What is the derivative of an angle? [closed]

Question

What is the derivative of an angle? I don't understand

Answer 1

Let's consider a 2-dimensional case. In a simple circular motion, the angular velocity is $$v_{\theta} \equiv \omega = r \frac{d\theta}{dt} = r \dot{\theta}~.$$ I called this velocity $v_{\theta}$ because it's a velocity connected to the change in the coordinate $\theta$.

OK, so that's a simple circular motion, like that of a bead at the end of a string. Now imagine I can (smoothly) make the string longer or shorter. Then the bead has not only a velocity connected with the motion described by coordinate $\theta$, but also a velocity connected with the motion along the radial direction, $$v_r = \frac{dr}{dt} = \dot{r}~.$$

The total velocity is the sum of these two, $$\vec{v} = v_r \hat{r} + v_{\theta} \hat{\theta}~,$$ and consequently $$v^2 = v_r^2 + v_{\theta}^2 = \dot{r}^2 + r^2\dot{\theta}^2~.$$

Your original problem is 3-dimensional and therefore also velocity connected with changes in the coordinate $\phi$ is needed. You should remember from, say, integration in spherical coordinates that the length element connected to an infinitesimal change $d\phi$ is given by $$dl_{\phi} = r \sin \theta ~d \phi~.$$ Then it follows that $$v_{\phi} = \frac{d l_{\phi}}{dt} = r \sin \theta \frac{d \phi}{dt} = r \sin \theta ~\dot{\phi}~.$$ Then $v^2 = v_r^2 + v_{\theta}^2 + v_{\phi}^2$ as shown in your textbook follows.

Answer 2

$\dot{\phi}\equiv\frac{d\phi}{dt}$ (same thing for $\dot{\theta}$) is only the time derivative of the angle $\phi$ (or $\theta$). The coordinates of a particle can be described in cartesiant, spherical or cylindrical coordinates. In spherical or cylindrical coordinates, the evolution of the particle's position is given by some angle(s). If the particle is moving, then those derivative tells how fast. For example, if a particle is moving on a circle of radius $R$, aligned on the $XY$ plane centered at the origin, its position at an instant $t$ can be described by $(x(t), y(t), 0)$ in cartesian coordinates where $\sqrt{x(t)^2 + y(t)^2} = R$, or can be expressed in cylindrical coordinates by $(R, \phi(t), 0)$ where $\phi(t) = \arctan{\frac{y(t)}{x(t)}}$ or in spherical coordinates by $(R, \frac{\pi}{2}, \phi(t))$. In either case, the derivative of the angle $\phi$ will not be zero if the particle is moving and this derivative will be the angular speed at which the particle is rotating on the circle.