course Mth 174 I intend to take test #3 this coming Thursday (5/3) and the final on the following Wednesday (5/9). ????·????????·????~?xassignment #021
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09:51:32 Query 11.9.7 (was page 576 #6) x' = x(1-y-x/3), y' = y(1-y/2-x)
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10:11:07 describe the phase plane, including nullclines, direction of solution trajectory in each region, and equilibrium points
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RESPONSE --> dx/dt = 0 has 2 solutions: x = 0 and y = -(1/3)x + 1. Therefore, x = 0 (y-axis) is a nullcline with vertical slopes, and the line segment (for x>=0 and y>=0) y = -(1/3)x + 1 is a nullcline with vertical slopes (endpoints at (0,1) and (3,0). dy/dt = 0 has 2 solutions: y = 0 and y = -2x+2. Therefore, y = 0 is a nullcline with vertical slopes, and the line segment (for x>=0 and y>=0) y = -2x + 2 is a nullcline with horizontal trajectories (endpoints at (0,2) and (1,0). The are 4 equilibrium points where the nullclines (with perpendicular trajectories) intersect: (0,0), (0,2), (0.6,0.8), and (3,0). Using test points in each region, the trajectories are: Region bounded by (0,1), (0,0), (1,0), and (0.6,0.8): x and y both increasing. Region bounded by (0.6,0.8), (1,0), and (3,0): x increasing and y decreasing. Region bounded by (0,1), (0,2), and (0.6,0.8): x decreasing and y increasing. Region outside (above and to the right of) segments from (0,2) to (0.6,0.8) and (0.6,0.8) to (3,0): x and y both decreasing.
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10:32:28 describe the trajectories that result
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RESPONSE --> Depending upon where the starting point at t=0 is in each region, all the trajectories will eventually move towards either y approaching 0 and x increasing without limit or towards x approaching 0 and y increasing without limit as t tends towards infinity. In other words for any starting point not at an equilibrium point, either x or y will go towards 0 while the other goes towards infinity.
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11:28:16 Query problem 11.10.22 (3d edition 11.10.19) (was 10.8.10) d^2 x / dt^2 = - g / L * x
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11:41:09 what is your solution assuming x(0) = 0 and x'(0) = v0?
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RESPONSE --> d^2x/dt^2 = (-g/l)x d^2x/dt^2 - (-g/l)x = 0 d^2x/dt^2 + (g/l)x = 0 x(t) = C1*cos[sqrt (g/l)*t] + C2*sin[sqrt (g/l)*t] x(0) = 0 = C1*cos[sqrt (g/l)*0 + C2*[sqrt (g/l)*0]] 0 = C1*cos(0) + C2*sin(0) 0 = C1 x(t) = C2*sin[sqrt (g/l)*t] x' = sqrt(g/l)*C2*cos[sqrt (g/l)*t] x'(0) = v0 = sqrt(g/l)*C2*cos[sqrt (g/l)*0] v0 = sqrt(g/l)*C2*cos(0) v0 = sqrt(g/l)*C2 C2 = v0/sqrt(g/l) x(t) = [v0/sqrt(g/l)]*sin[sqrt (g/l)*t]
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11:51:18 What is your solution if the pendulum is released from rest at x = x0?
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RESPONSE --> Using general solution calculated in previous response: x(t) = C1*cos[sqrt (g/l)*t] + C2*sin[sqrt (g/l)*t] x(0) = x0 = C1*cos(0) + C2*sin(0) x0 = C1 x(t) = x0*cos[sqrt (g/l)*t] + C2*sin[sqrt (g/l)*t] x' = -[sqrt(g/l)]*x0*sin[sqrt (g/l)*t] + [sqrt(g/l)]*C2*cos[sqrt (g/l)*t] x'(0) = v0 = 0 = -[sqrt(g/l)]*x0*sin(0) + [sqrt(g/l)]*C2*cos(0) 0 = [sqrt(g/l)]*C2 C2 = 0 x(t) = x0*cos[sqrt (g/l)*t]
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11:55:12 Query problem 11.10.25 (3d edition 11.10.24) (was 10.8.18) LC circuit L = 36 henry, C = 9 farad
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12:06:36 what is Q(t) if Q(0) = 0 and *(0) = 2?
