Let $v=ln(u+\sqrt{u^2+1})+C$ be the curve given on the right helicoid $x=u\cos(v),y=u\sin(v),z=2v$. Calculate the arc lengths of this curve between the points $M_1(0,0)$ and $M_2(1,ln(1+\sqrt{2}))$

So I am trying to find $r_u(u,v)$ and $r_v(u,v)$ to construct the first fundamental form, but I am having trouble with $v$ being reliant on $u$. It could be the case that I am forgetting some of my differentiation rules. Thanks for any help

(If some of the numbers seem odd, my professor changed the points to these since he said the others were misprints in the book. And this is just an extra problem not homework, so any steps after would also be helpful; I couldn't find many direct examples online elsewhere)


We usually use partial derivatives to get the metric $$g_{uu}=(\cos v,\sin v,0)\cdot (\cos v,\sin v,0)=1, \\ g_{vv}=(-u\sin v,u\cos v,2)\cdot (-u\sin v,u\cos v,2)=u^2+4 \\ g_{uu}=(\cos v,\sin v,0)\cdot (-u\sin v,u\cos v,2)=0$$ The the metric is given by $$(g)= \begin{pmatrix} 1 & 0\\ 0 & u^2+4 \end{pmatrix} $$ the curve is given by $$\alpha(u)=(u,\ln(u+\sqrt{u^2+1})\implies \alpha'(u)=(1,\frac{1}{\sqrt{u^2+1}})$$ Then $$||\alpha'(u)||=\sqrt{1\cdot 1^2+(u^2+4)\cdot (\frac{1}{\sqrt{u^2+1}})^2}$$ Finally, you can now integrate.

  • $\begingroup$ I understand that, but doesn't v's dependance on u factor in at all? $\endgroup$ – Jimmy Feb 18 '14 at 20:54
  • $\begingroup$ evaluating the metric is independent on the curve you want to find its arc length $\endgroup$ – Semsem Feb 18 '14 at 20:56
  • $\begingroup$ okay, so is the defined v used for the dv in the first fundamental form? $\endgroup$ – Jimmy Feb 18 '14 at 21:23
  • $\begingroup$ @Jimmy i complete it, sorry for inactivity $\endgroup$ – Semsem Feb 18 '14 at 22:50

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