| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Spherical geometry differential equations |
| Difficulty | Standard +0.3 This is a standard separable differential equations problem with clear setup. Students must form dr/dt = k/r², separate variables, integrate (giving r³ terms), apply two boundary conditions to find constants, then solve for time when r=0. While it requires multiple steps and careful algebra, the method is routine for Further Maths students and follows a well-practiced template with no geometric insight or novel problem-solving required. |
| 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 |
| \(\frac{dr}{dt} \propto \pm\frac{1}{r^2}\) or \(\frac{dr}{dt} = \pm\frac{k}{r^2}\) | M1 | AO3.3 — Translates description into mathematics |
| \(\int r^2\,dr = \int \pm k\,dt \Rightarrow \ldots\) | M1 | AO2.1 — Separates variables and attempts integration of at least one side |
| \(\frac{1}{3}r^3 = \pm kt\ \{+c\}\) | A1 | AO1.1b — Correct integration |
| Using \(t=0, r=5\) and \(t=4, r=3\) gives \(\frac{1}{3}r^3 = -\frac{49}{6}t + \frac{125}{3}\), where \(r\) (mm) is radius and \(t\) (minutes) is time from when mint was placed in mouth | M1 | AO3.1a — Complete process using both boundary conditions |
| OR using \(t=0, r=5\) and \(t=240, r=3\) gives \(\frac{1}{3}r^3 = -\frac{49}{360}t + \frac{125}{3}\), where \(r\) (mm) is radius and \(t\) (seconds) | A1 | AO1.1b — Correct equation with correct reference to units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r=0 \Rightarrow 0 = -\frac{49}{6}t + \frac{125}{3} \Rightarrow 0 = -49t + 250 \Rightarrow t = \ldots\) | M1 | AO3.4 — Sets \(r=0\) and rearranges |
| time \(= 5\) minutes 6 seconds | A1 | AO1.1b — Note: 306 seconds with no reference to minutes/seconds is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Suggests a suitable limitation, e.g.: Model does not consider how mint is sucked; Model does not consider whether mint is bitten; Model limited for times up to 5 min 6 sec; Mint may not retain spherical shape; Model indicates negative radius after dissolving; Does not consider temperature/saliva/swallowing | B1 | AO3.5b — Do not accept: mint may not dissolve at constant rate; rate of decrease must be constant; \(0\leq t < \frac{250}{49}\), \(r\geq 0\) without explanation; reference to \(r>5\) |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dr}{dt} \propto \pm\frac{1}{r^2}$ or $\frac{dr}{dt} = \pm\frac{k}{r^2}$ | M1 | AO3.3 — Translates description into mathematics |
| $\int r^2\,dr = \int \pm k\,dt \Rightarrow \ldots$ | M1 | AO2.1 — Separates variables and attempts integration of at least one side |
| $\frac{1}{3}r^3 = \pm kt\ \{+c\}$ | A1 | AO1.1b — Correct integration |
| Using $t=0, r=5$ and $t=4, r=3$ gives $\frac{1}{3}r^3 = -\frac{49}{6}t + \frac{125}{3}$, where $r$ (mm) is radius and $t$ (minutes) is time from when mint was placed in mouth | M1 | AO3.1a — Complete process using both boundary conditions |
| OR using $t=0, r=5$ and $t=240, r=3$ gives $\frac{1}{3}r^3 = -\frac{49}{360}t + \frac{125}{3}$, where $r$ (mm) is radius and $t$ (seconds) | A1 | AO1.1b — Correct equation with correct reference to units |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r=0 \Rightarrow 0 = -\frac{49}{6}t + \frac{125}{3} \Rightarrow 0 = -49t + 250 \Rightarrow t = \ldots$ | M1 | AO3.4 — Sets $r=0$ and rearranges |
| time $= 5$ minutes 6 seconds | A1 | AO1.1b — Note: 306 seconds with no reference to minutes/seconds is A0 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Suggests a suitable limitation, e.g.: Model does not consider how mint is sucked; Model does not consider whether mint is bitten; Model limited for times up to 5 min 6 sec; Mint may not retain spherical shape; Model indicates negative radius after dissolving; Does not consider temperature/saliva/swallowing | B1 | AO3.5b — Do not accept: mint may not dissolve at constant rate; rate of decrease must be constant; $0\leq t < \frac{250}{49}$, $r\geq 0$ without explanation; reference to $r>5$ |
---\begin{enumerate}
\item A spherical mint of radius 5 mm is placed in the mouth and sucked. Four minutes later, the radius of the mint is 3 mm .
\end{enumerate}
In a simple model, the rate of decrease of the radius of the mint is inversely proportional to the square of the radius.
Using this model and all the information given,\\
(a) find an equation linking the radius of the mint and the time.\\
(You should define the variables that you use.)\\
(b) Hence find the total time taken for the mint to completely dissolve. Give your answer in minutes and seconds to the nearest second.\\
(c) Suggest a limitation of the model.
\hfill \mbox{\textit{Edexcel Paper 2 2018 Q10 [8]}}