5. Multivariable function: A function f of two variable x and y is a
rule that assigns a unique real number f(x , y) in same domain D
in the x y-plane .
Example : z = x^2 + y^2
Polynomial function : A polynomial function in x and y is a sum
of function of the form Cx^m y^n with nonnegative integers m
and n and C is a constant .
Example:
Level curves : The level curves of a function f of two variables
are the curves with equation f(x , y)=k , where k is a constant (in
the range of f)
113y2xy7xy3x 235
5
6. 6
Application of multivariable function : Techniques of multivariable calculus are used
to study many objects of interest in the material world. In particular
Domain and
codomain
Application Techniques
Curves Length of curves line integrals and
curvature
Surface Areas of surfaces, surface integrals,
flux through surfaces, and curvature.
Scalar field Maxima and minima, Lagrange
multipliers, directional derivatives,
level sets.
Vector field Any of the operations of vector
calculus including gradient,
divergence, and curl.
7. 7
Example 01 : Find the domain and range of g(x , y) = 9 − 𝑥2 − 𝑦2
Solve: Now the domain of this function
D ={(x , y) | 9 − 𝑥2
− 𝑦2
≥ 0 }
= {(x , y) | 𝑥2
+ 𝑦2
≤ 9 }
Which is the disk with center (0 , 0) and radius 3
The range of g is
{ z | z = 9 − 𝑥2 − 𝑦2 , (x , y) ∈ D }
Since z is positive square root , z ≥ 0
So the range is = { z | 0 ≤ z ≤ 3 }
= [0 , 3]
8. 8
Example 02 : Find the domain and range of g(x , y) = 9 − 𝑥2 − 4𝑦2
Solve: Now the domain of this function
D ={(x , y) | 9 − 𝑥2
− 4𝑦2
≥ 0 }
= {(x , y) | 𝑥2
+ 4𝑦2
≤ 9 }
= {(x , y) | 𝑥2/32 + 𝑦2/
3
2
2
= 1 }
Which is the disk with center (3 , 3/2 )
The range of g is
{ z | z = 9 − 𝑥2 − 4𝑦2 , (x , y) ∈ D }
Since z is positive square root , z ≥ 0
So the range is = { z | 0 ≤ z ≤ 3 }
= [0 , 3]
9. 9
Example 03: Sketch the level curves of the function g(x , y) = 9 − 𝑥2 − 𝑦2 𝑓𝑜𝑟 𝑘 =
0 , 1, 2, 3
Solve: The level curves are
9 − 𝑥2 − 𝑦2 = k
Or, 9 − 𝑥2
− 𝑦2
= 𝑘2
Or, 𝑥2
+ 𝑦2
= 9 ─ 𝑘2
This is family of concentric circle with center (0 , 0) and radius 9 − 𝑘2
The cases k = 0 , 1 , 2 , 3
If k = 0 then the equation of the circle 𝑥2
+ 𝑦2
= 3^2 center (0 , 0) radius 3
If k = 1 then the equation of the circle 𝑥2
+ 𝑦2
= 8 center (0 , 0) radius 8
If k = 2 then the equation of the circle 𝑥2
+ 𝑦2
= 5 center (0 , 0) radius 5
If k = 3 then the equation of the circle 𝑥2
+ 𝑦2
= 0 center (0 , 0) radius 0
10. 10
Example 04: Sketch the level curves of the function f(x , y) = 10 ─ 𝑥2
− 𝑦2
Solve: The level curves are
10 ─ 𝑥2
− 𝑦2
= k
Or 𝑥2
+ 𝑦2
= 10 ─ k
This is family of concentric circle with center (0 , 0) and radius 10 − 𝑘
The cases k = 0 , 1 , 2 , 3 are
If k = 0 then the equation of the circle 𝑥2 + 𝑦2 = 10 center (0 , 0) radius 10
If k = 1 then the equation of the circle 𝑥2
+ 𝑦2
= 3^2 center (0 , 0) radius 3
If k = 2 then the equation of the circle 𝑥2
+ 𝑦2
= 8 center (0 , 0) radius 8
If k = 3 then the equation of the circle 𝑥2 + 𝑦2 = 7 center (0 , 0) radius 7
12. 12
Partial derivatives
The partial derivative of f(x , y) with respect to x at the point (xₒ , yₒ) is ,
𝜕𝑓
𝜕𝑥
⃒ 𝑥ₒ , 𝑦ₒ =
𝑑
𝑑𝑥
f(x , yₒ)⃒x=xₒ =lim
ℎ→0
𝑓 𝑥ₒ+ℎ ,𝑦ₒ −𝑓(𝑥ₒ ,𝑦ₒ)
ℎ
Provided limit exist
The partial derivative of f(x , y) with respect to x at the point (xₒ , yₒ) is ,
𝜕𝑓
𝜕𝑦
⃒ 𝑥ₒ , 𝑦ₒ =
𝑑
𝑑𝑦
f(xₒ , y)⃒y=yₒ =lim
ℎ→0
𝑓 𝑥ₒ ,𝑦ₒ+ℎ −𝑓(𝑥ₒ ,𝑦ₒ)
ℎ
Provided limit exist
13. 13
Application of partial derivatives in our real life:
• Derivatives are constantly used in everyday life to help measure how much
something is changing.
