| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths polar coordinates question requiring: (a) finding conditions on tangent perpendicularity using dy/dx = 0, involving differentiation and trigonometric manipulation; (b) curve sketching; (c) volume calculation using integration. While it requires several FM techniques, the individual steps are relatively standard for CP2 level—the tangent condition follows a known method, and the volume integral is straightforward. The conceptual demand is moderate for FM students who have practiced polar curves. |
| 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)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = r\cos\theta = a(p+2\cos\theta)\cos\theta\), leading to \(\frac{dx}{d\theta} = \alpha\sin\theta\cos\theta + \beta\sin\theta(p+2\cos\theta)\), or \(x = a(\cos 2\theta + p\cos\theta + 1)\) leading to \(\frac{dx}{d\theta} = \alpha\sin 2\theta + \beta\sin\theta\) | M1 | 3.1a — Complete method to find correct form for \(\frac{dx}{d\theta}\) |
| \(\frac{dx}{d\theta} = a\left[-2\sin\theta\cos\theta - \sin\theta(p+2\cos\theta)\right]\) or \(\frac{dx}{d\theta} = -4a\sin\theta\cos\theta - ap\sin\theta\) or \(\frac{dx}{d\theta} = -2a\sin 2\theta - ap\sin\theta\) | A1 | 1.1b — Correct \(\frac{dx}{d\theta}\) |
| \(a\left[-2\sin\theta\cos\theta - \sin\theta(p+2\cos\theta)\right] = 0\), leading to \(a\sin\theta(4\cos\theta + p) = 0\). Either \(\sin\theta = 0\) or \(\cos\theta = -\frac{p}{4}\) | M1 | 3.1a — Sets \(\frac{dx}{d\theta}=0\) and factorises to find values of \(\sin\theta\) or \(\cos\theta\) |
| \(\sin\theta = 0\) implies 2 solutions (tangents perpendicular to initial line) e.g. \(\theta = 0, \pi\) | B1 | 2.2a — Deduces \(\sin\theta=0\) provides two tangents |
| Therefore two solutions to \(\cos\theta = -\frac{p}{4}\) are required. \(-\frac{p}{4} > -1 \Rightarrow p < 4\) as \(p\) is a positive constant \(\therefore\ 2 < p < 4\) | A1* | 2.4 — Concludes \(\cos\theta = -\frac{p}{4} > -1 \Rightarrow p<4\) and \(p\) positive \(\therefore\ 0 < p < 4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape and position (condone cusp) — closed curve with inner loop shape | B1 | 2.2a — Correct shape and position |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Area \(= 2\times\frac{1}{2}\int_0^{\pi}\left[20(3+2\cos\theta)\right]^2 d\theta = 400\int_0^{\pi}(9+12\cos\theta+4\cos^2\theta)\,d\theta\) or \(= \frac{1}{2}\int_0^{2\pi}\left[20(3+2\cos\theta)\right]^2 d\theta\) | M1 | 3.4 — Uses model to find area of cross section with correct limits |
| \(\cos^2\theta = \frac{1}{2}+\frac{1}{2}\cos 2\theta\), leading to \(A = \ldots\int(9+12\cos\theta+2+2\cos 2\theta)\,d\theta = \alpha\theta \pm \beta\sin\theta \pm \lambda\sin 2\theta\) | M1 | 3.1a — Uses identity \(\cos 2\theta = 2\cos^2\theta - 1\) to integrate |
| \(= 400\left[11\theta + 12\sin\theta + \sin 2\theta\right]\) or \(= 200\left[11\theta+12\sin\theta+\sin 2\theta\right]\) | A1 | 1.1b — Correct integration |
| Using limits \(\theta=0\) and \(\theta=\pi\) (or \(\theta=0\) and \(\theta=2\pi\)) and subtracting correctly: \(= 400[11\pi - 0] = 4400\pi = 13823.0\,(\text{cm}^2)\) or \(= 200[11(2\pi)-0] = 4400\pi = 13823.0\,(\text{cm}^2)\) | M1 | 1.1b — Applies limits correctly |
| Volume \(= \text{area} \times 90 = 396\,000\pi = 1\,244\,070.691\,(\text{cm}^3)\) | M1 | 3.4 — Multiplies area by 90 (cm) |
| \(\text{time} = \frac{1\,244\,070.691}{50\,000} = \ldots\) or volume \(= 1244\) litres \(\therefore\) time \(= \frac{1244}{50} = \ldots\) | M1 | 2.2b — Divides volume by 50000 |
| \(25\) (minutes) | A1 | 3.2a — Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. Polar equation is not likely to be accurate. Sides will not be smooth. The hole may not be uniform depth. The pond may leak/ground may absorb some water. | B1 | 3.5b — Any valid limitation of the model (note: any reference to flow of water is B0) |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = r\cos\theta = a(p+2\cos\theta)\cos\theta$, leading to $\frac{dx}{d\theta} = \alpha\sin\theta\cos\theta + \beta\sin\theta(p+2\cos\theta)$, or $x = a(\cos 2\theta + p\cos\theta + 1)$ leading to $\frac{dx}{d\theta} = \alpha\sin 2\theta + \beta\sin\theta$ | M1 | 3.1a — Complete method to find correct form for $\frac{dx}{d\theta}$ |
