What is the significance of the constant k in the brachistochrone problem?

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In summary, the conversation discusses the brachistochrone problem and the constant k in the parametric equations. The value of k represents the height of the humps in the equation of the cycloid, and by finding a value for k using the bisection method, it can accurately draw the arc based on the initial point. The parametric equations given are actually the parametrisation of the cycloid, and the value of k can be calculated to make the equation equal zero and accurately draw the arc. The conversation also references a helpful Wolfram resource for further understanding of the cycloid.
  • #1
diffrac
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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!
 
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  • #2
Where is the symbol "k" in any of those expressions you wrote out? >_>
 
  • #3
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
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
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!
 

1. What does the "k" represent in a D.E. problem?

The "k" in a D.E. problem represents a constant or coefficient that is used in the differential equation to describe the relationship between the dependent and independent variables. It is typically a numerical value that is determined through experimentation or theoretical calculations.

2. How is the value of "k" determined in a D.E. problem?

The value of "k" can be determined through various methods such as integrating the differential equation, solving for the value that makes the equation balanced, or using initial or boundary conditions. In some cases, the value of "k" may also be assumed based on experimental data.

3. Is the value of "k" unique for each D.E. problem?

Yes, the value of "k" is unique for each D.E. problem as it is dependent on the specific variables and conditions described in the equation. It cannot be generalized or applied to other D.E. problems.

4. Can the value of "k" change over time in a D.E. problem?

Yes, the value of "k" can change over time in a D.E. problem if the variables or conditions involved in the equation also change. This is often seen in dynamic systems or processes that are subject to external factors.

5. How does the value of "k" affect the solution of a D.E. problem?

The value of "k" can greatly affect the solution of a D.E. problem as it can alter the shape and behavior of the solution curve. A larger or smaller value of "k" can result in a steeper or shallower curve, respectively, and can also impact the stability of the solution.

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