Explanation:
To find the directional derivative of the function h(r, s, t) = ln(3r + 6s + 9t) at the point (2,2,2) in the direction of the vector v = 8i + 24j + 12k, we can use the formula:
Dv(h) = ∇h · v
where ∇h is the gradient vector of the function h.
To find the gradient vector ∇h, we take the partial derivatives of h with respect to each variable r, s, and t:
∂h/∂r = 3/(3r + 6s + 9t)
∂h/∂s = 6/(3r + 6s + 9t)
∂h/∂t = 9/(3r + 6s + 9t)
Thus, the gradient vector ∇h is:
∇h = (3/(3r + 6s + 9t))i + (6/(3r + 6s + 9t))j + (9/(3r + 6s + 9t))k
At the point (2,2,2), the gradient vector ∇h is:
∇h(2,2,2) = (3/24)i + (6/24)j + (9/24)k
= (1/8)i + (1/4)j + (3/8)k
Now, we can find the directional derivative Dv(h) in the direction of the vector v as follows:
Dv(h) = ∇h · v
= ((1/8)i + (1/4)j + (3/8)k) · (8i + 24j + 12k)
= (1/8)(8) + (1/4)(24) + (3/8)(12)
= 3
Therefore, the directional derivative of the function h(r, s, t) = ln(3r + 6s + 9t) at the point (2,2,2) in the direction of the vector v = 8i + 24j + 12k is 3.