4. Some Applications of first order linear differential

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3 August 30, Some Applications of fist ode linea diffeential Equations The modeling poblem Thee ae seveal steps equied fo modeling scientific phenomena 1. Data collection (expeimentation) Given a cetain physical system, one has to un expeiments and get some idea of how the obseved data depend on time. 2. Setting up scientific law to descibe the time dependence This may involve diffeential o diffeence equations. The idea is to find the coect equations whose solutions give the obseved time dependence. 3. Analysis of solutions of appopiate equations to descibe obseved phenomena. We will descibe seveal known applications involving this pocess. Radioactive Decay It is known that cetain adioactive substances exhibit spontaneous decay. That is, if Q(t) epesents the amount of the substance at time t, then Q(t) satisfies the diffeential equation dq = Q(t) (1) dt whee is a positive eal numbe. This simply means that the ate of decay of the quantity at time t is popotional to the amount pesent at time t. We know that the geneal solution to (1) is Q(t) = Q(0)e t whee Q(0) is the amount pesent at time 0. We can use this to solve vaious questions elated to adio-active decay. 1. The element Thoium-234 (Th-234) exhibits adio-active decay. If 100 mg of Th-234 decays to mg in one week, find an expession fo

4 August 30, the amount at any time t. Also, find the half-life of the element (the amount of time it takes to decay to half its oiginal value). Let Q(t) denote the amount at time t. Let Q 0 = Q(0). Then, Q(t) = Q 0 e t. If t is measued in units of days, and Q(t) is measued in units of milligams (mg), then Q 0 = 100, Q(7) = 100e 7 = 82.04, e 7 = 82.04/100, So, = log(82.04/100) 7 = Q(t) = 100e t = 100e 0.028t. Let t h denote the half-life. Then, So, Q(t h ) = Q 0 e t h = Q = e t h, 2 = e t h. t h = log(2). (2)

5 August 30, Cabon Dating All living beings contain oughly the same concentation in cells of a cetain amount of the adio-active element Cabon-14, C 14. This substance decays at a cetain ate, but gets eplenished by living beings which beathe fom the atmosphee. When a being dies, it does not eplenish its Cabon-14, so the concentation which emains in the cells is smalle than that which was oiginally thee. Since the concentation satisfies Q(t) = Q 0 e t (3) fo some constants Q 0, > 0, and the half-life of C 14 is about 5730 yeas, one can use this fo dating achealogical objects. Let us take an example. Suppose that the emains of a cetain substance contains 20 % of its oiginal amount. How old is the substance? Fom fomula (2), we have o, 5730 = log(2) = log(2) 5730 We seek the time t 1 such that o o Q(t 1 ) = Q 0 5 = Q 0e t = e t 1 log(5) = t 1, t 1 t /(1.21 (10) 4 ) yeas

6 August 30, Compound inteest If inteest is compounded continuously, this gives an example of a model exhibiting exponential gowth. Let us eview inteest calculations. Suppose we ae given a cetain inital amount of money, called the pincipal amount P (0). If this is compounded annually at a ate of 5 % and P (t) denotes the amount of money available afte t yeas, we have P (1) = P (0) +.05 P (0) = P (0)(1 +.05) P (2) = P (1) +.05 P (1) = P (1)(1 +.05) = P (0)(1 +.05) 2 P (t) = P (0)(1 +.05) t Now, suppose the inteest is 5 % pe yea, but compounded monthly. The inteest pe month is 5/12 %. In t yeas, we compound 12t times. So, we get P (t) = P (0)(1 + (.05/12)) 12t. If the inteest is at the ate of %, we get P (t) = P (0)( )12t If the inteest is compounded n times pe yea, we get P (t) = P (0)( n )nt When n, we say we have inteest compounded continuously. What is P (t)? We take H def = n lim ( n )nt Taking logs we get log H = lim n nt log( n ) = lim n log(1 + 1 nt 100 n )

