| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find normal equation at point |
| Difficulty | Moderate -0.3 This is a straightforward implicit differentiation question with standard steps: verify a point lies on a curve, find dy/dx using the chain rule (dx/dy = 2cos y, so dy/dx = 1/(2cos y)), then find the normal line. All techniques are routine C3 material with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(y = \frac{\pi}{4} \Rightarrow x = 2\sin\frac{\pi}{4} = 2x\frac{1}{\sqrt{2}} = \sqrt{2} \Rightarrow P \in C\); Accept equivalent (reversed) arguments. In any method it must be clear that \(\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}\) or exact equivalent is used. | B1 | (1 mark) |
| (b) \(\frac{dx}{dy} = 2\cos y\) or \(1 = 2\cos y\frac{dy}{dx}\); \(\frac{dy}{dx} = \frac{1}{2\cos y}\); May be awarded after substitution; \(y = \frac{\pi}{4} \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{2}}\) | M1, A1, M1, A1 | cso (4 marks) |
| (c) \(m' = -\sqrt{2}\); \(y - \frac{\pi}{4} = -\sqrt{2}\left(x - \sqrt{2}\right)\); \(y = -\sqrt{2}x + 2 + \frac{\pi}{4}\) | B1, M1, A1 | (4 marks) |
(a) $y = \frac{\pi}{4} \Rightarrow x = 2\sin\frac{\pi}{4} = 2x\frac{1}{\sqrt{2}} = \sqrt{2} \Rightarrow P \in C$; Accept equivalent (reversed) arguments. In any method it must be clear that $\sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$ or exact equivalent is used. | B1 | (1 mark)
(b) $\frac{dx}{dy} = 2\cos y$ or $1 = 2\cos y\frac{dy}{dx}$; $\frac{dy}{dx} = \frac{1}{2\cos y}$; May be awarded after substitution; $y = \frac{\pi}{4} \Rightarrow \frac{dy}{dx} = \frac{1}{\sqrt{2}}$ | M1, A1, M1, A1 | cso (4 marks)
(c) $m' = -\sqrt{2}$; $y - \frac{\pi}{4} = -\sqrt{2}\left(x - \sqrt{2}\right)$; $y = -\sqrt{2}x + 2 + \frac{\pi}{4}$ | B1, M1, A1 | (4 marks)
---3. The curve $C$ has equation
$$x = 2 \sin y .$$
\begin{enumerate}[label=(\alph*)]
\item Show that the point $P \left( \sqrt { } 2 , \frac { \pi } { 4 } \right)$ lies on $C$.
\item Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { \sqrt { 2 } }$ at $P$.
\item Find an equation of the normal to $C$ at $P$. Give your answer in the form $y = m x + c$, where $m$ and $c$ are exact constants.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 2007 Q3 [9]}}