| Exam Board | OCR |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (polynomial/rational) |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring standard techniques: finding dy/dx using the chain rule (dy/dt รท dx/dt), finding the parameter value at a given point, and eliminating the parameter by algebraic manipulation. The hint to consider y/x makes part (ii) routine. Slightly above average due to the algebraic manipulation required, but all techniques are standard C4 material. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\) aef used | M1 | |
| \(=\frac{4t+3t^2}{2+2t}\) | A1 | |
| Attempt to find \(t\) from one/both equations | M1 | or diff (ii) cartesian eqn \(\to\) M1 |
| State/imply \(t=-3\) is only solution of both equations | A1 | subst\((3,-9)\), solve for \(\frac{dy}{dx}\to\) M1 |
| Gradient of curve \(=-\frac{15}{4}\) or \(\frac{-15}{4}\) or \(\frac{15}{-4}\) | A1 5 | grad of curve \(=-\frac{15}{4}\to\) A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{y}{x}=t\) | B1 | |
| Substitute into either parametric eqn | M1 | |
| Final answer \(x^3=2xy+y^2\) | A2 4 | [SR Any correct unsimplified form \(\to\) A1] |
# Question 5:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ aef used | M1 | |
| $=\frac{4t+3t^2}{2+2t}$ | A1 | |
| Attempt to find $t$ from one/both equations | M1 | or diff (ii) cartesian eqn $\to$ M1 |
| State/imply $t=-3$ is only solution of both equations | A1 | subst$(3,-9)$, solve for $\frac{dy}{dx}\to$ M1 |
| Gradient of curve $=-\frac{15}{4}$ or $\frac{-15}{4}$ or $\frac{15}{-4}$ | A1 **5** | grad of curve $=-\frac{15}{4}\to$ A1 |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{y}{x}=t$ | B1 | |
| Substitute into either parametric eqn | M1 | |
| Final answer $x^3=2xy+y^2$ | A2 **4** | [SR Any correct unsimplified form $\to$ A1] |
---5 A curve has parametric equations
$$x = 2 t + t ^ { 2 } , \quad y = 2 t ^ { 2 } + t ^ { 3 }$$
(i) Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $t$ and find the gradient of the curve at the point $( 3 , - 9 )$.\\
(ii) By considering $\frac { y } { x }$, find a cartesian equation of the curve, giving your answer in a form not involving fractions.
\hfill \mbox{\textit{OCR C4 2009 Q5 [9]}}