Question

F(x,y)=ylnx+xy 2

,P(1,2),θ= 4 /π
Find the gradient of f. Evaluate the gradient at the point P. Use the gradient to find the directional derivative in the directron of u=⟨cosθi+sinθj⟩

Answer 1

To find the gradient of f(x,y) = yln(x) + xy^2, we take the partial derivative with respect to each variable. Evaluating the gradient at P(1,2) gives us (2,4). The directional derivative in the direction of u = <cos(theta)i + sin(theta)j> is 2cos(theta) + 4sin(theta).

Step-by-step explanation:

The gradient of a function f(x,y) can be found by taking the partial derivative with respect to each variable. In this case, f(x,y) = yln(x) + xy^2. Taking the partial derivative with respect to x gives us y/x, and taking the partial derivative with respect to y gives us ln(x) + 2xy.

To evaluate the gradient at the point P(1,2), we substitute x=1 and y=2 into the partial derivatives. So the gradient at P(1,2) is (2/1, ln(1) + 2*1*2) = (2, ln(1) + 4) = (2,4).

To find the directional derivative in the direction of u = <cos(theta)i + sin(theta)j>, we take the dot product of the gradient and u. So the directional derivative is (2,4) · (<cos(theta), sin(theta)>) = 2cos(theta) + 4sin(theta).