L
C'T
y
a p CIO p p y X
r
It 45 4
171 1
T T
AI the level curves are ellipses
f
i

690. Problemse far y xy
my
a
I l l l l I l X
he Cx y xy
c O gives the coordinates lines
o
Hawken.to
c f 0
are hyperbolas

691. Probl.cm
fCx yj
x
Y
YOxty
0nyy
xc
Ii Deo
i
y
x i y y
i
I l l l l
il l l l X
I y
c I I C 2
g O
in
ty X
E I excluded from
the domain
Gg
fy
c
xty
cyty _X ex

692. y
ctt xC1 c
cased c I x O
y
axis
All the level curves on the
level of 1 are
straight
lines through the
origin

693. c 2tc y EY Tc cto
x't y
2
12412
circle with the centre in
0 LT and the
radius r Tac

694. Problemts flay 72
ry any Cop
to
y
O
f
I l l l
X
FL 2
L i
n
Lo Gig y
IT
y cq4cy
ur In
cxhqcy2 2
LI
1
fo O
b 2c

695. c 2tc y EY Tc cto
x't y
2
12412
circle with the centre in
0 LT and the
radius r Tac

696. Level surfaces
Let where and The set
f : D → ℝ D ⊂ ℝ3
c ∈ ℝ .
is called the level surface of f on level c.
bB/2` MQr n = 3X :`T?b Q7 f : R3 → R `2 Q#D2+ib BM 7Qm`@/BK2MbBQMH bT+2 bQ r2 +M
HBb2 i?2KX :2M2`HHv r2 +M QMHv MHvb2 bvKK2i`B2b Q7 i?2 ;`T?b M/ i`v iQ bF2i+?
H bm`7+2b
Lc = {(x, y, z) ∈ D : f(x, y, z) = c}
KTH2 R3X G2p2H bm`7+2b Q7 i?2 7mM+iBQM f(x, y, z) = x2+2y2+3z2 `2 +QM+2Mi`B+H 2HHBTb
i?2 +2Mi`2 BM i?2 Q`B;BMX
Mbp `/2M Q+? FQMiBMmBi2i
F Mm /2}MB2` ;` Mbp `/2M 7ƺ` p2FiQ`p `/ 7mMFiBQM2` Q+? 7mMFiBQM2` p ~2` p`B
/2bb 7mMFiBQM2` FM ;2M2`2HHi bF`Bpb bQK 2M 7mMFiBQM 7`´M Rn iBHH Rp 7ƺ` M´;` TQb
*QMbB/2` MQr n = 3X :`T?b Q7 f : R3 → R `2 Q#D2+ib BM 7Qm`@/BK2MbBQMH bT+2 bQ r2 +MMQi
pBbmHBb2 i?2KX :2M2`HHv r2 +M QMHv MHvb2 bvKK2i`B2b Q7 i?2 ;`T?b M/ i`v iQ bF2i+? i?2
H2p2H bm`7+2b
Lc = {(x, y, z) ∈ D : f(x, y, z) = c}
1tKTH2 R3X G2p2H bm`7+2b Q7 i?2 7mM+iBQM f(x, y, z) = x2+2y2+3z2 `2 +QM+2Mi`B+H 2HHBTbQB/b
rBi? i?2 +2Mi`2 BM i?2 Q`B;BMX
*QMbB/2` MQr n = 3X :`T?b Q7 f : R3 → R `2 Q#D2+ib BM 7Qm`@/BK2MbBQMH bT+2 bQ r2 +MMQi
pBbmHBb2 i?2KX :2M2`HHv r2 +M QMHv MHvb2 bvKK2i`B2b Q7 i?2 ;`T?b M/ i`v iQ bF2i+? i?2
H2p2H bm`7+2b
Lc = {(x, y, z) ∈ D : f(x, y, z) = c}
1tKTH2 R3X G2p2H bm`7+2b Q7 i?2 7mM+iBQM f(x, y, z) = x2+2y2+3z2 `2 +QM+2Mi`B+H 2HHBTbQB/b
rBi? i?2 +2Mi`2 BM i?2 Q`B;BMX

697. Problem solving
Describe some level surfaces of the functions:
f(x, y, z) =
x2
+ y2
z2
f(x, y, z) = x + 2y + 3z
f(x, y, z) = x2
+ y2
f(x, y, z) = x2
+ y2
+ z2
*QMbB/2` MQr n = 3X :`T?b Q7 f : R3 → R `2 Q#D2+ib BM 7Qm`@/BK2MbBQMH bT+2 bQ r2 +MMQi
pBbmHBb2 i?2KX :2M2`HHv r2 +M QMHv MHvb2 bvKK2i`B2b Q7 i?2 ;`T?b M/ i`v iQ bF2i+? i?2
H2p2H bm`7+2b
Lc = {(x, y, z) ∈ D : f(x, y, z) = c}
1tKTH2 R3X G2p2H bm`7+2b Q7 i?2 7mM+iBQM f(x, y, z) = x2+2y2+3z2 `2 +QM+2Mi`B+H 2HHBTbQB/b
rBi? i?2 +2Mi`2 BM i?2 Q`B;BMX

698. Problem 1 far y z
xt2yt3Z
Cx y z
f Ef c
X
fy 3z
I
All the level surfaces are planes
parallel to each other normal
to rt 1,2 3
Problem 2 f Cx ye x2ty2tz2
X2ty4z2 c
CLO LEO
c O co 0,0
c O Lc is a sphere
centered in the
origin
with
radius R

699. Problem 3 far y x2ty2
y c
Ceo 4 0
c O Lce z axis
c O Lea a cylinder
circular along z axis
with radius F
Problem 4 f Cx y z
tz
2ty C
2
2
Cco 4 0
Ceo LE z axis
c O L G y z x2tyEcz
in
a I z
ty 9 xFy
2 Eye

