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Consider the curve $x=a(\cos \theta+\theta \sin \theta)$ and $y=a(\sin \theta-\theta \cos \theta)$.
What is $\frac{d y}{d x}$ equal to?
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What is $\frac{d y}{d x}$ equal to?
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Verified Answer
The correct answer is:
$\tan \theta$
Given, $\mathrm{x}=\mathrm{a}(\cos \theta+\theta \sin \theta)$ and
$y=a(\sin \theta-\theta \cos \theta)$
$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(-\sin \theta+\theta \cos \theta+\sin \theta)=\mathrm{a} \theta \cos \theta$
$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(\cos \theta+\theta \sin \theta-\cos \theta)=\mathrm{a} \theta \sin \theta$
Now $\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{a} \theta \sin \theta}{\mathrm{a} \theta \cos \theta}=\tan \theta$
$y=a(\sin \theta-\theta \cos \theta)$
$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(-\sin \theta+\theta \cos \theta+\sin \theta)=\mathrm{a} \theta \cos \theta$
$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(\cos \theta+\theta \sin \theta-\cos \theta)=\mathrm{a} \theta \sin \theta$
Now $\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{a} \theta \sin \theta}{\mathrm{a} \theta \cos \theta}=\tan \theta$
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