Doubt with chain rule in calculating the torsion

Question

I have a problem where I have to prove that in a central force system $\;\vec F=F(r)\frac{\vec r}{r}\;$, the motion is planar

I am trying to prove that the Torsion of the system $\;[r'(t),r''(t),r'''(t)]=0\;$ (box product) is zero.Because the force is central the equation of motion is

$mr''(t)=F(r)\frac{\vec r}{r}\;$ and we know that r(position) depends on time $\vec r=\vec r(t)$.

I am not able to calculate $\;mr'''(t)=\frac{d}{dt}\Bigl(F(r)\frac{\vec r}{r}\Bigr)\;$. I am not able to put the chain rule correctly .

Can someone help me calculate $\frac{d}{dt}\Bigl(F(r)\frac{\vec r}{r}\Bigr)$, where position(r) is a function of time(t) : $\vec r=\vec r(t)$

Answer 1

\begin{align*} \frac{d}{dt}(F(r) \frac{\vec{r}}{r}) &= \dot{r} \frac{d}{dr}(F(r) \frac{\vec{r}}{r}) \\ &= \dot{r} \big( \frac{d F(r)}{dr} \frac{\vec{r}}{r} - F(r)\frac{\vec{r}}{r^2} + \frac{F(r)}{r} \frac{d \vec{r}}{dt}\frac{dt}{dr}) \\ &= \dot{r} \big( \frac{d F(r)}{dr} \frac{\vec{r}}{r} - F(r)\frac{\vec{r}}{r^2} + \frac{F(r)}{r} \frac{\vec{v}}{\dot{r}}) \end{align*}