x = asin2t(1+cos2t) and y = bcos2t(1-cos2t) find d2y/dx2 at t=(pi)/4 Share with your friends Share 2 Mayur Pisode answered this Dear Student, x = a sin 2t 1+cos2t⇒dxdt = 2a cos2t 1+cos 2t + a sin 2t -2sin 2t⇒dxdt = 2 a cos 2t + 2 a cos22t - 2 asin2 2t⇒dxdt= 2a cos 2t + 2a cos 4t y = b cos 2t1-cos 2t⇒dydt = -2b sin 2t 1-cos 2t + bcos 2t 2sin 2t⇒dydt = -2b Sin2t + 2b sin 2t . cos 2t + 2b sin 2t . cos 2t⇒dydt = -2b Sin2t + 2b 2 sin 2t . cos 2t ⇒dydt = -2b sin2t + 2b sin4tNow, dydx = dydt×dtdx=-2b sin 2t + 2b sin 4t 2a cos 2t + 2a cos 4t⇒dydx = basin 4t - sin 2tcos 4t + cos 2t⇒dydx = ba2 cos 3t . sin t2 cos 3t . cos t⇒dydx = ba tan t⇒d2ydx2 = basec2t × dtdx = basec2t × 12acos 4t + cos 2t⇒d2ydx2 =b2a2×sec2tcos 4t + cos 2t⇒d2ydx2t=π4 = b2a2×sec2π/4cos π + cosπ/2⇒d2ydx2t=π4 = b2a2 × 2-1+0 = -ba2 Regards 7 View Full Answer