How Do You Calculate Time in the Brachistochrone Problem?

Click For Summary
SUMMARY

The Brachistochrone problem involves determining the curve along which a ball travels between two points in the least amount of time. The curve is defined as a cycloid, represented in parametric form by the equations x(θ) = C(θ - sin(θ)) and y(θ) = -C(1 - cos(θ)). To calculate the travel time, the integral t_{12} = ∫_{T_1}^{T_2} (√(1 + (y')²) / √(2gy)) dx must be evaluated, where y' = dy/dx. The discussion emphasizes the need to derive dy/dx using the chain rule, specifically dy/dx = (dy/dθ) / (dx/dθ).

PREREQUISITES
  • Understanding of parametric equations
  • Knowledge of calculus, specifically integration and derivatives
  • Familiarity with cycloid geometry
  • Basic physics principles, particularly gravitational acceleration (g)
NEXT STEPS
  • Study the derivation of cycloid equations in detail
  • Learn advanced integration techniques for evaluating definite integrals
  • Explore the application of the chain rule in calculus
  • Investigate the historical context and significance of the Brachistochrone problem in physics
USEFUL FOR

Mathematicians, physicists, engineering students, and anyone interested in optimization problems in physics and calculus.

hriby
Messages
3
Reaction score
0
I hope that you've heard about Brachistochrone problem: http://mathworld.wolfram.com/BrachistochroneProblem.html
Given two points, I can find (calculate) the courve, on which the ball needs minimum time to travel from point 1 to point 2.
I get the equation for the courve, which is cycloid, in parametric form, let say:
<br /> x(\theta)&amp;=&amp;C\left(\theta-\sin{\theta}\right),<br />

<br /> y(\theta)&amp;=&amp;-C\left(1-\cos{\theta}\right).<br />

Now I also need to calculate the time needed...
How could I calculate it out of formula below using the equation for the courve/cycloid in parametric form?
t_{12}=\int_{T_1}^{T_2} \frac{\sqrt{1+{y&#039;}^2}}{\sqrt{2g\,y}}dx, \quad y&#039;=\frac{dy}{dx}
Thanks for your answers!
Hriby
 
Last edited:
Physics news on Phys.org
how 'bout:

dy/dx = (dy/d_theta)/(dx/d_theta)
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K