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Math Help - Length of Cycloid

  1. #1
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    Length of Cycloid

    Okay everyone, I'm sure this is a simple fix but let's see if I'm on the right track:

    We have been given a "project" to "prove" that the length of one complete arc of a cycloid generated by a circle of radius a, equals the perimeter of a square that circumscribes the same circle. I know that the answer to my integral should be 8|a|, but it is not coming out that way. I know that I have made an error somewhere but I just can't seem to find it!

    Assume a is a constant:

    f(t)=x=at-a\sin{t}

    g(t)=y=a-a\cos{t}

    0\leq{t}\leq{2\pi}

    f'(t)=a-a\cos{t}

    g'(t)=a\sin{t}

    L=\int_{0}^{2\pi}\sqrt{(f'(t))^2+(g'(t))^2} dt

    L=\int_{0}^{2\pi}\sqrt{(a-a\cos{t})^2+(a\sin{t})^2} dt

    L=\int_{0}^{2\pi}\sqrt{a^2-2a^2\cos^2{t}+a^2\cos^2{t}+a^2\sin^2{t}} dt

    L=\int_{0}^{2\pi}\sqrt{2a^2-2a^2\cos^2{t}} dt

    L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos^2{t})} dt

    L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(1-\cos^2{t})} dt

    L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(sin^2{t})} dt

    L=\sqrt{2}|a|\int_{0}^{2\pi}|\sin{t}| dt

    |\sin{t}| = \sin{t} for  t \epsilon [0,\pi]

    |\sin{t}| = -\sin{t} for  t \epsilon [\pi,2\pi]

    L=\sqrt{2}|a|\int_{0}^{\pi}\sin{t} dt + \sqrt{2}|a|\int_{\pi}^{2\pi}-\sin{t} dt

    L=\sqrt{2}|a|(-\cos{t})]_0^\pi + \sqrt{2}|a|(\cos{t})]_\pi^{2\pi}

    L=\sqrt{2}|a|(-(-1)-(-1))+\sqrt{2}|a|(1-(-1))

    L=4\sqrt{2}|a|

    It's evident that I should not have a \sqrt{2}, instead I should have just 2 outside the integral. Right?
    Last edited by elizsimca; April 18th 2008 at 09:26 PM. Reason: Interval Error
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  2. #2
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    Quote Originally Posted by elizsimca View Post
    Okay everyone, I'm sure this is a simple fix but let's see if I'm on the right track:

    We have been given a "project" to "prove" that the length of one complete arc of a cycloid generated by a circle of radius a, equals the perimeter of a square that circumscribes the same circle. I know that the answer to my integral should be 8|a|, but it is not coming out that way. I know that I have made an error somewhere but I just can't seem to find it!

    Assume a is a constant:

    f(t)=x=at-a\sin{t}

    g(t)=y=a-a\cos{t}

    0\leq{t}\leq{2\pi}

    f'(t)=a-a\cos{t}

    g'(t)=a\sin{t}

    L=\int_{0}^{2\pi}\sqrt{(f'(t))^2+(g'(t))^2} dt

    L=\int_{0}^{2\pi}\sqrt{(a-a\cos{t})^2+(a\sin{t})^2} dt

    L=\int_{0}^{2\pi}\sqrt{a^2-2a^2\cos^2{t}+a^2\cos^2{t}+a^2\sin^2{t}} dt

    L=\int_{0}^{2\pi}\sqrt{2a^2-2a^2\cos^2{t}} dt

    L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos^2{t})} dt

    L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(1-\cos^2{t})} dt

    L=\sqrt{2}|a|\int_{0}^{2\pi}\sqrt{(sin^2{t})} dt

    L=\sqrt{2}|a|\int_{0}^{2\pi}|\sin{t}| dt

    |\sin{t}| = \sin{t} for  t \epsilon [0,\pi]

    |\sin{t}| = -\sin{t} for  t \epsilon [\pi,2\pi]

    L=\sqrt{2}|a|\int_{0}^{\pi}\sin{t} dt + \sqrt{2}|a|\int_{\pi}^{2\pi}-\sin{t} dt

    L=\sqrt{2}|a|(-\cos{t})]_0^\pi + \sqrt{2}|a|(\cos{t})]_\pi^{2\pi}

    L=\sqrt{2}|a|(-(-1)-(-1))+\sqrt{2}|a|(1-(-1))

    L=4\sqrt{2}|a|

    It's evident that I should not have a \sqrt{2}, instead I should have just 2 outside the integral. Right?
    No time now. If no-one else replies I'll take a look later.
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  3. #3
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    It's okay Mr. F.
    It would have been up 45 mins ago, but I decided to try my hand at LaTeX so everyone could see the work I had already done.
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  4. #4
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    Errors

    Elizsimca
    You were correct up to here:
    L=\int_{0}^{2\pi}\sqrt{(a-a\cos{t})^2+(a\sin{t})^2}
    but you made an error in the next line:
    L=\int_{0}^{2\pi}\sqrt{a^2-{\color{red}2a^2\cos{t}}+a^2\cos^2{t}+a^2\sin^2{t}  }
    So
    L=\int_{0}^{2\pi}\sqrt{2a^2-2a^2\cos{t}}
    L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos{t})}
    L=\int_{0}^{2\pi}\sqrt{2a^2(1-\cos{t})}
    L=|a|\int_{0}^{2\pi}\sqrt{2(1-\cos{t})}
    L=|a|\int_{0}^{2\pi}\sqrt{2(2\sin^2(\frac{t}{2}))}
    L=2|a|\int_{0}^{2\pi}\sqrt{\sin^2(\frac{t}{2})}
    And from there, you should be able to get the correct answer.

    --Kevin C.
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  5. #5
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    Oh what a dumb mistake!! Thank you I super appreciate it!
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  6. #6
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    Quote Originally Posted by elizsimca View Post
    Oh what a dumb mistake!! Thank you I super appreciate it!
    And we appreciate you being so dilligent in showing all your working. Latex can be tricky. Well done - as you can see, it made it very easy to get a good answer.

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