OCR H240/02 2018 March — Question 7 9 marks

Exam BoardOCR
ModuleH240/02 (Pure Mathematics and Statistics)
Year2018
SessionMarch
Marks9
TopicDifferential equations
TypeTank/container - constant cross-section (cuboid/cylinder)
DifficultyStandard +0.8 This is a differential equations problem requiring students to set up dV/dt = inflow - outflow, convert to dh/dt using the cross-sectional area, then solve a separable differential equation with a quadratic term. While the setup is moderately standard, solving the resulting equation ∫dh/(25-4h²) requires partial fractions or a non-trivial substitution, making it harder than typical A-level questions but not exceptionally difficult.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)3.02f Non-uniform acceleration: using differentiation and integration
7 A tank is shaped as a cuboid. The base has dimensions 10 cm by 10 cm . Initially the tank is empty. Water flows into the tank at \(25 \mathrm {~cm} ^ { 3 }\) per second. Water also leaks out of the tank at \(4 h ^ { 2 } \mathrm {~cm} ^ { 3 }\) per second, where \(h \mathrm {~cm}\) is the depth of the water after \(t\) seconds. Find the time taken for the water to reach a depth of 2 cm .
AnswerMarks Guidance
\(V = 100h \Rightarrow \frac{dv}{dh} = 100\)M1
\(\frac{dv}{dt} = \frac{dv}{dh} \cdot \frac{dh}{dt} = 100 \frac{dh}{dt}\) \([= 25 - 4h^2]\)A1
\(\Rightarrow 25 - 4h^2 = 100 \frac{dh}{dt}\) oeM1 Equate \(25 - 4h^2\) to their \(\frac{dv}{dh} \cdot \frac{dh}{dt}\)
\(\Rightarrow \int_0^t \frac{1}{25-4h^2} dh = \int_0^t \frac{1}{100} dt\)M1 Attempt integration with correct denominator on LHS
\(\Rightarrow \frac{1}{10}\int_0^t \frac{1}{5+2h} + \frac{1}{5-2h} dh = \int_0^t \frac{1}{100} dt\)M1 Attempt partial fractions with correct denominators on LHS
\(\Rightarrow \frac{1}{10} \cdot \frac{1}{2}\left[\ln(5+2h) - \ln(5-2h)\right]_0^t = \frac{t}{100}\)A1 Correct integral; ignore limits
\(\Rightarrow 5\ln 9 = t\) oe
AnswerMarks Guidance
Time when depth is 2 cm is \(11.0\) seconds (3 sf)A1, A1 Any correct numerical expression for \(t\) (\(10.9861\ldots\))
Allow 11 seconds
AnswerMarks
[9] 
$V = 100h \Rightarrow \frac{dv}{dh} = 100$ | M1 | 

$\frac{dv}{dt} = \frac{dv}{dh} \cdot \frac{dh}{dt} = 100 \frac{dh}{dt}$ $[= 25 - 4h^2]$ | A1 | 

$\Rightarrow 25 - 4h^2 = 100 \frac{dh}{dt}$ oe | M1 | Equate $25 - 4h^2$ to their $\frac{dv}{dh} \cdot \frac{dh}{dt}$

$\Rightarrow \int_0^t \frac{1}{25-4h^2} dh = \int_0^t \frac{1}{100} dt$ | M1 | Attempt integration with correct denominator on LHS

$\Rightarrow \frac{1}{10}\int_0^t \frac{1}{5+2h} + \frac{1}{5-2h} dh = \int_0^t \frac{1}{100} dt$ | M1 | Attempt partial fractions with correct denominators on LHS

$\Rightarrow \frac{1}{10} \cdot \frac{1}{2}\left[\ln(5+2h) - \ln(5-2h)\right]_0^t = \frac{t}{100}$ | A1 | Correct integral; ignore limits

$\Rightarrow 5\ln 9 = t$ oe
Time when depth is 2 cm is $11.0$ seconds (3 sf) | A1, A1 | Any correct numerical expression for $t$ ($10.9861\ldots$)
Allow 11 seconds
| [9]

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7 A tank is shaped as a cuboid. The base has dimensions 10 cm by 10 cm . Initially the tank is empty. Water flows into the tank at $25 \mathrm {~cm} ^ { 3 }$ per second. Water also leaks out of the tank at $4 h ^ { 2 } \mathrm {~cm} ^ { 3 }$ per second, where $h \mathrm {~cm}$ is the depth of the water after $t$ seconds. Find the time taken for the water to reach a depth of 2 cm .

\hfill \mbox{\textit{OCR H240/02 2018 Q7 [9]}}
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