Math Practice exam 2 - solutions

Transcription

1 C Roettger, Fall 13 Math Practice exam 2 - solutions Problem 1 A solution of 10% perchlorate in water flows at a rate of 8 L/min into a tank holding 200L pure water. The solution is kept well stirred and flows out of the tank at a rate of 6 L/min. a) Determine the volume of perchlorate in the tank after t minutes. b) When will the percentage of perchlorate in the tank reach 6%? Solution. a) Let V (t) and x(t) the volume of water and perchlorate in the tank at time t, respectively. So V (t) = t (time in minutes, volume in liters). Then the amount x(t) of perchlorate satisfies x (t) = x(t) V (t). This is a linear equation, written in standard form as so an integrating factor is x (t) t x = 0.8, µ(t) = e 3/(100+t) dt = e 3 ln(100+t) = (100 + t) 3. We multiply the entire equation by µ and integrate: µ(t)x(t) = 0.8 (100 + t) 3 dt = 0.2(100 + t) 4 + C. so after solving for x, we get the general solution x(t) = 0.2(100 + t) + Since x(0) = 0, we can solve for C to get C (100 + t) 3. C = = 20, 000,

2 The amount of perchlorate after t minutes is b) We put and solve for t, getting first x(t) = 0.2(100 + t) 0.06 = x(t) V (t) 20, 000, 000 (100 + t) 3. = , 000, 000 (100 + t) 4 and then (in minutes). (100 + t) 4 = t = / Problem 2 A garage with no heating or cooling has a time constant of 2 hours. The outside temperature varies as a sine wave, M(t) = sin(π(t 8)/12). Determine the time when the building reaches its lowest and its highest temperature, assuming that the exponential term has died off. 2

3 Note how the building temperature T (blue) lags behind the ambient temperature M! Solution. equation The temperature T (t) of the building satisfies the differential T (t) = k(m(t) T (t)). Rewritten in standard form for a linear equation, this reads T + kt = km so we have the integrating factor µ(t) = e kt, and the usual recipe gives e kt T (t) = ke kt ( sin(π(t 8)/12)) dt. Integrating by parts twice gives for the hard part on the right-hand side e kt sin(αt + β) dt = ekt [k sin(αt + β) α cos(αt + β)] α 2 + k 2 + C. Integrating 65ke kt gives another 65e kt. We substitute k = 1/2 (1 / time constant), α = π/12, β = 2π/3. Altogether, ( ) ( ) 15 sin π(t 8) 5π cos π(t 8) T (t) = Ce t/2. π 2 / /4 For t large enough so that the exponential term has died off (meaning we can neglect it), we can find the maximum / minimum temperature by setting the derivative of the first term equal to 0, like in Calculus I. It helps to introduce a new variable u = π(t 8)/12. So we differentiate which gives d du [ 15 4 sin u 5π 8 cos u u = arctan for any integer n. Going back to t, ] = 15 4 cos u + 5π 8 sin u = 0 ( ) 6 + nπ nπ π π(t 8) 12 = nπ 3

4 gives the times t = n for any integer n. A second-derivative test (no, we don t want that) or firstderivative test (still, lots of work) or simply a graph (best! see above) shows that even integers n give times with minimal temperature, odd integers n give maximal temperature. Problem 3 A parachutist whose mass is 75 kg drops from a helicopter hovering 2000 meters above the ground. She falls to the ground under the force of gravity and air resistance, which is proportional to her velocity, with proportionality constant b 1 = 30 N sec/m when the chute is closed, and b 2 = 90 N sec/m when the chute is open. The chute does not open until the velocity reaches 20 m/sec. At what time will she reach the ground? Use g = 10N/kg. Solution. Let y(t) be the parachutist s height above ground. So y(0) = 2000, y (0) = 0, and my + by = mg We write v = y, so we get a first-order equation for the velocity, v + bv m = g which has integrating factor µ = e bt/m, vµ = g e bt/m dt = gmebt/m b and the solution v = gm b + Ce bt/m. + C This analysis is valid both for open and closed parachute. Now use v(0) = 0 to get for the first, rapid fall C 1 = gm b 1 = 25, v = 25(e b 1t/m 1). Let us find the time t 1 when the parachute opens, so v = 20 (negative sign because the movement is down!!) 20 = gm b 1 (e b 1t/m 1) 4

