Homework 2 Solutions Math 307 Summer 17

Homework 2 Solutions Math 307 Summer 17 July 8, 2017 Section 2.3 Problem 4. A tank with capacity of 500 gallons originally contains 200 gallons of water with 100 pounds of salt in solution. Water containing 1 pound of salt per gallon is entering at a rate of 3 gallons/minute, and the mixture is allowed to flow out of the tank at 2 gallons/minute. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. Compare this concentration with the theoretical limiting concentration if the tank had infinite capacity. Variables: Let Q(t) be the amount of salt in the tank in pounds. Let W (t) be the amount of water in the tank in gallons. Let t be time in minutes. In terms of just water, the rate of water in is 3 gal/min and the rate out is 2 gal/minute, so the tank is filling at a rate of 3 gal/min 2 gal/minute = 1 gal/min, so after every minute, there is one more gallon of water in the tank. Since the amount of water starts at 200 gallons, The setup for this mixing problem is dq dt W (t) = 200 + t. = rate in of salt rate out of salt where the rate in of salt is the concentration of salt in the water coming in times the rate of water coming in, so rate in of salt = (concentration in)(rate of water in) = (1 lb/gal)(3 gal/min) = 3 lb/min For the rate out, notice that the concentration of the salt coming out is the same as the concentration currently in the tank. Therefore, so our DE is rate out of salt = (concentration out)(rate of water out) ( ) Q lb 2Q lb = (2 gal/min) = W gal (200 + t) gal dq dt = 3 2Q 200 + t. This is not a separable equation, but it is linear, so we can solve for Q using integrating factors. 1

Step 1: Standard form: ( ) 2 Q + Q = 3 200 + t so p(t) = 2/(200 + t) and g(t) = 3. Step 2: Compute one solution for the integrating factor µ(t) (you don t need a constant of integration): µ(t) = e p(t)dt = e 2/(200+t)dt = e 2 ln 200+t = (200 + t) 2 Step 3: Multiply both sides of the DE by µ(t): Step 4: Rewrite and integrate: (200 + t) 2 Q + 2(200 + t)q = 3(200 + t) 2 ((200 + t) 2 Q) = 3(200 + t) 2 so integrating with respect to t, (200 + t) 2 Q = 3(200 + t) 2 dt = (200 + t) 3 + C Step 5: Solve for Q and C. Q = (200 + t) + C (200 + t) 2 Pluggin in the initial condition Q(0) = 100 (the amount of salt starts at 100), so C = 4, 000, 000 and 100 = 200 + C 200 2 Q = (200 + t) 4, 000, 000 (200 + t) 2. The tank is full when the amount of water W (t) reaches the capacity of the tank, which is 500, so when 500 = W (t) = 200 + t then t = 300 min. The concentration in the tank at this time is Q(300) 500 On the other hand, the concentration at time t is = 484 500 = 0.968. Q(t) W (t) = Q(t) 4, 000, 000 = 1 200 + t (200 + t) 3, so as t, the fraction converges to zero and the concentration converges to lim Q(t) = 1 0 = 1. t Problem 9. A certain college graduate borrows $8000 to buy a car. The lender charges interest at an annual rate of 10%. Assuming that interest is compounded continuously and that the borrower makes payments continuously at a constant rate k, determine the payment rate k that is required to pay off the loan in 3 years. Also determine how much interest is paid during the 3 year period. 2

