University of Notre Dame, Spring 2019

1. CALCULUS Ill

2. 3D Space
AZ
JE
:#
tangent plane
. ,
×
'
I
•
v
'
Y
n a
L
Z
Z 2
7
y
L X
-
y
L
× y
'
×
>
h2
( X
, 0,2 )
X
-
z y
-
2 L O , y ,
2)
plane plane
x -
y plane
>
y
L ( X. y ,
O )
x

3. ^
Distance ix. y )
•
D= V ( X -
X
'
)2t (
y
-
y
'
)2
• ( X
'
, y
'
)
>
^
2
( X , y ,
2)
•
I
Iz -
z 't
'
d
'
= #I d ( X
-
X
'
)2 t
( y
-
y
'
32
- -
di-
-
-
-
- d =
FEZ '
)2
•• C X
'
, Y
'
'
Z
'
)
D= j , ×
.
x. yz +
( y
-
y' 72 t
( 2 -
2
'
)
Z
L )
× Y
Z =
f ( × , y ) →
surface in 3D space
equation involving x. y , 2
2=2
a
-
Z
X
L dy
In
y
=
I
I I
I l
L
S
X t
y t 2=1
Nz
•
L 0,0 ,
I )
o ( O
,
I ,
O )
X
L
@
s
( 1,0 ,
O ) Y

4. X
Z
t
y
2
t 22 =
4
( X -
O )
2 t
( y
-
O )
2 t ( 2
-
O )
2 = 2
↳ sphere w/ center at
origin w/ radius 2
( X
-
2)
Z
is
L y
-
3)
2 t
L 2 -
4) 2 = I
↳ sphere w/ center at ( 2.3 ,
4) with r = I
Xt y
= I X t
y
= I X t
y
= I
I
equation x =
y X =
y
Surface in 3D 2 equations 2=2
curve in 3D 3 equations
point Cs )
^
•
L
)
^
#
X
Z
t
y
2
=/
# X
2
t
y 2=4
) , ± ×z +
y
z
e 4
-
L S

5. Vectors
'
Vector : a
quantity that has both magnitude da direction
-
ex :
velocity ,
forces
.
Vector addition
→
a
7
b
→
•
a
a
→ → →
atb b
7
→
•
a
.
Subtraction
^5
-
of 5 -
a→ta→=b→
→
a
x•
→
b
.
scalar multiplication
v→ 7 A
→ →
•
→ Ov -
v
•
ZV
,
L
•
.
vector components
A
NZ
→ ,
( 5,4 )
✓
( a , , az.az ) 7
y
a→= a. Ttazjtazk^ n
a
<
'
n
,
> >
, µ,
=
sa . .az ,
as >
< → s x
V→=5Tt 45=45,4 > i
u e
✓
a→= La ,
.az ) b→=Lb , .bz >
=
a. Ttazj =
b. Ttbzj
A→tb→= A. Ttazjtb , Ttbzj =
( A. tb ,
) it ( Aztbz )j
= L A. tb , ,
Aztbz >
T = ( V , ,
Vz )
cP=Ccv .
,
Cuz >

6. ^
magnitude L length ) of of = I at I =
. 9,2 t
Azz
7 ( a. , Az )
Az
( )
:
^
→
>
L COS O
,
since )
Ts =/Tl Loso ,
Sino >
o )
=L 181050 ,
181 Sino >
L S
✓
.
Rules of Vectors ( look in textbook )
.
Geometry Application
Thm :
the diagonals of a
parallelogram bisect each other
>
→
→ →
→
a t bb -
of - 5 2
→
×
Is +
IF =
Etba
2
-
Physics Application
a ) C is TT =L 1171 cost ,
ITT Is in a >
Fir IT TT =L ITT I COS B ,
ITT Isin B >
✓
LO ,
-100 >
TT t
TT =
LO ,
1007
100 lbs .
1171 cos × t 1171 COS 13=0
ITT Isin x t
ITT Isin 13=100

7. Dot Product E Cross Product :
Geometry of Vectors
Dot Product
-
A- vectors → At . B- number
B
ZD : At =
Ca . . Az )
A- . B- = a ,
b ,
t
Azbz
B- =
( b , .bz )
-
3D A =
La , , Az .
As > At . B- = a ,
b ,
t
azbz t
As bs-
B =
S b ,
.bz ,
bs )
Properties :
A- .
B- = B- oft
At o
( B- t I ) =
ATB- t A- . I
At •At =
Eid ;
2
=
I A- 12
-
A 7
o
) At . B- = I All BI Cos O
-
Its -
At
proof : Law of CosinesB
'
1B -
A- 12=1 At 12+1 B- 12 -
21 All BICOSO
( B- -
F) L B- -
F) =
B- .
B- t At .
F -
2ft .
B- = I A 12 t I B12 -
2nF .
B-
Projection ,
-
= lil =
I
A
I
o ) I
A- . ( c B- ) =
CLIO B- I =
( CF ) OF
T
Y
,
IATCOSO i
I
IF Hill cos O =
IF I .
lil
dot product w/ a unit vector
gives projection
T = ( A .
i ) it
-
un =
-4
III
J = C Ao FF,
1¥,
=
F. T
I
IT 12

8. Example :
What is the force vector E on the box ?
-
( 2 ,
I ) F =
projection of G- onto I
u -
U
=L
-
2 ,
-
I 7
E a- a- = -
mai
< E =
FEIT =
-39 C -
z .
-
I > =
C ⇒ ,
-
⇒ >
( O ,
O )
A- .
B- =
O
↳
perpendicular L cos 909=0
or Orthogonal
A- =
O ,
15=0
AT . I =
a ,
At =
La , , Az ,
937
A- .
j =
Az I =
L I ,
O ,
O )
A- .
I =
as

9. Cross Product
Dot product
A- .
B- =
a ,b , tdzbztazbz - 7
A i
=
ITIIIBTCOSO '
- -
,
- O ) i
T= B
y
s
'
BIBI'
s
Q :
What is the area of a
parallelogram spanned by
A- and B- ?
ZD :
y
n
Yr
,
B- =Lb , .bz >
Area = LATH
a
=
IATIBTSINO
h
7
B- A- =
IATIBTICOSO ' ' >
a X
B-'
,
=
At o B-
'
~
s B-'
=Lbz ,
-
b .
>
×
=
La , .az ) -
Lbz ,
-
bi )
= a .bz -
azb ,
a Right hand a
Left-hand
+ -
I
>
7
= I
di 22 B
A-=L .bz
-
azb ,
•
- O
Bb
,
bae n A n
[
right hand or left
F-
a
- -
-
A I 2 B 34
-
= I (4) -
2131=-2 -
= 2 -
a
B 3 4 A I 2 -
I I I I
-
> -
Byz
-
-
-
.
3D :
^Z
-
'
-
Area ( Eff )=
-
-
-
- r B -
Ayz
) -
- XZ
B
7 -
-
-
-
T
T . 2 2 2
-
YZ t
XZ t
Xy
-
- -
-
<
A A I -
XZ
>
y
I =
At xD
-
> -
Bxy-
-
I
X -
-
L At xy
) -

10. ^
A' xD
such that thx B- I =
area L £57 )
F. =
IATIBI since
I
f)O B
I =
A' tB
Tt '
I =
ab:% i -
as:3; it
}:3:
k
+ - t
i J I
=
a , 22 23
b , bz bs
^
n
i
n
J a k
= t L 2223 -
J
d , 23 t K d , 22
bzbz b , bz b.bz
Example : in j kn
I 2 O
-
-
6T -
3J
-
II
O
-
I 3
Area ( parallelogram
)=T6¥k
^
A' xD
→
jfc
?
Volume -_lFxB- 1h
F.
=L At x B- IIEICOSQ
I >
JO B- = ( At xD ) .
I
-
,
A

