14 DIFFERENTIATION
INSEVERAL
VARIABLES
n this chapter, we extend the concepts and techniques of differential calculus to
functions of several variables. As we will see, a function f that depends on two or
more variables has not just one derivative but rather a set of partial derivatives, one for
each variable. The partial derivatives are the components of the gradient vector, which
provides valuable insight into the function's behavior. In the last two sections, we apply
the tools we have developed to optimization in several variables.
I
The circulation of weather system s around
areas of low pressure can be understood
using the gradient vector , an important tool
arising in multivariable differentiation.
14.1 Functions
ofTwoorMoreVariables
A familiar example of a function of two variables is the area A of a rectangle, equal to
the product xy of the base x and height y. We write
A(x, y)
= xy
or A= f(x, y), where f(x, y) = xy. An example in three variables is the distance from
a point P = (x, y, z) to the origin:
g(x,y,z)
= /x2 + y2 + z2
An important but less familiar example is the density of seawater, denoted p, which
is a function of salinity S and temperature T and is a key factor in the makeup of ocean
current systems (Figure 1). Although there is no simple formula for p(S, T), scientists
determine values of the function experimentally. According to Table 1, if S = 32 (in
parts per thousand or ppt) and T = l0 °C, then
p(32, 10) = 1.0246 kg/m 3
FIGURE
1 The global climate is influenced
by the ocean "conveyer belt," a system of
deep currents driven by variations in
seawater density.
TABLE1 SeawaterDensityp
(kg/m3 ) as a Function
of Temperature
andSalinity.
Salinity (ppt)
oc
32
32.5
33
5
10
15
20
1.0253
1.0246
1.0237
1.0224
1.0257
1.0250
1.0240
1.0229
1.0261
1.0254
1.0244
1.0232
A function of n variables is a function f that assigns a real number f(xi, ... , Xn) to
each n-tuple (xi, ... , Xn) in a domain in Rn. Sometimes we write f ( P) for the value of
f at a point P = (xi, ... , Xn). When f is defined by an algebraic expression involving
xi, ... , Xn, we usually take as the domain the set of all n-tuples for which f (xi, ... , Xn)
is defined. The range of f is the set of all values f (xi, ... , Xn) for (xi, ... , Xn) in the
domain. Since we focus on functions of two or three variables, we shall often use the
variables x, y, and z (rather than xi, x2, x3).
EXAMPLE
1 Sketch the domains of:
(a) f (x, y)
= J9- x 2 -
y
(b) g(x, y, z)
= x,jy + ln(z -
1)
What are the ranges of these functions?
789
790 C H A P T E R 14
DIFFERENTIATION IN SEVERALVARIABLES
Solution
Jg-
(a) f(x, y) =
x 2 - y is defined only when 9 - x 2 - y > 0, or y < 9 - x 2 . Thus,
the domain consists of all points (x, y) lying on or below the parabola y = 9 - x 2
[Figure 2(A)]:
V
=
{(x, y): y < 9 - x 2 }
To determine the range, note that f is a nonnegative function and that f (0, y) =
- y.
Since 9 - y can be any nonnegative number, f (0, y) takes on all nonnegative values.
Therefore, the range off is the infinite interval [0, oo ).
(b) g(x, y, z) =
+ ln(z - 1) is defined only when both
and ln(z - 1) are defined. Therefore, both y > 0 and z > 1 are required, so the domain of the function is
given by {(x, y, z) : y > 0, z > 1} [Figure 2(B)]. The range of g is the entire real line R.
Indeed, for the particular choices y = l and z = 2, we have g(x, l, 2) = x
+ In 1 = x,
and since x is arbitrary, we see that g takes on all values.
y
9
1
.··· y
(A) The domain of f(x, y) = 9 - x 2 - y
is the set of all points lying below
the parabola y = 9 - x2.
(B) The domain of g(x , y, z) = x{y + ln(z - 1)
is the set of points with y > 0 and z > l.
The domain continues out to infinity in
the directions indicated by the arrows.
