| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Applied differentiation |
| Type | Connected rates of change |
| Difficulty | Moderate -0.3 This is a straightforward related rates problem with clear setup and standard formulas provided. Students apply the chain rule to connect dV/dt to dr/dt, then use this to find dS/dt. The calculations are routine with no conceptual surprises, making it slightly easier than average for C4. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V = \dfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dr} = 4\pi r^2\) | B1 oe | Can be implied by later working |
| \(\dfrac{dV}{dr} \times \dfrac{dr}{dt} = \dfrac{dV}{dt} \Rightarrow (4\pi r^2)\dfrac{dr}{dt} = 3\) | M1 oe | Candidate's \(\dfrac{dV}{dr} \times \dfrac{dr}{dt} = 3\) or \(3 \div\) candidate's \(\dfrac{dV}{dr}\) |
| When \(r = 4\): \(\dfrac{dr}{dt} = \dfrac{3}{4\pi(4)^2} = \dfrac{3}{64\pi}\) | dM1 | Dependent on previous M1; substitutes \(r=4\) |
| \(\dfrac{dr}{dt} = 0.01492...\ (\text{cm}^2\text{s}^{-1})\) | A1 | Anything that rounds to \(0.0149\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dS}{dt} = \dfrac{dS}{dr} \times \dfrac{dr}{dt} \Rightarrow \dfrac{dS}{dt} = 8\pi r \times \dfrac{3}{4\pi r^2}\) | M1 oe | \(8\pi r \times\) candidate's \(\dfrac{dr}{dt}\) |
| When \(r=4\): \(\dfrac{dS}{dt} = 8\pi(4) \times \dfrac{3}{4\pi(4)^2} = \dfrac{6}{4} = 1.5\) | ||
| \(\dfrac{dS}{dt} = 1.5\ (\text{cm}^2\text{s}^{-1})\) | A1 cso | Anything that rounds to \(1.5\) |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = \dfrac{4}{3}\pi r^3 \Rightarrow \dfrac{dV}{dr} = 4\pi r^2$ | B1 oe | Can be implied by later working |
| $\dfrac{dV}{dr} \times \dfrac{dr}{dt} = \dfrac{dV}{dt} \Rightarrow (4\pi r^2)\dfrac{dr}{dt} = 3$ | M1 oe | Candidate's $\dfrac{dV}{dr} \times \dfrac{dr}{dt} = 3$ or $3 \div$ candidate's $\dfrac{dV}{dr}$ |
| When $r = 4$: $\dfrac{dr}{dt} = \dfrac{3}{4\pi(4)^2} = \dfrac{3}{64\pi}$ | dM1 | Dependent on previous M1; substitutes $r=4$ |
| $\dfrac{dr}{dt} = 0.01492...\ (\text{cm}^2\text{s}^{-1})$ | A1 | Anything that rounds to $0.0149$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dS}{dt} = \dfrac{dS}{dr} \times \dfrac{dr}{dt} \Rightarrow \dfrac{dS}{dt} = 8\pi r \times \dfrac{3}{4\pi r^2}$ | M1 oe | $8\pi r \times$ candidate's $\dfrac{dr}{dt}$ |
| When $r=4$: $\dfrac{dS}{dt} = 8\pi(4) \times \dfrac{3}{4\pi(4)^2} = \dfrac{6}{4} = 1.5$ | | |
| $\dfrac{dS}{dt} = 1.5\ (\text{cm}^2\text{s}^{-1})$ | A1 cso | Anything that rounds to $1.5$ |
---5. At time $t$ seconds the radius of a sphere is $r \mathrm {~cm}$, its volume is $V \mathrm {~cm} ^ { 3 }$ and its surface area is $S \mathrm {~cm} ^ { 2 }$. [You are given that $V = \frac { 4 } { 3 } \pi r ^ { 3 }$ and that $S = 4 \pi r ^ { 2 }$ ]
The volume of the sphere is increasing uniformly at a constant rate of $3 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } r } { \mathrm {~d} t }$ when the radius of the sphere is 4 cm , giving your answer to 3 significant figures.
\item Find the rate at which the surface area of the sphere is increasing when the radius is 4 cm .
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2014 Q5 [6]}}