Question

If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t) then find $\frac{dy}{dx}att=\frac{\pi }{4}$.

Answer 1

$x=asin2t(1+cos2t),y=bcos2t(1–cos2t)\frac{dy}{dx}=2acos2t\left(1cos2t\right)+asin2t(-2sin2t)=2acos2t+2aco{s}^{2}2t-2asi{n}^{2}2t=2acos2t+2acos4t\frac{dy}{dx}=-2bsin2t(1-cos2t)+bcos2t\left(2sin2t\right)=-2bsin2t+2bsin2tcos2t+2bcos2tsin2t\left(2sin2t\right)=-2bsin2t+4bsin2tcos2t=-2bsin2t+2bsin4t\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-2bsin2t+2bsin4t}{2acos2t+2acos4t}\Rightarrow \frac{dy}{dt}=\frac{-2bsin2t+2bsin4t}{2acos2t+2acos4t}\Rightarrow {\left|\frac{dy}{dt}\right|}_{t=\frac{\pi }{4}}=\frac{-2bsin\frac{2\pi }{4}+2bsin\frac{4\pi }{4}}{2acos\frac{2\pi }{4}+2acos\frac{4\pi }{4}}\Rightarrow {\left|\frac{dy}{dt}\right|}_{t=\frac{\pi }{4}}=\frac{-2bsin\frac{\pi }{4}+2bsin\pi }{2acos\frac{\pi }{2}+2acos\pi }\Rightarrow {\left|\frac{dy}{dt}\right|}_{t=\frac{\pi }{4}}=\frac{-2b}{-2a}=\frac{b}{a}\therefore \Rightarrow {\left|\frac{dy}{dt}\right|}_{t=\frac{\pi }{4}}=\frac{b}{a}$