OCR C4 — Question 7 9 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeFirst-order integration
DifficultyStandard +0.3 This is a straightforward C4 differential equations question requiring integration of an exponential function, substitution to find a constant, and solving for when y equals h/100. All techniques are standard and the multi-step nature is typical for a longer C4 question, making it slightly easier than average overall.
Spec1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y)1.08l Interpret differential equation solutions: in context

At time \(t = 0\), a tank of height 2 metres is completely filled with water. Water then leaks from a hole in the side of the tank such that the depth of water in the tank, \(y\) metres, after \(t\) hours satisfies the differential equation $$\frac{dy}{dt} = -ke^{-0.2t},$$ where \(k\) is a positive constant.
  1. Find an expression for \(y\) in terms of \(k\) and \(t\). [4]
Given that two hours after being filled the depth of water in the tank is 1.6 metres,
  1. find the value of \(k\) to 4 significant figures. [2]
Given also that the hole in the tank is \(h\) cm above the base of the tank,
  1. show that \(h = 79\) to 2 significant figures. [3]

(i)
AnswerMarks
\(\int dy = \int -ke^{-0.2t} dt\)M1
\(y = 5ke^{-0.2t} + c\)A1
\(t = 0, y = 2 \Rightarrow 2 = 5k + c, \quad c = 2 - 5k\)M1
\(\therefore y = 5ke^{-0.2t} - 5k + 2\)A1
(ii)
AnswerMarks
\(t = 2, y = 1.6 \Rightarrow 1.6 = 5ke^{-0.4} - 5k + 2\)M1 A1
\(k = \frac{-0.4}{5e^{-0.4} - 5} = 0.2427 \text{ (4sf)}\)M1 A1
(iii)
AnswerMarks Guidance
as \(t \to \infty, y \to h\) (in metres)M1
\(\therefore "h" = -5k + 2 = 0.787 \text{ m} = 78.7 \text{ cm} \quad \therefore h = 79 \text{ (2sf)}\)M1 A1 (9)
## (i)
$\int dy = \int -ke^{-0.2t} dt$ | M1 |
$y = 5ke^{-0.2t} + c$ | A1 |
$t = 0, y = 2 \Rightarrow 2 = 5k + c, \quad c = 2 - 5k$ | M1 |
$\therefore y = 5ke^{-0.2t} - 5k + 2$ | A1 |

## (ii)
$t = 2, y = 1.6 \Rightarrow 1.6 = 5ke^{-0.4} - 5k + 2$ | M1 A1 |
$k = \frac{-0.4}{5e^{-0.4} - 5} = 0.2427 \text{ (4sf)}$ | M1 A1 |

## (iii)
as $t \to \infty, y \to h$ (in metres) | M1 |
$\therefore "h" = -5k + 2 = 0.787 \text{ m} = 78.7 \text{ cm} \quad \therefore h = 79 \text{ (2sf)}$ | M1 A1 | (9) |

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At time $t = 0$, a tank of height 2 metres is completely filled with water. Water then leaks from a hole in the side of the tank such that the depth of water in the tank, $y$ metres, after $t$ hours satisfies the differential equation
$$\frac{dy}{dt} = -ke^{-0.2t},$$
where $k$ is a positive constant.

\begin{enumerate}[label=(\roman*)]
\item Find an expression for $y$ in terms of $k$ and $t$. [4]
\end{enumerate}

Given that two hours after being filled the depth of water in the tank is 1.6 metres,

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item find the value of $k$ to 4 significant figures. [2]
\end{enumerate}

Given also that the hole in the tank is $h$ cm above the base of the tank,

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item show that $h = 79$ to 2 significant figures. [3]
\end{enumerate}

\hfill \mbox{\textit{OCR C4  Q7 [9]}}