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# Class Note for MATH 3321 at UH

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Date Created: 02/06/15

23 Some Applications In this section we give some examples of applications of linear and separable differential equations 1 Orthogonal Trajectories The oneparameter family of curves ltz72gt2lty71gt20 020 a is a family of circles with center at the point 27 1 and radius If we differentiate this equation with respect to x we get 2z722y71y0 and 2 777 ye yil b This is the differential equation for the family of circles Note that if we choose a speci c point mo7 yo7 yo 7 1 on one of the circles7 then b gives the slope of the tangent line at 07110 Now consider the family of straight lines passing through the point 27 1 y71Kz72 C 36 The differential equation for this family is Z verify this 1 24 Comparing equations b and d we see that right side of b is the negative reciprocal of the right side of Therefore we can conclude that if P mo yo is a point of intersection of one of the circles and one of the lines then the line and the circle are perpendicular orthogonal to each other at P The following figure shows the two families drawn in the same coordinate system 2 A curve that intersects each member of a given family of curves at right angles orthog onally is called an orthogonal trajectory of the family Each line in c is an orthogonal trajectory of the family of circles a and conversely each circle in a is an orthogonal trajectory of the family of lines In general if Fmyc0 and GmyK0 are oneparameter families of curves such that each member of one family is an orthogonal trajectory of the other family then the two families are said to be orthogonal trajectories A procedure for nding a family of orthogonal trajectories Cz y K 0 for a given family of curves Fz y C 0 is as follows Step 1 Determine the differential equation for the given family Fz y C 0 Step 2 Replace y in that equation by ily the resulting equation is the differential equation for the family of orthogonal trajectories Step 3 Find the general solution of the new differential equation This is the family of orthogonal trajectories Example Find the orthogonal trajectories of the family of parabolas y sz SOLUTION You can verify that the differential equation for the family y Caz can be written as Replacing y by ily we get the equation 1 2 if i which simplifies to y if y z 2y a separable equation Separating the variables7 we get 2yy 7x or 2y dy imdm integrating with respect to m we have 1 2 y277z20 or y2C 2 2 This is a family of ellipses with center at the origin and major axis on the z axis I Exercises 231 Find the orthogonal trajectories for the family of curves 1 y 0mg 2 mCy4 3 y 0x2 2 4 y2 207 Find the orthogonal trajectories for the family of curves 5 The family of parabolas symmetric with respect to the y axis and vertex at the origin 6 The family of parabolas with vertical axis and vertex at the point 17 2 7 The family of circles that pass through the origin and have their center on the z axis 8 The family of circles tangent to the m axis at 37 O Show that the given family is self orthogonal 9 y 4035 C 38 2 2 y 71 0274 m 10 w 2 Exponential Growth and Decay Radioactive Decay It has been observed and veri ed experimentally that the rate of decay of a radioactive material at time t is proportional to the amount of material present at time t Mathematically this says that if A At is the amount of radioactive material present at time 257 then A TA where r the constant of proportionality is negative To emphasize the fact that A is decreasing7 this equation is often written A ikA or ikA k gt 0 constant This is the form we shall use The constant of proportionality k is called the decay constant Note that this equation is both linear and separable and so we can use either method to solve it It is easy to show that the general solution is At 05 If A0 A0 is amount of material present at time t 07 then C A0 and At A0 5 Note that At 0 HalfLife An important property of a radioactive material is the length of time T it takes to decay to one half the initial amount This is the so called half life of the material Physicists and chemists characterize radioactive materials by their half lives To nd T we solve the equation A0 A0 ETkT for T 1 ikT AO A05 713 5 2 ikT ln127ln2 ln2 k 39 Conversely if we know the half life T of a radioactive material then the decay constant k is given by T Example Cobalt 60 is a radioactive element that is used in medical radiology It has a half life of 53 years Suppose that an initial sample of cobalt 60 has a mass of 100 grams a Find the decay constant and determine an expression for the amount of the sample that will remain t years from now b How long will it take for 90 of the sample to decay SOLUTION a Since the half life T ln 2k we have ln2 ln2 kTEO131 With A0 100 the amount of material that will remain after t years is At 100 5013 b If 90 of the material decays then 10 which is 10 grams remains Therefore we solve the equation 100 5013 10 for t 1 0 1 701311 7 7 n 39 y e 701 0131t ln01 t 701317176 It will take approximately 176 years for 90 of the sample to decay I Population Growth Growth of an Investment It has been observed and veri ed experimentally that under ideal conditions a population eg bacteria fruit ies humans etc tends to increase at a rate proportional to the size of the population Therefore if P Pt is the size of a population at time