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.
Find the arc length parameterization of $r(t)=\langle e^t\sin t,e^t\cos t,6e^t\rangle$.
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multivariable-calculus
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0I've tried several different forms of the answer, including putting a constant 0.5 outside the logarithm and simplifying $\sqrt{s^2}$ to s. – 2012-10-02
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It turns out I was supposed to integrate from zero to t, instead of from negative infinity. How arbitrary.
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0It 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... – 2013-07-20