700. 12.7
Gradient

701. Gradient: the several-variable counterpart of the first derivative
C+Q#BM2M Q+? ;`/B2Mi2M, 7ƺ`bi BMi`Q/mFiBQM
` iDmpFBF` pB HBi2 T´ ip´ /2}MBiBQM2` bQK 2;2MiHB;2M bF FQKK b2M`2 B Fm`b2M U68ĜdVX J2M
/2i ` T`FiBbFi ii ? /2K `2/M B/;X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+iBQM Q7 n p`B#H2bX
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ /2MQi2/ b ∇f(!
x) Q` ∇f(t)V
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22FbX
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v = 7(t)V
` iDmpFBF` pB HBi2 T´ ip´ /2}MBiBQM2` bQK 2;2MiHB;2M bF FQKK b2M`2 B Fm`b2M U68Ĝ
/2i ` T`FiBbFi ii ? /2K `2/M B/;X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+iBQM Q7 n p`B#H2bX
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ /2MQi2/ b ∇f(!
x) Q`
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22FbX
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v = 7(t)V
#2 p2+iQ`@pHm2/ 7mM+iBQM rBi? +QKTQM2Mi 7mM+iBQMb f1, . . . fm- BX2X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+iBQM Q7 n p
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ /2MQi2/
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22FbX
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v = 7(t)V
C+Q#BM2M Q+? ;`/B2Mi2M, 7ƺ`bi BMi`Q/mFiBQM
` iDmpFBF` pB HBi2 T´ ip´ /2}MBiBQM2` bQK 2;2MiHB;2M bF FQKK b2M`2 B Fm`b2M U68ĜdVX J2M
/2i ` T`FiBbFi ii ? /2K `2/M B/;X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+iBQM Q7 n p`B#H2bX
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ /2MQi2/ b ∇f(!
x) Q` ∇f(t)V
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22FbX
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v = 7(t)V
C+Q#BM2M Q+? ;`/B2Mi2M, 7ƺ`bi BMi`Q/mFiBQM
` iDmpFBF` pB HBi2 T´ ip´ /2}MBiBQM2` bQK 2;2MiHB;2M bF FQKK
/2i ` T`FiBbFi ii ? /2K `2/M B/;X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+i
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v
C+Q#BM2M Q+? ;`/B2Mi2M, 7ƺ`bi BMi`Q/mFiBQM
` iDmpFBF` pB HBi2 T´ ip´ /2}MBiBQM2` bQK 2;2MiHB;2M bF FQKK b2M`2 B Fm`b2
/2i ` T`FiBbFi ii ? /2K `2/M B/;X
.2}MBiBQM ReX G2i f : Rn → R #2 T`iBHHv /Bz2`2MiB#H2 7mM+iBQM Q7 n p`B#
h?2 ;`/B2Mi Q7 f BM i?2 TQBMi !
x Bb i?2 7QHHQrBM; p2+iQ`
;`/ f(!
x) =
!
∂f
∂x1
(!
x),
∂f
∂x2
(!
x), . . . ,
∂f
∂xn
(!
x)
UHbQ /2MQi2/ b ∇f
h?2 bvK#QH ∇ T`QMQmM+2b M#Hc i?2 rQ`/ K2Mb i?2 ?`T BM ;`22FX
:`/B2Mi Bb p2+iQ` }2H/ U 7mM+iBQM 7`QK Rn iQ RnVX
.2}MBiBQM RdX G2i !
y = !
f(!
x) : Rn → Rm UHi2`MiBp2 MQiiBQM, v = 7(t)V

702. In the 3-space: the graph of f In the plane: the gradient (vector field)
f(x, y) = x2
+ y2
∂f
∂x
(x, y) = 2x,
∂f
∂y
(x, y) = 2y
⃗
F(x, y) = ∇f(x, y) = (2x,2y)

703. Normal line and tangent line to a circle
(x1, y1)
(−y1, x1)
(x2, y2)
(−y2, x2)
The position vector in each
point on the circle is orthogonal
to the tangent vector.
(a, b) ⋅ (−b, a) = ab − ba = 0
The position vector is normal to
the circle

704. f(x, y) = x2
+ y2
∂f
∂x
(x, y) = 2x,
∂f
∂y
(x, y) = 2y
⃗
F(x, y) = ∇f(x, y) = (2x,2y)
Level curves are circles. They are orthogonal to the gradient.
Not a coincidence!
In the 3-space: the graph of f In the plane: the gradient (vector field)

705. f : ℝ2
→ ℝ, (x, y) ∈ Df
f(x, y) = C
∇f(x, y)
(x, y)
⃗
r′(t) = (x′(t), y′(t)) ⃗
r(t) = (x(t), y(t))
The gradient in each point is orthogonal
to the level curve through this point
φ(t) = f(x(t), y(t))
is constant on the level curve
φ′(t) = 0
0 = φ′(t) =
∂f
∂x
x′(t) +
∂f
∂y
y′(t) =
(
∂f
∂x
,
∂f
∂y)
⋅ (x′(t), y′(t)) = ∇f ⋅ ⃗
r′(t)

706. Tangent plane to the level surface
through the point on this surface
has a normal vector
thus it has an equation so:
f(x, y, z) = C
(a, b, c)
⃗
n =
(
∂f
∂x
(a, b, c),
∂f
∂y
(a, b, c),
∂f
∂z
(a, b, c)
)
⃗
n ⋅ (x − a, y − b, z − c) = 0
∂f
∂x
(a, b, c)(x − a) +
∂f
∂y
(a, b, c)(y − b) +
∂f
∂z
(a, b, c)(z − c) = 0

707. Q3
1tKTH2 9dX .2i2`KBM2 MQ`KH 2[miBQM 7Q` i?2 iM;2Mi THM2 iQ i?2 H2p2H bm`7+2 7Q`
f(x, y, z) = x2 + y2 + z2 BM i?2 TQBMi (1, 3, 4)X
h?2 ;`/B2Mi BM i?2 TQBMi Bb Q`i?Q;QMH iQ i?2 H2p2H bm`7+2 i?`Qm;? i?2 TQBMi- bQ r2 ;2i i?2
7QHHQrBM; 2[miBQM 7Q` i?2 iM;2Mi THM2, ∇f(1, 3, 4) · (x − 1, y − 3, z − 4) = 0- Q`,
0 =
∂f
∂x
(1, 3, 4)(x − 1) +
∂f
∂y
(1, 3, 4)(y − 3) +
∂f
∂z
(1, 3, 4)(z − 4).
h?2 T`iBH /2`BpiBp2b `2,
∂f
∂x
(x, y, z) = 2x,
∂f
∂y
(x, y, z) = 2y,
∂f
∂z
(x, y, z) = 2z,
bQ
∂f
∂x
(1, 3, 4) = 2,
∂f
∂y
(1, 3, 4) = 6,
∂f
∂z
(1, 3, 4) = 8,
r?B+? ;Bp2b i?2 7QHHQrBM; 2[miBQM 7Q` i?2 iM;2Mi THM2,
2(x − 1) + 6(y − 3) + 8(z − 4) = 0 ⇔ 2x + 6y + 8z − 52 = 0.
z
h?2 T`iBH /2`BpiBp2b `2,
∂f
∂x
(x, y, z) = 2x,
∂f
∂y
(x, y, z) = 2y,
∂f
∂z
(x, y, z)
bQ
∂f
∂x
(1, 3, 4) = 2,
∂f
∂y
(1, 3, 4) = 6,
∂f
∂z
(1, 3, 4)
r?B+? ;Bp2b i?2 7QHHQrBM; 2[miBQM 7Q` i?2 iM;2Mi THM2,
2(x − 1) + 6(y − 3) + 8(z − 4) = 0 ⇔ 2x + 6y +
x y
z
(1,3,4)
6B;m` k3, hM;2MiTHM iBHH MBp´viM iBHH f(x, y, z) = x2 + y2 + z2 ;2MQ
N26X LBp´viM ` b7 `2M K2/ KBiiTmMFi2M B Q`B;Q Q+? K2/ `/B2M
√
viM ` HBF K2/ ;`/B2Mi2M ∇f(1, 3, 4) = (2, 6, 8)X UBH/, MB- K2/
3
o2`v BKTQ`iMi, HH i?2 MQ`KH p2+iQ`b iQ i?2 bT?2`2 +2Mi2`2/ BM i?2 Q`B;BM
p2+iQ` 2M/BM; BM i?2 TQBMi r?2`2 r2 Tmi i?2 MQ`KHX 2`2 !
n = (2, 6, 8) = 2(1, 3,
at

