# Weird arc length question

1. Mar 17, 2005

### JonF

I want to find two functions f(n,x) and g(n,x) such that f(n,x)sin(g(n,x)) always has a constant arc length over some interval [a,b]. Where n increases the amplitude but decreases the period.

Any suggestions?

Last edited: Mar 17, 2005
2. Mar 17, 2005

### arildno

Well, you've got the trivial choice $$f(n,x)=\frac{1}{\sin(g(n,x)}$$

In the general case, the arc-length will be a function of "n".
Can you set up a differential equation for f and g which ensures that the arc-length is just the constant function?

3. Mar 17, 2005

### JonF

I did set up a DE. That’s how I know they have to be f(n,x) not just f(n)

If they were…
$$f(x) = g(n)sin(x*p(n))$$
$$f’(x) = g(n)p(n)cos(x*p(n))$$

So arch length is:

$$\int_{0}^{1} \sqrt{{1+ (g(n)p(n)cos(p(n)) )^2} dx =c$$

Differentiate
1+ (g(n)p(n)cos(p(n)))^2 – (1+ (g(n)p(n))^2) = 0

(g(n)p(n)cos(p(n)))^2 = (g(n)p(n))^2)

cos(p(n)) = 1

p(n) = 0

So I know if they were just functions of just n it would be trivial. For functions of x and n (possibly I will need to put a and b in there also) I know how to set up the de, no clue how to solve it.

4. Mar 17, 2005

### Crosson

Have you already tried seperation of variables to transform your partial DE into an ordinary DE?

What I mean is assume g(x,n) = X(x)*N(n).