A wheel of radius $a$ rolls on the outside of a circle with radius $b$ (see figure).

Fig. 1

Find the parameterization for the curve a point on the wheel follows. You may choose freely how you position the coordinate system, and the starting point.

My attempt:

Fig. 2

I'm not very well versed in parametrizing curves, so I tried first determining the curve that the center of the small circle would follow, and came up with $$P_s = ((a+b)\cos t, (a+b)\sin t)$$

Then I added to that the curve that the point $m$ would have relative to the center $s$ of the tiny wheel, $$P_m = P_s + (a\cos t, a\sin t) \\ = ((2a+b)\cos t, (2a+b)\sin t)$$

That is my answer, but unless I'm missing some fundamental algebra, this is not equal to the real answer.

The real answer:

With the origin at the center of the large circle, and with starting point $(b, 0)$, we get: $$x(t) = (a+b)\cos t - a\cos(\frac{a+b}{a}t) \\ y(t) = (a+b)\sin t - a\sin(\frac{a+b}{a}t)$$


enter image description here

Imagine the smaller circle has rolled along the bigger circle as shown above. The distance it has rolled along the larger circle is equal to $bt$ (shown as the green path). The distance that point $m$ on the smaller circle has moved is equal to $as$ (shown as the pink path). These two distances must be equal, therefore:$$as=bt$$$$\therefore s=\frac{bt}{a}$$Now look at the magnified section below the larger circle. If we drop a vertical line down from the center of the smaller circle then the angle made between the red line and this vertical line must be $\frac{\pi}{2}-t$. Therefore the angle marked as $p$ is given by:$$p=s-\left(\frac{\pi}{2}-t\right)=\frac{bt}{a}-\left(\frac{\pi}{2}-t\right)=\frac{(a+b)t}{a}-\frac{\pi}{2}$$Hopefully this should give you enough information to solve from here...


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