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Tricky substitution

  1. Apr 27, 2012 #1
    Hi,

    Firstly: This is not a homework Q. Check my previous posts, you will see the stuff i ask is for my own genuine learning.

    Let:
    [itex]\vartheta[/itex] = ([itex]\Omega[/itex]*t)/2

    Now I have: d^2x/dt^2

    And I want to sub in for t.

    So:

    d/dt = d[itex]\vartheta[/itex]/dt * d/d[itex]\vartheta[/itex]
    (Basic chain rule)

    I can work out that: d[itex]\vartheta[/itex]/dt = [itex]\Omega[/itex]/2

    So:

    d/dt = [itex]\Omega[/itex]/2 * d/d[itex]\vartheta[/itex]

    Now for the second derivative:

    d^2/dt^2 = d[itex]\vartheta[/itex]/dt * d/d[itex]\vartheta[/itex] * (d/dt)

    Now, I already have an expression for d/dt = d[itex]\vartheta[/itex]/dt * d/d[itex]\vartheta[/itex]

    So I can sub this in and get:

    d^2/dt^2 = d[itex]\vartheta[/itex]/dt * d/d[itex]\vartheta[/itex] * d[itex]\vartheta[/itex]/dt * d/d[itex]\vartheta[/itex]
    Which is:
    (correct me if I am wrong):
    d^2/dt^2 = d[itex]\vartheta[/itex]^2/dt^2 * d^2/d[itex]\vartheta[/itex]^2

    This is where I need the help of the experts :-)

    The text I am trying to understand gives this:
    d^2x/dt^2 = [itex]\Omega[/itex]^2 / 4 * d^2x/d[itex]\vartheta[/itex]^2

    Any ideas how they have made that step... it seems like they have simply said:
    d[itex]\vartheta[/itex]/dt * d[itex]\vartheta[/itex]/dt = (d[itex]\vartheta[/itex]/dt)^2
    Is this acceptable?
    If [itex]\Omega[/itex] = 2[itex]\pi[/itex]*(1/t). Then its like saying:
    d^2/dt^2 (([itex]\Omega[/itex]*t)/2) = [itex]\Omega[/itex]^2 / 4


    Any ideas/help?

    thanks
     
  2. jcsd
  3. Apr 27, 2012 #2

    tiny-tim

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    hi strokebow! :smile:

    (try using the X2 button just above the Reply box :wink:)
    i don't really understand your equations :redface:

    the way they got that is

    d2x/dt2

    = d/dt (dx/dt)

    = {d/dθ (dx/dt)} dθ/dt

    = {d/dθ (dθ/dt dx/dθ))} dθ/dt

    = {(Ω/2 d2x/dθ2)} Ω/2 :wink:
     
  4. Apr 28, 2012 #3
    Hi,

    Thanks you for your reply! :-)

    The inbetweener steps would be . . . (?)

    = {d/dθ (dθ/dt dx/dθ))} dθ/dt

    = {d/dθ dx/dθ (dθ/dt )} dθ/dt

    = {d/dθ dx/dθ (Ω/2)} Ω/2

    = {d2x/(dθ)2 (Ω/2)} Ω/2

    = {(Ω/2 d2x/dθ2)} Ω/2

    yar?
     
  5. Apr 28, 2012 #4

    tiny-tim

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