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Question
If x = ecos2t and y = esin2t, prove that `"dy"/"dx" = (-y log x)/(xlogy)`
Solution
Given that: ecos2t and y = esin2t
⇒ cos 2t = log x and sin 2t = log y.
Differentiating both the parametric functions w.r.t. t
`"dx"/"dt" = "e"^(cos2"t") * "d"/"dt" (cos 2"t")`
= `"e"^(cos 2"t") (- sin 2"t") * "d"/"dt" (2"t")`
= `- "e"^(cos2"t") * sin 2"t" * 2`
= `2"e"^(cos2"t") * sin 2"t"`
Now y = esin2t
`"dy"/"dt" = "e"^(sin2"t") * "d"/"dt"(sin 2"t")`
= `"e"^(sin2"t") * cos 2"t" * "d"/"dt"(2"t")`
= `"e"^(sin2"t") * cos 2"t" * 2`
= `2"e"^(sin2"t") * cos 2"t"`
∴ `"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`
= `(2"e"^(sin2"t") * cos2"t")/(-2"e"^(cos2"t") * sin 2"t")`
= `("e"^(sin2"t") * cos2"t")/(-"e"^(cos2"t") * sin2"t")`
= `(y cos 2"t")/(-x sin 2"t")`
= `(y log x)/(-x log y)` ......`[(because cos 2"t" = log x),(sin 2"t" = log y)]`
Hence, `"dy"/"dx" = - (y log x)/(x log y)`.
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