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Question
If x = esin3t, y = ecos3t, then show that `dy/dx = -(ylogx)/(xlogy)`.
Solution
x = esin3t, y = ecos3t
∴ log x = logesin3t, logy = logecos3t
∴ log x = (sin 3t)(log e), log y = (cos 3t)(log e)
∴ log x = sin 3t, log y = cos 3t ...(1) ... [∵ log e = 1]
Differentiating both sides w.r.t. t, we get
`(1)/x.dx/dt = d/dt(sin3t) = cos3t.d/dt(3t)`
= cos 3t x 3
= 3 cos 3t
and
`(1)/y.dy/dt = d/dt(cos 3t) = -sin3t.d/dx(3t)`
= – sin 3t x 3
= – 3 sin 3t
∴ `dx/dt = 3x cos 3t and dy/dt"= -3y sin 3t`
∴ `dy/dx = ((dy/dt))/((dx/dt)`
= `(-3y sin 3t)/(3x cos 3t)`
= `(-y sin 3t)/(x cos 3t)`
= `(-y log x)/(x log y)`. ...[By (1)]
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