FIGURE2
Graphing
Functions
of TwoVariables
In single-variable calculus, we use graphs to visualize the important features of a function
[Figure 3(A)]. Graphs play a similar role for functions of two variables. The graph of a
function f of two variables consists of all points (a, b, f (a, b)) in R 3 for (a, b) in the
domain V off. Assuming that f is continuous (as defined in the next section), the graph
is a surface whose height above or below the xy-plane at (a, b) is the value of the function
f(a, b) [Figure 3(B)]. We often write z = f(x, y) to stress that the z-coordinate of a point
on the graph is a function of x and y.
EXAMPLE2 Sketch the graph off (x, y)
= 2x 2 + 5y 2 .
Solution The graph is a paraboloid (Figure 4), which we saw in Section 12.6. We sketch
the graph using the fact that the horizontal cross section at height z = c is the ellipse
2x 2 +5y 2 =c.
Plotting more complicated graphs by hand can be difficult. Fortunately, graphing
technology (e.g., graphing calculators, computer algebra systems) eliminates the labor
and greatly enhances our ability to explore functions graphically. Graphs can be rotated
and viewed from different perspectives (Figure 5).
S E C T I O N 14.1
Functions
of Twoor MoreVariables 791
z
z
y
------------x
a
y
y
X
(A) Graph of y
= f(x)
(B) Graph of z =f(x, y)
X
FIGURE
4 Graph of
f (x, y) = 2x 2 + 5y 2 .
FIGURE
3
z
z
z
---X
ml FIGURE5
Different views of the graph of g(x, y)
= e-x 2 -Y 2 -
e-(x-1)
2
-(y-1)
2
•
Traces
One way of analyzing the graph of a function f (x, y) is to freeze the x-coordinate by
setting x = a and examine the resulting curve given by z = f (a, y). Similarly, we may
set y = b and consider the curve z = f (x, b). Curves of this type are called vertical
traces. They are obtained by intersecting the graph with planes parallel to a vertical
coordinate plane (Figure 6):
z
z
Vertical trace
atx=a
Vertical trace
aty=b
y
x=a
X
(A) Vertical traces parallel to yz-plane
y
X
y=b
(B) Vertical traces parallel to xz-plane
[[i FIGURE6
• Vertical trace in the plane x = a: Intersection of the graph with the vertical
plane x = a, consisting of all points (a, y, f (a, y))
• Vertical trace in the plane y = b: Intersection of the graph with the vertical
plane y = b, consisting of all points (x, b, f(x, b))
792 C H A P T E R 14
DIFFERENTIATION IN SEVERALVARIABLES
EXAMPLE 3 Describe the vertical traces of f (x, y) = x (sin y).
Solution When we freeze the x-coordinate by setting x = a, we obtain the trace curve
z = a(siny) (see Figure 7). This is a sine curve with amplitude lal, located in the plane
x =a.When
we set y = b, we obtain a line z = x(sinb) of slope sinb, located in the
plane y = b.
z
z
z = x sin y
(A) The traces in the planes x = a
are the curves z = a(sin y).
ml FIGURE7 Vertical traces of
f (X, Y)
z = x sin y
= X ( Siny ).
(B) The traces in the planes y
are the lines z = x (sin b).
=b
EXAMPLE4 IdentifyingFeaturesof a Graph Match the graphs in Figure 8 with the
following functions:
(i) f (X, y)
=X-
y2
(ii) g(x, y)
= x2 -
y
Solution Let's compare vertical traces. The vertical trace of f(x, y) = x - y 2 in the
plane x = a is a downward parabola z = a - y 2 . This matches (B). On the other hand,
the vertical trace of g(x, y) in the plane y =bis an upward parabola z = x 2 - b. This
matches (A).
Notice also that f(x, y) = x - y 2 is an increasing function of x [i.e., f (x, y) increases as x increases] as in (B), whereas g(x, y) = x 2 - y is a decreasing function of y
as in (A).
z
z
Upward parabolas
y = b, z =x 2 -b
I
Downward parabolas
x=a, z=a-y 2
\
y
z
Horizontal trace
atz=c
X
y
Decreasing in
positive y-direction
(A)
y
X
Level curve f (x, y) = c
ml FIGURE9 The level curve consists of all
points (x, y) where the function takes on
the value c.
