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Work and Parameterization?
I'd appreciate help with questions 1 and 2
1. Compute the work done by the force
field
F = (sin(xz) + 3)i + (x - y^2 + z)j + (e^x + y^2)k
on a particle starting at P = (2,-1,-4) and going in a straight line to Q = (2, 3, 0).
I know the parameterized line is r(t) = (2, 4t - 1, 4t - 4) and dr = (0, 4, 4)
I know F = (sin(2*(4t-4))+3, 2 - (4t-1)^2 + (4t-4), e^2 + (4t-1)^2)
= (sin(8t-8)+3, 2 - (16t^2-8t+1) + (4t-4), e^2 + 16t-8t+1)
= (sin(8t-8)+3; -16t^2 + 12t - 3; e^2 + 16t - 8t +1)
I also know Work = Integral(F * dr) =
Integral of ((sin(8t-8)+3; -16t^2 + 12t - 3; e^2 + 16t - 8t +1) * (0, 4, 4))
What I'm confused about is do I have my bounds for "t" from the question I was given or is this just an indefinite integral? Also I'd like help solving the rest. Thank you for your time.
2. I'm pretty sure I can solve this since it's a conservative vector field. I just want to affirm that I can find the work by finding the potential difference from (1,pi) to (0,-1).
Thank you for your time.
1 Answer
- kbLv 77 years agoFavorite Answer
1) With the way you have parameterized the line, t is in [0, 1]
(since t = 0 yields P, and t = 1 yields Q).
So, ∫c F · dr
= ∫(t = 0 to 1) <sin(8t-8)+3, -16t^2 + 12t - 3, e^2 + 16t - 8t +1> · <0, 4, 4> dt, using your work
= ∫(t = 0 to 1) 4(-16t^2 + 20t + e^2 - 2) dt
= 4(-16t^3/3 + 10t^2 + te^2 - 2t) {for t = 0 to 1}
= 32/3 + 4e^2.
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2) F is conservative, since ∇(cos(xy) + x^2 + 2y) = F.
So, the integral equals
(cos(xy) + x^2 + 2y) {from (1, π) to (0, -1)}
= -1 - 2π.
------------
I hope this helps!
