ࡱ> b5@ 0@>bjbj22 >DXXllldFdFdF8F$HIv U"VVVWWWkmmmmmmRBmQl#[W@W#[#[mVVowww#[T 8VlVkw#[kwwGxTlVtI `X;dFwe kԑ0,Yr,0,lWXw!YtYWWWmmAdFgwdFChapter 4 Applications of First-order Differential Equations to Real World Systems Cooling/Warming Law Population Growth and Decay Radio-Active Decay and Carbon Dating Mixture of Two Salt Solutions Series Circuits Survivability with AIDS Draining a tank Economics and Finance Mathematics Police Women Drug Distribution in Human Body A Pursuit Problem Harvesting of Renewable Natural Resources Exercises In Section 1.4 we have seen that real world problems can be represented by first-order differential equations. In chapter 2 we have discussed few methods to solve first order differential equations. We solve in this chapter first-order differential equations modeling phenomena of cooling, population growth, radioactive decay, mixture of salt solutions, series circuits, survivability with AIDS, draining a tank, economics and finance, drug distribution, pursuit problem and harvesting of renewable natural resources. 4.1 Cooling/Warming law We have seen in Section 1.4 that the mathematical formulation of Newtons empirical law of cooling of an object in given by the linear first-order differential equation (1.17)  EMBED Equation.3  This is a separable differential equation. We have  EMBED Equation.3  or ln|T-Tm |=(t+c1 or T(t) = Tm+c2e(t (4.1) Example 4.1: When a chicken is removed from an oven, its temperature is measured at 3000F. Three minutes later its temperature is 200o F. How long will it take for the chicken to cool off to a room temperature of 70oF. Solution: In (4.1) we put Tm = 70 and T=300 at for t=0. T(0)=300=70+c2e(.0 This gives c2=230 For t=3, T(3)=200 Now we put t=3, T(3)=200 and c2=230 in (4.1) then 200=70 + 230 e(.3 or  EMBED Equation.3  or  EMBED Equation.3  or  EMBED Equation.3  Thus T(t)=70+230 e-0.19018t (4.2) We observe that (4.2) furnishes no finite solution to T(t)=70 since limit T(t) =70. t( ( The temperature variation is shown graphically in Figure 4.1. We observe that the limiting temperature is 700F. Figure 4.1 4.2 Population Growth and Decay We have seen in section 1.4.1 that the differential equation  EMBED Equation.3  where N(t) denotes population at time t and k is a constant of proportionality, serves as a model for population growth and decay of insects, animals and human population at certain places and duration. Solution of this equation is N(t)=Cekt, where C is the constant of integration:  EMBED Equation.3  Integrating both sides we get lnN(t)=kt+ln C or  EMBED Equation.3  or N(t)=Cekt C can be determined if N(t) is given at certain time. Example 4.2: The population of a community is known to increase at a rate proportional to the number of people present at a time t. If the population has doubled in 6 years, how long it will take to triple? Solution : Let N(t) denote the population at time t. Let N(0) denote the initial population (population at t=0).  EMBED Equation.3  Solution is N(t)=Aekt , where A=N(0) Ae6k=N(6) =2N(0) = 2A or e6k=2 or k = ln 2 Find t when N(t)=3A=3N(0) or N(0) ekt=3N(0) or  EMBED Equation.3  or ln 3=  EMBED Equation.3  or t=  EMBED Equation.3 (9.6 years (approximately 9 years 6 months) Example 4.3 Let population of country be decreasing at the rate proportional to its population. If the population has decreased to 25% in 10 years, how long will it take to be half? Solution: This phenomenon can be modeled by  EMBED Equation.3  Its solution is N(t)=N(0) ekt, where N(0) in the initial population For t=10, N(10)=N(0) N(0) = N(0) e10k or e10k= EMBED Equation.3  or k= EMBED Equation.3 ln  EMBED Equation.3  Set N(t)= EMBED Equation.3 N(0)  EMBED Equation.3  or t= EMBED Equation.3  ( 8.3 years approximately. Example 4.4 Let N(t) be the population at time t and Let N0 denote the initial population, that is, N(0)=N0. Find the solution of the model  EMBED Equation.3  with initial condition N(0)=No Solution: This is a separable differential equation, and its solution is  EMBED Equation.3   EMBED Equation.3  To find A and B, observe that  EMBED Equation.3  Therefore, Aa+(B-bA)N=1. Since this equation is true for all values of N, we see that Aa=1 and B-bA=0. Consequently, A= EMBED Equation.3 , B=b/a, and  EMBED Equation.3  =  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Thus at = ln EMBED Equation.3  It can be verified that  EMBED