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Have i done this parametric differentiation right?

  1. May 3, 2006 #1
    y=t+cost x=t+sint

    dy/dt=1-sint dx/dt=1+cost

    dy/dx= (dy/dt).(dt/dx)

    = (1-sint).1/(1+cost) = (1-sint)/(1+cost)
    = 1-tant

    and how do i get from there to the second order differential?
  2. jcsd
  3. May 3, 2006 #2


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    Homework Helper

    How did you go from:
    [tex]\frac{dy}{dx} = \frac{1 - \sin t}{1 + \cos t} \quad \mbox{to} \quad = 1 - \tan t[/tex]?
    Just stop at [tex]\frac{dy}{dx} = \frac{1 - \sin t}{1 + \cos t}[/tex]. It's okay, you don't have to simplify it any further.
    To find:
    [tex]\frac{d ^ 2 y}{dx ^ 2}[/tex], we use the Chain rule, and the derivative of inverse function:
    [tex]\frac{dk}{du} = \frac{1}{\frac{du}{dk}}[/tex]
    [tex]\frac{d ^ 2 y}{dx ^ 2} = \frac{d \left( \frac{dy}{dx} \right)}{dx} = \frac{d \left( \frac{dy}{dx} \right)}{dt} \times \frac{dt}{dx} = \frac{d \left( \frac{dy}{dx} \right)}{dt} \times \frac{1}{\frac{dx}{dt}}[/tex]
    Can you go from here? :)
  4. May 3, 2006 #3
    the second order derivitve is just the derivitive of dy/dx all divided by dx/dt
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