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if x=2 cos t - cos 2t, y=2 sin t – sin 2t find d2y/dx2 at t=pi/2
dx/dt=-2sint+2sin2tdy/dt=2cost-2cos2tdy/dx=(dy/dt)/(dx/dt)=(cost-cos2t)/(-sint+sin2t)dy/dx=2sin3t/2.sint/2/2cos3t/2sint/2=tan3t/2d2y/dx2=d/dx(dy/dx)=d/dt(dy/dx).dt/dx=(3/2).sec23t/2.1/(-2sint+2sin2t)=3/4.1/cos2(3t/2).1/(sin2t-sint)=3/4.1/cos2(3t/2).1/2cos3t/2.sint/2=(3/8).1/(cos33t/2.sint/2put t= pi/2d2y/dx2=3/8. 1/(cos3(3pi/4).sin(pi/4))=-3/2Thanks & RegardsRinkoo GuptaAskIITians Faculty
dx/dt=-2sint+2sin2t dy/dt=2cost-2cos2t dy/dx=(dy/dt)/(dx/dt)=(cost-cos2t)/(-sint+sin2t) dy/dx=2sin3t/2.sint/2/2cos3t/2sint/2 =tan3t/2 d2y/dx2=d/dx(dy/dx) =d/dt(dy/dx).dt/dx =(3/2).sec23t/2.1/(-2sint+2sin2t) =3/4.1/cos2(3t/2).1/(sin2t-sint) =3/4.1/cos2(3t/2).1/2cos3t/2.sint/2 =(3/8).1/(cos33t/2.sint/2 put t= pi/2 d2y/dx2=3/8. 1/(cos3(3pi/4).sin(pi/4)) =-3/2 Thanks & Regards
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