Calculus for the Life Sciences II - Joseph M. Mahaffy

IntroductionBlood Pressure

Radioactive DecaySolution of Linear Growth and Decay Models

Newton’s Law of CoolingSolution of General Linear Model

Pollution in a Lake

Calculus for the Life Sciences II

Lecture Notes – Linear Differential Equations

Joseph M. Mahaffy,〈[email protected]〉

Department of Mathematics and StatisticsDynamical Systems Group

Computational Sciences Research Center

San Diego State UniversitySan Diego, CA 92182-7720

http://www-rohan.sdsu.edu/∼jmahaffy

Fall 2012

Joseph M. Mahaffy, 〈[email protected]〉Lecture Notes – Linear Differential Equations— (1/71)

http://www-rohan.sdsu.edu/~jmahaffy




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Outline

1 Introduction

2 Blood PressureCardiac CycleArterial Blood PressureModeling Blood PressureDiagnosis with ModelExample of Athlete

3 Radioactive DecayCarbon RadiodatingHyperthyroidism

4 Solution of Linear Growth and Decay Models

5 Newton’s Law of CoolingMurder InvestigationCooling Tea

6 Solution of General Linear Model

7 Pollution in a LakeExample of Pollution with Evaporation





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Introduction

Introduction

Examples of linear first order differential equations





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Introduction

Introduction


Arterial blood pressure





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Introduction

Introduction


Arterial blood pressureRadioactive decay





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Introduction

Introduction


Arterial blood pressureRadioactive decayNewton’s law of cooling





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Introduction

Introduction


Arterial blood pressureRadioactive decayNewton’s law of coolingPollution in a Lake





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Introduction

Introduction


Arterial blood pressureRadioactive decayNewton’s law of coolingPollution in a Lake

Extend earlier techniques to find solutions





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Cardiac CycleArterial Blood PressureModeling Blood PressureDiagnosis with ModelExample of Athlete

Blood Pressure

Blood Pressure

Blood Pressure is divided into systolic and diastolicpressure





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Blood Pressure

Blood Pressure


Normal reading is 120/80 (in mm of Hg)





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Blood Pressure

Blood Pressure



How are those numbers generated and what can we inferfrom them?





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Blood Pressure

Blood Pressure




The numbers for blood pressure reflect the force on arterialwalls





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Blood Pressure

Blood Pressure




The numbers for blood pressure reflect the force on arterialwalls

This pressure is generated by the beating of the heart





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Blood Pressure

Diagram of Heart





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Cardiac Cycle 1

Cardiac Cycle

Pulmonary circulation





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Cardiac Cycle 1

Cardiac Cycle


Blood flows from the body into the right atrium





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Cardiac Cycle 1

Cardiac Cycle


Blood flows from the body into the right atriumFlows to the right ventricle





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Cardiac Cycle 1

Cardiac Cycle


Blood flows from the body into the right atriumFlows to the right ventricleBlood goes through the pulmonary artery to the lungs





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Cardiac Cycle 1

Cardiac Cycle


Blood flows from the body into the right atriumFlows to the right ventricleBlood goes through the pulmonary artery to the lungsBlood exchanges O2 and CO2 in the lungs





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Cardiac Cycle 1

Cardiac Cycle


Blood flows from the body into the right atriumFlows to the right ventricleBlood goes through the pulmonary artery to the lungsBlood exchanges O2 and CO2 in the lungsBlood returns through the pulmonary vein to the left atrium





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Cardiac Cycle 1

Cardiac Cycle


Blood flows from the body into the right atriumFlows to the right ventricleBlood goes through the pulmonary artery to the lungsBlood exchanges O2 and CO2 in the lungsBlood returns through the pulmonary vein to the left atriumPressure in the pulmonary vein and left atrium is between 5and 15 mm of Hg





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Cardiac Cycle 1

Cardiac Cycle



Blood flows into the left ventricle





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Cardiac Cycle 1

Cardiac Cycle




The heart is rigid, so pressure increases only slightly





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Cardiac Cycle 1

Cardiac Cycle




The heart is rigid, so pressure increases only slightly

The right atrium contracts, then the AV valve between theatrium and the ventricle closes





