Question

Suppose f(x,y)=xy, P=(−4,−4) and v=2i+3j. A. Find the gradient of f. ∇f= i+ j Note: Your answers should be expressions of x and y; e.g. "3x - 4y" B. Find the gradient of f at the point P. (∇f)(P)= i+ j Note: Your answers should be numbers C. Find the directional derivative of f at P in the direction of v. Duf= Note: Your answer should be a number D. Find the maximum rate of change of f at P. Note: Your answer should be a number E. Find the (unit) direction vector in which the maximum rate of change occurs at P. u= i+ j Note: Your answers should be numbers

Answer 1

Answer:

a) The gradient of a function is the vector of partial derivatives. Then

$\nabla f=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})=(y,x)=y\hat{i} + x\hat{j}$

b) It's enough evaluate P in the gradient.

$\nabla f(P)=(-4,-4)=-4\hat{i} - 4 \hat{j}$

c) The directional derivative of f at P in direction of V is the dot produtc of $\nabla f(P)$ and v.

$\nabla f(P) v=(-4,-4)\left[\begin{array}{ccc}2\\3\end{array}\right] =(-4)2+(-4)3=-20$

d) The maximum rate of change of f at P is the magnitude of the gradient vector at P.

$||\nabla f(P)||=\sqrt{(-4)^2+(-4)^2}=\sqrt{32}=4\sqrt{2}$

e) The maximum rate of change occurs in the direction of the gradient. Then

$v=\frac{1}{4\sqrt{2}}(-4,-4)=(\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}})= \frac{-1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}$

is the direction vector in which the maximum rate of change occurs at P.