| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Parametric form dy/dx |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring the chain rule and knowledge that d/dy(tan(y)) = sec²(y). Part (a) uses the identity sec²(θ) = 1 + tan²(θ) to simplify. Part (b) is routine tangent line calculation. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 3\tan\left(y - \frac{\pi}{6}\right) \Rightarrow \frac{dx}{dy} = 3\sec^2\left(y - \frac{\pi}{6}\right)\) | B1 | Correct derivative including correct lhs. Condone \(\frac{dx}{dy} = k\sec^2\left(y - \frac{\pi}{6}\right)\) where \(k\) is constant |
| \(\Rightarrow \frac{dy}{dx} = \frac{1}{3\sec^2\left(y - \frac{\pi}{6}\right)}\) | M1 | Either attempts reciprocal rule \(\frac{dy}{dx} = 1 \div \frac{dx}{dy}\), or attempts to apply \(\sec^2\left(y-\frac{\pi}{6}\right) = 1 + \tan^2\left(y-\frac{\pi}{6}\right)\) with \(\tan\left(y-\frac{\pi}{6}\right)\) replaced by \(\frac{x}{3}\) |
| \(\frac{1}{3\left(1+\tan^2\left(y-\frac{\pi}{6}\right)\right)} = \frac{1}{3\left(1+\left(\frac{x}{3}\right)^2\right)}\) | dM1 | Attempts both steps to obtain \(\frac{dy}{dx}\) in terms of \(x\) |
| \(= \frac{3}{x^2+9}\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{\pi}{3} \Rightarrow x = 3\tan\frac{\pi}{6} = \sqrt{3}\) | B1 | Correct value for \(x\). Condone awrt 1.73 |
| \(x = \sqrt{3} \Rightarrow \frac{dy}{dx} = \frac{3}{\left(\sqrt{3}\right)^2+9}\) or \(y=\frac{\pi}{3} \Rightarrow \frac{dy}{dx} = \frac{1}{3\sec^2\left(\frac{\pi}{3}-\frac{\pi}{6}\right)} = \ldots\) | M1 | Uses correct method to find value of \(\frac{dy}{dx}\), which may be decimal |
| \(y - \frac{\pi}{3} = \frac{1}{4}\left(x - \sqrt{3}\right)\) | dM1 | Correct straight line method for tangent at \(\left(\sqrt{3}, \frac{\pi}{3}\right)\) using correctly found \(m\). Dependent on having found gradient and \(x\) value using correct method |
| \(y = 0 \Rightarrow -\frac{\pi}{3} = \frac{1}{4}\left(x - \sqrt{3}\right) \Rightarrow x = \ldots\) | ddM1 | Uses \(y=0\) to find \(x\). Dependent on having scored previous mark |
| \(x = \sqrt{3} - \frac{4\pi}{3}\) | A1 | Correct value or exact equivalent e.g. \(\frac{3\sqrt{3}-4\pi}{3}\) |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 3\tan\left(y - \frac{\pi}{6}\right) \Rightarrow \frac{dx}{dy} = 3\sec^2\left(y - \frac{\pi}{6}\right)$ | B1 | Correct derivative including correct lhs. Condone $\frac{dx}{dy} = k\sec^2\left(y - \frac{\pi}{6}\right)$ where $k$ is constant |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{3\sec^2\left(y - \frac{\pi}{6}\right)}$ | M1 | Either attempts reciprocal rule $\frac{dy}{dx} = 1 \div \frac{dx}{dy}$, or attempts to apply $\sec^2\left(y-\frac{\pi}{6}\right) = 1 + \tan^2\left(y-\frac{\pi}{6}\right)$ with $\tan\left(y-\frac{\pi}{6}\right)$ replaced by $\frac{x}{3}$ |
| $\frac{1}{3\left(1+\tan^2\left(y-\frac{\pi}{6}\right)\right)} = \frac{1}{3\left(1+\left(\frac{x}{3}\right)^2\right)}$ | dM1 | Attempts both steps to obtain $\frac{dy}{dx}$ in terms of $x$ |
| $= \frac{3}{x^2+9}$ | A1 | Correct answer |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{\pi}{3} \Rightarrow x = 3\tan\frac{\pi}{6} = \sqrt{3}$ | B1 | Correct value for $x$. Condone awrt 1.73 |
| $x = \sqrt{3} \Rightarrow \frac{dy}{dx} = \frac{3}{\left(\sqrt{3}\right)^2+9}$ or $y=\frac{\pi}{3} \Rightarrow \frac{dy}{dx} = \frac{1}{3\sec^2\left(\frac{\pi}{3}-\frac{\pi}{6}\right)} = \ldots$ | M1 | Uses correct method to find value of $\frac{dy}{dx}$, which may be decimal |
| $y - \frac{\pi}{3} = \frac{1}{4}\left(x - \sqrt{3}\right)$ | dM1 | Correct straight line method for tangent at $\left(\sqrt{3}, \frac{\pi}{3}\right)$ using correctly found $m$. Dependent on having found gradient and $x$ value using correct method |
| $y = 0 \Rightarrow -\frac{\pi}{3} = \frac{1}{4}\left(x - \sqrt{3}\right) \Rightarrow x = \ldots$ | ddM1 | Uses $y=0$ to find $x$. Dependent on having scored previous mark |
| $x = \sqrt{3} - \frac{4\pi}{3}$ | A1 | Correct value or exact equivalent e.g. $\frac{3\sqrt{3}-4\pi}{3}$ |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$x = 3 \tan \left( y - \frac { \pi } { 6 } \right) \quad x \in \mathbb { R } \quad - \frac { \pi } { 3 } < y < \frac { 2 \pi } { 3 }$$
(a) Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { a } { x ^ { 2 } + b }$$
where $a$ and $b$ are integers to be found.
The point $P$ with $y$ coordinate $\frac { \pi } { 3 }$ lies on $C$.\\
Given that the tangent to $C$ at $P$ crosses the $x$-axis at the point $Q$.\\
(b) find, in simplest form, the exact $x$ coordinate of $Q$.
\hfill \mbox{\textit{Edexcel P3 2023 Q7 [9]}}