# The first fundamental form to compute the lenght of a curve and the angle between tw0 curves

In my case, we consider an helicoid $$\varphi(u,v)= (u \; \sin\, v, u \; \cos\, v,v).$$ Its first fundamental form is $$\mathit{I} = \begin{pmatrix} E & F\\ F & G \end{pmatrix} = \begin{pmatrix} 1 & 0\\ 0 & u^2 + 1 \end{pmatrix}.$$

Now, we have to compute the area, the length of the sides and the angles of the triangle, defined by $$0< u< \sinh\, v,\; 0< v< a.$$

For the area I used the formula $$A = \int_{0}^{a}\int_{0}^{\sinh\, v}\sqrt{EG-F^2}dudv$$. For the sides I have done the following parametrizations $$\alpha _1(t) = (0,t)\quad \text{ where }\quad 0< t< a$$ $$\quad\quad\alpha _2(t) = (t,a)\quad \text{ where }\quad 0< t< \sinh\, a$$ $$\quad\quad\alpha _3(t) = (\sinh\, t,t)\quad \text{ where }\quad 0< t< a.$$ For the length of the curves I have that if $$\alpha (t)=(u(t),v(t))$$ then $$L = \int_{a}^{b} \sqrt{E(u')^2+2Fu'v'+G(v')^2}dt.$$ And finally, if $$\alpha$$ and $$\beta$$ are two curves that intersect in the surface they form an angle $$\theta$$ such that $$\cos\, \theta=\frac{\mathit{I(\alpha',\beta')}}{\left | \alpha' \right |\left | \beta' \right |}.$$ The problem I have is that the $$G$$ has an $$u^2$$ and the lenght of the curves and their angles depends of $$u$$. So there is something I'm not understanding well. I appreciate any help. Thanks!

You've set this up well, except for one issue. All you have to do is realize where the vertices of the triangle are (in $(u,v)$ coordinates). One is at $(0,0)$, one is at $(0,a)$, and the last is at $(\sinh a,a)$. So are your curves the correct boundary curves of the triangle? (Hint: One is wrong.) But once you know where the vertices are, there's no problem computing the first fundamental form at this points to find angles.
• $\alpha_2(t) = (t,a)$ right? Commented Apr 25, 2018 at 20:37
• @TedShifrin So it means we only need to calculate the first fundamental form at the intersection point and that makes $u$ gone? In particular of the problem, if $\alpha (t_0)=\beta(s_0)=\varphi(u_0, v_0)$ then we have $G=u_0^2 +1$. Am I correct sir ? Commented Jun 15, 2021 at 15:11
• @RopuToran I don't know what your "and that makes $u$ gone" means. But, yes, you need $E,F,G$ at the intersection point(s). Commented Jun 15, 2021 at 15:33
• @TedShifrin Sorry for my confusing expression. What I mean was when we substitute $u$ with value $u_0$ on $G$, the cosine function becomes a number, not a function anymore. Thank you! Commented Jun 16, 2021 at 4:56