708. 12.7
Directional derivatives, the definition

709. t → 0 ⇒ (a + tv1, b + tv2) → (a, b)
along the line (x, y) = (a + tv1, b + tv2), t ∈ ℝ
(a, b)
(a + tv1, b + tv2)
⃗
v
(a + v1, b + v2)
v1
v2
∥ ⃗
v ∥ = 1
D ⃗
v f(a, b) = f′ ⃗
v (a, b) = lim
t→0
f(a + tv1, b + tv2) − f(a, b)
t

710. a
b
(a, b, f(a, b))
z = f(x, y)
x
y
z
(a, b)
graph: surface
Tangent to the curve in
(a, b, f (a, b)) in the blue plane
D ⃗
v f(a, b) = f′ ⃗
v (a, b) = lim
t→0
f(a + tv1, b + tv2) − f(a, b)
t

711. t → 0 ⇒ (a + tv1, b + tv2) → (a, b)
längs linjen (x, y) = (a + tv1, b + tv2), t ∈ R
v1 ≠ 0 ⇒ y =
v2
v1
x + c
6B;m` kd, AHHmbi`iBQM iBHH `BFiMBM;b/2`Bpi- p/ ? M/2` B /2}MBiBQMbK M;/2MX UBH/, MB- K2/
E2vLQi2XV
h?2 /B`2+iBQMH /2`BpiBp2 b?Qrb i?2 bHQT2 Q7 i?2 iM;2Mi HBM2 BM i?2 TQBMi (a, b, f(a, b)) iQ
i?2 +m`p2 Q7 BMi2`b2+iBQM #2ir22M i?2 ;`T? bm`7+2 z = f(x, y) M/ i?2 p2`iB+H UBX2X T`HH2H
rBi? i?2 z@tBbV THM2 i?`Qm;? i?2 TQBMi (a, b)- HQM; !
vX
6ƺHDM/2 bib iBHH´i2` Qbb #´/2 #2` FM `BFiMBM;b/2`BpiQ` T´ 2ii 2MF2Hi b ii Q+? ;ƺ` M´;`
;2QK2i`BbF Q#b2`piBQM2` QK 7mMFiBQM2Mb iBHHp ti B QHBF `BFiMBM;`,
h?2Q`2K RR U/Kb RkXdXdVX PK f ` 2M /Bz2`2MiB2`#` 7mMFiBQM Q+? !
v bXX |!
v| = 1 2M ;Bp2M
`BFiMBM;- b´ `
f!
!
v(!
a) = ∇f(!
a) · !
v.

712. `BFiMBM;bp2FiQ` ` MQ`K2`/X
.2}MBiBQM kyX h?2 /B`2+iBQMH /2`BpiBp2 Q7 i?2 7mM+iBQM f : Rn → R BM i?2 TQBMi !
a BM i?2
/B`2+iBQM !
v, |!
v| = 1- Bb /2}M2/ b
D!
vf(!
a) = f!
!
v(!
a) = HBK
t→0
f(!
a + t!
v) − f(!
a)
t
U!
v = !
ej ;Bp2b f!
j(!
a)V
PK f ` 2M ip´p`B#2H7mMFiBQM UpBHF2i Q7ibi FQKK2` ii p` 7HH2i 7ƺ` QbbV- /´ ` `BFiMBM;b@
p2FiQ`M !
v = (v1, v2) Q+? /2}MBiBQM2M b2` mi T´ 7ƺHDM/2 b ii,
D!
vf(a, b) = f!
!
v(a, b) = HBK
t→0
f(a + tv1, b + tv2) − f(a, b)
t
.

713. f Bb /Bz2`2MiB#H2 7m
f!
!
v(!
a) = ∇f(!
a) · !
v.
h?2Q`2K RR U/Kb RkXdXdVX A7 f Bb /Bz2`2MiB#H2 7mM+iBQM M/ !
v bXiX |!
v | = 1 Bb ;Bp2M
/B`2+iBQM- i?2M

714. Compute the directional derivative
of the function
at
in the direction
f(x, y) = x2
y
(x, y) = (2, 2)
⃗
v =
1
2
(1, 1)
b #´/2 #2` FM `BFiMBM;b/2`BpiQ` T´ 2ii 2M
` QK 7mMFiBQM2Mb iBHHp ti B QHBF `BFiMBM;`,
kXdXdVX A7 f Bb /Bz2`2MiB#H2 7mM+iBQM M/
f!
!
v(!
a) = ∇f(!
a) · !
v.
example

715. fCx y1 x2y
2xy fyf x
ay 2,2 Tr Fra
2,21 2 2.2 8
2,27 22 4
f'p 2,27 012,2 Fe E
8,4 Fr
mum
Et 6

716. = |∇f( ⃗
a ) ⋅ ⃗
v | =
f!
!
v(!
a) = ∇f(!
a) · !
v.
+ t!
v)- bQ i?i ϕ /2b+`B#2b ?Qr f HQQFb HQM; i
f!
!
v(!
a) = HBK
t→0
ϕ(t) − ϕ(0)
t
= ϕ!
(0).
∇f( ⃗
a )
⃗
v
α
⃗
a
| f′ ⃗
v ( ⃗
a )| = = ∥∇f( ⃗
a )∥ ⋅ ∥ ⃗
v ∥ ⋅ |cos α|
⩽ ∥∇f( ⃗
a )∥ ⋅ ∥ ⃗
v ∥ = ∥∇f( ⃗
a )∥ ⋅ 1 = ∥∇f( ⃗
a )∥
∇f( ⃗
a )
⃗
v
⃗
a
∇f( ⃗
a )
⃗
v
⃗
a
max if α = 0 min if α = π
Geometrical consequences of the theorem about directional derivatives

717. |f!
v(!
a)| = |∇f(!
a) · !
v| ⩽ |∇f(!
a)||!
v| = |∇f(!
a)|
T`2+Bb b´ !
v Q+? ∇f(!
a) ` `BFi/2 ´i bKK ?´HH U+Qb 0 = 1VX _BFiMBM;b/2`
biƺ`bi B ;`/B2Mi2Mb `BFiMBM;X oB FM /` 7ƺHDM/2 bHmibib2`,
Ç h?2 7bi2bi BM+`2b2 Bb BM i?2 /B`2+iBQM ∇f(!
a)
Ui?2`2 r2 ?p2 α = 0 M/ +Qb α = 1V
M/ i?2 `i2 Q7 i?Bb +?M;2 Bb ∇f(!
a)X
Ç h?2 7bi2bi /2+`2b2 Bb BM i?2 /B`2+iBQM −∇f(!
a)
Ui?2`2 r2 ?p2 α = π M/ +Qb α = −1V
M/ i?2 `i2 Q7 i?Bb +?M;2 Bb ∇f(!
a)X
98X G2i
f(x, y) =
3
1 + x2 + y2
.

