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Math Help - Logarithmic Spiral

  1. #1
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    Logarithmic Spiral

    Good afternoon once again,

    I hope it is acceptable keep posting these. It seems like I make some small mistake on each and every one of them, so the feedback is rather helpful.

    Find the arc length of the logarithmic spiral ae^{b\theta} for 0\leq\theta\leq\pi.

    I have the following:

    r=ae^{b\theta}

    \frac{r}{dr}=abe^{b\theta}

    So...

    \displaystyle\int_{0}^{\pi}\sqrt{(ae^{b\theta})^2+  (abe^{b\theta})^2}d\theta


    \displaystyle\int_{0}^{\pi}\sqrt{(a)^2(e^{b\theta}  )^2+(ab)^2(e^{b\theta})^2}d\theta


    \displaystyle\int_{0}^{\pi}a+ab\sqrt{(e^{b\theta})  ^2+(e^{b\theta})^2}d\theta


    \displaystyle2a+b\int_{0}^{\pi}\sqrt{2(e^{b\theta}  )^2}d\theta


    \displaystyle2a+b\int_{0}^{\pi}e^{b\theta}\sqrt{2}  d\theta


    \displaystyle(2a+b)\sqrt{2}\int_{0}^{\pi}e^{b\thet  a}d\theta


    \displaystyle(2a+b)\sqrt{2}(\frac{e^{b\theta}}{b}|  _{0}^{\pi})


    \displaystyle(2a+b)\sqrt{2}(\frac{e^{b\pi}-e^{b0}}{b})


    \displaystyle(2a+b)\sqrt{2}(\frac{e^{b\pi}-1}{b})


    How did I do this time around?
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  2. #2
    MHF Contributor
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    The problem is in the third step

    \sqrt{a^2+(ab)^2} \ne a + ab
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