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Brachistochrone travel time

  1. Apr 28, 2013 #1
    1. The problem statement, all variables and given/known data

    I have to calculate minimum travel time between two points. I already have cycloid equations in parametric form:

    [itex]x=r*(t-\sin t)[/itex]
    [itex]y=r*(1-\cos t)[/itex]​

    2. Relevant equations

    For calculating time i want to use following formula:

    [itex]\int_{0}^{a} \frac{\sqrt{1+{y'}^2}}{\sqrt{2g\,y}}dx[/itex]​

    My question is what should I substitute for y and y'?
     
  2. jcsd
  3. Apr 28, 2013 #2
    The given equations as well as the y-prime notation suggest that you would use your y(t) expression for y, and take the derivative wrt t for y'(t).

    Let me know if that helps.
     
  4. Apr 29, 2013 #3

    haruspex

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    I would think the y' in the integrand stands for dy/dx. Use dy/dx = (dy/dt)/(dx/dt).
     
  5. Apr 29, 2013 #4
    So for y should I substitute y equation or integrate dy/dx and substitute solution of that?
     
  6. Apr 29, 2013 #5

    haruspex

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    There are two possible paths. You could eliminate t from the parametric form and obtain expressions for y and y' in terms of x. But it's probably neater to go the other way and eliminate x and y: turn everything in the integral (including dx) into functions of t.
     
  7. Apr 29, 2013 #6

    [itex]y'=\frac{dy}{dx}=\frac{r*(t-\sin t)}{-r*(1-\cos t}=-\frac{\sin t}{1-\cos t}[/itex]

    [itex]\int -\frac{\sin t}{1-\cos t}\,dt=-\ln (1-\cos t)[/itex]


    Is that what you meant?
     
  8. Apr 29, 2013 #7

    haruspex

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    You left out the d/dt above and below the line in the middle step, but the final expression is right.
    No. Look at your integral. It mentions y, y' and dx, and in the range it implicitly mentions x. You have expressions for x, y and y' as functions of t. Next you need an expression for dx as a function of t and dt. Then you can substitute all those in the integral.
     
  9. Apr 29, 2013 #8
    So i don't have to integrate y' to get y? Should i use one of the parametric form equations, or all i have to do is to use integrated y' and then integrate by substitution whole expression?

    That kinda confuses me a little bit.
     
  10. Apr 29, 2013 #9

    haruspex

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    No, it isn't necessary.
    Yes, use them all. You want to replace all references in the integral to x, y and y' with references to t. What's stopping you?
     
  11. Apr 30, 2013 #10
    I think i have solution.

    What I did is I derivated x equation to get dx as function of t: [itex]dx=1-\cos t dt [/itex].
    Next I substituted this to my integral and I got [itex]\int_{0}^{a} \frac{\sqrt{1+{y'}^2}*(1-\cos t)}{\sqrt{2g\,y}}\,dt [/itex]

    Is that right?
     
  12. Apr 30, 2013 #11

    haruspex

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    Yes, that's one part of what I advised you to do. Now do all the other parts: replace the y', the y, and the x in the ranges (x=0 etc.) with their representations in terms of t. I don't understand why you haven't done this. Is what I'm saying unclear?
     
  13. Apr 30, 2013 #12
    Sorry, I've already done that but i posted only part of my solution here. Thank you for your help.
     
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