Arc Length with Parametric Equations — Question 5

Question 5

Problem

A quarter ellipse is $x=5\cos t$, $y=3\sin t$, $0\le t\le\pi/2$. Which simple quarter-circle lengths bound it?

Question 5 – Solution

Solution

1. Differentiate: $$x'(t)=-5\sin t, \qquad y'(t)=3\cos t.$$ Thus the speed is $$v(t)=\sqrt{25\sin^2t+9\cos^2t}.$$

2. Rewrite the expression under the square root in two useful ways: $$25\sin^2t+9\cos^2t =9+16\sin^2t =25-16\cos^2t.$$ Since $0\le\sin^2t,\cos^2t\le1$, it follows that $$9\le v(t)^2\le25, \qquad\text{so}\qquad 3\le v(t)\le5.$$

3. Integrate these bounds over an interval of width $\pi/2$: $$3\int_0^{\pi/2}dt \le L\le 5\int_0^{\pi/2}dt.$$ Therefore, $$\boxed{\frac{3\pi}{2}\le L\le\frac{5\pi}{2}}.$$ These are the quarter-circumferences of circles with radii $3$ and $5$.