| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Chain rule with three variables |
| Difficulty | Moderate -0.8 This is a straightforward application of basic differentiation and the chain rule. Part (i) requires differentiating a simple power function (rewriting as 25v^{-1}), and part (ii) is a direct substitution into the chain rule formula dF/dt = (dF/dv)(dv/dt). Both parts are routine C3 techniques with no problem-solving or conceptual challenges beyond recognizing the chain rule application. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dF}{dy} = -25v^{-2}\) | M1, A1 [2] | \(\frac{d}{dv}(v^{-1}) = -v^{-2}\) soi; \(-25v^{-2}\) o.e mark final ans |
| Answer | Marks | Guidance |
|---|---|---|
| When \(v = 50, \frac{dF}{dy} = -25/50^2 (= -0.01)\) | B1, M1, A1cao [3] | \(-25/50^2\); o.e.; e.g. \(\frac{dF}{dv} = \frac{dF}{dv} \cdot \frac{dv}{dt}\); o.e. e.g. \(-3/200\) isw |
| \(\frac{dF}{dt} = \frac{dF}{dv} \cdot \frac{dv}{dt}\) | ||
| \(= -0.01 \times 1.5 = -0.015\) |
### (i)
$\frac{dF}{dy} = -25v^{-2}$ | M1, A1 [2] | $\frac{d}{dv}(v^{-1}) = -v^{-2}$ soi; $-25v^{-2}$ o.e mark final ans
### (ii)
When $v = 50, \frac{dF}{dy} = -25/50^2 (= -0.01)$ | B1, M1, A1cao [3] | $-25/50^2$; o.e.; e.g. $\frac{dF}{dv} = \frac{dF}{dv} \cdot \frac{dv}{dt}$; o.e. e.g. $-3/200$ isw
$\frac{dF}{dt} = \frac{dF}{dv} \cdot \frac{dv}{dt}$ | |
$= -0.01 \times 1.5 = -0.015$ | |
---The driving force $F$ newtons and velocity $v$ km s$^{-1}$ of a car at time $t$ seconds are related by the equation $F = \frac{25}{v}$.
\begin{enumerate}[label=(\roman*)]
\item Find $\frac{dF}{dv}$. [2]
\item Find $\frac{dF}{dt}$ when $v = 50$ and $\frac{dv}{dt} = 1.5$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2013 Q5 [5]}}