Page 71 - Vector Analysis
P. 71
§2.8 Directional Derivatives and Gradient Vectors 67
Solution: Let f (x, y, z) = x2 ´ y2 + xyz. Then
S = ␣(x, y, z) P R3 | f (x, y, z) = f (1, 0, 1)( ;
that is, S is a level set of f . Since (∇f )(1, 0, 1) = (2, 1, 0) ‰ (0, 0, 0), (2, 1, 0) is normal to
S at (1, 0, 1); thus the tangent plane of S at (1, 0, 1) is 2(x ´ 1) + y = 0. ˝
Proposition 2.84. Let f : Rn Ñ R be differentiable. If (∇f )(x0) ‰ 0, then ˘ (∇f )(x0)
}(∇f )(x0)}Rn
is the direction in which the function f increases/decreases most rapidly（最速上升／下降
方向）at x0.
Proof. Let x0 P Rn be given. Suppose that f increases most rapidly in the direction v,
then (Dvf )(x0) = sup (Dwf )(x0). Since f is differentiable, (Dwf )(x0) = (Df )(x0)(w) =
}w}Rn =1
(∇f )(x0) ¨ w which is maximized in the direction (∇f )(x0) . ˝
}(∇f )(x0)}Rn
Example 2.85. Let f : R3 Ñ R be given by f (x, y, z) = x2y sin z. Find the direction of
the greatest rate of change at (3, 2, 0).
Solution: We compute the gradient of f at (3, 2, 0) as follows:
(∇f )(3, 2, 0) = ( Bf (3, 2, 0), Bf (3, 2, 0), Bf (3, 2, )
0)
Bx By Bz
= (2xy sin z, x2 sin z, x2y cos z)ˇˇ(x,y,z)=(3,,2,0) = (0, 0, 18).
Therefore, the direction of the greatest rate of change of f at (3, 2, 0) is (0, 0, 1).
