| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Session | Specimen |
| Marks | 13 |
| 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 in part (a) and separation of variables in part (b). Both techniques are standard C4 content with no novel insight required, though the multi-step nature and 6-mark allocation for part (a) makes it slightly above average difficulty for routine exercises. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\frac{dr}{dt} = \frac{k}{r^2}\) | B1 | |
| \(A = \pi r^2 \therefore \frac{dA}{dr} = 2\pi r\) | M1 A1 | |
| \(\therefore \frac{dA}{dt} = 2\pi r \cdot \frac{k}{r^2} = \frac{(2\pi k)}{r} = \frac{(2\pi k)}{\left(\frac{A}{\pi}\right)^{\frac{1}{2}}} = \frac{2\pi^{\frac{3}{2}}k}{\sqrt{A}}\) | M1; M1 | |
| \(\therefore \frac{dA}{dt} \propto \frac{1}{\sqrt{A}}\) | A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\int\sqrt{S}\,dS = \int 2e^{2t}\,dt\) | M1 | |
| \(\frac{2}{3}S^{\frac{3}{2}} = e^{2t} + C\) | M1 A1 | |
| \(t=0, S=9 \Rightarrow C = 17\) | B1 | |
| \(\therefore \frac{2}{3}S^{\frac{3}{2}} = e^{2t} + 17\) and use \(S=16\) | M1 | |
| \(\left(\frac{128}{3} - 17\right) = e^{2t} \Rightarrow t = \frac{1}{2}\ln\left[\frac{77}{3}\right]\) | M1 | |
| \(= 1.6\) | A1 | (7 marks) |
# Question 8:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $\frac{dr}{dt} = \frac{k}{r^2}$ | B1 | |
| $A = \pi r^2 \therefore \frac{dA}{dr} = 2\pi r$ | M1 A1 | |
| $\therefore \frac{dA}{dt} = 2\pi r \cdot \frac{k}{r^2} = \frac{(2\pi k)}{r} = \frac{(2\pi k)}{\left(\frac{A}{\pi}\right)^{\frac{1}{2}}} = \frac{2\pi^{\frac{3}{2}}k}{\sqrt{A}}$ | M1; M1 | |
| $\therefore \frac{dA}{dt} \propto \frac{1}{\sqrt{A}}$ | A1 | (6 marks) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $\int\sqrt{S}\,dS = \int 2e^{2t}\,dt$ | M1 | |
| $\frac{2}{3}S^{\frac{3}{2}} = e^{2t} + C$ | M1 A1 | |
| $t=0, S=9 \Rightarrow C = 17$ | B1 | |
| $\therefore \frac{2}{3}S^{\frac{3}{2}} = e^{2t} + 17$ and use $S=16$ | M1 | |
| $\left(\frac{128}{3} - 17\right) = e^{2t} \Rightarrow t = \frac{1}{2}\ln\left[\frac{77}{3}\right]$ | M1 | |
| $= 1.6$ | A1 | (7 marks) |8. A circular stain grows in such a way that the rate of increase of its radius is inversely proportional to the square of the radius. Given that the area of the stain at time $t$ seconds is $A \mathrm {~cm} ^ { 2 }$,
\begin{enumerate}[label=(\alph*)]
\item show that $\frac { \mathrm { d } A } { \mathrm {~d} t } \propto \frac { 1 } { \sqrt { A } }$.\\
(6)
Another stain, which is growing more quickly, has area $S \mathrm {~cm} ^ { 2 }$ at time $t$ seconds. It is given that
$$\frac { \mathrm { d } S } { \mathrm {~d} t } = \frac { 2 \mathrm { e } ^ { 2 t } } { \sqrt { S } }$$
Given that, for this second stain, $S = 9$ at time $t = 0$,
\item solve the differential equation to find the time at which $S = 16$. Give your answer to 2 significant figures.
\section*{END}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q8 [13]}}