Math , Spring 2010: Exam 2 Solutions 1. #1.) /5 #2.) /15 #3.) /20 #4.) /10 #5.) /10 #6.) /20 #7.) /20 Total: /100

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1 Math , Spring 2010: Exam 2 Solutions 1 NAME: Math Exam 2 Solutions #1.) /5 #2.) /15 #3.) /20 #4.) /10 #5.) /10 #6.) /20 #7.) /20 Total: /100 Instructions: There are 5 pages and a total of 100 points on the exam. You must show all necessary work to get credit. You may not use your book, notes, or calculator. Partial credit will only be given for progress toward a correct solution. 1.) (5 points) Solve y + 3y 18y = 0. Solution The characteristic equation is r 2 + 3r 18 = 0. Factoring, we have (r 3)(r + 6) = 0, so r = 3 and r = 6 are the solutions. Hence y 1 = e 3t and y 2 = e 6t satisfy the differential equation, and the general solution is for constants C 1 and C 2. y = C 1 e 3t + C 2 e 6t, 2.) (15 points) A population is growing unrestrictedly, according to the Malthusian model. If the population takes 2 years to triple, how long will it take to become 5 times its original size? You may express your answer in terms of logarithms. Solution Let P (t) denote the population at time t years from the start. The Malthusian model is that P (t) = kp (t). Solving this (which was done in class) gives P (t) = P (0)e kt. Because the time for the population to triple is 2 years, we have P (2) = 3P (0). Hence 3P (0) = P (2) = P (0)e k 2. Cancelling P (0) gives 3 = e 2k. Taking the natural logarithm of both sides gives ln 3 = 2k, or k = 1 ln 3. Substituting this value for k in the equation P (t) = P 2 (0)ekt gives P (t) = P (0)e ( 1 2 ln 3)t. To find the time when the population is five times its original size, we set P (t) = 5P (0) and solve for t: 5P (0) = P (0)e ( 1 2 ln 3)t. We divide by P (0) to get 5 = e ( 1 2 ln 3)t and then take the natural logarithm of both sides to get ( ) 1 ln 5 = 2 ln 3 t, hence t = 2 ln 5 ln years.

2 Math , Spring 2010: Exam 2 Solutions 2 3.) (20 points) 20 points) A swimming pool originally holds 200 gallons of water containing 100 pounds of chlorine. Pure water is pumped into the pool at a rate of 4 gal/min, but pool water is being splashed out of the pool by enthusiastic children at a rate of 2 gal/min. Find the total amount of chlorine left in the pool when the pool holds 500 gallons of water. Solution First notice that the volume of water in the pool is initially 200 gallons, and for each minute afterwards, 4 gallons of water are added and 2 are lost, for a net change of 2 gallons per minute. So the volume of the pool is V = t. Let A(t) = the number of pounds of chlorine in the pool at time t minutes after the start. Then A(0) = 100 by the information we are given, and A amount of chlorine in amount of chlorine out (t) =. min min However, since the water coming into the pool is pure, the amount of chlorine in is 0. Water is being splashed out at a rate of 2 gallon/minute, and the concentration of chlorine in the water going out is the total amount A(t) of chlorine in the water, divided by the total volume V (t) of water. So the amount of chlorine out per minute is Thus This equation is separable: we get A(t) lbs V (t) gal 2 gal min = 2A lbs t min = da = A = A da A = Integrating both sides, using 1 = ln(100 + t) + C, we get 100+t Hence ln A = ln(100 + t) + C. A lbs t min. A(t) = e ln A = e ln(100+t)+c = e ln(100+t) e C = e ln[(100+t) 1] B = B(100 + t) 1 = where we have set B = e C. Since A(0) = 100, we get 100 = B, or B = = 10, Substituting this value of B into the formula for A(t), we get A(t) = 10, 000 B t, Since the volume of the pool is V (t) = t, the pool holds 500 gallons at time t = 150 minutes. At that time, the amount of chlorine in the pool is A(150) = 10, , 000 = = 40 pounds.

3 Math , Spring 2010: Exam 2 Solutions 3 4.) (10 points) Solve y 6y + 9y = 0, y(0) = 1, y (0) = 2. Solution The characteristic equation is r 2 6r + 9 = 0, which factors as (r 3) 2 = 0. Hence r = 3 is a repeated root. So y 1 = 3 3t and y 2 = te 3t are solutions, and the general solution is From the given condition y(0) = 1, we get y = C 1 e 3t + C 2 te 3t. 1 = C 1 e 0 + C 2 0 e 0 = C = C 1. Hence C 1 = 1. Therefore we have Using the product rule, we obtain y = e 3t + C 2 te 3t. From the given condition y (0) = 2, we get y = 3e 3t + C 2 e 3t + 3C 2 te 3t. 2 = 3e 0 + C 2 e 0 + 3C 2 0 e 0 = C = 3 + C 2. Hence C 2 = 5, and therefore y = e 3t + 5te 3t. 5.) (10 points) Solve y 2y y + 2y = 0. Solution The characteristic equation is r 3 2r 2 r+2 = 0. Noticing that = 0, we see that 1 is a root of r 3 2r 2 r +2. By the root/factor theorem, r 1 is a factor of r 3 2r 2 r +2. Dividing r 1 into r 3 2r 2 r + 2 by long division of polynomials, we obtain r 3 2r 2 r + 2 = (r 1)(r 2 r 2) = (r 1)(r 2)(r + 1). Hence r 3 2r 2 r + 2 has three distinct real roots, r = 1, 2, and 1. Hence y 1 = e t, y 2 = e 2t, and y 3 = e t are solutions of the differential equation. Therefore the general solution is for arbitrary constants C 1, C 2, and C 3. y = C 1 e t + C 2 e 2t + C 3 e t,

4 Math , Spring 2010: Exam 2 Solutions 4 6.) (20 points) An arrow of mass 1/2 kg is shot upward with an initial velocity of 98 m/sec. Assume that the force of air resistance, in newtons, is 2.45v, where v is the velocity of the of the arrow. Find the time it takes after the arrow is shot for the arrow to reach its maximum height. The magnitude of the force of gravity is 9.8 m/sec 2. Your answer can be expressed using logarithms. Solution We use a coordinate system with origin at the point from which the arrow is shot, and with positive axis directly upward. By Newton s law, F = ma = mv, where F is the net force acting on the arrow. Here F consists of the force of gravity, equal to mg newtons (the sign is negative because gravity pulls the arrow down, which is the negative direction in our coordinate system), and the force due to air resistance, which is given to be 2.45v newtons. Hence mv = F = mg 2.45v. Substituting 1/2 for m and 9.8 for g, we get Multiplying by 2, we have v = v, or 1 2 v = 1 (9.8) 2.45v. 2 v + 4.9v = 9.8. This equation is linear, with integrating factor e 4.9 = e 4.9t. Multiplying through gives e 4.9t v + 4.9e 4.9t v = 9.8e 4.9t. The left side is the derivative by the product rule of e 4.9t v, so d ( e 4.9t v ) = 9.8e 4.9t. Integrating both sides and applying the fundamental theorem of calculus gives d ( e 4.9t v = e 4.9t v ) = 9.8e 4.9t = e4.9t + C = 2e 4.9t + C. Multiplying the last equation through by e 4.9t gives v = 2 + Ce 4.9t. We are given that the initial velocity v(0) = 98, so Hence C = 100, and we have 98 = 2 + Ce 0 = 2 + C. v = e 4.9t. The arrow reaches its maximum height when its velocity is 0, We want to find the time when that happens, so we set v equal to 0 and solve for t. We get 0 = e 4.9t, so 2 = 100e 4.9t, or 1 = 50 e 4.9t. Taking the natural logarithm of both sides gives ln 1 = 4.9t. 50 Since ln 1 = ln 50, we have ln 50 = 4.9t. Cancelling the minus sign and dividing by 4.9 gives 50 t = ln This is how your answer should be left, but just for curiosity, using a calculator we get t.798 seconds, which is not very realistic.

5 Math , Spring 2010: Exam 2 Solutions 5 7.) (20 points) At noon, a glass of cold water, with temperature 40 F, is placed in a room with constant temperature 75 F. At 12:30 p.m., the temperature of the water is 50 F. How many hours after noon will it be when the water reaches the temperature 60 F? Assume that Newton s Law of heating/cooling applies in this problem. Your answer should be expressed in terms of logarithms. Let T (t) denote the temperature of the body at time t hours after noon, measured in degrees Farenheit. Newton s Law of Cooling states that = k(75 T ), for some constant k, because the ambient temperature is 75 F. This equation is separable: Integrating both sides gives 75 T = k, or 75 T = k. ln 75 T = kt + C. Multiplying through the last equation by 1 and noting that the temperature of the water will be less than 75 and hence 75 T = 75 T, we get Taking the exponential of both sides, where A = e C is a constant. Hence ln(75 T ) = kt C. 75 T = e ln(75 T ) = e kt C = e kt e C = Ae kt, T = 75 Ae kt. Using T (0) = 40, we get 40 = 75 Ae 0 = 75 A, so A = = 35. Therefore we have T = 75 35e kt. Since the temperature at 12:30 is 50 F, or T (1/2) = 50, we have 50 = 75 35e k/2. Hence 25 = 35e k/2, or e k/2 = 25 = 5. Taking the natural logarithm of both sides gives 35 7 k/2 = ln(5/7) = ln(7/5), or k = 2 ln(7/5). Therefore T = 75 35e 2 ln(7/5) t. We want to find t so that T = 60; so we set T = 60 and solve for t: or 15 = 35e 2 ln(7/5) t, or e 2 ln(7/5) t = ln(7/5)t = ln(3/7) = ln(7/3). Hence t = ln(7/3) 2 ln(7/5) 60 = 75 35e 2 ln(7/5) t, which is about minutes, or at about 1:15 p.m. = 3. Taking the natural logarithm of both sides, hours after noon,