# Family of curves on a surface (Differential geometry)

• May 11th 2008, 04:09 PM
Tchakra
Family of curves on a surface (Differential geometry)
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.

http://img126.imageshack.us/img126/807/diffgeombs3.png

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:
$\displaystyle xz-hy=> v*sin(u)=h(1-cos(u))=> h= v*sin(u)/(1-cos(u))$ which is a constant.

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

And after that i am stuck ... any help would be appreciated.
• Oct 27th 2008, 12:18 PM
Rebesques
Ok, lets see...

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

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