Question

In which direction's û , if any, is the rate of change of f(x,y) = x2 + sin(xy) at the point (1,0) equal to 1? If any, find the direction/s. If none, then specify that.

Answer 1

To find the direction's û where the rate of change of f(x,y) = x^2 + sin(xy) at the point (1,0) is equal to 1, we need to find the gradient vector at that point and normalize it to get the direction vector.

Step-by-step explanation:

To find the direction's û where the rate of change of f(x,y) = x^2 + sin(xy) at the point (1,0) is equal to 1, we need to find the gradient vector at that point and normalize it to get the direction vector. The gradient vector is found by taking the partial derivatives of the function with respect to x and y. Evaluating the gradient vector at (1,0), we normalize it to get the direction vector.

First, we find the partial derivative of f(x,y) with respect to x:

∂f/∂x = 2x + ycos(xy)

Then, we find the partial derivative of f(x,y) with respect to y:

∂f/∂y = xcos(xy)

Evaluating the partial derivatives at (1,0), we get:

∂f/∂x = 2(1) + 0cos(0) = 2

∂f/∂y = 1cos(0) = 1

So, the gradient vector at (1,0) is (2, 1). Normalizing the gradient vector, we divide each component by the magnitude of the gradient vector:

û = (2/√(2^2+1^2), 1/√(2^2+1^2)) = (2/√5, 1/√5)

Therefore, the direction's û where the rate of change of f(x,y) is equal to 1 at the point (1,0) is (2/√5, 1/√5).