LECTURE 4: PARTIAL DERIVATIVES
MINGFENG ZHAO
January 12, 2015
Graphs of functions of two variables
Definition 1. The graph of a function f of two variables is the set of points (x, y, z) that satisfy the equation z = f (x, y),
where (x, y) varies over the domain of the function.
Level curves
Definition 2. Given a surface z = f (x, y), for any z0 , the curve f (x, y) = z0 in the xy-plane is called a level curve of
the surface z = f (x, y) with level z0 .
Remark 1. A level curve of a surface z = f (x, y) with level z0 is just the projection of the trace of a surface z = f (x, y)
in the plane z = z0 onto the xy-plane.
Remark 2. The level curves with different levels do not intersect.
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2
MINGFENG ZHAO
Definition 3. Given a surface F (x, y, z) = 0, for any z0 , the curve F (x, y, z0 ) = 0 in the xy-plane is called a level curve
of the surface F (x, y, z0 ) = 0 with level z0 .
Example 1. Find and sketch the level curves of the function f (x, y) = y − x2 − 1.
The level curve with level z0 is described by the equation y − x2 − 1 = z0 , that is, y = x2 + 1 + z0 , which is a parabola
in the xy-plane. For example,
• With z0 = 0, the level curve is the parabola y = x2 + 1.
• With z0 = −1, the level curve is the parabola y = x2 .
• With z0 = −2, the level curve is the parabola y = x2 − 1.
• With z0 = 1, the level curve is the parabola y = x2 + 2.
Partial derivatives
Definition 4. The partial derivative of f with respect to x at the point (a, b) is:
fx (a, b) = lim
h→0
f (a + h, b) − f (a, b)
.
h
The partial derivative of f with respect to y at the point (a, b) is:
fy (a, b) = lim
h→0
f (a, b + b) − f (a, b)
.
h
Remark 3. The partial derivatives evaluated at point (a, b) are denoted in any of the following ways:
∂f
∂f ∂f
∂f fx (a, b) =
(a, b) =
, fy (a, b) =
(a, b) =
.
∂x
∂x (a,b)
∂y
∂y (a,b)
LECTURE 4: PARTIAL DERIVATIVES
3
Example 2. Let f (x, y) = x2 y, use the definitions of partial derivatives to compute fx (x, y) and fy (x, y).
For fx (x, y), we have
fx (x, y)
=
lim
h→0
=
lim
h→0
=
lim
h→0
=
lim
h→0
=
lim
h→0
=
f (x + h, y) − f (x, y)
h
2
(x + h) y − x2 y
h
2
(x + 2hx + h2 )y − x2 y
h
2
x y + 2hxy + h2 y − x2 y
h
2
2hxy + h y
h
lim (2xy + hy)
h→0
=
2xy.
For fy (x, y), we have
fy (x, y)
=
=
=
=
=
=
f (x, y + h) − f (x, y)
h→0
h
x2 (y + h) − x2 h
lim
h→0
h
x2 y + x2 h − x2 h
lim
h→0
h
x2 h
lim
h→0
h
lim
lim x2
h→0
x2 .
In summary, we have
fx (x, y) = 2xy,
fy (x, y) = x2 .
Remark 4. In practical computations of partial derivatives, to compute fx (x, y) is just to compute the derivative of
f (x, y) with respect to x when treating y is a constant; to compute fy (x, y) is just to compute the derivative of f (x, y)
with respect to y when treating x is a constant.
For example, let f (x, y) = x2 y, then
fx (x, y) = 2xy,
fy (x, y) = x2 .
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MINGFENG ZHAO
Example 3. Let f (x, y) = sin(xy) and g(x, y) = x2 exy , compute the partial derivatives of f and g.
In fact, we have
fx (x, y)
=
cos(xy) · y
By the Chain rule
= y cos(xy)
fy (x, y)
=
cos(xy) · x
By the Chain rule
= x cos(xy)
gx (x, y)
gy (x, y)
=
2xexy + x2 exy · y
=
2xexy + yx2 exy
=
x2 exy · x
=
x3 exy
By the Product rule and the Chain rule
By the Chain rule
Example 4. Let f (x, y) = esin(xy) , compute the partial derivatives of f .
In fact, we have
fx (x, y)
fy (x, y)
=
esin(xy) · cos(xy) · y
=
y cos(xy)esin(xy)
=
esin(xy) · cos(xy) · x By the Chain rule
=
x cos(xy)esin(xy) .
By the Chain rule
Example 5. Let f (x, y) = xxy with x > 0 and y > 0, compute the partial derivatives of f .
First, we know that
f (x, y) = xxy = exy ln(x) .
Then we have
fx (x, y)
=
exy ln(x) · [y ln(x) + xy ·
=
exy ln(x) [y ln(x) + y].
1
]
x
By the Chain rule and the product rule
xy
Example 6. Let f (x, y) = (xy)(xy) with x > 0 and y > 0, compute the partial derivatives of f .
Hint:
(xy)xy = e(xy) ln(xy) ,
xy
(xy)(xy)
= (xy)e
(xy) ln(xy)
= ee
(xy) ln(xy)
·ln(xy)
LECTURE 4: PARTIAL DERIVATIVES
5
Remark 5. The meanings of partial derivatives:
a. fx (a, b) represents the rate of change of f at point (a, b) with respect to x when y is fixed to be b.
b. fy (a, b) represents the rate of change of f at point (a, b) with respect to y when x is fixed to be a.
Example 7 (Ideal Gas Law). The pressure P , volume V , and temperature T of an ideal gas are related by the equation
P V = kT , where k > 0 is a constant depending on the amount of gas.
a. Determine the rate of change of the pressure with respect to the volume at constant temperature. Interpret the
result.
b. Determine the rate of change of the pressure with respect to the temperature at constant volume.
c. Explain these results using level curves. Interpret the result.
Since P V = kT , then P =
kT
V
.
a. So we get
∂P
∂V
=
∂
∂V
= kT
∂
∂V
= −kT
= −
Since k, T, V > 0, then
∂P
∂V
kT
V
1
V
1
V2
kT
.
V2
< 0, that is, the pressure is a decreasing function of volume at a constant
temperature.
b. So we get
∂P
∂T
=
=
=
Since k, V > 0, then
∂P
∂T
∂
∂T
kT
V
k ∂
(T )
V ∂T
k
.
V
> 0, that is, the pressure is an increasing function of temperature at a constant
volume.
c. Look at the level curves of P , for each P0 > 0, we have
P0 =
kT
.
V
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MINGFENG ZHAO
Then
T =
P0
V.
k
That is, T is a linear function of V .
Figure 1. Some level cuves
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca