Chapter 9

Multivariable
Calculus

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9- 1
 Section 9.1

Functions of
Several
Variables
© 2012 Pearson Education, Inc.. All rights reserved.

9- 2
Example
Safe diving requires an understanding
of how the increased pressure below
the surface affects the body’s intake of
nitrogen (氮气). An exercise in Section
2 of this chapter investigates a formula
for nitrogen pressure as a function of
two variables, depth and dive time.
Partial derivatives (偏导数) tell us how
this function behaves when one
variable is held constant as the other
changes. Dive tables based on the
formula help divers to choose a safe
time for a given depth, or a safe depth
for a given time.
 Introductory examples
The amounts of labor and capital needed to produce a certain number of items related?

If a company produces x items at a cost of $10 per item, then

the total cost C(x) of producing the items is given by
C(x)=10x.
The cost is a function of one independent variable, the number
of items produced. If the company produces two products, with
x of one product at a cost of $10 each, and y of another product
at a cost of $15 each, then the total cost to the firm is a function
of two independent variables, x and y. By generalizing
f(x) notation, the total cost can be written as C(x, y), where
C(x, y)=10x+15y.
When x=5 and y=12 the total cost is written C(5, 12), with
C(5, 12)=10⋅5+15⋅12=230.
 Volume of a can
Let r and h represent the radius and height of a can in cm. The
volume of the can is then a function of the two variables r and h
given by
V(r,h)   r 2 h.
 Function of Two or More Variables
Definition: The expression
z=f(x, y)
is a function of two variables if a unique value of z is obtained
from each ordered pair of real numbers (x, y).
-The variables x and y are independent variables, and z is the
dependent variable.
-The set of all ordered pairs of real numbers (x, y) such that
f(x, y) exists is the domain of f;
-The set of all values of f(x, y) is the range.
-Similar definitions could be given for functions of three, four,
or more independent variables.
Exercise 1
In the xy-plane, give the domain for
 Evaluate or Find the value of the function

3
Let f ( x, y )  4 x  2 xy  , find f (2,3).
2

y
Solution: Replace x with 2 and y with 3.

3
f (2,3)  4(2)  2(2)(3) 
2

3
 16  12  1
 29
 3
Let f ( x, y )  4 x 2  2 xy  , compute
y
 Your Turn

Let f ( x, y, z )  4 xz  3x 2 y  2 z 2 , find f (1, 2,3).

Solution: Replace x with 1, y with 2, and z with 3.
f (1, 2,3)  4(1)(3)  3(1) 2 (2)  2(3) 2
 12  6  18
 24
Express the points in the coordinate system

Functions of one independent

variable
- x-axis, y-axis
- locate points in the xy-plane.

Functions of two independent

variables
- x-axis, y-axis, z-axis
- z-axis perpendicular to the
xy-plane.
Graphing a Plane
Solution: The graph of this equation is a plane.
Find x-, y- and z-intercepts （x-,y-,z- 轴的截点） (the point
where the graph crosses the axis) :
x-intercept, (3,0,0)
y-intercept, (0, 6, 0),
z-intercept, (0, 0, 6).
Graphing x + 2y + 3z = 12 in the first octant.
Solution: The graph of this equation is a plane.
x-intercept, the point (12,0,0).
y-intercept, the point (0,6,0).
z-intercept, the point (0,0,4).
 Any findings?

The graph of y=f(x) is a curve in the plane.
The graph of z=f(x,y) is a surface in the three dimensional space.
 Graphing a Function

Traces （截痕） - the intersection of the surface with

a plane. （曲面和平面的交点）
The xy-trace is the intersection of the surface with the xy-plane.
The yz-trace is the intersection of the surface with the yz-plane
The xz-trace is the intersection of the surface with the xz-plane.
Level curves （等高线）- the intersection of the
surface with planes parallel to the xy-plane. （曲面和
平行于xy平面的平面的交点）
Such planes are of the form z=k, where k is a constant, and the
curves that result when they cut the surface are called level
curves.
(a) Find the intersection of the surface with xy-plane, yz-plane and xz-plane. (Traces)
Point (0,0,0) and parabolas in the yz-plane (x=0) and xz-plane (y=0).
(b) Find the intersection of the surface with the plane z=1, z=2, z=3 etc. (Level curves)
Circles which are parallel to the xy-plane.
(c) Draw the traces and level curves on the same set of axes and get a paraboloid（抛物面）.
 Figure 10

The level curves of Example 7 plotted in the xy-plane.

