| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| 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 substitution to find intersection points, then applying the chain rule to differentiate implicitly. All steps are standard C3 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 2 \Rightarrow y^2 + y = 12\) | M1 | Substituting \(x = 2\) |
| \(y^2 + y - 12 = 0\) | ||
| \((y-3)(y+4) = 0\) | A1 | \(y = 3\) |
| \(y = 3\) or \(-4\) | A1 | \(y = -4\) |
| \(2y\frac{dy}{dx} + \frac{dy}{dx} = 3x^2 + 2\) | M1 | Implicit differentiation – LHS must be correct |
| \(\frac{dy}{dx}(2y+1) = 3x^2 + 2\) | ||
| \(\frac{dy}{dx} = \frac{3x^2+2}{2y+1}\) | A1cao | |
| At \((2,3)\): \(\frac{dy}{dx} = \frac{12+2}{6+1} = 2\) | M1 | Substituting \(x=2\), \(y=3\) into their \(\frac{dy}{dx}\); must require both \(x\) and one of their \(y\) to be substituted |
| At \((2,-4)\): \(\frac{dy}{dx} = \frac{12+2}{-8+1} = -2\) | A1cao | \(2\) |
| A1cao [8] | \(-2\) |
## Question 2:
$y^2 + y = x^3 + 2x$
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 2 \Rightarrow y^2 + y = 12$ | M1 | Substituting $x = 2$ |
| $y^2 + y - 12 = 0$ | | |
| $(y-3)(y+4) = 0$ | A1 | $y = 3$ |
| $y = 3$ or $-4$ | A1 | $y = -4$ |
| $2y\frac{dy}{dx} + \frac{dy}{dx} = 3x^2 + 2$ | M1 | Implicit differentiation – LHS must be correct |
| $\frac{dy}{dx}(2y+1) = 3x^2 + 2$ | | |
| $\frac{dy}{dx} = \frac{3x^2+2}{2y+1}$ | A1cao | |
| At $(2,3)$: $\frac{dy}{dx} = \frac{12+2}{6+1} = 2$ | M1 | Substituting $x=2$, $y=3$ into their $\frac{dy}{dx}$; must require both $x$ and one of their $y$ to be substituted |
| At $(2,-4)$: $\frac{dy}{dx} = \frac{12+2}{-8+1} = -2$ | A1cao | $2$ |
| | A1cao [8] | $-2$ |
---2 Fig. 7 shows the curve defined implicitly by the equation
$$y ^ { 2 } + y = x ^ { 9 } + 2 x$$
together with the line $x = 2$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e0636807-d5bf-43c2-a484-68245e639cee-2_462_385_657_858}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
Not to scale
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 Q2 [8]}}