Question 2 - A-Level Maths

Edexcel C34 2018 October — Question 2 7 marks

Exam Board	Edexcel
Module	C34 (Core Mathematics 3 & 4)
Year	2018
Session	October
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 students to differentiate both sides with respect to x, substitute the given point to find dy/dx, then find the perpendicular gradient and write the normal equation. While it involves multiple steps (implicit differentiation, substitution, finding perpendicular gradient, equation of line), each step is routine and follows standard procedures taught in C3/C4 with no novel problem-solving required.
Spec	1.07m Tangents and normals: gradient and equations 1.07s Parametric and implicit differentiation

2. A curve $C$ has equation $$x ^ { 3 } - 4 x y + 2 x + 3 y ^ { 2 } - 3 = 0$$ Find an equation of the normal to $C$ at the point ( $- 3,2$ ), giving your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers. \includegraphics[max width=\textwidth, alt={}, center]{c6bde466-61ec-437d-a3b4-84511a98d788-05_108_166_2612_1781}

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Question 2:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$x^3 - 4xy + 2x + 3y^2 - 3 = 0 \Rightarrow 3x^2 - 4x\frac{dy}{dx} - 4y + 2 + 6y\frac{dy}{dx} = 0$	B1	Applies product rule to $-4xy$: $-4xy \rightarrow -4x\frac{dy}{dx} - 4y$. Accept exact alternatives.
M1	Attempts chain rule to $3y^2 \rightarrow Ay\frac{dy}{dx}$
A1	Correct differentiation of $x^3 + 2x + 3y^2 - 3 \Rightarrow 3x^2 + 2 + 6y\frac{dy}{dx}$
Substitute $(-3, 2) \Rightarrow \frac{dy}{dx} = \left(-\frac{7}{8}\right)$	M1	Substitutes $(-3,2)$ into differentiated form. Dependent on form having exactly two terms in $\frac{dy}{dx}$, one from $-4xy$ and one from $3y^2$
Uses gradient of normal \(= -\frac{1}{\left.\frac{dy}{dx}\right\	_{x=-3}}\)	dM1
$y - 2 = \frac{8}{7}(x+3) \Rightarrow 8x - 7y + 38 = 0$	M1, A1	Correct attempt at equation of normal at $(-3,2)$. Allow $k(8x-7y+38=0)$ where $k$ is integer

# Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^3 - 4xy + 2x + 3y^2 - 3 = 0 \Rightarrow 3x^2 - 4x\frac{dy}{dx} - 4y + 2 + 6y\frac{dy}{dx} = 0$ | B1 | Applies product rule to $-4xy$: $-4xy \rightarrow -4x\frac{dy}{dx} - 4y$. Accept exact alternatives. |
| | M1 | Attempts chain rule to $3y^2 \rightarrow Ay\frac{dy}{dx}$ |
| | A1 | Correct differentiation of $x^3 + 2x + 3y^2 - 3 \Rightarrow 3x^2 + 2 + 6y\frac{dy}{dx}$ |
| Substitute $(-3, 2) \Rightarrow \frac{dy}{dx} = \left(-\frac{7}{8}\right)$ | M1 | Substitutes $(-3,2)$ into differentiated form. Dependent on form having **exactly** two terms in $\frac{dy}{dx}$, one from $-4xy$ and one from $3y^2$ |
| Uses gradient of normal $= -\frac{1}{\left.\frac{dy}{dx}\right\|_{x=-3}}$ | dM1 | Attempts numerical value of gradient of normal; negative reciprocal. Dependent on previous M |
| $y - 2 = \frac{8}{7}(x+3) \Rightarrow 8x - 7y + 38 = 0$ | M1, A1 | Correct attempt at equation of normal at $(-3,2)$. Allow $k(8x-7y+38=0)$ where $k$ is integer |

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2. A curve $C$ has equation

$$x ^ { 3 } - 4 x y + 2 x + 3 y ^ { 2 } - 3 = 0$$

Find an equation of the normal to $C$ at the point ( $- 3,2$ ), giving your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers.

\includegraphics[max width=\textwidth, alt={}, center]{c6bde466-61ec-437d-a3b4-84511a98d788-05_108_166_2612_1781}\\

\hfill \mbox{\textit{Edexcel C34 2018 Q2 [7]}}

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Q1 8 Q2 7 Q3 6 Q4 10 Q5 10 Q6 9 Q7 8 Q8 10 Q9 9 Q10 12 Q11 10 Q12 13 Q13 13