| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate both sides with respect to x (applying product rule to the middle term), substitute the given point to find dy/dx, then write the tangent equation. While it involves multiple steps, each is a standard technique taught in C3/C4 with no novel insight required, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3x^2 + 6xy + 3x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0\) | M1 A1 B1 | M1: Implicit differentiation including either \(3x^2\frac{dy}{dx}\) or \(3y^2\frac{dy}{dx}\) term. A1: Differentiates \(y^3 \to 3y^2\frac{dy}{dx}\) and \(x^3 \to 3x^2\) and \(37 \to 0\). B1: Product rule on \(3x^2y\) giving \(6xy + 3x^2\frac{dy}{dx}\) |
| Substitutes \((1,3)\) and rearranges to get \(\frac{dy}{dx} = -\frac{7}{10}\) | M1 | Substitutes \(x=1, y=3\); note \(\frac{dy}{dx} = \frac{-3x^2-6xy}{3x^2+3y^2}\) |
| \((y-3) = -\frac{7}{10}(x-1)\) so \(7x + 10y - 37 = 0\) | M1 A1 (6) | M1: Use of \((y-3)=m(x-1)\) where \(m\) is numerical value of \(\frac{dy}{dx}\). A1: Accept integer multiples e.g. \(21x+30y-111=0\) |
# Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2 + 6xy + 3x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$ | M1 A1 B1 | M1: Implicit differentiation including either $3x^2\frac{dy}{dx}$ or $3y^2\frac{dy}{dx}$ term. A1: Differentiates $y^3 \to 3y^2\frac{dy}{dx}$ and $x^3 \to 3x^2$ and $37 \to 0$. B1: Product rule on $3x^2y$ giving $6xy + 3x^2\frac{dy}{dx}$ |
| Substitutes $(1,3)$ and rearranges to get $\frac{dy}{dx} = -\frac{7}{10}$ | M1 | Substitutes $x=1, y=3$; note $\frac{dy}{dx} = \frac{-3x^2-6xy}{3x^2+3y^2}$ |
| $(y-3) = -\frac{7}{10}(x-1)$ so $7x + 10y - 37 = 0$ | M1 A1 (6) | M1: Use of $(y-3)=m(x-1)$ where $m$ is numerical value of $\frac{dy}{dx}$. A1: Accept integer multiples e.g. $21x+30y-111=0$ |
---\begin{enumerate}
\item Find an equation of the tangent to the curve
\end{enumerate}
$$x ^ { 3 } + 3 x ^ { 2 } y + y ^ { 3 } = 37$$
at the point $( 1,3 )$. Give your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\
(6)\\
\hfill \mbox{\textit{Edexcel C34 2017 Q1 [6]}}