| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Chain rule with three variables |
| Difficulty | Standard +0.3 This is a standard connected rates of change question with clear scaffolding through parts (a)-(d). Part (a) is routine differentiation, (b) applies the chain rule directly as prompted, (c) is straightforward integration with given initial condition, and (d) involves substitution into derived formulas. While it requires multiple techniques (differentiation, chain rule, integration, substitution), each step is clearly signposted and uses standard C4 methods without requiring problem-solving insight or novel approaches. Slightly easier than average due to the extensive scaffolding. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{dV}{dr} = 4\pi r^2\) | B1 | (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Uses \(\frac{dr}{dt} = \frac{dV}{dt} \cdot \frac{dr}{dV}\) in any form \(= \frac{1000}{4\pi r^2(2t+1)^2}\) | M1, A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(V = \int 1000(2t+1)^{-2}\,dt\) and integrate to \(p(2t+1)^{-1}\), \(= -500(2t+1)^{-1}(+c)\) | M1, A1 | |
| Using \(V=0\) when \(t=0\) to find \(c\): \((c = 500)\) or equivalent | M1 | |
| \(\therefore V = 500\!\left(1 - \frac{1}{2t+1}\right)\) (any form) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| (i) Substitute \(t=5\) to give \(V\), then use \(r = \sqrt[3]{\left(\frac{3V}{4\pi}\right)}\) to give \(r = 4.77\) | M1, M1, A1 | (3) |
| (ii) Substitutes \(t=5\) and \(r =\) 'their value' into 'their' part (b): \(\frac{dr}{dt} = 0.0289\) \((\approx 2.90\times10^{-2})\) (cm/s) AG | M1, A1 | (2) [12] |
## Question 7:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dr} = 4\pi r^2$ | B1 | **(1)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Uses $\frac{dr}{dt} = \frac{dV}{dt} \cdot \frac{dr}{dV}$ in any form $= \frac{1000}{4\pi r^2(2t+1)^2}$ | M1, A1 | **(2)** |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $V = \int 1000(2t+1)^{-2}\,dt$ and integrate to $p(2t+1)^{-1}$, $= -500(2t+1)^{-1}(+c)$ | M1, A1 | |
| Using $V=0$ when $t=0$ to find $c$: $(c = 500)$ or equivalent | M1 | |
| $\therefore V = 500\!\left(1 - \frac{1}{2t+1}\right)$ (any form) | A1 | **(4)** |
### Part (d):
| Working/Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute $t=5$ to give $V$, then use $r = \sqrt[3]{\left(\frac{3V}{4\pi}\right)}$ to give $r = 4.77$ | M1, M1, A1 | **(3)** |
| (ii) Substitutes $t=5$ and $r =$ 'their value' into 'their' part (b): $\frac{dr}{dt} = 0.0289$ $(\approx 2.90\times10^{-2})$ (cm/s) AG | M1, A1 | **(2) [12]** |7. The volume of a spherical balloon of radius $r \mathrm {~cm}$ is $V \mathrm {~cm} ^ { 3 }$, where $V = \frac { 4 } { 3 } \pi r ^ { 3 }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } V } { \mathrm {~d} r }$.
The volume of the balloon increases with time $t$ seconds according to the formula
$$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 1000 } { ( 2 t + 1 ) ^ { 2 } } , \quad t \geqslant 0$$
\item Using the chain rule, or otherwise, find an expression in terms of $r$ and $t$ for $\frac { \mathrm { d } r } { \mathrm {~d} t }$.
\item Given that $V = 0$ when $t = 0$, solve the differential equation $\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 1000 } { ( 2 t + 1 ) ^ { 2 } }$, to obtain $V$ in terms of $t$.
\item Hence, at time $t = 5$,
\begin{enumerate}[label=(\roman*)]
\item find the radius of the balloon, giving your answer to 3 significant figures,
\item show that the rate of increase of the radius of the balloon is approximately $2.90 \times 10 ^ { - 2 } \mathrm {~cm} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2006 Q7 [12]}}