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

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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"`

  Is there an error in this question or solution?
Solution If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), find the values of dy/dx Concept: Derivatives of Functions in Parametric Forms.
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