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Second derivative from parametric form

  1. Oct 3, 2009 #1
    If x and y are defined in terms of a third vatiable say t , then to find d2y/dx2 , we cannot find d2y/dt2 and d2x/dt2 and divide them to get d2y/dx2 , i am unable to fingure out the reason for this !!!!!!!
  2. jcsd
  3. Oct 3, 2009 #2


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    This is the reason for the notation:
    [tex] \frac{d^2 y}{dx^2}[/tex]
    instead of...
    [tex] \frac{d^2 y}{d^2 x}[/tex]

    Remember that the first derivate gives the linear approximation for changes in the dependent variable as a (linear) function of changes in the dependent variable:

    [tex] dy = \frac{dy}{dx} \cdot dx[/tex]
    [tex] y(x+dx) = y(x) + \frac{dy}{dx}\cdot dx[/tex]

    Likewise the second derivative gives the second order approximation for changes in the dependent variable as a (quadratic) function fo changes in the dependent variable. Together you get:

    [tex] y(x+dx) = y(x) + \frac{dy}{dx} \cdot dx + \frac{d^2 y}{dx^2} \cdot (dx)^2[/tex]

    Notice you are looking for the second order variation in the value of y and not in the quadratic of y but as a quadratic function of the variation dx in x. So you see it is not a symmetric treatment of x and y.

    Consider the implicit equation below. I'll use dotted variables to represent t-derivatives:
    [tex] y = f(x)[/tex]

    [tex]\frac{d}{dt}y = \frac{d}{dt}f(x)[/tex]
    [tex]\dot{y} = f'(x)\dot{x}[/tex]

    [tex]\frac{d}{dt} \dot{y} = \frac{d}{dt}\left( f'(x)\dot{x}[/tex]
    [tex] \ddot{y} = \left(\frac{d}{dt} f'(x) \right) \dot{x} + f'(x)\left(\frac{d}{dt}\dot{x}\right)[/tex]
    [tex] \ddot{y} = \left(f''(x)\dot{x}\right)\dot{x} + f'(x)\ddot{x}[/tex]
    Now replace [itex] f(x) = y[/itex] to get:
    [tex] \ddot{y} = \frac{d^2y}{dx^2} \dot{x}^2 + \frac{dy}{dx}\ddot{x}[/tex]

    Now solve for the second derivative of y w.r.t. x:
    [tex] \frac{d^2 y}{dx^2} = \frac{ \ddot{y} + dy/dx \cdot \ddot{x}}{\dot{x}^2}
    = \frac{\ddot{y}}{\dot{x}^2} + \frac{\dot{y}\ddot{x}}{\dot{x}^3}[/tex]

    Notice that if you take the ratio you are trying to use:
    you get:
    [tex] \frac{\ddot{y}}{\ddot{x}}= \frac{d^2y}{dx^2}\frac{\dot{x}^2}{\ddot{x}} + \frac{dy}{dx}[/tex]
    which is not what you are trying to get.

    Finally notice how the units work out. Suppose y is some quantity measured in say kg and x has units of meters while t units of seconds.

    [tex]\frac{d^2 y}{dx^2} = \frac{d^2 y}{dx\,dx}[/tex]
    has units of kg per meter ^2.
    Note however:
    [tex] \frac{d^2 y}{dt^2}[/tex]
    has units of kg/s^2 while
    [tex]\frac{d^2 x}{dt^2}[/tex]
    has units of m/s^2 and so the ratio has units of kg/m not kg/m^2 as the 2nd derivative should have.
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