Increasing in
positive x-direction
(B)
FIGURE8
LevelCurvesandContour
Maps
In addition to vertical traces, the graph of f (x, y) has horizontal traces. These traces and
their associated level curves are especially important in analyzing the behavior of the
function (Figure 9):
S E C T I O N 14.1
Functions
of Twoor MoreVariables 793
Level curves
close together
z
z
Steep part
of graph
z =f(x, y)
y
y
X
Flatter part
of graph
Level curves
farther apart
(B) Horizontal traces
(A)
(C) Contour map
FIGURE10
On contour maps, level curves are often
referred to as contourlines. When we refer
to level curves on a contour map, we mean
the curves that are actually displayed.
Keep in mind that between the displayed
level curves there are additional curves
associated with other values off.
z
• Horizontal trace at height c: Intersection of the graph with the horizontal plane
z = c, consisting of the points (x, y, f (x, y)) such that f (x, y) = c
• Level curve: The curve f(x, y) = c in the xy-plane
Thus, the level curve corresponding to c consists of all points (x, y) in the domain of f
in the xy-plane where the function takes the value c. Each level curve is the projection
onto the xy-plane of the horizontal trace on the graph that lies above it.
A contour map is a plot in the domain in the xy-plane that shows the level curves
f(x, y) = c for equally spaced values of c. The interval m between the values of c is
called the contour interval. When you move from one level curve to the next, the value
of f(x, y) (and hence the height of the graph) changes by ± m.
Figure 10 compares the graph of a function f (x, y) in (A) and its horizontal traces
in (B) with the contour map in (C). The contour map in (C) has contour interval
m = 100.
It is important to understand how the contour map indicates the steepness of the
graph. If the level curves are close together, then a small move from one level curve to
the next in the xy-plane leads to a large change in height. In other words, the level curves
are close together if the graph is steep (Figure 10). Similarly, the graph is flatter when
the level curves are farther apart.
EXAMPLE5 EllipticParaboloid Sketch the contour map of
f(x, y)
y
and comment on the spacing of the contour curves.
Solution The level curves have equation f(x, y)
X
x
y
X
FIGURE11 f(x, y)
interval m = l O.
= x 2 + 3y 2
= x 2 + 3y 2 . Contour
2
= c, or
+ 3y 2 = C
• For c > 0, the level curve is an ellipse.
• For c = 0, the level curve is just the point (0, 0) because x 2 + 3y 2 = 0 only for
(x, y) = (0, 0).
• There is no level curve for c < 0 because f (x, y) is never negative.
The graph off (x, y) is an elliptic paraboloid (Figure 11). As we move away from
the origin, f(x, y) increases more rapidly. The graph gets steeper, and the level curves
become closer together.
794 C H A P T E R 14
DIFFERENTIATION IN SEVERALVARIABLES
EXAMPLE6
+-' REMINDERThe hyperbolic
paraboloid
in Figure 12 is often called a "saddle" or
"saddle-shaped surface."
HyperbolicParaboloid Sketch the contour map of
g(x, y)
= x2 -
Solution The level curves have equation g(x, y)
x2
-
3y 2
3y 2
= c, or
=C
• For c # 0, the level curve is the hyperbola x 2 - 3y 2 = c.
• For c = 0, the level curve consists of the two lines x =
tion g(x, y) = 0 factors as follows:
+
x 2 - 3y 2 = (x z
because the equa-
= 0
The graph of g(x, y) is a hyperbolic paraboloid (Figure 12). When you stand at the origin,
g(x, y) increases as you move along the x-axis in either direction and decreases as you
move along the y-axis in either direction. Furthermore, the graph gets steeper as you
move out from the origin, so the level curves grow closer together.