Variables: Let S(t) be the amount owed. Let t be time in years. Our DE is ds dt = 0.1S k We also have the initial condition S(0) = 8000. Notice that the problem statement implicitly assumes that 0.1S k is negative for all t. Otherwise, if the student doesn t make large enough payments, then the student s loan will grow exponentially and they will never be able to pay it off. Separating variables and integrating, we get the solution or 0.1S k = Ce 0.1t. S = Ce 0.1t + 10k for some C (notice this C and the previous one are different). Pluggin in S(0) = 8000, we get C = 8000 10k. Therefore, S = (8000 10k)e 0.1t + 10k. In order for the student to pay off the loan in 3 years, that means S(3) = 0, so which can be rewritten as Solving for k, 0 = (8000 10k)e 0.3 + 10k. k(10e 0.3 10) = 8000e 0.3. k $3087 per year. Finally, the amount of interest paid is the total amount paid minus the initial amount loaned, so Interest paid = Total paid 8000 = (3 yr)(k dollars/year) 8000 $1260. Problem 16. Newton s Law of Cooling states that the temperature of an object changes at a rate proportional to the difference betweeen its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton s law of cooling. If the coffee has a temperature of 200 F when freshly poured, and 1 min later has cooled to 190 F in a room at 70 F, determine when the coffee reaches a temperature of 150 F. Variables: Let T be temperature and t be time in minutes. Then our DE is dt dt = k(70 T ) for some positive k > 0. We have the two conditions T (0) = 200 and T (1) = 190. The DE is linear first order, so using integrating factors we can solve to get T = 70 + Ce kt. Plugging in T (0) = 200 gives you C = 130, and plugging in T (1) = 190 gives you k = ln ( 12 13). Therefore, T = 70 + 130e ln(12/13)t. 3

Setting T = 150 and solving we get so 8 13 = eln(12/13)t t = ln(8/13) ln(12/13). Problem 20a. A ball with mass 0.15kg is thrown upward with initial velocity 20m/s from the roof of a building 30m high. Neglecting air resistance, find the max height above the ground that the ball reaches. We basically did this in class. If v(t) is the velocity and h(t) is the height, we saw that they are given by the equations v(t) = gt + v 0 = 9.8t + 20 h(t) = (g/2)t 2 + v 0 t + h 0 = 4.9t 2 + 20t + 30 (notice that these are both independent of the mass m = 0.15. In order to find the maximum height, that is when v(t) = 0, so t = 20/9.8 = 2.04. Plugging this into h(t) we get h(2.04) 50.41m, the maximum height of the ball. Problem 21a. Assume that the conditions are as in Problem 20, except that there is a force due to air resistance of magnitude v /30 directed opposite to the velocity, where the velocity v is measured in m/s. Find the max height above the ground that the ball reaches. As seen in class, in this case our differential equation for the velocity is m dv dt = mg v 30. We also have the initial condition v(0) = 20 and we know m = 0.15kg. Solving the differential equation with integrating factors we get v = 44.55 + Ce 0.22t Then we use the initial condition to solve for C: 20 = 44.55 + Ce 0.22 0 = 44.55 + C, so C = 64.55 and v = 44.55 + 64.55e 0.22t. The max height occurs when v = 0, so and solving this gives 0 = 44.55 + 64.55e 0.22t t = ln(0.68) 0.22 1.753s. Remember how I told you to compare the answer you get here with the answer you get in Problem 20? Comparing the times, we see that the ball peaks faster if there is air resistance, which makes sense. Ok, 4

so what about the max height? Let s solve for h(t) by integrating v(t) and then pluggin in the initial condition. ( ) 9.8 h(t) = v(t)dt = 0.22 + 65.5e 0.22t dt + C = 9.8 0.22 t 65.5 0.22 e 0.22t + C Plugging in the initial condition h(0) = 30, we can solve to get C = 327.7. Then since the ball peaks at t = 1.753, the max height is h(1.753) = 9.8 65.5 (1.753) 0.22 0.22 e 0.22(1.753) + 327.7 47.157m We see that this is slightly lower than the peak height in Problem 20, so our intuition served us well! Section 2.4 Problem 1. Determine an interval in which the solution of the given IVP is certain to exist First put it in standard form (t 3)y + (ln t)y = 2t, y(1) = 2 y + ln t t 3 y = 2t t 3. then p(t) = ln t/(t 3) and g(t) = 2t/(t 3). Since ln t is only defined for t > 0, both p(t) and g(t) are continuous on the intervals (0, 3) and (3, ). Since the initial condition t 0 = 1 is in (0, 3), Theorem 2.4.1 says there exists a unique solution to the IVP, and it is guaranteed to exist on the interval (0, 3). Problem 7. State where in the ty-plane the hypotheses of Theorem 2.4.2 are satisfied for the differential equation y = t y 2t + 5y. What they really mean is, where could the initial condition (t 0, y 0 ) be in the ty-plane so that there exists some rectangle around it on which both f and the partial derivative f/ y are continuous? First, note that f = t y 2t + 5y and f y = (2t + 5y)( 1) (5)(t y) (2t + 5y) 2 = 7t (2t + 5y) 2. Then both functions are continuous when 2t + 5y 0, meaning they are continuous everywhere but on the line y = 2t/5. If we draw this line. But given any initial condition (t 0, y 0 ) not on this line, you can find a rectangle containing the point that does not intersect the line of discontinuity (the rectangle might have to be really small if your point is close to the line). 5

