# Homework Help: Arc length and parametric function

1. Apr 12, 2006

### Kb1jij

I'm having trouble with the following:

The problem is to find the arc length of the following parametric function:

x=(e^-t)(cos t), y=(e^-t)(sin t) from 0 to $$\pi$$

I found that
$$\frac{\partial y}{\partial t} = e^{-t}(\cos{t}-\sin{t})$$,
$$\frac{\partial x}{\partial t} = -e^{-t}(\sin{t}+\cos{t})$$

Then setting up the integral:
$$\int_{0}^{\pi} \sqrt{(-e^{-t}(\sin{t}+\cos{t}))^2+(e^{-t}(\cos{t}-\sin{t}))^2} dt$$

I then simplified the square root to;
$$e^{-2t}(-4\cos{t}\sin{t}))=e^{-2t}*{-2\sin{2t}}$$

This makes the integral:
$$\int_{0}^{\pi} e^{-t}\sqrt{-2\sin{2t} dt$$

Can this integral even be solved?
I don't think this problem was ment to be that difficult, so I think I made a mistake somewhere, but I can figure what.
Thanks!
Tom

Last edited: Apr 12, 2006
2. Apr 12, 2006

### Kb1jij

I figured it out.

The square root simplifies to 2e^(-2t), I just drop a sign.

Nevermind!

3. Apr 12, 2006

### Ouabache

Good job Tom.. How's the fishin' over in Noank?

Last edited: Apr 12, 2006
4. Apr 12, 2006

### Kb1jij

Well I'm really big into fishing, but from what I hear its good...
Are you from around here?