EXAMPLE7 Contour
Mapof a LinearFunction Sketch the graph of
f (x, y)
=
12 - 2x - 3y
and the associated contour map with contour interval m
X
g(x , y)
decreasing
c
= 30
I
1
g(x, y)
1ncreas1ng
~
g(x, y)
.
1ncreas1ng
C
:
= -30
interval m
= 10.
How can we measure steepness of the graph of a function quantitatively? Let's imagine the surface given by z = f (x, y) as a mountain (Figure 14). We place the xy-plane at
sea level, so that f (a, b) is the height (also called altitude or elevation) of the mountain
above sea level at the point (a, b) in the plane.
Figure 14(A) shows two points P and Qin the xy-plane, together with the points P
and Q on the graph that lie above them. We define the average rate of change:
,-...,;
g(x, ~
y
decreasing
FIGURE12 g(x, y)
Solution Note that if we set z = f(x, y), we can write the equation as 2x + 3y + z =
12. As we discussed in Section 12.5, this is the equation of a plane. To plot the graph, we
find the intercepts of the plane with the axes (Figure 13). The graph intercepts the z-axis at
z = f (0, 0) = 12. To find the x-intercept, we set y = z = 0 to obtain 12 - 2x - 3(0) =
0, or x = 6. Similarly, solving 12 - 3y = 0 gives they-intercept y = 4. The graph is the
plane determined by the three intercepts.
In general, the level curves of a linear function f (x, y) = qx + ry + s are the
lines with equation qx + ry + s = c. Therefore, the contour map of a linear function
consists of equally spaced parallel lines. In our case, the level curves are the lines
12 - 2x - 3y = c, or 2x + 3y = 12 - c (Figure 13).
I
X
.
= 4.
=x2 -
,-...,;
3y 2 . Contour
average rate of change from P to Q
where .L\ altitude
from P to Q.
=
,-...,;
.L\ altitude
= ---. --
.L\horizontal
,-...,;
change in the height from P and Q, and .L\ horizontal
=
distance
CONCEPTUAL
INSIGHT We will discuss the idea that rates of change depend on direction when we come to directional derivatives in Section 14.5. In single-variable calculus, we measure the rate of change by the derivative f'(a). In the multivariable case,
there is no single rate of change because the change in f(x, y) depends on the direction:
The rate is zero along a level curve [because f (x, y) is constant along level curves],
and the rate is nonzero in directions pointing from one level curve to the next [Figure
14(B)].
S E C T I O N 14.1
~
-----------
Functions
of Twoor MoreVariables 795
c=20
-----
C
= 16
c= 12
c=8
c=4
----
I
I
I
horizontal
c=O
V:
I I
I I
I ~---
y
X
I
I
I
2100 2000
D
c=20
c= 16
c= 12
c=8
c=4
c=O
c=-4
y
(Interval m = 4)
Contour interval: 0.8 km
Horizontal scale: 2 km '------'
A contour map is like a topographic map
that hikers would use to help understand
the terrain that they encounter. They are
both two-dimensional representations of
the features of three-dimensional
structures.
A
along the level curve
A-----C400
200
B
Contour interval: 100 m
Horizontal scale: 200 m --
(A)
X
FIGURE13 Graph and contour
map off (x, y) = 12 - 2x - 3y.
Function does not change
(B)
FIGURE14
EXAMPLE8 AverageRateof ChangeDepends
on Direction Compute the average
rate of change from A to the points B, C, and Din Figure 14(B).
Solution The contour interval in Figure 14(B) ism = 100 m. Segments AB and AC both
span two level curves, so the change in altitude is 200 m in both cases. The horizontal
scale shows that AB corresponds to a horizontal change of 200 m, and AC corresponds
to a horizontal change of 400 m. On the other hand, there is no change in altitude from A
to D. Therefore,
altitude
200
= = l.O
average rate of change from A to B = ----horizontal
200
altitude
200
= = 0.5
average rate of change from A to C = ----horizontal
400
average rate of change from A to D
altitude
horizontal
= -----
=0
We see here explicitly that the average rate varies according to the direction.