Equation.3  is always positive for 0<t<". Hence at = ln  EMBED Equation.3  Taking exponentials of both sides of this equation gives eat= EMBED Equation.3  N0(a-bN)eat = (a-bN0)N Bringing all terms involving N to the left-hand side of this equation, we see that [a-bNo + bN0eat] N(t) = aN0eat or N(t)= EMBED Equation.3  4.3 Radio-active Decay and Carbon Dating As discussed in Section 1.4.2. a radioactive substance decomposes at a rate proportional to its mass. This rate is called the decay rate. If m(t) represents the mass of a substance at any time, then the decay rate  EMBED Equation.3 is proportional to m(t). Let us recall that the half-life of a substance is the amount of time for it to decay to one-half of its initial mass. Example 4.5. A radioactive isotope has an initial mass 200mg, which two years later is 50mg. Find the expression for the amount of the isotope remaining at any time. What is its half-life? Solution: Let m be the mass of the isotope remaining after t years, and let -k be the constant of proportionality. Then the rate of decomposition is modeled by  EMBED Equation.3 = - km, where minus sign indicates that the mass is decreasing. It is a separable equation. Separating the variables, integrating, and adding a constant in the form lnc, we get lnm+lnc = - kt Simplifying, lnmc = - kt (4.3) or mc = e-kt or m = c1e-kt, where c1= EMBED Equation.3  To find c1, recall that m =200 when t=0. Putting these values of m and t in (4.3) we get 200 = c1 e-ko = c1.1 or c1=200 and m = 200e-kt (4.4) The value of k may now be determined from (4.4) by substituting t=2, m=150. 150 = 200 e-2k or  EMBED Equation.3  or 2k=ln  EMBED Equation.3  This gives  EMBED Equation.3 (0.2877)= 0.1438 ( 0.14 The mass of the isotope remaining after t years is then given by m(t) =200e -.1438t The half-life th is the time corresponding to m=100mg. Thus 100 = 200 e-0.14th or  EMBED Equation.3 = e-0.14th or th= -  EMBED Equation.3  Carbon Dating: The key to the carbon dating of paintings and other materials such as fossils and rocks lies in the phenomenon of radioactivity discovered at the turn of the century. The physicist Rutherford and his colleagues showed that the atoms of certain radioactive elements are unstable and that within a given time period a fixed portion of the atoms spontaneously disintegrate to form atoms of a new element. Because radioactivity is a property of the atom, Rutherford showed that the radioactivity of a substance is directly proportional to the number of atoms of the substance present. Thus, if N(t) denotes the number of atoms present at time t, then  EMBED Equation.3 , the number of atoms that disintegrate per unit time, is proportional to N; that is,  EMBED Equation.3 (N (4.5) The constant (, which is positive, is known as the decay constant of the substance. The larger ( is, the faster the substance decays. To compute the half life of substance in terms of (, assume that at time t=t0, N(t0)=N0. The solution of the initial value problem  EMBED Equation.3 (N N(t0) = N0 (4.6) is N(t)=N0e-((t-to) or  EMBED Equation.3 e-((t-to) Taking logarithms of both sides we obtain -( (t-t0)=ln EMBED Equation.3  (4.7) If  EMBED Equation.3 =  EMBED Equation.3 , then -( (t-t0)=ln  EMBED Equation.3 , so that t-t0 =  EMBED Equation.3  Thus the half life of a substance is ln2 divided by the decay constant (. The half-life of many substances have been determined and are well published. For example, half-life of carbon-14 is 5568 years, and the half-life of uranium 238 is 4.5 billion years. Remark 4.3.1 a) In (4.5) ( is positive and is decay constant. We may write equation (4.5) in the form  EMBED Equation.3 (N, where ( is negative constant, that is, ( <0. b) The dimension of ( is reciprocal time. It t is measured in years, then ( has the dimension of reciprocal years, and if t is measured in minutes, then ( has the dimension of reciprocal minutes. c) From (4.7) we can solve for t-t0=  EMBED Equation.3  (4.8) If t0 is the time the substance was initially formed or manufactured, then the age of the substance is  EMBED Equation.3 . The decay constant ( is known or can be computed in most cases. N can be computed quite usually. Computation or pre-knowledge of N0 will yield the age of the substance. By the Libbys discovery discussed in Section 1.4.2. the present rate R(t) of disintegration of the C-14 in the sample is given by