OCR C4 Specimen — Question 9 11 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
SessionSpecimen
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeSeparable variables - standard (applied/contextual)
DifficultyStandard +0.3 This is a standard differential equations question requiring separation of variables with a square root term. The setup is clearly guided ('proportional to √x'), the integration is routine (∫x^(-1/2)dx), and students are given the target answer to verify. Slightly easier than average due to clear scaffolding and straightforward algebra.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context

9 \includegraphics[max width=\textwidth, alt={}, center]{798da17d-0af5-4aa6-b731-564642dc28d5-4_572_917_294_607} A cylindrical container has a height of 200 cm . The container was initially full of a chemical but there is a leak from a hole in the base. When the leak is noticed, the container is half-full and the level of the chemical is dropping at a rate of 1 cm per minute. It is required to find for how many minutes the container has been leaking. To model the situation it is assumed that, when the depth of the chemical remaining is \(x \mathrm {~cm}\), the rate at which the level is dropping is proportional to \(\sqrt { } x\). Set up and solve an appropriate differential equation, and hence show that the container has been leaking for about 80 minutes.

AnswerMarks Guidance
\(\frac{dx}{dt} = -k\sqrt{x}\)M1 For use of derivative for rate of change
A1For correct equation (neg sign optional here)
\(x = 100\) and \(\frac{dx}{dt} = -1 \Rightarrow k = 0.1\)M1 For use of data and their DE to find \(k\)
Hence equation is \(\frac{dx}{dt} = -0.1\sqrt{x}\)A1 For any form of correct DE
\(\int x^{-\frac{1}{2}}\,dx = -0.1\int dt \Rightarrow 2x^{\frac{1}{2}} = -0.1t + c\)M1 For separation and integration of both sides
A1For \(2x^{\frac{1}{2}}\) correct
A1For \((\pm)kt\) correct (the numerical evaluation of \(k\) may be delayed until after the DE is solved)
\(x = 200, t = 0 \Rightarrow c = 2\sqrt{200}\)B1 For one arbitrary constant included (or equivalent statement of both pairs of limits)
\(2\sqrt{100} = -0.1t + 2\sqrt{200}\)M1 For evaluation of \(c\)
i.e. \(t = 82.8\)M1 For evaluation of \(t\)
A1For correct value 82.8 (minutes)
11
$\frac{dx}{dt} = -k\sqrt{x}$ | M1 | For use of derivative for rate of change
| A1 | For correct equation (neg sign optional here)
$x = 100$ and $\frac{dx}{dt} = -1 \Rightarrow k = 0.1$ | M1 | For use of data and their DE to find $k$
Hence equation is $\frac{dx}{dt} = -0.1\sqrt{x}$ | A1 | For any form of correct DE
$\int x^{-\frac{1}{2}}\,dx = -0.1\int dt \Rightarrow 2x^{\frac{1}{2}} = -0.1t + c$ | M1 | For separation and integration of both sides
| A1 | For $2x^{\frac{1}{2}}$ correct
| A1 | For $(\pm)kt$ correct (the numerical evaluation of $k$ may be delayed until after the DE is solved)
$x = 200, t = 0 \Rightarrow c = 2\sqrt{200}$ | B1 | For one arbitrary constant included (or equivalent statement of both pairs of limits)
$2\sqrt{100} = -0.1t + 2\sqrt{200}$ | M1 | For evaluation of $c$
i.e. $t = 82.8$ | M1 | For evaluation of $t$
| A1 | For correct value 82.8 (minutes)
| **11** |
9\\
\includegraphics[max width=\textwidth, alt={}, center]{798da17d-0af5-4aa6-b731-564642dc28d5-4_572_917_294_607}

A cylindrical container has a height of 200 cm . The container was initially full of a chemical but there is a leak from a hole in the base. When the leak is noticed, the container is half-full and the level of the chemical is dropping at a rate of 1 cm per minute. It is required to find for how many minutes the container has been leaking. To model the situation it is assumed that, when the depth of the chemical remaining is $x \mathrm {~cm}$, the rate at which the level is dropping is proportional to $\sqrt { } x$.

Set up and solve an appropriate differential equation, and hence show that the container has been leaking for about 80 minutes.

\hfill \mbox{\textit{OCR C4  Q9 [11]}}