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Parameterization of a path to find work

  1. Feb 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Evaluate the work done by the two-dimensional force F = ( x2, 2xy ) along each of the following three paths joining the origin to the point P = (1, 1) :
    The first two are fine
    The last path is: the path given parametrically as x = t3, y = t2 with a parameter t

    2. Relevant equations


    3. The attempt at a solution
    I think you just take x = t3, y = t2 and you plug them into F = ( x2, 2xy ) and then you set up the integral to find work ∫F⋅ds in terms of t. ∫(t6,2t5)⋅(3t2, 2t)dt = ∫(3t8+4t6)dt. Is this correct?
     
  2. jcsd
  3. Feb 23, 2017 #2

    Simon Bridge

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    How would you check?

    1. what happens when you use the same method/reasoning on the first two paths?
    2. how did the work for the first two paths compare with each other? What do you expect for the third one? What do you get?
     
  4. Feb 23, 2017 #3

    haruspex

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    Yes. You don't spell it out, but it looks like you have understood that ##\vec{ds}=(dx, dy)=(\dot x.dt, \dot y.dt)=(\dot x, \dot y).dt##.
     
  5. Feb 23, 2017 #4
    Okay. Now that I understand this part - what about the limits of t? In the original it was a function of two variables going up to the point (1,1). Now that I have paramaterized this (if that is the right wording) what would the limits be in t as a single variable? As in how do I go about figuring it out? I integrated from 0 to 1 because intuitively that makes sense but I'm usually wrong.
     
  6. Feb 23, 2017 #5

    haruspex

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    What are the values of the parameter at the path endpoints?
     
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