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Cardiac Cycle 2

Cardiac Cycle (cont)

The heart receives an electrical signal, which causes ventricularcontraction, beginning systole





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Cardiac Cycle 2



The left ventricle contracts, and the pressure increases until it“blows” open the aortic valve





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Cardiac Cycle 2




Blood rapidly flows into the aorta under this high pressure(systolic pressure)





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Cardiac Cycle 2





As pressure rises in the aorta, the AV valve reopens, while theaortic valve closes





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Cardiac Cycle 2






Now there is high pressure in the aorta, which forces the bloodinto the other arteries of the body





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Cardiac Cycle 2







As the blood flows through the body, the aortic pressure dropsto its low pressure, the diastolic pressure





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Cardiac Cycle 2







As the blood flows through the body, the aortic pressure dropsto its low pressure, the diastolic pressure

Process repeatsJoseph M. Mahaffy, 〈[email protected]〉Lecture Notes – Linear Differential Equations— (7/71)




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Arterial Blood Pressure 1

Arterial Blood Pressure: Model the arterial pressure, Pa(t),during a single beat of the heart





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Determine the important modeling parameters in the system





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The cardiac output, Q, represents the average amount of bloodpumped by the heart (in liters/min)





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The stroke volume, V , is the amount of blood pumped by theheart during one beat (liters/beat)





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T is the duration of a heart beat





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Relate flow from the cardiac output to the stroke volume by therelationship

Cardiac Output = Stroke Volume / Duration of the Flow

Q = V/T (liters/min)





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Relate flow from the cardiac output to the stroke volume by therelationship

Cardiac Output = Stroke Volume / Duration of the Flow

Q = V/T (liters/min)

1T

is the heart rate in beats/minJoseph M. Mahaffy, 〈[email protected]〉Lecture Notes – Linear Differential Equations— (8/71)




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Arterial Blood Pressure:





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The left ventricle completes pumping the blood into theaorta and the aortic valve closes at the maximum pressure,Psys





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The blood pressure begins to fall as the blood flowsthrough the arteries





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The rate of flowing of the blood depends on the resistanceof a blood vessel





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Viscosity of the blood





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Viscosity of the bloodLength of the vessels





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Viscosity of the bloodLength of the vesselsRadius of the blood vessels





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The viscosity of the blood is relatively constant, exceptunder diseased states like erythrocytemia (or when athletestake erythropoietin or EPO to overstimulate theproduction of red blood cells)





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The viscosity of the blood is relatively constant, exceptunder diseased states like erythrocytemia (or when athletestake erythropoietin or EPO to overstimulate theproduction of red blood cells)The length of the blood vessels are relatively constant,except for when conditions like pregnancy or amputationoccur





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The viscosity of the blood is relatively constant, exceptunder diseased states like erythrocytemia (or when athletestake erythropoietin or EPO to overstimulate theproduction of red blood cells)The length of the blood vessels are relatively constant,except for when conditions like pregnancy or amputationoccurThe main factor that changes resistance of the blood flow ischange in the radius





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The viscosity of the blood is relatively constant, exceptunder diseased states like erythrocytemia (or when athletestake erythropoietin or EPO to overstimulate theproduction of red blood cells)The length of the blood vessels are relatively constant,except for when conditions like pregnancy or amputationoccurThe main factor that changes resistance of the blood flow ischange in the radiusBlood pressure becomes a valuable tool for detectingnarrowing of the blood vessels by hypertension oratherosclerosis





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Modeling Blood Pressure 1

Modeling Blood Pressure:





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Experimentally, it has been observed that systemic bloodflow, Qs, is proportional to the difference between the arterialand venous pressures (Pa(t) − Pv(t)) with the proportionalitydependent on the resistance





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If Rs is the systemic resistance (mm Hg/liter/min), then we havethe following equation:

Qs(t) =1

Rs

(Pa(t) − Pv(t))





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If Rs is the systemic resistance (mm Hg/liter/min), then we havethe following equation:

Qs(t) =1

Rs

(Pa(t) − Pv(t))

