| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Show specific gradient value |
| Difficulty | Standard +0.3 This is a standard implicit differentiation question requiring students to differentiate implicitly, substitute a point to find the gradient, then find the normal's equation. While it involves multiple steps (differentiate, solve for dy/dx, evaluate at a point, find perpendicular gradient, write equation), these are all routine C4 techniques with no novel problem-solving required. The 8 marks reflect the working steps rather than conceptual difficulty. Slightly above average due to the multi-step nature and implicit differentiation being less routine than explicit differentiation. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(3x^2 + 2y + 2c\frac{dy}{dx} - 2y\frac{dy}{dx} = 0\) | M1 A2 | |
| \((2, -4) \Rightarrow 12 - 8 + 4\frac{dy}{dx} + 8\frac{dy}{dx} = 0\) | M1 A1 | |
| \(\frac{dy}{dx} = -\frac{1}{3}\) | M1 | gradient of normal = 3 |
| \(\therefore y + 4 = 3(x - 2)\) | M1 | |
| \(y = 3x - 10\) | A1 | (8) |
$3x^2 + 2y + 2c\frac{dy}{dx} - 2y\frac{dy}{dx} = 0$ | M1 A2 |
$(2, -4) \Rightarrow 12 - 8 + 4\frac{dy}{dx} + 8\frac{dy}{dx} = 0$ | M1 A1 |
$\frac{dy}{dx} = -\frac{1}{3}$ | M1 | gradient of normal = 3
$\therefore y + 4 = 3(x - 2)$ | M1 |
$y = 3x - 10$ | A1 | (8)
---\begin{enumerate}
\item A curve has the equation
\end{enumerate}
$$x ^ { 3 } + 2 x y - y ^ { 2 } + 24 = 0$$
Show that the normal to the curve at the point $( 2 , - 4 )$ has the equation $y = 3 x - 10$. (8)\\
\hfill \mbox{\textit{Edexcel C4 Q1 [8]}}