June 2023 Paper 3 Q5
5 A mathematics department is designing a new emblem to place on the walls outside its classrooms. The design for the emblem is shown in the diagram below.
The emblem is modelled by the region between the \(x\)-axis and the curve with parametric equations
\(x = 1 + 0.2t - \cos t, \qquad y = k\sin^2 t,\)
where \(k\) is a positive constant and \(0 \leqslant t \leqslant \pi\).
Lengths are in metres and the area of the emblem must be \(1\,\mathrm{m}^2\).
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0.2 + \sin t\) | B1* | 1.1 |
| \(\displaystyle\int_0^{\pi} k\sin^2 t(0.2 + \sin t)\,\mathrm{d}t \Rightarrow\) \(\displaystyle\int_0^{\pi} k(1 - \cos^2 t)(0.2 + \sin t)\,\mathrm{d}t\) | M1dep* | 3.1a |
| \(\displaystyle k\int_0^{\pi} (0.2 + \sin t - 0.2\cos^2 t - \sin t\cos^2 t)\,\mathrm{d}t\ (= 1)\) | A1 | 2.2a |
| [3] |
Notes
B1*: B1 for \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0.2 \pm \sin t\)
M1dep*: Uses \(\displaystyle\int y\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t\) and replaces \(\sin^2 t\) with \(1 - \cos^2 t\) to obtain an expression involving \(\cos^2 t\) and \(\sin t\)
No limits required for this mark
A1: AG (so must be checked carefully for any errors e.g. must contain relevant brackets around the \(1 - \cos^2 t\) term(s)). A correct expression e.g. \(\displaystyle\int_0^{\pi} k(1 - \cos^2 t)(0.2 + \sin t)\,\mathrm{d}t\) followed by the correct given answer can score this mark – limits and d\(t\) must be seen at least once (but need not be on the final integral)
Do not need to see \(= 1\) anywhere in their solution (and condone lack of brackets around the integrand)
| Scheme | Marks | AO |
|---|---|---|
| \(\displaystyle\int \left(0.2 + \sin t - 0.2\left(\tfrac{1}{2}\right)(1 + \cos 2t) - \sin t\cos^2 t\right)\mathrm{d}t\) | B1 | 1.2 |
| \(= 0.2t - \cos t\) | B1 | 1.1 |
| M1* | 1.1 | |
| \(-0.1t - 0.05\sin 2t + \frac{1}{3}\cos^3 t\) | A1 | 1.1 |
| \(\left[0.1t - \cos t - 0.05\sin 2t + \frac{1}{3}\cos^3 t\right]_0^{\pi}\) \(= \left(0.1\pi - (-1) - 0 + \frac{1}{3}(-1)^3\right) - \left(0 - 1 - 0 + \frac{1}{3}\right)\) | M1dep* | 1.1 |
| \(k\left(0.1\pi + 2 - \frac{2}{3}\right) = 1 \Rightarrow k = \dfrac{30}{3\pi + 40}\) | A1 | 1.1 |
| [6] |
Notes
B1: Correctly applies \(2\cos^2 t \equiv 1 + \cos 2t\) (so not just stating this identity) - implied by seeing \(-0.1t - 0.05\sin 2t\) after integration
e.g. applies could be for an attempt to integrate \(\frac{1}{2}(1 + \cos 2t)\) or stating this identity in an integral
B1: First two terms integrated correctly (Look out for those that have \(0.1t\) only from combining \(0.2t\) with \(-0.1t\))
This mark should be awarded if 0.2 is combined with another constant term and integrated correctly
M1*: M1 for an answer of the form \(\pm pt \pm q\sin 2t \pm r\cos^3 t\) or an answer of the form \(\pm pt \pm q\sin 2t \pm r\sin t\sin 2t \pm u\cos t\cos 2t\) or an answer of the form \(\pm pt \pm q\cos t \pm r\sin 2t \pm u\sin t\sin 2t \pm v\cos t\cos 2t\) for non-zero \(p\), \(q\), \(r\) (and \(u\), \(v\)) from integrating \(-0.2\cos^2 t - \sin t\cos^2 t\)
A1: A1 for the correct remaining three/four terms
(Alternatives: \(-0.1t - 0.05\sin 2t + \frac{1}{6}\sin t\sin 2t + \frac{1}{3}\cos t\cos 2t\) or \(-0.1t + 0.5\cos t - 0.05\sin 2t - \frac{1}{3}\sin t\sin 2t - \frac{1}{6}\cos t\cos 2t\))
M1dep*: Uses correct limits correctly \(F(\pi) - F(0)\) - condone limits the wrong way round only if the sign of their answer is subsequently changed
Must be using exact values (so must include term(s) in \(\pi\))
A1: www oe e.g. \(k = \dfrac{1}{0.1\pi + \frac{4}{3}}\) - isw once a correct expression for \(k\) seen
Correct exact value must be seen at some stage for the final mark – check first A mark carefully