| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Chain rule with three variables |
| Difficulty | Moderate -0.3 This is a straightforward connected rates of change question requiring basic differentiation of a simple reciprocal function and direct application of the chain rule. Part (i) is routine differentiation, and part (ii) involves substituting given values into dF/dt = (dF/dv)(dv/dt). While it tests understanding of the chain rule with three variables, the calculations are simple and the method is standard, making it slightly easier than average. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dF}{dv} = -25v^{-2}\) | M1 | \(\frac{d}{dv}(v^{-1}) = -v^{-2}\) soi |
| A1 | \(-25v^{-2}\) o.e. mark final answer | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| When \(v = 50\), \(\frac{dF}{dv} = -\frac{25}{50^2}\) (= \(-0.01\)) | B1 | \(-25/50^2\) |
| \(\frac{dF}{dt} = \frac{dF}{dv} \cdot \frac{dv}{dt}\) | M1 | o.e. e.g. \(\frac{dF}{dv} = \frac{dF}{dt} / \frac{dv}{dt}\) |
| \(= -0.01 \times 1.5 = -0.015\) | A1cao | o.e. e.g. \(-3/200\) isw |
| [3] |
## Question 3:
### Part (i)
| $\frac{dF}{dv} = -25v^{-2}$ | M1 | $\frac{d}{dv}(v^{-1}) = -v^{-2}$ soi |
|---|---|---|
| | A1 | $-25v^{-2}$ o.e. mark final answer |
| **[2]** | | |
### Part (ii)
| When $v = 50$, $\frac{dF}{dv} = -\frac{25}{50^2}$ (= $-0.01$) | B1 | $-25/50^2$ |
|---|---|---|
| $\frac{dF}{dt} = \frac{dF}{dv} \cdot \frac{dv}{dt}$ | M1 | o.e. e.g. $\frac{dF}{dv} = \frac{dF}{dt} / \frac{dv}{dt}$ |
| $= -0.01 \times 1.5 = -0.015$ | A1cao | o.e. e.g. $-3/200$ isw |
| **[3]** | | |
---3 The driving force $F$ newtons and velocity $v \mathrm {~km} \mathrm {~s} ^ { - 1 }$ of a car at time $t$ seconds are related by the equation $F = \frac { 25 } { v }$.\\
(i) Find $\frac { \mathrm { d } F } { \mathrm {~d} v }$.\\
(ii) Find $\frac { \mathrm { d } F } { \mathrm {~d} t }$ when $v = 50$ and $\frac { \mathrm { d } v } { \mathrm {~d} t } = 1.5$.
\hfill \mbox{\textit{OCR MEI C3 Q3 [5]}}