Shane Trulson's Calc Homework: Arc Length of Parametric Curve

[MarkFL]

Here is the question:

Calculus Homework?

Calculate the length of the curve x(t) = cost+tsint, y(t) = sint-tcost, 0<_t<_pi/2

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Re: Shane Trulson's question at Yahoo! Answers regardinf arc-length of parametric curve

Hello Shane Trulson,

We are give the following theorem:

If \(\displaystyle x=f(t)\) and \(\displaystyle y=g(t)\), $a\le t\le b$, define a smooth curve $C$ that does not intersect itself for $a<b<t$ then the length $s$ of $C$ is given by:

\(\displaystyle s=\int_a^b\sqrt{\left(\frac{dx}{dt} \right)^2+\left(\frac{dx}{dt} \right)^2}\,dt\)

Now, for this problem we are given:

\(\displaystyle x(t)=\cos(t)+t\sin(t)\)

\(\displaystyle y(t)=\sin(t)-t\cos(t)\)

And so we find:

\(\displaystyle \frac{dx}{dt}=-\sin(t)+t\cos(t)+\sin(t)=t\cos(t)\)

\(\displaystyle \frac{dy}{dt}=\cos(t)-\left(-t\sin(t)+\cos(t) \right)=t\sin(t)\)

To answer the issue of whether $C$ crosses itself or not, consider:

\(\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}= \frac{t\sin(t)}{t\cos(t)}=\tan(t)\)

Now, we know that on \(\displaystyle 0<t<\frac{\pi}{2}\) we have:

\(\displaystyle 0<\frac{dy}{dx}\)

And so $C$ must be strictly increasing and thus cannot cross itself. So, we may now proceed to find the arc-length:

\(\displaystyle s=\int_0^{\frac{\pi}{2}} \sqrt{ \left(t \cos(t) \right)^2+ \left(t \sin(t) \right)^2}\,dt= \int_0^{ \frac{\pi}{2}} \sqrt{t^2 \left( \cos^2(t)+ \sin^2(t) \right)}\,dt\)

Observing that $t$ is non-negative on the given interval and applying a Pythagorean identity, we may state:

\(\displaystyle s=\int_0^{\frac{\pi}{2}} t\,dt=\left[\frac{1}{2}t^2 \right]_0^{\frac{\pi}{2}}=\frac{1}{2}\left(\frac{\pi}{2} \right)^2-\frac{1}{2}\left(0 \right)^2\)

Hence, we find:

\(\displaystyle \bbox[5px,border:2px solid red]{s=\frac{\pi^2}{8}}\)