This alinea is about the $$\cot(\alpha)= \dfrac{d}{d\theta}\cdot \ln(r) = \dfrac{1}{r} \dfrac{dr}{dt}.$$ Where does the $\ln (r)$ come from? How can you derive it from that picture? I want to use that for the $$\nabla f(r,\theta)= \dfrac {\partial f}{\partial r}\vec e_r + \dfrac {1}{r}\dfrac{\partial f}{\partial \theta}\vec e_\theta.$$ I know what the definition is of the gradient (i.e. it's a partial derivative of the function with respect to every component in a particular dimension for my case $x,y,z$). The only thing is that I don't know how to convert that to polar coordinates. Can someone please clarify this?
EDIT: I know that the derivative of $\ln(x)= 1/x$. So bringing in $\ln r$ would be convenient.
The “physics-y” way to do this is to consider $P$ as the position of a particle moving along the curve $K$. If $\mathbf{r}$ is its position vector from $O$ to $P$, then $\alpha$ is the angle between its velocity vector $\dot{\mathbf{r}}$ (along the tangent line $t$) and the position vector $\mathbf{r}$. So
$$\cos{\alpha}=\frac{\dot{\mathbf{r}}}{|\dot{\mathbf{r}}|}\cdot\frac{\mathbf{r}}{|\mathbf{r}|}$$
In polar coordinates, the position vector is
$$\mathbf{r}=r\hat{\mathbf{r}}.$$
Differentiating it with respect to time gives the velocity vector
$$\dot{\mathbf{r}}=\dot{r}\hat{\mathbf{r}}+r\dot{\hat{\mathbf{r}}}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta}\hat{\mathbf{\theta}}.$$
Thus we have
$$\cos{\alpha}=\frac{\dot{r}\hat{\mathbf{r}}+r\dot{\theta}\hat{\mathbf{\theta}}}{|\dot{r}\hat{\mathbf{r}}+r\dot{\theta}\hat{\mathbf{\theta}}|}\cdot\hat{\mathbf{r}}=\frac{\dot{r}}{\sqrt{\dot{r}^2+r^2\dot{\theta}^2}}$$
which implies
$$\cot{\alpha}=\frac{\dot{r}}{r\dot{\theta}}=\frac{1}{r}\frac{dr}{d\theta}=\frac{d}{d\theta}\ln{r}.$$