Find dydx, if x and y are connected parametrically by the equations given in questions without eliminating the parameter.
x=sin3t√cost 2t,y=cos3t√cos 2t
Given, x=sin3t√cost 2t,y=cos3t√cos 2t
Differentiating w.r.t. t, we get
∴ dxdt=ddt(sin3t√cos 2t)=√cos 2t(3 sin2t cos t)−sin3t(−2sin 2tx√cos 2t)(√cos 2t)2 (Using quotient rule,ddx(uv)=vdudx−ududxv2)=3(cos 2t)sin2tcos t+sin 2tsin3tcos 2t√cos 2t=3(1−2sin2t)sin2t cos t+(2sin t cos t)sin3tcos 2t√cos 2t (∵ cos 2t=1−2sin2t and sin 2t=2sin t cos t)=3sin2t cos t−4sin4t cos tcos 2t√cos 2tand dydt=ddt(cos3t√cos 2t)=√cos 2t(−3cos2 tsin t)−cos3t(−2sin 2t2√cos 2t)(√cos 2t)2 (Using quotient rule)=−3(cos 2t)cos2t sin t+sin 2t cos3tcos 2t√cos 2t=−3(2cos2t−1)cos2t sin t+cos3t(2sin t cos t)cos 2t√cos 2t [∵cos 2t=2cos2t−1 sin 2t=2sin t cos t]=3cos2t sin t−4cos4t sin tcos 2t√cos 2t⇒ dydx=dydtdxdt=dydt×dtdx=3cos2t sin t−4cos4t sin t3sin2t cos t−4sin4t cos t=cos2tsin t(3−4cos2t)sin2t cos t(3−4sin2t)=cos t(3−4cos2t)sin t(3−4sin2t)=3 cos t−4cos3t3 sin t−4 sin3t=−3cos 3tsin 3t=−cot 3t [∵cos 3t=4cos3t−3cos tsin 3t=3 sin t−4 sin3t]