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Consider the curve $x=a(\cos \theta+\theta \sin \theta)$ and $y=a(\sin \theta-\theta \cos \theta)$.
What is $\frac{d^{2} y}{d x^{2}}$ equal to?
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What is $\frac{d^{2} y}{d x^{2}}$ equal to?
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Verified Answer
The correct answer is:
$\frac{\sec ^{3} \theta}{a \theta}$
Given $\mathrm{x}=\mathrm{a}(\cos \theta+\theta \sin \theta)$
$\mathrm{y}=\mathrm{a}(\sin \theta-\theta \cos \theta)$ and we have $\frac{\mathrm{dy}}{\mathrm{dx}}=\tan \theta$
According to question $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=\sec ^{2} \theta \frac{\mathrm{d} \theta}{\mathrm{dx}}$
$=\sec ^{2} \theta\left(\frac{1}{a \theta \cos \theta}\right)$ $\quad \left[\because \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a} \theta \cos \theta\right]$
Hence, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\sec ^{3} \theta}{\mathrm{a} \theta}$
$\mathrm{y}=\mathrm{a}(\sin \theta-\theta \cos \theta)$ and we have $\frac{\mathrm{dy}}{\mathrm{dx}}=\tan \theta$
According to question $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}=\sec ^{2} \theta \frac{\mathrm{d} \theta}{\mathrm{dx}}$
$=\sec ^{2} \theta\left(\frac{1}{a \theta \cos \theta}\right)$ $\quad \left[\because \frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a} \theta \cos \theta\right]$
Hence, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\sec ^{3} \theta}{\mathrm{a} \theta}$
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