| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Related rates |
| Difficulty | Standard +0.3 This is a straightforward related rates problem requiring chain rule application (part a) and separation of variables (part b). The setup is clearly guided, the differential equation is simple to separate and integrate (leading to A^(3/2) terms), and the initial conditions are straightforward to apply. While it requires multiple techniques, each step follows standard procedures with no novel insight needed. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dA}{dt} = 2\pi\sqrt{\frac{A}{\pi}} \times \frac{c}{A\pi} = \frac{2\pi\sqrt{c}}{\sqrt{A}} = \frac{k}{\sqrt{A}}\) | B1 | Uses the model to state \(\frac{dr}{dt} = \frac{c}{r^2}\) |
| M1 | Attempts to use \(\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}\) with \(A = \pi r^2\) and \(\frac{dA}{dr} = 2\pi r\) | |
| M1 | Substitutes \(r = \sqrt{\frac{A}{\pi}} = \frac{\sqrt{A}}{\sqrt{\pi}}\) into \(\frac{dA}{dt}\) and proceeds to an expression in terms of \(r\) for \(\frac{dA}{dt}\) | |
| A1 | Proceeds to the given answer with accurate work showing all necessary steps |
| Answer | Marks | Guidance |
|---|---|---|
| Therefore, an equation linking \(A\) and \(t\) is \(A = \left(\frac{189t}{10000} + \frac{27}{1000}\right)^{\frac{2}{3}}\) | B1 | Separates the variables \(\int\sqrt{A}\,dA = \int k\,dt\) |
| M1 | Integrating the lhs and rhs | |
| A1 | Correct integration | |
| B1 | Substitutes \(t = 0\) |
A circular stain is growing.
The rate of increase of its radius is inversely proportional to the square of the radius.
At time $t$ seconds the circular stain has radius $r$ cm and area $A$ cm².
**a. Show that $\frac{dA}{dt} = \frac{k}{\sqrt{A}}$**
$\frac{dr}{dt} \propto \frac{1}{r^2} \Rightarrow \frac{dr}{dt} = \frac{c}{r^2}$
Using the chain rule, $\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$. $A = \pi r^2 \Rightarrow \frac{dA}{dr} = 2\pi r$
$\frac{dA}{dt} = 2\pi r \times \frac{c}{r^2}$
Using $r = \sqrt{\frac{A}{\pi}}$ gives:
$\frac{dA}{dt} = 2\pi\sqrt{\frac{A}{\pi}} \times \frac{c}{A\pi} = \frac{2\pi\sqrt{c}}{\sqrt{A}} = \frac{k}{\sqrt{A}}$ | B1 | Uses the model to state $\frac{dr}{dt} = \frac{c}{r^2}$
| M1 | Attempts to use $\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}$ with $A = \pi r^2$ and $\frac{dA}{dr} = 2\pi r$
| M1 | Substitutes $r = \sqrt{\frac{A}{\pi}} = \frac{\sqrt{A}}{\sqrt{\pi}}$ into $\frac{dA}{dt}$ and proceeds to an expression in terms of $r$ for $\frac{dA}{dt}$
| A1 | Proceeds to the given answer with accurate work showing all necessary steps
**b. Solve the differential equation to find a complete equation linking $A$ and $t$.**
We have $\frac{dA}{dt} = \frac{k}{\sqrt{A}}$. Separation of variables gives $\int\sqrt{A}\,dA = \int k\,dt$.
Completing the integration gives: $\frac{2}{3}A^{\frac{3}{2}} = kt + c$.
From the question, at time $t = 0$, $A = 0.09$
$\frac{2}{3}(0.09)^{\frac{3}{2}} = k(0) + c \Rightarrow \frac{9}{500}$
At time $t = 10$, $A = 0.36$
$\frac{2}{3}(0.36)^{\frac{3}{2}} = k(10) + \frac{9}{500} \Rightarrow 10k = \frac{18}{125} - \frac{9}{500} \Rightarrow k = \frac{63}{5000}$
Therefore, an equation linking $A$ and $t$ is $A = \left(\frac{189t}{10000} + \frac{27}{1000}\right)^{\frac{2}{3}}$ | B1 | Separates the variables $\int\sqrt{A}\,dA = \int k\,dt$
| M1 | Integrating the lhs and rhs
| A1 | Correct integration
| B1 | Substitutes $t = 0$\begin{enumerate}
\item A circular stain is growing.
\end{enumerate}
The rate of increase of its radius is inversely proportional to the square of the radius. At time $t$ seconds the circular stain has radius $r \mathrm {~cm}$ and area $A \mathrm {~cm} ^ { 2 }$.\\
a. Show that $\frac { \mathrm { d } A } { \mathrm {~d} t } = \frac { k } { \sqrt { A } }$.
Given that
\begin{itemize}
\item the initial area of the circular stain is $0.09 \mathrm {~cm} ^ { 2 }$.
\item after 10 seconds the area of the circular stain is $0.36 \mathrm {~cm} ^ { 2 }$.\\
b. Solve the differential equation to find a complete equation linking $A$ and $t$.
\end{itemize}
\hfill \mbox{\textit{Edexcel PMT Mocks Q14 [10]}}