Question:
Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when
$x=a \cos \theta$ and $y=b \sin \theta$
Solution:
as $x=a \cos \theta$ and $y=b \sin \theta$
Then,
$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\frac{\mathrm{d}(\operatorname{acos} \theta)}{\mathrm{d} \theta}=-\mathrm{a} \sin \theta$
$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\frac{\mathrm{d}(\mathrm{b} \sin \theta)}{\mathrm{d} \theta}=\mathrm{b} \cos \theta$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{d} \theta}}{\frac{\mathrm{dx}}{\mathrm{d} \theta}}=\frac{\mathrm{b} \cos \theta}{-\mathrm{a} \sin \theta}=-\frac{\mathrm{b}}{\mathrm{a}} \cot \theta$
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