Edexcel FP2 2007 June — Question 8 14 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2007
SessionJune
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolar coordinates
TypeTangent parallel/perpendicular to initial line
DifficultyChallenging +1.2 This is a Further Maths FP2 polar coordinates question requiring sketching a limaçon, finding tangent conditions using dy/dx = 0, and computing area with polar integration. While these are standard FP2 techniques, the multi-part nature and requirement to work with polar calculus formulas places it moderately above average A-level difficulty.
Spec1.07r 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)4.09c Area enclosed: by polar curve

8. (a) Sketch the curve \(C\) with polar equation $$r = 5 + \sqrt { 3 } \cos \theta , \quad 0 \leq \theta \leq 2 \pi$$ (b) Find the polar coordinates of the points where the tangents to \(C\) are parallel to the initial line \(\theta = 0\). Give your answers to 3 significant figures where appropriate.
(c) Using integration, find the area enclosed by the curve \(C\), giving your answer in terms of \(\pi\).

AnswerMarks Guidance
(a) Shape (close curve, approx. symmetrical about the initial line, in all 'quadrants' and 'centred' to the right of the pole/origin)B1
Shape (at least one correct 'intercept' r value... shown on sketch or perhaps seen in a table)B12
(Also allow awrt 3.27 or awrt 6.73)
(b) \(x = r\sin\theta = 5\sin\theta + \sqrt{3}\sin\theta\cos\theta\)M1
\(\frac{d\theta}{d\theta} = 5\cos\theta - \sqrt{3}\sin^2\theta + \sqrt{3}\cos^2\theta(= 5\cos\theta + \sqrt{3}\cos 2\theta)\)A1
\(5\cos\theta - \sqrt{3}(1 - \cos^2\theta) + \sqrt{3}\cos^2\theta = 0\)M1
\(2\sqrt{3}\cos^2\theta + 5\cos\theta - \sqrt{3} = 0\)
\((2\sqrt{3}\cos\theta - 1)(\cos\theta + \sqrt{3}) = 0\)M1
\(\cos\theta = \ldots (0.288...)\) Also allow \(\pm \arccos\frac{1}{2\sqrt{3}}\)
\(\theta = 1.28\) and 5.01 (awrt) (Allow \(\pm 1.28\) awrt)A1
\(r = 5 + \sqrt{3}\left(\frac{2\sqrt{3}}{2}\right) = \frac{11}{2}\) (Allow awrt 5.50)A16
2nd M: Forming a quadratic in \(\cos\theta\)
3rd M: Solving a 3 term quadratic to find a value of \(\cos\theta\) (even if called \(\theta\))
Special case: Working with \(r\cos\theta = 5\sin\theta + \sqrt{3}\cos^2\theta\). 1st M1 for \(r\cos\theta = 5\sin\theta + \sqrt{3}\cos^2\theta\). 1st A1 for derivative \(= -5\sin\theta - 2\sqrt{3}\sin\theta\cos\theta\), then no further marks
(c) \(r^2 = 25 + 10\sqrt{3}\cos\theta + 3\cos^2\theta\)B1
\(\int(25 + 10\sqrt{3}\cos\theta + 3\cos^2\theta) d\theta = \frac{53\theta}{2} + 10\sqrt{3}\sin\theta + 3\left(\frac{\sin 2\theta}{4}\right)\)M1 A1ft A1ft
(It for integration of \((a + b\cos\theta)\) and \(c\cos 2\theta\) respectively)
\(\frac{1}{2}\left[25\theta + 10\sqrt{3}\sin\theta + \frac{3\sin 2\theta}{4} + \frac{3\theta}{2}\right]_0^{\pi/4} = \ldots\)M1
\(= \frac{1}{2}(50\pi + 3\pi) = \frac{53\pi}{2}\) or equiv. in terms of \(\pi\)A16
1st M: Attempt to integrate at least one term
2nd M: Requires use of the \(\frac{1}{2}\), correct limits (which could be 0 to \(2\pi\), or \(-\pi\) to \(\pi\), or 'double' 0 to \(\pi\)), and subtraction (which could be implied)
Total: [14]
**(a)** Shape (close curve, approx. symmetrical about the initial line, in all 'quadrants' and 'centred' to the right of the pole/origin) | B1 |
Shape (at least one correct 'intercept' r value... shown on sketch or perhaps seen in a table) | B12 |
(Also allow awrt 3.27 or awrt 6.73) | |

