<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\begin{document}


{\bf Question}

A particle is projected along the positive y-axis so that
initially, at time $t=0$, its position is $y=0$ and its velocity
is $\frac{dy}{dx}=2$.  The equation of motion for the particle is
described by

$$\frac{d^2y}{dx^2}= -8\cos^3(y)\sin(y).$$

\begin{description}
\item[(a)]
Calculate $\frac{df}{dy}$ where $f(y)=\cos^4(y)$
\item[(b)]
Find expressions for the velocity of the particle as a function of
y, and its position as a function of t.
\item[(c)]
Sketch the graph of the position $y$ as a function of time $t$ for
$t \geq 0$.  Comment in the time taken for the particle to reach
the point $y=\frac{\pi}{2}$
\end{description}


{\bf Answer}

\begin{description}
\item[(a)]
Find $\frac{df}{dy}$ where $f(y) = \cos^4 y$.

Let $u = \cos y$ so that $\frac{du}{dy} = -\sin y$

Then  $f(u) = u^4$   and $ \frac{df}{du}  =  4u^3$
\begin{eqnarray*}
{\rm using\ the\ chain\ rule:\  } \frac{df}{dy}  =  \frac{df}{du}
\times \frac{du}{dy}  &amp; = &amp; 4u^3 (\sin y) \\ &amp; = &amp; - \cos ^3 y
\sin y
\end{eqnarray*}
\item[(b)]
Let $v = \frac{dy}{dt}$ so that

$\displaystyle \frac{d^2y}{dt^2} = \frac{dV}{dt} = \frac{dV}{dy}
\times \frac{dy}{dt} = \frac{dV}{dy}V$
\begin{eqnarray*}
{\rm Equation\ of\ motion\ becomes\ } V \frac{dV}{dy} &amp; = &amp; -8
\cos^3y \sin y \\ {\rm which\ is\ separable:\ } \int V dV &amp; = &amp; 2
\int -4 \cos^3 y \sin y dy + C \\ {\rm by\ part\ (a)\ :}
\frac{1}{2} V^2 &amp; = &amp; 2 (\cos^4 y ) + C
\end{eqnarray*}
\begin{eqnarray*}
{\rm initial\ conditions: \ } V &amp; = &amp; 2 {\rm \ and\ } y = 0 {\rm
when\ } t = 0 \\ {\rm so\ } \frac{1}{2}(4) &amp; = &amp; 2 \cos^4(0) +C
\Rightarrow c = 0 \\ {\rm hence\ } V^2 &amp; = &amp; 4 \cos^4 y \\ {\rm
velocity\ as\ a\ function\ of\ y\ } v &amp; = &amp; \sqrt{4 \cos^4 y} = 2
\cos^ y
\end{eqnarray*}
(Take the positive square root to staify initial condition V = 2
when y = 0)
\begin{eqnarray*}
{\rm Now,\ } V = \frac{dy}{dt} {\rm so\ } \frac{dy}{dt} &amp; = &amp; 2
\cos^2 y \\ {\rm which\ is\ separable:\ } \int \frac{dy}{\cos^2 y}
&amp; = &amp; 2 \int dt + K \\ \int \frac{dy}{\cos^ y} &amp; = &amp; \tan y {\rm \
standard\ integral\ } \\ {\rm so\ } \tan y &amp; = &amp; 2t + K \\ {\rm
using\ initial\ conditions\ } y  =  0 {\rm \ when\ } t = 0 :
\tan(0) &amp; = &amp; 0 + K \\ \Rightarrow k &amp; = &amp; 0 \\{\rm therefore\ }
\tan y &amp; = &amp; 2t {\rm \ and\ } \\ {\rm position\ as\ a\ function\
of\ time:\ } y &amp; = &amp; \tan^{-1}(2t)
\end{eqnarray*}

\item[(c)]

Plot position $y$ against time $t$:

\begin{center}
\epsfig{file=158-3-1.eps, width=70mm}
\end{center}

The graph is asymptotic to the line $y=\frac{\pi}{2}.$

Hence the particle {\bf never} reaches $y=\frac{\pi}{2}.$

\end{description}


\end{document}


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