| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Related rates |
| Difficulty | Standard +0.3 This is a straightforward related rates problem requiring students to form a differential equation from a verbal description (dr/dt = k/√r), separate variables, integrate (standard power rule), apply initial conditions to find constants, then substitute into the sphere volume formula. All steps are routine C4 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dr}{dt}=\frac{k}{\sqrt{r}}\) oe | B2 | B1 for \(\frac{dr}{dt}=\); B1 for \(\frac{k}{\sqrt{r}}\). SR: B1 for \(\frac{dr}{dt}\propto\frac{1}{\sqrt{r}}\) |
| Sep variables of their diff eqn (or invert) & integrate each side, increasing powers by 1 (\(\frac{1}{r}\rightarrow\ln r\)) | *M1 | their d.e. must be \(\frac{dr}{dt}\) (or \(\frac{dt}{dr}\)) = f(r). Ignore absence of '+c' after integration |
| Subst \(\frac{dr}{dt}=1.08, r=9\) into their diff eqn to find \(k\) | M1 | their d.e. must include \(\frac{dr}{dt}\) (or \(\frac{dt}{dr}\)), \(r\) & \(k\). (\(\checkmark k=3.24\) but M mark, not A) |
| Substitute \(t=5, r=9\) to find 'c' | dep*M1 | Must involve '+c' here |
| Correct value of c (probably \(= 1.8\) or \(-1.8\)) | A1 | Check other values |
| \(r=(4.86t+2.7)^{\frac{2}{3}}\) ISW | A1 | Answer required in form \(r=f(t)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| subst \(t=0\) into any version of (i) result to find finite \(r\) | M1 | (\(\checkmark r\approx1.938991...\) but M mark, not A) |
| Any \(V\) in range \(30.5\leq V < 30.55\), but not fortuitously | A1 | Accept \(9.72\pi\) or \(\frac{243}{25}\pi\) |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dr}{dt}=\frac{k}{\sqrt{r}}$ oe | B2 | B1 for $\frac{dr}{dt}=$; B1 for $\frac{k}{\sqrt{r}}$. SR: B1 for $\frac{dr}{dt}\propto\frac{1}{\sqrt{r}}$ |
| Sep variables of their diff eqn (or invert) & integrate each side, increasing powers by 1 ($\frac{1}{r}\rightarrow\ln r$) | *M1 | their d.e. must be $\frac{dr}{dt}$ (or $\frac{dt}{dr}$) = f(r). Ignore absence of '+c' after integration |
| Subst $\frac{dr}{dt}=1.08, r=9$ into their diff eqn to find $k$ | M1 | their d.e. must include $\frac{dr}{dt}$ (or $\frac{dt}{dr}$), $r$ & $k$. ($\checkmark k=3.24$ but M mark, not A) |
| Substitute $t=5, r=9$ to find 'c' | dep*M1 | Must involve '+c' here |
| Correct value of c (probably $= 1.8$ or $-1.8$) | A1 | Check other values |
| $r=(4.86t+2.7)^{\frac{2}{3}}$ ISW | A1 | Answer required in form $r=f(t)$ |
---
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| subst $t=0$ into any version of (i) result to find finite $r$ | M1 | ($\checkmark r\approx1.938991...$ but M mark, not A) |
| Any $V$ in range $30.5\leq V < 30.55$, but not fortuitously | A1 | Accept $9.72\pi$ or $\frac{243}{25}\pi$ |
---8 At time $t$ seconds, the radius of a spherical balloon is $r \mathrm {~cm}$. The balloon is being inflated so that the rate of increase of its radius is inversely proportional to the square root of its radius. When $t = 5 , r = 9$ and, at this instant, the radius is increasing at $1.08 \mathrm {~cm} \mathrm {~s} ^ { - 1 }$.\\
(i) Write down a differential equation to model this situation, and solve it to express $r$ in terms of $t$.\\
(ii) How much air is in the balloon initially?\\[0pt]
[The volume of a sphere is $V = \frac { 4 } { 3 } \pi r ^ { 3 }$.]
\hfill \mbox{\textit{OCR C4 2013 Q8 [9]}}