| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Separable variables |
| Difficulty | Standard +0.3 This is a straightforward separable differential equation requiring standard integration technique (partial fractions after separation), followed by applying two boundary conditions to find constants, then algebraic manipulation to reach the given form. While it involves multiple steps, each is routine for P4 level with no novel insight required—slightly easier than average due to clear structure and standard method. |
| Spec | 1.06a Exponential function: a^x and e^x graphs and properties1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{d\theta}{dt} = -k(\theta-15)^2 \Rightarrow \int \frac{d\theta}{(\theta-15)^2} = \int -k\,dt\) | B1 | Correct separation of variables; \(d\theta\) and \(dt\) must be in correct positions |
| \(\int \frac{d\theta}{(\theta-15)^2} = -(\theta-15)^{-1}\) | M1 | Integrates \(\int \frac{1}{(\theta-15)^2}d\theta\) to \(\frac{\alpha}{\theta-15}\) |
| \(-\frac{1}{\theta-15} = -kt + c\) | A1 | Correct equation including \(k\) and a different constant |
| \(t=0, \theta=85 \Rightarrow -\frac{1}{70} = c\) | M1 | Uses \(t=0, \theta=85\) to find \(c\); may be awarded after incorrect integration |
| \(t=10, \theta=40 \Rightarrow \frac{1}{25} = 10k + \frac{1}{70} \Rightarrow k = \frac{9}{3500}\) | M1 | Uses \(t=10, \theta=40\) with their value of \(c\) to find \(k\) |
| \(\frac{1}{\theta-15} = \frac{9t}{3500} + \frac{1}{70} \Rightarrow \theta = \ldots\) | M1 | Rearranges using correct algebra to obtain \(\theta\) in terms of \(t\); integral must be correct form and both constants found by correct method |
| \(\theta = \frac{135t + 4250}{9t + 50}\) | A1 | Correct expression; allow integer multiples e.g. \(\theta = \frac{297500+9450t}{3500+630t}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(20 = \frac{135t+4250}{9t+50} \Rightarrow t = \ldots\) | M1 | Uses answer to part (a) or equivalent to find \(t\) when \(\theta=20\); must follow their work in (a) |
| \(t = \text{awrt } 72\) | A1 | Exact value is \(\frac{650}{9}\); allow awrt 72 |
## Question 6(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{d\theta}{dt} = -k(\theta-15)^2 \Rightarrow \int \frac{d\theta}{(\theta-15)^2} = \int -k\,dt$ | B1 | Correct separation of variables; $d\theta$ and $dt$ must be in correct positions |
| $\int \frac{d\theta}{(\theta-15)^2} = -(\theta-15)^{-1}$ | M1 | Integrates $\int \frac{1}{(\theta-15)^2}d\theta$ to $\frac{\alpha}{\theta-15}$ |
| $-\frac{1}{\theta-15} = -kt + c$ | A1 | Correct equation including $k$ and a different constant |
| $t=0, \theta=85 \Rightarrow -\frac{1}{70} = c$ | M1 | Uses $t=0, \theta=85$ to find $c$; may be awarded after incorrect integration |
| $t=10, \theta=40 \Rightarrow \frac{1}{25} = 10k + \frac{1}{70} \Rightarrow k = \frac{9}{3500}$ | M1 | Uses $t=10, \theta=40$ with their value of $c$ to find $k$ |
| $\frac{1}{\theta-15} = \frac{9t}{3500} + \frac{1}{70} \Rightarrow \theta = \ldots$ | M1 | Rearranges using correct algebra to obtain $\theta$ in terms of $t$; integral must be correct form and both constants found by correct method |
| $\theta = \frac{135t + 4250}{9t + 50}$ | A1 | Correct expression; allow integer multiples e.g. $\theta = \frac{297500+9450t}{3500+630t}$ |
---
## Question 6(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $20 = \frac{135t+4250}{9t+50} \Rightarrow t = \ldots$ | M1 | Uses answer to part (a) or equivalent to find $t$ when $\theta=20$; must follow their work in (a) |
| $t = \text{awrt } 72$ | A1 | Exact value is $\frac{650}{9}$; allow awrt 72 |
---\begin{enumerate}
\item In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
\end{enumerate}
The temperature, $\theta ^ { \circ } \mathrm { C }$, of a car engine, $t$ minutes after the engine is turned off, is modelled by the differential equation
$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - 15 ) ^ { 2 }$$
where $k$ is a constant.\\
Given that the temperature of the car engine
\begin{itemize}
\item is $85 ^ { \circ } \mathrm { C }$ at the instant the engine is turned off
\item is $40 ^ { \circ } \mathrm { C }$ exactly 10 minutes after the engine is turned off\\
(a) solve the differential equation to show that, according to the model
\end{itemize}
$$\theta = \frac { a t + b } { c t + d }$$
where $a , b , c$ and $d$ are integers to be found.\\
(b) Hence find, according to the model, the time taken for the temperature of the car engine to reach $20 ^ { \circ } \mathrm { C }$. Give your answer to the nearest minute.
\hfill \mbox{\textit{Edexcel P4 2023 Q6 [9]}}