Problems

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Solution: Points always maintain a constant fraction of the distance from the start, so the point distance \(x\) from the start at time \(t\) is moving with speed \(\frac{x}{a+ut} u\) The point is moving with speed \(v+\frac{x}{a+ut} u\) or in other words \begin{align*} && \frac{\d x}{\d t} &= v + \frac{x}{a+ut}u \\ \Rightarrow && \frac{\d x }{\d t} - \frac{u}{a+ut} x &= v \\ \Rightarrow && \frac{1}{a+ut} \frac{\d x}{\d t} - \frac{u}{(a+ut)^2} x &= \frac{1}{a+ut} v\\ \Rightarrow && \frac{\d}{\d x} \left ( \frac{x}{a+ut} \right) &= \frac{v}{a+ut} \\ \Rightarrow && \frac{x}{a+ut} &=\frac{v}{u} \ln (a + ut) + C \\ t = 0, x = 0: && 0 &= \frac{v}{u} \ln a + C \\ \Rightarrow && x &= (a+ut) \frac{v}{u} \ln \left ( \frac{a+ut}{a} \right) \\ \\ \Rightarrow && 1 &= \frac{v}{u} \ln \left ( \frac{a+uT}{a} \right) \\ \Rightarrow && e^k &= 1 + \frac{uT}{a} \\ \Rightarrow && uT &= a(e^k-1) \end{align*} On the return journey, we have \begin{align*} && \frac{\d x}{\d t} &= \frac{x}{a+ut}u - v \\ \Rightarrow && \frac{\d x}{\d t} - \frac{u}{a+ut} x &= - v \\ \Rightarrow && \frac{\d }{\d x} \left ( \frac{x}{a+ut} \right) &= -\frac{v}{a+ut} \\ \Rightarrow &&f &= -\frac{v}{u} \ln(a+ut) + K \\ t = T, f = 1: && 1 &= -\frac{v}{u}\ln(a+uT) + K \\ \Rightarrow && f &= 1+\frac{v}{u}\ln \left ( \frac{a+uT}{a+ut} \right) \\ \Rightarrow && 0 &= 1+\frac{v}{u} \ln \left ( \frac{a+uT}{a+uT_2} \right) \\ \Rightarrow && e^k &= \frac{a+uT_2}{a+uT}\\ \Rightarrow && uT_2 &= (a+uT)e^k - a \\ \Rightarrow && T_2 - T &= \frac{1}{u} \left ( (a+uT)e^k - a - uT\right) \\ &&&= \frac{1}{u} \left ((a+a(e^k-1))e^k-a-a(e^k-1) \right) \\ &&&= \frac{1}{u} \left (ae^{2k} -ae^k \right) \\ &&&= \frac{ae^k}{u} \left ( e^k-1 \right) \\ &&&= Te^k \end{align*}

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Solution: \begin{align*} && {\bf r} &=\left( \e^{t}\cos t \right) {\bf i}+ \left(\e^t \sin t\right) {\bf j} \\ \Rightarrow && \dot{\bf r} &= \left( \e^{t}\cos t -\e^t \sin t\right) {\bf i}+ \left(\e^t \sin t+\e^t \cos t\right) {\bf j} \\ \Rightarrow && \mathbf{r}\cdot\dot{ \mathbf{r}} &= e^{2t}(\cos^2 t - \sin t \cos t) + e^{2t}(\sin^2 t+ \sin t \cos t) \\ &&&= e^{2t} (\cos^2 t + \sin ^2 t)\\ &&&= e^{2t} \\ \\ && | {\bf r}| &= e^{t} \\ && |{\bf \dot{r}}| &= e^t \sqrt{(\cos t - \sin t)^2 + (\sin t + \cos t)^2} \\ &&&= e^t \sqrt{2 \cos^2 t + 2 \sin^2 t} \\ &&&= \sqrt{2} e^t \\ \\ \Rightarrow && \frac{\mathbf{r}\cdot\dot{ \mathbf{r}}}{ |{\bf {r}}| |{\bf \dot{r}}|} &= \frac{e^{2t}}{\sqrt{2}e^te^t} \\ &&&= \frac{1}{\sqrt{2}} \end{align*} Therefore the angle between the velocity and displacement is \(\frac{\pi}{4}\). \begin{align*} && \ddot{\bf{r}} &= \left( \e^{t}(\cos t - \sin t) - \e^t (\sin t + \cos t)\right) {\bf i}+ \left(\e^t (\sin t + \cos t) + \e^t(\cos t - \sin t)\right) {\bf j} \\ &&&= \left ( -2\e^{t} \sin t \right) {\bf i}+ \left ( 2\e^{t} \cos t \right) {\bf j} \\ \Rightarrow && {\bf r} \cdot \ddot{\bf{r}} &= 2e^{2t} \left ( -\sin t \cos t + \sin t \cos t \right) \\ &&&= 0 \end{align*} Therefore the acceleration is perpendicular.

\(Q\) has position $\mathbf{r}' = \left( \e^{t-T}\cos (t-T) \right) {\bf i}+ \left(\e^{t-T} \sin (t-T)\right) {\bf j}\( for \)t > T$. \begin{align*} && {\bf r' \cdot r} &= e^{2t-T} \left (\cos t \cos (t-T) + \sin t \sin(t - T) \right) \\ &&&= e^{2t-T} \cos (t - (t-T)) \\ &&&= e^{2t-T} \cos T \\ \\ && |{\bf r'}- {\bf r} |^2 &= |{\bf r}|^2 + |{\bf r}'|^2 - 2 {\bf r' \cdot r} \\ &&&= e^{2t} + e^{2(t-T)} - 2e^{2t-T} \cos T \\ &&&= e^{2t} \left (1 - 2e^{-T} \cos T + e^{-2T} \right) \\ \Rightarrow && |{\bf r'}- {\bf r} | &= e^{t} \sqrt{1 - 2e^{-T} \cos T + e^{-2T} } \end{align*} as required

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Solution:

1. 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 \(t = \frac{\pi}{2}, \frac{3\pi}{2}\)
2. \(\,\) \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 \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\)
3. 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.
4. \(\frac{\d }{\d t} (\mathbf{r} \cdot \mathbf{r}) = 2 \mathbf{r} \cdot \mathbf{\dot{r}}\) 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 \(2\).
   

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