Year: 2018
Paper: 2
Question Number: 10
Course: UFM Pure
Section: First order differential equations (integrating factor)
The pure questions were again the most popular of the paper, with only two of those questions being attempted by fewer than half of the candidates (none of the other questions was attempted by more than half of the candidates). Good responses were seen to all of the questions, but in many cases, explanations lacked sufficient detail to be awarded full marks. Candidates should ensure that they are demonstrating that the results that they are attempting to apply are valid in the cases being considered. In several of the questions, later parts involve finding solutions to situations that are similar to earlier parts of the question. In general candidates struggled to recognise these similarities and therefore spent a lot of time repeating work that had already been done, rather than simply observing what the result must be.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A uniform elastic string lies on a smooth horizontal table.
One end of the string
is attached to a fixed peg,
and the other
end is pulled at constant speed $u$. At time
$t=0$, the string is
taut and its length is $a$. Obtain an expression for
the speed, at time $t$,
of the point on the string
which is a distance
$x$ from the peg at time $t$.
An ant walks along the string starting at $t=0$ at the peg.
The ant walks at constant speed $v$ along the string (so that
its speed relative to the peg is the sum of $v$ and the speed of the
point on the string beneath the ant).
At time $t$, the ant is a distance $x$ from the
peg.
Write down
a first order differential equation
for $x$, and
verify
that
\[
\frac{\d }{\d t} \left( \frac x {a+ut}\right) = \frac v {a+ut} \,.
\]
Show that the time $T$ taken for the ant to
reach the end of the string is given by
\[uT = a(\e^k-1)\,,\]
where $k=u/v$.
On reaching the end of the string, the ant turns round and walks back to the
peg. Find in terms of $T$ and $k$
the time taken for the journey
back.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*}
The qualities of responses for this question were quite varied. Many candidates who struggled with this question failed to set up the speed of a point on the string in the first part of the question and were therefore unable to make any significant progress. In the part of the question considering the return journey a common error was to "restart the clock" once the ant had reached the endpoint, however, since the speed of the string is dependent on the time from the start of the problem this led to an error. A particularly elegant solution to the final part involved changing the frame of reference such that the endpoint was stationary and the peg was moving at constant speed. Stronger candidates were then able to use the symmetry of the problem to reduce the amount of algebra needed considerably.