# Finding arc length of vector valued function

1. Jul 9, 2012

### Yitin

1. The problem statement, all variables and given/known data

Find arc length of the graph of
r(t) = ti + ( (t6/6) - (6/t4) )j + t√3 k
1≤t≤2

2. Relevant equations

Arc length = ∫ ||dr/dt|| dt
(Integral from t0 to t1 of norm of derivative of r)

3. The attempt at a solution

dr/dt = i + (t5 + 24/(t3) )j + √3 k

12 = 1
(√3)2 = 3
(t5 + 24/(t3) )2 = t10 + 48t2 + 576/t6

||dr/dt|| = √(1 + 3 + t10 + 48t2 + 576/t6)

I get stuck here because I do not know how to integrate that. It doesn't seem anywhere close to a perfect square.
Did I make a mistake somewhere, or is there some kind of property to integrate this?

2. Jul 9, 2012

### jbunniii

You did not differentiate the $6/t^4$ term correctly.

3. Jul 9, 2012

### Yitin

Ok, so now it should be

(t5 +24t-5)2
(t10 + 48 + 576t-10)

||dr/dt|| = √( 1 + 3 + t10 + 48 + 576t-10 )
||dr/dt|| = √( t10 + 52 + 576t-10 )

Which I am still not sure how to integrate, but it at least looks nicer.