R(t)= (N(t)= (N0e-(t and the original rate of disintegration is R(o)=(N0. Thus  EMBED Equation.3  so that t= EMBED Equation.3  (4.9) d) If we measure R(t), that present rate of disintegration of the C-14 in the charcoal and observe that R(o) must equal the rate of disintegration of the C-14 in the comparable amount of living wood then we can compute the age t of the charcoal. e) The process of estimating the age of an artifact is called carbon dating. Example 4.6 : Suppose that we have an artifact, say a piece of fossilized wood, and measurements show that the ratio of C-14 to carbon in the sample is 37% of the current ratio. Let us assume that the wood died at time 0, then compute the time T it would take for one gram of the radio active carbon to decay this amount. Solution: By model (1.10)  EMBED Equation.3  This is a separable differential equation. Write it in the form  EMBED Equation.3  Integrate it to obtain ln|m|=kt+c Since mass is positive, lml=m and ln(m)=kt+c. Then m(t) = e kt+c=Aekt, where A = ec is positive constant. Let at some time, designated at time zero, there are M grams present. This is called the initial mass. Then m(o) = A = M, so m(t) = Mekt. If at some later time T we find that there are MT grams, then m(T) = MT = MekT. Then  EMBED Equation.3  hence  EMBED Equation.3  This gives us k and determines the mass at any time: m(t) =  EMBED Equation.3  Let T=( be the time at which half of the mass has radiated away, that is, half-life. At this time, half of the mass remains, so MT=M/2 and MT/M =  EMBED Equation.3 . Now the expression for mass becomes m(t) =  EMBED Equation.3  or m(t) =  EMBED Equation.3  Half-life of C-14 is 5600 years approximately, that is, ( = 5600  EMBED Equation.3  ( means approximately equal (all decimal places are not listed). Therefore m(t)=Me -0.00012378t or  EMBED Equation.3  by the given condition that  EMBED Equation.3  is .37 during t. (T= -  EMBED Equation.3  8031 years approximately. Example 4.7 (a) A fossilized bone is found to contain one thousandth the original amount of C-14. Determine the age of fossil. (b) Use the information provided in part (a) to determine the approximate age of a piece of wood found in an archaeological excavation at the site to date prehistoric paintings and drawing on the walls and ceilings of a cave in Lascaux, France, provided 85.5% of the C-14 had decayed. Solution: a. The separable differential equation  EMBED Equation.3 , where k is the constant of proportionality of decay, models the phenomenon as discussed above. The solution is N(t) = N0ekt (say ( = -k, if we want to put in the form of the above discussion). Half-life of C-14 is approximately 5600 years  EMBED Equation.3 = N(5600) or  EMBED Equation.3  N0 = N0e5600k. By cancelling N0 and taking logarithm of both sides we get  EMBED Equation.3  or k= - EMBED Equation.3  = -0.00012378 Therefore N(t) = N0 e-0.00012378t With N(t) = EMBED Equation.3  we have  EMBED Equation.3 N0=N0e-0.00012378t -0.00012378t = ln EMBED Equation.3 = - ln 1000. Thus  EMBED Equation.3  Let N(t)= N0ekt where k= -0.00012378 by part (a). 85.5% of C-14 had decayed; that is, N(t) = 0.145 N0 or N0e - 0.00012378t = 0.145 No Taking logarithm of both sides and solving for t, we get t(15,600 years 4.4 Mixture of Two Salt Solutions Example. 4.8 A tank contains 300 litres of fluid in which 20 grams of salt is dissolved. Brine containing 1 gm of salt per litre is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number N(t) of grams of salt in the tank at time t. Solution: By the data given in this example we have P(t ) = N(t) n=4, p=1, m=300 in the model (1.23)  EMBED Equation.3  or  EMBED Equation.3  This is a linear differential of first order in P whose integrating factor is  EMBED Equation.3  (See Section 2.3) Solution is given by  EMBED Equation.3  P(t) = 300 + c EMBED Equation.3  Since P(0) = 20 is given we get 20=P(0) = 300+Ce0, that is c= -280 Thus P(t) = 300 - 280 EMBED Equation.3  4.5 Series Circuits Let a series circuit contain only a resistor and an inductor as shown in Figure 4.2 Figure 4.2 LR Series circuit By Kirchhoffs second law the sum of the voltage drop across the inductor  EMBED Equation.3  and the voltage drop across the resistor (iR) is the same as the impressed voltage (E(t)) on the circuit. Current at time t, i(t), is the solution of the differential equation.  