Question
If $\text{x}=\text{a}(\theta-\sin\theta)\text{ and},\text{y}=\text{a}(1+\cos\theta),$ find $\frac{\text{dy}}{\text{dx}}\text{ at }\theta=\frac{\pi}{3}$
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Answer
Here,
$\text{x}=\text{a}(\theta-\sin\theta)\text{ and y}=\text{a}(1+\cos\theta)$
Then,
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\big[\text{a}(\theta-\sin\theta)\big]=\text{a}(1-\cos\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\big[\text{a}(1+\sin\theta)\big]=\text{a}(1-\sin\theta)$
$\therefore\frac{\text{dy}}{\text{dx}}=\Bigg[\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}\Bigg]_{\theta=\frac{\pi}{3}}$
$=-\frac{\sin\frac{\pi}{2}}{1-\cos\frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}=-\sqrt{3}$
$\text{x}=\text{a}(\theta-\sin\theta)\text{ and y}=\text{a}(1+\cos\theta)$
Then,
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\big[\text{a}(\theta-\sin\theta)\big]=\text{a}(1-\cos\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\frac{\text{d}}{\text{d}\theta}\big[\text{a}(1+\sin\theta)\big]=\text{a}(1-\sin\theta)$
$\therefore\frac{\text{dy}}{\text{dx}}=\Bigg[\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1-\cos\theta)}\Bigg]_{\theta=\frac{\pi}{3}}$
$=-\frac{\sin\frac{\pi}{2}}{1-\cos\frac{\pi}{3}}=\frac{\frac{\sqrt{3}}{2}}{1-\frac{1}{2}}=-\sqrt{3}$
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