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

952 Chapter 14 Partial Derivatives
(b) At s1, 3, 0d we have =f s1, 3, 0d − k0, 0, 3 l . The unit vector in the direction of
v − i 1 2 j 2 k is
Therefore Equation 14 gives
u − 1
s6
i 1 2
s6
j 2 1
s6
k
D u f s1, 3, 0d − =f s1, 3, 0d ? u
− 3k ?S 1
s6
i 1 2
s6
j 2 1
s6
kD
− 3S2
1
s6D−2Î 3 2
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Maximizing the Directional Derivative
Suppose we have a function f of two or three variables and we consider all possible
directional derivatives of f at a given point. These give the rates of change of f in all
possible directions. We can then ask the questions: in which of these directions does f
change fastest and what is the maximum rate of change? The answers are provided by the
following theorem.
TEC Visual 14.6B provides visual
confirmation of Theorem 15.
15
Theorem Suppose f is a differentiable function of two or three variables.
The maximum value of the directional derivative D u f sxd is | =f sxd | and it
occurs when u has the same direction as the gradient vector =f sxd.
Proof From Equation 9 or 14 we have
D u f − =f ? u − | =f || u | cos − | =f | cos
where is the angle between =f and u. The maximum value of cos is 1 and this
occurs when − 0. Therefore the maximum value of D u f is | =f | and it occurs when
− 0, that is, when u has the same direction as =f .
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EXAMPLE 6
(a) If f sx, yd − xe y , find the rate of change of f at the point Ps2, 0d in the direction
from P to Qs 1 2 , 2d.
(b) In what direction does f have the maximum rate of change? What is this maximum
rate of change?
SOLUTION
(a) We first compute the gradient vector:
=f sx, yd − k f x , f y l − ke y , xe y l
=f s2, 0d − k1, 2 l
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