# Arc length of

1. Sep 28, 2006

### Rozenwyn

$$c(t)=t^2\cos{t}.i+t^2\sin{t}.j+2t.k$$, and $$0 \leq{t} \leq1$$

Firstly, $$L(c) = \int_{0}_{1} || c'(t) ||.dt$$

Then, $$=\int_{0}^{1} || (-t^{2} \sin{t} + 2t \cos{t} ) i + (t^{2} \cos{t} + 2t \sin{t}) j + 2k || dt$$

Skipping one step, this simplifies to;
$$\int_{0}^{1} \sqrt{t^4 + 4t^2 + 4} .dt$$

$$\int_{0}^1 t^2+2.dt$$

$$\left[ \frac{t^3}{3} + 2t \right]_{0}^{1} = 1/3 + 2 = 7/3$$

Well, yeah is this correct ?

Let's see, I hope I didn't mess up the latex code.

Last edited: Sep 28, 2006
2. Sep 28, 2006

### Galileo

Remember that |c'(t)| is the square root of <c(t),c(t)>, so your integrand should be the square root of what you have now.

3. Sep 28, 2006

### Rozenwyn

You're absolutely right. How did I miss that ? :P... It should be correct now.