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Math Help - Archimedes Spiral

  1. #1
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    Archimedes Spiral

    Good afternoon,

    I am working on finding the length of Archimedes Spiral r=\theta for 0\leq\theta\leq2\pi. I think I may be doing something wrong. This is what I'm getting:

    r=\theta=f(x)

    f'(x)=1

    \displaystyle\int_{0}^{2\pi}\sqrt{\theta^2-1^2}d\theta

    I have evaluated this to:

    \displaystyle\frac{\theta}{2}\sqrt{\theta^2-1}-\frac{ln(\theta+\sqrt{\theta^2-1}}{2}|_{0}^{2\pi}

    When I try to wrap this up by plugging in my terminals... I get a mess. Is this right?

    \displaystyle\pi\sqrt{(2\pi)^2-1}-\frac{ln(\2\pi+\sqrt{(2\pi)^2-1})}{2}
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  2. #2
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    Quote Originally Posted by MechEng View Post
    Good afternoon,

    I am working on finding the length of Archimedes Spiral

    r=\theta

    for... 0\leq\theta\leq2\pi.

    I think I may be doing something wrong. This is what I'm getting:

    r=\theta

    \displaystyle\huge\frac{dr}{d\theta}=1

    \displaystyle\int_{0}^{2\pi}\sqrt{r^2+\left(\frac{  dr}{d\theta}\right)^2}\ d\theta

    Shouldn't you have been working from the above ?

    I have evaluated this to:

    \displaystyle\frac{\theta}{2}\sqrt{\theta^2-1}-\frac{ln(\theta+\sqrt{\theta^2-1}}{2}|_{0}^{2\pi}

    When I try to wrap this up by plugging in my terminals... I get a mess. Is this right?

    \displaystyle\pi\sqrt{(2\pi)^2-1}-\frac{ln(\2\pi+\sqrt{(2\pi)^2-1})}{2}
    \displaystyle\huge\int_{0}^{2{\pi}}\sqrt{r^2+1^2}\ d\theta=\int_{0}^{2{\pi}}\sqrt{1+\theta^2}\ d\theta

    Working through the integration....

    if you draw a right-angled triangle, perpendicular sides x and 1,
    then the hypotenuse length is

    \sqrt{1+x^2}

    Hence.... cos\theta=\frac{1}{\sqrt{1+x^2}}\ \Rightarrow\ \sqrt{1+x^2}=sec\theta

    x=tan\theta

    dx=sec^2\theta\ d\theta

    \int{\sqrt{1+x^2}}dx=\int{\sqrt{1+tan^2\theta}sec^  2\theta\ d\theta

    =\int{sec\theta\ sec^2\theta\ d\theta=\int{sec^3\theta}d\theta

    Integration by parts...

    u=sec\theta,\ dv=sec^2\theta\ d\theta

    \Rightarrow\ du=sec\theta\ tan\theta,\ v=tan\theta

    uv-\int{v}du=sec\theta\ tan\theta-\int{sec\theta\ tan^2\theta}d\theta=sec\theta\ tan\theta-\int{sec\theta\left(sec^2\theta-1\right)d\theta

    =sec\theta\ tan\theta-\int{sec^3\theta}d\theta+\int{sec\theta}d\theta

    2\int{sec^3\theta}d\theta=sec\theta\ tan\theta+\int{sec\theta}d\theta

    =sec\theta\ tan\theta+ln|sec\theta+tan\theta|

    \int{sec^3\theta}=\frac{1}{2}\left(sec\theta\ tan\theta+ln|sec\theta+tan\theta|\right)

    =\frac{1}{2}\left(x\sqrt{1+x^2}+ln|\sqrt{1+x^2}+x|  \right)

    Rewrite using \theta instead of x and evaluate using the limits.
    Last edited by Archie Meade; August 10th 2010 at 02:58 PM.
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  3. #3
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    ...yep. I'm not sure how I wound up with

    \displaystyle\int_{0}^{2\pi}\sqrt{r^2-\left(\frac{dr}{d\theta}\right)^2}\ d\theta

    VS.

    \displaystyle\int_{0}^{2\pi}\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\ d\theta


    I think I may have been looking at two different formulas at once, as they are all listed in the chapter summary. Since area formulas are genereally subtracting something from something else, that is the only explanation I can produce.

    I thank you once again, Sir.
    Last edited by MechEng; August 11th 2010 at 06:47 AM. Reason: Corrected Sign Placement and Formatting
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  4. #4
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    Anyway... Finishing this one up...

    For 0\leq\theta\leq2\pi...

    \displaystyle\frac{\left(2\pi\sqrt{1+2\pi^2}+ln|\s  qrt{1+2\pi^2}+2\pi|\right)}{2}-\frac{\left(0\sqrt{1+0^2}+ln|\sqrt{1+0^2}+0|\right  )}{2}


    \displaystyle\frac{\left(2\pi\sqrt{1+2\pi^2}+ln|\s  qrt{1+2\pi^2}+2\pi|\right)}{2}

    Is this a reasonable answer for this problem?
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  5. #5
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    Quote Originally Posted by MechEng View Post
    Anyway... Finishing this one up...

    For 0\leq\theta\leq2\pi...

    \displaystyle\frac{\left(2\pi\sqrt{1+(2\pi)^2}+ln|  \sqrt{1+(2\pi)^2}+2\pi|\right)}{2}-\frac{\left(0\sqrt{1+0^2}+ln|\sqrt{1+0^2}+0|\right  )}{2}


    \displaystyle\frac{\left(2\pi\sqrt{1+4\pi^2}+ln|\s  qrt{1+4\pi^2}+2\pi|\right)}{2}

    Is this a reasonable answer for this problem?
    Looks good to me.
    Be careful when squaring, parentheses helps.
    Lastly, you may evaluate the final expression.
    Last edited by Archie Meade; August 12th 2010 at 03:31 AM.
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