| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Moderate -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 the gradient, then write the tangent equation. While it involves the product rule and implicit differentiation technique, it's a standard textbook exercise with clear steps and no conceptual surprises, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3x^2 + 2xy - 2y^2 + 4 = 0 \Rightarrow 6x + 2x\frac{dy}{dx} + 2y - 4y\frac{dy}{dx} = 0\) | B1 | \(2xy\) differentiated correctly to give \(2x\frac{dy}{dx} + 2y\) or equivalent |
| M1 | Attempts chain rule on \(-2y^2\) to give form \(Ay\frac{dy}{dx}\) | |
| A1 | Fully correct differentiation of \(3x^2 - 2y^2 + 4\) giving \(6x - 4y\frac{dy}{dx}\) and "\(= 0\)" | |
| Sets \(x=2, y=4 \Rightarrow 12 + 4\frac{dy}{dx} + 8 - 16\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{5}{3}\) | M1 | Substitutes \(x=2\), \(y=4\) and attempts to find \(\frac{dy}{dx}\) |
| Uses \(x=2\), \(y=4\) and \(\frac{dy}{dx} = \frac{5}{3} \Rightarrow (y-4) = \frac{5}{3}(x-2)\) | M1 | Uses point and gradient to find tangent equation |
| \(5x - 3y + 2 = 0\) | A1 | Accept \(5x - 3y + 2 = 0\) or any integer multiple |
## Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3x^2 + 2xy - 2y^2 + 4 = 0 \Rightarrow 6x + 2x\frac{dy}{dx} + 2y - 4y\frac{dy}{dx} = 0$ | B1 | $2xy$ differentiated correctly to give $2x\frac{dy}{dx} + 2y$ or equivalent |
| | M1 | Attempts chain rule on $-2y^2$ to give form $Ay\frac{dy}{dx}$ |
| | A1 | Fully correct differentiation of $3x^2 - 2y^2 + 4$ giving $6x - 4y\frac{dy}{dx}$ and "$= 0$" |
| Sets $x=2, y=4 \Rightarrow 12 + 4\frac{dy}{dx} + 8 - 16\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{5}{3}$ | M1 | Substitutes $x=2$, $y=4$ and attempts to find $\frac{dy}{dx}$ |
| Uses $x=2$, $y=4$ and $\frac{dy}{dx} = \frac{5}{3} \Rightarrow (y-4) = \frac{5}{3}(x-2)$ | M1 | Uses point and gradient to find tangent equation |
| $5x - 3y + 2 = 0$ | A1 | Accept $5x - 3y + 2 = 0$ or any integer multiple |
---\begin{enumerate}
\item A curve $C$ has equation
\end{enumerate}
$$3 x ^ { 2 } + 2 x y - 2 y ^ { 2 } + 4 = 0$$
Find an equation for the tangent to $C$ at the point ( 2,4 ), giving 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]}}