HW 2.3.pdf - Shondel Wright Vaz MAT 275 ONLINE B Summer...

Unformatted text preview:Shondel Wright Vaz_MAT_275_ONLINE_B_Summer_2018 Assignment Section_2.3Modeling_with_First_Order due 07/08/2018 at 11:59pm MST 1. (1 point) Newton's law of cooling says that the rate of cooling of an object is proportional to the difference between the temperature of the object and that of its surroundings (provided the differ- ence is not too large). If T = T (t) represents the temperature of a (warm) object at time t, A represents the ambient (cool) temperature, and k is a negative constant of proportionality, which equation(s) accu- rately characterize Newton's law? . A. 27; = k(A — T) 2' _ : £895") ' dt _ D. 'fl—f = kT(T —A) B. All of the above 0 F. None of the above Answer(s) submitted: 0 C (correct) 2. (1 point) Water leaks from a vertical cylindrical tank through a small hole in its base at a rate proportional to the square root of the vol- ume of water remaining. The tank initially contains 350 liters and 18 liters leak out during the first day. A. When will the tank be half empty? I = days B. How much water will remain in the tank after 5 days? volume = Liters Solution: SOLUTION Let V(t) be the volume of water in the tank at time t, then dV — = kW. dt This is a separable equation which has the solution V(t) = (% +c)2. Since V(0) = 350 this gives 350 = C2 so V(t) = (g + M)? However, V(l) = 332, and so 332 = (g + my, so that k = 2(\/ 332 — \/ 350) = —0.9748. Therefore, V(t) = (—0.4874z + V350)? The tank will be half-empty when V(t) = 175, so we solve 175 = (—0.4874t + V350)2 to obtain t = 11.242 days. The tank will be half empty in 11.242 days. The volume after 5 days is V(5) which is approximately 264.755 liters. Answer(s) submitted: 0 11.24 0 264.75 (correct) 3. (1 point) A tank contains 2400 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 9 L/min, and is thoroughly mixed into it. The new solution drains out of the tank at the same rate. (a) How much sugar is in the tank at the begining? y(0) = — (kg) (b) Find the amount of sugar after t minutes. y(t) = — (kg) (c) As t becomes large, what value is y(t) approaching ? In other words, calculate the following limit. r11in y(t) = _ (kg) Solution: (a) Since the tank originally contains pure water, the initial amount of sugar is y(0) = 0. (b) First note that, since the incoming and outgoing flows of water are the same, the amount of water in the pool remains constant at 2400 L. We have dy . — 2 rate in — rate out dt where "rate in" and "rate out" refer to the rates at which the sugar enters and exits the tank, respectively. The rate at which the sugar enters the tank is given by rate in = (9 L/min) (0.07 kg/L) = 0.63 kg/min The concentation of sugar in the tank is 24% kg/L, so the rate of flow out is rate out = (9 L/min) [L kg/L] = {my kg/min 2400 Thus we obtain the differential equation dy 3 where each term has the units kg/min. We rewrite the differen- tial equation as dy E = _8(3)o(y_168) Separating the variables yields dy _ 3 /y—168_ /8°°dt 1n|y—168|=—fit+K SO and 3 y = 168 + Ce—m' The initial condition y(0) = 0 gives C = — 168, so i y =168—168e—800t (C) 3 1im<168—168e_mt> =168 t—)°° Answer(s) submitted: 0 0 o —((e'(—(t—266.66666ln(.63))/266.66666)—.63)/.00375) o 168 (correct) 4. (1 point) A tank contains 90 kg of salt and 1000 L of water. A solution of a concentration 0.045 kg of salt per liter enters a tank at the rate 7 L/min. The solution is mixed and drains from the tank at the same rate. (a) What is the concentration of our solution in the tank ini- tially? concentration = (kg/L) (b) Set up an initial value problem for the quantity y, in kg, of salt in the tank at time t minutes. "—3 = (kg/mix» y(0) = — (kg) (c) Solve the initial value problem in part (b). N) = ((1) Find the amount of salt in the tank after 4.5 hours. amount = — (kg) (e) Find the concentration of salt in the solution in the tank as time approaches infinity. concentration = (kg/L) Solution: (a) Since the tank originally contains 90 L, the initial concentration is % = 0.09kg/L. (b) First note that, since the incoming and outgoing flows of water are the same, the amount of water in the pool remains constant at 1000 L. We have