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Math Help - cycloid generated by a circle

  1. #1
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    cycloid generated by a circle

    See attachment for figure.
    Attached Thumbnails Attached Thumbnails cycloid generated by a circle-untitled.jpg  
    Last edited by rmpatel5; November 29th 2008 at 09:04 AM.
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    Hello,

    Consider the line from A to the y-axis. Its length is \pi a

    It is also equal to x+s+AP
    But x=a(t-sin(t))

    So we have \pi a=a(t-\sin(t))+s+AP \Leftrightarrow \boxed{s=\pi a-a(t-\sin(t))-AP}

    Consider the triangle ABP. It's a right angle triangle.
    Length of BP is a.
    Angle ABP is pi-t.
    So since sin(ABP)=AP/BP, we get :
    \sin(\pi-t)=\frac{AP}{a}

    by simple trigonometry, \sin(\pi-t)=\sin(t)
    Hence \boxed{AP=a \sin(t)}

    Finally :
    \begin{aligned}<br />
s&=\pi a-a(t-\sin(t))-AP \\<br />
&=\pi a-at+a \sin(t)-a \sin(t) \\<br />
&=\boxed{a(\pi-t)} \end{aligned}
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