The arclength parametrization of $\gamma(t)=(2t,(t-1)^2,3t^3)$

Question

Find the arclength paramztrization of the curve $$\gamma(t)=(2t,(t-1)^2,3t^3), t\in \mathbb R$$

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We have $$s(t)=\int_{t_0}^t\|\dot\gamma(u)\|du=\int_{t_0}^t\sqrt{4+4(u-1)^2+(9u^2)^2}du$$ I couldn't calculate it.

Any help is appreciated !

Answer 1

I am almost sure that there is a typo in the question. It could be $$\gamma(t)=(2t,(t-1)^2,3t^\color{red}{2})$$ If this is the case, then $$L=\int \sqrt{40 t^2-8 t+8}\,dt$$ $$t=\frac {u+1}{10}\implies L=\frac{1}{5 \sqrt{10}}\int\sqrt{u^2+19}\,du$$ $$u=\sqrt{19} \sinh (v)\implies L=\frac{19}{5 \sqrt{10}}\int \cosh^2(v)\,dv$$

Now, this one is very simple. When done, go back to $u$ and then to $t$.