EMBED Equation.3  (4.10) where ( and R are constants known as the inductance and the resistance respectively. The voltage drop across a capacitor with capacitance C is given by  EMBED Equation.3 , where q is the charge on the capacitor. Hence, for the series circuit shown in Figure 4.3 we get the following equation by applying Kirchhoffs second law  EMBED Equation.3  (4.11) Figure 4.3 RC Series Circuit Since  EMBED Equation.3 , (4.11) can be written as  EMBED Equation.3  (4.12) Example 4.9 Find the current in a series RL circuit in which the resistance, inductance, and voltage are constant. Assume that i(o)=0; that is initial current is zero. Solution: It is modeled by (4.10)  EMBED Equation.3  or  EMBED Equation.3  (4.13) Since (, R and E are constant (4.13) is linear equation of first-order in i with integrating factor  EMBED Equation.3  The solution of (4.13) is  EMBED Equation.3  or i(t) =  EMBED Equation.3  (4.14) Since i(0) = 0, c = - EMBED Equation.3  Putting this value of c in (4.14) we get  EMBED Equation.3  Example 4.10 A 100-volt electromotive force is applied to an RC series circuit in which the resistance is 200 ohms and the capacitance is 10-4 farads. Find the charge q(t) on the capacitor if q (0)=0. Find the current i(t). Solution: The phenomenon is modeled by (4.12):  EMBED Equation.3  , where R=200, C=10-4, E(t) = 100 Thus  EMBED Equation.3  (4.15) This is a linear differential equation of first-order The integrating factor is  EMBED Equation.3  and so the solution of (4.15) is q(t)e50t= EMBED Equation.3  or q(t)= EMBED Equation.3  q(0)=0= EMBED Equation.3  or  EMBED Equation.3  and so  EMBED Equation.3   EMBED Equation.3  But  EMBED Equation.3  and so i= EMBED Equation.3  4.6 Survivability with AIDS Equation (1.31) provides survival fraction S(t). It is a separable equation and its solution is S(t) =Si+(1-Si)e-kt: Given equation is  EMBED Equation.3   EMBED Equation.3  Integrating both sides, we get ln|S(t)-Si|=-kt+lnc  EMBED Equation.3  or  EMBED Equation.3  S(t)=Si+ce-kt Let S(0)=1 then c=1-Si. Therefore S(t) =Si +(1-Si)e-kt We can rewrite this equation in the equivalent form. S(t)=Si+(1-Si)e-t/T where, in analogy to radioactive nuclear decay, T is the time required for half of the mortal part of the cohort to die-that is, the survival half life. Example 4.11 Consider the initial-value problem  EMBED Equation.3  as the survivability with AIDS. Show that, in general, the half-life T for the mortal part of the cohort to die is  EMBED Equation.3  (b) Show that the solution of the initial value problem can be written as S(t)=Si+(1-Si)2-t/T (4.17) Solution: The solution of the separable differential equation in (4.16) is S(t) = (1-Si)e-kt+Si (4.18) Let S(t) = EMBED Equation.3 S(0), and solving for t we obtain the half=life T = EMBED Equation.3  (b) Putting  EMBED Equation.3  in (4.18) we obtain  EMBED Equation.3  4.7 Draining a Tank In Section 1.4.8 modeling of draining a tank is discussed. Equation (1.26) models the rate at which the water level is dropping. Example 4.12 A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. Find the height h of water in the tank at any time t if the initial height of the water is H. Solution: As discussed in Section 1.4.8, h(t) is the solution of the equation (1.26); that is,  EMBED Equation.3  (4.19) where A is the cross section area of the cylinder and B is the cross sectional area of the orifice at the base of the container. (4.19) can be written as  EMBED Equation.3  or  EMBED Equation.3  where  EMBED Equation.3  By integrating this equation we get  EMBED Equation.3  For t=0 h=H and so  EMBED Equation.3  Therefore h(t) = EMBED Equation.3  4.8 Economics and Finance We have presented models of supply, demand and compounding interest in Section 1.4.3. We solve those models, namely equations (1.11) and (1.16). (1.11), that is equation  EMBED Equation.3  is a separable differential equation of first-order. We can write it as dP=k(D-S) dt. Integrating both sides, we get P(t)=k(D-S)t+A where A is a constant of integration. Solution of (1.16), which is also a separable equation, is S(t)=S(0) ert (4.20) where S(0) is the initial money in the account Example 4.13 Find solution of the model of Example 1.21 with no initial demand (D(0)=0). Solution: The model is  EMBED Equation.3  This can be written as D1/2dD=k tdt Integrating both sides we get  EMBED Equation.3  where A is a constant integration. If