**(b)** $x = r\sin\theta = 5\sin\theta + \sqrt{3}\sin\theta\cos\theta$ | M1 |
$\frac{d\theta}{d\theta} = 5\cos\theta - \sqrt{3}\sin^2\theta + \sqrt{3}\cos^2\theta(= 5\cos\theta + \sqrt{3}\cos 2\theta)$ | A1 |

$5\cos\theta - \sqrt{3}(1 - \cos^2\theta) + \sqrt{3}\cos^2\theta = 0$ | M1 |
$2\sqrt{3}\cos^2\theta + 5\cos\theta - \sqrt{3} = 0$ | |
$(2\sqrt{3}\cos\theta - 1)(\cos\theta + \sqrt{3}) = 0$ | M1 |
$\cos\theta = \ldots (0.288...)$ | | Also allow $\pm \arccos\frac{1}{2\sqrt{3}}$
$\theta = 1.28$ and 5.01 (awrt) (Allow $\pm 1.28$ awrt) | A1 |
$r = 5 + \sqrt{3}\left(\frac{2\sqrt{3}}{2}\right) = \frac{11}{2}$ (Allow awrt 5.50) | A16 |

**2nd M:** Forming a quadratic in $\cos\theta$ | |

**3rd M:** Solving a 3 term quadratic to find a value of $\cos\theta$ (even if called $\theta$) | |

**Special case:** Working with $r\cos\theta = 5\sin\theta + \sqrt{3}\cos^2\theta$. 1st M1 for $r\cos\theta = 5\sin\theta + \sqrt{3}\cos^2\theta$. 1st A1 for derivative $= -5\sin\theta - 2\sqrt{3}\sin\theta\cos\theta$, then no further marks | |

**(c)** $r^2 = 25 + 10\sqrt{3}\cos\theta + 3\cos^2\theta$ | B1 |
$\int(25 + 10\sqrt{3}\cos\theta + 3\cos^2\theta) d\theta = \frac{53\theta}{2} + 10\sqrt{3}\sin\theta + 3\left(\frac{\sin 2\theta}{4}\right)$ | M1 A1ft A1ft |

(It for integration of $(a + b\cos\theta)$ and $c\cos 2\theta$ respectively) | |

$\frac{1}{2}\left[25\theta + 10\sqrt{3}\sin\theta + \frac{3\sin 2\theta}{4} + \frac{3\theta}{2}\right]_0^{\pi/4} = \ldots$ | M1 |
$= \frac{1}{2}(50\pi + 3\pi) = \frac{53\pi}{2}$ or equiv. in terms of $\pi$ | A16 |

**1st M:** Attempt to integrate at least one term | |

**2nd M:** Requires use of the $\frac{1}{2}$, correct limits (which could be 0 to $2\pi$, or $-\pi$ to $\pi$, or 'double' 0 to $\pi$), and subtraction (which could be implied) | |

**Total: [14]**

---
8. (a) Sketch the curve $C$ with polar equation

$$r = 5 + \sqrt { 3 } \cos \theta , \quad 0 \leq \theta \leq 2 \pi$$

(b) Find the polar coordinates of the points where the tangents to $C$ are parallel to the initial line $\theta = 0$. Give your answers to 3 significant figures where appropriate.\\
(c) Using integration, find the area enclosed by the curve $C$, giving your answer in terms of $\pi$.\\

\hfill \mbox{\textit{Edexcel FP2 2007 Q8 [14]}}