t then rP r gt 0 constant 1 In this case the constant of proportionality r is called the growth constant Similarly in a bank that compounds interest continuously the rate ofincrease of funds at time t is proportional to the amount of funds in the account at time t Thus equation 1 also represents the growth of a principal amount under continuous compounding Since the two cases are identical we ll focus on the population growth case 40 The general solution of equation 1 is Pt Cequot lf P0 P0 is the size of the population at time t 0 then Pt P0 5 is the size of the population at time 25 Note that Pt 00 In reality the rate of increase of a population does not continue to be proportional to the size of the population After some time has passed factors such as limitations on space or food supply introduc tion of diseases and so forth affect the growth rate the mathematical model is not valid indefinitely In contrast the model does hold inde nitely in the case of the growth of an investment under continuous compounding Doubling time The analog of the half life of a radioactive material is the so called doubling time the length of time T that it takes for a population to double in size Using the same analysis as above we have 2 A0 A0 ETT erT 2 TT ln 2 T M 7 In the banking investment and real estate communities there is a standard measure called the rule of 72 which states that the length of time approximately for a principal invested at r compounded continuously to double in value is 72r We know that the doubling time is ln2 069 7 69 N 72 r r TT39 This is the origin of the rule of 7277 72 is used rather than 69 because it has more H H l 22 divisors I Example Scientists have observed that a small colony of penguins on a remote Antarctic island obeys the population growth law There were 2000 penguins initially and 3000 penguins 4 years later a How many penguins will there be after 10 years b How long will it take for the number of penguins to double SOLUTION Let Pt denote the number of penguins at time 25 Since P0 2000 we have Pt 2000 e 41 We use the fact that P4 3000 to determine the growth constant T 3000 2000 5 5 15 47 ln 15 and so 1 15 r 11439 g0101 Therefore7 the number of penguins in the colony at any time t is Pt 2000 5010 a The number of penguins in the colony after 10 years is approximately P10 2000 503910110 2000 5101 e 5491 b To nd out how long it will take the number of penguins in the colony to double7 we need to solve 2000 5010 4000 for t ln 2 0101t 7 7 7 m e 7 27 0101t 7 ln 27 t 7 01017 686 years Note There is another way of expressing P that uses the exact value of r From the equation 3000 2000 5 we get r ln Thus t t t4 Pt 200051 111W 200051nl321 2000 l Exercises 232 1 A certain radioactive material is decaying at a rate proportional to the amount present If a sample of 50 grams of the material was present initially and after 2 hours the sample lost 10 of its mass7 find a An expression for the mass of the material remaining at any time t b The mass of the material after 4 hours c The half life of the material 2 What is the half life of a radioactive substance it takes 5 years for onethird of the material to decay 3 The size of a certain bacterial colony increases at a rate proportional to the size of the colony Suppose the colony occupied an area of 025 square centimeters initially7 and after 8 hours it occupied an area of 035 square centimeters 42 a Estimate the size of the colony 25 hours after the initial measurement b What is the expected size of the colony after 12 hours c Find the doubling time of the colony 4 A biologist observes that a certain bacterial colony triples every 4 hours and after 12 hours occupies 1 square centimeter a How much area did the colony occupy when rst observed b What is the doubling time for the colony 5 In 1980 the world population was approximately 45 billion and in the year 2000 it was approximately 6 billion Assume that the world population at each time 25 increases at a rate proportional to the population at time t Measure 25 in years after 1980 a Find the growth constant and give the world population at any time t b How long will it take for the world population to reach 9 billion double the 1980 population c The world population for 2002 was reported to be about 62 billion What population does the formula in a predict for the year 2002 6 It is estimated that the arable land on earth can support a maximum of 30 billion people Extrapolate from the data given in Exercise 5 to estimate the year when the food supply becomes insufficient to support the world population 3 Newton s Law of CoolingHeating Newton s Law of Cooling states that the rate of change of the temperature u of an object is proportional to the difference between u and the constant temperature a of the surrounding medium eg7 air or water7 called the ambient temperature The mathematical formulation of this statement is du a mu 7 a m constant The constant of proportionality7 m7 in this model must be negative for if the object is warmer than the ambient temperature u 7 a gt 07 then its temperature will decrease dudt lt 07 which implies m lt 0 if the object is cooler than the ambient temperature u 7 a lt 07 then its temperature will increase dudt gt 07 which again implies m lt 0 To emphasize that the constant of proportionality is negative7 we write Newton s Law of Cooling as du a 7ku 7 a k gt 0 constant 1 43 This differential equation is both linear and separable so either