Final answer:
a. The gradient of f(x, y) is (-8x, -32y). b. The equation of the projection of the path of steepest descent on the xy-plane is x = 1 - 8t, y = 1 - 32t. c. The parametric equations for the path C' on the surface are x = 1 - 8t, y = 1 - 32t, z = 21 - 4(1 - 8t)² - 16(1 - 32t)².
Step-by-step explanation:
a. To find the gradient of f(x, y), we need to find the partial derivatives of f with respect to x and y.
The gradient is the vector formed by these derivatives. Partial derivative with respect to x: ∂f/∂x = -8x. Partial derivative with respect to y: ∂f/∂y = -32y. Therefore, the gradient of f(x, y) is (-8x, -32y).
b. The path of steepest descent on the surface is in the direction opposite to the gradient vector.
So, starting from point P(1, 1, 1), we move in the direction (-8, -32) on the xy-plane to find the equation of C.
The equation of C is x = 1 - 8t, y = 1 - 32t, where t is a parameter.
c. To find parametric equations for the path C' on the surface, we substitute the equations of C into the equation of the surface f(x, y). f(x, y) = 21 - 4x² - 16y².
Substituting x = 1 - 8t and y = 1 - 32t into f(x, y), we get f(t) = 21 - 4(1 - 8t)² - 16(1 - 32t)².
Therefore, the parametric equations for the path C' are x = 1 - 8t, y = 1 - 32t, and z = 21 - 4(1 - 8t)² - 16(1 - 32t)².