To simplify the model, we take advantage of the fact that venouspressures are very low, so we approximate the systemic flow bythe equation:

Qs(t) =1

Rs

Pa(t)





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Compliance:





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Compliance:

Compliance is the stretchability of a vessel, which is aproperty that allows a vessel to change the volume inresponse to pressure changes





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Compliance:


The higher the compliance the easier it is for a vessel toexpand in response to increased pressure





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Compliance:



Resistance and compliance have a roughly inverserelationship





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Compliance:



Resistance and compliance have a roughly inverserelationship

Experimentally, the arterial volume, Va, is roughly equal tothe compliance, Ca, times the arterial pressure

Va(t) = CaPa(t)





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Differential Equation for Blood Flow:





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The flow representing the change in the arterial volume isgiven by the difference between the rate of flow enteringthe aorta and the rate of flow from the aorta





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Since the aortic valve is closed during systole, no blood isentering the aorta





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The differential equation is

dVa(t)

dt= flow rate in − flow rate out = 0 − Qs(t)





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dVa(t)

dt= flow rate in − flow rate out = 0 − Qs(t)

Thus,dVa(t)

dt= − 1

RsPa(t)





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Differential Equation for Blood Flow: SinceVa(t) = CaPa(t),

dVa(t)

dt= Ca

dPa(t)

dt





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dVa(t)

dt= Ca

dPa(t)

dt

This gives the initial value problem

dPa(t)

dt= − 1

CaRsPa(t) with Pa(0) = Psys





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dVa(t)

dt= Ca

dPa(t)

dt

This gives the initial value problem

dPa(t)

dt= − 1

CaRsPa(t) with Pa(0) = Psys

The solution is

Pa(t) = Psyse−

tCaRs for t ∈ [0, T ]





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Diagnosis with Model 1

Diagnosis with Model: How can this model be used toprovide a non-invasive method for estimating the physiologicalparameters for compliance, Ca , and resistance, Rs





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Measurable physiological quantities are





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Measurable physiological quantities areThe heart rate or pulse, 1

T





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T

Cardiac output, Q, using a doppler sonogram





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T

Cardiac output, Q, using a doppler sonogramThe systolic and diastolic pressures, Psys and Pdia





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T


Compliance comes from the stroke volume, V ,

V = Vsys − Vdia = CaPsys − CaPdia





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T


Compliance comes from the stroke volume, V ,

V = Vsys − Vdia = CaPsys − CaPdia

But V = QT , so compliance satisfies

Ca =QT

Psys − Pdia





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Resistance: The model gives the diastolic pressure just before thenext heart beat

Pdia = Psyse−

T

CaRs

Solve this equation for the resistance, Rs

Rs =T

Ca (ln(Psys) − ln(Pdia))





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Pdia = Psyse−

T

CaRs


Rs =T


Normal Person





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Pdia = Psyse−

T

CaRs


Rs =T


Normal Person

Pulse of approximately 70 beats/min ( 1T

)





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Pdia = Psyse−

T

CaRs


Rs =T


Normal Person


)Cardiac output of Q = 5.6 (liters/min)





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Pdia = Psyse−

T

CaRs


Rs =T


Normal Person


)Cardiac output of Q = 5.6 (liters/min)Systolic and diastolic pressures of Psys = 120 mm Hg andPdia = 80 mm Hg





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Pdia = Psyse−

T

CaRs


Rs =T


Normal Person


)Cardiac output of Q = 5.6 (liters/min)Systolic and diastolic pressures of Psys = 120 mm Hg andPdia = 80 mm Hg

Compute the compliance and resistance for a normal person

Ca = 0.002 (liters/mm Hg) and Rs = 17.6 (mm Hg/liter/min)Joseph M. Mahaffy, 〈[email protected]〉

Lecture Notes – Linear Differential Equations— (16/71)




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Example of Athlete 1

Example of an Athlete: Consider a trained athleteconsidered in very good condition





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Suppose an athlete has





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A pulse of 60 beats/min (at rest)





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A pulse of 60 beats/min (at rest)A blood pressure of 120/75





