I need to compute the arc length of the curve in $\mathbb{R}^3$ for $x(t)=(3t \cos t, 2t \sin t, 4t)$

I differentiated each component and I wanted to calculate the norm. However, I get a complicated expression which seems unlikely to integrate.

Please help

We have $v(t)=(3 \cos t -3t \sin t)\mathbf{i}+(2 \sin t+ 2t \cos t)\mathbf{j} +4\mathbf{k}$

Hence: the norm is $\sqrt{(3 \cos t -3t \sin t)^2+(2 \sin t+ 2t \cos t)^2 +4^2}$

I need to compute the arc length such that $0\leq t \leq a$

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    $\begingroup$ Show us your work. $\endgroup$ – Git Gud Mar 9 '13 at 22:56
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    $\begingroup$ What are your starting and ending values of $t$? That's pretty important to know. $\endgroup$ – Cameron Buie Mar 9 '13 at 23:00
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    $\begingroup$ Your expression for $v(t)$ is incorrect. It should be $v(t)=(3\cos t-3t\sin t)\hat{i}+(2\sin t+2t\cos t)\hat{j}+4\hat{k}$. $\endgroup$ – Zilliput Mar 9 '13 at 23:07
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    $\begingroup$ @Carpediem: If you knew the starting and ending value of t, how would you show your result to find the actual length? You don't know those values, so what is a way you can show the length without those limits? $\endgroup$ – Amzoti Mar 9 '13 at 23:13
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    $\begingroup$ @Carpediem Assuming the domain of $x$ is $[0,2\pi \textbf{]}$, Amzoti meant for you to just let it be $$s_x(t)=\int _0 ^{2\pi} \sqrt{(3 \cos t -3t\ sin t)^2+(2 \sin t+ 2t \cos t)^2 +4^2}dt$$ without further simplification. $\endgroup$ – Git Gud Mar 9 '13 at 23:14

It is quite rare to find curves that have a closed-form formula for arclength -- you usually end up with a big mess for which you can't find an anti-derivative. So, giving the answer as an integral seems reasonable. Alternatively, you could find the value of the integral by numerical methods, but that's probably not what you're being asked to do.


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