\fEXAMPLE 4 Consider the electric circuit shown in the figure and modeled by the differential equation below. [! + RI = E(t) dt E L Find an expression for the current in a circuit where the resistance is 9 0, the inductance is 3 H, a battery gives a constant voltage of 9 V, and the switch is turned on when t = 0. What is the limiting value of the current? SOLUTION With L = 3, R = 9, and E(t) = 9, the equation becomes switch 3-! + 91 = 9 or 01 = 3 - 31 Video Example () dt dt and the initial value problem is 3- 31 I(0) = 0. We recognize this equation as being separable, and we solve it as follows: 3 - 37 = at ( 3 - 31 = 0 ) Lin |3 - 31| = t + C er+ C 13 - 311 = 3 X 3 - 31 = 1 = Ae-3t I= 1 - Ae- 3t Since I(0) = 0, we have 1 - = 0, SO A = and the solution is I(t ) = The limiting current, in amperes, is t -+ co lim I(t) = lim (1 - 1e-3t) = 1 - 1 lim e-3t = t -+ co t - coEXAMPLE 5 Find the length of one arch of the cycloid x = r(0 - sin(0)), y = r(1 - cos(@)) SOLUTION From this example we see that one arch is described by the parameter interval 0 & 0 $ 27. Since dx and ay = de de 2 ar x we have Video Example () dx 2 L = + de de de Op V12(1 - 2 cos(8) + cos2(0) + sin?(0)) de = Vz(1 - cos (8) de. To evaluate this integral we use the identity sin?(x) = (1 - cos(2x)) with 0 = 2x, which gives 1 - cos(0) = 2 sin2( Since 0 5 0 S 2n, we have 0 S s n and so sin(0/2) 2 0. Therefore V 2(1 - cos(0)) = \\ 4 sin2(0/2) = 2 and so L = 2r sin(0/2) de = 2r or = 2r 86 X