Multivariate Calculus - Bruce E. Shapiro

LECTURE 11. GRADIENTS AND THE DIRECTIONAL DERIVATIVE 79
Solution. The gradient vector is
( ∂
∇f = ∇(e y sin x) =
∂x ey sin x, ∂ )
∂y ey sin x = (e y cos x, e y sin x)
= e y (cos x, sin x)
At p = (5π/6, 0), we find that the direction of steepest increase is
(
∇f(p) = e 0 (cos(5π/6), sin(5π/6)) = − √ )
3/2, 1/2
and the magnitude of the increase is
√
‖∇f(p)‖ = ( √ 3/2) 2 + (1/2) 2 = √ 3/4 + 1/4 = 1
Since u = ∇f(p) has a magnitude of 1, it is a unit vector in the direction that f
increases most rapidly.
Example 11.7 Suppose that the temperature T (x, y, z) of a ball of some material
centered at the origin is given by the function
T (x, y, z) =
100
10 + x 2 + y 2 + z 2
Starting at the point (1, 1, 1), in what direction must you move to obtain the greatest
increase in temperature?
Solution. We need to move in the direction of the gradient.
∇T (x, y, z) = i ∂T
∂x + j∂T ∂y + k∂T ∂z
= i ∂ 100
∂x 10 + x 2 + y 2 + z 2 + j ∂ ∂y
100
10 + x 2 + y 2 + z 2
+k ∂ ∂z
The partial derivative with respect to x is
100
10 + x 2 + y 2 + z 2
∂ 100
∂x 10 + x 2 + y 2 + z 2 = 100 ∂
∂x (10 + x2 + y 2 + z 2 ) −1
= 100(−1)(10 + x 2 + y 2 + z 2 ) −2 ∂
∂x (10 + x2 + y 2 + z 2 )
=
−200x
(10 + x 2 + y 2 + z 2 ) 2
By symmetry,
∂ 100
∂y 10 + x 2 + y 2 + z 2 = −200y
(10 + x 2 + y 2 + z 2 ) 2
∂ 100
∂z 10 + x 2 + y 2 + z 2 = −200z
(10 + x 2 + y 2 + z 2 ) 2
Math 250, Fall 2006 Revised December 6, 2006.