# How to obtain a 2D-coordinate system from two family of curves?

1. Aug 10, 2012

### mnb96

Hello,

it is known that if we have a curvilinear coordinate system in ℝ2 like $x=x(u,v)$, $y=y(u,v)$, and we keep one coordinate fixed, say $v=\lambda$, we obtain a family of one-dimensional curves $C_{\lambda}(u)=\left( x(u,\lambda),y(u,\lambda) \right)$. The analogous argument holds for the other coordinate u. These family of curves are sometimes called coordinate lines, or level curves.

My question is: if I am given two family of curves $C_v(u)$ and $C_u(v)$ is it possible to obtain the system of curvilinear coordinates $x(u,v)$, $y(u,v)$ that generated them?

2. Aug 10, 2012

### Staff: Mentor

Consider the family of radial curves and a family of nested circles. Together they create polar coordinates.

Can you determine the curvilinear coordinate system from that knowledge?

3. Aug 10, 2012

### mnb96

Yes, if we have a family of circles $C_r(\theta)=\left( r\cos\theta, r\sin\theta \right)$ for some $r\in \mathbb{R}^+$, and a family of straight lines passing through the origin $C_\theta(r)=\left( r\cos\theta, r\sin\theta \right)$ for some $\theta\in[0,2\pi)$ the solution is quite trivial.

I was interested more in a general procedure or simply a strategy that I could follow to solve this kind of problem.

If we cannot answer the general question then let's try at least a less trivial example I was unable to solve like this one: we have two families of "parallel" exponential curves, the first family is $C_\lambda(u) = (u, \; e^u +\lambda)$ for some fixed real scalars v, and the other family is $C_k(v) = (e^{-v} + k, \; v)$ for some real k.
I was unable to obtain two functions x(u,v) , y(u,v) such that $C_\lambda(u) = (x(u,\lambda), \; y(u,\lambda))$ and $C_k(v)=(x(k,v),\; y(k,v))$

Last edited: Aug 10, 2012
4. Aug 10, 2012

### Staff: Mentor

so i guess a way to investigate this is to determine if you lost any info when generating the two sets of curves such that you would find multiple different answers when you reverse the problem.

5. Aug 12, 2012

### GarageDweller

well I imagine you'd have to check the curves, not all curves form coordinates systems.