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Question
If X = f(t) and Y = g(t) Are Differentiable Functions of t , then prove that y is a differentiable function of x and
`"dy"/"dx" =("dy"/"dt")/("dx"/"dt" ) , "where" "dx"/"dt" ≠ 0`
Hence find `"dy"/"dx"` if x = a cos2 t and y = a sin2 t.
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Solution
x = a cos2 t and y = a sin2 t
Differentiating x and y with respect to t, we get
`"dx"/"dt" = "a cos"^2"t", "dy"/"dt" = "a sin"^2 "t"`
`=> "dx"/"dt" = "2a" "cos t" "d"/"dt" "cos t", "dy"/"dt" = "2a sin t" "d"/"dt" "sin t"`
`=> "dx"/"dt" = "2a" "cos t" xx (- "sin t"), "dy"/"dt" = "a" (2"sin t" xx "cos t")("d"/"dx" ("at"^2) = 2"at" "d"/"dx" "t")`
`=> "dx"/"dt" = "- 2a" "cos t" "sin t", "dy"/"dt" = "2a" "sin t cos" ("d"/"dx" ("sin" theta) = "cos" theta "and" "d"/"dx" ("cos" theta) = - "sin" theta)`
`=> "dx"/"dt" = - "a" "sin 2t", "dy"/"dt" = "a" "sin" "2t" ("sin" 2theta = 2 "sin" theta "cos" theta)`
Therefore
`"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`, where `"dx"/"dt" ≠ 0`
`"dy"/"dx" = ("a sin 2t")/(- "a sin 2t")`
`=> "dy"/"dx" = - 1`
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