dy . — I rate 111 — rate Ollt dt where "rate in" and "rate out" refer to the rates at which the salt enters and exits the tank, respectively. The rate at which the salt enters the tank is given by rate in = (7 L/n1in)(0.045 kg/L) = 0.315 kg/min The concentation of salt in the tank is 10% kg/L, so the rate of flow out is rate out = (7 L/min) [L kg/L] = 0.007y kg/min 1000 Thus we obtain the differential equation dy — = 0.315 — 0.007 dt y where each term has the units kg/min. The initial condition is y(0) = 90. (c) We rewrite the differential equation as dy _ E— 0.007(y 45) Separating the variables yields dy /y_—45——/0.007dt 1n |y — 45| = —0.007t +K SO and y = 45 + Ce—0.0071 The initial condition y(0) = 90 gives C = 45, so y(t) = 45 +4Se_°'°07' (d) The amount of salt after 4.5 hours is the given by y(4.5) = 45 + 45e—°-°°7<4-5) = 88.6046 (6) The concentration in the tank at any time t is given by 17% = 0.045 + 0.045e—0'007'so lim (0.045 + 004550-00") = 0.045 t—)°° Note that this limiting concentration equals the incoming con- centration. Answer(s) submitted: 9/100 .315-.007y 9o (.315+.315e"(—.007t))/.007 51.7982 45/1000 (correct) 5. (1 point) A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 8 L/min. The solution is mixed and drains from the tank at the rate 4 L/min. (a) Write an initial value problem for the amount of salt, y, in kilograms, at time t in minutes: d . d—f = <kg/nun)y(0) =—kg. (b) Solve the initial value problem in part (a) y(t) = — kg. (0) Find the amount of salt in the tank after 1 hours. amount = — fig) ((1) Find the concentration of salt in the solution in the tank as time approaches infinity. (Assume your tank is large enough to hold all the solution.) concentration = (kg/L) Solution: (a) First note that, since the incoming and outgo- ing flows of water are not the same, the amount of water in the tank is not constant. In fact the amount of water in the tank in- creases at a rate of 4 L/min and therefore the amount of water at any given time time t is given by 2000 + 4t. Let y(t) be the amount of salt (in kg) at time t minutes. We have dy . — = rate 1n — rate out dt where "rate in" and "rate out" refer to the rates at which the salt enters and exits the tank, respectively. Since pure water enters the tank, the rate at which the salt enters the tank is given by rate in = (8 L/min) (O kg/L) = 0 kg/min The concentation of salt in the tank is m kg/L, so the rate of flow out 1s . y y . t t:: 4 —————___ k 2:4________.k raw" ( um) [2000+4t 3'1"] 2000+4t gm" Thus we obtain the differential equation dy _ _ 4y dt _ 2000 + 4t where each term has the units kg/min. The initial condition is y(0) = 90. (b) Separating the variables gives dy 4 7 = _2000+4z dt Integrating ln|y| = —]n|2000+4t| +K and _ C y _ 2000+4t The initial condition y(0) = 90 gives C = 180000, so (t) _ 180000 y _ 2000+4t (d) The amount of salt after 1 hours, i.e. the amount of salt after 60 minutes, is the given by 1 80000 — = 80.3571 2000+4(60) y(60) = (d) The concentration in the tank at any time t is given by 2 180000 2000+4t — (2000+4t)2' lim 180000 _ 0 t—m (2000 + 4t)2 _ The limiting concentration is zero, as expected, since pure water is entering the tank. Answer(s) submitted: -y/(500+t) 9o 45000/(500+t) 1125/14 0 (correct) 6. (1 point) Newton's law of cooling states that the tempera- ture of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. Let k > 0 be the constant of proportionality. Assume the coffee has a temperature of 190 degrees Fahrenheit when freshly poured, and 1.5 minutes later has cooled to 177 degrees in a room at 62 degrees. (a) Write an initial value problem for the temperature T of the coffee, in Fahrenheit, at time t in minutes. Your answer will contain the uknown constant k: 6% = T(0) = (b) Solve the initial value problem in part (a). Your answer will contain the unknown constant k. T0): (0) Determine the value of the constant k k = _ minutes. ((1) Determine when the coffee reaches a temperature of 127 degrees. minutes. Solution: (a) T(t) satisfies the initial value problem d_T_ dz'_ —k(T — 62) T(0) = 190 (b) Solving the differential equation gives T(t) = 62 + Ce_kt Since T(0) = 190, we have T(t) = 62 + 128?," (c) After 1.5 minutes the temperature is 177, thus 177 2 62+ 128e—1-5k, which gives k = ——ln (E) % 0.0713988. (d) The temperature is given by To.) : 62+ 128e—0.0713988t To find when the coffee will reach a temperature of 127, we solve the equation 127 = 62 + 128e—0.0713988t This gives t = 9.49096 hours. Answer(s) submitted: —k(T—62) 190 62+(128)e' (—kt) .071399 9. 