Consider F and C below.
F(x, y, z) = (2xz + y2) i + 2xy j + (x2 + 15z2) k
C: x = t2, y = t + 2, z = 3t − 1, 0 ≤ t ≤ 1
(a) Find a function f such that F = ∇f.
f(x, y, z) =
x2z+xy2+5z3+C
Correct: Your answer is correct.
(b) Use part (a) to evaluate
C ∇f · dr
along the given curve C.
To evaluate the line integral, we need to evaluate C ∇f · dr along the curve C.
First, let's find ∇f by taking the partial derivatives of f with respect to x, y, and z.
∂f/∂x = 2xz + y^2,
∂f/∂y = 2xy,
∂f/∂z = x^2 + 15z^2.
Now that we have ∂f/∂x, ∂f/∂y, and ∂f/∂z, we can express the gradient ∇f = ∂f/∂x i + ∂f/∂y j + ∂f/∂z k.
So, the gradient vector ∇f is (2xz + y^2) i + 2xy j + (x^2 + 15z^2) k.
Now, we can evaluate C ∇f · dr by substituting the given parameterization of the curve C into the gradient vector ∇f.
C: x = t^2, y = t + 2, z = 3t - 1, 0 ≤ t ≤ 1.
dr = dx i + dy j + dz k = (2t dt) i + (dt) j + (3 dt) k.
Substituting the values of x, y, and z into ∇f, we get:
C ∇f · dr = (2(t^2)(3t - 1) + (t + 2)^2)(2t dt) + 2(t^2)(t + 2)(dt) + ((t^2)^2 + 15(3t - 1)^2)(3 dt).
Now, we can simplify the expression and integrate over the given range of t (0 to 1) to get the final value of C ∇f · dr.