 Figure 11

The topographical map is the surface of the land in a part of New York State.
Exercise: Matching the function with the level curves
 Example 8 Cobb-Douglas Production Function
Find the level curve at a production of 100 items for the Cobb-Douglas
production function
2 1
zx y3 3
 Figure 12a and 12b

The level curve of height 100 found in Example 8 is shown graphed in

three dimensions in Figure 12(a) and on the xy-plane in Figure 12(b).
Some common equations and their graphs
Some common equations and their graphs
Quadratic surfaces (二次曲面quadrics)

• 𝑎𝑥 2 + 𝑏𝑦 2 + 𝑐𝑧 2 + 2𝑓𝑦𝑧 + 2𝑔𝑧𝑥 + 2ℎ𝑥𝑦 +

2𝑝𝑥 + 2𝑞𝑦 + 2𝑟𝑧 + 𝑑 = 0

 Graphs in the subsequent pages are adopted from Wikipedia.

 𝑥2 𝑦2 𝑧2
Ellipsoid + + = 1 ,椭球
𝑎2 𝑏2 𝑐2
spheroid 𝑎 = 𝑏 类球体, sphere 𝑎 = 𝑏 = 𝑐
 𝑥2 𝑦2
Elliptic paraboloid + − 𝑧 = 0 ,椭圆抛物面
𝑎2 𝑏2
circular paraboloid 𝑎 = 𝑏 ，圆抛物面
 𝑥2 𝑦2
Hyperbolic paraboloid − − 𝑧 = 0 , 双曲
𝑎2 𝑏2
抛物面
 𝑥2 𝑦2
Elliptic hyperboloid of one sheet + 2 −
𝑎2 𝑏
 𝑥2 𝑦2
Elliptic hyperboloid of two sheets + −
𝑎2 𝑏2
 𝑥2 𝑦2 𝑧2
Elliptic cone + − = 0 , 椭圆锥面
𝑎2 𝑏2 𝑐2
circular cone 𝑎 = 𝑏 ，圆锥面
 𝑥2 𝑦2
Elliptic cylinder + = 1 , 椭圆柱面
𝑎2 𝑏2
circular cylinder 𝑎 = 𝑏 ，圆柱面
 𝑥2 𝑦2
Hyperbolic cylinder 2 − 2 = 1 , 双曲柱面
𝑎 𝑏
parabolic cylinder 𝑥 2 + 2𝑎𝑦 = 0 ， 抛物柱面
 Section 9.2

Partial
Derivatives

© 2012 Pearson Education, Inc.. All rights reserved.

9 - 45
Earlier, we found that the derivative
dy
dx
gives the rate of change of y with respect to x.
In this section, we will show how derivatives are found and
interpreted for multivariable functions.
z  f ( x, y)
 Marginal Profit
(profit for producing one additional unit of production)
A small firm makes only two products, radios and CD players. The
profits of the firm are given by

where x is the number of radios sold and y is the number of CD players sold.
How will a change in x or y affect P?
Example 1: partial derivatives
 Your Turn 1

Let f ( x, y )  2 x 2 y 3  6 x5 y 4 . Find f x ( x, y ) and f y ( x, y ).

Solution: To find fx(x, y), treat y as a constant and x as a

variable.
f ( x, y)  (2 y 3 ) x 2  (6 y 4 ) x5
f x ( x, y)  (2 y 3 )  2 x  (6 y 4 )  5 x 4  4 xy 3  30 x 4 y 4
To find fy(x, y), treat x as a constant and y as a variable.
f ( x, y)  (2 x 2 ) y 3  (6 x5 ) y 4
f y ( x, y)  (2 x 2 )  3 y 2  (6 x5 )  4 y 3  6 x 2 y 2  24 x5 y 3
 Example 2: Chain rule
The chain rule is a formula for computing the derivative of the
composite function.
 Your Turn 2

Let f ( x, y )  e 3 x2 y
. Find f x ( x, y) and f y ( x, y).