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A pulse of 60 beats/min (at rest)A blood pressure of 120/75A measured cardiac output of 6 liters/min





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A pulse of 60 beats/min (at rest)A blood pressure of 120/75A measured cardiac output of 6 liters/min

Find the compliance, Ca, and systemic resistance, Rs,of the arteries for this individual





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Solution: From the formula, compliance, Ca

Ca =QT

Psys − Pdia=

6.0/60

120 − 75= 0.00222 (liters/mm Hg)





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Ca =QT

Psys − Pdia=

6.0/60

120 − 75= 0.00222 (liters/mm Hg)

This is slightly larger than for a normal person





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Ca =QT

Psys − Pdia=

6.0/60

120 − 75= 0.00222 (liters/mm Hg)


The systemic resistance, Rs, satisfies

Rs =T


=1/60

0.00222(ln(120) − ln(75))= 15.96 (mm Hg/liter/min)





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Ca =QT

Psys − Pdia=

6.0/60

120 − 75= 0.00222 (liters/mm Hg)


The systemic resistance, Rs, satisfies

Rs =T


=1/60

0.00222(ln(120) − ln(75))= 15.96 (mm Hg/liter/min)

This is lower than for a normal person, which is what we wouldexpect for someone in better conditionJoseph M. Mahaffy, 〈[email protected]〉





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Carbon RadiodatingHyperthyroidism

Radioactive Decay

Radioactive Decay: Radioactive elements are important inmany biological applications





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Radioactive Decay


3H (tritium) is used to tag certain DNA base pairs





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Radioactive Decay


3H (tritium) is used to tag certain DNA base pairsAdd to mutant strains of E. coli that are unable tomanufacture one particular DNA base





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Radioactive Decay


3H (tritium) is used to tag certain DNA base pairsAdd to mutant strains of E. coli that are unable tomanufacture one particular DNA baseUsing antibiotics, one uses the radioactive signal todetermine how much DNA is replicated under a particularset of experimental conditions





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Radioactive Decay



Radioactive iodine is often used to detect or treat thyroidproblems





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Radioactive Decay



Radioactive iodine is often used to detect or treat thyroidproblemsMost experiments are run so that radioactive decay is notan issue





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Radioactive Decay



Radioactive iodine is often used to detect or treat thyroidproblemsMost experiments are run so that radioactive decay is notan issue

3H has a half-life of 12.5 yrs131I has a half-life of 8 days





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Carbon Radiodating 1

Carbon Radiodating: One important application ofradioactive decay is the dating of biological specimens





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A living organism is continually changing its carbon withthe environment





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Plants directly absorb CO2 from the atmosphere





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Plants directly absorb CO2 from the atmosphereAnimals get their carbon either directly or indirectly fromplants





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Gamma radiation that bombards the Earth keeps the ratioof 14C to 12C fairly constant in the atmospheric CO2





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Gamma radiation that bombards the Earth keeps the ratioof 14C to 12C fairly constant in the atmospheric CO2

14C stays at a constant concentration until the organismdies





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Modeling Carbon Radiodating: Radioactive carbon, 14C, decayswith a half-life of 5730 yr





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Living tissue shows a radioactivity of about 15.3 disintegrationsper minute (dpm) per gram of carbon





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The loss of 14C from a sample at any time t is proportional tothe amount of 14C remaining





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Let R(t) be the dpm per gram of 14C from an ancient object





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The differential equation for a gram of 14C

dR(t)

dt= −kR(t) with R(0) = 15.3





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The differential equation for a gram of 14C

dR(t)

dt= −kR(t) with R(0) = 15.3

This differential equation has the solution

R(t) = 15.3 e−kt, where k = ln(2)5730 = 0.000121





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Example: Carbon Radiodating

Example Carbon Radiodating: Suppose that an object isfound to have a radioactive count of 5.2 dpm per g of carbon





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Find the age of this object





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Solution: From above

5.2 = 15.3 e−kt





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5.2 = 15.3 e−kt

ekt = 15.35.2 = 2.94





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5.2 = 15.3 e−kt

ekt = 15.35.2 = 2.94

kt = ln(2.94)





