| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Spherical geometry differential equations |
| Difficulty | Moderate -0.3 This is a straightforward differential equations question requiring standard techniques: forming a DE from a verbal description, separating variables, using initial conditions to find constants, and substituting a value. Part (b) guides students to the answer, and part (d) is a standard modelling critique. Slightly easier than average due to the scaffolding and routine nature of all steps. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dr}{dt} = \frac{k}{\sqrt{r}}\) | B1 | Correctly sets up model; \(\frac{dr}{dt} \propto \frac{1}{\sqrt{r}}\); any letter except \(t\) or \(r\) for \(k\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.9 = \frac{k}{\sqrt{16}} \Rightarrow k = 3.6\) | B1 | \(k=3.6\) from valid method; condone \(k=-3.6\) |
| \(\int \sqrt{r} \, dr = \int \text{"3.6"} \, dt \Rightarrow \ldots\) | M1 | Separates variables correctly and attempts to integrate both sides |
| \(\frac{2}{3}r^{\frac{3}{2}} = \text{"3.6"}t \quad \{+c\}\) | A1 | Correct integration; \(\frac{2}{3}\) must be evident |
| \(t=10, r=16 \Rightarrow \frac{2}{3} \times 16^{\frac{3}{2}} = 3.6 \times 10 + c \Rightarrow c = \ldots\) | dM1 | Uses \(t=10\), \(r=16\) to find constant; dependent on first M1 |
| \(r^{\frac{3}{2}} = 5.4t + 10\) | A1* | Correct equation from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = 20 \Rightarrow r = (5.4(20)+10)^{\frac{2}{3}} = \ldots\) | M1 | Substitutes \(t=20\) into equation; index work must be correct |
| \(r = 24.1\) cm | A1 | cao 241mm or 24.1cm; correct answer with units implies both marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The model will not hold indefinitely as the balloon may burst | B1 | Must relate to model in context; e.g. balloon may burst/pop, unlikely to be perfectly spherical, predicts increase without limit |
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dr}{dt} = \frac{k}{\sqrt{r}}$ | B1 | Correctly sets up model; $\frac{dr}{dt} \propto \frac{1}{\sqrt{r}}$; any letter except $t$ or $r$ for $k$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.9 = \frac{k}{\sqrt{16}} \Rightarrow k = 3.6$ | B1 | $k=3.6$ from valid method; condone $k=-3.6$ |
| $\int \sqrt{r} \, dr = \int \text{"3.6"} \, dt \Rightarrow \ldots$ | M1 | Separates variables correctly and attempts to integrate both sides |
| $\frac{2}{3}r^{\frac{3}{2}} = \text{"3.6"}t \quad \{+c\}$ | A1 | Correct integration; $\frac{2}{3}$ must be evident |
| $t=10, r=16 \Rightarrow \frac{2}{3} \times 16^{\frac{3}{2}} = 3.6 \times 10 + c \Rightarrow c = \ldots$ | dM1 | Uses $t=10$, $r=16$ to find constant; dependent on first M1 |
| $r^{\frac{3}{2}} = 5.4t + 10$ | A1* | Correct equation from correct working |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = 20 \Rightarrow r = (5.4(20)+10)^{\frac{2}{3}} = \ldots$ | M1 | Substitutes $t=20$ into equation; index work must be correct |
| $r = 24.1$ cm | A1 | cao 241mm or 24.1cm; correct answer with units implies both marks |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The model will not hold indefinitely as the balloon may burst | B1 | Must relate to model in context; e.g. balloon may burst/pop, unlikely to be perfectly spherical, predicts increase without limit |
---\begin{enumerate}
\item A balloon is being inflated.
\end{enumerate}
In a simple model,
\begin{itemize}
\item the balloon is modelled as a sphere
\item the rate of increase of the radius of the balloon is inversely proportional to the square root of the radius of the balloon
\end{itemize}
At time $t$ seconds, the radius of the balloon is $r \mathrm {~cm}$.\\
(a) Write down a differential equation to model this situation.
At the instant when $t = 10$
\begin{itemize}
\item the radius is 16 cm
\item the radius is increasing at a rate of $0.9 \mathrm {~cm} \mathrm {~s} ^ { - 1 }$\\
(b) Solve the differential equation to show that
\end{itemize}
$$r ^ { \frac { 3 } { 2 } } = 5.4 t + 10$$
(c) Hence find the radius of the balloon when $t = 20$
Give your answer to the nearest millimetre.\\
(d) Suggest a limitation of the model.
\hfill \mbox{\textit{Edexcel Paper 1 2024 Q14 [9]}}