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.
∇f · dr = 2t3 + (t+2)2(3t-1) + t4 + 15(3t-1)2 = 24t3 - 18t2 + 8t + 15
Correct: Your answer is correct.
To evaluate ∇f · dr along the curve C, we can first calculate the gradient vector ∇f and the differential vector dr, and then take their dot product.
From part (a), we found that f(x, y, z) = x^2z + xy^2 + 5z^3 + C.
Now, let's find the differential vector dr. Given the parameterization of curve C: x = t^2, y = t + 2, z = 3t - 1, where 0 ≤ t ≤ 1, we can calculate the derivatives of x, y, and z with respect to t:
dx/dt = 2t,
dy/dt = 1,
dz/dt = 3.
Therefore, the differential vector dr is given by:
dr = (dx, dy, dz) = (2t, 1, 3).
Next, let's calculate the gradient vector ∇f. The gradient of f is the vector of partial derivatives of f with respect to each variable:
∂f/∂x = 2xz + y^2 = 2(t^2)(3t-1) + (t+2)^2 = 6t^3 - 2t^2 + t^2 + 4t + 4 = 6t^3 - t^2 + 4t + 4,
∂f/∂y = 2xy = 2(t^2)(t+2) = 2t^3 + 4t^2,
∂f/∂z = x^2 + 15z^2 = (t^2)^2 + 15(3t-1)^2 = t^4 + 45t^2 - 30t + 15.
Therefore, the gradient vector ∇f is given by:
∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z) = (6t^3 - t^2 + 4t + 4, 2t^3 + 4t^2, t^4 + 45t^2 - 30t + 15).
Finally, we can calculate ∇f · dr by taking their dot product:
∇f · dr = (6t^3 - t^2 + 4t + 4)(2t) + (2t^3 + 4t^2)(1) + (t^4 + 45t^2 - 30t + 15)(3).
To evaluate this expression along the curve C, we substitute the values of t from 0 to 1 into the expression and compute the result.
To evaluate ∇f · dr along the curve C, we need to find the gradient vector ∇f and the tangent vector dr.
From part (a), we found that ∇f = (2xz + y^2)i + (2xy)j + (x^2 + 15z^2)k.
The tangent vector dr is given by dr = dx/dt i + dy/dt j + dz/dt k.
Let's find dx/dt, dy/dt, and dz/dt:
dx/dt = d(t^2)/dt = 2t
dy/dt = d(t + 2)/dt = 1
dz/dt = d(3t - 1)/dt = 3
So, dr = 2ti + j + 3tk.
Now, let's calculate ∇f · dr:
∇f · dr = (2xz + y^2)i + (2xy)j + (x^2 + 15z^2)k · (2ti + j + 3tk)
= (2x(3t - 1) + (t + 2)^2)(2t) + 2(2t)(t + 2) + (t^2 + 15(3t - 1)^2)(3t)
= (6xt - 2x + t^2 + 4t + 4)(2t) + 4t(t + 2) + (t^2 + 45t^2 - 30t + 15)(3t)
= (12t^2x + 4xt - 4x + 2t^3 + 8t^2 + 8t + 8t + 8 + 36t^3 - 36t^2 + 45t^3 - 45t^2 + 15t)(2t + 2)
= (93t^3 + 56t^2 + 12tx - 4x + 12t)(2t + 2)
Therefore, ∇f · dr = (93t^3 + 56t^2 + 12tx - 4x + 12t)(2t + 2).