Mathlab Coding and Finding the Directional Derivative of the Function

PROBLEM C2
Solution
The function is, f (x, y) = x2-xy
At (x, y) = (2, -3) in the direction of vector v= (35, 45)
First, you find the function's gradient and evaluate it at the specified point.
You differentiate the function with respect to each variable to determine the gradient of the function, which is a vector.
∇f (x, y) =( ∂f∂x, ∂f∂y)
∂f∂x = 2x – y , ∂f∂y = -x
The gradient at point (2, -3) is;
∇ (x2-xy) (x, y) = (2x-y, -x)
Substitute x = 2 and y = -3 in the equations for ∂f∂x and ∂f∂y,
that is ∂f∂x = 2x – y = 2(2) – (-3) = 7 and ∂f∂y = -x = -(2) = -2
= (7, -2)
Length of the vector ū =352+452=1 Since the length of the vector is 1 then it’s already normalized.
Finally, the directional derivative is the dot product of the gradient and the normalized vector. This is, D= (7, -2). ( 35, 45)
=135
PROBLEM C3
* The function is, f (x, y) = x2 - xy +3y2
You differentiate the function with respect to each variable to determine the gradient of the function, which is a vector.
∇f (x, y) =( ∂f∂x, ∂f∂y)
∂f∂x = 2x – y , ∂f∂y = -x + 6y
The gradient = (2x-y, -x + 6y)
* f (x, y...