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.

Answer 1

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`