Final answer:
Yes. The directional derivative of the function at point P in the direction of point Q is 0.
Step-by-step explanation:
To find the directional derivative of the function at P in the direction of Q, we first need to find the gradient of the function.
The gradient of a function f(x, y) is given by ∇f = (∂f/∂x, ∂f/∂y).
In this case, f(x, y) = cos(x + y), so ∂f/∂x = -sin(x + y) and ∂f/∂y = -sin(x + y).
Now, we substitute the coordinates of point P(0, π) into the gradient, resulting in (-sin(0 + π), -sin(0 + π)) = (-sin(π), -sin(π)) = (0, 0).
Next, we substitute the coordinates of point Q(π/2, 0) into the direction vector, resulting in (π/2, 0).
To find the directional derivative in the direction of Q, we dot product the gradient with the direction vector: D = ∇f . Q = (0, 0) . (π/2, 0) = 0.
Therefore, the directional derivative of the function at P in the direction of Q is 0.