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RESPONSE --> L*(d^2Q/dt^2) + Q/C = 0 Dividing by L: d^2Q/dt^2 + Q/(L*C) = 0 Substituting for L and C: d^2Q/dt^2 + Q/(36*9) = 0 d^2Q/dt^2 + [(1/18)^2]*Q = 0 Q(t) = C1*cos[(1/18)t] + C2*sin[(1/18)t] Q(0) = 0 = C1*cos[(1/18)*0] + C2*sin[(1/18)*0] 0 = C1 Q(t) = C2*sin[(1/18)t] dQ/dt = (1/18)*C2*cos[(1/18)t] dQ/dt = I dQ/dt (0) = I(0) = 2 = (1/18)*C2*cos(0) 2 = (1/18)*C2 C2 = 36 Q(t) = 36*sin[(1/18)t]
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12:54:09 what is Q(t) if Q(0) = 6 and I(0) = 0?
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RESPONSE --> Using the general solution from the previous response: Q(t) = C1*cos[(1/18)t] + C2*sin[(1/18)t] Q(0) = 6 = C1*cos[(1/18)*0] + C2*sin[(1/18)*0] 6 = C1 Q(t) = 6cos[(1/18)t] + C2*sin[(1/18)t] Q'(t) = -(1/3)sin[(1/18)t] + (1/18)*C2*cos[(1/18)t] Q'(0) = I(0) = 0 = -(1/3)sin[(1/18)*0] + (1/18)*C2*cos[(1/18)*0] 0 = -(1/3)sin(0) + (1/18)*C2*cos(0) C2 = 0 Q(t) = 6cos[(1/18)t]
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12:54:21 What differential equation did you solve and what was its general solution? And
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RESPONSE --> See previous response.
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12:54:33 how did you evaluate your integration constants?
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RESPONSE --> See previous response.
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12:54:47 Query problem 11.11.12 (was 10.9.12)general solution of P'' + 2 P' + P = 0
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13:04:55 what is your general solution and how did you obtain it?
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RESPONSE --> P"" + 2P' + P = 0 r = -(1/2)b +/- (1/2)sqrt(b^2-4c) r = -(1/2)(2) +/- (1/2)sqrt[2^2-4(1)] r = -1 +/- sqrt(0) r = -1 Since b^2-4c = 0: y = (C1*t + C2)e^(-bt/2) y = (C1*t + C2)e^(-2t/2) y = (C1*t + C2)e^(-t)
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13:13:05 If not already explained, explain how the assumption that P = e^(rt) yields a quadratic equation.
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RESPONSE --> Using the second-order differential equation for P with respect to t, and substituting e^(rt) for P and r for dP/dt, we get[e^(rt)]*(r^2 + br + c) = 0. Since e^(rt) cannot be 0, r^2 + br + c = 0, and we have our quadratic equation.
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13:25:45 Explain how the solution(s) to to your quadratic equation is(are) used to obtain your general solution.
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RESPONSE --> Since b^2-4c = 0, the solution to the quadratic is r = -b/2 = -1. The two solutions for P are P = C2*e^(-bt/2) and P = C1*t*e^(-bt/2). Since the general solution is the sum of the two individual solutions, we get P = (C1t + C2)*e^(-bt/2). Since b = 2, we get P = (C1t + C2)*e^(-t).
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13:26:41 Query problem 11.11.36 (was 10.9.30) s'' + 6 s' + c s NO LONGER HERE. USE 11.11.30 s '' + b s ' - 16 s = 0
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13:31:29 for what values of c is the general solution underdamped?
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RESPONSE --> s"" + bs' -16s = 0 b^2 - 4c = b^2 - 4(-16) = b^2 + 64. Since b^2 is always positive for real b, b^2 - 4c is always > 0. Therefore, therefore there is no value of b for which the general solution is underdamped.
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13:36:46 for what values of c is the general solution overdamped?
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RESPONSE --> Since b^2-4c > 0 for all b, we check the values of r1 and r2. Let r1 = -(1/2)b + sqrt(b^2 - 4c). This solution results in r1 > 0 for all b. Let r2 = -(1/2)b - sqrt(b^2 - 4c). This solution results in r2 < 0 for all b. Since r1 and r2 are never both < 0, there is no value of b for which the general solution is overdamped.
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13:37:54 for what values of c is the general solution critically damped?
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RESPONSE --> Since b^2-4c is never = 0, there is no value of b for which the general solution is critically damped.
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13:57:17 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Question 11.9.10 involves analyzing the phase plane for two first-order differential equations (A' = 2A - 2B and B' = B - AB). The last part of the problem asks for a sketch of the slope field--using a computer or calculator. Finding dB/dA by dividing dB/dt by dA/dt I got an equation in B and A (dB/dA = (B-AB)/(2A-2B). The only way that I found to use the calculator to find the slope field was to pick a value for A, substitute it into the equation for dB/dA, graph it on the calculator, and pick off the values of y (dB/dA) for various values of x (B). I then repeated this with other values for A. Is there a better way to do this on the calculator? I really enjoyed the math in this chapter. Differential equations are very interesting.
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