| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Tank/container - constant cross-section (cuboid/cylinder) |
| Difficulty | Standard +0.3 This is a straightforward differential equations question requiring standard separation of variables. Part (a) is a simple 'show that' using the chain rule and given proportionality. Part (b) involves routine integration and finding constants from initial conditions. Part (c) is direct substitution. The question follows a very standard template for tank problems with clear scaffolding and no novel insights required—slightly easier than average. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dV}{dh}=200\) (o.e. e.g. \(\frac{dh}{dV}=\frac{1}{200}\)) | B1 | |
| \(\left(\frac{dh}{dt}=\right)\frac{dh}{dV}\times\frac{dV}{dt}=\frac{1}{200}\times\frac{k}{\sqrt{h}}\) | M1 | Chain rule applied |
| \(\frac{dh}{dt}=\frac{\lambda}{\sqrt{h}}\) * | A1\* | Given answer — must be derived correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dh}{dt}=\frac{\lambda}{\sqrt{h}} \Rightarrow \int h^{\frac{1}{2}}\,dh = \int\lambda\,dt \Rightarrow \ldots h^{\frac{3}{2}}=\lambda t\{+c\}\) | M1 | Separates variables and integrates both sides |
| \(\frac{2}{3}h^{\frac{3}{2}}=\lambda t\{+c\}\) | A1 | |
| \(\frac{2}{3}(1.44)^{\frac{3}{2}}=\lambda\times 0+c \Rightarrow c=1.152\left(=\frac{144}{125}\right)\) | dM1 | Substitutes \(t=0\), \(h=1.44\) to find \(c\) |
| \(\frac{2}{3}(3.24)^{\frac{3}{2}}=\lambda\times 8+\text{"1.152"} \Rightarrow \lambda=0.342\left(=\frac{171}{500}\right)\) | ddM1 | Substitutes \(t=8\), \(h=3.24\) and their \(c\) to find \(\lambda\) |
| \(h^{\frac{3}{2}}=0.513t+1.728\) o.e. e.g. \(h^{\frac{3}{2}}=\frac{513}{1000}t+\frac{216}{125}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(5^{\frac{3}{2}}=0.513t+1.728 \Rightarrow t=\ldots\) | M1 | Substitutes \(h=5\) into their equation |
| \((t=)\) awrt \(18.4\) min | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dV}{dh} = 200\) stated or used | B1 | May be implied by chain rule attempt |
| \(\frac{dV}{dt} = \pm\frac{k}{\sqrt{h}}\) or e.g. \(\frac{dV}{dt} = \pm\frac{1}{k\sqrt{h}}\) | M1 | Requires \(\frac{dV}{dh} = p\), \(p>1\); \(k\) may be \(\lambda\) but must not be a number |
| Application of chain rule \(\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}\) with correct placement | M1 | Any equivalent form with \(\frac{dV}{dt} = \pm\frac{k}{\sqrt{h}}\) or \(\pm\frac{1}{k\sqrt{h}}\) |
| \(\frac{dh}{dt} = \frac{\lambda}{\sqrt{h}}\) with no errors | A1* | Rigorous argument with all steps shown; do not allow \(\lambda\) used for both constants |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Separates variables and integrates to obtain \(\ldots h^{\frac{3}{2}} = \lambda t\{+c\}\) | M1 | Constant of integration not needed for this mark |
| \(\frac{2}{3}h^{\frac{3}{2}} = \lambda t (+c)\) | A1 | Constant of integration not needed |
| Substitutes \(t=0\), \(h=1.44\) and attempts to find \(c\) | dM1 | Dependent on previous method mark |
| Substitutes \(t=8\), \(h=3.24\) and their \(c\), attempts to find \(\lambda\) | ddM1 | Dependent on both previous method marks |
| \(h^{\frac{3}{2}} = 0.513t + 1.728\) or \(h^{\frac{3}{2}} = \frac{513}{1000}t + \frac{216}{125}\) | A1 | Allow 1.73 for 1.728; must follow A1 earlier |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Finds reciprocal of both sides and integrates to obtain \(t = \ldots h^{\frac{3}{2}}(+c)\) | M1 | |
