| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Moderate -0.3 This is a straightforward parametric equations question requiring standard techniques: showing a derivative is positive (simple calculation with rational functions) and eliminating the parameter to find Cartesian form. 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Differentiate \(x\) as a quotient, \(\frac{v\,du - u\,dv}{v^2}\) or \(\frac{u\,dv - v\,du}{v^2}\) | M1 | or product clearly defined |
| \(\frac{dx}{dt} = -\frac{1}{(t+1)^2}\) or \(\frac{-1}{(t+1)^2}\) or \(-(t+1)^{-2}\) | A1 | WWW \(\rightarrow 2\) |
| \(\frac{dy}{dt} = -\frac{2}{(t+3)^2}\) or \(\frac{-2}{(t+3)^2}\) or \(-2(t+3)^{-2}\) | B1 | |
| \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) | M1 | quoted/implied and used |
| \(\frac{dy}{dx} = \frac{2(t+1)^2}{(t+3)^2}\) or \(\frac{2(t+3)^{-2}}{(t+1)^{-2}}\) | *A1 | dep 1st 4 marks; ignore ref \(t=-1, t=-3\) |
| State squares +ve or \((t+1)^2\) & \((t+3)^2 +\) ve \(\therefore \frac{dy}{dx}\) +ve | dep*A1 | 6 marks or \(\left(\frac{t+1}{t+3}\right)^2 +\) ve. Ignore \(\geq 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempt to obtain \(t\) from either the \(x\) or \(y\) equation | M1 | No accuracy required |
| \(t = \frac{2-x}{x-1}\) AEF or \(t = \frac{2}{y}-3\) AEF | A1 | |
| Substitute in the equation not yet used in this part | M1 | or equate the 2 values of \(t\) |
| Use correct method to eliminate ('double-decker') fractions | M1 | |
| Obtain \(2x + y = 2xy + 2\) ISW AEF | A1 | 5 marks but not involving fractions |
# Question 7(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Differentiate $x$ as a quotient, $\frac{v\,du - u\,dv}{v^2}$ or $\frac{u\,dv - v\,du}{v^2}$ | M1 | or product clearly defined |
| $\frac{dx}{dt} = -\frac{1}{(t+1)^2}$ or $\frac{-1}{(t+1)^2}$ or $-(t+1)^{-2}$ | A1 | WWW $\rightarrow 2$ |
| $\frac{dy}{dt} = -\frac{2}{(t+3)^2}$ or $\frac{-2}{(t+3)^2}$ or $-2(t+3)^{-2}$ | B1 | |
| $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ | M1 | quoted/implied and used |
| $\frac{dy}{dx} = \frac{2(t+1)^2}{(t+3)^2}$ or $\frac{2(t+3)^{-2}}{(t+1)^{-2}}$ | *A1 | dep 1st 4 marks; ignore ref $t=-1, t=-3$ |
| State squares +ve or $(t+1)^2$ & $(t+3)^2 +$ ve $\therefore \frac{dy}{dx}$ +ve | dep*A1 | **6 marks** or $\left(\frac{t+1}{t+3}\right)^2 +$ ve. Ignore $\geq 0$ |
# Question 7(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempt to obtain $t$ from either the $x$ or $y$ equation | M1 | No accuracy required |
| $t = \frac{2-x}{x-1}$ AEF or $t = \frac{2}{y}-3$ AEF | A1 | |
| Substitute in the equation not yet used in this part | M1 | or equate the 2 values of $t$ |
| Use correct method to eliminate ('double-decker') fractions | M1 | |
| Obtain $2x + y = 2xy + 2$ ISW AEF | A1 | **5 marks** but not involving fractions |
**Total: 11 marks**
---7 The parametric equations of a curve are $x = \frac { t + 2 } { t + 1 } , y = \frac { 2 } { t + 3 }$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } > 0$.\\
(ii) Find the cartesian equation of the curve, giving your answer in a form not involving fractions.
\hfill \mbox{\textit{OCR C4 2010 Q7 [11]}}