718. Compute the rate of change
of
at
measured in each of the following directions:
a)
b)
c)
d)
f(x, y) = y4
+ 2xy3
+ x2
y2
(x, y) = (0, 1)
⃗
v = (1, 2)
⃗
v = (−2, 1)
⃗
v = (3, 0)
⃗
v = (1, 1)
/2`BpiBp2 b?Qrb i?2 bHQT2 Q7 i?2 iM;2Mi HBM2 BM i?2 TQBMi
2`b2+iBQM #2ir22M i?2 ;`T? bm`7+2 z = f(x, y) M/ i?2 p2`i
V THM2 i?`Qm;? i?2 TQBMi (a, b)- HQM; !
vX
H´i2` Qbb #´/2 #2` FM `BFiMBM;b/2`BpiQ` T´ 2ii 2MF2Hi b i
`piBQM2` QK 7mMFiBQM2Mb iBHHp ti B QHBF `BFiMBM;`,
/Kb RkXdXdVX A7 f Bb /Bz2`2MiB#H2 7mM+iBQM M/ !
v bXiX |
f!
!
v(!
a) = ∇f(!
a) · !
v.

719. fCxiyky4t2xy3tx2y2gCoI
fffG.y
21 Cxiy 4y3t6xy42x
Pf 0,1 54
167 526
I 11TH _rt4T gyhe
F 4,2
fujfqp pffqp.ua grad
2,4 IE 2 8 1051051 255
8 1 2,1 Hitler
E Iftar faicon 2,4 C 2,1
f 4 45 0
8 13,0 11711 3 i i
E 1,0 f'a 0,11 12,4 1,0
Con
F 11811 52
E Fallin
f an 2,4 Gink
l h
2 4 fz
6
3
ej

720. A directional vector of the normal line:
v1 = f′
1(a, b), v2 = f′
2(a, b)
f(x, y) = C
∇f(a, b) = (v1, v2)
(a, b)
A directional vector for the tangent line is any
vector which is perpendicular to the normal line.
The easiest one to find is (dot product zero):
(−v2, v1)
(x, y) = (a, b) + t(v1, v2), t ∈ ℝ
(x, y) = (a, b) + t(−v2, v1), t ∈ ℝ
Both can be converted to the
(m, b)-form / slope-intercept
form. I show you in the next
video how you do it.
TANGENT
NORMAL
A parametric equation for:
The normal line:
The tangent line:

721. Ç h?2 7bi2bi BM+`2b2 Bb BM i?2 /B`2+iBQM ∇f(!
a) M/ i?2 `i2 Q7 i?Bb +?M;2 Bb |∇f(!
a)|X
Ç h?2 7bi2bi /2+`2b2 Bb BM i?2 /B`2+iBQM −∇f(!
a) Ui?2`2 r2 ?p2 α = π M/ +Qb α = −1V
M/ i?2 `i2 Q7 i?Bb +?M;2 Bb |∇f(!
a)|X
1tKTH2 98X G2i
f(x, y) =
3
1 + x2 + y2
.
V AM r?i /B`2+iBQM M/ ?Qr 7bi /Q2b f(x, y) BM+`2b2 i i?2 ?B;?2bi `i2 i TQBMi (1, 1)
#V *QKTmi2 i?2 /B`2+iBQMH /2`BpiBp2 Q7 f BM (1, 1) BM i?2 /B`2+iBQM /2}M2/ #v B +
√
3DX
+V .2i2`KBM2 2[miBQMb Q7 i?2 iM;2Mi HBM2 M/ i?2 MQ`KH HBM2 iQ i?2 H2p2H +m`p2 i?`Qm;?
i?2 TQBMi (1, 1)X
/V .2i2`KBM2 M 2[miBQM Q7 i?2 iM;2Mi THM2 iQ i?2 bm`7+2 z = f(x, y) BM TQBMi (1, 1, 1)X
aQHmiBQM,
q2 Kmbi +QKTmi2 i?2 ;`/B2Mi- #2+mb2 Bi Bb M22/2/ 7Q` HH i?2 T`ib Q7 i?2 T`Q#H2KX h?2
7mM+iBQM Bb +QKTQbBi2 BM i?2 rv r?B+? M22/b i?2 *?BM _mH2 o2`bBQM k,
!

722. +V .2i2`KBM2 2[miBQMb Q7 i?2 iM;2Mi HBM2 M/ i?2 MQ`KH HBM2 iQ i?2 H2p2H +m`p2 i?`Qm;?
i?2 TQBMi (1, 1)X
/V .2i2`KBM2 M 2[miBQM Q7 i?2 iM;2Mi THM2 iQ i?2 bm`7+2 z = f(x, y) BM TQBMi (1, 1, 1)X
aQHmiBQM,
q2 Kmbi +QKTmi2 i?2 ;`/B2Mi- #2+mb2 Bi Bb M22/2/ 7Q` HH i?2 T`ib Q7 i?2 T`Q#H2KX h?2
7mM+iBQM Bb +QKTQbBi2 BM i?2 rv r?B+? M22/b i?2 *?BM _mH2 o`BMi k,
∇f(x, y) =
!
−
6x
(1 + x2 + y2)2
, −
6y
(1 + x2 + y2)2
=
6x
(1 + x2 + y2)2
· (−x, −y).
V h?2 7bi2bi BM+`2b2 Bb BM i?2 /B`2+iBQM Q7 i?2 ;`/B2Mi ∇f(1, 1)- #2BM; 6
9 (−1, −1) =
(−2
3 , −2
3 )X h?2 bT22/ Ui?2 `i2 Q7 +?M;2V Bb 2[mH iQ i?2 H2M;i? Q7 i?2 ;`/B2Mi,
|∇f(1, 1)| =
#!
−
2
3
2
+
!
−
2
3
2
=
2
3
√
2.
#V q2 M22/ iQ MQ`KHBx2 i?2 /B`2+iBQM, |(1,
√
3)| =
√
1 + 3 = 2X h?2 /B`2+iBQM Bb !
u = (1
2 ,
√
3
2 )X
h?2 /B`2+iBQMH /2`BpiBp2,
! $ √ % √ √
f(x, y) =
1 + x2 + y2
.
V AM r?i /B`2+iBQM M/ ?Qr 7bi /Q2b f(x, y) BM+`2b2 i i?2 ?B;?2bi `i2 i TQBMi (1, 1)
#V *QKTmi2 i?2 /B`2+iBQMH /2`BpiBp2 Q7 f BM (1, 1) BM i?2 /B`2+iBQM /2}M2/ #v B +
√
3DX
+V .2i2`KBM2 2[miBQMb Q7 i?2 iM;2Mi HBM2 M/ i?2 MQ`KH HBM2 iQ i?2 H2p2H +m`p2 i?`Qm;?
i?2 TQBMi (1, 1)X
/V .2i2`KBM2 M 2[miBQM Q7 i?2 iM;2Mi THM2 iQ i?2 bm`7+2 z = f(x, y) BM TQBMi (1, 1, 1)X
aQHmiBQM,
q2 Kmbi +QKTmi2 i?2 ;`/B2Mi- #2+mb2 Bi Bb M22/2/ 7Q` HH i?2 T`ib Q7 i?2 T`Q#H2KX h?2
7mM+iBQM Bb +QKTQbBi2 BM i?2 rv r?B+? M22/b i?2 *?BM _mH2 o`BMi k,
∇f(x, y) =
!
−
6x
(1 + x2 + y2)2
, −
6y
(1 + x2 + y2)2
=
6x
(1 + x2 + y2)2
· (−x, −y).
V h?2 7bi2bi BM+`2b2 Bb BM i?2 /B`2+iBQM Q7 i?2 ;`/B2Mi ∇f(1, 1)- #2BM; 6
9 (−1, −1) =
(−2
3 , −2
3 )X h?2 bT22/ Ui?2 `i2 Q7 +?M;2V Bb 2[mH iQ i?2 H2M;i? Q7 i?2 ;`/B2Mi,
|∇f(1, 1)| =
#!
−
2
3
2
+
!
−
2
3
2
=
2
3
√
2.