Solution: To find fx(x, y), treat y as a constant and x as a

variable.

f x ( x, y )  e  (3x 2 y )  e3 x y  3  2 x  y  6 xye3 x y
3 x2 y 2 2

x
To find fy(x, y), treat x as a constant and y as a variable.

f y ( x, y )  e 3 x2 y
 (3x 2 y )  e3 x2 y  3x 2 1  3x 2e3 x2 y
y
Partial derivative at a given point
 Example 3 Evaluating partial derivatives


such that
(d)
 Example 4
x

z  arctan
y
1 1 y
Solution: zx   2 ,
x y x  y
2 2

1  
 y
1  x  x
zy  2 
 2 2 .
x  y  x y
2

1  
 y
 Geometric interpretation

Tangent plane
In any particular direction,
there is only one tangent line.

The derivative of a function of one variable is the slope of the tangent line at that
point. At a point on the graph of a function of two variables, there may be many
tangent lines, all of which lie in the same tangent plane.
 Rate of Change: Figure 16

Partial derivatives to find the slope of the tangent lines in the x- and y-directions.

Curve z  f ( x, b)
 Example: Production Function


P 2
 0.1y [ln( 2 x  3 y  2)  x 
2
] ( product and chain rule)
x 2x  3y  2
P 2
(50,20)  0.1  20 [ln( 2  50  3  20  2)  50 
2
]  228
x 2  50  3  20  2
Thus, if the capital investment is held constant at $20,000 and
labor is increased from 50 to 51 work-hours per week,
production will increase by about 228 units.

In the same way, the marginal productivity of capital is

P 3
 0.1x[2 y  ln( 2 x  3 y  2)  y 
2
] ( product and chain rule)
y 2x  3y  2
P 3
(50,20)  0.1  50[2  20 ln( 2  50  3  20  2)  20 
2
]  1055
y 2  50  3  20  2

If work-hours are held constant at 50 hours per week and the

capital investment is increased from $20,000 to $21,000,
production will increase by about 1055 units.
Second-Order Partial Derivatives

Example 6 Second order partial derivatives
Example 7 Second-order partial derivatives
 Your Turn
Find all second partial derivatives for
f ( x, y)  x e  x y 2 7y 4 5

Solution :
Here f ( x, y )  2 xe  4 x y , and
x
7y 3 5
f ( x, y )  7 x e  5 x y
y
2 7y 4 4

Now find the sec ond  order partial derivatives :

f ( x, y )  2e  12 x y ,
xx
7y 2 5
f ( x, y )  49 x e  20 x y ,
yy
2 7y 4 3

f ( x, y )  14 xe  20 x y ,
xy
7y 3 4
f ( x, y )  14 xe  20 x y.
yx
7y 3 4
Example 8 Second-order partial derivatives
 Exercise 1--Laplace equation
 Laplace equation
Exercise 2 --Wave equation
Section 9.3

Maxima and
Minima
 Introduction

What amounts of sugar and flavoring produce the minimum

cost per batch of a soft drink? What is the minimum cost?

One of the most important applications of calculus is finding

maxima and minima of functions.
Earlier, we studied this idea extensively for functions of a single
independent variable; now we will see that extrema can be found
for functions of two variables and how to identify maxima or
minima.

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9 - 74
 Figure 17

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 Figure 18

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 Figure 19

The tangent line parallel to xz-plane

has a slope of 0 at the relative
maximum.
The tangent line parallel to yz-plane
has a slope of 0 at the relative
maximum.

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Extremum is used for either a relative maximum or a relative minimum.
The points satisfying the equations above are called critical points.

