Modeling with First-Order Equations

Modeling with First-Order Equations MATH 365 Ordinary Differential Equations J. Robert Buchanan Department of Mathematics Spring 2018

Radioactive Decay Radioactive decay takes place continuously. The number of atoms decaying at any instant is proportional to the number of un-decayed atoms present. Let Q represent the number of un-decayed atoms and t represent time. A first-order ODE describing radioactive decay is dq dt = k Q. The proportionality constant k is called the decay constant and is assumed to be positive.

Half-Life Definition The half-life (denoted t 1/2 ) of the radioactive substance is the time required for half of the initial amount to decay.

Half-Life Definition The half-life (denoted t 1/2 ) of the radioactive substance is the time required for half of the initial amount to decay. Find the relationship between the half-life t 1/2 and the decay constant k. Assuming t 0 = 0 then Q 0 2 = Q 0 e k t 1/2 1 2 = e k t 1/2 ln 1 2 = k t 1/2 ln 2 = k t 1/2 t 1/2 = ln 2 k.

Continuous Infusion Suppose that a chemical is eliminated from the blood stream of an animal at a rate proportional to the amount of the chemical present (similar to radioactive decay). The chemical enters the animal s blood stream at a rate given by the function F(t). A first order differential equation describing the amount of chemical in the blood stream is then dq dt = k Q + F(t).

Example (1 of 2) Suppose the chemical has a half-life of 2.5 hours in the blood stream and enters the blood stream at a constant rate C. If the chemical is a beneficial medicine whose effective dose is 50 ± 2.5 mg, what should the infusion rate be?

Example (1 of 2) If the infusion is constant and never ceases, then the ODE in addition to being first-order and linear is autonomous. The beneficial equilibrium solution is an amount of 50 mg. We can find the value of C for which Q(t) = 50 is an equilibrium solution. dq dt + k Q = C 0 + ln 2 2.5 (50) = C C = 20 ln 2 13.8629 mg/hr.

Example (1 of 2) (ln 2)t The integrating factor is µ(t) = e 2.5. d [µ(t)q(t)] = (20 ln 2)µ(t) dt d [µ(t)q(t)] dt = (20 ln 2) µ(t) dt dt µ(t)q(t) = (20 ln 2)µ(t) 2.5 ln 2 + D (ln 2)t Q(t) = 50 + De 2.5 0 = 50 + D D = 50 (ln 2)t Q(t) = 50 50e 2.5

Graph Q t 50 40 30 20 10 5 10 15 20 t

Graph Q t 50 40 30 20 10 5 10 15 20 t How long after the infusion is started will it take until the effective dose is achieved?

Effective Dose The medicine becomes effective when the dose is 50 2.5 = 47.5 mg.

Effective Dose The medicine becomes effective when the dose is 50 2.5 = 47.5 mg. Thus we must solve the equation: Q(t) = 47.5 (ln 2)t 50 50e 2.5 = 47.5 (ln 2)t 50e 2.5 = 2.5 (ln 2)t e (ln 2)t 2.5 2.5 = 0.05 = ln 0.05 t = 2.5 ln 0.05 ln 2

Example (2 of 2) Suppose the chemical has a half-life of 3 hours in the blood stream and enters the blood stream at a rate given by 10(1 + sin(2πt)). Initially 50 mg of the chemical is in the blood stream. How much of the chemical is in the blood stream at time t > 0? We must solve the IVP: dq dt = kq + 10(1 + sin(2πt)) Q(0) = 50.

Solution The integrating factor is µ(t) = e kt.

Solution The integrating factor is µ(t) = e kt. t 0 dq + kq = 10(1 + sin(2πt)) dt d [ ] e kt Q(t) = 10e kt (1 + sin(2πt)) dt d [ ] t e ks Q(s) ds = 10 e ku (1 + sin(2πu)) du ds e kt Q(t) Q(0) = 0 20π k 2 + 4π 2 + 10(ekt 1) k + 10ekt k 2 (k sin(2πt) 2π cos(2πt)) + 4π2 Q(t) = 50e kt + 20πe kt k 2 + 4π 2 + 10(1 e kt ) k 10 + k 2 (k sin(2πt) 2π cos(2πt)) + 4π2

Graph Q t 52 50 48 46 44 42 5 10 15 20 25 30 t

Stirred/Mixed Tanks (1 of 3) Suppose a tank (or some other vessel) has a constant volume V and is being fed a solution of water and a chemical in concentration γ g/l at a rate of r L/min. The tank is well stirred so that the chemical is evenly distributed throughout the tank. The well-stirred mixture flows out of the tank also at rate r L/min. If the tank is initially filled completely with pure water, find the concentration of the chemical in the well-stirred tank as a function of time.