49093 (correct) 7. (1 point) A curve passes through the point (0,2) and has the property that the slope of the curve at every point P is five times the y- coordinate of P. What is the equation of the curve? y(X) = Solution: y(x) satisfies the differential equation dy _ = 5 dx y with solution y(X) = C635" Since the curve passes through the point (0,2), we have 2 = C so y(x) = 265'" Answer(s) submitted: 0 2e " (5x) (correct) 8. (1 point) A thermometer is taken from a room where the temperature is 22°C to the outdoors, where the temperature is 2°C. After one minute the thermometer reads 17°C. (a) What will the reading on the thermometer be after 2 more minutes? (b) When will the thermometer read 3°C? minutes after it was taken to the outdoors. Solution: Let T(t) be the temperature at time t, measured in minutes. T(t) satisfies the differential equation dT — = —k T —2 dt ( ) with solution T(t) = 2+Ce-kt Since T(0) = 22, we have T(t) = 2 + 20e'k' After one minute the temperature is 17, thus 17 = 2 + 20e—k, which gives k = —1n(%) z 0.287682. Hence TU) = 2 + 20e—0.287682t (a) The reading on the thermometer after 2 more minutes will be T(3) = 2+20e_0'287682(3) m 10.4375 (a) To find when the termomether will read 3, we solve the equa- tion 3 = 2 + 20e—0.287682t This gives t = 10.4133 minutes. Answer(s) submitted: 0 10 .4375 o 10 . 4133 (correct) 9. (1 point) Dead leaves accumulate on the ground in a forest at a rate of 3 grams per square centimeter per year. At the same time, these leaves decompose at a continuous rate of 85 percent per year. A. Write a differential equation for the total quantity Q of dead leaves (per square centimeter) at time t: 019 _ W — — B. Sketch a solution to your differential equation showing that the quantity of dead leaves tends toward an equilibrium level. Assume that initially (t = 0) there are no leaves on the ground. What is the initial quantity of leaves? Q(0) = _ What is the equilibrium level? gm 2 Does the equilibrium value attained depend on the initial condi- tion? 0 A. yes 0 B. no Solution: SOLUTION Let Q(t) be the quantity of dead leaves, in grams per square centimeter. Then 7%— — 3 — 0.85 Q, where t is in years. If we suppose that the initial quantity is Q(0) = 0, then we can guess what the solution will look like without solving the differential equation: initially 11% = 3, so the quantity is in- creasing. However, as it increases the rate of decay —0.85Q increases, so Q will increase until the decay rate is equal to the accumulation rate. This occurs when 3 = 0.85Q, or Q m 3.53. Note that Q~ ~ 3. 53 is the equilibrium solution, which is where Q 1s constant: when Q 1s constant, fl—g: 0, which gives the same condition as we solved above. Therefore we know that the solution will look like the figure below, which shows the solution in blue and the equilibrium in red. (Click on the graph for a larger version.) We can also solve the differential equation using separation of variables. d_Q =3— 085 d—t Qa SO 3— O—QS.5Q ='/dt and 1 —0—851n|3—0.85Q|=t+C Solving for Q, we get Q = 3.53 —Ae—°-85', With 9(0) = Q = 3.53(1—e083). Note that no matter what the initial condition is, the same equilibrium solution will be attained. —0. 85C e where A: :|:— 0— 85 Answer(s) submitted: 0 3- . 