| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (exponential/logarithmic) |
| Difficulty | Standard +0.3 This is a straightforward parametric question requiring standard techniques: finding dy/dx using the chain rule (dy/dt รท dx/dt) and eliminating the parameter by taking logarithms. Both parts are routine C4 exercises with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dt} = \frac{(1+t)\cdot 2 - 2t\cdot 1}{(1+t)^2} = \frac{2}{(1+t)^2}\) | M1 A1 | |
| \(\frac{dx}{dt} = 2e^{2t}\) | B1 | |
| \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\) | M1 | |
| \(\Rightarrow \frac{dy}{dx} = \frac{2/(1+t)^2}{2e^{2t}} = \frac{1}{e^{2t}(1+t)^2}\) | A1 | |
| \(t = 0 \Rightarrow dy/dx = 1\) | B1ft [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2t = \ln x \Rightarrow t = \frac{1}{2}\ln x\) | M1 | or \(t\) in terms of \(y\) |
| \(\Rightarrow y = \frac{\ln x}{1 + \frac{1}{2}\ln x} = \frac{2\ln x}{2 + \ln x}\) | A1 [2] |
## Question 6:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dt} = \frac{(1+t)\cdot 2 - 2t\cdot 1}{(1+t)^2} = \frac{2}{(1+t)^2}$ | M1 A1 | |
| $\frac{dx}{dt} = 2e^{2t}$ | B1 | |
| $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ | M1 | |
| $\Rightarrow \frac{dy}{dx} = \frac{2/(1+t)^2}{2e^{2t}} = \frac{1}{e^{2t}(1+t)^2}$ | A1 | |
| $t = 0 \Rightarrow dy/dx = 1$ | B1ft [6] | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2t = \ln x \Rightarrow t = \frac{1}{2}\ln x$ | M1 | or $t$ in terms of $y$ |
| $\Rightarrow y = \frac{\ln x}{1 + \frac{1}{2}\ln x} = \frac{2\ln x}{2 + \ln x}$ | A1 [2] | |6 A curve has parametric equations
$$x = \mathrm { e } ^ { 2 t } , \quad y = \frac { 2 t } { 1 + t }$$
(i) Find the gradient of the curve at the point where $t = 0$.\\
(ii) Find $y$ in terms of $x$.
\hfill \mbox{\textit{OCR MEI C4 Q6 [8]}}