A path of steepest descent is the same as a
path of steepest ascent traversed in the
opposite direction. Water flowing down a
mountain approximately follows a path of
steepest descent.
When we walk up a mountain, the incline at each moment depends on the path we
choose. If we walk around the mountain, our altitude does not change at all. On the
other hand, at each point there is a steepest direction in which the altitude increases most
rapidly. On a contour map, the steepest direction is approximately the direction that takes
us to the closest point on the next highest level curve [Figure 15(A)]. We say "approximately" because the terrain may vary between level curves. A path of steepest ascent
is a path that begins at a point P and, everywhere along the way, points in the steepest
direction. We can approximate the path of steepest ascent by drawing a sequence of
796 C H A P T E R 14
DIFFERENTIATION IN SEVERALVARIABLES
segments that move as directly as possible from one level curve to the next. Figure 15(B)
shows two paths from P to Q. The solid path is a path of steepest ascent, but the dashed
path is not, because it does not move from one level curve to the next along the shortest
possible segment.
10
Approximate path
of steepest ascent
starting at P
(A) Vectors pointing approximately
in the direction of steepest ascent
steepest ascent
(B)
FIGURE15
MoreThanTwoVariables
z
x2 + Y2 + z2 = 1
x2 + Y2 + z2 = 4
y
There are many modeling situations where it is necessary to use a function of more than
two variables. For instance, we might want to keep track of temperature at the various
points in a room using a function T(x, y, z) that depends on the three variables corresponding to the coordinates of each point. In making quantitative models of the economy,
functions often depend on more than 100 variables.
Unfortunately, it is not possible to draw the graph of a function of more than
two variables. The graph of a function f (x, y, z) would consist of the set of points
(x, y, z, f (x, y, z)) in four-dimensional space R4 . However, just as we can use contour
maps to visualize a three-dimensional mountain using curves on a two-dimensional
plane, it is possible to draw the level surfaces of a function of three variables f (x, y, z).
These are the surfaces with equation f(x, y, z) = c for different values of c. For example,
the level surf aces of
f(x,y,z)
x2 + Y2 + z2 = 9
FIGURE16 The level surfaces of
f(x, y, z) = x 2 + y 2 + z 2 are spheres.
= x2 + y2 + z2
are the spheres with equation x 2 + y 2 + z 2 = c (Figure 16). In the case of a function
T (x, y, z) that represents temperature of points in space, we call the level surfaces corresponding to T (x, y, z) = k the isotherms. These are the collections of points, all of which
have the same temperature k.
For functions of four or more variables, we can no longer visualize the graph or the
level surfaces. We must rely on intuition developed through the study of functions of two
and three variables.
EXAMPLE9 Describe the level surfaces of g(x, y, z) = x 2 + y 2 - z 2.
Solution The level surface for c = 0 is the cone x 2 + y 2 - z 2 = 0. For c #-0, the level
surfaces are the hyperboloids x 2 + y 2 - z 2 = c. The hyperboloid has one sheet if c > 0
and it lies outside the cone. The hyperboloid has two sheets if c < 0, one sheet lies inside
the upper part of the cone, the other lies inside the lower part (Figure 17).
Functions
of Twoor MoreVariables 797
S E C T I O N 14.1
z
z
z
y
g(x , y, z)
=c
z
y
(c > 0)
g(x,y , z)
FIGURE17 Level surfaces of g(x, y, z)
= x 2 + y2 -
y
=0
g(x , y, z)
=c
y
(c < 0)
z2 .
14.1 SUMMARY
• The domain V of a function f (x 1, ... , Xn) of n variables is the set of n-tuples
(a1, ... , an) in Rn for which f (a1, ... , an) is defined. The range of f is the set
of values taken on by f.
• The graph of a continuous real-valued function f(x, y) is the surface in R3 consisting
of the points (a, b, f(a, b)) for (a, b) in the domain V off.
• A vertical trace is a curve obtained by intersecting the graph with a vertical plane
x = a or y = b.
