Question

Suppose f(x,y)=1x^2+3xy−2y^2, P=(3,2), and u=(−5/13,12/13).

A. Compute the gradient of f.
∇f=___i+___j
Note: Your answers should be expressions of x and y; e.g. "3x - 4y"

B. Evaluate the gradient at the point P.
(∇f)(3,2)=___i+___j
Note: Your answers should be numbers

C. Compute the directional derivative of f at P in the direction u .
(Duf)(P)=

Note: Your answer should be a number

Answer 1

A. The gradient of $f(x,y)$ is given by $∇f(x,y)=f_x(x,y) \vec{i}+f_y(x,y) \vec{j}.$

$f(x,y)=1x^2+3xy−2y^2,$

$f_x(x,y)=2x+3y,$

$f_y(x,y)=3x-4y.$

So we have:

$∇f(x,y)=(2x+3y) \vec{i}+(3x-4y) \vec{j}.$

Answer. $∇f=(2x+3y) \vec{i}+(3x-4y) \vec{j}.$

B. $∇f(x_0,y_0)=f_x(x_0,y_0) \vec{i}+f_y(x_0,y_0) \vec{j}.$ Hence at the point $P(3,2)$ we have:

$∇f(3,2)=(2 \cdot 3+3 \cdot 2) \vec{i}+ (3 \cdot 3-4 \cdot 2) \vec{j}= 12 \vec{i} +1 \vec{j}.$

Answer. $∇f(3,2)= 12 \vec{i} +1 \vec{j}.$

C. $D_u f(P)= ∇f(3,2) \cdot \vec{u},$ where $\vec{u}$ is a unit vector. Let's check it.

$||u||=\sqrt{(-\frac{5}{13})^2+(\frac{12}{13})^2}=\sqrt{\frac{25}{169}+\frac{144}{169}}=1.$

So $\vec{u}=-\frac{5}{13} \vec{i}+\frac{12}{13} \vec{j}$ is a unit vector.

$D_u f(3, 2)=(12 \vec{i} +1 \vec{j}) \cdot (-\frac{5}{13} \vec{i}+\frac{12}{13} \vec{j})= 12 \cdot (-\frac{5}{13})+1 \cdot (\frac{12}{13})=$

$=-\frac{60}{13}+ \frac{12}{13}= - \frac{48}{13}=-3 \frac{9}{13}.$

Answer. $D_u f(P) = -3 \frac{9}{13}.$