11. 12.5 Lines and Planes
Line in 3D Space
I .
Parametric Equation II. Solution to 2 Equations
2¥
! Y ( t ) ,
2 C t ) )
#←
Lines
Given
P= ( Xo , Yo ,
Zo )
J = La ,
b ,
C >
Line going through P in T direction
T a
( Xo , yo , Zo )
#
r Lt ) -
To = to
Taco
L -
' '
r Lt ) -
To II T ' '
ro Tht )
r Lt ) =
ro t tv
( Xo , yo ,
Zo )
•
X Lt ) =
Xo t at
O
y Ltt =
yo tbt
Z Lt ) =
Zo tbt
If ( Xo ,
yo ,
Zo ) =
LO ,
O .
O ) → L passes through origin
Xlt ) = I t t
y
Lt 1=2+2 t
} t = -
I → x Lt ) =
y Lt ) = 2 ( t 1=0
2 L t ) =3 +3T
Solution to 2 Equations
X =
Xo t at →
t =
y
=
yo tbt → t =
eye
b
2 =
Zo
t
Ct → t =
X
-
Xo
y
-
yo 2 -
Zo
a b C

12. Example :
Find line perpendicular to A- = 4. 2,0 > G 15=50,3 ,
47
passing through
F- ( S , 6,7 ) -
na
i j I
r
-
←
I L I 2 O At xD =L 8 ,
-4,37
B •
T=AtxBL 5. 6,7 ) O 3 4
rlt ) =
rot tv
=
L 5. 6,7 ) t
tL8 ,
-4,3 )
=
L St 8T ,
6- 4T ,7t3t )
P,
=
( X , , y , ,2 ,
)
- ooo
Po =
L Xo , yo , Zo 8=7,
-
to
• r Lt )=FottJ
^
-
t-rottlr.ro )
- r ,
% Lx , ,y , ,z , >
= l I -
t )Fottr
( Xo ,
yo ,2o )
Given pair of lines L ,
& Lz
3 possibilities
:{ infested
Skew -
-
L ,
: Xlt )=X ,
t
a.
t Lill Lz # La , ,
?, ,
C ,
> =L Laz , bz.cz >
y Lt )=y ,
+bit I k€0 vz
Ztt ) =
2.
tat L ,HLzc= > L , intersects ↳ or L ,
4 Lzskew
Lz :
X L s ) =
Xztazs ↳ Solve for L .
du Lz simultaneously ,
G determine if there
y ( s )=yztbzS is a solution
Z (5) =
Zztczs
Xlt
)=lt2t=S
y Ctl
=2t3t=2s
=) 2t3t=2t4t
zLtI=3t4t=3S
3T -
4t=O
X ( s
)=S a t=OYLSKZS
(
2151=35 t
-
-
O
3=35--3 ✓ to
5=1

13. Planes
Equation of a plane : Line
-
^ Ttt ) = tvtro
D= Fx B- J direction of line
- -
n B 7
To point on line
•7
A- >
a
-
n =
C a. b. C 7
T -
ro
a-*
( F -
To ) .rT=O
E
t.n-ro.rs
ro
=
axtbytcztd =O
L Xo , yo ,
Zo ) D= O iff plane passes through the
origin
F =
( X , y ,
27
Example :
What is the equation for a plane containing p ,
( I ,
2,3 ) , pz ( 3. 4,3 ) , Ps ( 1. 3,4 )
At = L2 ,
2,07
B- =L 0,1 ,
17
RT = Atx B- =
L2 ,
-2 ,
27
a b c
D=
-
To opt =
-
(1.2-2.2+3.2)=-4
2X -
24+22-4=0
Example :
What is line perpendicular to the plane :
Xt2y +32+4=0
passing through the point ( S ,
6,7 ) ?
Fit) = tvtro
=
L t +5
,
Ztt 6,3T +77
What is the distance between a point p ,
= L X , ,
y , ,
2 ,
) and a plane axtbytcztd
F. -
rT=
PIP
= C x , Y ,
z )
T -
n =
La ,
b ,
C)
To= L Xo . yo .
20£.
nee- -
-
E.
T
projnv-
i
-
'
it ,
D= lprojnvt -
Y rt =
'
Yon?
'
in , =
ly-n-j-l-KF-roi.nlInt
O
•
=
Irion - troon , =
tax ,
t
by,tC2 ,
+ d I
Int
22 t
BZ t
(
Z

14. Example :
L ,
:FLtI=Llt2t,2t3t,4t4t7
skew :
Lz :
g- G) =
LS ,
25,357
Distance btw them ?
D= I XVI

15. (€ty
) ,
2 C t ) )
Tct )
TLtI=LXLt ) ,
yet ) ,zCt ) )
drat
→ at tangent
Fctttvtro IF - T.at ftp..atDF -
at -
= v
lim ~v
dt
at → oat
drat =
Speed
Velocity
J=ddI=r
'
a- = IF =
r
"
Speed 181=1*71
Plane X
p
T Tht ) =
C Xlt ) ,
yet ) >
,
I =L cos Ltl ,
sink ) >
p
-
×
I l
.
I I I
I l
147I -
T
.
Lt ) =L -
yet ,
Xlt ) >
Sint -
it
Fits
t
=L -
sink ) ,
Cost ) )
-
Y 7
cost Y F' L t ) =
L X
'
Lt ) , y
'
Lt ) ) I l
-
y x
3- D
I
Flt )=LXLt ) , ylt ) ,
Ztt ) >
an
7
=L cos Lt ) ,
sink ) ,
7
B -
F' Lt ) =L -
sina.cosltl.IT )
o%Ya'heating IIEI its ,
-
sina.co >
3{
'
T -
E unit tangent vector *
osculating plane
^ TXT plane spanned by
B =
Txt unit binormal Vector tanda
=
T '
xrT ,
IT 'xT" I Normal plane
:
plane
N^=Bx 'T spanned by
A. BL -1T )

16. at
:
tangential acceleration
at =
lprojgal-T.tt
'
IT 't
lat =
rattan
an
=
-V 1212 -
af
Example :
compute normal plane for
Flt ) =
cost ,
Sint ,
3¥ ) att -
21T
F' (f) It -2 it
=L -
Sint ,
cost ,
It ) lzit
=L 0,1 ,
It 7
axtbytcztd-OL-rooca.b.cl
O .
Xt I .
yt
It 2
-
( 1. 0,3740 ,
I ,
7=0
Remark : N -
^
7 V
-
Class :N^=B^×F Book :
= ,
T
>
I =
-V
B T 'T
D= -1
> ,
g
ITI
18×81 IT 't J J=T .
D= xaTxT
a- =p
Ihtxalxvl
B

17. Integration of Vectors
Flt ) = L Xlt ) , ylt ) ,
zlt ) )
[ Differentiation ] ( Xlt ) ,yLt ) .
bit ) )
F' Lt ) =
points in
tangent (F) direction
t=b
with magnitude Ir
'
tht speed
J →=L X' Lt ) ,
y
'
( t ) ,
Z' C t ) > ( X ( b ) , ylb ) ,
2lb ) )
Integration : Flt )
•
gba-rthdt-e.LI?so..IrTtiktit=acxca).ycas.zcaD=LSbaXLt
Idt ,
sbayltldt.SE zltldt )
ftarltsdt Sbaxutdt
Stay
# dt
SEZ
# dt
= f-
avg
"
average
b -
a b -
a b -
a b -
a position
' '
? l
S :* Lost ,
Sint .
Et ) at
b- a
=
C SE
't
costdt.SZotsintdt.SI
't
at )
211=20,0,
EFFIE 7=20,0 ,
31T ) = C 0,0 , Z >
ZIT 21T
fbalr.lt/dt=arclength-
speed
Example :
arclength of our helix from t= Oto t .
-
21T
STILL -
Sint ,
cost ,
It >
ldt-fftvsinzttcosti-44itzdt-fztffahtitzd.li=
21T #/4I2
Flt ) =
( cos Lt ) ,
sin L ta ) ,
3¥ >
What is the trajectory ?
' '
re parameterization
"
U=t2
F =
Lcosu ,
sina.FI )

18. Canonical Parameterization :
"
parameterization by a
rclength
' '
⇐ IT '
I =
I
( X L S ) ,
y CS ) ,
2 ( S ) )
Foo
'
€.
✓s ¥
Parameterization of our helix by arc length
:
( cos
(¥%z) ,
sin ( FEW) ,
¥tz )

19. Newton 's Law of Motion
F--
mainpractice
:
particle object
know F
derive trajectory
ZD MY parabola
-
-
-
-
,
lxcthyctyyt
gravity
,
- F= -
mgjy
-
Vo ( Vocosx ,
'
,
ground
la
Vo sin >
>
know a- letter "
Lt )
×
then
get
JLtl-r.lt ) by anti -
differentiating
a-
Jlt )=rTt ) t
Vo
then
get Flt ) =
-
ttvotv ,
F=ma- =
-
mgj
a- =rT= -
gj
F.
Hi-Fi-
gtj-vocosxitlvosinx-g.tl
J
@ to
Fltkrotlvocosxtitlcvosinxlt -
Igt )j
f- O →
To =
FLO ) =L 0,07
FLtI= ( Vocosxltitllvosinx )t
-
Igt )j
Tangential and normal acceleration
An :
how quickly you
're
turning
at n
At :
how much your speed is
changing
at
.
-
-
-
*⇐
- -
-
-
say
sa'
^
,
i
An -
-

20. Z
I
¥¥i÷¥¥¥n¥¥¥. .
-
Sint ,
cost ,
3121T )
Tz'
Lt ) =L-
Ztsint
'
,
Zt cost ?
341T
)
n
n n T n
OT r ,
' '
Ltl =L -
cost ,
-
Sint ,
O )
'
B
Este
inssinn?iii.ftp.tj-4#sintZ
.
" it '
Jz x
g- z
=
(
' H'
ht Sint
2
,
-
I
ZTYIT Cost
2
,
8t3 )
×
Y

21. Functions of Several Variables I 14 .
I )
.
Single variable calculus :
.
I variable
.
graph in ZD space
.
Domain in ID space
.
Function of 2 Variables :
.
graph in 3D space
.
domain in 2b space
Contour Plot
^
'
÷÷÷i÷÷"
" " " " "
"
×
.
↳
paraboloid
⇐ O
.
f ( x , y ) = XZ -
y
Z
2
my
C =
0×2-
y 2=0
¥:÷:÷÷¥↳÷¥÷÷÷.
↳
pringle
N
y
✓
C = I C= -
I
✓
y
.
a Xt
by t CZ t D= O
'
-
2=1 -
ax -
by
-
d) 1C • 3
×
E.I:{yt.goyi.co/-/ C =
2
C = I
⇐ O

22. .
Function of 3 Variables
.
3D domain
.
4 D
graph
↳ t =
f L x ,
y ,
2 ) → level surfaces in 3D

23. Limits da
Continuity ( 14.2 ) and Partial Derivatives L 14.3 )
Example : sink )
iim
×→ot sin ( I ) undefined
Iim
µ
'' → O
-
sin ( I ) undefined
lim
IX. y ) → l Xo , yo
) FIX ,
y ) =L for LX ,
y ) Sufficiently close to ( Xo ,
yo ) ,
fl X. y ) is close to L
distance ( l x , y
) ,
( Xo , yo ) ) is small
Iim
S -
•
X →
Xo ft X ) =3
joy
f ( Xo ) ¥3 →
discontinuous
I
Xo
lim
Def :
f L X ,
y ) is continuous if for all ( Xo , yo ) in domain ,
IX. y ) → L Xo , yo )
=
f ( Xo , yo )
Iim
sincxy )
( X ,
y ) → ( Xo , yo ) X2tyZ
Partial Derivatives :
slopes of the
tangent line in the X direction FI L Xo , yo )=f× ( Xo , yo )
↳
treaty as a constant )
tangent line in the y direction
f- ( X , y
) =
XZTYZ +3C (
Xy )
f- ×
=
2x -
3 ysinlxy )
f- y
=
Zy
-
3 xsinlxy )
If× )×=2
-
3yZcosLxy ) → at
fxy =-3
xycoscxy ) -
3sinLxy ) →
Tatya
↳ f × first

24. Z
/•#f
yl Y
iv.×
f ( x ,
y
) = XEZY t sin (
Xy )
f ×
=
e
29 t
y COS (
Xy )
f-
y
= ZXEZY t
x cos ( Xy )
f××= -
y 25in L Xy )
f-
y y
=
4 XEZY -
XZ sin (
Xy )
f y ×
= IT =
2e2y t
cos (
Xy ) -
Xy sin ( Xy ) =
fxy
Theorem :
Suppose fxy and f-
y × exist near ( Xo ,
yo ) and fxy , fyx are continuous at ( X o , yo )
Then fxy ( Xo .
yo ) =
fyx ( Xo , yo )
↳ Mixed partials commute

25. Chain Rule
Single Variable Chain Rule
adz
Hg ( x )) =
f
'
( g L x ) ) g
'
( x )
or
W =
flu )
u =
g L x )
off = days .
dat
.
What does it mean ?
.
Q :
How much does w
change as I
change X ?
.
know :
4W I dadu
LIU
dx DX
U = -81 a ×
→
4W = -84 due
a = Fx ax
Multi -
Variable Chain Rule
W = f L X , y )
X =
x Lt )
y
-
-
Y Lt )
W Lt ) = f I X L t ) ,
y Lt ) )
What is IF in terms of IQ ,
?
aw = Fx ax t
IF ay
Also know :
DX = Etat ay = IF at
4W = FIEF at t
IF dat at
=
I FIFI t
Fwy date lat
IF =
# eat .
-38 It
More
generally
W =
w ( X , y )
× =
XLS.tt
y
=
yes ,
t )
W =
w L X ( s ,
t ) , y I s , t ) )
ETI
Ee ⇒ +
⇒ ⇒
-273ft # IF
Rule :
as
many summand S as variables that w depends on

26. de
Example : x Lt ) = t t
2
y Lt ) =3 t +4 WLX ,
y
) =
X
2 t
2yZ at
?
I
.
Could substitute E
"
brute force it
"
W Lt ) =
( t t
2)
2
t
213 t t
4 )
=
t 2+4 t t
4 t
18 t2 t
48T t 32
=
19 EZ
+52T
+36
W
'
Lt 1=38 t t 52
2. Use chain rule
Fat =
IT at +
Efta
=
2x .
I t
4
y
.
3
=
2 ( t t
2) t
12 ( 3 t t
4)
=
Ztt 4 +3Gt t
48=38t t
52
Implicit Functions :
I .
e . X
2
t
y
2
=/
dy
Tx ?
Implicit Differentiation
↳ chain rule
.
Apply Ex to equation
-3×1×2 t
y 2)
.
-
Ix LI )
2x t
2 y FL =
O
8¥ =
-
f-
W
=
w ( X , y ,
2. a ,
b )
x =
x Lt )
y
=
y Lt )
:
b =
b Lt )
Ldf =
IT Eft -37 It t
# at at

27. Review
Contour Plots
.
Set z
equal to constant E solve
TNB Frames
f= J=r
.
Lt )
A a- =tT=r "
vq
a
ta 's, N=BxT
Fcc )
Tian^
an
= a- oN^
- - - -
; a-
or =
Eat
an at-compta-a.T-a.IE ,
I
AT
Old Exam I # to
FLtI=L4t ,
costs 'll ,sint3t ) )
Jlt ) =
C 4 ,
-35in t ) ,
-30553T
) )
a- HI =
SO ,
-90513T) ,
-95hL -3T ) )
At
=
OT OFF,
=
( Ot 27sihbtkosbtlt27sihl-3tkosl-3.tl ) .
( 16+9 sin
'
Gt ) tacos
'
Est))
-
I
An =
✓ 8105434+81 sink -3T ) -
16+9517213+1+9054 -3T )
t →
at
9

28. Practice Exam I # 8
concentric circles ?
8- x -
y=c
y=8
-
X -
C ←
line
e-
1×2+427=(-1×2+42)
=
Inc
X2ty2=
-
Inc OCCLI →
circles
e
4×2 -
y
4×2 -
y
= Inc
4=4×2
-
Inc ←
parabola
sihlxty ) -
I > c > I
Xty=0 ,
IT ,
-
IT ,
21T ,
. . .
y=
-
Xt KIT Kisan
integer
Xz -
y2 saddle point
rltl-54t.cosbtl.sinl-3.tl )
r 'LtI=S4 ,
-35in
t ) ,
-305L -3T ) )
r "LtI=L0 , -905Gt) ,
-951hL -347
At --
A- °
I =O
NT
An =
19-1=9
Old Exam I # 6
r Lt )=( t2 -
I ,
1h42 ) ,t4 -
t
's -
Htt )
Slt ) =
( 2ft -
4. coscittltl.tt -
I )
Fios
'=Ir7l5tcosO
F' =L Zt ,
E. 4T
'
-
3tZ -
Zttl )
I
'
=
( ( tt3 )
-
' ' 2
,
Hsin Litt ) ,
It
-
" 2
)
t =
I
r 'Ll ) -_ l 2. 2,0 )
S 'Ll ) = CE ,
O ,
I )
r 'Ll )oS' (1) =L
IF
't-
Too 15'I=fz
1=2050
COS @
= I

