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

SectION 14.6 Directional Derivatives and the Gradient Vector 949
On the other hand, we can write tshd − f sx, yd, where x − x 0 1 ha, y − y 0 1 hb, so
the Chain Rule (Theorem 14.5.2) gives
t9shd − −f
−x
dx
dh 1 −f
−y
If we now put h − 0, then x − x 0 , y − y 0 , and
dy
dh − f xsx, yd a 1 f y sx, yd b
5 t9s0d − f x sx 0 , y 0 d a 1 f y sx 0 , y 0 d b
Comparing Equations 4 and 5, we see that
D u f sx 0 , y 0 d − f x sx 0 , y 0 d a 1 f y sx 0 , y 0 d b
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If the unit vector u makes an angle with the positive x-axis (as in Figure 2), then we
can write u − kcos , sin l and the formula in Theorem 3 becomes
6 D u f sx, yd − f x sx, yd cos 1 f y sx, yd sin
The directional derivative D u f s1, 2d
in Example 2 represents the rate of
change of z in the direction of u. This
is the slope of the tangent line to the
curve of intersection of the surface
z − x 3 2 3xy 1 4y 2 and the vertical
plane through s1, 2, 0d in the direction
of u shown in Figure 5.
z
EXAMPLE 2 Find the directional derivative D u f sx, yd if
f sx, yd − x 3 2 3xy 1 4y 2
and u is the unit vector given by angle − y6. What is D u f s1, 2d?
SOLUTION Formula 6 gives
D u f sx, yd − f x sx, yd cos 6 1 f ysx, yd sin 6
− s3x 2 2 3yd s3
2 1 s23x 1 8yd 1 2
− 1 2 f3 s3 x 2 2 3x 1 s8 2 3s3 dyg
0
x
FIGURE 55
(1, 2, 0)
π
6
u
y
Therefore
D u f s1, 2d − 1 2 f3s3 s1d 2 2 3s1d 1 s8 2 3s3 ds2dg −
13 2 3s3
2
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7et140605
05/04/10
MasterID: 01605
The Gradient Vector
Notice from Theorem 3 that the directional derivative of a differentiable function can be
written as the dot product of two vectors:
7 D u f sx, yd − f x sx, yd a 1 f y sx, yd b
− k f x sx, yd, f y sx, yd l ? ka, b l
− k f x sx, yd, f y sx, yd l ? u
The first vector in this dot product occurs not only in computing directional deriv atives
but in many other contexts as well. So we give it a special name (the gradient of f ) and
a special notation (grad f or =f , which is read “del f ”).
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