| \(t = \frac{2h^{\frac{3}{2}}}{3\lambda}(+c)\) | A1 | Constant of integration not needed |
| Substitutes \(t=0\), \(h=1.44\) AND \(t=8\), \(h=3.24\), attempts to find \(\lambda\) or \(c\) | dM1 | Dependent on previous method mark |
| Complete attempt to find both \(\lambda\) and \(c\) | ddM1 | Dependent on both previous method marks |
| \(h^{\frac{3}{2}} = 0.513t + 1.728\) or \(h^{\frac{3}{2}} = \frac{513}{1000}t + \frac{216}{125}\) | A1 | Allow 1.73 for 1.728 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| No attempt to integrate | M0A0 | |
| Substitutes \(t=0\), \(h=1.44\) to find \(B\) | M1 | |
| Substitutes \(t=8\), \(h=3.24\) with \(B\) to find \(A\) | dM1 | |
| Since given model not used | A0 | Maximum 00110; allow full recovery in (c) if equation is correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts to substitute \(h=5\) into equation of form \(h^{\frac{3}{2}} = At + B\) or rearranged form | M1 | Must use \(h=5\); obtain value for \(t\) even if negative |
| Awrt 18.4 minutes | A1 | Following correct equation in (b); units required (min/mins); allow 18 minutes 25 seconds or 18 mins 26 secs |
# Question 11(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dh}=200$ (o.e. e.g. $\frac{dh}{dV}=\frac{1}{200}$) | **B1** | |
| $\left(\frac{dh}{dt}=\right)\frac{dh}{dV}\times\frac{dV}{dt}=\frac{1}{200}\times\frac{k}{\sqrt{h}}$ | **M1** | Chain rule applied |
| $\frac{dh}{dt}=\frac{\lambda}{\sqrt{h}}$ * | **A1\*** | Given answer — must be derived correctly |
**(3 marks)**
---
# Question 11(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dh}{dt}=\frac{\lambda}{\sqrt{h}} \Rightarrow \int h^{\frac{1}{2}}\,dh = \int\lambda\,dt \Rightarrow \ldots h^{\frac{3}{2}}=\lambda t\{+c\}$ | **M1** | Separates variables and integrates both sides |
| $\frac{2}{3}h^{\frac{3}{2}}=\lambda t\{+c\}$ | **A1** | |
| $\frac{2}{3}(1.44)^{\frac{3}{2}}=\lambda\times 0+c \Rightarrow c=1.152\left(=\frac{144}{125}\right)$ | **dM1** | Substitutes $t=0$, $h=1.44$ to find $c$ |
| $\frac{2}{3}(3.24)^{\frac{3}{2}}=\lambda\times 8+\text{"1.152"} \Rightarrow \lambda=0.342\left(=\frac{171}{500}\right)$ | **ddM1** | Substitutes $t=8$, $h=3.24$ and their $c$ to find $\lambda$ |
| $h^{\frac{3}{2}}=0.513t+1.728$ o.e. e.g. $h^{\frac{3}{2}}=\frac{513}{1000}t+\frac{216}{125}$ | **A1** | |
**(5 marks)**
---
# Question 11(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $5^{\frac{3}{2}}=0.513t+1.728 \Rightarrow t=\ldots$ | **M1** | Substitutes $h=5$ into their equation |
| $(t=)$ awrt $18.4$ min | **A1** | |
**(2 marks)**
# Question (Differential Equation - Water Tank):
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dh} = 200$ stated or used | **B1** | May be implied by chain rule attempt |
| $\frac{dV}{dt} = \pm\frac{k}{\sqrt{h}}$ or e.g. $\frac{dV}{dt} = \pm\frac{1}{k\sqrt{h}}$ | **M1** | Requires $\frac{dV}{dh} = p$, $p>1$; $k$ may be $\lambda$ but must not be a number |
| Application of chain rule $\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}$ with correct placement | **M1** | Any equivalent form with $\frac{dV}{dt} = \pm\frac{k}{\sqrt{h}}$ or $\pm\frac{1}{k\sqrt{h}}$ |
| $\frac{dh}{dt} = \frac{\lambda}{\sqrt{h}}$ with no errors | **A1*** | Rigorous argument with all steps shown; do not allow $\lambda$ used for both constants |
---
