| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| 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 tangent condition formula (dr/dθ = r tan φ) and area integration. Part (a) involves routine differentiation and solving, while part (b) requires setting up and evaluating a polar area integral with some algebraic manipulation. The techniques are well-practiced in FM courses, though the integration requires care with trigonometric identities and logarithms. |
| 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = r\cos\theta = (1+\tan\theta)\cos\theta = \cos\theta + \sin\theta\); \(\frac{dx}{d\theta} = \alpha(1+\tan\theta)\sin\theta + \beta\sec^2\theta\cos\theta\) or \(\frac{dx}{d\theta} = \alpha\sin\theta + \beta\cos\theta\) | M1 | Substitutes equation of \(C\) into \(x = r\cos\theta\) and differentiates |
| \(\frac{dx}{d\theta} = -(1+\tan\theta)\sin\theta + \sec^2\theta\cos\theta\) or \(\frac{dx}{d\theta} = -\sin\theta + \cos\theta\); \(\frac{dx}{d\theta} = -\sin\theta + \sec^2\theta\cos\theta - \tan\theta\sin\theta\) or \(\frac{dx}{d\theta} = -\sin\theta + \sec\theta - \tan\theta\sin\theta\) | A1 | Fully correct differentiation |
| Setting \(\frac{dx}{d\theta} = 0\): \(-\sin\theta + \cos\theta = 0 \Rightarrow \tan\theta = 1 \Rightarrow \theta = \ldots\) (various equivalent approaches shown) | dM1 | Dependent on M1; sets \(\frac{dx}{d\theta}=0\) and uses correct trig identities to find \(\theta\); alternatively substitutes \(\theta=\frac{\pi}{4}\) and shows equals 0 |
| \(r = 1 + \tan\left(\frac{\pi}{4}\right) = 2\), therefore \(A\left(2, \frac{\pi}{4}\right)\) | A1* | Shows \(r=2\) hence polar coordinates \(\left(2, \frac{\pi}{4}\right)\) from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Area bounded by curve \(= \frac{1}{2}\int(1+\tan\theta)^2\, d\theta\) | M1 | Applies area \(= \frac{1}{2}\int r^2\,d\theta\), multiplies out, uses identity \(\pm 1 \pm \tan^2\theta = \sec^2\theta\) to get integrable form and integrates |
| \(= \frac{1}{2}\int(1 + 2\tan\theta + \tan^2\theta)\,d\theta = \frac{1}{2}\int(1 + 2\tan\theta + [\sec^2\theta - 1])\,d\theta\) | ||
| \(= \frac{1}{2}[2\ln | \sec\theta | + \tan\theta]\) or \(\ln |
| \(= \frac{1}{2}\left[2\ln\left | \sec\frac{\pi}{4}\right | + \tan\frac{\pi}{4}\right] - \frac{1}{2}[2\ln |
| Area of triangle \(= \frac{1}{2}xy = \frac{1}{2}\left(2\cos\frac{\pi}{4}\right)\left(2\sin\frac{\pi}{4}\right) = \ldots = 1\) | M1 | Correct method to find area of triangle; area \(= 1\) is M0 only |
| Finds required area \(=\) area of triangle \(-\) area bounded by curve \(= 1 - \left[\ln\sqrt{2} + \frac{1}{2}\right]\) | M1 | Finds required area = area of triangle minus area bounded by curve |
| \(= \frac{1}{2}(1 - \ln 2)\) | A1* | Correct answer, no errors or omissions, cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2}\int(1+\tan\theta)^2\,d\theta = \frac{1}{2}\int(1+2\tan\theta+\tan^2\theta)\,d\theta\); let \(u=\tan\theta \Rightarrow \frac{du}{d\theta}=\sec^2\theta\); leading to \(\frac{1}{2}\int_0^1\frac{1+2u+u^2}{1+u^2}\,du = \frac{1}{2}\int_0^1 1 + \frac{2u}{1+u^2}\,du\) | M1 | Uses substitution \(u=\tan\theta\) to get integrable form and integrates |
| \(\frac{1}{2}[u + \ln(1+u^2)]\) | A1 | Correct integration |
| \(\frac{1}{2}[(1+\ln(1+1^2))-(0+\ln 1)]\) or \(\frac{1}{2}\left[\left(\tan\frac{\pi}{4}+\ln\left(1+\tan^2\frac{\pi}{4}\right)\right)-(\tan 0 + \ln(1+\tan^2 0))\right]\) \(\left\{= \frac{1}{2}\ln 2 + \frac{1}{2}\right\}\) | dM1 | Applies limits \(u=0\) and \(u=1\) or substitutes back and uses \(\theta\) limits |
| Area of triangle \(= \frac{1}{2}xy = \frac{1}{2}\left(2\cos\frac{\pi}{4}\right)\left(2\sin\frac{\pi}{4}\right) = 1\) | M1 | Correct method to find area of triangle |
| Finds required area \(= 1 - \left[\ln\sqrt{2}+\frac{1}{2}\right]\) | M1 | Finds required area = area of triangle minus area bounded by curve |
| \(= \frac{1}{2}(1-\ln 2)\) | A1* | Correct answer, cso |
