A force can be expressed as $F=f(r)e_r$ is called a a central force. Show that the angular momentum $J(mr^2d \theta /dt$ in this case) is conserved.
NOTE: $a(t) = [\frac{d^2r}{dt^2} - r(\frac{d\theta }{dt})^2]e_r + [r (\frac{d^2 \theta}{d^2}) + 2 (\frac{dr}{dt})(\frac{d \theta}{dt})]e_\theta$
Solution: Newton's equations, $ma=F$ can be written as....
$ \tag{1} m[\frac{d^2r}{dt} - r(\frac{d \theta}{dt})^2] = f(r)$
and
$ \tag{1} m(r (\frac{ d^2 \theta}{ dt^2}) + 2 (\frac{dr}{dt})(\frac {d \theta}{dt}))=0$
But this equation can be written as
$\tag{2} (\frac{1}{r})(\frac{d}{dt})(mr^2 (\frac{d \theta}{dt})) = 0$
which gives $J = mr^2 \frac{d \theta}{dt}$ to be a constant.
So I am having trouble figuring out how to get from step 1 to step 2. How can (1) be written as (2) when there are: $\frac{d^2 \theta}{dt^2}$, and $\frac{dr}{dt}$ in equation 1?
Work backward: The critical point is that
\begin{align} \frac{d}{dt} \left(r^2 \frac{d\theta}{dt}\right) & = \frac{d}{dt}r^2 \frac{d\theta}{dt} + r^2 \frac{d}{dt} \frac{d\theta}{dt} \\ & = 2r\frac{dr}{dt} \frac{d\theta}{dt} + r^2 \frac{d^2\theta}{dt^2} \\ & = r \left(2\frac{dr}{dt} \frac{d\theta}{dt} + r \frac{d^2\theta}{dt^2}\right) \end{align}
Just an application of the product rule for the derivative.