Let x(t)=2√2cost√sin2t and y(t)=2√2sint√sin2t,t∈(0,π2). Then 1+(dydx)2d2ydx2 at t=π4 is equal to
A
13
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B
−23
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C
−2√23
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D
23
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Solution
The correct option is B−23 x=2√2cost√sin2t,y=2√2sint√sin2t ∴dxdt=2√2cos3t√sin2t,dydt=2√2sin3t√sin2t ∴dydx=tan3t,( at t=π4,dydx=−1)
and d2ydx2=3sec23t.dtdx=3sec23t.√sin2t2√2cos3t (At t=π4,d2ydx2=−3) ∴1+(dydx)2d2ydx2=2−3=−23