I need to find the arc lenght of this curve (in cylindrical coordinates):
$$\boldsymbol\Psi(t)=(t-1)\mathbf{\hat{h_\rho}}+(t^2-1)\mathbf{\hat{h_z}}$$
With $\varphi=t$
I already tried to use the formula:
$$L=\int_{t_o}^{t_1}{\sqrt{ \left( \frac{\partial\rho}{\partial t} \right)^2 +\rho^2\left( \frac{\partial\varphi}{\partial t} \right)^2 +\left( \frac{\partial z}{\partial t} \right)^2 }} \space\space dt$$
However, with this formula I'm getting irrational numbers inside the square root and I'm quite confused.
Any hint? Thanks in advance.
PS. Sorry for the small integral, I didn't figure how to fix it.
Assuming $\phi(t) = t$ just as you mention. Then,
The parametrized equations are: $\rho(t) = t - 1, \phi(t) = t, z(t) = t^2 -1 $.
Thus, $\frac{d}{dt}\rho(t) = 1, \frac{d}{dt}\phi(t) = 1, \frac{d}{dt}z(t) = 2t $.
Therefore, $\sqrt{\left(\frac{d}{dt}\rho(t)\right)^2 + \rho^2 \left(\frac{d}{dt}\phi(t)\right)^2 + \left(\frac{d}{dt}z(t)\right)^2} = \sqrt{1 + (t-1)^2 + 4t^2 } = \sqrt{5t^2 - 2t + 2} $.
Finally, completing the square and a substitution $u=\sqrt{5}x-\frac{1}{\sqrt{5}}$ gives: \begin{align*} L = \int^{t_1}_{t_0}\sqrt{5t^2 - 2t + 2}dt &= \frac {1}{\sqrt{5}}\int^{t_1}_{t_0}\sqrt{u^2 + \frac 95}du\\ \end{align*} Now this requires one more substitution, namely $s = \frac{3\tan(u)}{\sqrt{5}}$,to solve. Hopefully this helps.