Question

 

 If x=a sin 2t(1+cos 2t) and y=b cos 2t(1−cos 2t), find `dy/dx `

 

Answer 1

`x=asin 2t(1+cos2t) `

`y=bcos2t(1−cos2t)`

We know that

`dy/dx=dy/dt xx dt/dx`

`y=bcos2t(1−cos2t)`

`⇒dy/dt=−2bsin2t(1−cos2t)+ (2bcos2t sin2t)`

`⇒dy/dt=−2bsin2t+2bsin2t cos2t+2bcos2t sin2t`

`⇒dy/dt=−2bsin2t+4bsin2t cos2t`

`⇒dy/dt=2b(sin4t−sin2t)`

 

`x=asin2t(1+cos2t)`

`⇒dx/dt=2acos2t(1+cos2t)−2asin2t sin2t`

`⇒dx/dt=2acos2t+2acos^2 2y−2asin^2 2t`

`⇒dx/dt=2a(cos2t+cos4t)`

`∴(dy/dx)_(t=π/4)=b/a ((sin4(π/4)−sin2(π/4))/(cos2(π/4)+cos4(π/4)))=b/axx((0−1)/(0−1))=b/a`