# Question 8:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $SA = 2\pi \int r\sin\theta\sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta = 2\pi\int r\sin\theta\sqrt{25\cos2\theta + \ldots}\,d\theta$ or equivalent with $\cos\theta$ | M1 | 2.1 — Applies surface area formula about $x$ or $y$ axis with substitution of at least $r^2$ and attempt at $\left(\frac{dr}{d\theta}\right)^2$. The $2\pi$ must appear at some stage. |
| $r^2 = 25\cos2\theta \Rightarrow 2r\frac{dr}{d\theta} = k\sin2\theta$, or $r = 5\cos^{\frac{1}{2}}2\theta \Rightarrow \frac{dr}{d\theta} = A\cos^{-\frac{1}{2}}2\theta \times B\sin2\theta$ | M1 | 2.1 — Attempts to find $\frac{dr}{d\theta}$ via implicit differentiation or first square rooting then chain rule. |
| $2r\frac{dr}{d\theta} = -50\sin2\theta$ or $\frac{dr}{d\theta} = \frac{-50\sin2\theta}{2r}$, or $\frac{dr}{d\theta} = \frac{5}{2}\cos^{-\frac{1}{2}}2\theta \times -2\sin2\theta$ | A1 | 1.1b — Correct expression for $\frac{dr}{d\theta}$; need not be simplified. |
| $SA = 2\pi\int 5\sqrt{\cos2\theta}\sin\theta\sqrt{25\cos2\theta + \frac{25\sin^22\theta}{\cos2\theta}}\,d\theta = 2\pi\int 5\sqrt{\cos2\theta}\sin\theta\cdot\frac{5}{\sqrt{\cos2\theta}}\,d\theta = k\pi\int\sin\theta\,d\theta$ | M1 | 2.1 — Complete substitution into SA formula; applies trig identities to simplify to form $k\pi\int\sin\theta\,d\theta$ or $k\pi\int\cos\theta\,d\theta$. |
| $= 50\pi\int\sin\theta\,d\theta$ or $50\pi\int\cos\theta\,d\theta$ | A1 | 1.1b — Correct simplified integral. |
| $= 50\pi\int_0^{\frac{\pi}{4}}\sin\theta\,d\theta = 50\pi\left[-\cos\theta\right]_0^{\frac{\pi}{4}}$ | M1 | 3.4 — Uses model with appropriate limits to determine the surface area; for $x$-axis rotation limits likely $0$ and $\frac{\pi}{4}$ (or $-\frac{\pi}{4}$ and $0$); for $y$-axis allow $\frac{\pi}{4}$ and $\frac{\pi}{2}$. |
| $= 25\pi\left(2-\sqrt{2}\right)\ (\text{cm}^2)$ | A1 | 2.2a — Correct expression with no errors. If rotation about $y$-axis used, must have made clear reference to $r^2 = -25\cos2\theta$. |

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## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Adopts correct strategy: attempting $\frac{dy}{d\theta}$, finding $\theta$ when $\frac{dy}{d\theta}=0$, and using that $\theta$ to find $CD$ | M1 | 3.1a — Complete method for finding $CD$; maximum must be identified and length $CD$ calculated. |
| $y = r\sin\theta = 5\sqrt{\cos2\theta}\sin\theta \Rightarrow \frac{dy}{d\theta} = -\frac{5\sin2\theta\sin\theta}{\sqrt{\cos2\theta}} + 5\sqrt{\cos2\theta}\cos\theta$ | M1 | 1.1b — Uses product rule correctly to differentiate $r\sin\theta$; expect form $\alpha\sin2\theta\sin\theta(\cos2\theta)^{-\frac{1}{2}} + \beta\cos\theta(\cos2\theta)^{\frac{1}{2}}$ |
| $\frac{dy}{d\theta} = 0 \Rightarrow 5\cos\theta - 20\cos\theta\sin^2\theta = 0 \Rightarrow \theta = \ldots$ | M1 | 2.1 — Sets derivative to $0$ and proceeds via correct trig work to reach a value for $\theta$. Various routes possible, e.g. $\cos\theta(1-4\sin^2\theta)=0$, or $\cos3\theta=0$, or $\tan2\theta = \frac{1}{\tan\theta}$. |
| E.g. $\sin^2\theta = \frac{1}{4} \Rightarrow \theta = \frac{\pi}{6}$, or $\cos3\theta = 0 \Rightarrow 3\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{6}$ | A1 | 1.1b — Correct value of $\theta$ from correct working; derivative must have been correct. May be implied by correct $\sin$ and $\cos$ values used. SC award for $\frac{\pi}{3}$ if using $x = r\cos\theta$. |
| $CD = 2r\sin\frac{\pi}{6} = 2\times5\times\sqrt{\cos\frac{\pi}{3}}\times\frac{1}{2} = \frac{5\sqrt{2}}{2}\ (\text{cm})$* | M1; A1* | 3.4; 2.1 — M1: Uses value of $\theta$ in model to find $CD$, i.e. $CD = 2\times5\sqrt{\cos"2\theta"}\times\sin"\theta"$. Allow $2r\cos\theta$ for attempts from $\frac{dx}{d\theta}=0$. A1*: cso correct proof. NB for $x=r\cos\theta$ used, max M0M0M1A1M1A0 unless $r^2=-25\cos2\theta$ used. |