Problem source

A particle moves so that ${\bf r}$, 
its displacement from a fixed origin at time $t$, 
is given by 
\[{\bf r} = \l \sin{2t} \r {\bf i} + \l 2\cos t \r \bf{j}\,,\] 
where $0 \le t < 2\pi$.
\begin{questionparts}
\item  Show that the particle passes through the origin exactly twice.
\item  Determine the times when the velocity of the particle is perpendicular to its displacement. 
\item  Show that, when the particle is not at the origin,
its velocity is never parallel to its displacement.
\item  Determine the maximum distance of the particle from the origin, and sketch the path of the particle.
\end{questionparts}

Solution source

\begin{questionparts}
\item It is at the origin when both $\sin 2t$ and $\cos t = 0$, but this $\sin 2t = 2 \sin t \cos t$ so this happens precisely when $\cos t = 0$, ie when $t = \frac{\pi}{2}, \frac{3\pi}{2}$

\item $\,$ \begin{align*}
&& \dot{\mathbf{r}} &= 2 \cos 2t \mathbf{i} - 2 \sin t \mathbf{j} \\
&& \mathbf{r} \cdot \dot{\mathbf{r}} &= 2\cos 2t \sin 2t - 2 \sin t 2 \cos t  \\
&&&= \sin 2t \left (2\cos 2t - 2 \right) 
\end{align*}
Therefore they are perpendicular when $\sin 2t = 0 \Rightarrow t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ and when $\cos 2t = 1 \Rightarrow 2t = 0, 2\pi, 4\pi \Rightarrow t = 0, \pi, 2\pi$, therefore all solutions are $ t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$

\item For $\mathbf{r}$ and $\dot{\mathbf{r}}$ to be parallel, we would need 
\begin{align*}
&& \frac{2 \cos 2t}{\sin 2t} &= \frac{-2 \sin t}{2 \cos t}\\
&& 2 \cos 2t \cos t  &= - \sin t \sin 2t \\
&& 0 &= 2\cos t (\cos 2t + \sin ^2 t) \\
&&&= 2 \cos t (\cos^2 t) \\
&&&= 2 \cos^3 t
\end{align*}

Therefore the only time we can be parallel is when $\cos t = 0$, which is when we are at the origin.

\item $\frac{\d }{\d t} (\mathbf{r} \cdot \mathbf{r}) = 2 \mathbf{r} \cdot \mathbf{\dot{r}}$ so we should check the values when velocity and displacement are perpendicular, ie $ t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ which have values $\mathbf{r} = \binom{0}{2}, \binom{0}{0}, \binom{0}{-2}, \binom{0}{0}, \binom{0}{2}$. Therefore the maximum distance is $2$.

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sqrt(1/((#1)^2+1))};
    \def\xl{-3};
    \def\xu{3};
    \def\yl{-3};
    \def\yu{3};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);


        \draw[thick, blue, smooth, domain=0:400, samples=400] 
            plot ({sin(2*\x)}, {2*cos(\x)});

        \filldraw (0, 2) circle (1.5pt) node[above right] {$2$};
        \filldraw (0, -2) circle (1.5pt) node[above right] {$-2$};
        % \filldraw (0, -2) circle (1.5pt) {$-2$};

%         \foreach \x in {0,1,2,3,4,5} {
% % %     \draw (\x,0.1) -- (\x,-0.1) node[below] {$x_{\x}$};
% % %   }
        % \node[blue, above, rotate=-20] at (3, {4*\functionf(3)}) {\tiny $y^2(x^2+a^2) = c^2$}; 
    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

\end{questionparts}