• They're used by the government in population censuses, various types of
sciences, and even in economics .
• Partial Derivatives are used in basic laws of Physics for example- Newton's
law of Linear motion , Maxwell’s equations of electromagnetism and
Einstein’s equation in general relativity .
• Derivatives in chemistry. One use of derivatives in chemistry is when you
want to find the concentration of an element in a product.
• In economics we use Partial Derivative to check what happens to other
variables while keeping one variable constant.
14. 14
Example 01 : Find
𝜕𝑓
𝜕𝑥
where f(x , y) =
2𝑦
𝑦+𝑐𝑜𝑠𝑥
Solve: Differentiating equation partially w . r to x , we have
)cos(
sin2
)cos(
)sin0(20).cos(
)cos(
)cos(2)2()cos(
)
cos
2
(
2
2
xy
xy
xy
xyxy
xy
xy
dx
d
yy
dx
d
xy
x
f
xy
y
dxdx
f
15. 15
Example 02 : Find
𝜕𝑓
𝜕𝑦
where f(x , y) = 𝑦2 𝑒 𝑥 + 𝑥𝑦 sin(𝑥 + 𝑦)
Solve: Differentiating equation partially w . r to y , we have
)sin()cos(2
).sin()cos(2
)sin()(
)sin(
2
2
yxxyxxyye
xyxyxxyye
yxxy
y
ey
y
yxxyey
yy
f
x
x
x
x
16. 16
Example 03 :
Find 𝑓𝑥 , 𝑓𝑦 , 𝑓𝑥𝑦 , 𝑓𝑦𝑥 , 𝑓𝑥𝑥𝑥 , 𝑓𝑦𝑥𝑦 , 𝑓𝑥𝑦𝑥 𝑖𝑓 𝑓 𝑥 , 𝑦 =
𝑎𝑥2 + 2ℎ𝑥𝑦 + 𝑏𝑦2
Solve:
hhyax
yx
f
y
fxy
hbyhx
xy
f
x
fyx
byhxbyhxyax
yy
f
fy
hyaxbyhxyax
xx
f
fx
2)22()(
2)22()(
22)2(
22)2(
22
22
18. 18
Statement of Chain rule : If x=x(t) , y=y(t) are differentiable at t
and if z = f(x , y) is differentiable at the point (x , y) = (x(t) , y(t)
then z=f(x(t) ,Y(t)) is differentiable at t and we write ,
∂𝑧
∂𝑡
=
∂𝑧
∂𝑥
∂𝑥
∂𝑡
+
∂𝑧
∂𝑦
∂𝑦
∂𝑡
19. Example 04 : If 𝑧 = 𝑒 𝑥𝑦 , 𝑥 = 2𝑢 = 𝑣 , 𝑦 =
𝑢
𝑣
, 𝐹𝑖𝑛𝑑
𝑑𝑧
𝑑𝑢
solve:
v
u
yvux
putting
v
vu
v
u
e
v
x
ye
v
xeye
u
y
y
z
u
x
x
z
xy
xy
xyxy
,2
)
2
2(
)2(
1
2
u
z