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Parametric equation confusion of d2y/dx2

  1. Apr 1, 2012 #1
    x=t2 and y=t3-3t, find dy/dx and d2y/dx2

    I understand how to get to dy/dx but an confused on how to get to d2y/dx2 can someone please explain in depth, I know the formula is (d/dt dy/dx)/(dx/dt) I dont understand where d/dt comes from and what it is. why do we not just derive it twice? I really do not understand the 2nd part at all.
  2. jcsd
  3. Apr 1, 2012 #2


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    It's just several applications of the chain rule

    [tex] \frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \left(\frac{dy}{dt} / \frac{dx}{dt} \right) [/tex]
    from what your OP says you're familiar with this calculation. We're just going to do it again
    [tex] \frac{d}{dx} \left( \frac{dy}{dx} \right)[/tex][tex] = \frac{d}{dt} \left( \frac{dy}{dx} \right) \frac{dt}{dx}[/tex][tex] = \frac{d}{dt} \left( \frac{dy}{dt}/\frac{dx}{dt} right) / \frac{dx}{dt} [/tex]
    Last edited by a moderator: Apr 2, 2012
  4. Apr 1, 2012 #3
    Im sorry but i still have no idea : \.
  5. Apr 1, 2012 #4
    It will perhaps become more clear if you switch your notation slightly. How about we denote dy/dx instead by ω. Okay, so ω is a function, and what we are looking for is dω/dx. We will apply the chain rule:

    dω/dx=dω/dt * dt/dx=d/dt(ω)*dt/dx

    Now we simply put back dy/dx for ω to get

    d2y/dx2=d/dt(dy/dx)*dt/dx as claimed. Hope that clears it up :smile:
  6. Apr 1, 2012 #5


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    Only one note of caution with this approach: make sure dt/dx is defined which means dx/dt is non-zero (i.e. t changes with respect to x at the point you are considering and hence x also changes with respect to y)
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