723. ∇f(x, y) = −
(1 + x2 + y2)2
, −
(1 + x2 + y2)2
=
(1 + x2 + y2)2
· (−x, −y).
V h?2 7bi2bi BM+`2b2 Bb BM i?2 /B`2+iBQM Q7 i?2 ;`/B2Mi ∇f(1, 1)- #2BM; 6
9 (−1, −1) =
(−2
3 , −2
3 )X h?2 bT22/ Ui?2 `i2 Q7 +?M;2V Bb 2[mH iQ i?2 H2M;i? Q7 i?2 ;`/B2Mi,
|∇f(1, 1)| =
#!
−
2
3
2
+
!
−
2
3
2
=
2
3
√
2.
#V q2 M22/ iQ MQ`KHBx2 i?2 /B`2+iBQM, |(1,
√
3)| =
√
1 + 3 = 2X h?2 /B`2+iBQM Bb !
u = (1
2 ,
√
3
2 )X
h?2 /B`2+iBQMH /2`BpiBp2,
D!
uf(1, 1) = ∇f(1, 1) · !
u =
!
−
2
3
, −
2
3
·
$
1
2
,
√
3
2
%
= −
2
3
·
1
2
−
2
3
·
√
3
2
= −
1 +
√
3
3
.
+V /B`2+iBQMH p2+iQ` Q7 i?2 MQ`KH HBM2 Bb i?2 ;`/B2Mi BM i?2 TQBMi UQ` Mv b+HBM; Q7
i?Bb p2+iQ`- HBF2 7Q` 2tKTH2 (1, 1)VX h?2 MQ`KH HBM2 ?b i?mb i?2 7QHHQrBM; T`K2i`B+
2[miBQM,
(x, y) = (1, 1) + t(1, 1), t ∈ R
Q`- BM i?2 xy@THM2, y = xX
V AM r?i /B`2+iBQM M/ ?Qr 7bi /Q2b f(x, y) BM+`2b2 i i?2 ?B;?2bi `i2 i TQBMi (1, 1)
#V *QKTmi2 i?2 /B`2+iBQMH /2`BpiBp2 Q7 f BM (1, 1) BM i?2 /B`2+iBQM /2}M2/ #v B +
√
3DX
+V .2i2`KBM2 2[miBQMb Q7 i?2 iM;2Mi HBM2 M/ i?2 MQ`KH HBM2 iQ i?2 H2p2H +m`p2 i?`Qm;?
i?2 TQBMi (1, 1)X
/V .2i2`KBM2 M 2[miBQM Q7 i?2 iM;2Mi THM2 iQ i?2 bm`7+2 z = f(x, y) BM TQBMi (1, 1, 1)X
aQHmiBQM,
q2 Kmbi +QKTmi2 i?2 ;`/B2Mi- #2+mb2 Bi Bb M22/2/ 7Q` HH i?2 T`ib Q7 i?2 T`Q#H2KX h?2
7mM+iBQM Bb +QKTQbBi2 BM i?2 rv r?B+? M22/b i?2 *?BM _mH2 o`BMi k,
∇f(x, y) =
!
−
6x
(1 + x2 + y2)2
, −
6y
(1 + x2 + y2)2
=
6x
(1 + x2 + y2)2
· (−x, −y).
V h?2 7bi2bi BM+`2b2 Bb BM i?2 /B`2+iBQM Q7 i?2 ;`/B2Mi ∇f(1, 1)- #2BM; 6
9 (−1, −1) =
(−2
3 , −2
3 )X h?2 bT22/ Ui?2 `i2 Q7 +?M;2V Bb 2[mH iQ i?2 H2M;i? Q7 i?2 ;`/B2Mi,
|∇f(1, 1)| =
#!
−
2
3
2
+
!
−
2
3
2
=
2
3
√
2.
#V q2 M22/ iQ MQ`KHBx2 i?2 /B`2+iBQM, |(1,
√
3)| =
√
1 + 3 = 2X h?2 /B`2+iBQM Bb !
u = (1
,
√
3
)X

724. 7mM+iBQM Bb +QKTQbBi2 BM i?2 rv r?B+? M22/b i?2 *?BM _mH2 o`BMi k,
∇f(x, y) =
!
−
6x
(1 + x2 + y2)2
, −
6y
(1 + x2 + y2)2
=
6x
(1 + x2 + y2)2
· (−x, −y).
V h?2 7bi2bi BM+`2b2 Bb BM i?2 /B`2+iBQM Q7 i?2 ;`/B2Mi ∇f(1, 1)- #2BM; 6
9 (−1, −1) =
(−2
3 , −2
3 )X h?2 bT22/ Ui?2 `i2 Q7 +?M;2V Bb 2[mH iQ i?2 H2M;i? Q7 i?2 ;`/B2Mi,
|∇f(1, 1)| =
#!
−
2
3
2
+
!
−
2
3
2
=
2
3
√
2.
#V q2 M22/ iQ MQ`KHBx2 i?2 /B`2+iBQM, |(1,
√
3)| =
√
1 + 3 = 2X h?2 /B`2+iBQM Bb !
u = (1
2 ,
√
3
2 )X
h?2 /B`2+iBQMH /2`BpiBp2,
D!
uf(1, 1) = ∇f(1, 1) · !
u =
!
−
2
3
, −
2
3
·
$
1
2
,
√
3
2
%
= −
2
3
·
1
2
−
2
3
·
√
3
2
= −
1 +
√
3
3
.
+V /B`2+iBQMH p2+iQ` Q7 i?2 MQ`KH HBM2 Bb i?2 ;`/B2Mi BM i?2 TQBMi UQ` Mv b+HBM; Q7
i?Bb p2+iQ`- HBF2 7Q` 2tKTH2 (1, 1)VX h?2 MQ`KH HBM2 ?b i?mb i?2 7QHHQrBM; T`K2i`B+
2[miBQM,
(x, y) = (1, 1) + t(1, 1), t ∈ R
Q`- BM i?2 xy@THM2, y = xX
f(x, y) = C
∇f(a, b) = (v1, v2)
(a, b)
TANGENT
NORMAL
M/ i?2 `i2 Q7 i?Bb +?M;2 Bb |∇f(!
a)|X
1tKTH2 98X G2i
f(x, y) =
3
1 + x2 + y2
.
V AM r?i /B`2+iBQM M/ ?Qr 7bi /Q2b f(x, y) BM+`2b2 i i?2 ?B;?2bi `i2 i TQBMi (1, 1)
#V *QKTmi2 i?2 /B`2+iBQMH /2`BpiBp2 Q7 f BM (1, 1) BM i?2 /B`2+iBQM /2}M2/ #v B +
√
3DX
+V .2i2`KBM2 2[miBQMb Q7 i?2 iM;2Mi HBM2 M/ i?2 MQ`KH HBM2 iQ i?2 H2p2H +m`p2 i?`Qm;?
i?2 TQBMi (1, 1)X
/V .2i2`KBM2 M 2[miBQM Q7 i?2 iM;2Mi THM2 iQ i?2 bm`7+2 z = f(x, y) BM TQBMi (1, 1, 1)X
aQHmiBQM,
q2 Kmbi +QKTmi2 i?2 ;`/B2Mi- #2+mb2 Bi Bb M22/2/ 7Q` HH i?2 T`ib Q7 i?2 T`Q#H2KX h?2
7mM+iBQM Bb +QKTQbBi2 BM i?2 rv r?B+? M22/b i?2 *?BM _mH2 o`BMi k,
∇f(x, y) =
!
−
6x
(1 + x2 + y2)2
, −
6y
(1 + x2 + y2)2
=
6x
(1 + x2 + y2)2
· (−x, −y).
V h?2 7bi2bi BM+`2b2 Bb BM i?2 /B`2+iBQM Q7 i?2 ;`/B2Mi ∇f(1, 1)- #2BM; 6
9 (−1, −1) =
(−2
3 , −2
3 )X h?2 bT22/ Ui?2 `i2 Q7 +?M;2V Bb 2[mH iQ i?2 H2M;i? Q7 i?2 ;`/B2Mi,
|∇f(1, 1)| =
#!
−
2
3
2
+
!
−
2
3
2
=
2
3
√
2.
√ √ 1
√
3

725. ∇f(a, b) = (v1, v2)
(a, b)
TANGENT
NORMAL
M/ i?2 `i2 Q7 i?Bb +?M;2 Bb |∇f(!
a)|X
1tKTH2 98X G2i
f(x, y) =
3
1 + x2 + y2
.
V AM r?i /B`2+iBQM M/ ?Qr 7bi /Q2b f(x, y) BM+`2b2 i i?2 ?B;?2bi `i2 i TQBMi (1, 1)
#V *QKTmi2 i?2 /B`2+iBQMH /2`BpiBp2 Q7 f BM (1, 1) BM i?2 /B`2+iBQM /2}M2/ #v B +
√
3DX
+V .2i2`KBM2 2[miBQMb Q7 i?2 iM;2Mi HBM2 M/ i?2 MQ`KH HBM2 iQ i?2 H2p2H +m`p2 i?`Qm;?
i?2 TQBMi (1, 1)X
/V .2i2`KBM2 M 2[miBQM Q7 i?2 iM;2Mi THM2 iQ i?2 bm`7+2 z = f(x, y) BM TQBMi (1, 1, 1)X
aQHmiBQM,
q2 Kmbi +QKTmi2 i?2 ;`/B2Mi- #2+mb2 Bi Bb M22/2/ 7Q` HH i?2 T`ib Q7 i?2 T`Q#H2KX h?2
7mM+iBQM Bb +QKTQbBi2 BM i?2 rv r?B+? M22/b i?2 *?BM _mH2 o`BMi k,
∇f(x, y) =
!
−
6x
(1 + x2 + y2)2
, −
6y
(1 + x2 + y2)2
=
6x
(1 + x2 + y2)2
· (−x, −y).
V h?2 7bi2bi BM+`2b2 Bb BM i?2 /B`2+iBQM Q7 i?2 ;`/B2Mi ∇f(1, 1)- #2BM; 6
9 (−1, −1) =
(−2
3 , −2
3 )X h?2 bT22/ Ui?2 `i2 Q7 +?M;2V Bb 2[mH iQ i?2 H2M;i? Q7 i?2 ;`/B2Mi,
|∇f(1, 1)| =
#!
−
2
3
2
+
!
−
2
3
2
=
2
3
√
2.
√ √ 1
√
3
h?2 iM;2Mi HBM2 Bb T2`T2M/B+mH` iQ i?2 MQ`KH HBM2- r?B+? K2Mb i?i Bib /B`2+iBQMH
p2+iQ` Bb Mv p2+iQ` T2`T2M/B+mH` iQ (1, 1)- bQ 7Q` 2tKTH2 (1, −1)X h?2 iM;2Mi HBM2 ?b
i?mb i?2 7QHHQrBM; T`K2i`B+ 2[miBQM,
(x, y) = (1, 1) + t(1, −1), t ∈ R.
AM Q`/2` iQ i`Mb7Q`K i?Bb 2[miBQM BMiQ M (m, b)@2[miBQM BM i?2 xy@THM2- r2 bQHp2 #Qi?
2[miBQMb 7Q` t M/ ;2i M 2[miBQM rBi? Dmbi x M/ y, x = 1 + t M/ y = 1 − t ;Bp2
x − 1 = 1 − y- bQ i?2 (m, b)@2[miBQM Q7 i?2 iM;2Mi HBM2 BM i?2 xy@THM2 Bb y = −x + 2X
q2 MQiB+2 i?i i?Bb HBM2 `2HHv Bb T2`T2M/B+mH` iQ i?2 HBM2 y = x BM i?2 xy@THM2X

