| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2023 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric differentiation |
| Type | Find tangent equation at parameter |
| Difficulty | Standard +0.3 This is a standard parametric equations question covering routine techniques: finding intercepts, parametric differentiation using the chain rule, finding a tangent equation, and volume of revolution. All parts follow textbook methods with no novel insight required. Part (b) is a 'show that' which guides students to the answer. The trigonometric identities needed (double angle formulas, cosec) are standard P4 content. Slightly easier than average due to the structured, multi-part nature that scaffolds the solution. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3\pi\) | B1 | Condone \((3\pi, 0)\) and allow \(x = 3\pi\). Do not allow decimals. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-2\sin t}{6 - 6\cos 2t}\) | M1 A1 | Attempts to differentiate \(x(t)\) and \(y(t)\) and calculates \(\frac{\mathrm{d}y}{\mathrm{d}x}\) by using \(\frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t}\). Condone poor differentiation but one term must be correct form; requires \(x \to \alpha + \beta\cos 2t\) or \(y \to A\sin t\) |
| \(= \dfrac{-2\sin t}{6-6(1-2\sin^2 t)} = \dfrac{-2\sin t}{12\sin^2 t} = -\dfrac{1}{6}\cosec t\) | M1 A1* | Uses appropriate trig to reach \(\frac{\mathrm{d}y}{\mathrm{d}x} = \alpha\cosec t\). Condone sign errors only. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P = \left(\dfrac{3\pi}{2}-3,\ \sqrt{2}\right)\) | B1 | At \(P\), \(x=\frac{3\pi}{2}-3\), \(y=\sqrt{2}\) or exact equivalents. May be implied by tangent equation. |
| Uses point and gradient to form tangent equation | M1 | Requires: substitution of \(t=\frac{\pi}{4}\) into \(\frac{\mathrm{d}y}{\mathrm{d}x} = \alpha\cosec t\) to find gradient; use of \(\left(\frac{3\pi}{2}-3, \sqrt{2}\right)\) in tangent equation; setting \(x=0\) to find \(y\) or finding \(c\) |
| \(\dfrac{\pi\sqrt{2}}{4} + \dfrac{\sqrt{2}}{2}\) o.e. | A1 | E.g. \(\dfrac{\pi}{2\sqrt{2}}+\dfrac{1}{\sqrt{2}}\). Does not need to be identified as \(y\) value. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Attempts \(y^2\dfrac{\mathrm{d}x}{\mathrm{d}t} = 4\cos^2 t\left(6-6\cos 2t\right)\) | M1 A1 | Condone poor notation e.g. \(\cos t^2\) for \(\cos^2 t\) as long as recovered. |
| \(= (2\cos 2t+2)(6-6\cos 2t) = 6\!\left(2-2\cos^2 2t\right) = 6\!\left(2-(1+\cos 4t)\right)\) | dM1 | Full method to express in terms of \(\cos 4t\). Condone sign errors. Depends on first M mark. Many valid approaches. |
| Volume \(= \displaystyle\int_0^{\pi/2} \pi y^2\,\dfrac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t = \int_0^{\pi/2} 6\pi(1-\cos 4t)\,\mathrm{d}t\) | A1 | All correct with correct limits and including "\(\mathrm{d}t\)". |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left[6\pi t - \dfrac{6\pi\sin 4t}{4}\right]_0^{\pi/2} = 3\pi^2\) | M1, A1 | M1: \(\int\beta(1-\cos 4t)\,\mathrm{d}t \to \beta\!\left(t - \dfrac{\sin 4t}{4}\right)\), allow made-up \(\beta\). A1: \(3\pi^2\) following full marks in (d)(i), correct integration and correct limits. |
