694
CHAPTER 15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
50.
Find a function
f
(
x
,
y
,
z
)
such that
∇
f
= h
z
,
2
y
,
x
i
.
SOLUTION
f
(
x
,
y
,
z
)
=
xz
+
y
2
is a good choice.
51.
Find a function
f
(
x
,
y
)
such that
∇
f
= h
y
,
x
i
.
We must ±nd a function
f
(
x
,
y
)
such that
∇
f
=
¿
∂
f
∂
x
,
∂
f
∂
y
À
= h
y
,
x
i
That is,
∂
f
∂
x
=
y
,
∂
f
∂
y
=
x
We integrate the ±rst equation with respect to
x
.Since
y
is treated as a constant, the constant of integration is a function
of
y
.Weget
f
(
x
,
y
)
=
Z
ydx
=
yx
+
g
(
y
)
(1)
We differentiate
f
with respect to
y
and substitute in the second equation. This gives
∂
f
∂
y
=
∂
∂
y
(
+
g
(
y
))
=
x
+
g
±
(
y
)
Hence,
x
+
g
±
(
y
)
=
x
⇒
g
±
(
y
)
=
0
⇒
g
(
y
)
=
C
Substituting in (1) gives
f
(
x
,
y
)
=
+
C
One of the solutions is
f
(
x
,
y
)
=
(obtained for
C
=
0).
52.
Show that there does not exist a function
f
(
x
,
y
)
such that
∇
f
=

y
2
,
x
®
.
Hint:
Use Clairaut’s Theorem
f
xy
=
f
.
Suppose that for some differentiable function
f
(
x
,
y
)
,
∇
f
=

f
x
,
f
y
®
=
D
y
2
,
x
E
That is,
f
x
=
y
2
and
f
y
=
x
. Therefore,
f
=
∂
∂
y
f
x
=
∂
∂
y
y
2
=
2
y
and
f
=
∂
∂
x
f
y
=
∂
∂
x
x
=
1
Since
f
and
f
are both continuous, they must be equal by Clairaut’s Theorem. Since