method can be used to solve it As you can check the general solution is ut a 05 If the initial temperature of the object is 740 uo then u0a050a0 and 071070 Thus the temperature of the object at any time t is given by w 7 a 7 770 7 UV 2 The graphs of ut in the cases uo lt a and uo gt a are given below Note that limtnoo ut a in each case In the rst case u is increasing and its graph is concave down in the second case u is decreasing and its graph is concave up Sigma Example A metal bar with initial temperature 25 C is dropped into a container of boiling water 1000 C After 5 seconds the temperature of the bar is 350 C a What will the temperature of the bar be after 1 minute b How long will it take for the temperature of the bar to be Within 050 C of the boiling water SOLUTION Applying equation 2 the temperature of the bar at any time t is Tt 100 25 7100e t 100 7 75 5 The rst step is to determine the constant k Since T5 35 we have 35 100 7 75 a 75 a 65 75k ln6575 k 7 00286 Therefore Tt 100 7 75 570028675 44 a The temperature of the bar after 1 minute is7 approximately T60 100 7 75 50028660 g 100 7 75 51717 g 86530 b We want to calculate how long it will take for the temperature of the bar to reach 9950 Thus7 we solve the equation 995 100 7 75 500286t for t 995 100775 50028 775 50028 705 7008262 ln 0575 Exercises 233 1 A thermometer is taken from a room where the temperature is 72 F to the outside where the temperature is 32quot F After 12 minute7 the thermometer reads 50 F a What will the thermometer read after it has been outside for 1 minute b How many minutes does the thermometer have to be outside for it to read 350 F 2 A metal ball at room temperature 200 C is dropped into a container of boiling water 1000 C given that the temperature of the ball increases 20 in 2 seconds7 find a The temperature of the ball after 6 seconds in the boiling water b How long it will take for the temperature of the ball to reach 900 C 3 Suppose that a corpse is discovered at 10 pm and its temperature is determined to be 85 F Two hours later7 its temperature is 740 F If the ambient temperature is 680 F7 estimate the time of death 4 Falling Objects With Air Resistance Consider an object with mass m in free fall near the surface of the earth The object experiences the downward force of gravity its weight as well as air resistance7 which may be modeled as a force that is proportional to velocity and acts in a direction opposite to the motion By Newton s second law of motion7 we have 1 771077mgikv where g is the gravitational acceleration constant and k gt 0 is the dmg coe cient that depends upon the density of the atmosphere and aerodynamic properties of the object Note if time is measured in seconds and distance in feet7 then 9 is approximately 32 feet 45 t 6066 seconds per second per second if time is measured in seconds and distance in meters7 then 9 is approximately 98 meters per second per second Rearrangement gives 11 dt m 9 where r km Once this equation is solved for the velocity v the height of the object is obtained by simple integration Exercises 234 1 a Solve the initial value problem all ETv97v010 in terms of r 97 and 120 b Show that vt a 7mgk as t a 00 This is called the terminal velocity of the object c lntegrate v to obtain the height y assuming an initial height y0 yo 2 An object with mass 10kg is dropped from a height of 200m Given that its drag coefficient is k 25 Nms7 after how many seconds does the object hit the ground 3 An object with mass 50 kg is dropped from a height of 200m lt hits the ground 10 seconds later Find the object s drag coefficient k 4 An object with mass 10 kg is projected upward from ground level with initial velocity 60 ms lt hits the ground 84 seconds later a Find the object s drag coefficient k b Find the maximum height c Find the velocity with which the object hits the ground 5 Mixing Problems Here we consider a tank7 or other type of container7 that contains a volume V of water in which some amount of impurity eg7 salt is dissolved Water containing the dissolved impurity at a known concentration ows or is pumped into the tank at a given volume flow mite7 and water ows out of the tank also at a given volume ow rate We assume that the water in the tank remains thoroughly mixed at all times Let At be the amount of impurity in the tank at time t The impurity concentration in the tank is then 1425V7 and At will satisfy a differential equation of the generic form dA E in ow rate 7 out ow rate 46 These ow rates are products of the form concentration x volume ow rate So if we let Rm and Rout denote the volume ow rates and let kin denote the impurity concentration of the in ow7 then we have dA A E kinRin 7 V When Rm Rout the volume V is constant If Rm 7 Rout but each is constant7 then Rout V V0 l Rim 7 Rout t Exercises 235 1 A tank with a capacity of 2m3 2000 liters is initially full of pure water At time t 07 salt water with salt concentration 5 gramsliter begins to ow into the tank at a rate of 10 litersminute The well mixed solution in the tank is pumped out at the same rate a Set up7 and then solve7 the initial value problem for the amount of salt in the tank at time 25 minutes b Find the time when the salt concentration in the tank becomes 4 gramsliter 2 A 100 gallon tank is initially full of water At time t 07 a 20 hydrochloric acid solution begins to ow into the tank at a rate of 2 gallonsminute The well mixed solution in the tank is pumped out at the same rate