| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Convert Cartesian to polar equation |
| Difficulty | Standard +0.3 This is a straightforward Further Maths polar coordinates question requiring algebraic manipulation and standard substitutions (x=r cos θ, y=r sin θ). Part (a) is simple algebra, and part (b) follows a routine procedure taught in FP3. While it's a Further Maths topic, the execution is mechanical with no novel insight required, making it slightly easier than average overall. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x^2 + y^2 = 1 - 2y + y^2 \Rightarrow x^2 + y^2 = (1-y)^2\) | B1 | AG |
| Total | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x^2 + y^2 = r^2\) | M1 | Or \(x = r\cos\theta\) |
| \(y = r\sin\theta\) | M1 | |
| \(x^2 = 1-2y\) so \(x^2 + y^2 = (1-y)^2 \Rightarrow r^2 = (1 - r\sin\theta)^2\) | A1 | OE e.g. \(r^2\cos^2\theta = 1 - 2r\sin\theta\); PI by next line |
| \(r = 1 - r\sin\theta\) or \(r = -(1 - r\sin\theta)\); \(r(1+\sin\theta)=1\) or \(r(1-\sin\theta)=-1\) | m1 | Either |
| \(r > 0\) so \(r = \dfrac{1}{1+\sin\theta}\) | A1 | CSO |
| Total | 6 |
## Question 3(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x^2 + y^2 = 1 - 2y + y^2 \Rightarrow x^2 + y^2 = (1-y)^2$ | B1 | AG |
| **Total** | **1** | |
## Question 3(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x^2 + y^2 = r^2$ | M1 | Or $x = r\cos\theta$ |
| $y = r\sin\theta$ | M1 | |
| $x^2 = 1-2y$ so $x^2 + y^2 = (1-y)^2 \Rightarrow r^2 = (1 - r\sin\theta)^2$ | A1 | OE e.g. $r^2\cos^2\theta = 1 - 2r\sin\theta$; PI by next line |
| $r = 1 - r\sin\theta$ or $r = -(1 - r\sin\theta)$; $r(1+\sin\theta)=1$ or $r(1-\sin\theta)=-1$ | m1 | Either |
| $r > 0$ so $r = \dfrac{1}{1+\sin\theta}$ | A1 | CSO |
| **Total** | **6** | |
---3
\begin{enumerate}[label=(\alph*)]
\item Show that $x ^ { 2 } = 1 - 2 y$ can be written in the form $x ^ { 2 } + y ^ { 2 } = ( 1 - y ) ^ { 2 }$.
\item A curve has cartesian equation $x ^ { 2 } = 1 - 2 y$.
Find its polar equation in the form $r = \mathrm { f } ( \theta )$, given that $r > 0$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2008 Q3 [6]}}