Area with Parametric Equations — Question 7

Question 7 – Solution

Solution

1. Compared with the standard cycloid $x=t-\sin t$, $y=1-\cos t$, the $x$-coordinates are unchanged while every $y$-coordinate is multiplied by $2$. A vertical stretch by a factor of $2$ doubles area, so the predicted arch area is $2(3\pi)=6\pi$.

2. Verify this prediction directly on $0\le t\le2\pi$. Differentiate $x$: $$x'(t)=1-\cos t.$$ Since $x'(t)\ge0$, the arch is traced from left to right.

3. Substitute $y=2(1-\cos t)$: $$A=\int_0^{2\pi}y(t)x'(t)\,dt =2\int_0^{2\pi}(1-\cos t)^2\,dt.$$

4. From $$\int_0^{2\pi}(1-\cos t)^2\,dt =2\pi+\pi=3\pi,$$ it follows that $$\boxed{A=2(3\pi)=6\pi}.$$ The computation agrees with the scaling prediction.