Note:
1. The fact that the slopes of the tangent lines are 0 is no guarantee that a relative
extremum has been located.
2. For functions of more than one variable, to avoid complications, we will only
consider cases in which the function is differentiable, i.e, critical points exclude
points from the domain where the derivative does not exist.
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 Figure 20： Saddle point

Saddle point: it is a minimum when approached from one direction

but a maximum when approached from another direction. A saddle
point is neither a maximum nor a minimum.
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 Example 1 Critical Points


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 Your Turn 1
Find all critical points for f (x,y) = 4x3+3xy+4y3.
Solution: Find all points (a,b) such that
f x (a, b)  0 and f y (a, b)  0.

f x ( x, y )  12 x 2  3 y  0 f y ( x, y )  3x  12 y 2  0
3 y  12 x 2 3 x  12(4 x 2 ) 2  0
y  4 x 2 3 x  192 x 4  0
3x(1  64 x 3 )  0
x  0 or 1  64 x3  0
1 1
x 
3
implies x   .
Continued 64 4

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 Your Turn 1 Continued

y  4 x 2
When x  0, then y  0
2
1  1 1 1
and when x   , then y  4     4     .
4  4  16  4
 1 1
The solution of the system of equations is (0,0) and   ,   .
 4 4
Since these are the only solutions of the system,
(0,0) and  1 1 
 , 
 4 4
are the critical points for the given function.

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 Test for relative extrema

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 Test for relative extrema

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 Example 2: Relative Extrema

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 Example 2: Relative Extrema

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 Figure 21

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 Example 3 Saddle Point

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 Example 3 Saddle Point


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 Figure 22

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 Your Turn 2
Identify each of the critical points in Your Turn 1 as a relative
maximum, relative minimum, or saddle point.
Solution: Critical points for f (x,y) = 4x3+3xy+4y3 are (0, 0)
and (−1/4, −1/4).
f x  12 x 2  3 y f y  3x  12 y 2
f xx  24 x f yy  24 y
f xy  3

For (0, 0) : f xx (0,0)  24(0)  0 f yy (0,0)  24(0)  0

f xy (0,0)  3 D  0  0  (3) 2  9  0.
(0,0) is a saddle point.
Continued
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 Your Turn 2 Continued

For (−1/4, −1/4) :

f xx (1/ 4, 1/ 4)  24(1/ 4)  6  0
f xy (1/ 4, 1/ 4)  3
f yy (1/ 4, 1/ 4)  24(1/ 4)  6
Here D  (6)  (6)  (3) 2  36  9  27  0;
there is a relative maximum at (  1/ 4, 1/ 4) .

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 Example 4: Production Costs


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 Example 4: Production Costs

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 Example 4: Production Costs
when x=4, y=18.
Now check to see whether the critical point (4,18) leads to a
relative minimum. Here

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 Figure 23

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 More Examples

1. z  x  y
2 2

For  0, 0  fx 0,0   f y 0,0   0,

f xx  0 , 0   2 , f xy  0 , 0   0 , f yy  0 , 0   2 ,
D  2  2  0  4  0 and f (0,0)  0
2
xx

there is a relative minimum at (0,0). This relative minimum is f  0, 0   0

2. z  x y
For  0, 0  fx  0,0   f y  0,0   0,
f x x  0 , 0   0 , f x y  0 , 0   1, f y y  0 , 0   0 ,
D  0  0  1  1  0
2

there is a saddle point at (0,0).

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 More Examples

3. z  x 4 y 4

For  0, 0  fx 0,0   f y 0,0   0,

f xx  0 , 0   0 , f xy  0 , 0   0 , f yy  0 , 0   0 ,
D  00  0  0 2

D=0 gives no information, but it is obvious f  0, 0   0 is a minimum.

4. z  x 3  y 3
For  0, 0  f x  0 , 0   f y  0 , 0   0 ,
f x x  0 , 0   0 , f x y  0 , 0   1, f y y  0 , 0   0 ,
D  00  0  0 2

D=0 gives no information, and f  0 , 0   0 is neither a maximum or a

minimum.
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