726. AM Q`/2` iQ i`Mb7Q`K i?Bb 2[miBQM BMiQ M (m, b)@2[miBQM BM i?2 xy@THM2- r2 bQHp2 #Qi?
2[miBQMb 7Q` t M/ ;2i M 2[miBQM rBi? Dmbi x M/ y, x = 1 + t M/ y = 1 − t ;Bp2
x − 1 = 1 − y- bQ i?2 (m, b)@2[miBQM Q7 i?2 iM;2Mi HBM2 BM i?2 xy@THM2 Bb y = −x + 2X
q2 MQiB+2 i?i i?Bb HBM2 `2HHv Bb T2`T2M/B+mH` iQ i?2 HBM2 y = x BM i?2 xy@THM2X
/V h?2 2[miBQM Q7 i?2 iM;2Mi THM2i Bb z = f(1, 1) + ∇f(1, 1) · (x − 1, y − 1)- bQ
z = 1 + (−
2
3
, −
2
3
) · (x − 1, y − 1).
1[miBQM Q7 i?2 iM;2Mi THM2 iQ z = f(x, y) BM i?2 TQBMi (a, b, f(a, b))
z = f(a, b) + f!
x(a, b)(x − a) + f!
y(a, b)(y − b)
+M #2 r`Bii2M BM b?Q`i rv mbBM; i?2 ;`/B2Mi M/ /Qi T`Q/m+i,
z = f(a, b) + ∇(a, b) · (x − a, y − b).
bQ z = 1− 2
3 (x−1)− 2
3 (y −1)X JmHiBTHB+iBQM Q7 #Qi? bB/2b rBi? j M/ bQK2 bBKTHB}+iBQMb
H2/ iQ i?2 7QHHQrBM; 2[miBQM Q7 i?2 iM;2Mi THM2 iQ i?2 ;`T? bm`7+2 i?`Qm;? (1, 1, 1),
2x + 2y + 3z − 7 = 0.
i?2 TQBMi (1, 1)X
/V .2i2`KBM2 M 2[miBQM Q7 i?2 iM;2Mi THM2 iQ i?2 bm`7+2 z = f(x, y) BM TQBMi (1, 1, 1)X
aQHmiBQM,
q2 Kmbi +QKTmi2 i?2 ;`/B2Mi- #2+mb2 Bi Bb M22/2/ 7Q` HH i?2 T`ib Q7 i?2 T`Q#H2KX h?2
7mM+iBQM Bb +QKTQbBi2 BM i?2 rv r?B+? M22/b i?2 *?BM _mH2 o`BMi k,
∇f(x, y) =
!
−
6x
(1 + x2 + y2)2
, −
6y
(1 + x2 + y2)2
=
6x
(1 + x2 + y2)2
· (−x, −y).
V h?2 7bi2bi BM+`2b2 Bb BM i?2 /B`2+iBQM Q7 i?2 ;`/B2Mi ∇f(1, 1)- #2BM; 6
9 (−1, −1) =
(−2
3 , −2
3 )X h?2 bT22/ Ui?2 `i2 Q7 +?M;2V Bb 2[mH iQ i?2 H2M;i? Q7 i?2 ;`/B2Mi,
|∇f(1, 1)| =
#!
−
2
3
2
+
!
−
2
3
2
=
2
3
√
2.
#V q2 M22/ iQ MQ`KHBx2 i?2 /B`2+iBQM, |(1,
√
3)| =
√
1 + 3 = 2X h?2 /B`2+iBQM Bb !
u = (1
2 ,
√
3
2 )X
h?2 /B`2+iBQMH /2`BpiBp2,
D!
uf(1, 1) = ∇f(1, 1) · !
u =
!
−
2
3
, −
2
3
·
$
1
2
,
√
3
2
%
= −
2
3
·
1
2
−
2
3
·
√
3
2
= −
1 +
√
3
3
.
+V /B`2+iBQMH p2+iQ` Q7 i?2 MQ`KH HBM2 Bb i?2 ;`/B2Mi BM i?2 TQBMi UQ` Mv b+HBM; Q7

727. ProblemI Find an
equation to
the tangent plane to the levee
surface of
fCny E
x2ytyztz2x
at G
mm
gradient in the point G 1,1
is nominal to the level
surface through this
joint
OfCengiz 8 8
2xyt Et2yZ y72zx
a
Tfa 1,1 1 1,3
I lay Tz Num
Eq of the plane through 1 1,1
with normal vector
mmmm

728. M
Axt By 1 Cz D 0
JT x
y
3z Q
G 1 1 EST
E Ty E
1 3 0 0
D 3
The tangent plane has following
equation
x y 3z 3 O
xty 32 13 0

729. Quadric surfaces, an introduction
10.5

730. x2
+ y2
+ z2
= r2
x
y
z
1
1
1
Sphere
(x − x0)2
+ (y − y0)2
+ (z − z0)2
= r2

731. Ellipsoid
Elliptic paraboloid
Hyperbolic paraboloid
Elliptic hyperboloid of one sheet
Elliptic hyperboloid of two sheets
if one of the variables is
not squared
If there is one MINUS
If there is a MINUS, but all three variables squared
One sheet: just one MINUS
Two sheets: two times MINUS (if +1 on the RHS)
(only PLUS)

732. Elliptic cone
Elliptic cylinder
Hyperbolic cylinder
Parabolic cylinder
Cylinder: if one of the variables is MISSING
if there is a MINUS
(all variables squared and the RHS=0)
if one of the variables is
not squared
(only PLUS)

733. When two or more of the parameters of the canonical equation are equal, one gets a quadric of revolution, which
remains invariant when rotated around an axis (or infinitely many axes, in the case of the sphere).
Quadrics of revolution
Oblate and prolate spheroids (special cases of
ellipsoid)
Sphere (special case of spheroid)
Circular paraboloid (special case of elliptic paraboloid)
Circular hyperboloid of one sheet (special case of
elliptic hyperboloid of one sheet)
Circular hyperboloid of two sheets (special case of
elliptic hyperboloid of two sheets)
Circular cone (special case of cone)

734. The parabola y = x2 (a curve) is a graph to the (single-variable) function f (x) = x2
The paraboloid z = x2 + y2 + 1 (a surface) is a graph to the (two-variable) function f (x, y) = x2 + y2 + 1
x
y
TmMFi2`- b´ FM HQFH 2ti`2KTmMFi2` #
2tBbi2`` 2DV Q+? `M/TmMFi2` iBHH Df X
.2i ` BMi2 b F2`i ii 2M F`BiBbF TmMFi
ivT2` p mib22M/2 F`BM; 2M F`BiBbF TmMF
x
y
f(x, y)
1

735. The graph to the (single-variable) function f (x) = |x|
The cone (a surface) is a graph to the (two-variable) function
z = x2
+ y2
f(x, y) = x2
+ y2
x
y

736. x
y
y0
x0
y = y0
x = x0

737. y
z
x
(x0, y, z)
(x0, 0, 0)
x = x0
Plane x = x0

738. y
z
x
(x, y0, z)
(0, y0, 0)
y = y0
Plane y = y0

739. y
z
x
(x, y, z0)
(0, 0, z0)
z = z0
Plane z = z0
(x, y,0)

740. y
z
r
x
x2
+ y2
= r2
cylinder

741. z2
= x2
+ y2
z = x2
+ y2
z = − x2
+ y2

742. Non-degenerate real quadric surfaces
Ellipsoid
Elliptic paraboloid
(only PLUS)

743. (0, 0) a
b
F2 F1
c
c
a
a
x2
a2
+
y2
b2
= 1
x2
+ y2
= r2
r
r
−r
−r
x2
a2
+
y2
b2
+
z2
c2
= 1
x2
+ y2
+ z2
= r2
x x
y
y

744. x2
4
+
y2
9
+
z2
16
= 1
x
y
z

745. a
−a
b
c
x2
a2
−
y2
b2
= 1
hyperbola
x ⩽ − a ∨ x ⩾ a

746. (x − x0)2
a2
−
(y − y0)2
b2
= 1
(x0, y0)
a
a
−
(x − x0)2
a2
+
(y − y0)2
b2
= 1
(x0, y0)
b
b
hyperbola