# Question 8:
**(a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $3\pi$ | B1 | Condone $(3\pi, 0)$ and allow $x = 3\pi$. Do not allow decimals. |
**(b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-2\sin t}{6 - 6\cos 2t}$ | M1 A1 | Attempts to differentiate $x(t)$ and $y(t)$ and calculates $\frac{\mathrm{d}y}{\mathrm{d}x}$ by using $\frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t}$. Condone poor differentiation but one term must be correct form; requires $x \to \alpha + \beta\cos 2t$ or $y \to A\sin t$ |
| $= \dfrac{-2\sin t}{6-6(1-2\sin^2 t)} = \dfrac{-2\sin t}{12\sin^2 t} = -\dfrac{1}{6}\cosec t$ | M1 A1* | Uses appropriate trig to reach $\frac{\mathrm{d}y}{\mathrm{d}x} = \alpha\cosec t$. Condone sign errors only. |
**(c)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $P = \left(\dfrac{3\pi}{2}-3,\ \sqrt{2}\right)$ | B1 | At $P$, $x=\frac{3\pi}{2}-3$, $y=\sqrt{2}$ or exact equivalents. May be implied by tangent equation. |
| Uses point and gradient to form tangent equation | M1 | Requires: substitution of $t=\frac{\pi}{4}$ into $\frac{\mathrm{d}y}{\mathrm{d}x} = \alpha\cosec t$ to find gradient; use of $\left(\frac{3\pi}{2}-3, \sqrt{2}\right)$ in tangent equation; setting $x=0$ to find $y$ or finding $c$ |
| $\dfrac{\pi\sqrt{2}}{4} + \dfrac{\sqrt{2}}{2}$ o.e. | A1 | E.g. $\dfrac{\pi}{2\sqrt{2}}+\dfrac{1}{\sqrt{2}}$. Does not need to be identified as $y$ value. |
**(d)(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Attempts $y^2\dfrac{\mathrm{d}x}{\mathrm{d}t} = 4\cos^2 t\left(6-6\cos 2t\right)$ | M1 A1 | Condone poor notation e.g. $\cos t^2$ for $\cos^2 t$ as long as recovered. |
| $= (2\cos 2t+2)(6-6\cos 2t) = 6\!\left(2-2\cos^2 2t\right) = 6\!\left(2-(1+\cos 4t)\right)$ | dM1 | Full method to express in terms of $\cos 4t$. Condone sign errors. Depends on first M mark. Many valid approaches. |
| Volume $= \displaystyle\int_0^{\pi/2} \pi y^2\,\dfrac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t = \int_0^{\pi/2} 6\pi(1-\cos 4t)\,\mathrm{d}t$ | A1 | All correct with correct limits and including "$\mathrm{d}t$". |
**(d)(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[6\pi t - \dfrac{6\pi\sin 4t}{4}\right]_0^{\pi/2} = 3\pi^2$ | M1, A1 | M1: $\int\beta(1-\cos 4t)\,\mathrm{d}t \to \beta\!\left(t - \dfrac{\sin 4t}{4}\right)$, allow made-up $\beta$. A1: $3\pi^2$ following full marks in (d)(i), correct integration and correct limits. |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7f5fc83d-ab7c-4edb-a2c6-7a58f1357d5a-24_579_642_251_715}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a sketch of the curve $C$ with parametric equations
$$x = 6 t - 3 \sin 2 t \quad y = 2 \cos t \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$
The curve meets the $y$-axis at 2 and the $x$-axis at $k$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item State the value of $k$.
\item Use parametric differentiation to show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \lambda \operatorname { cosec } t$$
where $\lambda$ is a constant to be found.
The point $P$ with parameter $\mathrm { t } = \frac { \pi } { 4 }$ lies on $C$.\\
The tangent to $C$ at the point $P$ cuts the $y$-axis at the point $N$.
\item Find the exact $y$ coordinate of $N$, giving your answer in simplest form.
The region bounded by the curve, the $x$-axis and the $y$-axis is rotated through $2 \pi$ radians about the $x$-axis to form a solid of revolution.
\item \begin{enumerate}[label=(\roman*)]
\item Show that the volume of this solid is given by
$$\int _ { 0 } ^ { \alpha } \beta ( 1 - \cos 4 t ) d t$$
where $\alpha$ and $\beta$ are constants to be found.
\item Hence, using algebraic integration, find the exact volume of this solid.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2023 Q8 [14]}}