Therefore, the hypotheses are satisfied at all points not on the line y = 2t/5. Section 2.5 Problem 3. Determine the equilibrium points and classify each one as asymptotically stable or unstable. Draw the phase line, and sketch several graphs of solutions in the ty-plane. dy dt = y(y 1)(y 2), y 0 0 The equilibrium solutions are y = 0, y = 1 and y = 2. For 0 < y < 1, the derivative is positive dy/dt > 0. For 1 < y < 2, the derivative is negative dy/dt < 0. For y > 2, the derivative is positive dy/dt > 0. In order to figure out the signs of the derivatives in each interval, you can simply plug in any y value in that interval into f(y) to check the sign. For example, f(1.5) = (1.5)(0.5)( 0.5) < 0. Therefore, all values of f(y) are negative in the interval 1 < y < 2. It follows that y = 0 is unstable, y = 1 is stable, and y = 2 is unstable. Problem 18. A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k and is lost through evaporation at a rate proportional to the surface area. (a) Show that the volume V (t) of water in the pond at time t satisfies the DE where α is the coefficient of evaporation. dv dt = k απ ( ) 2/3 3a V 2/3 πh 6

(b) Find the equilibrium depth of water in the pond. Is the equilibrium asymptotically stable? (c) Find a condition that must be satisfied if the pond is not to overflow. (a) As mentioned in the hint, let L(t) be the height of the water from the bottom of the cone. The surface of the water forms a circle of radius r(t), so the surface area is πr 2. Notice that the ratio of r to L is the same as the ratio of the radius of the whole cone a to the height of the whole cone h, so r L = a h. (0.1) Furthermore, since the body of water forms a cone of height L and radius r, its volume is V = 1 3 πr2 L. (0.2) from the volume formula of a right circular cone. Now let s write the differential equation. Since the rate in of water is k and the rate out of water is proportional to the surface area of the water, which is πr 2, we have dv dt = k απr2 where α is the coefficient of evaporation. So we see that we need to write r in terms of V. Let s solve for L in equation (0.1) and plug it into equation (0.2), then solve for r in terms of V : Plugging this back into the DE we get dv dt = k απ L = rh a V = 1 ( ) rh 3 πr2 = rhπ a ( ) 1/3 3V a r = πh ( ) 2/3 3V a = k απ πh 3a r3 ( ) 2/3 3a V 2/3 πh (b) Notice that they asking for you to solve for the value of L at equilibrium. Let s go back to the equation and set this quantity equal to zero. Then dv dt = k απr2 r = k απ (we take the positive square root because the radius can t be negative). Then we can plug back into equation (0.1) to get L. L = rh ( ) h k a = a απ First, notice that V, L and r all increase together, or they all decrease together. You can see that this equilibrium solution is indeed stable because as L increases, r increases as well, and the right hand side of the DE decreases. So as you increase r the rate of change must go from positive, to 0 at the above equilibrium, and then to negative. This means that our equilibrium is asymptotically stable. 7

