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QUESTION

\begin{description}

\item[(b)]
The position of a particle $P$ at a time $t$ is given by

$$\mathbf{r}=\sin(2t)\mathbf{i}+\cos(2t)\mathbf{j}+t\mathbf{k}.$$

\begin{description}

\item[(i)]
Find the velocity and speed of $P$ at time $t$.

\item[(ii)]
Find the acceleration of $P$ at time $t$, and deduce that the
acceleration is perpendicular to the velocity.

\item[(iii)]
Determine the angle between the particle's position \textbf{r} and
the velocity at time $t$, and find the limiting value if this
angle at large time.

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ANSWER

\begin{description}

\item[(b)]
$\mathbf{r}=\sin(2t)\mathbf{i}+\cos(2t)\mathbf{j}+t\mathbf{k}$

\begin{description}

\item[(i)]
Velocity
$\ds=\frac{d\mathbf{r}}{dt}=2\cos(2t)\mathbf{i}-2\sin(2t)\mathbf{j}+\mathbf{k}$\\
Speed
$\ds=\left|\frac{d\mathbf{r}}{dt}\right|=\sqrt{(2\cos(2t))^2+(-2\sin(2t))^2+1^2}\\
=\sqrt{(4(\cos^2(2t)+\sin^2(2t))+1)}=\sqrt{5}$

\item[(ii)]
Acceleration
$\ds=\frac{d^2\mathbf{r}}{dt^2}=-4\sin(2t)\mathbf{i}-4\cos(2t)\mathbf{j}$

$\ds\frac{d\mathbf{r}}{dt}\cdot\frac{d^2\mathbf{r}}{dt^2}
=(2\cos(2t),-2\sin(2t),1)\cdot(-4\sin(2t,-4\cos(2t),0)\\
=-8\cos(2t)\sin(2t)+8\sin(2t)\cos(2t)+0=0$

Therefore the velocity is perpendicular to the acceleration.

\item[(iii)]
$\ds\cos(\theta)=\frac{\mathbf{r}.\frac{d\mathbf{r}}{dt}}{|\mathbf{r}||\frac{d\mathbf{r}}{dt}|}
\\ =\frac{2\sin(2t)\cos(2t)-2\cos(2t)\sin(2t)+t}{\sqrt{\sin^2(2t)+\cos^2(2t)+t^2}\sqrt{5}}
=\frac{t}{\sqrt{5(1+t^2)}}$
\\Therefore
$\ds\theta=\cos^{-1}\left(\frac{t}{\sqrt{5(1+t^2)}}\right)$\\ As
$\ds t\to\infty,\
\theta\to\cos^{-1}\left(\frac{1}{\sqrt{5}}\right)=63.4^{\circ}$

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