| $\frac{dx}{d\theta} = a\left[-2\sin\theta\cos\theta - \sin\theta(p+2\cos\theta)\right]$ or $\frac{dx}{d\theta} = -4a\sin\theta\cos\theta - ap\sin\theta$ or $\frac{dx}{d\theta} = -2a\sin 2\theta - ap\sin\theta$ | A1 | 1.1b — Correct $\frac{dx}{d\theta}$ |
| $a\left[-2\sin\theta\cos\theta - \sin\theta(p+2\cos\theta)\right] = 0$, leading to $a\sin\theta(4\cos\theta + p) = 0$. Either $\sin\theta = 0$ or $\cos\theta = -\frac{p}{4}$ | M1 | 3.1a — Sets $\frac{dx}{d\theta}=0$ and factorises to find values of $\sin\theta$ or $\cos\theta$ |
| $\sin\theta = 0$ implies 2 solutions (tangents perpendicular to initial line) e.g. $\theta = 0, \pi$ | B1 | 2.2a — Deduces $\sin\theta=0$ provides two tangents |
| Therefore two solutions to $\cos\theta = -\frac{p}{4}$ are required. $-\frac{p}{4} > -1 \Rightarrow p < 4$ as $p$ is a positive constant $\therefore\ 2 < p < 4$ | A1* | 2.4 — Concludes $\cos\theta = -\frac{p}{4} > -1 \Rightarrow p<4$ and $p$ positive $\therefore\ 0 < p < 4$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape and position (condone cusp) — closed curve with inner loop shape | B1 | 2.2a — Correct shape and position |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Area $= 2\times\frac{1}{2}\int_0^{\pi}\left[20(3+2\cos\theta)\right]^2 d\theta = 400\int_0^{\pi}(9+12\cos\theta+4\cos^2\theta)\,d\theta$ or $= \frac{1}{2}\int_0^{2\pi}\left[20(3+2\cos\theta)\right]^2 d\theta$ | M1 | 3.4 — Uses model to find area of cross section with correct limits |
| $\cos^2\theta = \frac{1}{2}+\frac{1}{2}\cos 2\theta$, leading to $A = \ldots\int(9+12\cos\theta+2+2\cos 2\theta)\,d\theta = \alpha\theta \pm \beta\sin\theta \pm \lambda\sin 2\theta$ | M1 | 3.1a — Uses identity $\cos 2\theta = 2\cos^2\theta - 1$ to integrate |
| $= 400\left[11\theta + 12\sin\theta + \sin 2\theta\right]$ or $= 200\left[11\theta+12\sin\theta+\sin 2\theta\right]$ | A1 | 1.1b — Correct integration |
| Using limits $\theta=0$ and $\theta=\pi$ (or $\theta=0$ and $\theta=2\pi$) and subtracting correctly: $= 400[11\pi - 0] = 4400\pi = 13823.0\,(\text{cm}^2)$ or $= 200[11(2\pi)-0] = 4400\pi = 13823.0\,(\text{cm}^2)$ | M1 | 1.1b — Applies limits correctly |
| Volume $= \text{area} \times 90 = 396\,000\pi = 1\,244\,070.691\,(\text{cm}^3)$ | M1 | 3.4 — Multiplies area by 90 (cm) |
| $\text{time} = \frac{1\,244\,070.691}{50\,000} = \ldots$ or volume $= 1244$ litres $\therefore$ time $= \frac{1244}{50} = \ldots$ | M1 | 2.2b — Divides volume by 50000 |
| $25$ (minutes) | A1 | 3.2a — Correct answer |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. Polar equation is not likely to be accurate. Sides will not be smooth. The hole may not be uniform depth. The pond may leak/ground may absorb some water. | B1 | 3.5b — Any valid limitation of the model (note: any reference to flow of water is B0) |\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$r = a ( p + 2 \cos \theta ) \quad 0 \leqslant \theta < 2 \pi$$
where $a$ and $p$ are positive constants and $p > 2$\\
There are exactly four points on $C$ where the tangent is perpendicular to the initial line.\\
(a) Show that the range of possible values for $p$ is
$$2 < p < 4$$
(b) Sketch the curve with equation
$$r = a ( 3 + 2 \cos \theta ) \quad 0 \leqslant \theta < 2 \pi \quad \text { where } a > 0$$
John digs a hole in his garden in order to make a pond.\\
The pond has a uniform horizontal cross section that is modelled by the curve with equation
$$r = 20 ( 3 + 2 \cos \theta ) \quad 0 \leqslant \theta < 2 \pi$$
where $r$ is measured in centimetres.
The depth of the pond is 90 centimetres.\\
Water flows through a hosepipe into the pond at a rate of 50 litres per minute.\\
Given that the pond is initially empty,\\
(c) determine how long it will take to completely fill the pond with water using the hosepipe, according to the model. Give your answer to the nearest minute.\\
(d) State a limitation of the model.
\hfill \mbox{\textit{Edexcel CP2 2021 Q6 [14]}}