Question 7: If x and y are connected parametrically by the equation, without eliminating the parameter, find. x = sin³/√cos2t ,y= cos³t/√cos2t
Class 12 - Math - Continuity and Differentiability
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here x = sin³t/√cos2t and y = cos³t/√cos2t
dy/dt =[ √cos2t.3sin²t.cost -sin³t.1/2.1/√cos2t.(-sin2t).2 ]/cos2t
= cos2t.3sin²t.cost+sin2t.sin³t/(cos2t)³/²
and dy/dt = [√cos2t.3cos²t(-sint)-cos³t.1/2.1/√cos2t.(-sin2t).2]/cos2t
=>[ -(cos2t).(3sintcos²t) + sin2t.cos³t]/(sin2t)³/²
dy/dx = (dy/dt)/dx/dt
= -3cos²tsintcos2t+sin2tcos³t/3sin²tcostcos2t+sin2tsin³t
= -6sintcos⁴t+3cos²tsint+2sintcos⁴t/3sin²tcost-6sin⁴tcost+2sin⁴tcost
= -4sintcos⁴t + 3cos²tsint/-4sin⁴tcost+3sin²tcost
= sintcost(3cost-4cos³t)/-sintcost4sin³t-3sint)
= cos3t/-sin3t = -cot3t.
dy/dt =[ √cos2t.3sin²t.cost -sin³t.1/2.1/√cos2t.(-sin2t).2 ]/cos2t
= cos2t.3sin²t.cost+sin2t.sin³t/(cos2t)³/²
and dy/dt = [√cos2t.3cos²t(-sint)-cos³t.1/2.1/√cos2t.(-sin2t).2]/cos2t
=>[ -(cos2t).(3sintcos²t) + sin2t.cos³t]/(sin2t)³/²
dy/dx = (dy/dt)/dx/dt
= -3cos²tsintcos2t+sin2tcos³t/3sin²tcostcos2t+sin2tsin³t
= -6sintcos⁴t+3cos²tsint+2sintcos⁴t/3sin²tcost-6sin⁴tcost+2sin⁴tcost
= -4sintcos⁴t + 3cos²tsint/-4sin⁴tcost+3sin²tcost
= sintcost(3cost-4cos³t)/-sintcost4sin³t-3sint)
= cos3t/-sin3t = -cot3t.
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