Question

I am walking on the elliptic paraboloid: z=16-2x²-y². The projection of my path on the x-y plane follows the ellipse (at right) 4x²+y²=4. At what point is my path the steepest (going up or going down) and what is the slope of the climb at that point?

Answer 1

The point on the ellipse where the path is the steepest is (1/2, 2), and the slope of the climb at that point is -4.

Step-by-step explanation:

To find the point on the elliptic paraboloid where the path is the steepest, we need to determine the gradient of the surface function z=16-2x²-y². The gradient vector will point in the direction of maximum increase, so the negative of the gradient vector will point in the direction of maximum decrease or steepest climb. So, we need to find the negative gradient vector and evaluate it at the point on the ellipse where 4x²+y²=4.

The gradient of the surface function is ∇z = (-4x, -2y).

Substituting the equation of the ellipse into the surface function, we have z=16-2(1/4)(4-y²)-y² = 16-2-y²-y²=12-2y².

To find the point on the ellipse where the path is the steepest, we need to evaluate -∇z at that point, which is (-4(1/2), -2(2)) = (-2,-4).

So, the point on the ellipse where the path is the steepest is (1/2, 2), and the slope of the climb at that point is -4.