I have a question regarding the following forms in the case that both angular velocity and radius change per $dt$. Wiki defines angular velocity in two dimensions as:
$$ \omega = \frac {d \phi} {dt} $$
then linear velocity $v$ as:
$$ \mathcal{l} = r \phi, v(t) = \frac {d \mathcal{l}}{d t} = r \omega(t) $$
What is the equation for linear velocity if both $r$ and $\theta$ change per $dt$? Is $r$ parametrized to $r(t)$? I'm confused because $r$ is a constant in the above forms, where I would expect it to be part of the change $d \mathcal{l}$.
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$\begingroup$We have $$(x(t),y(t))=(r(t)\cos \theta(t),r(t) \sin\theta(t)).$$ Then\begin{align} |\mathbf{v}(t)|^2&=(x')^2+(y')^2\\ &=(r'\cos \theta-r(\sin \theta) \theta')^2+(r'\sin \theta+ r(\cos \theta) \theta')\\ &=(r')^2+r^2(\theta')^2\\ &=(\frac{dr}{dt})^2+r^2\omega^2. \end{align}So$$ |\mathbf{v}(t)|=\sqrt{(\frac{dr}{dt})^2+r^2\omega^2}. $$
$\endgroup$ $\begingroup$It explains in that Wikipedia article that for a general case of a point moving in the plane (so not specifically with a fixed distance from the origin), that the derivative of its position vector $\mathbf{r}$ with respect to $t$ (i.e. its velocity vector) is given by
$$\frac{d \mathbf{r}}{dt} = \dot{r} \hat{\mathbf{r}} + r \dot{\varphi}\hat{\varphi}$$
The linear speed is the magnitude of the velocity vector, so
$$\begin{eqnarray} \frac{dl}{dt} & = & \left | \frac{d \mathbf{r}}{dt} \right | \\ & = & \left| \dot{r} \hat{r} + r \dot{\varphi}\hat{\varphi} \right | \\ & = & \sqrt{|\dot{r} \hat{r}|^2 + |r \dot{\varphi}\hat{\varphi}|^2} & \text{since } \hat{r} \text{ and } \hat{\varphi} \text{ are orthogonal} \\ & = & \sqrt{\dot{r}^2 + r^2 \dot{\varphi}^2} & \text{since } \hat{r} \text{ and } \hat{\varphi} \text{ are unit vectors} \end{eqnarray}$$
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