Optimizing Travel Time: Calculating the Brachistochrone Curve

[Divh]

Homework Statement

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

$x=r*(t-\sin t)$
$y=r*(1-\cos t)$​

Homework Equations

For calculating time i want to use following formula:

$\int_{0}^{a} \frac{\sqrt{1+{y'}^2}}{\sqrt{2g\,y}}dx$​

My question is what should I substitute for y and y'?

 

[Ocifer]

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.

 

[haruspex]

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).

I would think the y' in the integrand stands for dy/dx. Use dy/dx = (dy/dt)/(dx/dt).

 

[Divh]

I would think the y' in the integrand stands for dy/dx. Use dy/dx = (dy/dt)/(dx/dt).

So for y should I substitute y equation or integrate dy/dx and substitute solution of that?

 

[haruspex]

So for y should I substitute y equation or integrate dy/dx and substitute solution of that?

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.

 

[Divh]

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.

$y'=\frac{dy}{dx}=\frac{r*(t-\sin t)}{-r*(1-\cos t}=-\frac{\sin t}{1-\cos t}$

$\int -\frac{\sin t}{1-\cos t}\,dt=-\ln (1-\cos t)$


Is that what you meant?

 

[haruspex]

$y'=\frac{dy}{dx}=\frac{r*(t-\sin t)}{-r*(1-\cos t}=-\frac{\sin t}{1-\cos t}$

You left out the d/dt above and below the line in the middle step, but the final expression is right.

$\int -\frac{\sin t}{1-\cos t}\,dt=-\ln (1-\cos t)$
Is that what you meant?

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.

 

[Divh]

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.

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.

 

[haruspex]

So i don't have to integrate y' to get y?

No, it isn't necessary.

Should i use one of the parametric form equations

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?

 

[Divh]

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?

I think i have solution.

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

Is that right?

 

[haruspex]

I think i have solution.

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

Is that right?

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?

 

[Divh]

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?

Sorry, I've already done that but i posted only part of my solution here. Thank you for your help.