The discussion revolves around calculating the minimum travel time between two points using the Brachistochrone problem, specifically employing cycloid equations in parametric form. Participants are exploring the appropriate substitutions for variables in an integral that involves the derivative of the vertical position with respect to horizontal position.
The conversation is active, with participants providing guidance on how to approach the substitutions needed for the integral. There is a recognition of multiple paths to the solution, and some participants express confusion about the necessary steps, indicating a collaborative effort to clarify the process.
Participants are navigating the complexities of the integral and the relationships between the variables, with some expressing uncertainty about the integration process and the need for clarity on the use of parametric forms.
I would think the y' in the integrand stands for dy/dx. Use dy/dx = (dy/dt)/(dx/dt).Ocifer said: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).
haruspex said:I would think the y' in the integrand stands for dy/dx. Use dy/dx = (dy/dt)/(dx/dt).
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 said:So for y should I substitute y equation or integrate dy/dx and substitute solution of that?
haruspex said: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.
You left out the d/dt above and below the line in the middle step, but the final expression is right.Divh said:y'=\frac{dy}{dx}=\frac{r*(t-\sin t)}{-r*(1-\cos t}=-\frac{\sin t}{1-\cos t}
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.\int -\frac{\sin t}{1-\cos t}\,dt=-\ln (1-\cos t)
Is that what you meant?
haruspex said: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.
No, it isn't necessary.Divh said:So i don't have to integrate y' to get y?
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?Should i use one of the parametric form equations
haruspex said: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?
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 said: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 said: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?