29. Practice Exam I # 6
f- =
e
XYZ
f ×
=
y Ze
XYZ
f×y
=
2ye×Y
'
t
y
'
( 2x y ) ex Y
'
=
2ye×Y
'
t
2×y3e×y2
Practice Exam # 8
htt =L cost ,
Sint ,
t ) Osculating plane @ L
1,0
,
O )
plane that contains TE I
✓ Lt ) =L -
Sint ,
cost ,
17
a Ltt =
C -
cost ,
-
Sint ,
O ) t =
I
I
!I! I so .
-
ins
O X
-
y
t 2 =
d
d =
OLI
) -
L O ) t
O
-
y
t 2=0
Practice Exam I # 9
r Lt ) =
L2 cost ,
et ,
t ) point in
tangent line at 12.1.01
r
'
Ltt =L -
Zs int ,
et ,
I ) t =
O
r
'
( O ) =
SO ,
I ,
I )
eqn for line : 5ft I =
tvtro
5 Lt ) =L 2 ,
t t I ,
t )
541=52 ,
2 ,
I )

30. Directional Derivative E Gradient
it =
unit vector
Question : What is the rate of change of f in the un direction at ( Xo ,
yo ) ?
this number Dun f I Xo .
yo )
"
directional derivative of f at ( Xo ,
yo )
"
h ( s ) = f ( Xo t
Sa ,
yo
t
Sb )
Dun f L Xo , yo ) =
Is Is =
o h L S )
=
Is Is = o f ( Xo tsa , yo
t sb )
=
f × ( Xo ,
yo la t
fy ( Xo ,
yo ) b
Dun f ( Xo , yo ) =
( f × , fy ) •
( a ,
b )
- -
gradient of f I
grid f =
If
⑦ = L at , Ey )
TT =
C Ix , Fy ) f =
tax ,
If )
Dj f =
TT f o
is
What is the
meaning of TT f ?
I fo un = I If I I ill COS 0
= I If I COSO for which Un is Dun f
biggest ? 0=0
If points in the direction of
greatest increase
left =
Slope in this direction
f L X ,
y ) = Ii ( X
2
t
y 2)
If =
Lf × ,
y x )
=
'T ( 2x , Zy )
=L E .
¥7
Vector field → a different vector at each point
TT I level curves

31. Level curves for functions of 2 variables
The
' '
C -
level curve
"
of f L x , y ) is defined to be the set { ( x ,
y
) I f ( X , y ) =
C }
f ( X , y ) =
X
2
t
y
2
C =
O ,
X
2
t
y
2
=
O
C =
I
,
X
2
t
y
Z
=
I
C = r
2
,
XZ t
y
2
=p
2
↳
equation for the circle centered at 0 of radius r
' '
Gradient of f
"
or If ,
which is in the field in the domain of f whose
value at p
is Dflp ) = tax ( p ) ,
L p ) >
Duff L x , y )
=
at ( Xt tu , , y
t
tu z
) I
t
-
-
o
¥ L X , y ) U ,
t
I x , y ) Uz =
If L x , y ) out =/ If I x ,
y ) I I it I COS O
17 f I Level curves of f
Given w =
f ( X , y ,
2 )
Dw L X , y ,
2) =
( If L X , y . 2) ,
L x , y ,
2) ,
It ( X , y ,
2 ) )
Dw is I to level surfaces
Tangent Plane Problem
Given a surface
' '
2 = f I x ,
y )
"
,
find eqn for a
tangent plane a point ( X , y ,
f- L X ,
y
) )
need a normal Vector for the tangent plane ;Surface
' '
2 =
f ( x , y )
"
is level surface of
a new function Wl X , y ,
2 )
W ( X , y ,
2) =
2
-
f L X , y ) →
the O -
level surface of w is
"
z =
f L x , y
)
"
17W L X , y ,
2) =L -
f × ( X , y ) ,
-
fy ( X , y
I ,
I )
-
fx ( X , y ) l X
-
Xo ) -
fy L X , y ) (
y
-
yo ) t 2 -
Zo
=
O
Z =
Zo
-
fx L X ,
y ) ( X -
Xo ) -
f
y
L X ,
y
) L
y
-
yo )

32. Minimal Maximal critical points of f- L X , y
)
•
I -
var :
-
local minimax
'
absolute minimax
.
f
'
( X ) =
O
↳ find critical points
.
Places of potential minimax :
critical point ,
end points
-
f
' '
L Xo ) > O → local min
.
f
' '
( Xo ) ( O →
local Max
-
f
' '
( Xo 1=0 →
no conclusion
Multi -
Variable
.
z =
f I X , y )
Ivorry about local maxima on bound ry
=
( f × ,
f-
y
) =
O →
critical point
↳ horizontal tangent planes
.
places of potential Maximin :
critical point ,
bound ry
EX : L X -
1) ( y
-
1) ( Xtlkytl ) 2yZ
-
2 4
Xy
4
Xy 2×2 -
z
=
{4 l 0,01
=
1×2 -
1) ( yz
-
I )
-
16 ( 1,1111 ,
-
Ill -
I ,
Ill -
I ,
-
I )
=
Xzyz
-
Xz -
y
-
t I ↳
saddle point
If =L 2X y
'
-
2x ,
2×2
y
-
Zy > =
SO ,
07
ZXYZ -
2x = O =
2x L
y
2 -
11=0
2×2 y
-
24=0
=
2yLX2
-
I 1=0
I O ,
O )
( I ,
1) l I .
-
Ill -
I ,
1) Hi -
I )
The
"
second derivative test
' '
f xx
fy×
iffyYy =
fxxfyy -
f Ly =D
d) O →
local min or Max
do →
saddle point
D= O →
no conclusion

33. f ( X , y ) =
XZy2
-
xz
-
yz t I
-
22×22
-
22
y 22
critical points :L X. yl
=
10,01L -
I .
-
II. C- I. 17 ( I ,
-
1) L 1,11
mint #
Max saddle points
f xx ( Xo , yo )
concave down in x direction
concave up in x direction
Checking the Boundary
(a) X =
2
-
22422
f C
2. y )=4y2
-
4 -
y
' -
1=342-3
Tdy f L 2. y ) =
by =O
y=o
→
minimum
f L 2. y ) =-3
f ( 2 ,
2) =f( 2 ,
-27=9
min :L O ,
-
2) ( 0,21L -
2. 0712,0 )
Max :
(
-
2 ,
-
2) L 2 .
-
2) L -
2. 2) ( 2,21
Find minimax of flx , y )=X2t2y2 on unit disk
I. interior
•
t =
The
TT =
( 2X ,
4y ) =
SO ,
O ) f=2
m.in
Lt ) -
- cost
2X=0 44=0 , yet ) -
- sint
• • •
f =/
X =
O y
=
O t =
IT f- =L f=O t =O
Max Max
( 0,0 )
f=z
} 40
= 8 → minimax
•
t= 3172
min
f××= 270 → min
fLtl= cost t 2 Sint =
ltsinzt
f-
'
( t ) =
Zsintcost =
O
t=O ,
IT ,
The ,
3172
f
' '
Lt ) = -
Zsinzt +2057
to →
2 Checking boundary for absolute minimax ,
not local Maximin
t=T/z→ -2
t =
IT -
2
t =3 → -
2

