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Line integral to calculate work

  1. Apr 11, 2012 #1
    Hi. I have a concrete doubt with this problem. Here's the pic.

    attachment.php?attachmentid=46109&stc=1&d=1334149747.jpg

    It asks me to calculate the work done by force P (the ball moves with constant speed). So the solution is in the book and I understood everything, but the problem comes here,

    the force P in axis y is zero so the work of P should be:

    [tex]\int\limits_{{x_o}}^{{x_f}} {{P_x}dx} [/tex]

    And we know that Px is equal to the tension in axis x so:

    [tex]\int\limits_{}^{} {T\sin \theta dx} [/tex]

    But we need to convert the variable dx into theta variable. And the books states that as:
    [tex]x = L\sin \theta [/tex]
    then:
    [tex]dx = L\cos \theta d\theta [/tex]

    But shouldn't it be:
    [tex]dx = L\sin\theta d\theta [/tex]??

    Why did it take the derivative of sine and not of x on the other side? Thank you!
     

    Attached Files:

  2. jcsd
  3. Apr 11, 2012 #2

    gneill

    User Avatar

    Staff: Mentor

    Given:
    ##~~~x = L sin(\theta)##

    Differentiate w.r.t. θ:
    ##~~~\frac{dx}{d \theta} = L cos(\theta) ##

    Rearrange:
    ##~~~dx = L cos(\theta) d\theta##
     
  4. Apr 11, 2012 #3
    Sorry but what did you do in the second step (wrt?)? I think I'm understaning now but I still don't get it.

    Thanks!
     
  5. Apr 11, 2012 #4
    Differentiate both sides with respect to theta.
     
  6. Apr 11, 2012 #5
    Aham! So it would be an intermediate step there:

    [tex]\frac{{dx}}{{d\theta }} = \frac{d}{{d\theta }}L\sin \theta [/tex]

    I didn't know that you could take dO to the other side as a product. Thanks!
     
  7. Apr 11, 2012 #6
    Think about it. If two functions are constantly the same, shouldn't their derivatives also be equal?
     
  8. Apr 13, 2012 #7
    And could this exercise be solved easily using the concepts of energy and conservative forces? Then, how could I begin?

    Thanks!

    Edit. I'll create a new thread.
     
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