| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Balloon or expanding shape |
| Difficulty | Standard +0.3 This is a standard C3 related rates question with two routine parts: (a) requires differentiating A=πr² implicitly with respect to time (dA/dt given, find dr/dt), and (b) requires differentiating an exponential function and solving dm/dt=-3. Both are textbook applications of differentiation with no novel insight required, making it slightly easier than average. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.07b Gradient as rate of change: dy/dx notation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Either: State or imply either \(\frac{dA}{dr}=2\pi r\) or \(\frac{dA}{dt}=250\) | B1 | or both |
| Attempt manipulation of derivatives to find \(\frac{dr}{dt}\) | M1 | using multiplication / division |
| Obtain correct \(\frac{250}{2\pi r}\) | A1 | or equiv |
| Obtain 1.6 | A1 | 4 or equiv; allow greater accuracy |
| Or: Attempt to express \(r\) in terms of \(t\) | M1 | using \(A=250r\) |
| Obtain \(r=\sqrt{\frac{250t}{\pi}}\) | A1 | or equiv |
| Differentiate \(kt^{\frac{1}{2}}\) to produce \(\frac{1}{2}kt^{-\frac{1}{2}}\) | M1 | any constant k |
| Substitute \(t=7.6\) to obtain 1.6 | A1 | (4) allow greater accuracy |
| (b) State \(\frac{dm}{dt}=-150ke^{-kt}\) | B1 | |
| Equate to \((\pm 3)\) and attempt value for \(t\) | M1 | using valid process; condone sign confusion |
| Obtain \(-\frac{1}{k}\ln(\frac{1}{50k})\) or \(\frac{1}{k}\ln(50k)\) or \(\frac{\ln 50+\ln k}{k}\) | A1 | 3 or equiv but with correct treatment of signs |
(a) Either: State or imply either $\frac{dA}{dr}=2\pi r$ or $\frac{dA}{dt}=250$ | B1 | or both
Attempt manipulation of derivatives to find $\frac{dr}{dt}$ | M1 | using multiplication / division
Obtain correct $\frac{250}{2\pi r}$ | A1 | or equiv
Obtain 1.6 | A1 | 4 or equiv; allow greater accuracy
Or: Attempt to express $r$ in terms of $t$ | M1 | using $A=250r$
Obtain $r=\sqrt{\frac{250t}{\pi}}$ | A1 | or equiv
Differentiate $kt^{\frac{1}{2}}$ to produce $\frac{1}{2}kt^{-\frac{1}{2}}$ | M1 | any constant k
Substitute $t=7.6$ to obtain 1.6 | A1 | (4) allow greater accuracy
(b) State $\frac{dm}{dt}=-150ke^{-kt}$ | B1
Equate to $(\pm 3)$ and attempt value for $t$ | M1 | using valid process; condone sign confusion
Obtain $-\frac{1}{k}\ln(\frac{1}{50k})$ or $\frac{1}{k}\ln(50k)$ or $\frac{\ln 50+\ln k}{k}$ | A1 | 3 or equiv but with correct treatment of signs
**Total: 7**
---\begin{enumerate}[label=(\alph*)]
\item Leaking oil is forming a circular patch on the surface of the sea. The area of the patch is increasing at a rate of 250 square metres per hour. Find the rate at which the radius of the patch is increasing at the instant when the area of the patch is 1900 square metres. Give your answer correct to 2 significant figures. [4]
\item The mass of a substance is decreasing exponentially. Its mass now is 150 grams and its mass, $m$ grams, at a time $t$ years from now is given by
$$m = 150e^{-kt},$$
where $k$ is a positive constant. Find, in terms of $k$, the number of years from now at which the mass will be decreasing at a rate of 3 grams per year. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR C3 2010 Q7 [7]}}