Demand D=0 at the initial time t=0, then A=0 and demand D(t) at any time t is given by  EMBED Equation.3  4.9 Mathematics Police Women The time of death of a murdered person can be determined with the help of modeling through differential equation. A police personnel discovers the body of a dead person presumably murdered and the problem is to estimate the time of death. The body is located in a room that is kept at a constant 70 degree F. For some time after the death, the body will radiate heat into the cooler room, causing the bodys temperature to decrease assuming that the victims temperature was normal 98.6F at the time of death. Forensic expert will try to estimate this time from bodys current temperature and calculating how long it would have had to lose heat to reach this point. According to Newtons law of cooling, the body will radiate heat energy into the room at a rate proportional to the difference in temperature between the body and the room. If T(t) is the body temperature at time t, then for some constant of proportionality k, T'(t)=k[T(t)-70] This is a separable differential equation and is written as  EMBED Equation.3  Upon integrating both sides, one gets ln|T-70|=kt+c Taking exponential, one gets |T-70|=ekt+C=Aekt where A = eC. Then T-70= ( Aekt= Bekt Then T(t)=70 + Bekt Constants k and B can be determined provided the following information is available: Time of arrival of the police personnel, the temperature of the body just after his arrival, temperature of the body after certain interval of time. Let the officer arrived at 10.40 p.m. and the body temperature was 94.4 degrees. This means that if the officer considers 10:40 p.m. as t=0 then T(0)=94.4=70+B and so B=24.4 giving T(t)=70 + 24.4 ekt. Let the officer makes another measurement of the temperature say after 90 minutes, that is, at 12.10 a.m. and temperature was 89 degrees. This means that T(90)=89=70+24.4 e90k Then  EMBED Equation.3  so  EMBED Equation.3  and  EMBED Equation.3  The officer has now temperature function  EMBED Equation.3  In order to find when the last time the body was 98.6 (presumably the time of death), one has to solve for time the equation  EMBED Equation.3  To do this, the officer writes  EMBED Equation.3  and takes logarithms of both sides to obtain  EMBED Equation.3  Therefore, the time of death, according to this mathematical model, was  EMBED Equation.3  which is approximately 57.0.7 minutes. The death occurred approximately 57.07 minutes before the first measurement at 10.40 p.m. , that is at 9.43 p.m. approximately 4.10 Drug Distribution (Concentration) in Human Body To combat the infection to human a body appropriate dose of medicine is essential. Because the amount of the drug in the human body decreases with time medicine must be given in multiple doses. The rate at which the level y of the drug in a patients blood decays can be modeled by the decay equation  EMBED Equation.3  where k is a constant to be experimentally determined for each drug. If initially, that is, at t=0 a patient is given an initial dose yp, then the drug level y at any time t is the solution of the above differential equations, that is, y(t)=yp e-kt Remark: 4.10.1. In this model it is assumed that the ingested drug is absorbed immediately which is not usually the case. However, the time of absorption is small compared with the time between doses. Example 4.14: A representative of a pharmaceutical company recommends that a new drug of his company be given every T hours in doses of quantity y0, for an extended period of time. Find the steady state drug in the patient s body. Solution: Since the initial dose is y0, the drug concentration at any time t e"o is found by the equation y=y0e-kt, the solution of the equation  EMBED Equation.3  At t=T the second dose of y0 is taken, which increases the drug level to y(T)=y0+y0 e-kT = y0(1+e-kT) The drug level immediately begins to decay. To find its mathematical expression we solve the initial-value problem:  EMBED Equation.3  y(T)=y0(1+e-kT) Solving this initial value problem we get y=y0(1+e-kT)e-k(t-T) This equation gives the drug level for t>T. The third dose of y0 is to be taken at t=2T and the drug just before this dose is taken is given by  EMBED Equation.3  The dosage y0 taken at t=2T raises the drug level to y(2T) = y0 + y0(1+e-kT)e-kT = y0(1+e-kT+e-2kt) Continuing in this way, we find after (n+1)th dose is taken that the drug level is y(nT)=y0(1+e-kT+e-2kT+..