## Question 7:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = r\cos\theta = (1+\tan\theta)\cos\theta = \cos\theta + \sin\theta$; $\frac{dx}{d\theta} = \alpha(1+\tan\theta)\sin\theta + \beta\sec^2\theta\cos\theta$ or $\frac{dx}{d\theta} = \alpha\sin\theta + \beta\cos\theta$ | M1 | Substitutes equation of $C$ into $x = r\cos\theta$ and differentiates |
| $\frac{dx}{d\theta} = -(1+\tan\theta)\sin\theta + \sec^2\theta\cos\theta$ or $\frac{dx}{d\theta} = -\sin\theta + \cos\theta$; $\frac{dx}{d\theta} = -\sin\theta + \sec^2\theta\cos\theta - \tan\theta\sin\theta$ or $\frac{dx}{d\theta} = -\sin\theta + \sec\theta - \tan\theta\sin\theta$ | A1 | Fully correct differentiation |
| Setting $\frac{dx}{d\theta} = 0$: $-\sin\theta + \cos\theta = 0 \Rightarrow \tan\theta = 1 \Rightarrow \theta = \ldots$ (various equivalent approaches shown) | dM1 | Dependent on M1; sets $\frac{dx}{d\theta}=0$ and uses correct trig identities to find $\theta$; alternatively substitutes $\theta=\frac{\pi}{4}$ and shows equals 0 |
| $r = 1 + \tan\left(\frac{\pi}{4}\right) = 2$, therefore $A\left(2, \frac{\pi}{4}\right)$ | A1* | Shows $r=2$ hence polar coordinates $\left(2, \frac{\pi}{4}\right)$ from correct working |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Area bounded by curve $= \frac{1}{2}\int(1+\tan\theta)^2\, d\theta$ | M1 | Applies area $= \frac{1}{2}\int r^2\,d\theta$, multiplies out, uses identity $\pm 1 \pm \tan^2\theta = \sec^2\theta$ to get integrable form and integrates |
| $= \frac{1}{2}\int(1 + 2\tan\theta + \tan^2\theta)\,d\theta = \frac{1}{2}\int(1 + 2\tan\theta + [\sec^2\theta - 1])\,d\theta$ | | |
| $= \frac{1}{2}[2\ln|\sec\theta| + \tan\theta]$ or $\ln|\sec\theta| + \frac{1}{2}\tan\theta$ or $-\ln\cos\theta + \frac{1}{2}\tan\theta$ | A1 | Correct integration |
| $= \frac{1}{2}\left[2\ln\left|\sec\frac{\pi}{4}\right| + \tan\frac{\pi}{4}\right] - \frac{1}{2}[2\ln|\sec 0| + \tan 0]$ $\left\{= \ln\sqrt{2} + \frac{1}{2}\right\}$ | dM1 | Applies limits $\theta=0$ and $\theta=\frac{\pi}{4}$, subtracts correct way |
| Area of triangle $= \frac{1}{2}xy = \frac{1}{2}\left(2\cos\frac{\pi}{4}\right)\left(2\sin\frac{\pi}{4}\right) = \ldots = 1$ | M1 | Correct method to find area of triangle; area $= 1$ is M0 only |
| Finds required area $=$ area of triangle $-$ area bounded by curve $= 1 - \left[\ln\sqrt{2} + \frac{1}{2}\right]$ | M1 | Finds required area = area of triangle minus area bounded by curve |
| $= \frac{1}{2}(1 - \ln 2)$ | A1* | Correct answer, no errors or omissions, cso |
**Alternative:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\int(1+\tan\theta)^2\,d\theta = \frac{1}{2}\int(1+2\tan\theta+\tan^2\theta)\,d\theta$; let $u=\tan\theta \Rightarrow \frac{du}{d\theta}=\sec^2\theta$; leading to $\frac{1}{2}\int_0^1\frac{1+2u+u^2}{1+u^2}\,du = \frac{1}{2}\int_0^1 1 + \frac{2u}{1+u^2}\,du$ | M1 | Uses substitution $u=\tan\theta$ to get integrable form and integrates |
| $\frac{1}{2}[u + \ln(1+u^2)]$ | A1 | Correct integration |
| $\frac{1}{2}[(1+\ln(1+1^2))-(0+\ln 1)]$ or $\frac{1}{2}\left[\left(\tan\frac{\pi}{4}+\ln\left(1+\tan^2\frac{\pi}{4}\right)\right)-(\tan 0 + \ln(1+\tan^2 0))\right]$ $\left\{= \frac{1}{2}\ln 2 + \frac{1}{2}\right\}$ | dM1 | Applies limits $u=0$ and $u=1$ or substitutes back and uses $\theta$ limits |
| Area of triangle $= \frac{1}{2}xy = \frac{1}{2}\left(2\cos\frac{\pi}{4}\right)\left(2\sin\frac{\pi}{4}\right) = 1$ | M1 | Correct method to find area of triangle |
| Finds required area $= 1 - \left[\ln\sqrt{2}+\frac{1}{2}\right]$ | M1 | Finds required area = area of triangle minus area bounded by curve |
| $= \frac{1}{2}(1-\ln 2)$ | A1* | Correct answer, cso |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{33292670-3ad0-4125-a3bb-e4b7b21ed5f4-22_678_776_248_639}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve $C$ with equation
$$r = 1 + \tan \theta \quad 0 \leqslant \theta < \frac { \pi } { 3 }$$
Figure 1 also shows the tangent to $C$ at the point $A$.\\
This tangent is perpendicular to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Use differentiation to prove that the polar coordinates of $A$ are $\left( 2 , \frac { \pi } { 4 } \right)$
The finite region $R$, shown shaded in Figure 1, is bounded by $C$, the tangent at $A$ and the initial line.
\item Use calculus to show that the exact area of $R$ is $\frac { 1 } { 2 } ( 1 - \ln 2 )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP2 2022 Q7 [10]}}