5 gives with g = 10N/kg t 1 = m b 1 ln (in seconds). After that time, we get ( 1 20b ) gm v = gm b 2 + C 2 e b 2t/m. and the initial condition v(t 1 ) = 20 gives C 2 = Finally, find the position y(t) by integrating v. In both cases, y(t) = gmt b For the first case, y(0) = 2000 gives Cm b e bt/m + D. D 1 = C 1m b 1 = so the position at time t 1 is y(t 1 ) meters. With the parachute open, we get y(t) = gmt b 2 C 2m b 2 e b 2t/m + D 2 and from the initial condition y(t 1 ) = , we can determine D2 = Finally, the parachutist reaches the ground when so at time 0 = y(t) gmt b 2 + D 2 t = D 2b 2 gm (in seconds). The exponential term at around this time is completely negligible. 5

6 Note. This is 3.4 # 7, and the book s answer is 241 seconds, because they used the more accurate value g = 9.81N/kg. Problem 4 Find the general solution of y 2y + 5y = 0. Solution. The auxiliary equation is r 2 2r + 5 = 0 which has roots 1 ± 2i. So we can write the general solution as y(t) = e t (c 1 cos 2t + c 2 sin 2t). Problem 5 Consider the differential equation x (t) 2x (t) + x(t) = cos t. (1) a) Find a particular solution of equation (1). b) Find the complimentary function (general solution of the associated homogeneous equation). c) Find the unique solution satisfying both (1) and the initial conditions x(0) = 0, x (0) = 1. Solution. a) We try x p (t) = A cos t + B sin t. This leads to A = 0, B = 1/2. b) The characteristic equation is (r 1) 2 = 0, with double root r = 1, so the complimentary function is x c (t) = (c 1 + c 2 t)e t c) We get c 1 = 0, c 2 = 3/2 after substituting the general solution x p + x c for x. Problem 6 A mass of 2 kg is traveling horizontally on wheels without friction. It is hooked up to a spring with spring constant k = 162 N/m. Let x(t) be the position of the mass with x = 0 being the equilibrium position. 6

7 a) Find the second-order linear differential equation governing x. b) What is the natural frequency ω of this system? c) Find the position function x(t) if x(0) = 12 and x (0) = 45. d) Suppose an external force F = 2 cos(10t) starts acting on the mass, and x(0) = x (0) = 0. Find the resulting position function x(t). Solution. a) 2x + 162x = 0. b) ω = 162/2 = 9 (in radians per second). c) The general solution here is x c (t) = c 1 cos 9t + c 2 sin 9t. The initial conditions give c 1 = 12 and c 2 = 5, so x(t) = 12 cos 9t 5 sin 9t. d) Now we have to find a particular solution x p for 2x + 162x = 2 cos 10t. (2) There is no overlap between cos 10t, sin 10t and cos 9t, sin 9t., So we try x p (t) = A cos 10t + B sin 10t, and substituting this into (2) yields equations B = 0, 200A + 162A = 2, so A = 1/19. The general solution is now x(t) = 1 19 cos 10t + c 1 cos 9t + c 2 sin 9t. The initial conditions give x (0) = 9c 2 = 0, so c 2 = 0 and c 1 = 1/19. Problem 7 Consider the Euler-Cauchy equation t 2 y + 3ty + y = 0. a) Find the general solution. b) Then find the solution satisfying y(1) = 4, y (1) = 3. Solution. The characteristic equation here is r(r 1) + 3r + 1 = 0 7

8 which has a double root r = 1, so solutions are y = 1/t and y = ln(t)/t. The general solution is y = c 1 + c 2 ln t. t b) So c 1 = 4 and from we get c 2 c 1 = 3, so c 2 = 7. y (t) = c 2/t (c 1 + c 2 ln t) t 2, 8