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