Second derivative from parametric form

[vikcool812]

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 !

 

[jambaugh]

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

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:

$$dy = \frac{dy}{dx} \cdot dx$$
or
$$y(x+dx) = y(x) + \frac{dy}{dx}\cdot dx$$

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:

$$y(x+dx) = y(x) + \frac{dy}{dx} \cdot dx + \frac{d^2 y}{dx^2} \cdot (dx)^2$$

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:
$$y = f(x)$$

$$\frac{d}{dt}y = \frac{d}{dt}f(x)$$
$$\dot{y} = f'(x)\dot{x}$$

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

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

Notice that if you take the ratio you are trying to use:
$$\frac{\ddot{y}}{\ddot{x}}$$
you get:
$$\frac{\ddot{y}}{\ddot{x}}= \frac{d^2y}{dx^2}\frac{\dot{x}^2}{\ddot{x}} + \frac{dy}{dx}$$
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.

$$\frac{d^2 y}{dx^2} = \frac{d^2 y}{dx\,dx}$$
has units of kg per meter ^2.
Note however:
$$\frac{d^2 y}{dt^2}$$
has units of kg/s^2 while
$$\frac{d^2 x}{dt^2}$$
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.