# Contradiction of a scalar product

Can anyone resolve this contradiction:

$$\vec{r}\cdot\dot{\vec{r}}=\frac{1}{2}\frac{d}{dt}\left(\vec{r}^2\right)=\frac{1}{2}\frac{d}{dt}\left(\left|\vec{r}\right|^2\right)\equiv\frac{1}{2}\frac{d}{dt}\left(r^2\right)=r\dot{r}, \qquad r=|\vec{r}|.$$

But the velocity $\vec{v}=\dot{\vec{r}}$ has not to be parallel to $\vec{r}$, so actually:

$$\vec{r}\cdot\dot{\vec{r}}=r \dot{r} \cos{\angle\left(\vec{r},\dot{\vec{r}}\right)}$$

What am I doing wrong? Has anyone an idea?

P.S. I have this problem from the book "Electromagnetic Theory" from Ferraro (p. 543).

The issue here is that $\dot{r}$ is not the magnitude of $\dot{\vec{r}}$, but rather the rate of change of $r$, the magnitude of $\vec{r}$. Think about a particle moving in a circle. Since $r$ is constant, $\dot{r}=0$, but the velocity is certainly not zero!
The formula $\vec{r}\cdot\dot{\vec{r}} = r\dot{r}$ makes perfect sense, because $\dot{r}$ is the component of the velocity in the radial direction, i.e., in the direction of $\vec{r}$.
• Thanks! The assumption $\dot{r}=\dot{\vec{r}}$ was my error of reasoning. – Andy May 6 '14 at 9:07