AQA Further Paper 2 2021 June — Question 7 7 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2021
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVolumes of Revolution
TypeSurface area of revolution: parametric curve
DifficultyChallenging +1.8 This is a Further Maths surface area of revolution question requiring parametric differentiation, trigonometric manipulation (including multiple applications of sin²t + cos²t = 1 and double angle formulas), and careful integration. While the setup is standard, the algebraic manipulation of the astroid's derivatives and the subsequent integration demand sustained technical skill beyond typical A-level questions, placing it well above average difficulty.
Spec1.03g Parametric equations: of curves and conversion to cartesian1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=14.08d Volumes of revolution: about x and y axes
7 \includegraphics[max width=\textwidth, alt={}, center]{13abb93f-2fef-465c-980c-3b412de06618-10_854_1027_264_520} The diagram shows a curve known as an astroid.
The curve has parametric equations $$\begin{aligned} & x = 4 \cos ^ { 3 } t \\ & y = 4 \sin ^ { 3 } t \\ & ( 0 \leq t < 2 \pi ) \end{aligned}$$ The section of the curve from \(t = 0\) to \(t = \frac { \pi } { 2 }\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Show that the curved surface area of the shape formed is equal to \(\frac { b \pi } { c }\), where \(b\) and \(c\) are integers.
Question 7:
AnswerMarks Guidance
\(\dot{x} = -12\cos^2 t \sin t\), \(\dot{y} = 12\sin^2 t \cos t\)M1 Obtains derivatives of \(x\) and \(y\)
\(\dot{x}^2 + \dot{y}^2 = 144\cos^4 t \sin^2 t + 144\sin^4 t \cos^2 t\)A1 Correct expression for \(\dot{x}^2 + \dot{y}^2\)
\(= 144\cos^2 t \sin^2 t(\cos^2 t + \sin^2 t) = 144\cos^2 t \sin^2 t\)B1 Uses trig identity to simplify
\(\sqrt{\dot{x}^2 + \dot{y}^2} = 12\cos t \sin t\)
AnswerMarks Guidance
\[S = 2\pi \int_0^{\pi/2} y\sqrt{\dot{x}^2+\dot{y}^2}\,dt\]M1 Substitutes into surface area formula; condone missing limits and missing \(2\pi\)
\[= 2\pi \int_0^{\pi/2} 4\sin^3 t (12\cos t \sin t)\,dt = 96\pi\int_0^{\pi/2}\sin^4 t \cos t\,dt\]A1 Correct expression including correct limits
\[= 96\pi\left[\frac{\sin^5 t}{5}\right]_0^{\pi/2} = \frac{96\pi}{5}\]A1 Obtains \(k\sin^5 t\) by integration; condone missing limits
Completes rigorous argument to show required resultR1 
## Question 7:

$\dot{x} = -12\cos^2 t \sin t$, $\dot{y} = 12\sin^2 t \cos t$ | M1 | Obtains derivatives of $x$ and $y$

$\dot{x}^2 + \dot{y}^2 = 144\cos^4 t \sin^2 t + 144\sin^4 t \cos^2 t$ | A1 | Correct expression for $\dot{x}^2 + \dot{y}^2$

$= 144\cos^2 t \sin^2 t(\cos^2 t + \sin^2 t) = 144\cos^2 t \sin^2 t$ | B1 | Uses trig identity to simplify

$\sqrt{\dot{x}^2 + \dot{y}^2} = 12\cos t \sin t$

$$S = 2\pi \int_0^{\pi/2} y\sqrt{\dot{x}^2+\dot{y}^2}\,dt$$ | M1 | Substitutes into surface area formula; condone missing limits and missing $2\pi$

$$= 2\pi \int_0^{\pi/2} 4\sin^3 t (12\cos t \sin t)\,dt = 96\pi\int_0^{\pi/2}\sin^4 t \cos t\,dt$$ | A1 | Correct expression including correct limits

$$= 96\pi\left[\frac{\sin^5 t}{5}\right]_0^{\pi/2} = \frac{96\pi}{5}$$ | A1 | Obtains $k\sin^5 t$ by integration; condone missing limits

Completes rigorous argument to show required result | R1 |

---
7\\
\includegraphics[max width=\textwidth, alt={}, center]{13abb93f-2fef-465c-980c-3b412de06618-10_854_1027_264_520}

The diagram shows a curve known as an astroid.\\
The curve has parametric equations

$$\begin{aligned}
& x = 4 \cos ^ { 3 } t \\
& y = 4 \sin ^ { 3 } t \\
& ( 0 \leq t < 2 \pi )
\end{aligned}$$

The section of the curve from $t = 0$ to $t = \frac { \pi } { 2 }$ is rotated through $2 \pi$ radians about the $x$-axis.

Show that the curved surface area of the shape formed is equal to $\frac { b \pi } { c }$, where $b$ and $c$ are integers.\\

\hfill \mbox{\textit{AQA Further Paper 2 2021 Q7 [7]}}
This paper (13 questions)
View full paper
Q1 1 Q2 1 Q3 1 Q4 7 Q5 5 Q6 8 Q7 7 Q8 6 Q9 14 Q10 13 Q11 9 Q12 12 Q13 16