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Math Help - arc length

  1. #1
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    arc length

     y = sin^-1(x)+\sqrt{1-x^2} starting at point (0,1)

    I have worked this down and obtained the integral

     \int \frac{2}{(x+1)}dx

    but I do not know how to set up my limits on the integral. I know the arcsin function is bound between -1 and 1 on the x axis and -pi/2 and pi/2 on the y axis
    Last edited by silencecloak; April 2nd 2009 at 02:24 PM. Reason: typos
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  2. #2
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    Quote Originally Posted by silencecloak View Post
     y = sin^-1(x)+\sqrt{1-x^2} starting at point (0,1) ... and ending where?

    I have worked this down and obtained the integral

     \int \frac{2}{(x+1)}dx

    but I do not know how to set up my limits on the integral. I know the arcsin function is bound between -1 and 1 on the x axis and -pi/2 and pi/2 on the y axis
    .
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  3. #3
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    Quote Originally Posted by skeeter View Post
    .
    It doesnt tell me where it ends

    "Find the arc length function for the curve below with starting point (0, 1)."
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  4. #4
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    Quote Originally Posted by silencecloak View Post
    It doesnt tell me where it ends

    "Find the arc length function for the curve below with starting point (0, 1)."
    the water clears ... the question is asking for a function.

    note that I reworked and corrected your arc length integral ...

    S(x) = \sqrt{2} \int_0^x \frac{1}{\sqrt{t+1}} \, dt

    S(x) = 2\sqrt{2}\left[\sqrt{t+1}\right]_0^x

    S(x) = 2\sqrt{2}\left[\sqrt{x+1} - 1\right]
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    how did you know to start the integral at 0?
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  6. #6
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    Quote Originally Posted by silencecloak View Post
    how did you know to start the integral at 0?
    what's the given starting point?
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  7. #7
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    Quote Originally Posted by skeeter View Post
    what's the given starting point?

    Mmmm ok, didn't know it was that straightforward

    Gracias Senor
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