# Meaning of k in D. E. problem

1. Aug 20, 2011

### diffrac

Hello!

I am currently studying the brachistochrone problem in the Boyce-Di Prima D. E. book. While following their indications on the resolution of the problem for the curve as it passes through the point (x0; y0), i find myself wondering what the constant k really means (physically speaking).

Using the bisection method i was told can find a value for k (replacing values in the function f(θ)=2θ-2"sin" θ+"cos" θ-1) but yet again what does that value represent and how is that a solution to the brach. problem?

I know this sounds a little murky, but that's the state of my understanding...so thanks for any help!

2. Aug 20, 2011

### Dr. Seafood

Where is the symbol "k" in any of those expressions you wrote out? >_>

3. Aug 21, 2011

### diffrac

it's actually in the parametric equations: x= (k^2 (θ-"sin" θ))/2 and
y=(k^2 (1-"cos" θ))/2
for the position in function of time of the particle as it slides along the curve (i think).

4. Aug 22, 2011

### kdbnlin78

The parametric representation you give above is actually the parametrisation of the cycloid. The value of $k$ here (or $k^2$ if you will) represents the "height" of the humps in the equation of the cylcoid itself.

If you are unfamiliar with the cycloid, check Wolfram.

http://mathworld.wolfram.com/Cycloid.html

5. Aug 22, 2011

### diffrac

So if i say i want the cycloid to pass through the point x0=1 and y0=2 and i resolve the parametric equations until i get f(θ)=2θ-2"sin" θ+"cos" θ-1 and then i replace random values into the function until i get one that will make it equal zero, then k equals it as well (k≅2.193)...and so if i understand correctly 2.193 is what 'makes' the equation 'draw' the arc accurately based on the first point (x0=1 and y0=2) it starts from?

Thanks a lot for the informative wolfram reference!