Finding L(c) when c(t) = t^2cos(t).i + t^2sin(t).j + 2t.k

[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.

 

[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.

 

[Rozenwyn]

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