15
DIFFERENTIATION IN
SEVERAL VARIABLES
15.1 Functions of Two or More Variables
(ET Section 14.1)
Preliminary Questions
1.
What is the difference between a horizontal trace and a level curve? How are they related?
SOLUTION
A horizontal trace at height
c
consists of all points
(
x
,
y
,
c
)
such that
f
(
x
,
y
)
=
c
. A level curve is the
curve
f
(
x
,
y
)
=
c
in the
xy
plane. The horizontal trace is in the
z
=
c
plane. The two curves are related in the sense that
the level curve is the projection of the horizontal trace on the
xy
plane. The two curves have the same shape but they are
located in parallel planes.
2.
Describe the trace of
f
(
x
,
y
)
=
x
2

sin
(
x
3
y
)
in the
xz
plane.
SOLUTION
The intersection of the graph of
f
(
x
,
y
)
=
x
2

sin
(
x
3
y
)
with the
xz
plane is obtained by setting
y
=
0 in
the equation
z
=
x
2

sin
(
x
3
y
)
. We get the equation
z
=
x
2

sin 0
=
x
2
. This is the parabola
z
=
x
2
in the
xz
plane.
3.
Is it possible for two different level curves of a function to intersect? Explain.
SOLUTION
Two different level curves of
f
(
x
,
y
)
are the curves in the
xy
plane defined by equations
f
(
x
,
y
)
=
c
1
and
f
(
x
,
y
)
=
c
2
for
c
1
=
c
2
. If the curves intersect at a point
(
x
0
,
y
0
)
, then
f
(
x
0
,
y
0
)
=
c
1
and
f
(
x
0
,
y
0
)
=
c
2
, which
implies that
c
1
=
c
2
. Therefore, two different level curves of a function do not intersect.
4.
Describe the contour map of
f
(
x
,
y
)
=
x
with contour interval 1.
SOLUTION
The level curves of the function
f
(
x
,
y
)
=
x
are the vertical lines
x
=
c
. Therefore, the contour map of
f
with contour interval 1 consists of vertical lines so that every two adjacent lines are distanced one unit from another.
5.
How will the contour maps of
f
(
x
,
y
)
=
x
and
g
(
x
,
y
)
=
2
x
with contour interval 1 look different?
SOLUTION
The level curves of
f
(
x
,
y
)
=
x
are the vertical lines
x
=
c
, and the level curves of
g
(
x
,
y
)
=
2
x
are the
vertical lines 2
x
=
c
or
x
=
c
2
. Therefore, the contour map of
f
(
x
,
y
)
=
x
with contour interval 1 consists of vertical
lines with distance one unit between adjacent lines, whereas in the contour map of
g
(
x
,
y
)
=
2
x
(with contour interval
1) the distance between two adjacent vertical lines is
1
2
.
Exercises
In Exercises 1–4, evaluate the function at the specified points.
1.
f
(
x
,
y
)
=
x
+
yx
3
,
(
2
,
2
)
,
(

1
,
4
)
,
(
6
,
1
2
)
SOLUTION
We substitute the values for
x
and
y
in
f
(
x
,
y
)
and compute the values of
f
at the given points. This gives
f
(
2
,
2
)
=
2
+
2
·
2
3
=
18
f
(

1
,
4
)
=

1
+
4
·
(

1
)
3
=

5
f
6
,
1
2
=
6
+
1
2
·
6
3
=
114
2.
g
(
x
,
y
)
=
y
x
2
+
y
2
,
(
1
,
3
)
,
(
3
,

2
)
SOLUTION
We substitute
(
x
,
y
)
=
(
1
,
3
)
and
(
x
,
y
)
=
(
3
,

2
)
in the function to obtain
g
(
1
,
3
)
=
3
1
2
+
3
2
=
3
10
;
g
(
3
,

2
)
=

2
3
2
+
(

2
)
2
=

2
13
3.
h
(
x
,
y
,
z
)
=
xyz

2
,
(
3
,
8
,
2
)
,
(
3
,

2
,

6
)
SOLUTION
Substituting
(
x
,
y
,
z
)
=
(
3
,
8
,
2
)
and
(
x
,
y
,
z
)
=
(
3
,

2
,

6
)
in the function, we obtain
h
(
3
,
8
,
2
)
=
3
·
8
·
2

2
=
3
·
8
·
1
4
=
6
h
(
3
,

2
,

6
)
=
3
·
(

2
)
·
(

6
)

2
=

6
·
1
36
=

1
6
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C H A P T E R
15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
4.
Q
(
y
,
z
)
=
y
2
+
y
sin
z
,
(
y
,
z
)
=
(
2
,
π
2
), (

2
,
π
6
)
SOLUTION
We have
Q
2
,
π
2
=
2
2
+
2 sin
π
2
=
4
+
2
·
1
=
6
Q

2
,
π
6
=
(

2
)
2

2 sin
π
6
=
4

2
·
1
2
=
3
In Exercises 5–16, sketch the domain of the function.
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 Spring '08
 Hubscher
 Level set, F E R E N T, S E V E R A L VA R

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