| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find normal equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring product rule and chain rule, followed by finding a normal line equation. The algebra is manageable, the point is given (no need to find it), and the techniques are standard C4 content with no novel problem-solving required. Slightly above average difficulty due to the exponential term and implicit differentiation, but still a routine textbook exercise. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(e^{-2x}\frac{dy}{dx}-2ye^{-2x} = 2+2y\frac{dy}{dx}\) | M1 A1 | A1 correct RHS |
| \(\frac{d}{dx}(ye^{-2x}) = e^{-2x}\frac{dy}{dx}-2ye^{-2x}\) | B1 | |
| \(\left(e^{-2x}-2y\right)\frac{dy}{dx} = 2+2ye^{-2x}\) | M1 | |
| \(\frac{dy}{dx} = \frac{2+2ye^{-2x}}{e^{-2x}-2y}\) | A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \(P\): \(\frac{dy}{dx} = \frac{2+2e^0}{e^0-2} = -4\) | M1 | |
| Using \(mm'=-1\): \(m'=\frac{1}{4}\) | M1 | |
| \(y-1=\frac{1}{4}(x-0)\) | M1 | |
| \(x-4y+4=0\) | A1 (4) | or any integer multiple |
# Question 4:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $e^{-2x}\frac{dy}{dx}-2ye^{-2x} = 2+2y\frac{dy}{dx}$ | M1 A1 | A1 correct RHS |
| $\frac{d}{dx}(ye^{-2x}) = e^{-2x}\frac{dy}{dx}-2ye^{-2x}$ | B1 | |
| $\left(e^{-2x}-2y\right)\frac{dy}{dx} = 2+2ye^{-2x}$ | M1 | |
| $\frac{dy}{dx} = \frac{2+2ye^{-2x}}{e^{-2x}-2y}$ | A1 (5) | |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $P$: $\frac{dy}{dx} = \frac{2+2e^0}{e^0-2} = -4$ | M1 | |
| Using $mm'=-1$: $m'=\frac{1}{4}$ | M1 | |
| $y-1=\frac{1}{4}(x-0)$ | M1 | |
| $x-4y+4=0$ | A1 (4) | or any integer multiple |
---4. The curve $C$ has the equation $y \mathrm { e } ^ { - 2 x } = 2 x + y ^ { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.
The point $P$ on $C$ has coordinates $( 0,1 )$.
\item Find the equation of the normal to $C$ at $P$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2009 Q4 [9]}}