| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2020 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Separable variables - standard (applied/contextual) |
| Difficulty | Moderate -0.3 This is a standard separable differential equation with straightforward integration (power rule and partial fractions), followed by applying initial conditions and finding a limit. While it requires multiple steps and careful algebra, it follows a completely routine procedure for P4/Further Maths with no novel insight required—slightly easier than average due to its predictable structure. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| VIIV SIHI NI JIIIM ION OC | VIAV SIHI NI I II M I I O N OC | VAYV SIHI NI JIIIM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Separates variables: \(\displaystyle\int \dfrac{dA}{A^{\frac{3}{2}}}=\int\dfrac{dt}{5t^2}\) | M1 | \(A^{\frac{3}{2}}\), \(dA\), \(t^2\) and \(dt\) must be in correct positions; condone without integral signs; don't be too concerned about position of "5" |
| \(-2A^{-\frac{1}{2}}=-\dfrac{1}{5}t^{-1}\) | M1 | Integrates one side correctly: look for \(pA^{-\frac{1}{2}}\) or \(qt^{-1}\) |
| \(-2A^{-\frac{1}{2}}=-\dfrac{1}{5}t^{-1}(+c)\) | dM1 | Integrates both sides correctly: look for \(pA^{-\frac{1}{2}}\) and \(qt^{-1}\) |
| \(-2A^{-\frac{1}{2}}=-\dfrac{1}{5}t^{-1}(+c)\) correct (without \(+c\)) | A1 | e.g. \(-10A^{-\frac{1}{2}}=-t^{-1}(+c)\) or \(-\dfrac{A^{-0.5}}{0.5}=-\dfrac{1}{5}t^{-1}(+c)\) |
| Substitutes \(t=3\), \(A=2.25 \Rightarrow c=\ldots\left(-\dfrac{19}{15}\right)\) | M1 | Must have "\(+c\)"; some evidence for award; one index must have started correct; may be awarded after incorrect change of subject |
| Proceeds to \(A=\left(\dfrac{pt}{qt+r}\right)^2\) from \(\dfrac{p}{\sqrt{A}}=\dfrac{q}{t}+r\) with correct operations | M1 | Must use common factor on RHS, invert both sides (not each term), then square both sides to reach \(A=\ldots\); candidates do not need a value for \(r\) |
| \(A=\left(\dfrac{30t}{19t+3}\right)^2\) | A1 | cso; this mark can be awarded independent of first M1 |
| (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| As \(t\to\infty\), \(A\to\left(\dfrac{30}{19}\right)^2=\dfrac{900}{361}\) or awrt \(2.49\ \text{cm}^2\) | M1 A1 | |
| (2) | ||
| (9 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \left(\dfrac{300t}{190t + 30}\right)^2\) is also correct | Be aware this alternative form is acceptable |
| Answer | Marks | Guidance |
|---|---|---|
| As \(t \to \infty\), \(A \to \left(\dfrac{a}{b}\right)^2\) | M1 | The form of part (a) must be \(A = \left(\dfrac{at}{bt+c}\right)^2\) where \(a\), \(b\) and \(c\) are all positive. The reason is that if we had "\(bt - c\)", the area would be infinite at \(t = \dfrac{c}{b}\) and a limit would not exist |
| \(\dfrac{900}{361}\) or awrt \(2.49 \text{ cm}^2\) following correct work | A1 | Condone for both marks \(A < \dfrac{900}{361}\) following correct work |
## Question 9:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Separates variables: $\displaystyle\int \dfrac{dA}{A^{\frac{3}{2}}}=\int\dfrac{dt}{5t^2}$ | M1 | $A^{\frac{3}{2}}$, $dA$, $t^2$ and $dt$ must be in correct positions; condone without integral signs; don't be too concerned about position of "5" |
| $-2A^{-\frac{1}{2}}=-\dfrac{1}{5}t^{-1}$ | M1 | Integrates one side correctly: look for $pA^{-\frac{1}{2}}$ or $qt^{-1}$ |
| $-2A^{-\frac{1}{2}}=-\dfrac{1}{5}t^{-1}(+c)$ | dM1 | Integrates both sides correctly: look for $pA^{-\frac{1}{2}}$ **and** $qt^{-1}$ |
| $-2A^{-\frac{1}{2}}=-\dfrac{1}{5}t^{-1}(+c)$ correct (without $+c$) | A1 | e.g. $-10A^{-\frac{1}{2}}=-t^{-1}(+c)$ or $-\dfrac{A^{-0.5}}{0.5}=-\dfrac{1}{5}t^{-1}(+c)$ |
| Substitutes $t=3$, $A=2.25 \Rightarrow c=\ldots\left(-\dfrac{19}{15}\right)$ | M1 | Must have "$+c$"; some evidence for award; one index must have started correct; may be awarded after incorrect change of subject |
| Proceeds to $A=\left(\dfrac{pt}{qt+r}\right)^2$ from $\dfrac{p}{\sqrt{A}}=\dfrac{q}{t}+r$ with correct operations | M1 | Must use common factor on RHS, invert both sides (not each term), then square both sides to reach $A=\ldots$; candidates do not need a **value** for $r$ |
| $A=\left(\dfrac{30t}{19t+3}\right)^2$ | A1 | cso; this mark can be awarded independent of first M1 |
| | **(7)** | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| As $t\to\infty$, $A\to\left(\dfrac{30}{19}\right)^2=\dfrac{900}{361}$ or awrt $2.49\ \text{cm}^2$ | M1 A1 | |
| | **(2)** | |
| | **(9 marks total)** | |
## Part (a) - Additional Note:
| $A = \left(\dfrac{300t}{190t + 30}\right)^2$ is also correct | | Be aware this alternative form is acceptable |
---
## Part (b):
| As $t \to \infty$, $A \to \left(\dfrac{a}{b}\right)^2$ | M1 | The form of part (a) must be $A = \left(\dfrac{at}{bt+c}\right)^2$ where $a$, $b$ and $c$ are all positive. The reason is that if we had "$bt - c$", the area would be infinite at $t = \dfrac{c}{b}$ and a limit would not exist |
| $\dfrac{900}{361}$ or awrt $2.49 \text{ cm}^2$ following correct work | A1 | Condone for both marks $A < \dfrac{900}{361}$ following correct work |9. Bacteria are growing on the surface of a dish in a laboratory.
The area of the dish, $A \mathrm {~cm} ^ { 2 }$, covered by the bacteria, $t$ days after the bacteria start to grow, is modelled by the differential equation
$$\frac { \mathrm { d } A } { \mathrm {~d} t } = \frac { A ^ { \frac { 3 } { 2 } } } { 5 t ^ { 2 } } \quad t > 0$$
Given that $A = 2.25$ when $t = 3$
\begin{enumerate}[label=(\alph*)]
\item show that
$$A = \left( \frac { p t } { q t + r } \right) ^ { 2 }$$
where $p , q$ and $r$ are integers to be found.
According to the model, there is a limit to the area that will be covered by the bacteria.
\item Find the value of this limit.\\
\includegraphics[max width=\textwidth, alt={}, center]{79ac81c3-cd05-4f28-8840-3c8a6960e7b7-31_2255_50_314_34}\\
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\hfill \mbox{\textit{Edexcel P4 2020 Q9 [9]}}