Stirred/Mixed Tanks (2 of 3) The amount of chemical in the incoming stream is the product of the incoming concentration and the incoming flow rate, γ r g/min.

Stirred/Mixed Tanks (2 of 3) The amount of chemical in the incoming stream is the product of the incoming concentration and the incoming flow rate, γ r g/min. We can express the concentration of the chemical in the tank C(t) as the amount of chemical in the tank Q(t) divided by the volume of the tank V, C(t) = 1 Q(t) g/l. V Since the concentration of the chemical in the outflow is the same as the concentration of the chemical in the tank (recall that the tank is well-stirred), the amount of chemical leaving the tank per minute is the product of the concentration in the tank and the outgoing flow rate, r Q(t) V = r C(t) g/min.

Stirred/Mixed Tanks (3 of 3) The rate of change in the amount of the chemical in the tank Q(t) is the difference between the inflow rate and the outflow rate, then an ODE describing the amount of chemical in the tank is dq = γ r r Q dt V. The concentration of the chemical in the tank C(t) is just the amount of chemical in the tank divided by the volume of the tank. An ODE for the concentration is 1 dq V dt dc dt = γ r V r V Q V = γ r V r V C.

Example Suppose a 500 L tank initially containing 500 L of pure water is fed with a solution of saline and water at a rate of 2 L/min with a concentration of 1/2 g/l of saline. Set up an IVP describing the concentration of saline in the tank as a function of time. Solve the IVP for t > 0. Find the lim t C(t). The appropriate IVP: dc dt dc dt = (1/2)(2) 500 + 1 250 C = 1 500 C(0) = 0. 2 500 C The ODE is first-order, linear with an integrating factor of µ(t) = e t/250.

Solution t 0 d ds dc + 1 dt 250 C = 1 500 d [ ] e t/250 1 C(t) = dt [ ] t e s/250 C(s) ds = 500 et/250 0 1 500 eu/250 du e t/250 C(t) C(0) = 1 2 (et/250 1) C(t) = 1 2 (1 e t/250 )

Solution t 0 d ds dc + 1 dt 250 C = 1 500 d [ ] e t/250 1 C(t) = dt [ ] t e s/250 C(s) ds = 500 et/250 0 1 500 eu/250 du e t/250 C(t) C(0) = 1 2 (et/250 1) C(t) = 1 2 (1 e t/250 ) ( ) 1 lim C(t) = lim t t 2 (1 e t/250 ) = 1 2

Graph C t 0.5 0.4 0.3 0.2 0.1 200 400 600 800 1000 t

Newton s Law of Cooling Suppose the temperature of an object is θ and the temperature of the environment in which the object sits is T. Furthermore assume the temperature of the environment is constant. Newton s Law of Cooling states that the rate of change in the temperature of the object is proportional to the difference between the temperature of the object and the temperature of the environment. A first-order, linear ODE describing the change in θ is dθ dt = k(θ T ).

Example Suppose a cup of coffee is initially 200 degrees Fahrenheit and one minute later is at 190 degrees Fahrenheit in a room whose temperature is 70 degrees Fahrenheit. When will the temperature of the coffee be 150 degrees Fahrenheit? An IVP describing this situation is dθ dt = k(θ 70) θ(0) = 200. We should also use the observation that θ(1) = 190 degrees Fahrenheit.

Solution Re-writing the ODE in standard form as dθ dt + k θ = 70k we see that the integrating factor is µ(t) = e k t.

