# Multivariable calculus: Length of curve

1. Oct 26, 2013

### Minusu

1. The problem statement, all variables and given/known data
Find the length of the curve traced by the given vector function on the indicated interval.
r(t)=<t, tcost, tsint> ; 0<t<pi

2. Relevant equations

s= ∫||r'(t)||dt

3. The attempt at a solution

r'(t) = <1, -tsint + cost, tcost + sint>
s= ∫||r'(t)||dt

||r'(t)|| = sqrt(1^2+(-tsint+cost)^2+(tcost+sint)^2)
||r'(t)|| = sqrt(1+t^2sin^2t-2tsintcost+cos^2t+t^2cos^t+2tsintcost+sin^2t)
||r'(t)|| = sqrt(1 + t^2sin^2t+t^2cos^2t+sin^2t+cos^2t)

I'm not sure where to go from here. I was thinking of using the trig identity sin^2x+cos^2x= 1, but I don't think I can do that.

2. Oct 26, 2013

### haruspex

Why not?

3. Oct 26, 2013

### Minusu

I'd end up with sqrt(2+t^2), but then I don't know how to integrate that from 0 to pi. At first, I tried u-substitution, so u =2+t^2, du=2tdt, du/2=tdt, but there is no tdt outside the sqrt so that I can substitute.

4. Oct 26, 2013

### CAF123

Try integration by parts.

5. Oct 26, 2013

### Minusu

:S isn't what I'm doing integration by parts? Or do you mean separate it into two integrals? I don't think you can do that with a sqrt sign

6. Oct 26, 2013

### CAF123

No, what you did before was substitution. Your integral is $$I= \int_0^\pi \sqrt{t^2 + 2}\,dt$$ Let $u = \sqrt{t^2+ 2}$ and $dv = dt$. With one application of integration of parts, you will get a form that appears to not be any simpler: $$t \sqrt{t^2 + 2}|_0^\pi - \int_0^\pi t \frac{t}{\sqrt{t^2 + 2}}dt$$What I thought you could then do was use integ by parts again, but this only leads to the completely trivial statement $I= I$
I'll think over it some more - maybe a hyperbolic substitution would be useful.

7. Oct 26, 2013

### CAF123

Try the substitution t = √2 tan x.

8. Oct 26, 2013

### haruspex

I think √2 sinh x works out a little more easily.

9. Oct 26, 2013

### CAF123

Yes, I agree.