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Question

# $\mathrm{If}x=a\left(\mathrm{cos}t+t\mathrm{sin}t\right)\mathrm{and}y=a\left(\mathrm{sin}t-t\mathrm{cos}t\right),\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\frac{{d}^{2}y}{d{x}^{2}}\mathrm{at}t=\frac{\mathrm{\pi }}{4}.$

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Solution

## $\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}x=a\left(\mathrm{cos}t+t\mathrm{sin}t\right)\mathrm{and}y=a\left(\mathrm{sin}t-t\mathrm{cos}t\right)\phantom{\rule{0ex}{0ex}}\mathrm{On}\mathrm{differentiating}\mathrm{with}\mathrm{respect}\mathrm{to}t,\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{dx}{dt}=\frac{d}{dt}\left[a\left(\mathrm{cos}t+t\mathrm{sin}t\right)\right]=-a\mathrm{sin}t+a\mathrm{sin}t+at\mathrm{cos}t\phantom{\rule{0ex}{0ex}}=at\mathrm{cos}t\phantom{\rule{0ex}{0ex}}\mathrm{and}\phantom{\rule{0ex}{0ex}}\frac{dy}{dt}=\frac{d}{dt}\left[a\left(\mathrm{sin}t-t\mathrm{cos}t\right)\right]=a\mathrm{cos}t-a\mathrm{cos}t+at\mathrm{sin}t\phantom{\rule{0ex}{0ex}}=at\mathrm{sin}t\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\left(\frac{dy}{dx}\right)=\frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{at\mathrm{sin}t}{at\mathrm{cos}t}=\mathrm{tan}t\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{{d}^{2}y}{d{x}^{2}}=\frac{\mathrm{d}}{dx}\left(\frac{dy}{dx}\right)=\frac{\mathrm{d}}{dx}\left(\mathrm{tan}t\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{d}}{dt}\left(\mathrm{tan}t\right)×\frac{\mathrm{d}t}{\mathrm{d}x}={\mathrm{sec}}^{2}t×\frac{1}{at\mathrm{cos}t}\phantom{\rule{0ex}{0ex}}=\frac{1}{at{\mathrm{cos}}^{3}t}\phantom{\rule{0ex}{0ex}}{\left(\frac{{d}^{2}y}{d{x}^{2}}\right)}_{t=\frac{\mathrm{\pi }}{4}}=\frac{1}{a\left(\frac{\mathrm{\pi }}{4}\right){\mathrm{cos}}^{3}\left(\frac{\mathrm{\pi }}{4}\right)}=\frac{8\sqrt{2}}{a\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{at}t=\frac{\mathrm{\pi }}{4},\frac{{d}^{2}y}{d{x}^{2}}=\frac{8\sqrt{2}}{a\mathrm{\pi }}.$

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