| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 8 |
| 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 with standard techniques. Part (a) requires routine application of implicit differentiation rules for trigonometric functions, and part (b) is a standard tangent line calculation. The algebra is clean with nice values (π/3, π/6), making it slightly easier than average but still requiring proper technique. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(-4 \sin x + (2 \cos y)\frac{dy}{dx} = 0\) | M1 A2 | |
| \(\frac{dy}{dx} = \frac{4\sin x}{2 \cos y} = \frac{2\sin x}{\cos y} = 2 \sin x \sec y\) | M1 A1 | |
| (b) \(\text{grad} = 2 \times \frac{\sqrt{2}}{2} \times \frac{2}{\sqrt{3}} = 2\) | B1 | |
| \(\therefore y - \frac{\pi}{6} = 2(x - \frac{\pi}{3})\) | M1 | |
| \(6y - \pi = 12x - 4\pi\) or \(4x - 2y = \pi\) | A1 | (8) |
**(a)** $-4 \sin x + (2 \cos y)\frac{dy}{dx} = 0$ | M1 A2 |
$\frac{dy}{dx} = \frac{4\sin x}{2 \cos y} = \frac{2\sin x}{\cos y} = 2 \sin x \sec y$ | M1 A1 |
**(b)** $\text{grad} = 2 \times \frac{\sqrt{2}}{2} \times \frac{2}{\sqrt{3}} = 2$ | B1 |
$\therefore y - \frac{\pi}{6} = 2(x - \frac{\pi}{3})$ | M1 |
$6y - \pi = 12x - 4\pi$ or $4x - 2y = \pi$ | A1 | (8)\begin{enumerate}
\item A curve has the equation
\end{enumerate}
$$4 \cos x + 2 \sin y = 3$$
(a) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \sin x \sec y$.\\
(b) Find an equation for the tangent to the curve at the point $\left( \frac { \pi } { 3 } , \frac { \pi } { 6 } \right)$, giving your answer in the form $a x + b y = c$, where $a$ and $b$ are integers.\\
\hfill \mbox{\textit{Edexcel C4 Q2 [8]}}