Re: Shane Trulson's question at Yahoo! Answers regardinf arc-length of parametric curve
Hello Shane Trulson,
We are give the following theorem:
If $$x=f(t)$$ and $$y=g(t)$$, $a\le t\le b$, define a smooth curve $C$ that does not intersect itself for $a<b<t$ then the length $s$ of $C$ is given by:
$$s=\int_a^b\sqrt{\left(\frac{dx}{dt} \right)^2+\left(\frac{dx}{dt} \right)^2}\,dt$$
Now, for this problem we are given:
$$x(t)=\cos(t)+t\sin(t)$$
$$y(t)=\sin(t)-t\cos(t)$$
And so we find:
$$\frac{dx}{dt}=-\sin(t)+t\cos(t)+\sin(t)=t\cos(t)$$
$$\frac{dy}{dt}=\cos(t)-\left(-t\sin(t)+\cos(t) \right)=t\sin(t)$$
To answer the issue of whether $C$ crosses itself or not, consider:
$$\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}= \frac{t\sin(t)}{t\cos(t)}=\tan(t)$$
Now, we know that on $$0<t<\frac{\pi}{2}$$ we have:
$$0<\frac{dy}{dx}$$
And so $C$ must be strictly increasing and thus cannot cross itself. So, we may now proceed to find the arc-length:
$$s=\int_0^{\frac{\pi}{2}} \sqrt{ \left(t \cos(t) \right)^2+ \left(t \sin(t) \right)^2}\,dt= \int_0^{ \frac{\pi}{2}} \sqrt{t^2 \left( \cos^2(t)+ \sin^2(t) \right)}\,dt$$
Observing that $t$ is non-negative on the given interval and applying a Pythagorean identity, we may state:
$$s=\int_0^{\frac{\pi}{2}} t\,dt=\left[\frac{1}{2}t^2 \right]_0^{\frac{\pi}{2}}=\frac{1}{2}\left(\frac{\pi}{2} \right)^2-\frac{1}{2}\left(0 \right)^2$$
Hence, we find:
$$\bbox[5px,border:2px solid red]{s=\frac{\pi^2}{8}}$$
