Applications of integration: TJC 2023 Promo Q11

(a)

$C$ meets the $y$-axis where $x = 0$: $$\begin{aligned} \cos 2\theta &= 0\\ 2\theta &= \frac{\pi}{2},\ \frac{3\pi}{2},\ \frac{5\pi}{2},\ \ldots\\ \theta &= \frac{\pi}{4},\ \frac{3\pi}{4},\ \frac{5\pi}{4},\ \ldots \end{aligned}$$ For $\dfrac{\pi}{2} \leq \theta \leq \dfrac{3\pi}{2}$, $\theta = \dfrac{3\pi}{4}$ or $\dfrac{5\pi}{4}$.

(b)

When $\theta = \dfrac{\pi}{2}$, $(x, y) = (-1, 2)$ (end-point). When $\theta = \dfrac{3\pi}{2}$, $(x, y) = (-1, 0)$ (end-point).

(c)

$$\begin{aligned} x = \cos 2\theta \;&\Rightarrow\; \frac{\,\mathrm{d} x}{\,\mathrm{d}\theta} = -2\sin 2\theta\\ y = 1 + \sin\theta \;&\Rightarrow\; \frac{\,\mathrm{d} y}{\,\mathrm{d}\theta} = \cos\theta\\ \frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= \frac{\cos\theta}{-2\sin 2\theta}. \end{aligned}$$ When $\theta = \dfrac{3\pi}{4}$, $y = 1 + \sin\dfrac{3\pi}{4} = 1 + \dfrac{1}{\sqrt{2}}$ and $$\begin{aligned} \frac{\,\mathrm{d} y}{\,\mathrm{d} x} &= \frac{\cos\frac{3\pi}{4}}{-2\sin\frac{3\pi}{2}} = -\frac{1}{2\sqrt{2}}. \end{aligned}$$ The tangent passes through $\left(0,\ 1 + \dfrac{1}{\sqrt{2}}\right)$: $$\begin{aligned} y - \left(1 + \frac{1}{\sqrt{2}}\right) &= -\frac{1}{2\sqrt{2}}(x - 0)\\ y &= -\frac{1}{2\sqrt{2}}x + 1 + \frac{1}{\sqrt{2}}. \end{aligned}$$

(d)

$$\begin{aligned} \text{Area} &= \int_{y_1}^{y_2} x \,\mathrm{d} y = \int_{\frac{5\pi}{4}}^{\frac{3\pi}{4}} \cos 2\theta\cos\theta \,\mathrm{d}\theta\\ &= \frac{1}{2}\int_{\frac{5\pi}{4}}^{\frac{3\pi}{4}} \cos 3\theta + \cos\theta \,\mathrm{d}\theta\\ &= \frac{1}{2}\left[\frac{\sin 3\theta}{3} + \sin\theta\right]_{\frac{5\pi}{4}}^{\frac{3\pi}{4}}\\ &= \frac{1}{2}\left[\frac{1}{3}\!\left(\frac{1}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} - \left(-\frac{1}{3}\!\left(\frac{1}{\sqrt{2}}\right) - \frac{1}{\sqrt{2}}\right)\right]\\ &= \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3} \text{ units}^2. \end{aligned}$$ Hence $\theta_1 = \dfrac{5\pi}{4}$, $\theta_2 = \dfrac{3\pi}{4}$ and $\mathrm{f}(\theta) = \cos 2\theta\cos\theta$.

(e)

$$\begin{aligned} \frac{x}{y^2} &= \frac{\cos 2\theta}{(1 + \sin\theta)^2}\\ &\approx \frac{1 - \frac{(2\theta)^2}{2}}{(1 + \theta)^2} \qquad (\theta \text{ sufficiently small})\\ &= \left(1 - 2\theta^2\right)(1 + \theta)^{-2}\\ &= \left(1 - 2\theta^2\right)\left(1 - 2\theta + 3\theta^2 - \cdots\right)\\ &= 1 - 2\theta + 3\theta^2 - 2\theta^2 + \cdots\\ &\approx 1 - 2\theta + \theta^2 \qquad (\theta \text{ sufficiently small})\\ &= (1 - \theta)^2. \quad\text{((shown))} \end{aligned}$$