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

SectION 14.6 Directional Derivatives and the Gradient Vector 955
so Equation 19 becomes
f x sx 0 , y 0 dsx 2 x 0 d 1 f y sx 0 , y 0 dsy 2 y 0 d 2 sz 2 z 0 d − 0
which is equivalent to Equation 14.4.2. Thus our new, more general, definition of a tangent
plane is consistent with the definition that was given for the special case of Section 14.4.
EXAMPLE 8 Find the equations of the tangent plane and normal line at the point
s22, 1, 23d to the ellipsoid
x 2
4 1 y 2 1 z2
9 − 3
SOLUTION The ellipsoid is the level surface (with k − 3) of the function
Figure 10 shows the ellipsoid, tangent
plane, and normal line in Example 8.
4
2
0
z 2
4
Therefore we have
F x sx, y, zd − x 2
Fsx, y, zd − x 2
4 1 y 2 1 z2
9
F y sx, y, zd − 2y
F z sx, y, zd − 2z
9
F x s22, 1, 23d − 21 F y s22, 1, 23d − 2 F z s22, 1, 23d − 2 2 3
Then Equation 19 gives the equation of the tangent plane at s22, 1, 23d as
21sx 1 2d 1 2sy 2 1d 2 2 3 sz 1 3d − 0
6
0 2 2 0 2
y
x
FIGURE 10 10
7et140610
05/04/10
MasterID: 01610
y
level curve
f(x, y)=k
0 x
7et140611–12
P(x¸, y¸)
±f(x¸, y¸)
FIGURE 11 11
FIGURE 12
which simplifies to 3x 2 6y 1 2z 1 18 − 0.
By Equation 20, symmetric equations of the normal line are
x 1 2
21 − y 2 1
2
Significance of the Gradient Vector
− z 1 3
2 2
3
We now summarize the ways in which the gradient vector is significant. We first consider
a function f of three variables and a point Psx 0 , y 0 , z 0 d in its domain. On the one hand,
we know from Theorem 15 that the gradient vector =f sx 0 , y 0 , z 0 d gives the direction of
fastest increase of f. On the other hand, we know that =f sx 0 , y 0 , z 0 d is orthogonal to the
level surface S of f through P. (Refer to Figure 9.) These two properties are quite compatible
intu itively because as we move away from P on the level surface S, the value of
f does not change at all. So it seems reasonable that if we move in the perpendicular
direction, we get the maximum increase.
In like manner we consider a function f of two variables and a point Psx 0 , y 0 d in its
domain. Again the gradient vector =f sx 0 , y 0 d gives the direction of fastest increase of f.
Also, by considerations similar to our discussion of tangent planes, it can be shown that
=f sx 0 , y 0 d is perpendicular to the 300 level curve f sx, yd − k that passes through P. Again
curve of
this is intuitively plausible because 200
steepest
the values of f remain constant as we move along the
curve. (See ascent Figure 11.)
100
If we consider a topographical map of a hill and let f sx, yd represent the height above
sea level at a point with coordinates sx, yd, then a curve of steepest ascent can be drawn
Copyright 2016 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
■