Year: 2011
Paper: 1
Question Number: 7
Course: LFM Pure
Section: Differential equations
There were again significantly more candidates attempting this paper than last year (just over 1100), but the scores were significantly lower than last year: fewer than 2% of candidates scored above 100 marks, and the median mark was only 44, compared to 61 last year. It is not clear why this was the case. One possibility is that Questions 2 and 3, which superficially looked straightforward, turned out to be both popular and far harder than candidates anticipated. The only popular and well-answered questions were 1 and 4. The pure questions were the most popular as usual, though there was noticeable variation: questions 1–4 were the most popular, while question 7 (on differential equations) was fairly unpopular. Just over half of all candidates attempted at least one mechanics question, which one-third attempted at least one probability question, an increase on last year. The marks were surprising, though: the two best-answered questions were the pure questions 1 and 4, but the next best were statistics question 12 and mechanics question 9. The remainder of the questions were fairly similar in their marks. A number of candidates ignored the advice on the front cover and attempted more than six questions, with a fifth of candidates trying eight or more questions. A good number of those extra attempts were little more than failed starts, but still suggest that some candidates are not very effective at question-picking. This is an important skill to develop during STEP preparation. Nevertheless, the good marks and the paucity of candidates who attempted the questions in numerical order does suggest that the majority are being wise in their choices. Because of the abortive starts, I have generally restricted my attention to those attempts which counted as one of the six highest-scoring answers in the detailed comments. On occasions, candidates spent far longer on some questions than was wise. Often, this was due to an algebraic slip early on, and they then used time which could have been far better spent tackling another question. It is important to balance the desire to finish a question with an appreciation of when it is better to stop and move on. Many candidates realised that for some questions, it was possible to attempt a later part without a complete (or any) solution to an earlier part. An awareness of this could have helped some of the weaker students to gain vital marks when they were stuck; it is generally better to do more of one question than to start another question, in particular if one has already attempted six questions. It is also fine to write "continued later" at the end of a partial attempt and then to continue the answer later in the answer booklet. As usual, though, some candidates ignored explicit instructions to use the previous work, such as "Hence", or "Deduce". They will get no credit if they do not do what they are asked to! (Of course, a question which has the phrase "or otherwise" gives them the freedom to use any method of their choosing; often the "hence" will be the easiest, though.) It is wise to remember that STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom. There were a number of common errors and issues which appeared across the whole paper. The first was a lack of fluency in algebraic manipulations. STEP questions often use more variables than A-level questions (which tend to be more numerical), and therefore require candidates to be comfortable engaging in extended sequences of algebraic manipulations with determination and, crucially, accuracy. This is a skill which requires plenty of practice to master. Another area of weakness is logic. A lack of confidence in this area showed up several times. In particular, a candidate cannot possibly gain full marks on a question which reads "Show that X if and only if Y" unless they provide an argument which shows that Y follows from X and vice versa. Along with this comes the need for explanations in English: a sequence of formulæ or equations with no explicit connections between them can leave the reader (and writer) confused as to the meaning. Brief connectives or explanations would help, and sometimes longer sentences are necessary. Another related issue continues to be legibility. Many candidates at some point in the paper lost marks through misreading their own writing. One frequent error was dividing by zero. On several occasions, an equation of the form xy = xz appeared, and candidates blithely divided by x to reach the conclusion y = z. Again, I give a strong reminder that it is vital to draw appropriate, clear, accurate diagrams when attempting some questions, mechanics questions in particular.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, you may assume that $\ln (1+x) \approx x -\frac12 x^2$ when $\vert x \vert $ is small.
The height of the water in a tank at time $t$ is $h$.
The initial height of the water is $H$ and water flows into the tank at a constant rate. The cross-sectional area of the tank is constant.
\begin{questionparts}
\item Suppose that water leaks out at a rate proportional to the height of the water in the tank, and that when the height reaches $\alpha^2 H$, where $\alpha$ is a constant greater than 1, the height remains constant.
Show that
\[
\frac {\d h}{\d t } = k( \alpha^2 H -h)\,,
\]
for some positive constant $k$. Deduce
that the time $T$ taken for the water to reach height $\alpha H$ is given by
\[
kT = \ln \left(1+\frac1\alpha\right)\,,
\]
and that $kT\approx \alpha^{-1}$ for large values of $\alpha$.
\item Suppose that the rate at which water leaks out of the tank is proportional to $\sqrt h$ (instead of $h$), and that when the height reaches $\alpha^2 H$, where $\alpha$ is a constant greater than 1, the height remains constant.
Show that the time $T'$ taken for the water to reach height $\alpha H$ is given by
\[
cT'=2\sqrt H \left( 1 - \sqrt\alpha +\alpha \ln
\left(1+\frac1 {\sqrt\alpha} \right)\right)\,
\]
for some positive constant $c$, and that $cT'\approx \sqrt H$ for large values of $\alpha$.
