S E C T I O N
15.5
The Gradient and Directional Derivatives
(ET Section 14.5)
677
(d)
If
f
(
x
,
y
)
were differentiable at
(
0
,
0
)
, then there would exist a function
(
h
,
k
)
satisfying
lim
(
h
,
k
)
→
(
0
,
0
)
(
h
,
k
)
=
0
such that
f
(
h
,
k
)
=
L
(
h
,
k
)
+
(
h
,
k
)
h
2
+
k
2
in a disk containing
(
0
,
0
)
. By part (c),
f
x
(
0
,
0
)
=
f
y
(
0
,
0
)
=
0, therefore
L
(
h
,
k
)
=
f
(
0
,
0
)
+
f
x
(
0
,
0
)
h
+
f
y
(
0
,
0
)
k
=
0
+
0
h
+
0
k
=
0
Therefore we get
f
(
h
,
k
)
=
(
h
,
k
)
h
2
+
k
2
(1)
We define the following functions:
R
1
(
h
,
k
)
=
(
h
,
k
)
·
h
h
2
+
k
2
;
R
2
(
h
,
k
)
=
(
h
,
k
)
·
k
h
2
+
k
2
We now show that
R
1
(
h
,
k
)
and
R
2
(
h
,
k
)
satisfy the required properties. First notice that
h
h
2
+
k
2
≤
h
√
h
2
=
1
and
k
h
2
+
k
2
≤
k
√
k
2
=
1
Therefore,
0
≤

R
1
(
h
,
k
)

≤

(
h
,
k
)

and
0
≤

R
2
(
h
,
k
)

≤

(
h
,
k
)

.
Since
lim
(
h
,
k
)
→
(
0
,
0
)
(
h
,
k
)
=
0, the Squeeze Theorem implies that also
lim
(
h
,
k
)
→
(
0
,
0
)
R
1
(
h
,
k
)
=
lim
(
h
,
k
)
→
(
0
,
0
)
R
2
(
h
,
k
)
=
0
We show that the conditions are satisfied:
hR
1
(
h
,
k
)
+
kR
2
(
h
,
k
)
=
h
2
(
h
,
k
)
h
2
+
k
2
+
k
2
(
h
,
k
)
h
2
+
k
2
=
h
2
+
k
2
(
h
,
k
)
h
2
+
k
2
=
(
h
,
k
)
h
2
+
k
2
Combining with (1) we get
f
(
h
,
k
)
=
hR
1
(
h
,
k
)
+
kR
2
(
h
,
k
).
(e)
Setting
h
=
k
, we get
f
(
h
,
h
)
=
hR
1
(
h
,
h
)
+
hR
2
(
h
,
k
)
or
f
(
h
,
h
)
h
=
R
1
(
h
,
h
)
+
R
2
(
h
,
h
)
Taking the limit as
h
→
0 gives
lim
h
→
0
f
(
h
,
h
)
h
=
lim
h
→
0
R
1
(
h
,
h
)
+
lim
h
→
0
R
2
(
h
,
h
)
=
0
+
0
=
0
However by the definition of
f
, we have for
h
=
0
f
(
h
,
h
)
=
2
h
2
(
h
+
h
)
h
2
+
h
2
=
2
h
⇒
f
(
h
,
h
)
h
=
2
So obviously lim
h
→
0
f
(
h
,
h
)
h
=
2
=
0. We arrive at a contradiction and conclude that
f
is not differentiable at
(
0
,
0
)
.
(f)
By the Criterion for Differentiability, the continuity of
f
x
and
f
y
at
(
0
,
0
)
implies that
f
(
x
,
y
)
is differentiable at
(
0
,
0
)
. In part (e) we showed that
f
is not differentiable at
(
0
,
0
)
, hence
f
x
and
f
y
are not continuous at this point.
15.5 The Gradient and Directional Derivatives
(ET Section 14.5)
Preliminary Questions
1.
Which of the following is a possible value of the gradient
∇
f
of a function
f
(
x
,
y
)
of two variables?
(a)
5
(b)
3
,
4
(c)
3
,
4
,
5
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678
C H A P T E R
15
DIFFERENTIATION IN SEVERAL VARIABLES
(ET CHAPTER 14)
SOLUTION
The gradient of
f
(
x
,
y
)
is a vector with two components, hence the possible value of the gradient
∇
f
=
∂
f
∂
x
,
∂
f
∂
y
is (b).
2.
True or false: A differentiable function increases at the rate
∇
f
P
in the direction of
∇
f
P
?
SOLUTION
The statement is true. The value
∇
f
P
is the rate of increase of
f
in the direction
∇
f
P
.
3.
Describe the two main geometric properties of the gradient
∇
f
.
SOLUTION
The gradient of
f
points in the direction of maximum rate of increase of
f
and is normal to the level curve
(or surface) of
f
.
4.
Express the partial derivative
∂
f
∂
x
as a directional derivative
D
u
f
for some unit vector
u
.
SOLUTION
The partial derivative
∂
f
∂
x
is the following limit:
∂
f
∂
x
(
a
,
b
)
=
lim
t
→
0
f
(
a
+
t
,
b
)

f
(
a
,
b
)
t
Considering the unit vector
i
=
1
,
0 , we find that
∂
f
∂
x
(
a
,
b
)
=
lim
t
→
0
f
(
a
+
t
·
1
,
b
+
t
·
0
)

f
(
a
,
b
)
t
=
D
i
f
(
a
,
b
)
We see that the partial derivative
∂
f
∂
x
can be viewed as a directional derivative in the direction of
i
.
5.
You are standing at point where the temperature gradient vector is pointing in the northeast (NE) direction. In which
direction(s) should you walk to avoid a change in temperature?
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 Spring '08
 Hubscher
 Chain Rule, Derivative, Gradient, F E R E N T

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