• A level curve is a curve in the xy-plane defined by an equation f (x, y) = c. The level
curve f (x, y) = c is the projection onto the xy-plane of the horizontal trace curve,
obtained by intersecting the graph with the horizontal plane z = c.
• A contour map shows the level curves f (x, y) = c for equally spaced values of c. The
spacing m is called the contour interval.
• When reading a contour map, keep in mind:
- Your altitude does not change when you hike along a level curve.
- Your altitude increases or decreases by m (the contour interval) when you hike
from one level curve to the next.
• The spacing of the level curves indicates steepness: They are closer together where
the graph is steeper.
~altitude
• The average rate of change from P to Q is the ratio
.
.
~honzontal
• A direction of steepest ascent at a point P is a direction along which f (x, y) increases
most rapidly. The steepest direction is obtained (approximately) by drawing the segment from P to the nearest point on the next level curve.
• Level surfaces can be used to understand a function f(x, y, z). In the case where the
function represents temperature, we call the level surfaces isotherms.
14.1 EXERCISES
Preliminary
Questions
1. What is the difference between a horizontal trace and a level curve?
How are they related?
2.
Describe the trace of f(x, y)
=x
2
-
3
sin(x y) in the xz-plane .
3. Is it possible for two different level curves of a function to intersect?
Explain.
4.
Describe the contour map of f (x, y)
5.
How will the contour maps of
f(x, y)
=x
and
with contour interval 1 look different?
=x
with contour interval 1.
g(x, y)
= 2x
798 C H A P T E R 14
DIFFERENTIATION IN SEVERALVARIABLES
Exercises
z
In Exercises 1-6, at each point evaluate the function or indicate that the
function is undefined there.
1.
f(x,y)=x+yx3,
2.
g(x, y) =
z
(1, 2), (-1, 6), (e, rr)
y
x2 _ y2'
(1,3), (3,-3),
(v"2,2)
y
3.
h(x, y) =
4.
k(x, y)
5.
h(x, y, z)
6.
Jx-y2
'
x-y
= xe-Y,
(A)
(1, 0), (3, -3), (0, 12)
= xyzr-s
= .
w(r,s,t)
X
(20, 2), (1, -2), (1, 1)
t
Sln
¼),(4, -4,
2,
(3, 7, -2), (3, 2,
,
(2,2, ;), (rr,rr,rr),
0)
( -2,2,
FIGURE19
21. Match the functions (a)-(f) with their graphs (A)-(F) in Figure 20.
~)
(a) f(x, y)
(b) f(x, y)
In Exercises 7-14, sketch the domain of the function.
7.
9.
f(x, y)
f(x, y)
11. g(y, z)
=
= ln(4x 2 =
8.
12x - 5y
1
12. f(x, y)
2
z+y
f(x, y)
10. h(x, t)
y)
(B)
=
- x2
1
x +t
= --
= sin
= lxl + IYI
= cos(x - y)
-1
(c) f(x, y)
= 1 + 9x2 + y2
(d) f(x, y)
= cos(y2)e-O.l(x2+y2)
-1
(e) f(x, y)
= 1 + 9x2 + 9y2
(f)
= cos(x2 + y2)e-O.l(x2+y2)
f(x, y)
y
X
14. f(x, y)
13. F(I,R)=ffi
= cos-
z
1(x
z
+ y)
In Exercises 15-18, describe the domain and range of the function.
15. f(x, y, z)
= xz + eY
16. f(x, y, z)
17. P(r,s, t)
= J16
18. g(r,s)
- r 2s2t 2
=
= cos-
+ zezfx
19. Match graphs (A) and (B) in Figure 18 with the functions:
(i)
f(x, y)
(ii) g(x, y)
X
1 (rs)
= -x + y 2
= X + y2
(A)
(B)
z
z
z
X
(C)
y
X
(D)
z
z
(E)
(F)
X
(A)
(B)
FIGURE18
20. Match each of graphs (A) and (B) in Figure 19 with one of the following functions:
(i)
f(x,y)=(cosx)(cosy)
(ii) g(x, y) = cos(x 2 + y 2)
FIGURE20
S E C T I O N 14.1
y
22. Match the functions (a)-(d) with their contour maps (A)-(D) m
Figure 21.