Solution Re-writing the ODE in standard form as dθ dt + k θ = 70k we see that the integrating factor is µ(t) = e k t. t 0 d [ ] e k t θ(t) = 70ke k t dt d [ ] t e k s θ(s) ds = 70ke k u du ds e k t θ(t) θ(0) = 70(e k t 1) 0 θ(t) = 200e k t + 70(1 e k t ) θ(t) = 70 + 130e k t

Solution Re-writing the ODE in standard form as dθ dt + k θ = 70k we see that the integrating factor is µ(t) = e k t. t 0 d [ ] e k t θ(t) = 70ke k t dt d [ ] t e k s θ(s) ds = 70ke k u du ds e k t θ(t) θ(0) = 70(e k t 1) Question: how do we find k? 0 θ(t) = 200e k t + 70(1 e k t ) θ(t) = 70 + 130e k t

Determining the Value of k Recall that θ(1) = 190 so that we have the equation, θ(1) = 70 + 130e k(1)

Determining the Value of k Recall that θ(1) = 190 so that we have the equation, θ(1) = 70 + 130e k(1) 190 = 70 + 130e k e k = 12 13 k = ln 12 13 k = ln 13 12

Determining the Value of k Recall that θ(1) = 190 so that we have the equation, θ(1) = 70 + 130e k(1) 190 = 70 + 130e k e k = 12 13 k = ln 12 13 k = ln 13 12 0.0800427

Determining the Value of k Recall that θ(1) = 190 so that we have the equation, θ(1) = 70 + 130e k(1) 190 = 70 + 130e k e k = 12 13 k = ln 12 13 k = ln 13 12 0.0800427 t ln(13/12) θ(t) = 70 + 130e

When is the Coffee s Temperature 150 F? 150 = t ln(13/12) 70 + 130e 80 = t ln(13/12) 130e e t ln(13/12) = 8 ( ) 13 13 t ln = ln 12 ( 8 13 ) t = ln(8/13) ln(12/13)

When is the Coffee s Temperature 150 F? 150 = t ln(13/12) 70 + 130e 80 = t ln(13/12) 130e e t ln(13/12) = 8 ( ) 13 13 t ln = ln 12 ( 8 13 ) t = ln(8/13) 6.06561 min. ln(12/13)

Graph 200 Θ t 180 160 140 120 100 80 10 20 30 40 50 60 t

Escape Velocity (1 of 2) An object of mass m is launched perpendicularly from the surface of the earth with initial velocity v 0. Question: what is the minimum initial velocity that will allow the object to escape the earth s gravitational attraction? Assumptions: The only force on the object is gravity which varies inversely with the square of the distance separating the centers of mass of the object and the earth. The radius of the earth is R. The distance from the object to the surface of the earth is x.

Escape Velocity (2 of 2) Recall Newton s Second Law of Motion, F = m a, then Comments: a = dv dt = d 2 x dt 2 m a = m g R 2 (R + x) 2. The right-hand side of the equation captures the effect that the gravitational attraction of the earth decreases as x increases. We can treat x as the independent variable and v as the dependent variable using relationship (justified by the Chain Rule for Derivatives) a = dv dt = dv dx dx dt = dv dx v.

Solution (1 of 3) We now have the IVP: m v dv R 2 = m g dx (R + x) 2 v(0) = v 0. The ODE is first-order, separable.

Solution (1 of 3) We now have the IVP: m v dv R 2 = m g dx (R + x) 2 v(0) = v 0. The ODE is first-order, separable. v v dv dx = g R 2 (R + x) 2 R 2 v dv = g (R + x) 2 dx v 0 u du = x 1 2 (v 2 v 2 0 ) = g R2 R + x g R 0 g R 2 (R + z) 2 dz

Solution (2 of 3) Solving for v(x) yields v(x) = v 2 0 2g R + 2g R R + x.

Solution (2 of 3) Solving for v(x) yields v(x) = v 2 0 2g R + 2g R R + x. As the object travels upward, its velocity decreases toward 0 at its maximum altitude (apogee). Solving the equation v(x) = 0 produces 2g R 2g R R + x 2g R R + x R + x = = v 2 0 = 2g R v 2 0 2g R 2g R v 2 0 x = R v 2 0 2g R v 2 0

Solution (3 of 3) R v0 2 x max = 2g R v0 2 If the object has achieved escape velocity, the maximum altitude is +, so we need an initial velocity v 0 for which 2g R v 2 0 = 0 v 0 = 2g R. If we use R 6, 378, 140 m and g 9.80665 m/s 2 then v 0 11184.6 m/s 6.95 mi/s!

Homework Read Section 2.3 Exercises: 1, 3, 15, 19, 31, 32