34. Rectangular box w/ surface area 12
2X
y
t
2×2 i
Zyz
=
12
V =
Xyz
z = z ( X ,
y ) =
?
V ( X
, y I =
Xyz C X ,
y l

35. Lagrange Multipliers
Idea :
f C x ,
y ) or f C X , y ,
2 ) Wrt a constraint g ( X , y 7=2 Or
g l X , y ,
2) = C
f ( x , y ) = X
2 t
2y2 ,
find Maximin on unit disc
constraint :
x
2 t
y 2=1
I
.
An extrema point Off With constraint g is where level curve for f first touches level
curve for
g
2.
Tangents to level curves for f and
g are same
3. If I I TTg
TT f = HTT
g
( X , y ) is extremal iff If I X , y ) =
XTTGLX , y )
g ( X , y ) =
c
In above example :
f- ( X , y ) =
X
2 t 2
y
2
g ( X , y ) .
-
X
2
t
y
2 =
I
If =L 2X ,
4
y
)
TT
g =L 2X .
2y7
If =
X #
g
( 2X ,
4
y
> =
X C 2X , Zy )
2X
.
-
X 2x
4y=X2y
X
2 t
y
2
=/
Case I :
X to → X =
I →
y
=
O → x = I I
Case 2 :
X = O →
y
= I I
✓ =
Xyz
2X
y
t
2×2+242=12
yz
=
X I
Ly
t
22 )
X 2 = X ( 2X t 22 )
yz
= X ( 2
y
t 22 )
2X y
t
2×2 t
Zyz
=
12

36. Given fl X , y ,
2) want to Maximin
subject to constraint g L X , y ,
2) =
C
'
I
Volume =3
2 I
Xyz =3
. - - - -
!
212×21+242+3 l
Xy ) = f ( X , Y ,
2)
'
IZ i
42+3 y
=
XYZX
'
'
n 22+3 X = XXZ
3 Y 4×+2
y
=
X
Xy
Xyz =3
x=4 =
Ey ''
I
gq=¥→z×=y → x -
Ey
22+3 X 2 3
X =
Tz =
I+
I
; -32 4g →
34=42 →
2=-34Y
X =
-4×+24=-4+-2
Xy
IEy )
y
I I
y
) =3 = I
y
's
y
3=8
y
=
2
X =L (2) =
I
z
-
-
I, (2) = I
2 constraints
f L X ,
y ,
2) subject to
g L X , Y ,
21 =
C and h L X ,
y .
2) =
k
intersection of g G h satisfy both Constraints
T
tangent to Level surface ,
i. e .
If IT
If is in normal plane of the curve ( Ig x ⑦ h ) .
TTV =
O
i. e. If is in the plane spanned by Ig t
Ih g ( x , y ,
27 =C
If = XTTF t
mtg h ( X , y . 2) = K
Example : f ( X , y ,
2) =
Xt2y +32 .
g ( X ,
y ,
2) =
X
-
y -12=1 ,
h L X ,
y ,
2) =
XZ t
y
'
=
I
I =
X t
M 2X
}
x +
may } FIFE; } Ey

37. Integration
Sdc Sba
Hx . yldxdy →
volume
a ex Eb C Eyed
Compute volume using ring method
ith
Volume ( slab ) =
( area of face ) ay
-
Sba f- L x. yldx
Volume = -2 (Sba ft x. yildx ) ay J !( Sba f ( x , y ) dx ) dy
Ex : SIS : lxztyzldxdy →
SISI XZ t
y
Z
dydx
If 3×3 t
XYZ I
'
o
dy
go I +
y
'
dy
I y
t
Ty 318
It's =
¥
Area of Slab at fixed X :
Shg, fix , y ) dy
Volume : Sba SITE's fix ,
y
)
dydx
Ex : f ( x , y ) =
y
R :
y =
fix
=
hlx )
9"%
.
× y = I -
x =
g ( X )
fo
'
Sixy dydx
Sj Eye IIE dx
=
SIEH -
N -
u -
xp ) DX = fo 2x DX
×2 I
'
o
=
I
x= ray =
hey ) SIS
'
y dxdy
S
'
oxy IFI
'
dy
"%%.
y -
. >
S
'
oyvfyi -
yu
-
y
)
dy
×=I -
y
=
gly )

38. Sometimes the integrand prefers a certain order of integration
Ex : SS r sing dxdyIT 12 Y
S
× sings dydx
S :
"
So
dxdy V
R R
Tk a The
>
I l
The The
Sometimes a region of integration prefers one order over another
x= -
REF -
4
)
Z
! I
"
fix . ysdydx v
or
p y
Sit Siri
"
fix , yidxdy
t
fifty , f l x , yldxdytfof-Ttyflx.gl dxdy
x=
-
Ty try
y
= E t
2
y
-
z
=
×=±T2y-4 x =
try
Volume Between 2 Surfaces
IS r
f ( x. yldxdy
-
SIR glx ,
y
)
dxdy
=
Sfr I f L X , y ) -
g ( x , Y ) ) dxdy
EX :
Volume of a sphere of radius I
-
Fye
z =
TEY z =
-
T¥yd ,
x =
iffy
fi,
S
2M¥42 dxdy
Can use double integrals to compute area
Volume =
SIR I
dxdy =
area of region
Volume of Paraboloid X 't
y
Z -
I under Xy plane
IS R 1×2 t
y
' -
1) dxdy →
neg .
Volume
↳ Sfp I I
-
XZ -
y 2)
dxdy multiply by
-
I

39. Mass of Bar = I Slxldxdy () )
Where 8 l X ) is
density function T
SIX )
Or SIR 81×1 dxdy depending on shape

40. Polar Coordinates
OEOEZIT
y
-
←
•
RIO
rsin0{
/ ×=rcoso rt-VXZ.ly '
10 , y=rsinO 07am
'
( E )#rcoso
z=T¥y
Sphere of Radius a
Volume ?
SSDZTE-yzdxdyze-rafz.gr
,⑧ "
"
→
←
Ira .
If
ftx.yldxdy~Eflrijcosoij.rijsinoijlrijdraodxdy-rdrdof.fi
, 2T¥yZdxdy
SITS.az#rzrdrdOt
Example :
DD X2tyZd×dy
Zcoso
S Too rzrdrdoo %
-1 2

41. Exam 2 8:00am
What did we do ?
Chain Rule
*
f L X , y ) X =
X ( U ,
V ) y
=
y ( u .
V )
Ef -
-
If .
+
Ef
*
Implicit Differentiation
a
Constraint :
K2
t
y
2
t 22=1 ) Ty
What is -38 ? ↳ 2=2 l x , y )
O t 2
y
t 22 toy =
O
Ey =
-
24/22 =
-
Y
12
Directional Derivatives and Gradients
*
TT f =
C f× , fy ) Dj f =
slope in u direction =
If oil
↳ direction of
largest increase
length
=
Slope in that direction
f ( X , y .
2) → If =
( f × ,
f
y ,
f- z >
Tangent Plane
TT f =
54 ,
5 ,
67
4 X t
Sy t 62 t d =
O
Normal Line
Flt ) =
t C 4,5 ,
67 t C
p , , Pz . Ps )
gradient ( X , y ,
2)
Local Maximin
Check interior
, boundary ,
E corners
TT f =
O
f ×
.
-
O f
y
=
O →
crit .
points
given crit .
point
Iffffyyy I
}
If 70 minimax → If fxx or fyy > O :
min ,
CO :
Max
If CO saddle point

42. Lagrange Multipliers
One constraint :
Max f- ( X ,
y ,
2 ) subject to
glx , y ,
2) =L
If =
x Fg
two constraint :
Max f L X , y , 2 ) subject to
g
L X , y ,
2) and h L X , y .
2)
If =
XTTG t MTT h

43. Applications of Double Integral
e.
g .
D is a lamina l thin stab plate )
f C x , y ) =
density ex , y )
Mass =
SSD f IX. y ) dxdy
Center of Mass : I I , 4- I
I = th SSD XC ( X. yldxdy
g-
= Im SSD
yet x. yldxdy
where m= SSD Ctx ,
yldxdy
Ex : ELX . Y ) =3 for [0,23×50,2]
F-tmf:S } x. 3 dxdy=L .
12=1
5=1
Why does this work ?
Averages : the Xi
Assume uniform
density
→
e is constant
÷÷÷.
not ' values :
Fifties y
'
fidgety
↳
¥,
Laxey
More
generally
: E Wi Xi EXisjelxi.j.yin.la/Ly
→
Axe L x. yldxdy
E Wi E e L Xi , j , Yi .
j ) IX
dy Sf e ( x. yldxdy
Moment of Inertia
Solid
e= I
Si
"
So
'
r2 .
I rdrdo
rot
•
R SI
"
Er "
to do
Io =
ZIT (E) =
I
M =
IT
bigger
same mass
smaller
R= TIME =
VII =
÷
Io
.
-
Sf ( X 't
y
'
) e ( x. yldxdy
p2
R2 =
Io
m