7 August 30, Fo small x, log(1 + x) x, so we get So, and we have log H = lim n 100 n 1 nt = t 100 H = exp( t 100 ), P (t) = P (0)e t 100. Mixing Poblems Suppose a tank contains a solution of Q 0 lbs. of salt dissolved in 100 gallons of wate. Assume that a solution containing 1 lb of salt/gal is poued 4 into the tank at a ate of gal/min. Assume that the solution mixes instantaneously and that the combined solution is dained fom the tank at the same ate of gal/min. 1. How much salt is thee in the tank at time t > 0? 2. Find the limiting amount Q L as t. 3. If = 3, and Q 0 = 2Q L, find the time T fo Q(T ) to be within 2% of Q L. 4. What must be fo T to be no lage than 45 minutes? Solution: Let Q(t) be the amount of salt in the tank at time t. We fist find Q(t). Then we will see that the othe questions can be answeed simply. Let Q in (t) denote the amount of salt that has flowed into the tank at time t, and Q out (t) denote the amount that has flowed out of the tank at time t. Since the numbe of gallons flowing into the tank equals the numbe of gallons flowing out of the tank. The total numbe of gallons emains fixed at 100.

8 August 30, Then, Q(t) = Q 0 + Q in Q out and Q (t) = Q in(t) Q out(t) Now, Q in = 4 lb/min and Q out = (amount of salt pe gallon) Hence, we get the d.e. (numbe of gallons flowing out pe minute) = Q(t) 100 o Q = 4 Q(t) 100, Q + Q(t) 100 = 4. This is a linea d.e., with solution obtained fom µ = e t 100 [ ] Q(t) = e t e t C Q(t) = 25 + Ce t 100 Q 0 = 25 + C, C = Q 0 25

9 August 30, Q L = 25. Q 0 = 2Q L Q 0 = 50, C = 25. Find T such that Q(T ) Q L < (.02)Q L. Plug into above and get Q(T ) < (1.02)25 Q(T ) = e 3T 100 < (1.02)25 Then, solve fo T. Obseve that if we had diffeent ates in of input and out of output, and we let V (t) be the volume in the tank at time t, then we would get the elations and V (t) = V (0) + t( in out ), and So, Q in = (amount of salt pe gal coming in) (numbe of gallons pe unit time coming in), Q out = (amount of salt pe gal going out) (numbe of gallons pe unit time going out), = Q(t) V (t) out.

10 August 30, Q = Q in Q out = (amount of salt pe gal coming in) in Q(t) V (t) out. Newton s Law of Cooling: Assume a solid body B with initial tempeatue Θ 0 (at time t = 0) is immesed in an ambient fluid whose tempeatue is kept at the constant value T. Let Θ(t) denote the tempeatue of the body at time t. Newton s law of cooling states that Θ (t) = k(θ(t) T ) fo some constant k. That is, the ate of change of the tempeatue of B at time t > 0 is popotional to the diffeence of the tempeatue of B and the tempeatue T of the ambient fluid. Let us solve this d.e. We have dθ dt = k(θ T ) dθ Θ T = kdt log(θ T ) = kt + c Θ T = Ce kt Θ = T + Ce kt = T + (Θ 0 T )e kt Have 3 paametes T, θ 0, k to detemine. Typical Poblem Suppose that an object whose tempeatue is 40 degees Celsius (40 C) is placed in a oom whose tempeatue is maintained at

11 August 30, degees Celsius (20 C). One minute late, the tempeatue of the object is 36 C. Assuming that Newton s law of cooling holds, what is the tempeatue of the object 10 minutes late? Hee we have Θ 0 = 40, T = 20. Then, letting t = 1, we have Now, plugging in T = 11 gives Θ(t) = Θ(1) = 20 + (40 20)e k 36 = e k = ek k = ln 4 5 Θ = e 11(ln 4 5 )