Backlinks
1 We Begin at Polar 2D Motion
Let's think about a particle with some velocity \(\vec{v}\), some acceleration \(\vec{a}\), and some location \(\vec{r}\).
Parameterize \(\vec{r}\), we figure that:
\begin{equation} \vec{r} = (x(t),y(t)) \end{equation}Therefore, the velocity would be:
\begin{equation} \dot{\vec{r}} = \vec{v} = (\dot{x}, \dot{y}) = \dot{x}\hat{i} + \dot{y}\hat{j} \end{equation}By the same token:
\begin{equation} \vec{a} = \ddot{\vec{r}} \end{equation}1.1 A weird coordinate
We create a new coordinate system
\begin{equation} \hat{r} = \hat{r}(t) \end{equation}and
\begin{equation} \hat{\theta} = \hat{\theta}(t) \end{equation}and also
\begin{equation} \hat{r} \perp \hat{\theta} \end{equation}1.2 Relating the new systems
\begin{equation}
\vec{r} = r \hat{r}
\end{equation}
"the value of the vector \(r\) is the direction of \(r\) times the magnitude of \(r\)."
\begin{equation} \vec{v} = \frac{d\vec{r}}{dt} = r \frac{d \hat{r}}{dt} + \hat{r} \frac{dr}{dt} \end{equation}Now, we note that because \(\hat{r} \perp \hat{\theta}\), \(\frac{d\hat{r}}{dt}\), the tiny instant change of the direction must be in the \(\hat{\theta}\) direction.
Our job today is to write a function \(\vec{v} = f_1(\hat{r}, \hat{\theta}, \dot{r}, \dot{\theta}, r)\), \(\vec{v} = f_2(\hat{r}, \hat{\theta}, \dot{r}, \dot{\theta}, \ddot{r}, \ddot{\theta}, r)\)