the core of a Central Forces problem is that F(r) = F(r)$ \hat{r}$. Due to this, the choice of spherical coordinates is pretty obvious.
So now, $$\mathbf{\dot{r}} = r\,\mathbf{\hat{r}} + r\dot{\varphi} \,\boldsymbol{\hat{\varphi}} \tag{*}$$
How do we get this? I would suggest checking out a classical mechanics book, like Taylor or Goldstein. Basically, this equation is a vector representation of radial and transverse velocities.
Differentiating (*) you would easily yield both the radial and transverse accelerations. I would highly suggest you try doing it yourself. Once you are done differentiating you would be getting two components, one of them would be the radial acceleration and the other the transverse acceleration.
hint on how to get equation (*).
$\mathbf{r} = (x, y) = (r\sin{\varphi}, r\cos{\varphi}) $
$\mathbf{\dot{r}}=\frac{d\mathbf{r}}{dt}= {\mathbf {v}}={\dot {r}}(\cos \varphi ,\ \sin \varphi )+r{\dot {\varphi }}(-\sin \varphi ,\cos \varphi ) $
From here on hopefully you can clearly see what the unit vectors are.
Best Answer
Close. This is what you want: $$ \frac{dx}{dt}(t)=\alpha x(t)^{4}\text{ where }\alpha\text{ is a constant}. $$ $x(t)$ is the position of the particle at time $t$, from which it follows that $\frac{dx}{dt}(t)$ is its velocity at time $t$. $\alpha$ appears in the above since the question asks the velocity to be proportional to $x(t)^{4}$.