| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar coordinates question requiring the formula dy/dx = 0 for tangents parallel to the initial line, followed by straightforward algebraic manipulation and substitution. While it involves multiple steps and the polar tangent formula, it's a textbook application with no novel insight required—moderately above average difficulty due to the Further Maths content and multi-part nature. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Cardioid shape with dimple, \(3+\sqrt{5}\) marked | B1 | Recalls correct shape for the type of curve, including 'dimple' |
| Correct position with pole, initial line and point labelled | B1 | Correct position with labelling of pole, initial line and point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{d}{d\theta}(r\sin\theta) = \frac{d}{d\theta}(3\sin\theta + \sqrt{5}\sin\theta\cos\theta) = A\cos\theta + B\cos 2\theta\) | M1 | Uses \(y = r\sin\theta\) with the curve and attempts to differentiate. Accept any correct form, slips in coefficients allowed |
| \(\frac{d}{d\theta}(r\sin\theta) = 3\cos\theta + \sqrt{5}\cos 2\theta\) | A1 | Correct differentiation. Accept equivalents e.g. \(3\cos\theta + \sqrt{5}\cos^2\theta - \sqrt{5}\sin^2\theta\) |
| \(\frac{dy}{dx} = 0 \Rightarrow 3\cos\theta + \sqrt{5}(2\cos^2\theta - 1) = 0\) leading to \(\{2\sqrt{5}\cos^2\theta + 3\cos\theta - \sqrt{5} = 0\}\) | M1 | Sets derivative equal to zero, using trig identities to form a quadratic in \(\cos\theta\) |
| \(\cos\theta = \frac{1}{\sqrt{5}}\) from \(\cos\theta = \frac{-3\pm 7}{4\sqrt{5}}\), quadrant 1 needs \(\cos\theta > 0\) | A1 | Solves quadratic and selects correct value for \(\cos\theta\). If other value given it is A0 unless clearly rejected |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = 4\) | B1 | Correct value for \(r\) |
# Question 4:
**(a)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Cardioid shape with dimple, $3+\sqrt{5}$ marked | B1 | Recalls correct shape for the type of curve, including 'dimple' |
| Correct position with pole, initial line and point labelled | B1 | Correct position with labelling of pole, initial line and point |
**(b)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{d\theta}(r\sin\theta) = \frac{d}{d\theta}(3\sin\theta + \sqrt{5}\sin\theta\cos\theta) = A\cos\theta + B\cos 2\theta$ | M1 | Uses $y = r\sin\theta$ with the curve and attempts to differentiate. Accept any correct form, slips in coefficients allowed |
| $\frac{d}{d\theta}(r\sin\theta) = 3\cos\theta + \sqrt{5}\cos 2\theta$ | A1 | Correct differentiation. Accept equivalents e.g. $3\cos\theta + \sqrt{5}\cos^2\theta - \sqrt{5}\sin^2\theta$ |
| $\frac{dy}{dx} = 0 \Rightarrow 3\cos\theta + \sqrt{5}(2\cos^2\theta - 1) = 0$ leading to $\{2\sqrt{5}\cos^2\theta + 3\cos\theta - \sqrt{5} = 0\}$ | M1 | Sets derivative equal to zero, using trig identities to form a quadratic in $\cos\theta$ |
| $\cos\theta = \frac{1}{\sqrt{5}}$ from $\cos\theta = \frac{-3\pm 7}{4\sqrt{5}}$, quadrant 1 needs $\cos\theta > 0$ | A1 | Solves quadratic and selects correct value for $\cos\theta$. If other value given it is A0 unless clearly rejected |
**(c)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = 4$ | B1 | Correct value for $r$ |
---\begin{enumerate}
\item (a) Sketch the polar curve $C$, with equation
\end{enumerate}
$$r = 3 + \sqrt { 5 } \cos \theta \quad 0 \leqslant \theta \leqslant 2 \pi$$
On your sketch clearly label the pole, the initial line and the value of $r$ at the point where the curve intersects the initial line.
The tangent to $C$ at the point $A$, where $0 < \theta < \frac { \pi } { 2 }$, is parallel to the initial line.\\
(b) Use calculus to show that at $A$
$$\cos \theta = \frac { 1 } { \sqrt { 5 } }$$
(c) Hence determine the value of $r$ at $A$.
\hfill \mbox{\textit{Edexcel CP2 2023 Q4 [7]}}