| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| 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 standard technique (product rule, chain rule) followed by finding a normal equation. While it involves multiple steps, each is routine for A-level Further Maths students, making it slightly easier than average overall difficulty. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^3 \rightarrow \ldots x^2\) and \(3y^2 \rightarrow \ldots y\frac{dy}{dx}\) | M1 | Attempts to differentiate \(x^3 \rightarrow \ldots x^2\) and \(3y^2 \rightarrow \ldots y\frac{dy}{dx}\) where … are constants |
| \(2xy \rightarrow 2y + 2x\frac{dy}{dx}\) | B1 | Correct application of product rule on \(2xy\) |
| \(3x^2 + 2x\frac{dy}{dx} + 2y + 6y\frac{dy}{dx} = \ldots \Rightarrow \frac{dy}{dx} = \ldots\) | M1 | Valid attempt to make \(\frac{dy}{dx}\) subject, with exactly 2 different terms in \(\frac{dy}{dx}\) coming from \(3y^2\) and \(2xy\) |
| \(\frac{dy}{dx} = -\frac{2y+3x^2}{2x+6y}\) | A1 | Accept equivalent forms e.g. \(\frac{-2y-3x^2}{2x+6y}\), \(\frac{2y+3x^2}{-2x-6y}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = -\frac{2(5)+3(-2)^2}{2(-2)+6(5)}\) or \(3(-2)^2+2(-2)\frac{dy}{dx}+2\times5+6\times5\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx} = \ldots \left(-\frac{11}{13}\right)\) | M1 | Substitutes \(x=-2\) and \(y=5\); must have \(x\)'s and \(y\)'s in their \(\frac{dy}{dx}\) |
| \(y-5 = \frac{13}{11}(x+2)\) | dM1 | Equation of normal using negative reciprocal of tangent gradient, with \(x=-2\), \(y=5\) correctly placed |
| \(13x - 11y + 81 = 0\) | A1 | Integer multiple of this equation acceptable |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^3 \rightarrow \ldots x^2$ and $3y^2 \rightarrow \ldots y\frac{dy}{dx}$ | M1 | Attempts to differentiate $x^3 \rightarrow \ldots x^2$ **and** $3y^2 \rightarrow \ldots y\frac{dy}{dx}$ where … are constants |
| $2xy \rightarrow 2y + 2x\frac{dy}{dx}$ | B1 | Correct application of product rule on $2xy$ |
| $3x^2 + 2x\frac{dy}{dx} + 2y + 6y\frac{dy}{dx} = \ldots \Rightarrow \frac{dy}{dx} = \ldots$ | M1 | Valid attempt to make $\frac{dy}{dx}$ subject, with exactly 2 different terms in $\frac{dy}{dx}$ coming from $3y^2$ and $2xy$ |
| $\frac{dy}{dx} = -\frac{2y+3x^2}{2x+6y}$ | A1 | Accept equivalent forms e.g. $\frac{-2y-3x^2}{2x+6y}$, $\frac{2y+3x^2}{-2x-6y}$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = -\frac{2(5)+3(-2)^2}{2(-2)+6(5)}$ or $3(-2)^2+2(-2)\frac{dy}{dx}+2\times5+6\times5\frac{dy}{dx}=0 \Rightarrow \frac{dy}{dx} = \ldots \left(-\frac{11}{13}\right)$ | M1 | Substitutes $x=-2$ and $y=5$; must have $x$'s and $y$'s in their $\frac{dy}{dx}$ |
| $y-5 = \frac{13}{11}(x+2)$ | dM1 | Equation of normal using negative reciprocal of tangent gradient, with $x=-2$, $y=5$ correctly placed |
| $13x - 11y + 81 = 0$ | A1 | Integer multiple of this equation acceptable |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying on calculator technology are not acceptable.\\
A curve has equation
$$x ^ { 3 } + 2 x y + 3 y ^ { 2 } = 47$$
(a) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$
The point $P ( - 2,5 )$ lies on the curve.\\
(b) Find the equation of the normal to the curve at $P$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers to be found.
\hfill \mbox{\textit{Edexcel Paper 2 2023 Q7 [7]}}