747. https://faculty.math.illinois.edu/~nmd/quadrics/
with thanks to Jonathan Rogness and Nathan Dunfield
To understand quadratic surfaces

748. The intersections between the surface and the planes z = c, and the
intersections between the surface and the planes y = c (for each number
c such that |c| ≠ 2) are hyperbolas. This explains the name hyperboloid.
What kind of hyperbolas we get (meaning how they lie in the space and
what equations they have) depends of whether |c| 2 or |c| 2. What
happens if |c| = 2? We get a pair of diagonal lines in a plane.
In the yz-plane we have: x = 0. The equation of the surface there is thus y2 + z2 = 4,
meaning that the intersection between the surface and the yz-plane is a circle with
radius 2. The intersections between the surface and the planes x = c (which are parallel
to the yz-plane) are even larger circles, because there we have y2 + z2 = 4 + c2.
This shows that the surface forms a kind of pipe which is narrow in the middle and
broadens up in both directions (positive and negative) along the x-axis. The surface
consists of one entire piece and this is why it is called hyperboloid of one sheet.
hyperboloid of one sheet
only one minus
Fig. 10.5.7
8. −x2 + y2 + z2 = 4 represents a hyperboloid of one sheet,
with circular cross-sections in all planes perpendicular to
the x-axis.
x
y
z
y
−x2+y2+z2=4
Fig. 10.5.8
9. z = xy represents a hyperbolic paraboloid containing the
x- and y-axes.
y
z
Fig. 10.5.10
11. x2 − 4z2 = 4 represents a hyperbolic cylinder with axis
along the y-axis.
x
y
z
x
x2−4z2=4
Fig. 10.5.11
12. y = z2 represents a parabolic cylinder with vertex line
along the x-axis.
z
y=z2
x
−x2
+ y2
+ z2
= 4

749. The surface exists only there where |x| ≥ 2; this explains the gap.
The intersection between the surface and the plane z = c (for
any number c) gives the hyperbola x2 - y2 = 4 + c2 and the
intersection between the surface and the plane y = c gives the
hyperbola x2 - z2 = 4 + c2. This explains the name hyperboloid.
On the x-axis we have: y = 0 och z = 0. The equation of the surface reduces there to x2 = 4,
which means that the intersection between the surface and the x-axis consists of two points:
(2, 0, 0) and (-2, 0, 0). Note that no points (x, y, z) with -2 x 2 can belong to the surface,
because for such points x2 4, while obviously - y2 - z2 ≤ 0 which makes x2 - y2 - z2 = 4
impossible. This explains why the surface consists of two disjoint pieces and why it is
called hyperboloid of two sheets.
The intersection between the surface and the plane x = c for such c that |c| 2 are circles,
because there it holds y2 + z2 = c2 - 4.
hyperboloid of two sheets
-2
2 times minus
INSTRUCTOR’S SOLUTIONS MANUAL SECTION 10.5 (PAGE 598)
x
y
z
x2−y2−z2=4
Fig. 10.5.7
8. −x2 + y2 + z2 = 4 represents a hyperboloid of one sheet,
with circular cross-sections in all planes perpendicular to
the x-axis.
z
−x2+y2+z2=4
x
y
z
x2+4z2=4
Fig. 10.5.10
11. x2 − 4z2 = 4 represents a hyperbolic cylinder with axis
along the y-axis.
z
2
x
y
x2
− y2
− z2
= 4

750. F ;2HbMBii bQK FQKK2` ii p` F`ƺFic /2i }MMb 7v` QHBF ivT2` p M/`;`/bFm`pQ`X 1M
B``2/m+B#2H M/`;`/b2FpiBQM B p`B#H2`M x Q+? y KQibp`` M´;QM p 7ƺHDM/2 7v` ivT2` p
Fm`pQ`, +B`F2H- 2HHBTb- ?vT2`#2H Q+? T`#2H U2;2MiHB;2M b´ ` /2i i`2 ivT2`- 27i2`bQK p`D2
+B`F2H ` 2M 2HHBTbVX a2 #BH/2`M B TT2M/Bt T´ bHmi2i p /2bb Mi2+FMBM;`X
1M HHK M M/`;`/b2FpiBQM B p`B#H2`M x- y Q+? z FM bF`Bpb
Ax2
+ By2
+ Cz2
+ Dxy + Exz + Fyz + Gx + Hy + Iz + J = 0.
A mM/Mi;b7HH FM /2ii mii`v+F 7FiQ`Bb2`b iBHH T`Q/mFi2M
(A1x + B1y + C1z + D1)(A2x + B2y + C2z + D2) = 0.
p`pB/ HƺbMBM;`M ;2b p ip´ THM, A1x + B1y + C1z + D1 = 0 Q+? A2x + B2y + C2z + D2 = 0
UQK A2
1 + B2
1 + C2
1 != 0 Q+? A2
2 + B2
2 + C2
2 != 0VX A HH M/` 7HH 7´` pB 2M b´ FHH/ Fp/`iBbF
vi bQK FQKK2` ii p` F`ƺFiX .2i }MMb b2t QHBF ivT2` p Fp/`iBbF viQ`X 1M B``2/m+B#2H
M/`;`/b2FpiBQM B p`B#H2`M x- y Q+? z KQibp`` M´;QM p 7ƺHDM/2 b2t ivT2` Ub7 ` ` 2ii
bT2+BH7HH p 2M 2HHBTbQB/ K2/ a2 = b2 = c2 = r2V p F`ƺFi viQ`,

751. (x − x0)2
+ (y − y0)2
= r2
(x − x0)2
a2
+
(y − y0)2
b2
= 1
(x0, y0)
r
(x0, y0) a
b
(x − x0)2
a2
−
(y − y0)2
b2
= 1
(x0, y0)
a
a
−
(x − x0)2
a2
+
(y − y0)2
b2
= 1
(x0, y0)
b
b
Ax2
+ Bxy + Cy2
+ Dx + Ey + F = 0
A2
+ B2
+ C2
0
= B2
4AC discriminant
0 gives a hyperbola
0 gives an ellipse (circle)
= 0 gives a parabola
(a + b)2
= a2
+ 2ab + b2
(a − b)2
= a2
− 2ab + b2

752. y = x2
− 8x + 11
= x2
− 8x + 16 − 5
= (x − 4)2
− 5
New vertex in
(4, − 5)
(4, − 5)

753. Identify surfaces defined by the following equations
x2
+ y2
+ z2
= 2z
x2
+ y2
− 6x + 4y − z + 10 = 0
x2
+ 4y2
+ 9z2
= 36
y2
= 4x2
+ 16z2
x2
+ y2
− z2
− 4x − 6y + 2z + 12 = 0
x = z2
+ z

754. x
y
x
y
(3, − 2)
x2
+ y2
− 6x + 4y − z + 10 = 0
z = (x − 3)2
+ (y + 2)2
− 3