Find the arc length of r(t)= <tsin(t), tcost(t), 3t> from 0 to t to 2pi (inclusive)
Integral from 0 to 2pi of the magnitude of r'(t) dt
The Attempt at a Solution
1. Must find the derivative of the function.
Using the product rule a few times, the derivative becomes
<tcos(t)+sin(t), -tsin(t)+cos(t), 3>
2. Now take the magnitude
Well, this is a bit lengthy, but it simplifies pretty nice with the double angle trig identity. (I hope what I did is correct, please check my work in this step.
(tcos(t)+sin(t))^2 = t^2cos^2t+2tsin(t)*cos(t)+sin^2t
Simplifies to: t^2cos^2t+tsin(2t)+sin^2t
(-tsin(t)+cos(t))^2 = t^2sin^2t+(-2tsin(t)*cos*(t))+cos^2t
simplifies to: t^2sin^2t-tsin(2t)+cos^2t
3^2 = 9
So, when we add everything the double angles cancel out. The cos^2t+sin^2t is equal to 1. We factor a t^2 out of the t^2cos^2t and t^2sin^2t equations and this becomes t^2*1.
We are left with 9+1+t^2
3. Integrate from 0 to 2pi the square root of (t^2+10) dt
Well, here's where I am stuck.
I feel like I've done everything properly up to this point. But this integral cannot be taken easily (at least with calc 3 tools).
Did I do something wrong? Any hints on the integral?