| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Tank/reservoir mixing problems |
| Difficulty | Challenging +1.2 This is a multi-part question combining coordinate geometry, volumes of revolution, and related rates. Parts (a)-(c) involve standard techniques (finding constants, integration for volume), while part (d) requires implicit differentiation of the volume formula—a straightforward application of the chain rule. The context is applied but the mathematical steps are routine for Further Maths students who have practiced integrating factor problems and related rates. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k = 2.6\) | B1 | 3.4 — Uses the model to obtain a correct value for \(k\). Must be 2.6 not −2.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 1.18 \Rightarrow \ln(3.6 \times 1.18 - \text{"2.6"}) = \ldots\) | M1 | 1.1b — Substitutes their value of \(k\) and \(x=1.18\) into the given model to find a value for \(y\) |
| \(h = 0.4995\ldots\) m | A1 | 2.2b — Infers that the depth of the pool could be awrt 0.5 m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \ln(3.6x - 2.6) \Rightarrow x = \dfrac{e^y + 2.6}{3.6}\) or \(\dfrac{5e^y+13}{18}\) | B1ft | 1.1a — Uses the model to obtain \(x\) correctly in terms of \(y\) (follow through their \(k\)) |
| \(V = \pi\displaystyle\int\left(\dfrac{e^y+2.6}{3.6}\right)^2 dy = \dfrac{\pi}{3.6^2}\displaystyle\int\left(e^{2y}+5.2e^y+6.76\right)dy\) or \(\dfrac{\pi}{324}\displaystyle\int\left(25e^{2y}+130e^y+169\right)dy\) | M1 | 3.3 — Uses the model to obtain an expression for the volume using \(\pi\int(f(y))^2\,dy\); must expand to reach integrable form |
| \(= \dfrac{\pi}{3.6^2}\left[\dfrac{1}{2}e^{2y}+5.2e^y+6.76y\right]\) or \(\dfrac{\pi}{324}\left[\dfrac{25}{2}e^{2y}+130e^y+169y\right]\) | A1 | 1.1b — Correct integration |
| Substitutes limits \(h\) and \(0\), clearly showing use of both limits | M1 | 2.1 — Selects limits appropriate to the model (\(h\) and 0) |
| \(= \dfrac{\pi}{3.6^2}\left(\dfrac{1}{2}e^{2h}+5.2e^h+6.76h - 5.7\right)\) | A1 | 1.1b — Correct expression (allow unsimplified and isw if necessary) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dV}{dh} = \dfrac{\pi}{3.6^2}\left(e^{2h}+5.2e^h+6.76\right) = \dfrac{\pi}{3.6^2}\left(e^{0.4}+5.2e^{0.2}+6.76\right)\) | M1 | 3.1a — Recognises that \(\dfrac{dV}{dh}\) is required and attempts to find \(\dfrac{dV}{dh}\) or \(\dfrac{dh}{dV}\) from their integration |
| \(\dfrac{dh}{dt} = \dfrac{dh}{dV}\cdot\dfrac{dV}{dt} = \dfrac{1}{3.539\ldots}\times 0.015 \times 60\) | M1 | 1.1b — Evidence of correct use of chain rule; attempts to divide 15 (or converted 15) by their \(\dfrac{dV}{dh}\) |
| \(\dfrac{dh}{dt} = 25.4\ \text{cm h}^{-1}\) | A1 | 3.2a — Correct answer awrt 25.4 with correct units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=0.2 \Rightarrow x = \dfrac{2.6+e^{0.2}}{3.6} \Rightarrow A = \pi\!\left(\dfrac{2.6+e^{0.2}}{3.6}\right)^2\ (=3.54)\) | M1 | 3.1a — Uses \(y=0.2\) to find \(x\) and the surface area of the water at that instant |
| \(\dfrac{dh}{dt} = \dfrac{0.015\times 60}{3.54}\) | M1 | 1.1b — Attempts to divide the rate by their area |
| \(\dfrac{dh}{dt} = 25.4\ \text{cm h}^{-1}\) | A1 | 3.2a — Correct answer awrt 25.4 with correct units |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = 2.6$ | B1 | 3.4 — Uses the model to obtain a correct value for $k$. Must be 2.6 not −2.6 |
## Question 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 1.18 \Rightarrow \ln(3.6 \times 1.18 - \text{"2.6"}) = \ldots$ | M1 | 1.1b — Substitutes their value of $k$ and $x=1.18$ into the given model to find a value for $y$ |
| $h = 0.4995\ldots$ m | A1 | 2.2b — Infers that the depth of the pool could be awrt 0.5 m |
## Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \ln(3.6x - 2.6) \Rightarrow x = \dfrac{e^y + 2.6}{3.6}$ or $\dfrac{5e^y+13}{18}$ | B1ft | 1.1a — Uses the model to obtain $x$ correctly in terms of $y$ (follow through their $k$) |
| $V = \pi\displaystyle\int\left(\dfrac{e^y+2.6}{3.6}\right)^2 dy = \dfrac{\pi}{3.6^2}\displaystyle\int\left(e^{2y}+5.2e^y+6.76\right)dy$ or $\dfrac{\pi}{324}\displaystyle\int\left(25e^{2y}+130e^y+169\right)dy$ | M1 | 3.3 — Uses the model to obtain an expression for the volume using $\pi\int(f(y))^2\,dy$; must expand to reach integrable form |
| $= \dfrac{\pi}{3.6^2}\left[\dfrac{1}{2}e^{2y}+5.2e^y+6.76y\right]$ or $\dfrac{\pi}{324}\left[\dfrac{25}{2}e^{2y}+130e^y+169y\right]$ | A1 | 1.1b — Correct integration |
| Substitutes limits $h$ and $0$, clearly showing use of both limits | M1 | 2.1 — Selects limits appropriate to the model ($h$ and 0) |
| $= \dfrac{\pi}{3.6^2}\left(\dfrac{1}{2}e^{2h}+5.2e^h+6.76h - 5.7\right)$ | A1 | 1.1b — Correct expression (allow unsimplified and isw if necessary) |
## Question 8(d) — Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dV}{dh} = \dfrac{\pi}{3.6^2}\left(e^{2h}+5.2e^h+6.76\right) = \dfrac{\pi}{3.6^2}\left(e^{0.4}+5.2e^{0.2}+6.76\right)$ | M1 | 3.1a — Recognises that $\dfrac{dV}{dh}$ is required and attempts to find $\dfrac{dV}{dh}$ or $\dfrac{dh}{dV}$ from their integration |
| $\dfrac{dh}{dt} = \dfrac{dh}{dV}\cdot\dfrac{dV}{dt} = \dfrac{1}{3.539\ldots}\times 0.015 \times 60$ | M1 | 1.1b — Evidence of correct use of chain rule; attempts to divide 15 (or converted 15) by their $\dfrac{dV}{dh}$ |
| $\dfrac{dh}{dt} = 25.4\ \text{cm h}^{-1}$ | A1 | 3.2a — Correct answer awrt 25.4 **with correct units** |
## Question 8(d) — Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=0.2 \Rightarrow x = \dfrac{2.6+e^{0.2}}{3.6} \Rightarrow A = \pi\!\left(\dfrac{2.6+e^{0.2}}{3.6}\right)^2\ (=3.54)$ | M1 | 3.1a — Uses $y=0.2$ to find $x$ and the surface area of the water at that instant |
| $\dfrac{dh}{dt} = \dfrac{0.015\times 60}{3.54}$ | M1 | 1.1b — Attempts to divide the rate by their area |
| $\dfrac{dh}{dt} = 25.4\ \text{cm h}^{-1}$ | A1 | 3.2a — Correct answer awrt 25.4 **with correct units** |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e63eca0e-e2e9-4660-855f-0696c680a8f9-24_359_552_274_438}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e63eca0e-e2e9-4660-855f-0696c680a8f9-24_323_387_283_1238}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 1 shows the central vertical cross section $A B C D$ of a paddling pool that has a circular horizontal cross section. Measurements of the diameters of the top and bottom of the paddling pool have been taken in order to estimate the volume of water that the paddling pool can contain.
Using these measurements, the curve $B D$ is modelled by the equation
$$y = \ln ( 3.6 x - k ) \quad 1 \leqslant x \leqslant 1.18$$
as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find the depth of the paddling pool according to this model.
The pool is being filled with water from a tap.
\item Find, in terms of $h$, the volume of water in the pool when the pool is filled to a depth of $h \mathrm {~m}$.
Given that the pool is being filled at a constant rate of 15 litres every minute,
\item find, in $\mathrm { cmh } ^ { - 1 }$, the rate at which the water level is rising in the pool when the depth of the water is 0.2 m .
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP2 2019 Q8 [11]}}