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Math Help - Arc length using integrals

  1. #1
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    Arc length using integrals

    Ok, I've got this equation here,

    y=ln((e^x+1)/(e^x-1)

    and from this equation I need to find the arc length where p< x < q, p > 0.

    I know the formula for the length is the integral from p to q of [1+(dy/dx)^2]^2dx, and I've tried u substitution eventually winding up with the integral from e^p to e^q [(u^2-1)/(u^2+1)*du/u, but here I'm stuck. Does anybody know where to go from here, or if I've just messed it up entirely?

    Thanks
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  2. #2
    MHF Contributor matheagle's Avatar
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    I assume you want y=ln((e^x+1)/(e^x-1))

    and

    integral from p to q of [1+(dy/dx)^2]^{1/2}dx
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  3. #3
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    Yep, its pretty much that integral that's stumped me
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  4. #4
    MHF Contributor matheagle's Avatar
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    You had left out the square root and the ).
    BUT after some algebra I got

    {dy\over dx}= {-2e^x\over e^{2x}-1}, then the integrand becomes  {e^{2x}+1\over e^{2x}-1}.

    So you need to evaluate \int_q^p {e^{2x}+1\over e^{2x}-1}dx=\int_q^p 1+{2\over e^{2x}-1} dx.
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  5. #5
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    Hello, bnay!

    I believe your final function is inverted . . .


    y\:=\:\ln\left(\frac{e^x+1}{e^x-1}\right)

    I need to find the arc length for:  0 < p< x < q
    We have: . y \;=\;\ln(e^x+1) - \ln(e^x-1)

    . . Then: . \frac{dy}{dx} \;=\;\frac{e^x}{e^x+1} + \frac{e^x}{e^x-1} \;=\;\frac{-2e^x}{(e^{2x}-1)^2}

    And: . 1 + \left(\frac{dy}{dx}\right)^2 \;=\; 1 + \frac{4e^{2x}}{(e^{2x}-1)^2} \;=\;\frac{(e^{2x}-1)^2 + 4e^{2x}}{(e^{2x}-1)^2} \;=\;\frac{e^{4x} + 2x^{2x} + 1}{(e^{2x} - 1)^2} . = \;\frac{(e^{2x} + 1)^2}{(e^{2x}-1)^2}

    . . Hence: . \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \;=\;\frac{e^{2x}+1}{e^{2x}-1}

    And we must integrate: . L \;=\;\int^p_q\frac{e^{2x}+1}{e^{2x}-1}\,dx


    Let: u = e^x\quad\Rightarrow\quad x = \ln u \quad\Rightarrow\quad dx = \tfrac{du}{u}

    Substitute: . L \;=\;\int \frac{u^2+1}{u^2-1}\,\frac{du}{u} \;=\;\int\frac{u^2+1}{u(u-1)(u+1)}\,du

    Partial Fractions: . \int\left[-\frac{1}{u} + \frac{1}{u-1} + \frac{1}{u+1}\right]\,du

    I'll let you finish it . . .

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