Answer:
Explanation:
(a) To find the rate of change of V at P in the direction of vector v, we need to compute the directional derivative of V at P in the direction of v, denoted by ∇v V(P).
The directional derivative of V in the direction of v is given by the dot product of the gradient of V at P with v:
∇v V(P) = ∇V(P) · v
where ∇V(P) is the gradient of V at P, which is a vector that points in the direction of the maximum rate of change of V at P, and its magnitude is the maximum rate of change at P.
To compute ∇V(P), we take the partial derivatives of V with respect to x, y, and z:
∂V/∂x = 10x - 2y + yz
∂V/∂y = -2x + xz
∂V/∂z = xy
So,
∇V(P) = (10(3) - 2(6) + (6)(4))i + (-2(3) + (3)(4))j + (3)(6)k
= 34i - 6j + 18k
The unit vector in the direction of v is
|v| = √(1^2 + 1^2 + (-1)^2) = √3
so the direction of v is
v/|v| = (1/√3)i + (1/√3)j - (1/√3)k
Therefore, the rate of change of V at P in the direction of v is
∇v V(P) = ∇V(P) · v/|v| = (34i - 6j + 18k) · ((1/√3)i + (1/√3)j - (1/√3)k)/√3
= (28√3)/3
Hence, the rate of change of V at P in the direction of v is (28√3)/3.
(b) The direction in which V changes most rapidly at P is given by the direction of the gradient vector ∇V(P). This is because the gradient vector points in the direction of the maximum rate of change of V at P, and its magnitude is the maximum rate of change.
We have already computed ∇V(P) in part (a):
∇V(P) = 34i - 6j + 18k
So the direction in which V changes most rapidly at P is the direction of the vector (34i - 6j + 18k)/|∇V(P)|, which simplifies to
(2i - (1/3)j + (2/3)k)/√3
(c) The maximum rate of change of V at P is given by the magnitude of the gradient vector ∇V(P).
We have already computed ∇V(P) in part (a):
|∇V(P)| = √(34^2 + (-6)^2 + 18^2) = 2√466
So the maximum rate of change of V at P is 2√466.