Answer:
To find the directional derivative of the function f(x, y) = x/y² at the point p(3, −1) in the direction of the point q(−9, 15), we need to follow these steps:
First, we need to find the gradient of f at p, which is a vector that points in the direction of the greatest rate of change of f. The gradient is given by ∇f(x, y) = ⟨fx(x, y), fy(x, y)⟩, where fx and fy are the partial derivatives of f with respect to x and y respectively. To find the partial derivatives, we use the quotient rule:
fx(x, y) = (y² - 2xy(1))/(y²)² = (y - 2x)/(y³)
fy(x, y) = (-2x(y)(-1))/(y²)² = (2x)/(y³)
Next, we plug in the coordinates of p into the gradient to get ∇f(3, −1) = ⟨(−1 - 2(3))/((−1)³), (2(3))/((−1)³)⟩ = ⟨5, −6⟩.
Then, we need to find the unit vector that points in the direction from p to q. To do this, we subtract the coordinates of p from q to get a vector v = ⟨−9 - 3, 15 - (−1)⟩ = ⟨−12, 16⟩. Then, we divide this vector by its magnitude to get a unit vector u = v/∥v∥. The magnitude of v is given by ∥v∥ = √((-12)² + 16²) = √400 = 20. So,
u = v/∥v∥ = ⟨−12/20, 16/20⟩ = ⟨−0.6, 0.8⟩.
Finally, we use the formula for the directional derivative, which is
Explanation: