Name: October 24, 2014 ID Number: Fall Midterm I. Number Total Points Points Obtained Total 40

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1 Math 307O: Introduction to Differential Equations Name: October 24, 204 ID Number: Fall 204 Midterm I Number Total Points Points Obtained Total 40 Instructions.. Show all your work and box the final answer. 2. You are allowed a handwritten cheat sheet 8.5 inches, double-sided 3. You are allowed a scientific calculator with trigonometric functions, however the calculator should not have calculus operations (i.e., differentiation, integration, etc).

2 Problem. A tank with a capacity of 320 gal originally contains 200 gal of water with 00 lb of salt in solution. Water containing 2 lb of salt per gallon is entering at a rate of 2 gal/min, and the mixture is allowed to flow out of the tank at a rate of 4 gal/min. (a) Find an explicit formula for the amount of salt Q(t) in the tank at any time. Sketch a graph of Q(t). (5 points) (b) Find the interval of time where the solution Q(t) obtained in part (a) accurately describes the amount of salt in the tank. ( point) (c) Find the concentration (in pounds per gallon) of salt in the tank when it is filled half its capacity. (2 points) (d) Find the maximum amount of salt in the tank at any moment. (2 points) Solution. (a) We have then V 0 = 200, Q 0 = 00, α = 2, r = 2, r 2 = 4 dq(t) dt This is a linear DE with integral factor Using the solution formula for + 2 Q(t) = 4, Q(0) = t 2 µ(t) = e 00 t dt = e 2 ln 00 t = (00 t) 2. Q(t) = (00 t) 2 4 (00 t) ( 2 ) = (00 t) 2 4 (00 t) + C = 4(00 t) + C(00 t) 2 Using the initial condition Q(0) = 00 we obtain C = 3/00. Finally,we obtain Q(t) = 4(00 t) 3 00 (00 t)2 () or equivalently, Q(t) = t 3t2 00 (2) 2

3 Scratch Paper (b) The formula for Q(t) in Eq. (2) represents the quantity of salt in the tank for t 0 and as long as Q(t) 0. Since Q(t) = 0 t = 40 or t = 00, then the interval is I = [0, 00] (c) The tank if half its capacity when V (t) = 200 2t = 320/2 = 60, then t = 20. The concentration at t = 20 is Q(20) V (20) = 28 = 4/5 = (d) The maximum amount of salt is attain at time t = t 0 such that Q (t 0 ) = 0, i.e., Q (t 0 ) = 3 50 t t 0 = 00/3 = Then the maximum amount of salt int he tank at any moment is Q(00/3) = 400/3 = 33.3 In particular the graph of Q(t) is shown below. 3

4 Problem 2. Two substance P and Q react to form a substance X. The rate at which X is produced is proportional to the product of the amount of substance P and Q that is present at time t. To form X we need 2 lb of P for each pound of Q. If initially there are 00 lb of P and 50 lb of Q and after 0 minutes there are 0 lb of X. (a) Find x(t) the amount of the substance X after t minutes. (7 points) (b) What is the maximum amount of X that can be formed? ( point) (c) How much will remain of substances P and Q after a long period of time? (2 points) Solution. (a) We have p = 00, q = 50 and n = 2/3 then x(t) satisfies the following DE dx (00 dt = k 23 ) (50 x 3 ) x, x(0) = 0, x(0) = 0. Simplifying the DE we get Using separation of variables dx dt = 2k 9 (50 x)2. dx 2k (50 x) 2 = 9 dt then 50 x = 2k 9 t + C. Using x(0) = 0 we obtain C = /50, similarly using x(0) = 0 we get k = 3/4, 000, then 50 x = 2, 000 t Solving for x(t) we obtain x(t) = 50 t 2, or alternatively x(t) = 50t t

5 Scratch Paper (b) Taking limit as t of x(t) we get the maximum amount of X that can be formed, 50t lim x(t) = lim t t t + 40 = 50 (c) The amount of P and Q at any time t is given by P (t) = x(t), Q(t) = 50 3 x(t). When x = 50 what remains of P and Q is lim P (t) = 00 2 t 3 50 = 0 lim Q(t) = 50 t 3 50 = 0 5

6 Problem 3. Let y(t) denote the population of passenger pigeons on a small forest at any given time t, where time is measured in months. Assume that the rate of growth or decline of the pigeons is model by the differential equation dy dt = 2y y2, 000 (a) Draw a direction field of the differential equation and use it to indicate the environmental carrying capacity (i.e., the amount of pigeons expected after a long period of time) of the forest. (3 points) (b) Assume that at time t = 0 there are,500 pigeons. Use Euler s method with step size h = 0.5 to approximate the amount of pigeons after one month. (2 points) (c) Assume that we want to introduce a threshold T =, 000 to the model, below which grow does not occur. Modify the differential equation to include the desired threshold and graph the direction field for this new differential equation. You do not need to solve the differential. (5 points) Solution. (a) The equilibrium solutions of the DE are given by dy dt = 2y y2 = 0 y = 0 and y = 2, 000., 000 Since dy/dt is negative for y < 0 and y > 2, 000; and positive when 0 < y < 2, 000 then the direction field looks like Then the environmental carrying capacity is given by y = 2, 000 6

7 Scratch Paper (b) Since time is measured in months, we need to approximate y() using steps of h = 0.5. Denote f(y) = dy dt = 2y y2,000. From Euler s Method we have y(0.5) = y(0) + f(, 500) 0.5 = 875. and similarly y() = y(0.5) + f(, 875) , 992 (c) We notice that the DE dy dt = 2y y2, 000 is a logistic equation written in standard form as ( dy dt = 2 y ) y 2, 000 To introduce a threshold T =, 500 we consider the logistic equation with threshold given by ( dy dt = 2 y ) ( y ) y (3) 2, 000, 000 The direction field of this DE is given by Figure : Graph of the direction field of DE (3). The environmental carrying capacity K = 2, 000 is in red color while the threshold T =, 000 in green color. 7

8 Problem 4. A college graduate borrows $0,000 to buy a car to a the lender that charges interest at an variable annual rate of r(t) = percent. Assume 0+t the student make payments continuously at a variable annual rate k(t) = 200t + 2, 000. If S(t) denotes the amount due by the student after t years, then S(t) satisfies the following differential equation, ds(t) dt = r(t)s(t) k(t). (a) Find an explicit formula for S(t). (6 points) (b) Find the time t 0 that it takes the student to pay off and the final amount of interest paid. (3 points) (c) Assume that initially, the student was given the option changing his variable annual rate r(t) for a constant annual rate r 0 equals to the average rate along the period of the loan, i.e., r 0 = r(0) + r(t 0). 2 Give an heuristic reason of whether or not he made a good decision by taking the variable annual rate r(t). ( point) [Hint: Consider the deposit function] Solution. (a) We have that for t > 0, S(t) satisfies the DE ds dt = S(t) 200t 2, 000, S(0) = 0, t This is a linear DE with integral factor µ(t) = e 0+t dt = e ln 0+t = 0 + t Then S(t) = (0 + t) = (0 + t) 200t + 2, 000 dt 0 + t 200dt = 200t(0 + t) + C(0 + t) Using the initial condition S(0) = 0, 000 we get C =, 000 8

9 Scratch Paper Hence or equivalently S(t) = 200t(0 + t) +, 000(0 + t) (4) S(t) = 0, 000, 000t 200t 2 (b) The debt is paid when S(t) = 0, using Eq.(4) we get 200t(0 + t) +, 000(0 + t) = (0 + t)( 200t +, 000) = 0 t = 5 Notice that t must be a positive value so t = 0 is not considered. The deposit function is given by D(t) = t Hence after 5 year he has paid in interests 0 200s + 2, 000sds = 00t 2 + 2, 000t. D(5) 0, 000 = 2, 500 0, 000 = 2, 500 (c) We have r 0 = 2 (/0 + /5) = /2. By looking that the graphs of r(t) = 0+t vs r 0 = /2 we notice that at the beginning of the loan he is charged more interest if choosing the variable rate, while at the end the end of the loan the opposite happens. Figure 2: Comparison of interest rates during a 5 years period. 9

10 Figure 3: Graph of deposit function D(t) = 00t 2 + 2, 000t. Since the deposit function is increasing as shown in Fig. 3, his best strategy is to pay less interest at the begging of the loan. In other words, he made a bad decision by choosing the variable rate. In fact, if using the constant rate r 0 = /2 he would have paid the loan in approximate 4, 95 years saving about $50 in total. 0