a Set up7 and then solve7 the initial value problem for the amount of hydrochloric acid in the tank at time 25 minutes b Find the time when the hydrochloric acid concentration becomes 10 3 A room measuring 10 m x 5 m x 3 m initially contains air that is free of carbon monox ide At time t 07 air containing 3 carbon monoxide enters the room at a rate of 1 milminute7 and the well circulated air in the room leaves at the same rate a Set up7 and then solve7 the initial value problem for the amount of carbon monox ide in the room at time 25 minutes b Find the time when the carbon monoxide concentration in the room reaches 2 4 A tank with a capacity of 1m3 1000 liters is initially half full of pure water At time t 07 4 salt solution begins to ow into the tank at a rate of 30 litersminute The well mixed solution in the tank is pumped out at a rate of 20 litersminute 47 a Set up and then solve the initial value problem for the amount of salt in the tank between time t 0 and the time when the tank becomes full b Find the salt concentration of the solution in the tank during this process 5 A 100 gallon tank is initially full of pure water At time t 0 water containing salt at concentration 15 gramsgallon begins to ow into the tank at a rate of 1 gallonminute while the well mixed solution in the tank is pumped out at a rate of 2 gallonsminute a Set up and then solve the initial value problem for the amount of salt in the tank between time t 0 and the time when the tank becomes empty b Find the maximum amount of salt in the tank during this process 6 The Logistic Equation In the mid nineteenth century the Belgian mathematician RF Verhulst used the differential equation dy a IiiM7 i lt1 where k and M are positive constants to study the population growth of various countries This equation is now known as the logistic equation and its solutions are called logistic functions Life scientists have used this equation to model the spread of an infectious disease through a population and social scientists have used it to study the ow of information In the case of an infectious disease if M denotes the number of people in the population and yt is the number of infected people at time t then the differential equation states that the rate of change of infected people is proportional to the product of the number of people who have the disease and the number of people who do not The constant M is called the carrying capacity of the enVironment Note that dydt gt 0 when 0 lt y lt M dydt 0 when y M and dydt lt 0 when y gt M The constant k is the intrinsic growth rate The differential equation 1 is separable It is also a Bernoulli equation We write the equation as 7y 7k0 MM y mdyikdt 01 1 M 1 M dy 7 kt Cl partial fraction decomposition y M i y and integrate 1 1 Mlnlyli lnlMigl kt01 48 We can solve this equation for y as follows 1 y MIn Miy 7 kt01 In My MktM01Mkt02 02M01 29 y 7 Mkt027 02 Mkti Mkt 7 02 Miy 7 e 75 e 705 075 Now7 in the context of this discussion7 y yt satis es 0 lt yt lt M Therefore7 yMi y gt 0 and we have Mg y C 5M1 Solving this equation for y we get CM Mt m Finally7 if y0 R7 R lt M7 then CM R of which implies C and M R 2405 W39 2 The graph of this particular solution is shown below Note that y is an increasing function In the Exercises you are asked to show that the graph is concave up on 07 a and concave down on 01700 This means that the disease is spreading at an increasing rate up to time a after at7 the disease is still spreading7 but at a decreasing rate Note7 also7 that limtnoo yt M Example An in uenza virus is spreading through a small city with a population of 50000 people Assume that the virus spreads at a rate proportional to the product of the number of people who have been infected and the number of people who have not been infected lf 100 people were infected initially and 1000 were infected after 10 days7 find 49 a The number of people infected at any time t b How long it will take for half the population to be infected SOLUTION a Substituting the given data into equation b7 we have 10050 000 50 000 t 7 M l 100 49 900 5750900195 1 499 5750900 We can determine the constant k by applying the condition y10 1000 We have 50000 1000 149957500000k 4995750090019 49 75000009 ln49499 k g 00000046 Thus7 the number of people infected at time t is approximately D7 50000 y 1499e0v23t39 b To nd how long it will take for half the population to be infected7 we solve yt 257 000 for t 50 000 25 000 7 1 499570231 1 499 70231 7 5 499 t w E 27 days I Exercises 236 1 A rumor spreads through a small town with a population of 5000 at a rate propor tional to the product of the number of people who have heard the rumor and the number who have not heard it Suppose that 100 people initiated the rumor and that 500 people heard it after 3 days a How many people will have heard the rumor after 8 days b How long will it take for half the population to hear the rumor 2 Let y be the logistic function Show that dydt increases for y lt M2 and decreases for y gt M2 What can you conclude about dydt when y M2 50 3 Solve the logistic equation by means of the change of variables W U05 Mt vt 2v t Express the constant of integration in terms of the initial value y0 yo 4 Suppose that a population governed by a logistic model exists in an environment with carrying capacity of 800 If an initial population of 100 grows to 300 in 3 years7 nd the intrinsic growth rate k 51

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