See attachment for figure.

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- Nov 22nd 2008, 06:10 PMrmpatel5cycloid generated by a circle
See attachment for figure.

- Nov 23rd 2008, 12:25 AMMoo
Hello,

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

It is also equal to $\displaystyle x+s+AP$

But $\displaystyle x=a(t-sin(t))$

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

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 :

$\displaystyle \sin(\pi-t)=\frac{AP}{a}$

by simple trigonometry, $\displaystyle \sin(\pi-t)=\sin(t)$

Hence $\displaystyle \boxed{AP=a \sin(t)}$

Finally :

$\displaystyle \begin{aligned}

s&=\pi a-a(t-\sin(t))-AP \\

&=\pi a-at+a \sin(t)-a \sin(t) \\

&=\boxed{a(\pi-t)} \end{aligned}$