James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

SectION 14.6 Directional Derivatives and the Gradient Vector 953
The unit vector in the direction of PQ l − k2 3 2 , 2l is u − k2 3 5 , 4 5 l, so the rate of change
of f in the direction from P to Q is
D u f s2, 0d − =f s2, 0d ? u − k1, 2l ? k2 3 5 , 4 5 l
− 1(2 3 5) 1 2( 4 5) − 1
(b) According to Theorem 15, f increases fastest in the direction of the gradient vector
=f s2, 0d − k1, 2 l. The maximum rate of change is
| =f s2, 0d | − | k1, 2l | − s5 ■
At s2, 0d the function in Example 6
increases fastest in the direction of the
gradient vector =f s2, 0d − k1, 2 l.
Notice from Figure 7 that this vector
appears to be perpendicular to the level
curve through s2, 0d. Figure 8 shows the
graph of f and the gradient vector.
y
Q
2
1
±f(2, 0)
0 1 P 3 x
20
15
z 10
5
0
0 1
x
2
3 0
1
y
2
FIGURE 77 FIGURE 88
7et140607
7et140608
EXAMPLE 7 Suppose that the temperature at a point sx, y, zd in space is given by
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Tsx, y, zd − 80ys1 1 x
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2 1 2y 2 1 3z 2 d, where T is measured in degrees Celsius and
MasterID: x, y, z in meters. 01607 In which direction does the temperature MasterID: increase 01608 fastest at the point
s1, 1, 22d? What is the maximum rate of increase?
SOLUTION The gradient of T is
=T − −T
−x i 1 −T
−y j 1 −T
−z k
160x
− 2
s1 1 x 2 1 2y 2 1 3z 2 d i 2 2
−
320y
s1 1 x 2 1 2y 2 1 3z 2 d 2 j 2
160
s1 1 x 2 1 2y 2 1 3z 2 2
s2x i 2 2y j 2 3z kd
d
At the point s1, 1, 22d the gradient vector is
=Ts1, 1, 22d − 160
256 s2i 2 2 j 1 6 kd − 5 8 s2i 2 2 j 1 6 kd
480z
s1 1 x 2 1 2y 2 1 3z 2 d 2 k
By Theorem 15 the temperature increases fastest in the direction of the gradient vector
=Ts1, 1, 22d − 5 8 s2i 2 2 j 1 6 kd or, equivalently, in the direction of 2i 2 2 j 1 6 k
or the unit vector s2i 2 2 j 1 6 kdys41. The maximum rate of increase is the length of
the gradient vector:
| =Ts1, 1, 22d | − 5 8 | 2i 2 2 j 1 6 k | − 5 8 s41
Therefore the maximum rate of increase of temperature is 5 8 s41 < 48Cym.
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