+e-nkT) We notice that the drug level after (n+1)th dose is the sum of the first n terms of a geometric series, with first term as yo and the common ratio e-kT. This sum can be written as  EMBED Equation.3  As n becomes large, the drug level approaches a steady state value, say ys given by ys = lim y(nT) n(( = EMBED Equation.3  The steady state value ys is called the saturation level of the drug. 4.11 A Pursuit Problem Figure 4.4 A dog chasing a rabbit is shown in Figure 4.4. The rabbit starts at the position (0,0) and runs at a constant speed vR along the y-axis. The dog starts chase at the position (1.0) and runs at a constant speed vD so that its line of sight is always directed at the rabbit. If vD>vR, the dog will catch the rabbit; otherwise the rabbit gets away. Finding the function representing the pursuit curve gives the path the dog follows. Since the dog always runs directly at the rabbit during the pursuit, the slope of the line of sight between the dog and the rabbit at any time t is given by  EMBED Equation.3  If we assume that the line of sight is tangent to the pursuit curve y=f(x), then m=  EMBED Equation.3  and therefore  EMBED Equation.3  (4.21) is the mathematical model of the Pursuit Problem. The solution of (4.21) will give the path taken by the dog. The position of the dog at any time t>0 is (x,y), and the y coordinate of the rabbit at the corresponding time is yR=0+vRt =vRt, so  EMBED Equation.3  or  EMBED Equation.3  Implicitly differentiating this expression with respect to x yields  EMBED Equation.3  where  EMBED Equation.3  This may be written as  EMBED Equation.3  (4.22) Finally, we note that the speed of the dog can be written as  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  Solving this for  EMBED Equation.3 , we have  EMBED Equation.3  Substituting this result into Equation (4.22) yields  EMBED Equation.3  Put y'=w, then this equation takes the form  EMBED Equation.3  or  EMBED Equation.3  Integrating both sides we get w and by integrating w we get y. The constant of integration can be found by using the initial conditions y(1)=0 and y' (1)=0. 4.12 Harvesting of Renewable Natural Resources There are many renewable natural resources that humans desire to use. Examples are fishes in rivers and sea and trees from our forests. It is desirable that a policy be developed that will allow a maximal harvest of a renewable natural resource and yet not deplete that resource below a sustainable level. We introduce a mathematical model providing some insights into the management of renewable resources. Let P(t) denote the size of a population at time t, the model for exponential growth begins with the assumption that  EMBED Equation.3  for some k>0. In this model the relative or specific, growth rate defined by  EMBED Equation.3  is assumed to be a constant. In many cases  EMBED Equation.3  is not constant but a function of P, let  EMBED Equation.3  = f(P) or  EMBED Equation.3  Suppose an environment is capable of sustaining no more than a fixed number K of individuals in its population. The quantity is called the carrying capacity of the environment. Special cases: (i) f (P)=c1P +c2 (ii) If f(0)=r and f(K)=0 then c2=r and c1= - EMBED Equation.3 , and so (i) takes the form f (P) = r-( EMBED Equation.3 )P. Simple Renewable natural resources model is  EMBED Equation.3  This equation can also be written as  EMBED Equation.3  Example 4.15: Find the solution of the following harvesting model  EMBED Equation.3  P(o)=Po Solution: 4.15 The differential equation can be written as  EMBED Equation.3  or  EMBED Equation.3  or  EMBED Equation.3  Integrating we get  EMBED Equation.3  or  EMBED Equation.3  Setting t=0 and P=P0 we find c1=(Po-4)/(Po-1). Solving for P we get  EMBED Equation.3  Exercises Newtons Law of Cooling/Warming A thermometer reading 1000 F is placed in a pan of oil maintained at 100 F. What is the temperature of the thermometer when t=20 sec, if its temperature is 600 F when t = 8 sec? A thermometer is removed from a room where the air temperature is 600 F and is taken outside, where the temperature is 100 F. After 1 minute the thermometer reads 500 F. What is the reading of the thermometer at t=2 minutes? How long will it take for the thermometer to reach 200 F . Water is heated to a boiling point