| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Related rates |
| Difficulty | Standard +0.3 This is a straightforward related rates problem with standard techniques: differentiation of a power function, separation of variables integration, finding a limit, and applying the chain rule. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2\) | B1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{900}{(2t+3)^2} \Rightarrow V = -\frac{450}{2t+3} + c\) | M1 A1 | M1: Integrates to form \(V = \frac{k}{2t+3}\) (o.e.) with or without \(+c\). Condone sign error in \(2t-3\). A1: \(V = -\frac{450}{2t+3}(+c)\) (o.e.); no need for \(+c\) |
| \(t=0,\ V=0 \Rightarrow 0 = -\frac{450}{3}+c \Rightarrow c = \ldots\) | M1 | Substitutes \(V=0,\ t=0\) and proceeds to find \(c\); must have attempted integration to achieve function in \(V\) and \(t\) with constant of integration |
| \(V = 150 - \frac{450}{2t+3} = \frac{300t+450-450}{2t+3} = \frac{300t}{2t+3}\) * | A1* | Correct integration and value for \(c\) with at least one intermediate step shown before reaching given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(150\ \text{cm}^3\) | B1 | Must include units |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(t=3 \Rightarrow V = \frac{300\times3}{2\times3+3} = (100)\) | M1 | Attempts to substitute \(t=3\) into equation for \(V\); allow if there is a slip in substitution |
| \(100 = \frac{4}{3}\pi r^3 \Rightarrow r = 2.88\ \text{cm}\) | dM1, A1cao | Uses their \(V\) in \(V = \frac{4}{3}\pi r^3\) to find \(r\). A1: cao \(r=2.88\) cm, to 3 s.f.; must include units unless already penalised in (b)(ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r}\times\frac{\mathrm{d}r}{\mathrm{d}t} \Rightarrow \frac{900}{(2t+3)^2} = 4\pi r^2 \times \frac{\mathrm{d}r}{\mathrm{d}t}\) | M1 | Attempts to use chain rule with given \(\frac{\mathrm{d}V}{\mathrm{d}t}\) and attempt at substituting their \(\frac{\mathrm{d}V}{\mathrm{d}r}\) |
| \(t=3,\ r=\text{"2.88"} \Rightarrow \frac{900}{81} = 4\pi\times2.88^2\times\frac{\mathrm{d}r}{\mathrm{d}t} \Rightarrow \frac{\mathrm{d}r}{\mathrm{d}t} = \ldots\) | dM1 | Substitutes both \(t=3\) and their value of \(r\); proceeds to find \(\frac{\mathrm{d}r}{\mathrm{d}t}\). If no substitution shown, answer must be correct for their \(r\) |
| \(\frac{\mathrm{d}r}{\mathrm{d}t} = \text{awrt}\ 0.11\ \text{cm s}^{-1}\) | A1 | Must include units unless already penalised in (b)(ii) or (c) |
## Question 7:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2$ | B1 | cao |
### Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{900}{(2t+3)^2} \Rightarrow V = -\frac{450}{2t+3} + c$ | M1 A1 | M1: Integrates to form $V = \frac{k}{2t+3}$ (o.e.) with or without $+c$. Condone sign error in $2t-3$. A1: $V = -\frac{450}{2t+3}(+c)$ (o.e.); no need for $+c$ |
| $t=0,\ V=0 \Rightarrow 0 = -\frac{450}{3}+c \Rightarrow c = \ldots$ | M1 | Substitutes $V=0,\ t=0$ and proceeds to find $c$; must have attempted integration to achieve function in $V$ and $t$ with constant of integration |
| $V = 150 - \frac{450}{2t+3} = \frac{300t+450-450}{2t+3} = \frac{300t}{2t+3}$ * | A1* | Correct integration and value for $c$ with at least one intermediate step shown before reaching given answer |
### Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $150\ \text{cm}^3$ | B1 | Must include units |
### Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $t=3 \Rightarrow V = \frac{300\times3}{2\times3+3} = (100)$ | M1 | Attempts to substitute $t=3$ into equation for $V$; allow if there is a slip in substitution |
| $100 = \frac{4}{3}\pi r^3 \Rightarrow r = 2.88\ \text{cm}$ | dM1, A1cao | Uses their $V$ in $V = \frac{4}{3}\pi r^3$ to find $r$. A1: cao $r=2.88$ cm, to 3 s.f.; must include units unless already penalised in (b)(ii) |
### Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}r}\times\frac{\mathrm{d}r}{\mathrm{d}t} \Rightarrow \frac{900}{(2t+3)^2} = 4\pi r^2 \times \frac{\mathrm{d}r}{\mathrm{d}t}$ | M1 | Attempts to use chain rule with given $\frac{\mathrm{d}V}{\mathrm{d}t}$ and attempt at substituting their $\frac{\mathrm{d}V}{\mathrm{d}r}$ |
| $t=3,\ r=\text{"2.88"} \Rightarrow \frac{900}{81} = 4\pi\times2.88^2\times\frac{\mathrm{d}r}{\mathrm{d}t} \Rightarrow \frac{\mathrm{d}r}{\mathrm{d}t} = \ldots$ | dM1 | Substitutes both $t=3$ and their value of $r$; proceeds to find $\frac{\mathrm{d}r}{\mathrm{d}t}$. If no substitution shown, answer must be correct for their $r$ |
| $\frac{\mathrm{d}r}{\mathrm{d}t} = \text{awrt}\ 0.11\ \text{cm s}^{-1}$ | A1 | Must include units unless already penalised in (b)(ii) or (c) |\begin{enumerate}
\item The volume $V \mathrm {~cm} ^ { 3 }$ of a spherical balloon with radius $r \mathrm {~cm}$ is given by the formula
\end{enumerate}
$$V = \frac { 4 } { 3 } \pi r ^ { 3 }$$
(a) Find $\frac { \mathrm { d } V } { \mathrm {~d} r }$ giving your answer in simplest form.
At time $t$ seconds, the volume of the balloon is increasing according to the differential equation
$$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 900 } { ( 2 t + 3 ) ^ { 2 } } \quad t \geqslant 0$$
Given that $V = 0$ when $t = 0$\\
(b) (i) solve this differential equation to show that
$$V = \frac { 300 t } { 2 t + 3 }$$
(ii) Hence find the upper limit to the volume of the balloon.\\
(c) Find the radius of the balloon at $t = 3$, giving your answer in cm to 3 significant figures.\\
(d) Find the rate of increase of the radius of the balloon at $t = 3$, giving your answer to 2 significant figures. Show your working and state the units of your answer.
\hfill \mbox{\textit{Edexcel P4 2023 Q7 [12]}}