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).