Edexcel C4 — Question 5 10 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeFirst-order integration
DifficultyModerate -0.3 This is a straightforward C4 differential equations question requiring direct integration of an exponential function, substitution of initial conditions, and solving for constants. The integration is standard (∫e^(-0.2t) dt), and all parts follow mechanically from the previous one with no conceptual challenges or novel problem-solving required. Slightly easier than average due to the explicit form of dy/dt and clear step-by-step structure.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)

5. 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 $$\frac { \mathrm { d } y } { \mathrm {~d} t } = - k \mathrm { e } ^ { - 0.2 t }$$ where \(k\) is a positive constant,
  1. Find an expression for \(y\) in terms of \(k\) and \(t\). Given that two hours after being filled the depth of water in the tank is 1.6 metres,
  2. find the value of \(k\) to 4 significant figures. Given also that the hole in the tank is \(h \mathrm {~cm}\) above the base of the tank,
  3. show that \(h = 79\) to 2 significant figures.
    5. continued

AnswerMarks Guidance
(a) \(\int dy = \int -ke^{-0.2t} dt\)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
(b) \(t = 2, y = 1.6 \Rightarrow 1.6 = 5ke^{-0.4} - 5k + 2\)M1
\(k = \frac{-0.4}{5e^{-0.4} - 5} = 0.2427\) (4sf)M1 A1
(c) as \(t \to \infty, y \to h\) (in metres)M1
\(\therefore "h" = -5k + 2 = 0.787\) m \(= 78.7\) cm ∴ \(h = 79\)M1 A1 (10)
**(a)** $\int dy = \int -ke^{-0.2t} dt$ | 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 |

**(b)** $t = 2, y = 1.6 \Rightarrow 1.6 = 5ke^{-0.4} - 5k + 2$ | M1 |

$k = \frac{-0.4}{5e^{-0.4} - 5} = 0.2427$ (4sf) | M1 A1 |

**(c)** as $t \to \infty, y \to h$ (in metres) | M1 |

$\therefore "h" = -5k + 2 = 0.787$ m $= 78.7$ cm ∴ $h = 79$ | M1 A1 | (10)
5. 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

$$\frac { \mathrm { d } y } { \mathrm {~d} t } = - k \mathrm { e } ^ { - 0.2 t }$$

where $k$ is a positive constant,
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $y$ in terms of $k$ and $t$.

Given that two hours after being filled the depth of water in the tank is 1.6 metres,
\item find the value of $k$ to 4 significant figures.

Given also that the hole in the tank is $h \mathrm {~cm}$ above the base of the tank,
\item show that $h = 79$ to 2 significant figures.\\
5. continued
\end{enumerate}

\hfill \mbox{\textit{Edexcel C4  Q5 [10]}}