Suppose f(x,y)=xy, P=(−3,−4) and v=−4i−2j. Find the (unit) direction vector in which the maximum rate of change occurs at P.

Answer 1

To find the direction in which the function f(x,y) = xy has its maximum rate of change at point P(-3,-4), we'll use the gradient of f.

1. First, compute the partial derivatives:

df/dx = y and

df/dy = x.

2. Evaluate the partial derivatives at the point P(-3,-4):

df/dx at (-3,-4) = -4 and

df/dy at (-3,-4) = -3.

So, the gradient of f at P(-3,-4) is given by the vector:

Gradient f = <-4,-3>.

3. The gradient gives the direction of maximum rate of change. To get the unit vector in this direction, we'll normalize the gradient:

Magnitude of the gradient at P = sqrt((-4)^2 + (-3)^2) = sqrt(16 + 9) = sqrt(25) = 5.

Unit direction vector of maximum rate of change = Gradient f / Magnitude = <-4/5, -3/5>.

So, the unit direction vector in which the maximum rate of change occurs at P is <-4/5, -3/5>.