# Closed curve and orthogonal curvilinear coordinate system

1. Mar 23, 2006

### traianus

Hello, I have a "simple" problem for you guys. I am not expert in math and so try to be simple.

I explain the problem by starting with one example. The polar coordinate system has the following main property: with two parameters, rho and theta, each point is described as the intersection of a circle and a straight line. These two curves are orthogonal and so the curvilinear coordinate system is orthogonal.

Another example is if we use ellipses and hyperbolas: they are orthogonal. The transformation is simple:

x = c*cosh(a)* cos(b) (1)
y = c*sinh(a) * sin(b) (2)

where a>0, and 0<=b<=2pi

Now my question. Suppose that we like to use a set of curvilinear coordinate system (plane x-y only) which is orthogonal. Suppose that instead of ellipses or circles one of the type of curves is a closed path defined by (for example) the equation

x^8/A^8 + y^8/B^8 = 1

which is closed curve similar to an ellipse. Ho do I find a similar trasnformation (like equations (1) and (2)) with the property that the curvilinear coordinate system is also orthogonal? Please help me!

2. Mar 23, 2006

### Hurkyl

Staff Emeritus
Did you mean the family of curves (parametrized by k > 0)

x^8/A^8 + y^8/B^8 = k

where A and B are fixed constants?

In the obvious way! Just look at what you want: you want functions of two variables a and b such that:

(1) The derivative WRT a is parallel to x^8/A^8 + y^8/B^8 = k
(2) The derivative WRT b is perpendicular to x^8/A^8 + y^8/B^8 = k

So you should be able to set up a system of partial differential equations. I guess it might not be very easy to solve, though.

3. Mar 23, 2006

### traianus

Thank you. Can you show the system of partial differential equations that solves the problem? Actually in my case I like to have the coordinate system of the "hyper-ellipse"

x^(2n)/A^(2n) + y^(2n)/B^(2n) = k

where n is a positive integer.

4. Mar 23, 2006

### Hurkyl

Staff Emeritus
If C(k) is the loop defined by:

x^8 / A^8 + y^8 / B^8 = k

and (p, q) is a point on C(k)

then can you find a tangent vector to C(k) at (p, q)? What about a vector perpendicular to C(k) at (p, q)?

5. Mar 23, 2006

### traianus

It is an implicit function. So I think that the tangent has equation

7*p^7/A^8*(x-p) + 7*q^7/B^8*(y-q) = 0 (1)

from (1) it follows that the vector perpendicular to the curve is

Vperp= q^7/B^8 * i - p^7/A^8*j (2)

where i,j are the unit vectors. The vector parallel to the curve is then

Vpar = p^7/A^8*i+ q^7/B^8*j (3)

Let me see if I get your point. Curvilinear coordinates means that there is a transformation of the type

x = f(a,b) (4)

y = g(a,b) (5)

Now I have to obtain the system. Please correct if am wrong. If I am not, can you write the system of equations?

6. Mar 24, 2006

### Hurkyl

Staff Emeritus
Your work is off. I think you just switched which is perpendicular and which is parallel.

You're right that it's an implicit equation: when you differentiate, you should get:

$$\frac{8 p^7}{A^8} \, dx + \frac{8 q^7}{B^8} \, dy = 0$$

is the equation for motion along your loop at the point (p, q).

Yep.

Since you wanted one of your parameters to go around the loop, that means when you partially differentiate, the result should be the tangent vector! i.e.

$\frac{\partial}{\partial a} \langle f(a, b), g(a, b) \rangle$ = the tangent vector at <f(a, b), g(a, b)>

and a similar condition for the other direction!

7. Mar 25, 2006

### HallsofIvy

Staff Emeritus
Finding "orthogonal trajectories" used to be a standard problem in ordinary d.e.s. If a family of curves is given by f(x,y)= c, then
$$\frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}y'= 0$$
so
$$y'= -\frac{\frac\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$

The "orthogonal trajectories" must satisfy the differential equation
$$y'= \frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}$$

In the case that $$\frac{x^{2n}}{A^{2n}}+ \frac{y^{2n}}{B^{2n}}= k$$ (you still haven't answered the question as to whether A and B are fixed constants while k determines the member of the family- actually, I suspect that there are some relations between A, B, k), assuming that A, B are constant while k varies over the family, the orthogonal trajectories satisfy
$$y'= \frac{A^{2n}}{B^{2n}}\frac{y^{2n-1}}{x^{2n-1}}$$

8. Mar 25, 2006

### traianus

Thank you for you answers. A, B are constant, as n. So, what is the parameteric representation

y = f(a,b)

x = g(a,b)

in which when a = constant I have the curve

$$\frac{x^{2n}}{A^{2n}}+ \frac{y^{2n}}{B^{2n}}= k$$

and when b = constant I have a family of curves orthogonal to it (like it happens for polar coordinates in which the lines are perpendicular to the circles)?

9. Mar 26, 2006

### traianus

A and B must be positive. If I move from a curve fto another one, A and B can change, but they can be function ONLY of the parameter a. Also, they have to be positive.

10. Mar 28, 2006

### traianus

Nobody knwos?