A) \[-\frac{1}{2}\]
B) \[-\frac{1}{4}\]
C) 0
D) \[\frac{1}{2}\]
Correct Answer: A
Solution :
At point \[(1,1),1={{e}^{t}}\sin t,1={{e}^{t}}\cos t\] \[\Rightarrow \] \[\tan t=1\Rightarrow t=\frac{\pi }{4}\] Now, \[\frac{dy}{dt}={{e}^{t}}(\cos t-\sin t)\] and \[\frac{dx}{dt}={{e}^{t}}(\sin t+\cos t)\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{\cos t-\sin t}{\cos t+\sin t}.\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dt}\left( \frac{\cos t-\sin t}{\cos t+\sin t} \right)\frac{dt}{dx}\] \[=\left[ \frac{\begin{align} & (\cos t+\sin t)(-\sin t-\cos t) \\ & -(\cos t-\sin t)(-\sin t+\cos t) \\ \end{align}}{{{(\cos t+\sin t)}^{2}}} \right]\frac{dt}{dx}\] \[=\frac{-2}{{{(\cos t+\sin t)}^{2}}}.\frac{1}{{{e}^{t}}(\sin t+\cos t)}\] \[=\frac{-2}{({{e}^{t}}\cos t+{{e}^{t}}\sin t)}.\frac{1}{{{(\cos t+\sin t)}^{2}}}\] \[=\frac{-2}{x+y}.\frac{1}{{{(\cos t+\sin t)}^{2}}}\] \[=\frac{-2}{1+1}.\frac{1}{\left( \cos \frac{\pi }{4}+\sin \frac{\pi }{4} \right)}=-\frac{1}{2}\]
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