| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Balloon or expanding shape |
| Difficulty | Moderate -0.3 This is a straightforward related rates problem requiring the chain rule. Part (i) involves differentiating A = πr² and applying dA/dt = (dA/dr)(dr/dt) with given dr/dt = 0.5, which is a standard technique. Part (ii) is simple substitution. The question is slightly easier than average because it's a textbook application of the chain rule with clear setup and no conceptual obstacles. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \pi r^2 \Rightarrow \frac{dA}{dr} = 2\pi r\) | M1, A1 | |
| \(\Rightarrow \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = \frac{1}{2} \times 2\pi r = \pi r\) | M1, A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| So when \(r = 6\), \(\frac{dA}{dt} = 6\pi (= 15.707...)\). The area increases at \(15.7 \text{ km}^2\text{h}^{-1}\), to 3sf. | M1, A1 | Substituting. 2 marks |
### (i)
$A = \pi r^2 \Rightarrow \frac{dA}{dr} = 2\pi r$ | M1, A1
$\Rightarrow \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = \frac{1}{2} \times 2\pi r = \pi r$ | M1, A1 | 4 marks
### (ii)
So when $r = 6$, $\frac{dA}{dt} = 6\pi (= 15.707...)$. The area increases at $15.7 \text{ km}^2\text{h}^{-1}$, to 3sf. | M1, A1 | Substituting. 2 marks
---An oil slick is circular with radius $r$ km and area $A$ km$^2$. The radius increases with time at a rate given by $\frac{dr}{dt} = 0.5$, in kilometres per hour.
\begin{enumerate}[label=(\roman*)]
\item Show that $\frac{dA}{dt} = \pi r$. [4]
\item Find the rate of increase of the area of the slick at a time when the radius is 6 km. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 Q7 [6]}}