| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Related rates |
| Difficulty | Standard +0.3 This is a straightforward differential equations question requiring standard techniques: setting up simple DEs from verbal descriptions, using given conditions to find constants, and solving by direct integration. The partial fractions in part (iv)-(v) are routine, and all steps follow predictable patterns with no novel problem-solving required. Slightly easier than average due to heavy scaffolding across six parts. |
| Spec | 4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dr}{dt}=\frac{k}{(t+1)}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t=3,\ \frac{dr}{dt}=0.5 \Rightarrow \frac{k}{(4)}=0.5 \Rightarrow k=2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dr}{dt}=\frac{2}{(t+1)} \Rightarrow r=2\ln(t+1)+c\) | M1, A1 | |
| \(t=0,\ r=0 \Rightarrow c=0\) | A1 | |
| When \(t=10,\ r=2\ln 11\approx 4.796\) cm | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dr}{dt}=\frac{k}{(t+1)(t+2)}\) | B1 | |
| \(\frac{dr}{dt}=0.5,\ t=3 \Rightarrow 0.5=\frac{k}{4\times 5} \Rightarrow k=10\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{(t+1)(t+2)}=\frac{A}{(t+1)}+\frac{B}{(t+2)}\) | M1 | Using partial fractions |
| \(\Rightarrow A(t+2)+B(t+1)\equiv 1 \Rightarrow A+B=0,\ 2A+B=1\) | ||
| \(\Rightarrow A=1,\ B=-1\) | A1 | both |
| \(\frac{dr}{dt}=\frac{10}{(t+1)(t+2)}=10\left(\frac{1}{t+1}-\frac{1}{t+2}\right)\) | M1 | Integrate to log functions |
| \(\Rightarrow r=10\ln\left(\frac{t+1}{t+2}\right)+c\) | A1 | condone omission of \(c\) |
| \(r=0,\ t=0 \Rightarrow c=-10\ln\frac{1}{2}\) | M1, A1 | Sub to find \(c\) |
| \(\Rightarrow r=10\ln 2\left(\frac{t+1}{t+2}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t=10 \Rightarrow r=10\ln\frac{22}{12}\approx 6.06\) metres | M1, A1 | Substituting |
## Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dr}{dt}=\frac{k}{(t+1)}$ | B1 | |
**Total: 1 mark**
## Question 7(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t=3,\ \frac{dr}{dt}=0.5 \Rightarrow \frac{k}{(4)}=0.5 \Rightarrow k=2$ | B1 | |
**Total: 1 mark**
## Question 7(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dr}{dt}=\frac{2}{(t+1)} \Rightarrow r=2\ln(t+1)+c$ | M1, A1 | |
| $t=0,\ r=0 \Rightarrow c=0$ | A1 | |
| When $t=10,\ r=2\ln 11\approx 4.796$ cm | A1 | |
**Total: 4 marks**
## Question 7(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dr}{dt}=\frac{k}{(t+1)(t+2)}$ | B1 | |
| $\frac{dr}{dt}=0.5,\ t=3 \Rightarrow 0.5=\frac{k}{4\times 5} \Rightarrow k=10$ | M1, A1 | |
**Total: 3 marks**
## Question 7(v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{(t+1)(t+2)}=\frac{A}{(t+1)}+\frac{B}{(t+2)}$ | M1 | Using partial fractions |
| $\Rightarrow A(t+2)+B(t+1)\equiv 1 \Rightarrow A+B=0,\ 2A+B=1$ | | |
| $\Rightarrow A=1,\ B=-1$ | A1 | both |
| $\frac{dr}{dt}=\frac{10}{(t+1)(t+2)}=10\left(\frac{1}{t+1}-\frac{1}{t+2}\right)$ | M1 | Integrate to log functions |
| $\Rightarrow r=10\ln\left(\frac{t+1}{t+2}\right)+c$ | A1 | condone omission of $c$ |
| $r=0,\ t=0 \Rightarrow c=-10\ln\frac{1}{2}$ | M1, A1 | Sub to find $c$ |
| $\Rightarrow r=10\ln 2\left(\frac{t+1}{t+2}\right)$ | A1 | |
**Total: 7 marks**
## Question 7(vi):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t=10 \Rightarrow r=10\ln\frac{22}{12}\approx 6.06$ metres | M1, A1 | Substituting |
**Total: 2 marks**
---7 A quantity of oil is dropped into water. When the oil hits the water it spreads out as a circle. The radius of the circle is $r \mathrm {~cm}$ after $t$ seconds and when $t = 3$ the radius of the circle is increasing at the rate of 0.5 centimetres per second.\\
One observer believes that the radius increases at a rate which is proportional to $\frac { 1 } { ( t + 1 ) }$.\\
(i) Write down a differential equation for this situation, using $k$ as a constant of proportionality.\\
(ii) Show that $k = 2$.\\
(iii) Calculate the radius of the circle after 10 seconds according to this model.
Another observer believes that the rate of increase of the radius of the circle is proportional to $\frac { 1 } { ( t + 1 ) ( t + 2 ) }$.\\
(iv) Write down a new differential equation for this new situation. Using the same initial conditions as before, find the value of the new constant of proportionality.\\
(v) Hence solve the differential equation.\\
(vi) Calculate the radius of the circle after 10 seconds according to this model.
\hfill \mbox{\textit{OCR MEI C4 Q7 [18]}}