Are $\frac{d(f(r(t),\theta(t)))}{dt}$ and $\frac{\bar{v}}{|\bar{v}|}\bar{\nabla}f$ same in spherical polar coordinates?
Where $\bar{v}$ defines the vector given by the derivative w.r.t t of the parametric curve $(r(t),\theta(t))$.
In words, I want to ask if the directional derivative is same as the derivative $\frac{d(f(r(t),\theta(t)))}{dt}$ in plane polar coordinates?
I am confused as I am not able to find them to be equal.
Note that from the chain rule we have
$$\frac{d f}{d t}=\frac{\partial f}{\partial r}\frac{dr}{dt}+\frac{\partial f}{\partial \theta}\frac{d\theta}{dt}+\frac{\partial f}{\partial \phi}\frac{d\phi}{dt}\tag1$$
With $\vec r=\hat r(t)r(t)$ and $\vec v=\frac{d\vec r}{dt}$ we can write
$$\vec v=\hat r\frac{d r}{dt}+\hat \theta r \frac{d\theta}{dt}+\hat \phi r\sin(\theta)\frac{d\phi}{dt}\tag2$$
In spherical coordinates $\nabla f(r,\theta)$ is
$$\nabla f=\hat r\frac{\partial f}{\partial r}+\hat \theta \frac1r \frac{\partial f}{\partial \theta}+\hat \phi\frac1{r\sin(\theta)}\frac{\partial f}{\partial \phi}\tag3$$
Using $(2)$ and $(3)$, we see that
$$\vec v\cdot \nabla f=\frac{\partial f}{\partial r}\frac{dr}{dt}+\frac{\partial f}{\partial \theta}\frac{d \theta}{d t}+\frac{\partial f}{\partial \phi}\frac{d\phi}{dt}\tag4$$
Comparing $(1)$ and $(4)$ reveals
$$\vec v\cdot \nabla f=\frac{d f}{d t}$$