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प्रश्न
If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.
`x = (sin^3t)/sqrt(cos 2t), y = (cos^3t)/sqrt(cos 2t)`
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उत्तर
Here x = `(sin^3t)/(sqrtcos 2t)` ....(1)
y = `(cos^3 t)/ (sqrtcos 2t)` .....(2)
Differentiating (1) and (2) w.r.t. t, we get,
`dx/dt = (sqrtcos2t d/dt sin^3 t - sin ^3 t d/dt (sqrt cos2t))/(cos2t)`
= `((sqrt cos2t) 3 sin^2 t cos t - sin^3 t. 1/(2 sqrtcos2t) . (-sin 2t).2)/(cos 2t)`
= `(sqrt cos 2t 3 sin^2 t cos t + (sin^3 t sin 2t)/(sqrtcos2t))/(cos 2t)`
= `(3 cos 2t sin^2 t cos t + sin^3 t sin 2t)/ ((cos 2t)^(3//2))`
`dy/dt = (sqrt cos 2t d/dt cos^3 t - cos^3 t d/dt sqrtcos2t)/(cos 2t)`
= `(sqrtcos2t.3 cos^2 t (- sint) - cos^3 t. 1/(2sqrtcos 2t).(-sin 2t).2)/(cos 2t)`
= `(-3 cos^2 t. sin t. sqrt cos2t + (cos^3 t sin 2t)/(sqrtcos2t))/(cos2t)`
= `(cos^3 t sin 2t - 3 cos^2 t. sin t cos 2t)/((cos2t)^(3//2))`
`dy/dx = (dy/dt)/(dx/dt)`
= `(cos^3 t sin 2t - 3 cos^2 t . sin t cos 2t)/(3 cos2t sin^2 t cos t + sin^3 t sin 2t)`
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