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# Directional Derivatives and Gradient Vectors - Exercises, Partial Derivatives Notes | EduRev

## : Directional Derivatives and Gradient Vectors - Exercises, Partial Derivatives Notes | EduRev

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14.5 Directional Derivatives and Gradient Vectors 1013
EXERCISES 14.5
In Exercises 1–4, find the gradient of the function at the given point.
Then sketch the gradient together with the level curve that passes
through the point.
1. 2. ƒsx, yd = ln sx
2
+ y
2
d, s1, 1d ƒsx, yd = y - x, s2, 1d
3. 4.
In Exercises 5–8, find at the given point.
5.
6. ƒsx, y, zd = 2z
3
- 3sx
2
+ y
2
dz + tan
-1
xz, s1, 1, 1d
ƒsx, y, zd = x
2
+ y
2
- 2z
2
+ z ln x, s1, 1, 1d
§f
gsx, yd =
x
2
2
-
y
2
2
, A22, 1B gsx, yd = y - x
2
, s -1, 0d
4100 AWL/Thomas_ch14p965-1066  8/25/04  2:53 PM  Page 1013
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14.5 Directional Derivatives and Gradient Vectors 1013
EXERCISES 14.5
In Exercises 1–4, find the gradient of the function at the given point.
Then sketch the gradient together with the level curve that passes
through the point.
1. 2. ƒsx, yd = ln sx
2
+ y
2
d, s1, 1d ƒsx, yd = y - x, s2, 1d
3. 4.
In Exercises 5–8, find at the given point.
5.
6. ƒsx, y, zd = 2z
3
- 3sx
2
+ y
2
dz + tan
-1
xz, s1, 1, 1d
ƒsx, y, zd = x
2
+ y
2
- 2z
2
+ z ln x, s1, 1, 1d
§f
gsx, yd =
x
2
2
-
y
2
2
, A22, 1B gsx, yd = y - x
2
, s -1, 0d
4100 AWL/Thomas_ch14p965-1066  8/25/04  2:53 PM  Page 1013
7.
8.
Finding Directional Derivatives
In Exercises 9–16, find the derivative of the function at in the
direction of A.
9.
10.
11.
12.
13.
14.
15.
16.
Directions of Most Rapid Increase and Decrease
In Exercises 17–22, find the directions in which the functions increase
and decrease most rapidly at Then find the derivatives of the func-
tions in these directions.
17.
18.
19.
20.
21.
22.
Tangent Lines to Curves
In Exercises 23–26, sketch the curve together with
and the tangent line at the given point. Then write an equation for the
tangent line.
23. 24.
25. 26.
Theory and Examples
27. Zero directional derivative In what direction is the derivative
of at P(3, 2) equal to zero?
28. Zero directional derivative In what directions is the derivative
of at P(1, 1) equal to zero?
29. Is there a direction u in which the rate of change of
2
- 3xy + 4y
2
ƒsx, yd =
ƒsx, yd = sx
2
- y
2
d>sx
2
+ y
2
d
ƒsx, yd = xy + y
2
x
2
- xy + y
2
= 7, s -1, 2d xy =-4, s2, -2d
x
2
- y = 1, A22, 1B x
2
+ y
2
= 4, A22, 22B
§f ƒsx, yd = c
hsx, y, zd = ln sx
2
+ y
2
- 1d + y + 6z, P
0
s1, 1, 0d
ƒsx, y, zd = ln xy + ln yz + ln xz, P
0
s1, 1, 1d
gsx, y, zd = xe
y
+ z
2
, P
0
s1, ln 2, 1>2d
ƒsx, y, zd = sx>yd - yz, P
0
s4, 1, 1d
ƒsx, yd = x
2
y + e
xy
sin y, P
0
s1, 0d
ƒsx, yd = x
2
+ xy + y
2
, P
0
s -1, 1d
P
0
.
A = i + 2j + 2k
hsx, y, zd = cos xy + e
yz
+ ln zx, P
0
s1, 0, 1>2d,
gsx, y, zd = 3e
x
cos yz, P
0
s0, 0, 0d, A = 2i + j - 2k
ƒsx, y, zd = x
2
+ 2y
2
- 3z
2
, P
0
s1, 1, 1d, A = i + j + k
ƒsx, y, zd = xy + yz + zx, P
0
s1, -1, 2d, A = 3i + 6j - 2k
A = 3i - 2j
hsx, yd = tan
-1
sy>xd +23 sin
-1
sxy>2d, P
0
s1, 1d,
A = 12i + 5j
gsx, yd = x - s y
2
>xd +23 sec
-1
s2xyd, P
0
s1, 1d,
ƒsx, yd = 2x
2
+ y
2
, P
0
s -1, 1d, A = 3i - 4j
ƒsx, yd = 2xy - 3y
2
, P
0
s5, 5d, A = 4i + 3j
P
0
ƒsx, y, zd = e
x + y
cos z + s y + 1d sin
-1
x, s0, 0, p>6d
ƒsx, y, zd = sx
2
+ y
2
+ z
2
d
-1>2
+ ln sxyzd, s -1, 2, -2d
30. Changing temperature along a circle Is there a direction u in
which the rate of change of the temperature function
(temperature in degrees Celsius, distance in feet) at
31. The derivative of ƒ(x, y) at in the direction of is
and in the direction of is What is the derivative of
32. The derivative of ƒ(x, y, z) at a point P is greatest in the direction
of In this direction, the value of the derivative is
b. What is the derivative of ƒ at P in the direction of
33. Directional derivatives and scalar components How is the
derivative of a differentiable function ƒ(x, y, z) at a point in the
direction of a unit vector u related to the scalar component of
34. Directional derivatives and partial derivatives Assuming that
the necessary derivatives of ƒ(x, y, z) are defined, how are
and related to and Give reasons for your
35. Lines in the xy-plane Show that
is an equation for the line in the xy-plane through the point
normal to the vector
36. The algebra rules for gradients Given a constant k and the
and
use the scalar equations
and so on, to establish the following rules.
a.
b.
c.
d.
e. §a
ƒ
g
b =
g§ƒ - ƒ§g
g
2
§sƒgd = ƒ§g + g§ƒ
§sƒ - gd =§ƒ -§g
§sƒ + gd =§ƒ +§g
§skƒd = k§ƒ

0
0x
sƒgd = ƒ
0g
0x
+ g
0ƒ
0x
,
0
0x
a
ƒ
g
b =
g
0ƒ
0x
- ƒ
0g
0x
g
2
,

0
0x
skƒd = k
0ƒ
0x
,
0
0x
sƒ ; gd =
0ƒ
0x
;
0g
0x
,
§g =
0g
0x
i +
0g
0y
j +
0g
0z
k,
§ƒ =
0ƒ
0x
i +
0ƒ
0y
j +
0ƒ
0z
k
N = Ai + Bj. sx
0
, y
0
d
Asx - x
0
d + Bsy - y
0
d = 0
ƒ
z
? ƒ
x
, ƒ
y
, D
k
ƒ D
j
ƒ,
D
i
ƒ,
s§ƒd
P 0
P
0
i + j?
§ƒ
213.
v = i + j - k.
-i - 2j?
-3. -2j 212
i + j P
0
s1, 2d
-3°C>ft ? Ps1, -1, 1d
2xy - yz
Tsx, y, zd =
1014 Chapter 14: Partial Derivatives
4100 AWL/Thomas_ch14p965-1066  8/25/04  2:53 PM  Page 1014
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