# Simple integral

1. Jul 24, 2004

### SummerPhysStudent

Ok, So i need to find the arc length of the curve r(t) = < 2t, e^t, e^(-t)> from 0<=t<=1
so L should be the integral from 0 to 1 over the magnitude of r'(t) dt. So what I'm getting is
S01 sqrt(4 + e2t + e-2t)dt
and this I'm not sure how to do this integral. Anyhelp?

Last edited: Jul 24, 2004
2. Jul 24, 2004

### e(ho0n3

You can use Latex to type up the integral instead of using that weird S. You might want to check what |r'(t)| dt is again. The integral you'll get wont be pretty though.

3. Jul 24, 2004

### homology

For whatever its worth, I chucked the integral into Mathematica and it didn't solve it.

Oh, wait, well if we screw around with the integral to get

$$\int_{0}^{1}\sqrt{4+2(\frac{e^2t + e^-2t}{2})}dt=\int_{0}^{1}\sqrt{4+2cosh(2t)}dt$$

and then chuck it into Mathematica, it says:

$$i\sqrt{6}EllipticE[i,\frac{2}{3}]$$

The EllipticE[ , ] is a complete elliptic integral of the second kind.

In the words of Gandalf in the mines of moria (movie) "This foe is beyond any of you. Run!" (beyond me anyway, elliptic integrals, yick! :yuck: )

kevin

Last edited: Jul 24, 2004