| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | First-order integration |
| Difficulty | Moderate -0.5 This is a straightforward differential equations question requiring basic integration of an exponential function, application of initial conditions, and solving for t. All steps are routine A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{H =\}\ 0.6e^{-0.2t} \{+c\}\) | M1 | Attempts integration to achieve \(\{H =\}\ ke^{-0.2t}\{+c\}\) with \(k\) a numerical constant \(\neq -0.12\) |
| \(t=0, H=1.5 \Rightarrow 1.5 = \text{"0.6"} + c \Rightarrow c = 0.9\) | dM1 | Uses \(t=0, H=1.5\) and model of form \(H = ke^{-0.2t}+c\) to find value of constant \(c\); cannot just "make up" value for \(k\) |
| \(\Rightarrow H = 0.6e^{-0.2t} + 0.9\) | A1 | Correct complete equation with \(H\) present. Allow exact equivalents in required form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1.2 = 0.6e^{-0.2t} + 0.9 \Rightarrow 0.6e^{-0.2t} = 0.3\) | M1 | Uses \(H=1.2\) in model of form \(H = Ae^{-0.2t}+B\), \(B\neq 0\), rearranges to make \(Ae^{\pm 0.2t}\) or \(e^{\pm 0.2t}\) the subject |
| \(e^{-0.2t} = \dfrac{1}{2} \Rightarrow t = -5\ln\!\left(\dfrac{1}{2}\right)\) | dM1 | Correct use of ln to make \(t\) the subject. Requires \(A>0\), \(0 |
| \(\{t =\}\ 3\ \text{hours}\ 28\ \text{minutes}\) | A1 | Correct time in hours and minutes; must come from correct \(A\) and \(B\) in (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{\text{As } t \text{ gets large } H \rightarrow\}\ 0.9\) | M1 | Identifies requirement to establish limit as \(t\to\infty\). Can be implied by stating \(H = Ae^{-0.2t}+B \to B\) or \(\lim_{t\to\infty}[0.6e^{-0.2t}+0.9]=0.9\) |
| \(0.9\ \text{m or}\ 90\ \text{cm}\) | A1ft | Correct height including units. Follow through on value of \(B\) where \(0 |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{H =\}\ 0.6e^{-0.2t} \{+c\}$ | M1 | Attempts integration to achieve $\{H =\}\ ke^{-0.2t}\{+c\}$ with $k$ a numerical constant $\neq -0.12$ |
| $t=0, H=1.5 \Rightarrow 1.5 = \text{"0.6"} + c \Rightarrow c = 0.9$ | dM1 | Uses $t=0, H=1.5$ and model of form $H = ke^{-0.2t}+c$ to find value of constant $c$; cannot just "make up" value for $k$ |
| $\Rightarrow H = 0.6e^{-0.2t} + 0.9$ | A1 | Correct complete equation with $H$ present. Allow exact equivalents in required form |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1.2 = 0.6e^{-0.2t} + 0.9 \Rightarrow 0.6e^{-0.2t} = 0.3$ | M1 | Uses $H=1.2$ in model of form $H = Ae^{-0.2t}+B$, $B\neq 0$, rearranges to make $Ae^{\pm 0.2t}$ or $e^{\pm 0.2t}$ the subject |
| $e^{-0.2t} = \dfrac{1}{2} \Rightarrow t = -5\ln\!\left(\dfrac{1}{2}\right)$ | dM1 | Correct use of ln to make $t$ the subject. Requires $A>0$, $0<B<1.2$ and $e^{0.2t}=\lambda>0$ |
| $\{t =\}\ 3\ \text{hours}\ 28\ \text{minutes}$ | A1 | Correct time in hours and minutes; must come from correct $A$ and $B$ in (a) |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{\text{As } t \text{ gets large } H \rightarrow\}\ 0.9$ | M1 | Identifies requirement to establish limit as $t\to\infty$. Can be implied by stating $H = Ae^{-0.2t}+B \to B$ or $\lim_{t\to\infty}[0.6e^{-0.2t}+0.9]=0.9$ |
| $0.9\ \text{m or}\ 90\ \text{cm}$ | A1ft | Correct height including units. Follow through on value of $B$ where $0<B<1.2$ |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e116a86f-63e0-4e80-b49c-d9f3c819ce15-14_495_711_243_641}
\captionsetup{labelformat=empty}
\caption{Diagram not drawn to scale.}
\end{center}
\end{figure}
Figure 2
Figure 2 shows a cylindrical tank of height 1.5 m .\\
Initially the tank is full of water.\\
The water starts to leak from a small hole, at a point $L$, in the side of the tank.\\
While the tank is leaking, the depth, $H$ metres, of the water in the tank is modelled by the differential equation
$$\frac { \mathrm { d } H } { \mathrm {~d} t } = - 0.12 \mathrm { e } ^ { - 0.2 t }$$
where $t$ hours is the time after the leak starts.\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item show that
$$H = A \mathrm { e } ^ { - 0.2 t } + B$$
where $A$ and $B$ are constants to be found,
\item find the time taken for the depth of the water to decrease to 1.2 m . Give your answer in hours and minutes, to the nearest minute.
In the long term, the water level in the tank falls to the same height as the hole.
\item Find, according to the model, the height of the hole from the bottom of the tank.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q7 [8]}}