# Calculating arc length

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.

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 $0\leq t \leq a$

• Show us your work. – Git Gud Mar 9 '13 at 22:56
• What are your starting and ending values of $t$? That's pretty important to know. – Cameron Buie Mar 9 '13 at 23:00
• Your expression for $v(t)$ is incorrect. It should be $v(t)=(3\cos t-3t\sin t)\hat{i}+(2\sin t+2t\cos t)\hat{j}+4\hat{k}$. – Zilliput Mar 9 '13 at 23:07
• @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? – Amzoti Mar 9 '13 at 23:13
• @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. – Git Gud Mar 9 '13 at 23:14