8 5Q o 0 0 3.52941 0 B (correct) 10. (1 point) According to a simple physiological model, an athletic adult male needs 20 calories per day per pound of body weight to maintain his weight. If he consumes more or fewer calories than those required to maintain his weight, his weight changes at a rate proportional to the difference between the number of calo- ries consumed and the number needed to maintain his current weight; the constant of proportionality is 1/3500 pounds per calorie. Suppose that a particular person has a constant caloric intake of H calories per day. Let W(t) be the person's weight in pounds at time t (measured in days). (a) What differential equation has solution W(t)? W W — (Your answer may involve W, H and values given in the prob- lem. ) (b) Solve this differential equation, if the person starts out weighing 165 pounds and consumes 3200 calories a day. W = — (c) What happens to the person's weight as t —> 0°? W —> Solution: SOLUTION (a) Since the rate of change of the weight is equal to l 3500 —(Intake — Amount to maintain weight) we have dW_ 1 H— 2 d—t 3500 ( 0W) (b) Starting off with the equation dW 1 H I: 1—75(W_%)' we separate variables and integrate: dW _ L... W—%_ 175 Thuswehave 1n _ __ 'w 10:] 175t+C sothat H —t/175 W 20=Ae or, in other words w_ _ E +Ae_t/175. 20 If the caloric intake is 3200 calories per day we know that H = 3200. Then, if the person initially weighs 165 pounds, we have 165 =160+A, so thatA = 5. Thus W =160+5e_'/175. (c) From our solution in (b), we know that as t —) 0°, W —> 160. This is seen in the graph of weight vs. time, shown below. Answer(s) submitted: 0 (l/3500)(H—20W) o l60+5e'(-t/l75) o 160 (correct) 11. (1 point) A 59kg skydiver jumps out of an airplane. We assume that the forces acting on the body are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with magnitude cv2 where c = 0. 15 25 1% and v(t) is the velocity of the skydiver at time t (and upward is positive velocity). The gravitational constant is g = 9.8m / s2. a) Find a differential equation for the velocity v : dv E— b) Determine the terminal velocity in meters per second for free-fall (no parachute). terminal velocity = — m/s Note: Answer should be negative for downward velocity. Solution: (a) The net force acting on the skydiver is given by the sum of the gravitational force and the air resistance: dv ma = —mg + cv2 Dividing by m yields dv _ c E _ _g 2 Substituting the given values of m, g and c yields the differential equation d 0.1525 —V = —9.8 + 2 dt 59 v (b) The terminal velocity is given by the equilibrium solution, which we can find by solving the equation 0.1525 2 59v 0 = —9.8+ This gives 578.2 V2 _ 0.1525" Since the velocity is downward, the limiting value is I 578.2 Answer(s) submitted: 0 -9.8+(.002585v'2) o —6l.575 (correct) 12. (1 point) A body of mass 6 kg is projected vertically upward with an initial velocity 20 meters per second. We assume that the forces acting on the body are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with magnitude c|v(t)| where c = 0.4% and v(t) is the velocity of the ball at time t. The gravitational constant is g = 9.8m / s2. a) Find a differential equation for the velocity v: dv E— b) Solve the differential equation in part a) and find a formula for the velocity at any time t: v(t) = Find a formula for the position function at any time t, if the initial position is s(O) = 0: s(t) = How does this compare with the solution to the equation for velocity when there is no air resistance? If c = 0, then v(t) = 20 — 9.8t, and if s(O) = 0, then s(t) = 20t—4.9t2. We then have that v(t) = 0 when t m 2.041, and s(2.041) x 20.408, and that the positive t solution to s(t) = 0 is t m 4.082, which leads to v(4.082) = —20 meters per second. Answer(s) submitted: 0 -9.8-(.4/6)v o 167e"(-.066667t)—l47 o —2505e'(—.06667t)—l47t+2505 (correct) When t = 0, Q = 0, 13. (1 point) A drug is administered intravenously at a con- r stant rate of r mg/hour and is excreted at a rate proportional to Q = g (1 — 6—,") the quantity present, with constant of proportionality k > 0. Th us, , r (a) Set up a differential equation for the quantity, Q, in mil- Q00 = tlgno Q = k- ligrams, of the drug in the body at time t hours. Your answer will contain the unknown constants r and k. Q'= This is shown in the following graph. (b) Solve this differential equation, assuming there is no drug in the body initially. Your answer will contain r and k. Q = (c) What is the limiting long-run value of Q? tlgg 90?) = Solution: SOLUTION The differential equation for Q is dQ _ k E _ r _ Q' (Click on the graph for a larger version.) so that d Answer(s) submitted: / Q = / dt, 0 r—kQ _ r—kQ . (r/k)*(l-e"(-kt)) and, after solvmg, r . r/k —k Q = g +149 t- (correct) Generated by ©WeBWorK, http:/Iwebworkmaa.org, Mathematical Association of America