Question
If $\text{x}=\text{a}(\theta+\sin\theta),\text{y}=\text{a}(1+\cos\theta),$ find $\frac{\text{dy}}{\text{dx}}.$
✓
Answer
Here,
$\text{x}=\text{a}(\theta+\sin\theta)$
Differentiating it with respect to $\theta$,
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big(\frac{\text{d}}{\text{d}\theta}(\theta)+\frac{\text{d}}{\text{d}\theta}(\sin\theta)\Big)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(1+\cos\theta)\ .....(\text{i})$
And, $\text{y}=\text{a}(1+\cos\theta)$
Differentiating it with respect to $\theta$,
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(0-\sin\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\sin\theta\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1+\cos\theta)}$
$=\frac{-\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}{\frac{2\cos^2\theta}{2}}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\tan\theta}{2}$
$\text{x}=\text{a}(\theta+\sin\theta)$
Differentiating it with respect to $\theta$,
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\Big(\frac{\text{d}}{\text{d}\theta}(\theta)+\frac{\text{d}}{\text{d}\theta}(\sin\theta)\Big)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(1+\cos\theta)\ .....(\text{i})$
And, $\text{y}=\text{a}(1+\cos\theta)$
Differentiating it with respect to $\theta$,
$\frac{\text{dx}}{\text{d}\theta}=\text{a}(0-\sin\theta)$
$\frac{\text{dx}}{\text{d}\theta}=\text{a}\sin\theta\ .....(\text{ii})$
Dividing equation (ii) by (i),
$\frac{\frac{\text{dy}}{\text{d}\theta}}{\frac{\text{dx}}{\text{d}\theta}}=\frac{-\text{a}\sin\theta}{\text{a}(1+\cos\theta)}$
$=\frac{-\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}{\frac{2\cos^2\theta}{2}}$
$\frac{\text{dy}}{\text{dx}}=-\frac{\tan\theta}{2}$
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