(a) f(x, y)
= 3x + 4y
(b) g(x, y)
(c) h(x,y)=4x-3y
(d) k(x, y)
10
10
5
5
0
0
-5
-5
-10
-10
-10
0
-5
5
10
= x3 = x2 -
Functions
of Twoor MoreVariables 799
y
y
c=6
c=0
FIGURE
22 Contour map with contour interval m
"-
-10
-5
(A)
0
40. Use the contour map in Figure 23 to calculate the average rate of
change:
(a) from A to B .
(b) from A to C.
10
5
= 6.
(B)
y
10
10
5
5
------
......................., . ....
··••i••········· ·····••:••···
:c
...........
..............
0
0
-5
-5
I
-10
-10
0
-5
5
!c=-3
················
-10
10
-10
-5
(C)
0
5
-6
10
12 - 3x - 4 y
= sin(x
24. f(x,y)=J4-x2-y2
- y)
28. f (x, y)
29. Sketch contour maps off (x, y) = x
and 2.
30. Sketch the contour map off
4, 8, 12, 16.
31. f (x, y)
=
33. f(x, y)
= y2
=
2
X
42. For each of A-C indicate in which of the four cardinal directions,
N, S, E, or W, pressure is increasing the greatest.
1
1020
2
+y + 1
+ y with contour intervals m =
10
1
;::___
1024
(x, y) = x 2
+
y2
with level curves c = 0,
/
= xy
35. f(x, y)
= x 2 + 4y 2
36. f(x, y)
= X + 2y
37. f(x, y)
=
38. f (x, y)
=
1012
l
~- --------
1012
(
1016
1008
y
34. f(x, y)
X
l
I
1020
=-
x2
1028
1
=
y
6
43. Rank the following states in order from greatest change in pressure
across the state to least: Arkansas, Colorado, North Dakota, Wisconsin.
32. f(x, y)
y
4
41. (a) At which of A-C is pressure increasing in the northern direction?
(b) At which of A-C is pressure increasing in the westerly direction?
In Exercises 31-38, draw a contour map of f(x, y) with an appropriate
contour interval, showing at least six level curves.
x2 -
2
Exercises 41-4-3 refer to the map in Figure 24.
26. f(x, y)
27. f(x, y)
-2
FIGURE
23
In Exercises 23-28, sketch the graph and draw several vertical and horizontal traces.
=
-4
(D)
FIGURE
21
23. f (x, y)
·················"·····
I
2
X
3x 2 -
- 1
y2
39.
Find the linear function whose contour map (with contour interval m = 6) is shown in Figure 22. What is the linear function if m = 3
(and the curve labeled c = 6 is relabeled c = 3)?
FIGURE24 Atmospheric pressure (in millibars) over North America
on March 26, 2009.
In Exercises 44-4-7, let T(x, y, z) denote temperature at each point in
space. Draw level suifaces (also called isotherms) corresponding to the
fixed temperatures given.
+ 3y - z, T = 0, 1, 2
- y + 2z, T = 0, 1,2
44. T (x, y, z) = 2x
45. T(x,y,z)
= x
800 C H A P T E R 14
46. T(x,y,z)
DIFFERENTIATION IN SEVERALVARIABLES
+ y 2 - z, T = 0, 1,2
x 2 - y 2 + z2 , T = 0, 1,2, -1,
= x2
47. T(x, y, z) =
In Exercises 52-55, refer to Figure 26.
52. Find the change in seawater density from A to B.
-2
In Exercises 48-51, p(S, T) is seawater density (kilograms per cubic meter) as a function of salinity S (parts per thousand) and temperature T
(degrees Celsius). Refer to the contour map in Figure 25.
48. Calculate the average rate of change of p with respect to T from B
to A.
