| 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 vertical tangent points |
| Difficulty | Standard +0.8 This question requires implicit differentiation (standard C3 technique) but then asks students to find where gradient is infinite (vertical tangent), requiring them to set the denominator to zero and solve a transcendental equation involving e^(2y). The conceptual leap to connect infinite gradient with denominator = 0, plus solving the resulting equation, elevates this above routine implicit differentiation exercises. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e^{2y} = x^2 + y \Rightarrow 2e^{2y}\frac{dy}{dx} = 2x + \frac{dy}{dx}\) | M1 | Implicit differentiation – allow one slip (but with \(\frac{dy}{dx}\) both sides) |
| A1 | ||
| \(\Rightarrow \left(2e^{2y} - 1\right)\frac{dy}{dx} = 2x\) | M1 | collecting terms |
| \(\Rightarrow \frac{dy}{dx} = \frac{2x}{2e^{2y}-1}\) | E1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Gradient is infinite when \(2e^{2y} - 1 = 0\) | M1 | |
| \(\Rightarrow e^{2y} = \frac{1}{2} \Rightarrow 2y = \ln\frac{1}{2}\) | A1 | must be to 3 s.f. |
| \(\Rightarrow y = \frac{1}{2}\ln\frac{1}{2} = -0.347\) (3 s.f.) | M1 | substituting their \(y\) and solving for \(x\) |
| \(x^2 = e^{2y} - y = \frac{1}{2} - (-0.347) = 0.8465\) | A1 | cao – must be to 3 s.f., but penalise accuracy once only |
| \(\Rightarrow x = 0.920\) | [4] |
# Question 5:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e^{2y} = x^2 + y \Rightarrow 2e^{2y}\frac{dy}{dx} = 2x + \frac{dy}{dx}$ | M1 | Implicit differentiation – allow one slip (but with $\frac{dy}{dx}$ both sides) |
| | A1 | |
| $\Rightarrow \left(2e^{2y} - 1\right)\frac{dy}{dx} = 2x$ | M1 | collecting terms |
| $\Rightarrow \frac{dy}{dx} = \frac{2x}{2e^{2y}-1}$ | E1 [4] | |
## Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Gradient is infinite when $2e^{2y} - 1 = 0$ | M1 | |
| $\Rightarrow e^{2y} = \frac{1}{2} \Rightarrow 2y = \ln\frac{1}{2}$ | A1 | must be to 3 s.f. |
| $\Rightarrow y = \frac{1}{2}\ln\frac{1}{2} = -0.347$ (3 s.f.) | M1 | substituting their $y$ and solving for $x$ |
| $x^2 = e^{2y} - y = \frac{1}{2} - (-0.347) = 0.8465$ | A1 | cao – must be to 3 s.f., but penalise accuracy once only |
| $\Rightarrow x = 0.920$ | [4] | |5 Fig. 6 shows the curve $\mathrm { e } ^ { 2 y } = x ^ { 2 } + y$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e3850377-bd1a-4e3c-8424-e3db7fd3c4db-3_736_1331_893_459}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x } { 2 \mathrm { e } ^ { 2 y } - 1 }$.\\
(ii) Hence find to 3 significant figures the coordinates of the point P , shown in Fig. 6, where the curve has infinite gradient.
\hfill \mbox{\textit{OCR MEI C3 Q5 [8]}}