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Question
If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t) then find `dy/dx `
Solution
x = a sin 2t (1 + cos 2t)
y = b cos 2t (1 – cos 2t)
`dx/dt=2acos2t(1+cos2t)+asin2t(-2sin2t)`
`=2acos2t+2acos^2 2t-2a sin^2 2t`
`=2a cos2t+2a cos4t`
`dy/dt=-2dsin2t(1-cos2t)+bcos2t(2sin2t)`
`=-2bsin2t+2b sin2tcos2t+2b cos2t sin2t`
`=-2b sin2t+4b sin2tcos2t`
`=-2bsin2t+2bsin4t`
`(dy/dt)/(dx/dt)=(-2bsin2t+2bsin4t)/(2a cos2t+2a cos4t)`
`dy/dx=(-2bsin2t+2bsin4t)/(2a cos2t+2a cos4t)`
`|dy/dx|_(t=pi/4)=(-2bsin((2pi)/(4))+2bsin((4pi)/4))/(2a cos((2pi)/4)+2a cos((4pi)/4))`
`=>|dy/dx|_(t=pi/4)=(-2bsin(pi/2)+2bsinpi)/(2a cos(pi/2)+2a cospi)`
`=>|dy/dx|_(t=pi/4)=(-2b)/(-2a)=b/a`
`therefore |dy/dx|_(t=pi/4)=b/a`
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