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Math Help - Calculate the length of the curve

  1. #1
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    Question Calculate the length of the curve

    x12 Calculate the length of the curve
    between x=0 och x=1.


    Calculate the length of the curve-picture-1.png

    I think i managed to solve until
    Calculate the length of the curve-whatever.jpg
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  2. #2
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    Quote Originally Posted by sf1903 View Post
    x12 Calculate the length of the curve
    between x=0 och x=1.


    Click image for larger version. 

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ID:	14468

    I think i managed to solve until Click image for larger version. 

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Views:	9 
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ID:	14467

    y=\frac{x^{3\slash 2}}{3}-x^{1\slash 2}\Longrightarrow y'=\frac{x^{1\slash 2}}{2}-\frac{1}{2x^{1\slash 2}}

    1+(y')^2=1+\frac{1}{4}x-\frac{1}{2}+\frac{1}{4x}=\frac{1}{4}\left(x+2+\fra  c{1}{x}\right)=\frac{1}{4}\left(x+\frac{1}{x}\righ  t)^2

    So the function in the integral is pretty simple: {1}{2}\left(x+\frac{1}{x}\right) ...unfortunately this improper integral diverges at zero since \ln x\xrightarrow[x\to 0]{}-\infty ...

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    y=\frac{x^{3\slash 2}}{3}-x^{1\slash 2}\Longrightarrow y'=\frac{x^{1\slash 2}}{2}-\frac{1}{2x^{1\slash 2}}

    1+(y')^2=1+\frac{1}{4}x-\frac{1}{2}+\frac{1}{4x}=\frac{1}{4}\left(x+2+\fra  c{1}{x}\right)=\frac{1}{4}\left(x+\frac{1}{x}\righ  t)^2
    No, tonio, that last term is
    \frac{1}{4}\left(x^{1/2}+ \frac{1}{x^{1/2}}\right)^2.

    The integral becomes \frac{1}{2}\int_0^1 \left(x^{1/2}+ x^{-1/2}\right)dx which, while an improper integral, does have a finite value.

    So the function in the integral is pretty simple: {1}{2}\left(x+\frac{1}{x}\right) ...unfortunately this improper integral diverges at zero since \ln x\xrightarrow[x\to 0]{}-\infty ...

    Tonio
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    No, tonio, that last term is
    \frac{1}{4}\left(x^{1/2}+ \frac{1}{x^{1/2}}\right)^2.

    The integral becomes \frac{1}{2}\int_0^1 \left(x^{1/2}+ x^{-1/2}\right)dx which, while an improper integral, does have a finite value.
    This is how far I have come, am I correct?
    Calculate the length of the curve-upp.jpg
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    No, tonio, that last term is
    \frac{1}{4}\left(x^{1/2}+ \frac{1}{x^{1/2}}\right)^2.

    The integral becomes \frac{1}{2}\int_0^1 \left(x^{1/2}+ x^{-1/2}\right)dx which, while an improper integral, does have a finite value.

    You are, of course, right. I copied, pretty stupidly, from the line before the last one instead of he alst one...and forgot.
    Thanx....oh, and then the integral converges (it still is an improper integral)

    Tonio
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