To find the work done in moving a particle along a curve in a force field, we can use the line integral of the force over the curve. The line integral is given by:
W = ∫C F · dr
where F is the force field, dr is the differential displacement vector along the curve, and the integral is taken over the curve C.
Given the force field F = 3xyi - 5zj + 10xk and the curve x = t^2 + 1, y = 2t^3, we can compute the line integral over the curve.
First, let's calculate the differential displacement vector dr. Since x = t^2 + 1 and y = 2t^3, we can express dr as follows:
dr = dx i + dy j
Now, we need to express dx and dy in terms of dt:
dx = (d/dt)(t^2 + 1) dt = 2t dt dy = (d/dt)(2t^3) dt = 6t^2 dt
Substituting dx and dy into dr, we have:
dr = (2t dt) i + (6t^2 dt) j
Next, we compute F · dr:
F · dr = (3xy) (2t dt) + (-5z) (6t^2 dt) + (10x) (2t dt) = 6t^3(t^2 + 1)(2t dt) - 30t^2(6t^2 dt) + 20t(t^2 + 1)(2t dt)
Simplifying, we get:
F · dr = 12t^4 dt + 12t^3 dt + 40t^4 dt = 52t^4 dt + 12t^3 dt
Now, we can integrate F · dr over the given range t = 1 to t = 2:
W = ∫[1 to 2] (52t^4 + 12t^3) dt = [13t^5/5 + 3t^4/2] [1 to 2] = [(13(2)^5/5 + 3(2)^4/2) - (13(1)^5/5 + 3(1)^4/2)] = [(13(32)/5 + 3(16)/2) - (13/5 + 3/2)] = [(416/5 + 48/2) - (13/5 + 3/2)] = (416/5 + 24) - (13/5 + 3/2) = 83.2 + 24 - 2.6 - 1.5 = 103.1
Therefore, the work done in moving the particle along the given curve from t = 1 to t = 2 is 103.1 units.