Work done by F for a particle moving along C

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SUMMARY

The discussion centers on calculating the work done by a radial vector field F on a particle moving along a specified curve C. The vector field is defined as F = r, and the derivative of the position vector is given by dr/dt = 2cos(t)i + 5sin(2t)j + 6cos(t)sin^2(t)k. The initial attempt to compute the work resulted in an incorrect value due to a misunderstanding of the integration process. The correct approach involves integrating the dot product of F and dr/dt over the interval from 0 to π/2, leading to the conclusion that the work done is W = ∫[0, π/2] F · dr/dt dt.

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glid02
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Here's the question:
If C is the curve given by http://ada.math.uga.edu/webwork2_files/tmp/equations/01/126d71403184689027c86bb27202771.png ,[/URL] http://ada.math.uga.edu/webwork2_files/tmp/equations/56/f3b3a7b9c3e5d804f53afad7a07bc71.png and F is the radial vector field http://ada.math.uga.edu/webwork2_files/tmp/equations/28/71b024e85380a1ae00268ee0af38721.png ,[/URL] compute the work done by F on a particle moving along C.

This is what I've done so far...
F = r because F is only x, y, z
dr/dt = 2cos(t)i+5sin(2t)j+6cos(t)sin^2(t)k

Then F dot dr/dt =
2cos(t)(1+sin(t))+5sin(2t)(1+5sin^2(t))+6cos(t)sin^2(t)(1+2sin^3(t))


Now cos(pi/2) = 0 and sin(pi) = 0 - from 5sin(2t) - so when plugging in pi/2 the answer should be zero for all.
sin(0) = 0 and cos(0) = 1, so when evaluated at 0 the only answer should be 2 (2cos(0)*1) so by my account the answer should be -2.

Apparently this isn't the right answer, if someone could tell me what I missed it'd be awesome.

Thanks a lot.
 
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F dot dr is the incremental work along the path... you need to integrate this over the entire path, not just plug in a value!

In this case, W = Int[0, pi/2] F dot dr/dt dt.

Also be careful... you lost a 2 in 2cos(t)(1 + 2sin(t))!
 
oh wow, that's a great point.

thanks.
 

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