53. Estimate the average rate of change from A to B and from A
to C.
54. Estimate the average rate of change from A to points i, ii, and iii.
55. Sketch the path of steepest ascent beginning at D.
49. Calculate the average rate of change of p with respect to S from B
to C.
50. At a fixed level of salinity, is seawater density an increasing or a decreasing function of temperature?
51. Does water density appear to be more sensitive to a change in temperature at point A or point B?
25
'
I
r1,')\)
I
....\~
!.,
,,-..._
u.._
20
N
a.>
I,'
15 ,
B
a
,,,,,,
,,,
10
5
,'
/
0
...,"
/
)'
V
/
,,,
,,V
,.,....
...
/
I
\
.
.
V
If
/
32.5 33.0 33.5
Salinity (ppt)
/
V
,/
Contour interval = 20 m
,..,.
.....
...
---
1
2km
FIGURE26
56. Let temperature in 3-space be given by T (x, y, z) = x 2 + y 2
Draw isotherms corresponding to temperatures T = - 2, -1, 0, 1, 2.
I I/
\,·
0
.....
_.....i--
a_,6':> ,,,
·,
\)
...,..,,
/
,,,,,.,.
~'
32.0
_......,,
,,,.,.
,/
...,,,,,..
·,
A
....
v
,,,,..-,.,,..
.....
: \ \)1.,~9
,,,,...
,,,,
l/
.....
,..,.
,..,.
__,I,.,-
~':>
4-
.,,, \ ·~
/"
V
•
I/
....,
\) I
L/
,,,.,.
,/
I
\ \)1.,'S
/
I,'
/
/
/
............ .\.I\')1.,~
I
..........
I
.-,..,.
,.,,,.. """' _,.,,,..
---
.
,.,,.. ·t
I/
..........
I ':>.....
\ ~1::' ')___
....
V
I/'
,.,,,..
L..,
..\. \)1.,')
, ~\)
........
.........
..........
,,,,,V
I/
31.5
,,,,..
V
,,, V
/
L..,, ...
.....
V
.,,,,,.
...
,,,,.,.
,,;"
/
-,..,.
L..,
...,.,,,..
...,.i--
/
V
,..,.
........
V
1-4
a.>
.,,
.,,
0
.,,
--
I
I
,..,.
...
I (\~ - ~l); ,- I'
V
{
/
34.0
-
34.5
FIGURE25 Contour map of seawater density p(S, T) (kilograms per
cubic meter).
57. Let temperature in 3-space be given by T(x, y, z) =
Draw isotherms corresponding to temperatures T = 0, 1, 2.
-
z.
x;+ Y:+ z
58. Let temperature in 3-space be given by T(x, y, z) = x 2
Draw isotherms corresponding to temperatures T = -1, 0, 1.
-
y2
59. Let temperature in 3-space be given by T(x, y, z) = x 2 - y 2
Draw isotherms corresponding to temperatures T = - 2, -1, 0, 1, 2.
-
-
2
.
z.
z2 .
FurtherInsights
andChallenges
60.
The function f(x, t) = t- 112 e-x ft, whose graph is shown in
Figure 27, models the temperature along a metal bar after an intense burst
of heat is applied at its center point.
2
(a) Sketch the vertical traces at times t
Temperature T
=
1, 2, 3. What do these traces tell
us about the way heat diffuses through the bar?
Timet
(b) Sketch the vertical traces x = c for c = ±0.2, ±0.4. Describe how
temperature varies in time at points near the center.
61. Let
-0.4_0.2 0
0.2 0.4
X
f(x,y)=
Write f as a function
level curves of f.
JX 2 +y 2
f (r, 0)
for (x, y) =I=(0, 0)
in polar coordinates, and use this to find the
FIGURE27 Graph of f (x, t)
t = 0.
= t- 112 e- x 2 It beginning
shortly after
14.2 LimitsandContinuity
in SeveralVariables
This section develops limits and continuity in the multivariable setting. We focus on
functions of two variables, but similar definitions and results apply to functions of three
or more variables.