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प्रश्न
If x = cos t (3 – 2 cos2 t) and y = sin t (3 – 2 sin2 t), find the value of dx/dy at t =4/π.
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उत्तर
x = cos t (3 – 2 cos2 t) and y = sin t (3 – 2 sin2 t)
We need to find dy/dx :
`dy/dx=(dy/dt)/(dx/dt)`
Let us find dx/dt:
x = cos t (3 – 2 cos2 t)
`dx/dt=cost(4costsint)+(3-2cos^2t)(-sint)`
`=>dx/dt=-3sint+4cos^2tsint+2cos^2tsint`
Let us find dy/dx:
y = sin t (3 – 2 sin2 t)
`dy/dt=sint(-4sintcost)+(3-2sin^2t)(cost)`
`=>dy/dt=3cost-4sin^2tcost-2sin^2tcost`
thus,
`dy/dx=(3cost-4sin^2tcost-2sin^2tcost)/(-3sint+4cos^2tsint+2cos^2tsint)`
`=>dy/dx=(3cost-6sin^2tcost)/(-3sint+6cos^2tsint)`
`=>dy/dx=(3cost(1-2sin^2t))/(-3sint(1-2cos^2t))`
`=>dy/dx=(3cost(1-2sin^2t))/(3sint(2cos^2t-1))`
`=>dy/dx=cost/sint [because 2cos^2t-1=1-2sin^2t]`
`=>dy/dx=cott`
`=>(dy/dx)_(t=pi/4)=cot(pi/4)=1`
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