| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Find normal equation |
| Difficulty | Standard +0.3 This is a straightforward parametric equations question requiring routine techniques: substitution to verify a point, differentiation of sin²y using chain rule, reciprocal for dy/dx, finding a normal equation, and calculating a triangle area. All steps are standard P3 procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((k=) \ 4\sin^2\left(\frac{\pi}{3}\right)-1=2\) | B1* | Verifies \(k=2\) with no errors seen. Condone \(x=2\). Do not allow if rounded numbers are seen, e.g. do not allow \(k=4\sin\left(\frac{\pi}{3}\right)^2-1=2\). Working in degrees substituting \(60°\) is acceptable. Must conclude \(k=2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x=4\sin^2 y-1 \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}y}=8\sin y\cos y\) | M1 A1 | M1: differentiates to form \(p\sin y\cos y\) or equivalent. A1: correct lhs and rhs. Note \(\frac{\mathrm{d}y}{\mathrm{d}x}=\ldots\) is A0. May use identity \(\cos 2y = \pm1\pm2\sin^2 y \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}y}=q\sin 2y\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempts \(\sin^2 y = \frac{\pm x\pm1}{4}\) or \(\cos^2 y=\pm1\pm\frac{x+1}{4}\) | M1 | Attempts to find \(\sin y\) or \(\sin^2 y\) or \(\cos y\) or \(\cos^2 y\) in terms of \(x\) |
| Both \(\sin^2 y\) and \(\cos^2 y\) found in terms of \(x\) | dM1 | Dependent on previous M1. Must find both |
| \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{8\sin y\cos y}=\frac{1}{8\times\sqrt{\frac{x+1}{4}}\times\sqrt{\pm1\pm\frac{x+1}{4}}}=\frac{1}{2\sqrt{x+1}\sqrt{3-x}}\) | ddM1 A1* | ddM1: full method to get \(\frac{\mathrm{d}y}{\mathrm{d}x}\) of form \(\frac{1}{p\sin y\cos y}\) in terms of \(x\). A1*: given answer, all previous marks scored, no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \(x=2\), gradient of curve \(=\frac{1}{2\sqrt{3}} \Rightarrow\) gradient of normal \(=-2\sqrt{3}\) | M1 | Full method for gradient of normal using \(x\) or \(y\) value. Must be negative reciprocal of tangent gradient |
| \(y-\frac{\pi}{3}=-2\sqrt{3}(x-2) \Rightarrow x=2+\frac{\pi}{6\sqrt{3}}\) \((=2.3\ldots)\) | dM1 | Dependent on previous M1. Uses point \(\left(2,\frac{\pi}{3}\right)\), sets \(y=0\), proceeds to find \(x\) |
| \(\text{Area}=\frac{1}{2}\times\left(2+\frac{\pi}{6\sqrt{3}}\right)\times\frac{\pi}{3}=\frac{\pi}{3}+\frac{\pi^2}{36\sqrt{3}}\) | A1 | Accept exact equivalent e.g. \(\frac{\pi}{3}+\frac{\pi^2\sqrt{3}}{108}\), in form \(a\pi+b\pi^2\) |
# Question 9:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(k=) \ 4\sin^2\left(\frac{\pi}{3}\right)-1=2$ | B1* | Verifies $k=2$ with no errors seen. Condone $x=2$. Do not allow if rounded numbers are seen, e.g. do not allow $k=4\sin\left(\frac{\pi}{3}\right)^2-1=2$. Working in degrees substituting $60°$ is acceptable. Must conclude $k=2$ |
**(1 mark)**
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x=4\sin^2 y-1 \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}y}=8\sin y\cos y$ | M1 A1 | M1: differentiates to form $p\sin y\cos y$ or equivalent. A1: correct lhs and rhs. Note $\frac{\mathrm{d}y}{\mathrm{d}x}=\ldots$ is A0. May use identity $\cos 2y = \pm1\pm2\sin^2 y \Rightarrow \frac{\mathrm{d}x}{\mathrm{d}y}=q\sin 2y$ |
**(2 of 6 marks)**
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempts $\sin^2 y = \frac{\pm x\pm1}{4}$ or $\cos^2 y=\pm1\pm\frac{x+1}{4}$ | M1 | Attempts to find $\sin y$ or $\sin^2 y$ **or** $\cos y$ or $\cos^2 y$ in terms of $x$ |
| Both $\sin^2 y$ and $\cos^2 y$ found in terms of $x$ | dM1 | Dependent on previous M1. Must find both |
| $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{8\sin y\cos y}=\frac{1}{8\times\sqrt{\frac{x+1}{4}}\times\sqrt{\pm1\pm\frac{x+1}{4}}}=\frac{1}{2\sqrt{x+1}\sqrt{3-x}}$ | ddM1 A1* | ddM1: full method to get $\frac{\mathrm{d}y}{\mathrm{d}x}$ of form $\frac{1}{p\sin y\cos y}$ in terms of $x$. A1*: given answer, all previous marks scored, no errors |
**(4 of 6 marks, total 6)**
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $x=2$, gradient of curve $=\frac{1}{2\sqrt{3}} \Rightarrow$ gradient of normal $=-2\sqrt{3}$ | M1 | Full method for gradient of normal using $x$ or $y$ value. Must be negative reciprocal of tangent gradient |
| $y-\frac{\pi}{3}=-2\sqrt{3}(x-2) \Rightarrow x=2+\frac{\pi}{6\sqrt{3}}$ $(=2.3\ldots)$ | dM1 | Dependent on previous M1. Uses point $\left(2,\frac{\pi}{3}\right)$, sets $y=0$, proceeds to find $x$ |
| $\text{Area}=\frac{1}{2}\times\left(2+\frac{\pi}{6\sqrt{3}}\right)\times\frac{\pi}{3}=\frac{\pi}{3}+\frac{\pi^2}{36\sqrt{3}}$ | A1 | Accept exact equivalent e.g. $\frac{\pi}{3}+\frac{\pi^2\sqrt{3}}{108}$, in form $a\pi+b\pi^2$ |
**(3 marks, total 10)**9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5a695b86-1660-4c06-ac96-4cdb07af9a2e-30_714_1079_251_495}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
The curve shown in Figure 5 has equation
$$x = 4 \sin ^ { 2 } y - 1 \quad 0 \leqslant y \leqslant \frac { \pi } { 2 }$$
The point $P \left( k , \frac { \pi } { 3 } \right)$ lies on the curve.
\begin{enumerate}[label=(\alph*)]
\item Verify that $k = 2$
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } x } { \mathrm {~d} y }$ in terms of $y$
\item Hence show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 1 } { 2 \sqrt { x + 1 } \sqrt { 3 - x } }$
The normal to the curve at $P$ cuts the $x$-axis at the point $N$.
\end{enumerate}\item Find the exact area of triangle $O P N$, where $O$ is the origin.
Give your answer in the form $a \pi + b \pi ^ { 2 }$ where $a$ and $b$ are constants.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2024 Q9 [10]}}