temperature 1200C. It is then removed from the burner and kept in a room of 300C temperature. Assuming that there is no change in the temperature of the room and the temperature of the hot water is 110oC after 3 minutes. (a) Find the temperature of water after 6 minutes (b) Find the duration in which water will cool down to the room temperature? Population Growth and decay A culture initially has Po number of bacteria. At t=1 hour, the number of bacteria is measured to be  EMBED Equation.3  P0. If the rate of growth is proportional to the number of bacteria P(t) present at time t, determine the time necessary for the number of bacteria to triple. Solve the logistic differential equation:  EMBED Equation.3  Insects in a tank increase at a rate proportional to the number present. If the number increases from 50,000 to 100,000 in one hour, how many insects are present at the end of two hours. It was estimated that the earths human population in 1961 was 3,060,000,000. Assuming the population increases at the rate of 2 percent, find the earths population in 1996 using model of population growth (1.8). Check this number with the actual population of the earth available from authentic sources. Radio-Active Decay and Carbon Dating A breeder reactor converts relatively stable uranium 238 into the isotope plutonium 239. After 30 years it is determined that 0.022% of the initial amount N0 of plutonium has disintegrated. Find the half-life of this isotope if the rate of disintegration in proportional to the amount remaining. The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 4 hours. If 1 gram of lead is present initially, how long will it take for 80% of the lead to decay? Solve the model obtained in Exercise 32 of chapter 1. In the 1950 excavation at Nippur, a city of Babylonia, charcoal from a roof beam gave a count of 4.09 dis/min/g. Living wood gave 6.68 disintegrations. Assuming that this charcoal was formed during the time of Hammurabis reign, find an estimate for the likely time of Hammurabis succession. Mixture of Two Salt Solutions A tank with a capacity of 600 litres initially contains 200 litres of pure water. A salt solution containing 3 Kg of salt per litre is allowed to run into the tank at a rate of 16 lit/min, and the mixture is then removed at a rate of 12 lit/min. Find the expression for the number of Kilograms of salt in the tank at any time t. A large tank is filled with 600 liters of pure water. Brine containing 2 Kg of salt per litre is pumped into the tank at a rate of 5 litre/min. The well-mixed solution is pumped out at the same rate. Find the number P(t) of kilograms of salt in the tank at time t. What is the concentration of the solution in the tank at t=10 min? A 250-litre tank contains 100 litres of pure water. Brine containing 4 kg of salt per litre flows into the tank at 5 lit/hr. If the well-stirred mixture flows out at 3 lit/hr, find the concentration of salt in the tank at the instant it is filled to the top. Series circuit A series RL circuit has a resistance 20 ohms, and an inductance of 1 henry, and an impressed voltage of 12 volts. Find the current i(t) if the initial current is zero. An electromotive force 120, 0 ( t ( 20 E(t)= 0 , t > 20 is applied to an LR series circuit in which the inductance is 20 henries and the resistance is 2 ohms. Find the current i(t) if i(0)=0 Survivability with AIDS Find survival fraction S(t) with aids after 2 years by applying model (1.31). Draining a Tank A tank in the form of a right-circular cylinder standing on end is leaking water through a circular hole in its bottom. Let us assume that the height of the tank is 10 ft. high and has radius 2 ft. and circular hole has radius inches. If the tank is initially full, how long it will take to empty? Economics and Finance What rate of interest payable annually is equivalent to 6% continuously compounded? Suppose a person deposits 10,000 Indian rupees in a bank account at the rate of 5% per annum compounded continuously. How much money will be in his bank account 18 months later? How much he has in the account if the interest were compounded monthly. Drug distribution(Concentration) in Human Body A drug with k=0.01 is administered every 12 hours in doses of 4 mg. Calculate the amount of the drug in the patients body after the 4th dose is taken.  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