44. Triple Integrals
Mass =
SSSDCLX.y.ztdxdydZI-tmSSSDXCCX.y.UA/dYd2y-=-mSSSDyecx.y
,
2) dxdydz
I = Tn SSS.pzecx.ly ,
2) dxdydz
Ex :
Z
Mass =L .
It .
ooh =
Z
C =
I
=
I =
SSSDXDV
±
.
.
" " "
x=V€ Y
=¥S'of
xdxdydz
- Y I M¥2 , y ,
z )
X
2×2+42+22=1=¥fIfF"
'
2×21
"
dydz y=o
=¥S'oSoF"
ELI -
YZ
-
22
)dyd2
-
) y
.
=
¥ S
'
o
y
-
Ey
'
-
zzyl dz
,
-
RE
✓ y
Switch order : S 'oST*SF⇒X dzdydx
'

45. Cylindrical Coordinates
( r , 0,2 )
I
c
421¥
X=rcosO
y - -
f -
I
# LIA
-
-
rooter y=rsino
§
}z
¥Tio> 2=2
Example :
Compute mass of density Elr ,
O
,
2) =p
Z
^
2 o rdrdoedz
.
}
Ksi 's
" .
v
.
Zr
x
y
Example :
Find center of mass bounded
by 2=X2ty2 and
2=24 Where C =L
I = Im SSSD Xdxdydz
↳
'
a is
rcosordxdydz=
fordz 2y=2rsinO X2ty2=rZZrsino
frz RCOSO r d2 fordo X2ty2=2

46. Spherical Coordinates
z r = es in Q
( x , y ,
2 ) ( e ,
O , 0 ) X =
rcoso = e sin ① cos O
e -
to = e Zo y
=
rsino =
esinosin
O
- =
-
y
020121T 2 =
Pcos Q
-
-
O r =
0202 IT
×
fffpflx ,
y ,
2) dxdydz rectangular
=
fffpflrcoso ,
rsino ,
2) rdrdodz cylindrical
=
SSS is f- I e sin locos O ,
esinosino ,
Pcos Q ) e
'
sin Oded 0 do spherical
EX :
Volume of a sphere of radius R
S
I.
ezsinodedodo
Sf
't
SI 7- sin 0 DODO
goat
-
RI cos lol 't do
12ft I R
's
cos 0 do
R
3
Ex :
mass of upside down cone in first octant
density
=
f ( X , y ,
2) =
xztyz
Mass =
SSS is XZTYZ dxdydz
=
SSS ( e sin 012 e sin 0 de DODO
=
Stones 't
'
4S go.TO
( e sinope Zsin Oded 0 do
Example
:
center of mass of part of sphere with radius I in first octant
constant
density
=
I
m =
volume = too .
=
If
Myz=
SE
"
SIT
"
SL e sin cos O e
'
sin Oded 0 do
=
Stones S
'
o essinzocoso.de DODO
=f to
"
Sto
"
IT sin 20 cos Od 0 do
=
f to
"
got
"
I .
I ( I -
cos ( 20 ) ) COSO DO DO
M×y=
I 't
"
SI
"
S
'
o
e cos 0 P2 sin Qdpd Q do
=
S I
"
SI
"
I, cos 0 sin dado
= S 't
"
too sin 2018
"
do

47. General
Change of Variables for Multiple Integrals
I .
-
La . .az )
b- =
Lb . .bz )
a .az =
LIA
b , bz
frandomvalueofv
Fi ( u )=LX( U.ll.ylu.lt )
RT ( UI =
L x ( 2. VI. yl2.ir ) )
T
a- =
-dF
Gu
random value of u
du =L Fu xlu.D.dauylu.lt > =L Fu )Liu
b-
=8LW=c£×cz.rs ,
IT ycz.us > = at
,
Lw
4A=
443¥04
ex aye
ZU 24
IF LIV au
=
ex aye
dully
OV OV
dXdy=
If
If pay
dudu
-
Jacobian f
( X. y )
( 4. V )
Ex :
ffp l 4×2 -
YZ )
' 00
dxdy
4×2-42=12×+4 ) ( 2X -
y )
- -
U v
100 24,41 y
Sf U V
' °°
one ,v , dudu 2
*
( u=2,V=
-
21
4=2×+4 4=2
-
2X
V =
2x
-
y R 2=2×+4
UtV=4X U
-
V =2y
*
2=4
114=2,V= 2) X
X = 'Ll ( Utv ) y=E( u
-
V ) X=O
U
goes up
4 I
=
-
I ¥( UTVKO
y=
-
2t2×
I -
I 4 U=
-
V
2=2×-4-
2x V goes upFfs' '
u9do!9ouf¥,auauu⇒v⇒
⇐ u .
3 Variables
Eu Ew
ex at ey
all OV ow
⇐ ez ⇐
OU OV 2W

48. Line
Integral
fit , L Xlt ) ,
ylt ) )
. I It
Integrate a function over c
art( Illntegratea vector
fieldoverc@IifLx.y
) fcflx ,y)dS
"
adding up function over curve
"
{ I :FL x. y
) ScFodF= ' '
work done by force field F "
I Ilfflx , yids siarclength
I -_
Tn
SXFCX.ytdsy-tmsyflx.ly/dS
Computing Line Integrals
Steph :
parameterize curve Steps :
do this I -
var
integral l integrate wrtt )
FLTKLXHI , ylt ) >
Step 2 computed
ds= ( EE ) !
⇒ Zdt =/ Idt
Compute mass of semicircle w/
density f L x. y )=y2
z -
Mass =Ly
'
IS =L ,
yzdstfczyds
J
C ,
irlt ) =
SO ,
t ) -
2EtE2
Cz :FHI=C2cost .
Zsint ) Etc
^ "
ds=TEdt=dt ds=FsinEI2dt=V4dt=2dt
c ,
SI tzdt SITE 4sinZLt ) .
Zdt-2-
C

49. Start with a vector field
F L x ,
y ) =
L M l x , y ) ,
N I X , y I )
.
wind
.
ocean currents
.
Tff
.
force field ,
e.
g .
gravity
Jc F .
dir =
integral of F- along C =
work done by F along C
Work
=
force .
distance
-
y
F
!
I
i n
.
> T-
F . I :
force in I direction
Sc F .
I ds =
Sc Food F
-
#
Sc F. Tds = SEES EL x Lt ) , yet I I .
¥¥{I
¥1Idt =
Sc F . dir
Step I :
parameterize r Lt I =L x Lt ) ,
y It I )
Step 2 : Write as
integral of t
- d F
I =
I at
IT I
=
TarIt
EX :
C =
part of a parabola
y
=
X
2
•
( I ,
I ) Sc F .
IF
✓ for F = -
yi t
XJ
( O ,
O )
Step I :
parameterize C :
X Lt I =
t
y L t ) = t
2
r Lt I = Ct ,
t 27
Step 2 :
Write as integral of t :
Sc F .
dir .
-
S
'
o
F . 8¥ at
=
I s -
t 2
,
t > .
C I ,
Zt > At
Step 3 :
Integrate
I
'
o
-
t
'
t
2E d t
=
f
'
o t
Z
at =
I

