# Find the arc length parameterization of $r(t)=\langle e^t\sin t,e^t\cos t,6e^t\rangle$.

I computed $s= \int_{-\infty}^t \! |r'(u)| \, \mathrm{d} u$ to be $s=\sqrt{38e^{2t}}$. Solving for t yields $t=\ln\left(\sqrt{\frac{s^2}{38}}\right)$. The online homework system I'm using isn't accepting the answer after substituting this value into the original vector function.

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Maybe the computer doesn't like $\sqrt{s^2}$ where $s$ will do. Though I would expect it to test correctness by substituting random values. – André Nicolas Oct 2 '12 at 3:38
I've tried several different forms of the answer, including putting a constant 0.5 outside the logarithm and simplifying $\sqrt{s^2}$ to s. – Max Lipton Oct 2 '12 at 3:45

It is not so arbitrary because if you integrate from $-\infty$, you always get $+\infty$ as the result no matter what $t$ is. Still, the number $0$ seems quite arbitrary because you can start from anywhere as long as it is a real number. But well, if you have to pick a number, $0$ seems not too arbitrary... – Tunococ Jul 20 '13 at 21:36