Question

Let \(x(t)=2\sqrt2costsin\sqrt2t \) and \(y(t)=2\sqrt2sintsin\sqrt2t\) ,\(t ∈(0,\frac{π}{2})\)

 Then \(\frac{1+(\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}\) at \(t=\frac{π}{4}\) is equal to

• A. \(\frac{-2\sqrt{2}}{3}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{1}{3}\)
• D. \(\frac{-2}{3}\)

Answer 1

The Correct Option is D

\(x=2\sqrt2cost\sqrt{sin2t}\), \(y=2\sqrt2sint\sqrt{sin2t}\)

∴ \(\frac{dx}{dt}=\frac{2\sqrt2cos3t}{\sqrt{sin2t}}\), \(\frac{dy}{dt}=\frac{2\sqrt2sin3t}{\sqrt{sin2t} }\)

∴\(\frac{dy}{dx}=tan3t\),( at \(t=\frac{π}{4},\frac{dy}{dx}=−1\)) 

and \(\frac{d^2y}{dx^2}=3sec^23t.\frac{dt}{dx}=\frac{3sec^23t.\sqrt{sin2t}}{2\sqrt2cos3t }\)

(At \(t=\frac{π}{4},\frac{d^2y}{dx^2}=−3\)) 

∴\(\frac{1+(\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}=\frac{2}{−3}=\frac{-2}{3}\)