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Integration and trig

  1. Dec 25, 2005 #1

    Alkatran

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    Let's say I wanted to find the length of the perimeter of a semi-circle using integration with x(t) = cos(t) and y(t) = sin(t). I would do like so:

    [tex]\int_{0}^{\pi} \sqrt{sin(t)^2 + cos(t)^2} dt
    = \int_{0}^{\pi} \sqrt{1} dt
    = \pi[/tex]

    now if I wanted to do it for the whole circle I would do it like so:

    [tex]\int_{0}^{\pi} \sqrt{sin(2t)^2 + cos(2t)^2} dt
    = \int_{0}^{\pi} \sqrt{1} dt
    = \pi[/tex]

    Now obviously I made a mistake: I didn't replace 2x with u (etc...). But my question is: why can't I do this? What rule stops me from eliminating the cos^2 + sin^2 and getting the wrong answer?
     
    Last edited: Dec 25, 2005
  2. jcsd
  3. Dec 25, 2005 #2
    Well even if you make the substitution u=2t you would still arrive at an answer of just pi, becuase you have to account for du = 2dt.
     
  4. Dec 25, 2005 #3

    Tide

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    If [itex]x = \cos(2t)[/itex] then [itex]dx^2 = 4 \sin^2(2t) dt^2[/itex] and similarly for y so your distance/perimeter formula is incorrect.
     
    Last edited: Dec 26, 2005
  5. Dec 25, 2005 #4

    Alkatran

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    No wonder I was having trouble doing the problem, I forgot to differentiate x(t) and y(t)! :rofl:
     
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