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anonymous
 5 years ago
By explicitly evaluating the line integrals, calculate the work done by the
force field F = (1,z,y) on a particle when it is moved from (1, 0, 0) to (1, 0, pi )
(i) along the helix (cos t, sin t, t), and
(ii) along the straight line
joining the two points. (iii) Do you expect your answers to (i) and (ii) to be
the same? Explain.
I'm not sure how to quite set it up. I'm confused with parametrization and how it may apply here.
Thanks in advance
anonymous
 5 years ago
By explicitly evaluating the line integrals, calculate the work done by the force field F = (1,z,y) on a particle when it is moved from (1, 0, 0) to (1, 0, pi ) (i) along the helix (cos t, sin t, t), and (ii) along the straight line joining the two points. (iii) Do you expect your answers to (i) and (ii) to be the same? Explain. I'm not sure how to quite set it up. I'm confused with parametrization and how it may apply here. Thanks in advance

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I can get you set up. The parametrization is\[\int\limits_{(1,0,0)}^{(1,0,\pi)}F.ds=\int\limits_{0}^{\pi}(1,t,\sin t).(\sin t, \cos t, 1)dt\]You take the dot product and evaluate the integral. You replace x, y, z in the force vector with their parametrizations (e.g. x=cos t), and ds as a vector represents the differentials of your parametrized curve, so for the curve s=(cos t, sin t , t), ds = (sin t, cos t, 1)dt. The limits are worked out by solving for t given your x, y, z coordinates.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0A straight line is parametrized as: s(t)=(initial vector) +t((final vector)(initial vector)) where t runs from 0 to 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By the way, I calculated the same value for the two arcs you were given, but you wouldn't expect this to be the case in general. A counter example you could use is a path given by s(t) = (cos t, 0, t). This curve starts and ends at the same points (here, t runs from 0 to pi), but yields a different line integral (i.e different work).
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