Question

Find the distance traveled by a particle with position (x, y) as t varies in the given time interval.
x = 5 sin2 t, y = 5 cos2 t, 0 ≤ t ≤ 3π

Answer 1

The traveled distance will be given by the arc length of the given curve. For parametric equations, the arc length \(L\) of a curve \(C\) defined over an interval \(a\le t\le b\) is
\[L=\int_CdS=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}~dt\]

Answer 2

Are the equations
\[\begin{cases}x=5\sin2t\\y=5\cos2t\end{cases}~~\text{or}~~\begin{cases}x=5\sin^2t\\y=5\cos^2t\end{cases}~~?\]

Answer 3

It is the squared version

Answer 4

\[\begin{cases}x=5\sin^2t\\y=5\cos^2t\end{cases}~~\implies~~\begin{cases}\dfrac{dx}{dt}=10\sin t\cos t\\\\\dfrac{dy}{dt}=-10\cos t\sin t\end{cases}\]
Substitute into the integral, you have
\[\large\begin{align*} L&=\int_0^{3\pi}\sqrt{\left(10\sin t\cos t\right)^2+\left(-10\cos t\sin t\right)^2}~dt\\\\
&=\int_0^{3\pi}\sqrt{100\sin^2 t\cos^2 t+100\cos^2 t\sin^2 t}~dt\\\\
&=10\sqrt2\int_0^{3\pi}\sqrt{\cos^2 t\sin^2 t}~dt\\\\
&=10\sqrt2\int_0^{3\pi}|\cos t\sin t|~dt\\\\
&=10\sqrt2\int_0^{3\pi}\left|\frac{1}{2}\sin 2t\right|~dt\\\\
&=10\int_0^{3\pi}\left|\sin 2t\right|~dt\\\\
&=10\left(3\int_0^{\pi/2}\sin 2t~dt-3\int_{\pi/2}^\pi\sin2t~dt\right)\end{align*}\]

Answer 5

So, what would the final answer be? I keep getting a negative number

Answer 6

I'm getting 60. I'll have to recheck later though, I would have expected an answer in terms of \(\pi\)...