(c) In order for the pond not to overflow, we would need the equilibrium L value to be at most the height of the cone h, so we need ( ) h k a απ h k απ a2 so we would need the inflow k to be at most απa 2. k απa 2 Problem 20. At a given level of effort, it is reasonable to assume that the rate at which fish are caught depends on the population y: the more fish there are, the easier it is to catch them. Thus we assume that the rate at which fish are caught is given by Ey where E is a positive constant, with units 1/time, that measures the total effort made to harvest the given species of fish. To include this effect, the logistic equation is replaced by dy = r(1 y/k)y Ey dt This equation is known as the Schaefer model. (a) Show that if E < r, then there are two equilibrium points, y 1 = 0 and y 2 = K(1 E/r) > 0. (b) Show that y = y 1 is unstable and y = y 2 is asymptotically stable. (c) A sustainable yield Y of the fishery is a rate at which fish can be caught indefinitely. It is the product of the effort E and the asymptotically stable population y 2. Find Y as a function of the effort E; the graph of this function is known as the yield-effort curve. (d) Determine E so as to maximize Y and thereby find the maximum sustainable yield Y m. (a) Let A be the expression A = (r E)K/r. Then the first thing to notice is that with a bit of rewriting we can factor the equation as dy = y(a y)(r/k) dt so the equilibrium points are y = 0 and y = A, which is the same as y 2 given in the problem statement. Then it is obvious that y 2 = A = (r E)K/r > 0 if E < r. (b) Since A, r and K are all positive, the graph of y versus y(a y)(r/k) looks like 8

so solutions below y = A increase up to y = A, and solutions above y = A decrease down to y = A. Therefore, y = y 2 = A is a stable equilibrium. Similarly, solutions near y = 0 diverge away from y = 0, so y = y 1 = 0 is unstable. (c) They literally just want you to write Y = Ey 2 = E(r E)K/r (d) In order to maximize Y with respect to E, we need to set dy/de = 0, so 0 = dy de = (r E)K/r EK/r by the product rule, and collecting and simplifying we get so E = r/2. Plugging back into Y we get E(2K/r) = K Y m = (r/2)(r r/2)k/r = rk 4 Section 2.6 Problem 7. Determine whether the equation is exact. If it is exact, find the solution. (e x sin y 2y sin x) + (e x cos y + 2 cos x)y = 0 9

Let M = e x sin y 2y sin x and N = e x cos y + 2 cos x. Since M y = ex cos y 2 sin x = N x, the ODE is exact. Therefore, in order to solve it, we must find a Φ(x, y) such that Φ x = M, Φ y = N. In order to find such a Φ, we integrate M and N with respect to x and y, respectively, so Φ = M dx = (e x sin y 2y sin x) dx = e x sin y + 2y cos x + C 1 (y) Φ = N dy = (e x cos y + 2 cos x) dy = e x sin y + 2y cos x + C 2 (x) where C 1 (y) is some function of y and C 2 (x) is some function of x. Since these expressions for Φ must match, then we see that both C 1 (y) and C 2 (x) must be constants. Hence, Φ = e x sin y + 2y cos x and the final solution is for any constant C. e x sin y + 2y cos x = C Additional Problems Problem A1. Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y 0, Rewrite the equation as y + t 2 y 2 = 0, y(0) = y 0. y = t 2 y 2, then we see that it is a separable equation. Also notice that since f(t, y) and f/ y are continous for all t and y, there exists a unique solution to the differential equation for any initial conditions. However, we cannot yet tell where those solutions are continuous. First, we should look for equilibrium solutions. Setting the right hand side of the equation equal to 0, we see the only equilibrium solution is y = 0. Therefore, when y 0 = 0 the unique solution is y = 0, which is continuous for all t. Second, we solve the separable equation, y 2 dy = t 2 dt We can solve for C using the initial conditions, y 1 = t 3 /3 + C 1 y = t 3 /3 + C. y 0 = 1 0 3 /3 + C = 1 C 10

so C = y 1 0, assuming y 0 0. So for y 0 0, the solution is y = 1 t 3 /3 + y0 1 = 3y 0 y 0 t 3 + 3 which is not defined when y 0 t 3 = 3 so when t = 3 3 y 0. The sign of this value depends on the sign of y 0, so there are two cases. If y 0 > 0, then 3 3/y 0 < 0 so the largest interval on which the solution is continuous is ( 3 3/y 0, ). Otherwise, if y 0 < 0, then 3 3/y 0 > 0 so the largest interval on which the solution is continuous is (, 3 3/y 0 ). In summary, Initial y 0 value Largest interval on which solution is continuous y 0 = 0 (, ) y 0 > 0 ( 3 3/y 0, ) y 0 < 0 (, 3 3/y 0 ) 11