1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Brachistochrone Differential Equation

  1. Jan 20, 2013 #1
    The first part of this problem asks me to solve the following for y' :
    [tex] \left( 1 + {y'}^2 \right)y = k^2 [/tex]
    So I have:
    [tex] 1 + {y'}^2 = \frac{k^2}{y} [/tex]
    [tex] {y'}^2 = \frac{k^2}{y} - 1 [/tex]
    [tex] y' = \sqrt{{\frac{k^2}{y} - 1}} [/tex]

    Then I am asked to show that if I introduce the following:
    [tex] y = k^2 \sin^2t [/tex]

    Then the equation for y' found above takes the form:
    [tex] 2k^2\sin^2t dt = dx [/tex]

    My attempt looks like this:
    [tex] y' = \sqrt{ \frac{k^2}{y} - 1} [/tex]
    [tex] y' = \sqrt{ \frac{k^2}{k^2 \sin^2t}-1} [/tex]
    [tex] y' = \sqrt{ \csc^2t - 1 } [/tex]
    [tex] y' = \sqrt{ \cot^2t } [/tex]
    [tex] \frac{dy}{dx} = \cot t [/tex]

    At this point I feel like I have gotten off track or that I am following the wrong line of argumentation. Either that or I'm out of steam and I can't see how keep this going and get what I'm being asked for. Any insight would be greatly appreciated.
  2. jcsd
  3. Jan 20, 2013 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You need to use the substitution for y in dy/dx also. What does that give you?
  4. Jan 20, 2013 #3
    Thank you!!! I get:

    [tex] \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} [/tex]
    [tex] \frac{dy}{dx} = 2k^2 \sin t \cos t \frac{dt}{dx} [/tex]
    [tex] \cot t = 2k^2 \sin t \cos t \frac{dt}{dx} [/tex]
    [tex] 1 = 2k^2 \sin^2 t \frac{dt}{dx} [/tex]
    [tex] dx = 2k^2 \sin^2 t dt [/tex]

    Your genius rivals that of Gauss himself!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook