Parametric Equations: Get Started Now

halvizo1031
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Can someone help me get started on number one please?
 

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It's not clear to me whether the wheel is rolling along the x-axis, or rotating in place. Which is it?
 
Well that's where I'm stuck. Since it says "rim of a wheel", my only guess is that it is moving along the x-axis. If this is the case, then it produces a cycloid right?
 
Strictly speaking, a cycloid is produced only when the pen lies on the rim of the rolling wheel; our pen is in the interior.

If you suppose the wheel to be rolling, then decompose the motion of the pen into two parts, the motion of the wheel, and the motion of the pen about the wheel's hub.
 
wow that's toughy
 
If P = (x,y) and if you call the center of the circle C you know that

\vec{OP}=\vec{OC} + \vec{CP}

Since the wheel isn't slipping, you know the x coordinate of the center is the same as the arc of the wheel a\theta and the y coordinate is a. So you can begin with

\langle x,y\rangle=\langle a\theta, a\rangle + \vec{CP}

Now figure out the components of CP in terms of \theta. It isn't difficult, especially if you write them in terms of the standard polar angle at the center C and use that go get it in terms of \theta.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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