| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2020 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Spherical geometry differential equations |
| Difficulty | Standard +0.3 This is a standard related rates problem requiring the chain rule to connect dV/dt to dr/dt, then solving a separable differential equation. The steps are routine: use V = (4/3)πr³, separate variables, integrate r² dr, and apply initial conditions. While multi-part, each step follows a predictable template with no novel insight required, making it slightly easier than average. |
| Spec | 1.07r 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/Answer | Mark | Guidance |
| \(\frac{dV}{dt} = -c\) (for positive constant \(c\)) | B1 | Any letter acceptable including \(k\). Note \(\frac{dV}{dt}=c\) is B0 unless \(c\) stated as negative |
| Uses \(\frac{dV}{dt} = \frac{dV}{dr}\times\frac{dr}{dt}\) with \(\frac{dV}{dt}=-c\) and \(\frac{dV}{dr}=4\pi r^2\) | M1 | Allow \(\frac{dV}{dr}=\lambda r^2\) for any constant. No requirement to use correct volume formula for this mark |
| \(-c = 4\pi r^2 \times \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = -\frac{c}{4\pi r^2} = -\frac{k}{r^2}\) | A1* | Intermediate line equivalent to \(\frac{dr}{dt}=-\frac{c}{4\pi r^2}\) required. If started with \(\frac{dV}{dt}=-k\), must explain how \(\frac{dr}{dt}=-\frac{k}{4\pi r^2}\to\frac{dr}{dt}=-\frac{k}{r^2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{dr}{dt}=-\frac{k}{r^2} \Rightarrow \int r^2\,dr = \int -k\,dt\), integrates with one side "correct" | M1 | Integral signs not required to be seen |
| \(\frac{r^3}{3} = -kt\ (+\alpha)\) | A1 | \(+\alpha\) not required for this mark. May be awarded if \(k\) given a value |
| Uses \(t=0,\ r=40 \Rightarrow \alpha = \ldots\), giving \(\alpha = \frac{64000}{3}\) | M1 | If no constant of integration present, or \(k\) predefined, only first two marks in (b) available |
| Uses \(t=5,\ r=20\) and \(\alpha\) to find \(k\) | M1 | May be awarded if equation adapted incorrectly, e.g. each term cube rooted |
| \(r^3 = 64000 - 1200t\) or exact equivalent | A1 | e.g. \(\frac{r^3}{3}=\frac{64000}{3}-\frac{11200}{3}t\). ISW after correct answer. Condone \(21333.\dot{3}\) for \(\frac{64000}{3}\). Do not award if only rounded/truncated decimal used |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Uses model, recognises valid when \(r\geq 0\), uses this to find \(t\). E.g. "\(64000-11200t\ldots 0 \Rightarrow t\ldots\)" | M1 | Award for attempt to find \(t\) when \(r=0\). Must be from form \(ar^n = b-ct\) with \(a,b,c>0\) giving positive values of \(t\) |
| Valid for times up to and including \(\frac{40}{7}\) seconds | A1ft | Allow \(t < \frac{40}{7}\) or \(t \leq \frac{40}{7}\). No requirement for left-hand side of inequality. Follow through: allow \(t <\) their \(\frac{64000}{11200}\) as long as this value is greater than 5 |
# Question 14 (Differential Equations):
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dt} = -c$ (for positive constant $c$) | B1 | Any letter acceptable including $k$. Note $\frac{dV}{dt}=c$ is B0 unless $c$ stated as negative |
| Uses $\frac{dV}{dt} = \frac{dV}{dr}\times\frac{dr}{dt}$ with $\frac{dV}{dt}=-c$ and $\frac{dV}{dr}=4\pi r^2$ | M1 | Allow $\frac{dV}{dr}=\lambda r^2$ for any constant. No requirement to use correct volume formula for this mark |
| $-c = 4\pi r^2 \times \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = -\frac{c}{4\pi r^2} = -\frac{k}{r^2}$ | A1* | Intermediate line equivalent to $\frac{dr}{dt}=-\frac{c}{4\pi r^2}$ required. If started with $\frac{dV}{dt}=-k$, must explain how $\frac{dr}{dt}=-\frac{k}{4\pi r^2}\to\frac{dr}{dt}=-\frac{k}{r^2}$ |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{dr}{dt}=-\frac{k}{r^2} \Rightarrow \int r^2\,dr = \int -k\,dt$, integrates with one side "correct" | M1 | Integral signs not required to be seen |
| $\frac{r^3}{3} = -kt\ (+\alpha)$ | A1 | $+\alpha$ not required for this mark. May be awarded if $k$ given a value |
| Uses $t=0,\ r=40 \Rightarrow \alpha = \ldots$, giving $\alpha = \frac{64000}{3}$ | M1 | If no constant of integration present, or $k$ predefined, only first two marks in (b) available |
| Uses $t=5,\ r=20$ and $\alpha$ to find $k$ | M1 | May be awarded if equation adapted incorrectly, e.g. each term cube rooted |
| $r^3 = 64000 - 1200t$ or exact equivalent | A1 | e.g. $\frac{r^3}{3}=\frac{64000}{3}-\frac{11200}{3}t$. ISW after correct answer. Condone $21333.\dot{3}$ for $\frac{64000}{3}$. Do not award if **only** rounded/truncated decimal used |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Uses model, recognises valid when $r\geq 0$, uses this to find $t$. E.g. "$64000-11200t\ldots 0 \Rightarrow t\ldots$" | M1 | Award for attempt to find $t$ when $r=0$. Must be from form $ar^n = b-ct$ with $a,b,c>0$ giving positive values of $t$ |
| Valid for times up to and including $\frac{40}{7}$ seconds | A1ft | Allow $t < \frac{40}{7}$ or $t \leq \frac{40}{7}$. No requirement for left-hand side of inequality. Follow through: allow $t <$ their $\frac{64000}{11200}$ as long as this value is greater than 5 |
---\begin{enumerate}
\item A large spherical balloon is deflating.
\end{enumerate}
At time $t$ seconds the balloon has radius $r \mathrm {~cm}$ and volume $V \mathrm {~cm} ^ { 3 }$\\
The volume of the balloon is modelled as decreasing at a constant rate.\\
(a) Using this model, show that
$$\frac { \mathrm { d } r } { \mathrm {~d} t } = - \frac { k } { r ^ { 2 } }$$
where $k$ is a positive constant.
Given that
\begin{itemize}
\item the initial radius of the balloon is 40 cm
\item after 5 seconds the radius of the balloon is 20 cm
\item the volume of the balloon continues to decrease at a constant rate until the balloon is empty\\
(b) solve the differential equation to find a complete equation linking $r$ and $t$.\\
(c) Find the limitation on the values of $t$ for which the equation in part (b) is valid.
\end{itemize}
\hfill \mbox{\textit{Edexcel Paper 1 2020 Q14 [10]}}