A) \[\frac{3}{2\sqrt{2}}\]
B) \[\frac{1}{6}\]
C) \[\frac{1}{6\sqrt{2}}\]
D) \[\frac{1}{3\sqrt{2}}\]
Correct Answer: C
Solution :
\[x=3\text{ }tan\text{ }t\text{ }and\text{ }y=3\text{ }sec\text{ }t\] So that \[\frac{dx}{dt}=3{{\sec }^{2}}t\] and \[\frac{dy}{dt}=3\sec t\,\,\tan \,\,t\] \[\frac{dy}{dx}\,\,=\,\,\frac{dy/dt}{dx/dt}\,\,=\,\sin \,t\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\,\,=\,\,(\cos \,\,t.)\,\,=\,\,\frac{dt}{dx}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\,\,=\,\,(\cos \,\,t)\,.\,\,\,\frac{1}{3\,{{\sec }^{2}}t}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}\,\,=\,\,\,\,\frac{1}{3}(co{{s}^{3}}\,t)\] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{t=\pi /4}}\,=\,\,\,\frac{1}{3}\times \,\,{{\left( \frac{1}{\sqrt{2}} \right)}^{3}}\,\,=\,\,\frac{1}{6\sqrt{2}}\]
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