#### Question

If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), find the values of `dy/dx `.

#### Solution

We have,

x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 - cos 2t)

`therefore "dx"/"dt" = "a"["sin" "2t" ."d"/"dt" (1 + "cos" "2t") + (1 + "cos" 2"t") "d"/"dt" "sin" "2t"]`

`= "a" ["sin" 2"t" . (-2 "sin" "2t") + (1 + "cos" "2t") . 2 "cos" "2t"]`

`= -2 "a" "sin"^2 "2t" + 2"a" "cos" 2"t" (1 + "cos" "2t")`

`=> "dx"/"dt" = -2"a" ["sin"^2 "2t" - "cos" "2t" (1 + "cos" "2t")]` .....(1)

and `"dy"/"dt" = "b" ["cos" "2t" . (2 "sin" "2t") + (1 - "cos" "2t") + (1 - "cos" "2t") . "d"/"dt" "cos" "2t" . "d"/"dt" "cos" "2t"]`

`= "b" ["cos" "2t" . (2 "sin" "2t") + (1 - "cos" "2t") (-2 "sin " "2t")]`

`= "2b" ["sin" "2t" . "cos" "2t" - (1 - "cos" "2t") "sin" "2t"]`

`therefore "dy"/"dx" = ("dy"/"dt")/("dx"/"dt") = ("2b" ["sin" "2t" . "cos" "2t" - (1 - "cos" "2t") "sin" "2t"])/(-2"a" ["sin"^2 "2t" - "cos" "2t" (1 + "cos" "2t")])`

`=> ("dy"/"dx")_("t" = pi/4) = - "b"/"a" ["sin" pi/2 "cos" pi/2 - (1 - "cos" pi/2) "sin" pi/2]/["sin"^2 pi/2 - "cos" pi/2 (1 + "cos" pi/2)]`

`= -"b"/"a" . (0-1)/(1 - 0) = "b"/"a"`