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Mathematics Question on Derivatives

Let x(t)=22costsin2tx(t)=2\sqrt2costsin\sqrt2t and y(t)=22sintsin2ty(t)=2\sqrt2sintsin\sqrt2t ,t∈(0,π2)t ∈(0,\frac{π}{2})

Then 1+(dydx)2d2ydx2\frac{1+(\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}} at t=π4t=\frac{π}{4} is equal to

A

−223\frac{-2\sqrt{2}}{3}

B

23\frac{2}{3}

C

13\frac{1}{3}

D

−23\frac{-2}{3}

Answer

−23\frac{-2}{3}

Explanation

Solution

x=22costsin2tx=2\sqrt2cost\sqrt{sin2t}, y=22sintsin2ty=2\sqrt2sint\sqrt{sin2t}

∴ dxdt=22cos3tsin2t\frac{dx}{dt}=\frac{2\sqrt2cos3t}{\sqrt{sin2t}}, dydt=22sin3tsin2t\frac{dy}{dt}=\frac{2\sqrt2sin3t}{\sqrt{sin2t} }

∴dydx=tan3t\frac{dy}{dx}=tan3t,( at t=π4,dydx=−1t=\frac{π}{4},\frac{dy}{dx}=−1)

and d2ydx2=3sec23t.dtdx=3sec23t.sin2t22cos3t\frac{d^2y}{dx^2}=3sec^23t.\frac{dt}{dx}=\frac{3sec^23t.\sqrt{sin2t}}{2\sqrt2cos3t }

(At t=π4,d2ydx2=−3t=\frac{π}{4},\frac{d^2y}{dx^2}=−3)

∴1+(dydx)2d2ydx2=2−3=−23\frac{1+(\frac{dy}{dx})^2}{\frac{d^2y}{dx^2}}=\frac{2}{−3}=\frac{-2}{3}