| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 7 |
| 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 standard parametric equations question requiring chain rule differentiation and elimination of parameter. Part (i) uses the standard formula dy/dx = (dy/dt)/(dx/dt) with straightforward exponential differentiation. Part (ii) requires recognizing that t = (1/3)ln(x) from x = e^(3t), then substituting into y—routine algebraic manipulation for C4 level. Slightly above average due to the logarithmic form required, but follows predictable patterns. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = e^{3t}\), \(y = te^{2t}\) | B1 | soi |
| \(dy/dt = 2te^{2t} + e^{2t}\) | M1 | Their \(\frac{dy}{dt} \div \frac{dx}{dt}\) in terms of \(t\) |
| \(\Rightarrow dy/dx = (2te^{2t} + e^{2t})/3e^{3t}\) | A1 | oe cao; allow unsimplified form even if subsequently cancelled incorrectly |
| When \(t=1\), \(dy/dx = 3e^2/3e^3 = 1/e\) | A1 | cao www; must be simplified to \(1/e\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3t = \ln x\), \(y = \frac{\ln x}{3} e^{2\ln x/3} = \frac{x^{2/3}\ln x}{3}\) | B1 | Any equivalent form of \(y\) in terms of \(x\) only |
| \(dy/dx = \frac{1}{3}x^{2/3}\cdot\frac{1}{x} + \ln x \cdot \frac{2}{9}x^{-1/3}\) | M1 | Differentiating their \(y\); need a product including \(\ln kx\) and \(x^p\); subst \(x = e^{3t}\) to obtain \(dy/dx\) in terms of \(t\) |
| \(= \frac{1}{3e^t} + \frac{2t}{3e^t}\) | A1 | oe cao |
| \(dy/dx = 1/3e + 2/3e = 1/e\) | A1 | www cao; exact only; must be simplified to \(1/e\) or \(e^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3t = \ln x \Rightarrow t = (\ln x)/3\) | B1 | Finding \(t\) correctly in terms of \(x\) |
| \(y = (\ln x)/3e^{(2\ln x)/3}\) | M1 | Substituting in \(y\) using their \(t\) |
| \(y = \frac{1}{3}x^{\frac{2}{3}}\ln x\) | A1 | Required form \(ax^b \ln x\) only |
## Question 2:
### Part (i)
**EITHER:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = e^{3t}$, $y = te^{2t}$ | B1 | soi |
| $dy/dt = 2te^{2t} + e^{2t}$ | M1 | Their $\frac{dy}{dt} \div \frac{dx}{dt}$ in terms of $t$ |
| $\Rightarrow dy/dx = (2te^{2t} + e^{2t})/3e^{3t}$ | A1 | oe cao; allow unsimplified form even if subsequently cancelled incorrectly |
| When $t=1$, $dy/dx = 3e^2/3e^3 = 1/e$ | A1 | cao www; must be simplified to $1/e$ oe |
**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3t = \ln x$, $y = \frac{\ln x}{3} e^{2\ln x/3} = \frac{x^{2/3}\ln x}{3}$ | B1 | Any equivalent form of $y$ in terms of $x$ only |
| $dy/dx = \frac{1}{3}x^{2/3}\cdot\frac{1}{x} + \ln x \cdot \frac{2}{9}x^{-1/3}$ | M1 | Differentiating their $y$; need a product including $\ln kx$ and $x^p$; subst $x = e^{3t}$ to obtain $dy/dx$ in terms of $t$ |
| $= \frac{1}{3e^t} + \frac{2t}{3e^t}$ | A1 | oe cao |
| $dy/dx = 1/3e + 2/3e = 1/e$ | A1 | www cao; **exact only**; must be simplified to $1/e$ or $e^{-1}$ |
**[4 marks]**
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3t = \ln x \Rightarrow t = (\ln x)/3$ | B1 | Finding $t$ correctly in terms of $x$ |
| $y = (\ln x)/3e^{(2\ln x)/3}$ | M1 | Substituting in $y$ using their $t$ |
| $y = \frac{1}{3}x^{\frac{2}{3}}\ln x$ | A1 | Required form $ax^b \ln x$ only |
**[3 marks]**
---2 A curve has parametric equations $x = \mathrm { e } ^ { 3 t } , y = t \mathrm { e } ^ { 2 t }$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$. Hence find the exact gradient of the curve at the point with parameter $t = 1$.\\
(ii) Find the cartesian equation of the curve in the form $y = a x ^ { b } \ln x$, where $a$ and $b$ are constants to be determined.
\hfill \mbox{\textit{OCR MEI C4 Q2 [7]}}