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Separates variables and integrates to obtain $\ldots h^{\frac{3}{2}} = \lambda t\{+c\}$ | **M1** | Constant of integration not needed for this mark |
| $\frac{2}{3}h^{\frac{3}{2}} = \lambda t (+c)$ | **A1** | Constant of integration not needed |
| Substitutes $t=0$, $h=1.44$ and attempts to find $c$ | **dM1** | Dependent on previous method mark |
| Substitutes $t=8$, $h=3.24$ and their $c$, attempts to find $\lambda$ | **ddM1** | Dependent on both previous method marks |
| $h^{\frac{3}{2}} = 0.513t + 1.728$ or $h^{\frac{3}{2}} = \frac{513}{1000}t + \frac{216}{125}$ | **A1** | Allow 1.73 for 1.728; must follow A1 earlier |
### Part (b) Alternative:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds reciprocal of both sides and integrates to obtain $t = \ldots h^{\frac{3}{2}}(+c)$ | **M1** | |
| $t = \frac{2h^{\frac{3}{2}}}{3\lambda}(+c)$ | **A1** | Constant of integration not needed |
| Substitutes $t=0$, $h=1.44$ AND $t=8$, $h=3.24$, attempts to find $\lambda$ or $c$ | **dM1** | Dependent on previous method mark |
| Complete attempt to find both $\lambda$ and $c$ | **ddM1** | Dependent on both previous method marks |
| $h^{\frac{3}{2}} = 0.513t + 1.728$ or $h^{\frac{3}{2}} = \frac{513}{1000}t + \frac{216}{125}$ | **A1** | Allow 1.73 for 1.728 |
### Special Case (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| No attempt to integrate | **M0A0** | |
| Substitutes $t=0$, $h=1.44$ to find $B$ | **M1** | |
| Substitutes $t=8$, $h=3.24$ with $B$ to find $A$ | **dM1** | |
| Since given model not used | **A0** | Maximum 00110; allow full recovery in (c) if equation is correct |
---
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts to substitute $h=5$ into equation of form $h^{\frac{3}{2}} = At + B$ or rearranged form | **M1** | Must use $h=5$; obtain value for $t$ even if negative |
| Awrt 18.4 minutes | **A1** | Following correct equation in (b); units required (min/mins); allow 18 minutes 25 seconds or 18 mins 26 secs |
---11.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3f6f3f19-a1d0-488b-a1a4-302cc4cf5a1e-30_455_997_210_552}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A tank in the shape of a cuboid is being filled with water.\\
The base of the tank measures 20 m by 10 m and the height of the tank is 5 m , as shown in Figure 1.
At time $t$ minutes after water started flowing into the tank the height of the water was $h \mathrm {~m}$ and the volume of water in the tank was $V \mathrm {~m} ^ { 3 }$
In a model of this situation
\begin{itemize}
\item the sides of the tank have negligible thickness
\item the rate of change of $V$ is inversely proportional to the square root of $h$
\begin{enumerate}[label=(\alph*)]
\item Show that
\end{itemize}
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { \lambda } { \sqrt { h } }$$
where $\lambda$ is a constant.
Given that
\begin{itemize}
\item initially the height of the water in the tank was 1.44 m
\item exactly 8 minutes after water started flowing into the tank the height of the water was 3.24 m
\item use the model to find an equation linking $h$ with $t$, giving your answer in the form
\end{itemize}
$$h ^ { \frac { 3 } { 2 } } = A t + B$$
where $A$ and $B$ are constants to be found.
\item Hence find the time taken, from when water started flowing into the tank, for the tank to be completely full.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2023 Q11 [10]}}