50. Ex : F = -
yi
-
xj
• ( I ,
I )
y=x
( 0,0 )
X L tht
yCtI=t
So
'
C -
t ,
t > o
( 1. I > dt=0

51. Fundamental Theorem of Calculus
Sba tax dx =
f ( b ) -
f La )
For line
integrals
:
Sc F .
IF =
f L X , , y ,
) -
f ( Xo .
yo )
Example : F =
( x
2
t
y
2) I t
2 Xy J
, .
X =
Cost
OE t E Iz
y
=
s int
,
SE
"
F . EE at
Sto
"
l cos 't t Sint ,
2 costs int ) .
C -
Sint ,
cost > at
SE
"
-
s int t
2 cos 't Sint dt
cost lot
"
-
E cos 't It
"
-
I t
I =
-
I
f- ( X , y ) = I ×
3
t
Xy
2
TT f =
L X
'
t
y
2
,
2X
y ) =
F
Using FTC :
f- ( O ,
I ) -
f ( I ,
O )
O
-
I =
-
I
Given E ,
how do
you get fix ,
y
) so that If = F
Note :
not
every F has an f
Physics Example :
x
E =
-
mg j
t t I I I
f- = -
mg y
= -
mgh
"
potential energy
"
✓ If =
( O ,
-
mg )
I Work : Sc E .
IF =
-
Mgh ,
+
Mgh o
y

52. Finding Anti -
Derivatives
Given Fix , y
I =
Mit Nj
Ql :
When does there exist fix , y ) so that If = F
TT f =
f ,
I t f
y Jw w
M N
My
=
fxy f exists →
My =
Nx
Nx =
f
y ×
"
F is conservative
"
ex : F =
L x
'
t
y 2) it Zxyj- -
M N
My
=
2y ✓
Nx =
2
y
ex : F = -
y it xj
My = -
I
×
Nx =
I
Q2 :
If My
=
Mx ,
does f exist ?
A lot of the time
F =
1×2 t
y
'
Ii +
Zxyj
Want f
f- ×
=
X
2
t
y
2
→
f =
I X
't
XYZ
t
C (
y )
f- y
=
2X
y
→
f =
X y
2
i
C L x )
f =
5×3 t
XYZ

53. Missed Class 415
-
a
L
,
←
. TF
t '
→
i r
we

54. Curl
Green 's Theorem
' >
Fundamental Thm of Line Integrals
SSD curl
"
(F) dxdy
=
!F. IF Sc TT fo DF =
FIX , , y ,
)
-
flxz.
yz )
F -
-
Pit Qj
Ex say
p Q
=
Qx -
Py =
curl
20
( F ,
-
ZD boundary of D
①
Orient
boundrysotha.to?.isuca!waysonuttC ,
D
F conservative →
curl
" >
(E) =
O
E =
x2ty2
Curl
ZD
F =
O
IFI =
5×+52=1×2ty
2
XZTYZ
Sc ,
Fo dF=§ ,
L Fo Tlds =
Soc
.
Ids =
21T
11
*
O
SSD
DFdxdy=O
O
↳
proof stuff
21T
F =
-
yitxj
Curl
20
E
' '
-2 2-
2X ay = +2
-
y x

55. curl
3D
F =
Vector =
curl F
↳
points in axis of rotation
I CUTI F I =
rate of rotation
F =
Pi t
n
Q j t
RE
i j k
a-
ee2x a
y 22
P Q R

56. 213 : TT =
JI I t
Tay j
3D : TT = IT t
Ey j
t
-2 I
n
i j k
curl E- Ex Ey IT =
TT x F
P Q R
F =
pit Qj t RE
3D :
E
TT x E =
curl F
I .
E =
div E
TT .
E =
taxi +
Ey j to II ) .
( Pi +
Qj t RE )
= t t
FF
F = -
yitxj t
OI
IT x F =
i J k
'
Ix Ey Iz =
Oi t
Oj t 2E
-
y x O
TT .
F = Ex L -
y ) t
Tay X t Iz O =
O
F- =
xityjtznk
Ex E =
i J I
Ex Ey of =
Oi t
Oj t
OI = O
X y 2
I . F =
Ex X t
Ey y
t
IT 2 .
-
I t I t I =3
Divergence measures creation of fluid .
Curl measures
swirling of fluid .
ZD :
curl
ZD
( Ff ) =
O
am
3D : for I rt If =
O for E div curl F =
O
I I I I
TT x
( I f ) =
O I .
(TT F ) =
O
Q : Is F conservative ? i. e .
F = If Q : Is F =
curl G- ?
First check is curl E =
O
Only possible if div F .
- O

57. Parameter icing Surfaces
sphere with radius I :
X ( O ,
0 ) =
sin COSO
y ( O ,
O ) =
sin since
Z ( O ,
O ) = cos 01
sphere with radius 2 :
X l O ,
01=2
sin locos O
E =
2
y ( O ,
01=2sin since
Z ( O ,
07=2cos Q
cylinder on side w/ center on y axis with radius 3
2 ( U ,
V ) =3 COS U
X Lu ,
v ) =3 sin u
y ( U ,
V ) =
v
Arc length (c) =
Sods = Sba HII Idt
Where THI =L Xlt ) ,
y
L t ) ,
z t t ) )
Italia 't day
'
+
¥
Surfaces :
Flu ,
V ) = C X L u ,
V ) , y Lu .
V ) ,
2 LU ,
V ) )
Tangent Vectors
- -
er er
OU
,
OV
Chunk of Area
→ -
or
aux
Or
IV
OU OV
→ -
NGA =
Or
Aux
Or
IV
ou OV
surface Area :
IS 3%37 dudu

58. EX :
Find Surface area of a paraboloid ,
z=X 't
y
'
L top disk of radius 2 )
( vi. V ) = ( r ,
@ ) or ( U ,
V ) =
( x , Y )
X=u
y=V
Z ( vi. V ) =
f ( U ,
V )
F = L X ( UN ) , y L UN ) ,
ZCU ,
V ) )
=
( U ,
v ,
flu ,
V ) )
n
n
I k
x -37 =
It
O f×C a. V ) =L -
f× ,
-
fy ,
I )
O I f
y
( UN )
f) Vf×2tfy 't
12 dudu where D is circle of radius 2:22 =X2tyZD
y
dxdy
SIS EEy. V 4×2+442+1 dxdy
X =
UCOSV
y=usinV
ZLU ,
V ) =
XZTYZ =p 2=42
F= LUCOSV ,
usinv ,
U2 )

59. Flux Integrals
Last Time :
DA =
III ldudv
Surface area : Ds DA
Iff .in#uxFIdUdV
Today : F =P I X , Y ,
2) it QLX , Y ,
21 j t R I x ,
y ,
2) I
Flux :
the amount of fluid that passes through surface per unit time
z → →
→ →
⇐ '
%¥##E- I→ →
y
×
→
→
→ →
Constant vector field :
Flux =
I E .
N I l areal
Generally
:
Flux = IS CF .
NIDA =
IS F. DE
=
SS E .
( Eru ''
E) dudu
Ex : F =
xityjtzk
Compute flux through a sphere of radius 2
u =
to
✓ = O
X LU ,
v ) =
Zsinucosv
y
LU ,
V ) =
2 sinus inv
2 I U ,
V ) =
2 cos U
DF = Eu ×
-37 dudu
^ n
n
L
j k
Zcosucosv Zcosusinv
-
Zsinu = ( 45in ZUCOSV ,
45in Zusinv ,
4 sihucosu ) dudu
-
2 sinus inv Zsinucosv O
F. d A=L Zsinucosv ,
Zsinusihv ,
ZCOSU > o
L 4S in ZUCOSV ,
4 sin Zusihv ,
4 sinucosu )
IT
I }
"
to C Zsinucosv ,
Zsinusihv ,
ZCOSU > 044 sin ZUCOSV ,
4 sin Zusihv ,
4 sinucosu ) dudu
13
"
SE8 sins ut 8 sin UCOSZU dudu
,
fo
" SE8 since dudu =
to"
-
8 cos u to du = SI
't
16 DV =
321T

60. Simplifications
I .
dV= e' sin ① d ed Odo
=
de DA
DA =
e2sin0dQdO
I Fux I =
Ve 's in
"
utesinhicos '
u
=
Tetsu =
Esinu

61. Review
Polar Coordinates
y X=rcosO
•
y=rsinO
r
RIO
O ) x 020121T
dXdy=rdrdO
Mass ,
Centers of Mass
,
E Moments
g.
m
:¥÷
.IE#i:7.iaax.aa
.
Triple Integrals
f L X , y ,
2) =
density
SSSD fix , g. 2) dxdydz
=
mass
Cylindrical Coordinates
×=rcosO RIO
•
y=rsinO 020221T
Z
2=2 2 El -
Cs
,
co )
F r
dxdydzt-rdrdo.dz
Spherical Coordinates
X= esinocoso r=esinQ
it •
y
=
esinosino EZO
e
z =
ECOSQ 020121T
g-
r
e =
Tty 020221T
dxdydztesinodedodo

