Multivariate Calculus - Bruce E. Shapiro

LECTURE 11. GRADIENTS AND THE DIRECTIONAL DERIVATIVE 75
Definition 11.5 The gradient of a function f(x, y) : R 2 ↦→ R of two variables is
given by
∇f(x, y) = gradf(x, y) = i ∂f
∂x + j∂f
(11.4)
∂y
The gradient of a function f(x, y, z) : R 3 ↦→ R of three variables is
∇f(x, y) = gradf(x, y) = i ∂f
∂x + j∂f ∂y + k∂f ∂z
(11.5)
Note that if we apply the definition in three-dimensions to a function of two variables,
we obtain the same result as the first definition, because the partial derivative
of f(x, y) with respect to z is zero (z does not appear in the equation).
Example 11.1 Find the gradient of f(x, y) = sin 3 (x 2 y)
Solution.
∇f = i ∂f
∂x + j∂f ∂y = i ∂
∂x sin3 (x 2 y) + j ∂ ∂y sin3 (x 2 y)
[
= i 3 sin 2 (x 2 y)) ∂ ] [
∂x sin(x2 y) + j 3 sin 2 (x 2 y)) ∂ ]
∂y sin(x2 y)
= i [ 3 sin 2 (x 2 y)) cos(x 2 y)(2xy) ] + j [ 3 sin 2 (x 2 y)) cos(x 2 y)(x 2 ) ]
= 3x sin 2 ( x 2 y ) cos ( x 2 y ) (2yi + xj)
Example 11.2 Find the gradient of f(x, y) = x 2 y + y 2 z + z 2 x
Solution.
∇f = i ∂f
∂x + j∂f ∂y + k∂f ∂z
= i ∂ (
x 2 y + y 2 z + z 2 x ) + j ∂ (
x 2 y + y 2 z + z 2 x ) +
∂x
∂y
k ∂ (
x 2 y + y 2 z + z 2 x )
∂z
= i(2xy + z 2 ) + j(x 2 + 2yz) + k(y 2 + 2zx)
Theorem 11.1 Properties of the Gradient VectorSuppose that f and g are
functions and c is a constant. Then the following are true:
∇[f + g] = ∇f + ∇g (11.6)
∇(cf) = c∇f (11.7)
∇(fg) = f∇g + g∇f (11.8)
Math 250, Fall 2006 Revised December 6, 2006.