\end{questionparts}\begin{questionparts}
\item \begin{align*}
\frac{\d h}{\d t} &= \underbrace{c}_{\text{flow in}} - \underbrace{kh}_{\text{flow out}}
\end{align*}.
We also know that when $h = \alpha^2 H$, $\frac{\d h}{\d t} = 0$, ie $c - k \alpha^2 H = 0$ therefore:
\[ \frac{\d h}{\d t} = k(\alpha^2 H - h) \]
\begin{align*}
&& \frac{\d h}{\d t} &= k(\alpha^2 H - h) \\
&& \int \frac{1}{\alpha^2 H - h} \d h &= \int k \d t \\
&& - \ln |\alpha^2H -h| &= kt + C \\
t = 0, h = H: && -\ln |(1-\alpha^2 )H| &= C \\
\Rightarrow && kt &= \ln \left | \frac{(\alpha^2-1)H}{h-\alpha^2 H }\right | \\
&& kT &= \ln \frac{(\alpha^2-1)H}{\alpha H - \alpha^2 H} \\
&&&= \ln \frac{1+\alpha}{\alpha} \\
&&&= \ln \left (1 + \frac1{\alpha} \right) \\
&&&\approx \frac1{\alpha} - \frac12 \frac1{\alpha^2}\\
&&&\approx \alpha^{-1}
\end{align*}
\item \begin{align*}
&& \frac{\d h}{\d t} &=c(\alpha \sqrt{H} - \sqrt{h}) \\
\Leftrightarrow && c \int_0^{T'} \d t&= \int_{H}^{\alpha H} \frac{1}{\alpha \sqrt{H}-\sqrt{h}} \d h \\
u = \sqrt{h/H}: && cT' &= \int_1^{\sqrt{\alpha}} \frac{1}{\alpha \sqrt{H} - \sqrt{H}u} 2\sqrt{H}u \d u \\
&&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u}{\alpha - u} \d u \\
&&&= 2\sqrt{H}\int_1^{\sqrt{\alpha}} \frac{u - \alpha + \alpha}{\alpha - u} \d u \\
&&&= 2\sqrt{H}\left [-u - \alpha \ln |\alpha - u| \right]_1^{\sqrt{\alpha}} \\
&&&= 2\sqrt{H}\left ( -\sqrt{\alpha} + 1- \alpha \ln (\alpha - \sqrt{\alpha}) + \alpha \ln |\alpha - 1| \right) \\
&&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\alpha-1}{\alpha - \sqrt{\alpha}}
\right)\right)\\
&&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}^2-1}{\sqrt{\alpha}(\sqrt{\alpha}-1)}
\right)\right)\\
&&&= 2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( \frac{\sqrt{\alpha}+1}{\sqrt{\alpha}}
\right)\right)\\
&&&= \boxed{2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \ln \left ( 1+\frac{1}{\sqrt{\alpha}}
\right)\right)}\\
&&&\approx2\sqrt{H}\left (1-\sqrt{\alpha} + \alpha \left ( \frac{1}{\sqrt{\alpha}}-\frac12 \frac{1}{\alpha}
\right)\right) \\
&&&=2\sqrt{H} \left ( 1 - \frac12 \right) \\
&&&= \sqrt{H}
\end{align*}
as required.
\end{questionparts}
This was by far the least popular Pure Mathematics question, attempted by only one-third of candidates. The marks achieved were poor; the median mark was 3. (i) For this first part of the question, very few candidates were able to justify the form of differential equation given. Many candidates talked about the water level decreasing until it reached α²H, which is not what actually happens (it continually increases until it reaches that point). This lack of understanding had a knock-on effect later on, too, in that they were unable to generate the required differential equation in part (ii). Nevertheless, from the given DE, a good number were able to separate the variables and at least make progress towards a solution, though only a minority were able to complete the task. A number of candidates muddled H and h; this sloppiness prevented them from getting the right answer. It was also unfortunate that the symbols for proportionality and the Greek letter alpha are similar; students had to take extra care to keep those distinct as well. Many candidates used the given result and approximation for ln(1 + x) to deduce the stated approximation. (ii) Few candidates were able to generate the correct differential equation for this part, with the offering of dh/dt = c(α²H − √h) being far more common. (We were generous and only deducted a few marks if they made progress with this incorrect equation.) Of those who attempted to solve either the correct or this variant differential equation, very few could work out how to integrate the resulting expression. Even when they used the substitution u = √h, they were usually unable to integrate the fractional linear expression resulting from it. Some were more successful, especially when they used the substitution v = α√H − √h. A number of students wisely picked up a couple of marks by demonstrating the validity of the final approximation on the basis of the solution given in the question.