Family of curves on a surface (Differential geometry)

Tchakra · #1 $Old$ 05-11-2008, 06:09 PM

Here is a question i am not sure how to tackle, I am not familiar with how to deal with family of curves and don't really have much time to look around for the definition as i am sitting the exam in two days.

The question is divided into three parts: Here is my attempt any help appreciated.
1) I am have no idea, i think it is a case of knowing the definition and i don't.

2) It is simply constraining the local parametrization to the given function so:
$xz-hy=> v*sin(u)=h(1-cos(u))=> h= v*sin(u)/(1-cos(u))$ which is a constant.

3) $\psi(u,v)=const$ is like phi therefore the tangent vectors to the family defined by the psi are of the multiples of $\psi_{v}x_{u}-\psi_{u}x_{v}$
So for the families to be orthogonal their tangent must be orthogonal and so $(\psi_{v}x_{u}-\psi_{u}x_{v}).(\phi_{v}x_{u}-\phi_{u}x_{v})=0$
Using the fundamental forms E=1=G and F=0 we get $\psi_{v}\phi_{v}+\psi_{u}\phi_{u}=0$ which after differentiating gives $\psi_{v}\sin(u)-\psi_{u}v=0$

And after that i am stuck ... any help would be appreciated.

Rebesques · #2 $Old$ 10-27-2008, 02:18 PM

Ok, lets see...

1) Consider a curve in the domain. The gradient of $\phi$ is normal to the tangent $(du,dv)$ of the curve. Since all directions of such surface curves are given by $x_{u}du+x_vdv$ , the functional dependence of $x$ and $\phi$ means that the determinant of their Jacobian matrix is zero, whence the required result.

2) We have $\psi_{v}\sin(u)-\psi_{u}v=0,$ so $\sin(u)du=vdv,$ from which $\psi(u,v)=\frac{v^2}{2}-\cos(u)=const.$