Exercise 8c

\[\vec{r}(t) = \left( \begin{matrix} c \cos^3 t \\ c \sin^3 t \end{matrix} \right)\] \[\vec{v}(t) = \left( \begin{matrix} -3c \cos^2 t \sin t \\ 3c \sin^2 t \cos t\end{matrix} \right)\] \[\left| \vec{v}(t) \right| = \sqrt{(-3c \cos^2 t \sin t )^2 + (3c \sin^2 t \cos t )^2 } \] \[ = 3c \sqrt{ (\cos^4 t \sin^2 t) + ( \sin^4 t \cos^2 t) } \] \[ = 3c \cos^2 t \sin^2 t \sqrt{ \cos^2 t + \sin^2 t} \] \[ = 3c \cos^2 t \sin^2 t\] \[\vec{a}(t) = \left( \begin{matrix} -3c \left( 2 \cos t (-\sin t) \sin t + \cos^2 t \cos t \right) \\ 3c \left( 2 \sin t \cos t \cos t - \sin^2 t \sin t \right) \end{matrix} \right)\]

Using Pythagoras \(\sin^2 t + \cos^2 t = 1\) we get:

\[\vec{a}(t) = \left(\begin{matrix} -3c \cos t \, (1-3 \sin^2 t) \\ 3c \sin t \, (3 \cos^2 t -1)\end{matrix} \right)\]

Animation

position vector: red
velocitiy vector: blue
accelaration vector: green

The animation presents two periods, time is dilated by factor 2. In the \(x\)-\(y\) diagram the trajectory of the position vector is shown in red. The velocity vector is always tangent to the trajectory, this is easy to see in the 2nd repetition, when the trajectory is already plotted. For clarification, the velocitiy vector was placed at the tip of the position vector. The magnitude of the velocitiy vector becomes zero at the tipping points.

In the \(x\prime\)-\(y\prime\) diagram the velocity vector is shown again, this time starting from the origin. In the velocitiy space the velocity vector moves on the blue trajectory. The accelaration is always tangent to the trajectory of the velocity.