Modeling with First Order ODEs (cont). Existence and Uniqueness of Solutions to First Order Linear IVP. Second Order ODEs

Transcription

1 Modeling with First Order ODEs (cont). Existence and Uniqueness of Solutions to First Order Linear IVP. Second Order ODEs September 18 22, 2017

2 Mixing Problem Yuliya Gorb Example: A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. 1 Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. 2 Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. t time, [min] (independent variable) Q amount of salt, Q = Q(t), [lb] (dependent variable) V volume of mixture, [gal] r 1 rate-in of mixture entering tank, r 1 = 3 [gal/min] c 1 concentration of salt in mixture entering tank, c 1 = 1 [lb/gal] r 2 rate-out of mixture leaving tank, r 2 = 2 [gal/min] V 0 initial volume of mixture, V 0 = 200 [gal] Q 0 initial amount of salt, Q 0 = 100 [lb]

3 Mixing Problem (cont.) Yuliya Gorb We assume that salt is neither created nor destroyed in the tank. Therefore, variations in the amount of salt are due solely to the flows in and out of the tank, i.e. the rate of change of salt in the tank is equal to the rate at which salt is flowing in minus the rate at which it is flowing out. That is, dq = Rate in (salt) Rate out (salt) (1) where Rate in (salt) = r 1c 1 and Rate out (salt) = r 2c 2 with c 2 being concentration of salt in mixture that flows out of the tank. It is equal c 2(t) = Q(t), where V (t) is the volume of mixture (2) V (t) This volume is variable due to different rates r 1 and r 2, and can be found from the IVP: { { dv dv = r 1 r 2 = 1 or V (0) = V 0 V (0) = 200 whose solution is V (t) = t [gal] (3)

4 Mixing Problem (cont.) Yuliya Gorb Now, collect info from (1), (2), (3) to obtain the model for the amount of salt: dq Q(t) dq = r 1c 1 r 2 = 3 2Q(t) V (t) or t Q(0) = Q 0 Q(0) = 100 The ODE: Q + 2Q(t) = 3 is linear and its integrating factor is t µ(t) = e t+200 = (t + 200) 2 d [ ] (t + 200) 2 Q(t) = 3(t + 200) 2 C Q(t) = t (t + 200). Using the IC we find C = , hence, 1 Q(t) = t (t + 200) 2 [lb] amount of salt at any time t 2 Now, we find the moment of time T when tank begins to overflow: V (T ) = T = 500 T = 300 min. Then the concentration of salt at that time is c(t ) = Q(300) V (300) = = 121 = [lb/gal] 125 So, the concentration of salt at the moment when the tank begins to overflow is lb/gal

5 Newton s Law of Cooling Example: Suppose that a murder victim was found at 8:30am and the temperature of the body at that time was 30 o C. The room in which the murder victim was discovered is a constant 22 o C. An hour later the body temperature became 28 o C. Determine the approximate time that the murder occurred? The physical principle that governs the process of body temperature change is Newton s Law of Cooling: The temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings, i.e. if T (t) is the current temperature of an object, T a is an ambient temperature, then dt where k is called the heat transfer coefficient that we will use to set the model = k (T (t) Ta), (4)

6 Newton s Law of Cooling (cont.) t time, [h] (independent variable) T current body temperature, T = T (t), [ o C] (dependent variable) T a room temperature, T a = 22[ o C] T 0 initial body temperature, T 0 = 30[ o C] T 1 body temperature in 1 hour, T 1 = 28[ o C] t 0 the initial time t 0 = 0 set at 8:30am k heat transfer coefficient, [1/h] Then the IVP describing our situation is dt = k (T (t) 22) (5) T (0) = 30 where the coefficient k will be determined from the condition T (1) = 28 later The ODE in (5) is separable dt = k. After integrating we have T (t) 22 log T (t) 22 = kt + C that yields T (t) = 22 + Ce kt o C, which is the body at any time t > 0. Using the IC we obtain T (t) = e kt o C

7 Newton s Law of Cooling (cont.) Now, use T (1) = 28 o C, we have 28 = e kt k = log 4 log 3, then T (t) = e (log 4 log 3)t o C We now need to find an instant of time T in hours when the body temperature was normal, i.e. 37 o C: (log 4 log 3)T 37 = e then log 15 log 8 T = log 4 log h Finally, to find the time of death, we note that 0.19 h 11 min, hence, the murder had occurred 2h11min prior 8:30am, that is, at 6:19am