| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Emblem or applied region area |
| Difficulty | Standard +0.3 This is a straightforward parametric area question requiring standard techniques: applying the parametric area formula, using the double angle identity sin(2t) = 2sin(t)cos(t), and integrating sin(t)cos²(t) by substitution. Part (b) involves simple coordinate substitution. All steps are routine A-level procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y \times \dfrac{dx}{dt} = 5\sin 2t \times 6\cos t\) or \(5\times 2\sin t\cos t \times 6\cos t\) | M1 | 1.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int 5\sin 2t \times 6\cos t \, dt = \int 5 \times 2\sin t\cos t \times 6\cos t \, dt\) or \(\int 5\sin 2t \times 6\cos t \, dt = \int 60\sin t\cos^2 t \, dt\) | dM1 | Attempts to use \(\sin 2t = 2\sin t\cos t\) within an integral |
| \((\text{Area} =) \int_0^{\frac{\pi}{2}} 60\sin t\cos^2 t \, dt\) | A1* | Fully correct work leading to given answer; must include \(\sin 2t = 2\sin t\cos t\) explicitly |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int 60\sin t\cos^2 t \, dt = -20\cos^3 t\) | M1 | Obtains \(k\cos^3 t\); may use substitution \(u = \cos t\) |
| \(-20\cos^3 t\) (correct integration) | A1 | Correct integration; equivalent e.g. \(-20u^3\) |
| \(\text{Area} = \left[-20\cos^3 t\right]_0^{\frac{\pi}{2}} = 0-(-20) = 20\) | A1* | Rigorous proof with correct limits and \(0-(-20)\); not just \(-20\cos^3\frac{\pi}{2}-(-20\cos^3 0)=20\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5\sin 2t = 4.2 \Rightarrow \sin 2t = \frac{4.2}{5}\) | M1 | Uses model to find values of \(t\) when \(\sin 2t = \frac{4.2}{5}\) |
| \(t = 0.4986\ldots, 1.072\ldots\) | A1 | At least one correct value to 2 d.p. |
| Attempts to find \(x\) values at both \(t\) values | dM1 | Condone poor trig work |
| \(t = 0.4986\ldots \Rightarrow x = 2.869\ldots\); \(t = 1.072 \Rightarrow x = 5.269\ldots\) | A1 | Both values correct to 2 d.p. |
| Width of path \(= 2.40\) metres | A1 | Allow awrt 2.40 m or 240 cm; units required |
# Question 12:
## Part (a)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $y \times \dfrac{dx}{dt} = 5\sin 2t \times 6\cos t$ or $5\times 2\sin t\cos t \times 6\cos t$ | M1 | 1.2 |
## Question (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 5\sin 2t \times 6\cos t \, dt = \int 5 \times 2\sin t\cos t \times 6\cos t \, dt$ or $\int 5\sin 2t \times 6\cos t \, dt = \int 60\sin t\cos^2 t \, dt$ | dM1 | Attempts to use $\sin 2t = 2\sin t\cos t$ within an integral |
| $(\text{Area} =) \int_0^{\frac{\pi}{2}} 60\sin t\cos^2 t \, dt$ | A1* | Fully correct work leading to given answer; must include $\sin 2t = 2\sin t\cos t$ explicitly |
## Question (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 60\sin t\cos^2 t \, dt = -20\cos^3 t$ | M1 | Obtains $k\cos^3 t$; may use substitution $u = \cos t$ |
| $-20\cos^3 t$ (correct integration) | A1 | Correct integration; equivalent e.g. $-20u^3$ |
| $\text{Area} = \left[-20\cos^3 t\right]_0^{\frac{\pi}{2}} = 0-(-20) = 20$ | A1* | Rigorous proof with correct limits and $0-(-20)$; not just $-20\cos^3\frac{\pi}{2}-(-20\cos^3 0)=20$ |
## Question (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5\sin 2t = 4.2 \Rightarrow \sin 2t = \frac{4.2}{5}$ | M1 | Uses model to find values of $t$ when $\sin 2t = \frac{4.2}{5}$ |
| $t = 0.4986\ldots, 1.072\ldots$ | A1 | At least one correct value to 2 d.p. |
| Attempts to find $x$ values at both $t$ values | dM1 | Condone poor trig work |
| $t = 0.4986\ldots \Rightarrow x = 2.869\ldots$; $t = 1.072 \Rightarrow x = 5.269\ldots$ | A1 | Both values correct to 2 d.p. |
| Width of path $= 2.40$ metres | A1 | Allow awrt 2.40 m or 240 cm; units required |
**Cartesian approach:** $5\sin 2t = 4.2 \Rightarrow 10\frac{x}{6}\sqrt{1-\frac{x^2}{36}} = 4.2 \Rightarrow x^4 - 36x^2 + 228.6144 = 0 \Rightarrow x = 2.869\ldots, 5.269\ldots$
---12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e28350e9-5090-4079-97da-e669ef9a5a7a-34_396_515_251_772}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The curve shown in Figure 3 has parametric equations
$$x = 6 \sin t \quad y = 5 \sin 2 t \quad 0 \leqslant t \leqslant \frac { \pi } { 2 }$$
The region $R$, shown shaded in Figure 3, is bounded by the curve and the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the area of $R$ is given by $\int _ { 0 } ^ { \frac { \pi } { 2 } } 60 \sin t \cos ^ { 2 } t \mathrm {~d} t$
\item Hence show, by algebraic integration, that the area of $R$ is exactly 20
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e28350e9-5090-4079-97da-e669ef9a5a7a-34_451_570_1416_742}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Part of the curve is used to model the profile of a small dam, shown shaded in Figure 4. Using the model and given that
\begin{itemize}
\end{enumerate}\item $x$ and $y$ are in metres
\item the vertical wall of the dam is 4.2 metres high
\item there is a horizontal walkway of width $M N$ along the top of the dam
\item calculate the width of the walkway.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2020 Q12 [11]}}