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Arc length and parametric function

  1. Apr 12, 2006 #1
    I'm having trouble with the following:

    The problem is to find the arc length of the following parametric function:

    x=(e^-t)(cos t), y=(e^-t)(sin t) from 0 to [tex] \pi [/tex]

    I found that
    [tex] \frac{\partial y}{\partial t} = e^{-t}(\cos{t}-\sin{t}) [/tex],
    [tex] \frac{\partial x}{\partial t} = -e^{-t}(\sin{t}+\cos{t})[/tex]

    Then setting up the integral:
    [tex]\int_{0}^{\pi} \sqrt{(-e^{-t}(\sin{t}+\cos{t}))^2+(e^{-t}(\cos{t}-\sin{t}))^2} dt [/tex]

    I then simplified the square root to;
    [tex] e^{-2t}(-4\cos{t}\sin{t}))=e^{-2t}*{-2\sin{2t}} [/tex]

    This makes the integral:
    [tex] \int_{0}^{\pi} e^{-t}\sqrt{-2\sin{2t} dt[/tex]

    Can this integral even be solved?
    I don't think this problem was ment to be that difficult, so I think I made a mistake somewhere, but I can figure what.
    Last edited: Apr 12, 2006
  2. jcsd
  3. Apr 12, 2006 #2
    I figured it out.

    The square root simplifies to 2e^(-2t), I just drop a sign.

  4. Apr 12, 2006 #3


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    Good job Tom.. How's the fishin' over in Noank?:smile:
    Last edited: Apr 12, 2006
  5. Apr 12, 2006 #4
    Well I'm really big into fishing, but from what I hear its good...
    Are you from around here?
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