| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line of intersection of planes |
| Difficulty | Challenging +1.2 This is part (c) of a multi-part vectors question asking for the line of intersection of two planes. While it requires understanding that two planes intersect in a line and involves finding direction vectors and a point on both planes, this is a standard Core Pure 1 technique. The question provides the planes explicitly (one from earlier parts), making it a methodical application of learned procedures rather than requiring novel insight. It's moderately above average difficulty due to the multi-step nature and being Further Maths content, but remains a textbook-style exercise. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dV}{dt}=3-\frac{4}{1+e^{0.8t}}\pm kV\) (where \(k\) is constant) | M1 | 3.3 |
| \(t=0, V=10, \frac{dV}{dt}=-3 \Rightarrow -3=3-\frac{4}{1+1}-10k \Rightarrow k=\ldots\) | dM1 | 3.4 |
| \(10k=4 \Rightarrow k=\frac{2}{5} \Rightarrow \frac{dV}{dt}=3-\frac{4}{1+e^{0.8t}}-0.4V\) | A1* | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{d}{dt}(\arctan e^{0.4t})=\frac{1}{1+(e^{0.4t})^2}\times ke^{0.4t}\) | M1 | 1.1b |
| \(\frac{d}{dt}(\arctan e^{0.4t})=\frac{2e^{0.4t}}{5(1+e^{0.8t})}\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(y=\arctan e^{0.4t} \Rightarrow \tan y = e^{0.4t} \Rightarrow \sec^2 y\frac{dy}{dx}=0.4e^{0.4t}\) | M1 | 1.1b |
| \(\frac{dy}{dx}=\frac{0.4e^{0.4t}}{\sec^2 y}=\frac{0.4e^{0.4t}}{1+\tan^2 y}=\frac{0.4e^{0.4t}}{1+(e^{0.4t})^2}\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dV}{dt}+0.4V=3-\frac{4}{1+e^{0.8t}} \Rightarrow I.F. = e^{\int 0.4\,dt}=e^{0.4t}\) | B1 | 2.2a |
| \(e^{0.4t}V=\int 3e^{0.4t}-\frac{4e^{0.4t}}{1+e^{0.8t}}\,dt\) | M1 | 1.1b |
| \(=Ae^{0.4t}-B\arctan(e^{0.4t})(+c)\) | M1 | 1.1b |
| \(e^{0.4t}V=7.5e^{0.4t}-10\arctan(e^{0.4t})\,(+c)\) | A1 | 1.1b |
| \(V=10, t=0 \Rightarrow 10=7.5-10\arctan 1+c \Rightarrow c=\ldots\) | M1 | 3.4 |
| \(V=7.5-10e^{-0.4t}\arctan(e^{0.4t})+2.5(\pi+1)e^{-0.4t}\) | A1 | 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| E.g. \(V(10)\approx 7.4\) litres so the model is not very accurate as it predicts approximately 7.5% below the actual level | B1ft | 3.5a |
## Question 6(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dt}=3-\frac{4}{1+e^{0.8t}}\pm kV$ (where $k$ is constant) | M1 | 3.3 |
| $t=0, V=10, \frac{dV}{dt}=-3 \Rightarrow -3=3-\frac{4}{1+1}-10k \Rightarrow k=\ldots$ | dM1 | 3.4 |
| $10k=4 \Rightarrow k=\frac{2}{5} \Rightarrow \frac{dV}{dt}=3-\frac{4}{1+e^{0.8t}}-0.4V$ | A1* | 2.1 |
---
## Question 6(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d}{dt}(\arctan e^{0.4t})=\frac{1}{1+(e^{0.4t})^2}\times ke^{0.4t}$ | M1 | 1.1b |
| $\frac{d}{dt}(\arctan e^{0.4t})=\frac{2e^{0.4t}}{5(1+e^{0.8t})}$ | A1 | 1.1b |
**Alternative:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $y=\arctan e^{0.4t} \Rightarrow \tan y = e^{0.4t} \Rightarrow \sec^2 y\frac{dy}{dx}=0.4e^{0.4t}$ | M1 | 1.1b |
| $\frac{dy}{dx}=\frac{0.4e^{0.4t}}{\sec^2 y}=\frac{0.4e^{0.4t}}{1+\tan^2 y}=\frac{0.4e^{0.4t}}{1+(e^{0.4t})^2}$ | A1 | 1.1b |
---
## Question 6(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dt}+0.4V=3-\frac{4}{1+e^{0.8t}} \Rightarrow I.F. = e^{\int 0.4\,dt}=e^{0.4t}$ | B1 | 2.2a |
| $e^{0.4t}V=\int 3e^{0.4t}-\frac{4e^{0.4t}}{1+e^{0.8t}}\,dt$ | M1 | 1.1b |
| $=Ae^{0.4t}-B\arctan(e^{0.4t})(+c)$ | M1 | 1.1b |
| $e^{0.4t}V=7.5e^{0.4t}-10\arctan(e^{0.4t})\,(+c)$ | A1 | 1.1b |
| $V=10, t=0 \Rightarrow 10=7.5-10\arctan 1+c \Rightarrow c=\ldots$ | M1 | 3.4 |
| $V=7.5-10e^{-0.4t}\arctan(e^{0.4t})+2.5(\pi+1)e^{-0.4t}$ | A1 | 2.1 |
---
## Question 6(d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| E.g. $V(10)\approx 7.4$ litres so the model is not very accurate as it predicts approximately 7.5% below the actual level | B1ft | 3.5a |6 \\
2
\end{array} \right)$$
where $\mu$ is a scalar parameter.
The plane $\Pi _ { 2 }$ contains the line $l _ { 1 }$ and the line $l _ { 2 }$\\
(c) Determine a vector equation for the line of intersection of $\Pi _ { 1 }$ and $\Pi _ { 2 }$
The plane $\Pi _ { 3 }$ has equation r. $\left( \begin{array} { l } 1 \\ 1 \\ a \end{array} \right) = b$ where $a$ and $b$ are constants.\\
Given that the planes $\Pi _ { 1 } , \Pi _ { 2 }$ and $\Pi _ { 3 }$ form a sheaf,\\
(d) determine the value of $a$ and the value of $b$.
\begin{enumerate}
\item Water is flowing into and out of a large tank.
\end{enumerate}
Initially the tank contains 10 litres of water.\\
The rate of flow of the water is modelled so that
\begin{itemize}
\item there are $V$ litres of water in the tank at time $t$ minutes after the water begins to flow
\item water enters the tank at a rate of $\left( 3 - \frac { 4 } { 1 + \mathrm { e } ^ { 0.8 t } } \right)$ litres per minute
\item water leaves the tank at a rate proportional to the volume of water remaining in the tank
\end{itemize}
Given that when $t = 0$ the volume of water in the tank is decreasing at a rate of 3 litres per minute, use the model to\\
(a) show that the volume of water in the tank at time $t$ satisfies
$$\frac { \mathrm { d } V } { \mathrm {~d} t } = 3 - \frac { 4 } { 1 + \mathrm { e } ^ { 0.8 t } } - 0.4 V$$
(b) Determine $\frac { \mathrm { d } } { \mathrm { d } t } \left( \arctan \mathrm { e } ^ { 0.4 t } \right)$
Hence, by solving the differential equation from part (a),\\
(c) determine an equation for the volume of water in the tank at time $t$.
Give your answer in simplest form as $V = \mathrm { f } ( t )$
After 10 minutes, the volume of water in the tank was 8 litres.\\
(d) Evaluate the model in light of this information.
\hfill \mbox{\textit{Edexcel CP1 2023 Q6 [12]}}