The directional derivative of g(x, y, z) at the point (4, 3, 3) in the direction of the vector v = 4j - k is 522/sqrt(17).
To find the directional derivative of the function:
1. Define the function and the direction vector:
Function: g(x, y, z) = (x⁴y³z)^(3/2)
Point: P = (4, 3, 3)
Direction vector: v = 4j - k
2. Calculate the gradient of the function:
The gradient of a scalar field represents the direction of the steepest ascent and its magnitude is the rate of change in that direction. It is calculated as:
∇g(x, y, z) = (∂g/∂x, ∂g/∂y, ∂g/∂z)
For our function:
∂g/∂x = 6x²y³z^(3/2)
∂g/∂y = 3x⁴y²z^(3/2)
∂g/∂z = 3x⁴y³z^(1/2)
Therefore, the gradient at point P:
∇g(P) = (144, 108, 54)
3. Normalize the direction vector:
To find the directional derivative, we need the unit vector in the direction of v.
u = v / ||v||
u = (4j - k) / ||4j - k||
||4j - k|| = sqrt(4^2 + 1^2) = sqrt(17)
u = (4/sqrt(17))j - (1/sqrt(17))k
4. Calculate the directional derivative:
The directional derivative is the dot product of the gradient and the normalized direction vector.
D_ug(P) = ∇g(P) · u
D_ug(P) = (144, 108, 54) · (4/sqrt(17))j - (1/sqrt(17))k
D_ug(P) = 576/sqrt(17) - 54/sqrt(17)
D_ug(P) = 522/sqrt(17)
Therefore, the directional derivative of g(x, y, z) at the point (4, 3, 3) in the direction of the vector v = 4j - k is 522/sqrt(17).