| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Deduce integral value from area |
| Difficulty | Challenging +1.8 This is a sophisticated multi-part question requiring conversion between Cartesian and polar coordinates, area calculations in both systems, and deducing an integral value by equating areas. Part (ii) requires algebraic manipulation to derive the polar form, and part (iii) demands understanding that the polar area formula ½∫r²dθ equals the Cartesian area from part (i). While systematic, it requires strong conceptual understanding of coordinate systems and multi-step reasoning across three connected parts, placing it well above average difficulty for Further Maths. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempt integral as \(k(2x+1)^{1.5}\) | M1 | |
| Get 9 | A1 | cao |
| Attempt subtraction of areas (triangle) | M1 | Their answer minus triangle |
| Get 3 | A1\(\checkmark\) | Their answer \(- 6\) (\(>0\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(r^2=x^2+y^2\) and \(x=r\cos\theta\), \(y=r\sin\theta\) | B1 | |
| Eliminate \(x\) and \(y\) to produce quadratic equation (\(=0\)) in \(r\) (or \(\cos\theta\)) | M1 | |
| Solve their quadratic to get \(r\) in terms of \(\theta\) | A1\(\checkmark\) | |
| Clearly get A.G. | A1 | \(r>0\) may be assumed |
| Clearly show \(\theta_1\)(at \(B\))\(=\tan^{-1}\frac{3}{4}\) and \(\theta_2\)(at \(A\))\(=\pi\) | B1 | |
| SC: Eliminate \(y\) to get \(r\) in terms of \(x\) only | M1 | |
| Get \(r = x+1\) | A1 | |
| SC: Start with \(r=1/(1-\cos\theta)\) and derive Cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use area \(= \frac{1}{2}\int r^2\,d\theta\) with correct \(r\) | B1 | cwo; ignore limits |
| Rewrite as \(k\csc^4(\frac{1}{2}\theta)\) | M1 | Not just quoted |
| Equate to their part (i) and tidy | M1 | To get \(\int =\) some constant |
| Get 24 | A1 | A.G. |
# Question 9:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempt integral as $k(2x+1)^{1.5}$ | M1 | |
| Get 9 | A1 | cao |
| Attempt subtraction of areas (triangle) | M1 | Their answer minus triangle |
| Get 3 | A1$\checkmark$ | Their answer $- 6$ ($>0$) |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $r^2=x^2+y^2$ and $x=r\cos\theta$, $y=r\sin\theta$ | B1 | |
| Eliminate $x$ and $y$ to produce quadratic equation ($=0$) in $r$ (or $\cos\theta$) | M1 | |
| Solve their quadratic to get $r$ in terms of $\theta$ | A1$\checkmark$ | |
| Clearly get A.G. | A1 | $r>0$ may be assumed |
| Clearly show $\theta_1$(at $B$)$=\tan^{-1}\frac{3}{4}$ and $\theta_2$(at $A$)$=\pi$ | B1 | |
| SC: Eliminate $y$ to get $r$ in terms of $x$ only | M1 | |
| Get $r = x+1$ | A1 | |
| SC: Start with $r=1/(1-\cos\theta)$ and derive Cartesian | | |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use area $= \frac{1}{2}\int r^2\,d\theta$ with correct $r$ | B1 | cwo; ignore limits |
| Rewrite as $k\csc^4(\frac{1}{2}\theta)$ | M1 | Not just quoted |
| Equate to their part (i) and tidy | M1 | To get $\int =$ some constant |
| Get 24 | A1 | A.G. |9\\
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The diagram shows the curve with equation $y = \sqrt { 2 x + 1 }$ between the points $A \left( - \frac { 1 } { 2 } , 0 \right)$ and $B ( 4,3 )$.\\
(i) Find the area of the region bounded by the curve, the $x$-axis and the line $x = 4$. Hence find the area of the region bounded by the curve and the lines $O A$ and $O B$, where $O$ is the origin.\\
(ii) Show that the curve between $B$ and $A$ can be expressed in polar coordinates as
$$r = \frac { 1 } { 1 - \cos \theta } , \quad \text { where } \tan ^ { - 1 } \left( \frac { 3 } { 4 } \right) \leqslant \theta \leqslant \pi$$
(iii) Deduce from parts (i) and (ii) that $\int _ { \tan ^ { - 1 } \left( \frac { 3 } { 4 } \right) } ^ { \pi } \operatorname { cosec } ^ { 4 } \left( \frac { 1 } { 2 } \theta \right) \mathrm { d } \theta = 24$.
www.ocr.org.uk after the live examination series.\\
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\hfill \mbox{\textit{OCR FP2 2010 Q9 [13]}}