62. General
Change of Variables
X =
X ( U ,
V )
y
=
y LU ,
V )
ffyf.nl#syldxdy=u.ffimitsxcuivl.ylu.vHFYu7lidudV
=/
Eu ⇐
⇒ ⇒
I
Line Integrals
Sc f L X , y ,
2) AS
SF L X , y ,
2) od F work done by
F
I .
Parameterize
F Lt ) =L X It ) , y l t ) ,
2 I t ) )
2 .
Sc f L x , y ,
2) ds
ds =
Ideaf ) 't
fffYt¥
'
at =
II at
Sc Fo DF = So ( F . N^ ) ds =
Sc Pdx t Q dy t
Rd z
F =
L P L X , y ,
2) ,
Q ( X , y ,
2) ,
R ( X , y ,
2 ) )
dried
at
dt
Sc F .
dir = Sba F .
at
Sc ds =
arc length
Fundamental Theorem of Calc for Line Integrals
I TT f od F =
f L X , , y , ,
2 ,
) -
f- ( X o , yo ,
Zo )
w
E
F is conservative if F =
If for some f
F =
LP ,
Q ,
R )
ZD :
conservative → Q ×
-
Py
=
O
3D :
conservative → curl F = I
Given F such that Q x
-
Py =
O
If = E
f ×
=
p f
y
=
Q
f- =
f Pdx t.ly) SQdy t
CfX )

63. Green 's Theorem
④
c
SSD ( Qx -
Py )dXdy=&cPdXtQdy=§cFodr
Curl
F=LP,Q ,
R )
n
Curie :FxE=a¥ Ey at
Q R
divE-TT.FI II +
If to
⑦ =
L II. Ey ,
-32 )
f- Unc →
rect
→
rect → fun
I Fx TT .
f If Ix f) =O
F TT xF TT . x E 1=0

64. Stoke 's Theorem
^
IN
→
#
Givens .
an orientation is a choice of D
↳→
Make sure
III points indirection of orientation←
If F. DF =
Sfs I F. in
)dA=ffFlxlu,
VI. ycu.vt.zcu.VH.IE °
-37 )dudv
u -
V
limits
Z
-
'
←
parabola in y
-
z
S= surface of revolution obtained by rotating
y=I
-
zz this parabola around z-axis
,
" "
y
l outwards orientation )
I
Compute flux of E -
xityjtzk throughs
- -
I
Think cylindrical
:
U=
"
O
' '
Check orientation V
X=rcosO 4=0
} →
X= ( I
-
VZ ) cos u
y=rsihO V=z
y= I I -
v2 )
since
2=2 r =/ -
22 Z= V
✓
O EOE ZIT
-
12221
I -
' ' ' '
I U U V
-
I
-
LIT
Do F.
CEE
Idudv
II =L -
( I -
✓ 2) sina.LI
-
VZICOSU ,
O )
=L -
Zvcosu ,
-
Zvsinu ,
17
n
k
⇒ FI =
I -
v 2) since L I 2) cosy O
=
L L I
-
V 2) cosy
,
( I -
v 2) since ,
ZVLI -
V2 ) )
-
ZVCOSU
-
Zvsinu I
Stiff
"
L ( I -
V 2) cosy ,
Ll -
v 2) since ,
V ) .
( ( I -
V 2) cosy ,
Ll -
V 2) since ,
ZVLI -
V2 ) ) dudu

65. N
SOE . dt-SSCUHE.DE
-
-
-
-
-
.
if
Example :
use Stoke 's theorem to evaluate §FodF
where F =
s 2. X. y >
C is unit circle ,
centered at
origin in the plane xtyt 2=0
↳ Nestor's .
't )
J= ftp.
-
tf ,
Oy
I Perpendicular
check To N=O
I. T=O
-
D= Ciro ,
%)
Parameterize c :
TH )= cost Jtsintw
=L ¥ -
s ,
t
-
sing ,
t ) . . .
Or using Stoke 's
theoryi j k
curl E =
Ix By Ez =
( 1.
1,1)
z X
y
ISCURIF .dF=fSkurIE .
NIDA =
554,1 ,
1) ol 't .
Is .
To I da
=SSr5dA=BSSdA=T3 IT
Some Interesting Examples :
f- =
-
xi -
yJ -
2K
where e=yxE
E3
C is unit circle in x
-
y plane
ft .dF=O
=
Note : I ( f) =L
x y z
( X2ty 2+2213/2 , 1×2+42+221312, 1×2+42+221312
)
curl E =
curl ( I ( et ) ) = I for all IX. y ,
2) not at LO
,
O.O )

66. Divergence Theorem
let D be a region in 3D
space
Where S = surface which forms the
boundary of D ( outward orientation )
and F- =
Vector field in 3D space
SSSDTT .
Edu =
Sss F. DE
EX :
D= hemisphere ,
X
2
t
y
2+22=4 ( radius = 2) ,
with
y 10 ( so D is a half hemisphere )
F =
C X . y ,
27
LHS :
SSS is TT .
Fav RHS :
Sss Fod At =
Sfs,
E .
DA t
Sss .
F. DA
TT .
F =
Ex Xt
Ey y to 2
.
-
I +
It I =3 where S ,
is the curved hemisphere part E
ISS D 3 DV =3 SSS DV Sz is the flat back
-
volume As ,
IF o
D) dat Sss ,
LF .
D) DA
- -
3 .
I ( 4J IT 23 ) =
161T IF 1=2 O
2. Area L S ,
)
2. I 41T 22
16 IT

67. Calc Review
Vector
Geometry :
.
NT :
normal Vector La ,
b ,
C >
↳
plane :
a x t
by t CZ =
d
.
Lines : II t ) =L Xo , yo ,
Zo ) t
t VT
→
r It I
- -
r
'
I t I =
V Lt )
T
"
Lt ) = T '
( t ) =
a Lt )
Arc length
Kds = Sc I at Idt =
So VEE 't
eat 4-
IT
'
at
Mass
Ic f ( x , y ,
z Ids =
Sba f ( x Lt I .
y I t ) ,
2 Lt ) )
/ dat Idt
Sc Eod F =
SLF .
F) ds = Sba F . FE at
TN B Frames
T T -
-
II
•
Bn
"
s N I =
Tv
Fits I J I
B =
T '
x F "
I F '
XT ' '
I
N = B ×
I
Chain Rule
f ( X , y ) X ( U ,
v I y l u ,
v )
tf
= If +
If
.eu
ZU all
Gradients
f-
I points in direction of
greatest increase
15ft I :
rate of
change in this direction
Directional derivative : If °
IT

68. Tangent Plane
r Lu ,
v ) → FI × = N- =
La , b. c) →
axtbytcz =D
z =
f ( X ,
y )
parameterize
U =
X ,
V =
y
→ FLU ,
V ) =
( y ,
V
,
fly ,
v ) )
D= FIX FI = -
fi -
f y J t I
a b C
-
f ×
X
-
f
y y
t
z =D
D= -
f , Xo
-
f
y yo
t f ( Xo , yo )
Z = d t f ×
X t
f y y
= f ( Xo , yo ) t
f ×
( X
-
Xo ) t
f y
( y
-
yo )
Maximal Minima
TT f =
O →
critical points
I
ft ffxyyy / =D
D > O min f xx ) O or fyy > O
Max f xx LO
D SO Saddle point
D= O ?
Max I min f ( X , y )
Subject to constraint g L x , y I =
C
If =
XTTG
Polar I Cylindrical I Spherical
dxdy =
rdrdo
dxdydz =
rdrdocdz
dxdydz =Psin Old ed ① do
× =
X Lu ,
V )
y
=
y Lu ,
V )
dxdy =
1¥71,
I dxdy
=/ III dudu

69. Surface Integrals
FLU ,V )
Asda
=
surface area
ffsflx ,y,z1dAd# I ×
FIldudv
Flux Integral
Is Fo DE dF=FI× # dudu
↳ =
( F. NIDA
.fSs3dA=3 .
surface area