| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find dy/dx at a point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring students to (1) solve a quadratic to find intersection points, (2) differentiate implicitly using standard rules, and (3) substitute values. All steps are routine C3 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| At \((2, -4)\): \(\frac{dy}{dx} = \frac{12 + 2}{-8 + 1} = -2\) | M1, A1, A1, M1, A1cao, M1, A1cao, A1cao [8] | Substituting \(x = 2\); \(y = 3\); \(y = -4\); Implicit differentiation – LHS must be correct; substituting \(x = 2, y = 3\) into their dy/dx, but must require both x and one of their y to be substituted; 2; -2 |
$y^2 + y = x^3 + 2x$ with $x = 2 \Rightarrow y^2 + y - 12 = 0 \Rightarrow (y-3)(y+4) = 0 \Rightarrow y = 3$ or $y = -4$
$2y\frac{dy}{dx} + \frac{dy}{dx} = 3x^2 + 2 \Rightarrow \frac{dy}{dx}(2y+1) = 3x^2 + 2 \Rightarrow \frac{dy}{dx} = \frac{3x^2 + 2}{2y + 1}$
At $(2, 3)$: $\frac{dy}{dx} = \frac{12 + 2}{6 + 1} = 2$
At $(2, -4)$: $\frac{dy}{dx} = \frac{12 + 2}{-8 + 1} = -2$ | M1, A1, A1, M1, A1cao, M1, A1cao, A1cao [8] | Substituting $x = 2$; $y = 3$; $y = -4$; Implicit differentiation – LHS must be correct; substituting $x = 2, y = 3$ into their dy/dx, but must require both x and one of their y to be substituted; 2; -27 Fig. 7 shows the curve defined implicitly by the equation
$$y ^ { 2 } + y = x ^ { 3 } + 2 x ,$$
together with the line $x = 2$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3efea8db-9fa1-47a8-89b8-e4888f87a313-3_465_378_534_808}
\captionsetup{labelformat=empty}
\caption{Not to scale}
\end{center}
\end{figure}
Fig. 7
Find the coordinates of the points of intersection of the line and the curve.\\
Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$. Hence find the gradient of the curve at each of these two points.
\hfill \mbox{\textit{OCR MEI C3 2005 Q7 [8]}}