| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Spherical geometry differential equations |
| Difficulty | Standard +0.3 This is a straightforward applied differential equations question requiring chain rule to establish the DE, separation of variables to solve it, and a simple modelling comment. All steps are standard A-level techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dV}{dt} = \frac{k}{x}\) | M1 | Expresses inverse proportionality with a constant |
| When \(t = 0\), \(x = 5\) and \(\frac{dV}{dt} = 21\), so \(\frac{dV}{dt} = \frac{105}{x}\) | A1 | Evaluating \(k\), oe (may be done later) |
| \(\frac{dV}{dx} = 4\pi x^2\) | B1 | May be embedded in chain rule |
| So the chain rule gives \(\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} = 4\pi x^2 \frac{dx}{dt}\) | M1 | Use of the chain rule |
| Hence \(\frac{dx}{dt} = \frac{1}{4\pi x^2} \times \frac{105}{x} = \frac{105}{4\pi x^3}\) AG | A1 [5] | Convincing argument |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int 4\pi x^3 \, dx = \int 105 \, dt\) | M1 | Separating the variables |
| \(\pi x^4 = 105t + c\) | A1 | Condone missing \(+c\) here |
| When \(t = 0\), \(x = 5\) so \(c = 625\pi\) | M1, A1 | Using initial conditions; correct value for \(c\) |
| When \(t = 120\): \(x = \sqrt[4]{\frac{105}{\pi} \times 120 + 625} = 8.25\) cm | A1 [5] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| As \(t\) gets very large, the volume gets very large so the balloon will get beyond the maximum it can be without bursting and so burst. | E1 [1] | Conveys the idea that \(t \to \infty \Rightarrow V \to \infty\) or \(x \to \infty\). Indicates a practical problem with very large volume |
## Question 11:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dV}{dt} = \frac{k}{x}$ | M1 | Expresses inverse proportionality with a constant |
| When $t = 0$, $x = 5$ and $\frac{dV}{dt} = 21$, so $\frac{dV}{dt} = \frac{105}{x}$ | A1 | Evaluating $k$, oe (may be done later) |
| $\frac{dV}{dx} = 4\pi x^2$ | B1 | May be embedded in chain rule |
| So the chain rule gives $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} = 4\pi x^2 \frac{dx}{dt}$ | M1 | Use of the chain rule |
| Hence $\frac{dx}{dt} = \frac{1}{4\pi x^2} \times \frac{105}{x} = \frac{105}{4\pi x^3}$ **AG** | A1 [5] | Convincing argument |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int 4\pi x^3 \, dx = \int 105 \, dt$ | M1 | Separating the variables |
| $\pi x^4 = 105t + c$ | A1 | Condone missing $+c$ here |
| When $t = 0$, $x = 5$ so $c = 625\pi$ | M1, A1 | Using initial conditions; correct value for $c$ |
| When $t = 120$: $x = \sqrt[4]{\frac{105}{\pi} \times 120 + 625} = 8.25$ cm | A1 [5] | cao |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| As $t$ gets very large, the volume gets very large so the balloon will get beyond the maximum it can be without bursting and so burst. | E1 [1] | Conveys the idea that $t \to \infty \Rightarrow V \to \infty$ or $x \to \infty$. Indicates a practical problem with very large volume |
---11 A balloon is being inflated. The balloon is modelled as a sphere with radius $x \mathrm {~cm}$ at time $t \mathrm {~s}$. The volume $V \mathrm {~cm} ^ { 3 }$ is given by $\mathrm { V } = \frac { 4 } { 3 } \pi \mathrm { x } ^ { 3 }$.
The rate of increase of volume is inversely proportional to the radius of the balloon. Initially, when $t = 0$, the radius of the balloon is 5 cm and the volume of the balloon is increasing at a rate of $21 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $x$ satisfies the differential equation $\frac { \mathrm { dx } } { \mathrm { dt } } = \frac { 105 } { 4 \pi \mathrm { x } ^ { 3 } }$.
\item Find the radius of the balloon after two minutes.
\item Explain why the model may not be suitable for very large values of $t$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2021 Q11 [11]}}