OCR MEI C3 2008 January — Question 6 8 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2008
SessionJanuary
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind vertical tangent points
DifficultyStandard +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 recognize this means the denominator equals zero and solve the resulting transcendental equation e^(2y) = 1/2, which involves logarithms and numerical work. The conceptual leap from 'infinite gradient' to 'denominator = 0' and solving the implicit equation elevates this above routine differentiation exercises.
Spec1.07s Parametric and implicit differentiation

6 Fig. 6 shows the curve \(\mathrm { e } ^ { 2 y } = x ^ { 2 } + y\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a66c2da5-46b4-4467-933e-179be04b03b1-3_739_1339_349_404} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure}
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x } { 2 \mathrm { e } ^ { 2 y } - 1 }\).
  2. Hence find to 3 significant figures the coordinates of the point P , shown in Fig. 6, where the curve has infinite gradient. Section B (36 marks)

6 Fig. 6 shows the curve $\mathrm { e } ^ { 2 y } = x ^ { 2 } + y$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{a66c2da5-46b4-4467-933e-179be04b03b1-3_739_1339_349_404}
\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.

Section B (36 marks)\\

\hfill \mbox{\textit{OCR MEI C3 2008 Q6 [8]}}