2 ar Find the length of one arch of the cycloid x = r(0 - sin(0)), y = r(1 - cos(@)) Solution From a previous
2 ar Find the length of one arch of the cycloid x = r(0 - sin(0)), y = r(1 - cos(@)) Solution From a previous example we see that one arch is described by the parameter interval 0 s / $ 2x. Since dx and ay de de we have L = dx \\ 2 de de de V/2 (1 - 2 cos(0) + cos?(0) + sin?(0)) de V 2 (1 - cos(#)) do. To evaluate this integral we use the identity sin (x) = _ (1 - cos(2x)) with 0 = 2x, which gives 1 - cos(0) = 2 sin 2 Since 0 S 0 S 2x, we have 0 s s x and so sin() 2 0. Therefore V 2(1 - cos(0)) = 4 sin() = 2 and so L = 2r/sin(5) de = zr = 2rDetermine whether the statement is true or false. If the parametric curve x = At), y = g(t), satisfies g (1) = 0, then it has a horizontal tangent when t = 1. True O FalseDetermine whether the statement is true or false. 2 Ifx = rm and y = gm are twice differentiable. then d_y = \"I . ax: i at:2 0 True 0 False Need Help? _ 13. [-IIJ.12 Points] DETAILS SCALCETB 102.048. [\"100 Submissions Used MY NOTES ASK YOUR TEACHER Find the exact length of the curve. X : e't, y=4e\"2, 05:53
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