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5.2 = 15.3 e−kt

ekt = 15.35.2 = 2.94

kt = ln(2.94)

Thus, t = ln(2.94)k = 8915 yr,





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5.2 = 15.3 e−kt

ekt = 15.35.2 = 2.94

kt = ln(2.94)

Thus, t = ln(2.94)k = 8915 yr, so the object is about 9000 yrs old





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Hyperthyroidism 1

Hyperthyroidism is a serious health problem caused by anoveractive thyroid





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Hyperthyroidism 1


The primary hormone released is thyroxine, whichstimulates the release of other hormones





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Hyperthyroidism 1


The primary hormone released is thyroxine, whichstimulates the release of other hormonesToo many other hormones, such as insulin and the sexhormones





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Hyperthyroidism 1


The primary hormone released is thyroxine, whichstimulates the release of other hormonesToo many other hormones, such as insulin and the sexhormonesResult is low blood sugar causing lethargy or mooddisorders and sexual dysfunction





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Hyperthyroidism 1


The primary hormone released is thyroxine, whichstimulates the release of other hormonesToo many other hormones, such as insulin and the sexhormonesResult is low blood sugar causing lethargy or mooddisorders and sexual dysfunctionOne treatment for hyperthyroidism is ablating thethyroid with a large dose of radioactive iodine, 131I





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Hyperthyroidism 1



The thyroid concentrates iodine brought into the body





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Hyperthyroidism 1



The thyroid concentrates iodine brought into the bodyThe 131I undergoes both β and γ radioactive decay, whichdestroys tissue





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Hyperthyroidism 1



The thyroid concentrates iodine brought into the bodyThe 131I undergoes both β and γ radioactive decay, whichdestroys tissuePatient is given medicine to supplement the loss of thyroxine





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Hyperthyroidism 2

Hyperthyroidism: Treatment





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Hyperthyroidism 2


Based upon the thyroid condition and body mass, astandard dose ranges from 110-150 mCi (milliCuries),given in a special “cocktail”





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Hyperthyroidism 2



It is assumed that almost 100% of the 131I is absorbed bythe blood from the gut





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Hyperthyroidism 2




The thyroid uptakes 30% of this isotope of iodine, peakingaround 3 days





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Hyperthyroidism 2





The remainder is excreted in the urine





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Hyperthyroidism 2






The half-life of 131I is 8 days, so this isotope rapidly decays





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Hyperthyroidism 2






The half-life of 131I is 8 days, so this isotope rapidly decays

Still the patient must remain in a designated room for 3-4days for this procedure, so that he or she does not irradiatethe public from his or her treatment





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Hyperthyroidism 3

Hyperthyroidism Example: Assume that a patient is givena 120 mCi cocktail of 131I and that 30% is absorbed by thethyroid





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Hyperthyroidism 3


Find the amount of 131I in the thyroid (in mCi), if thepatient is released four days after swallowing theradioactive cocktail





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Hyperthyroidism 3


Find the amount of 131I in the thyroid (in mCi), if thepatient is released four days after swallowing theradioactive cocktail

Calculate how many mCis the patient’s thyroid retainsafter 30 days, assuming that it was taken up by the thyroidand not excreted in the urine





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Hyperthyroidism 4

Solution:





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Hyperthyroidism 4

Solution:

Assume for simplicity of the model that the 131I isimmediately absorbed into the thyroid, then stays thereuntil it undergoes radioactive decay





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Hyperthyroidism 4

Solution:


Since the thyroid uptakes 30% of the 120 mCi, assumethat the thyroid has 36 mCi immediately after theprocedure





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Hyperthyroidism 4

Solution:



This is an oversimplification as it takes time for the 131I toaccumulate in the thyroid





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Hyperthyroidism 4

Solution:



This is an oversimplification as it takes time for the 131I toaccumulate in the thyroid

This allows the simple model

dR

dt= −k R(t) with R(0) = 36 mCi





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Hyperthyroidism 5

Solution (cont): The radioactive decay model is

dR






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Hyperthyroidism 5


dR


The solution isR(t) = 36 e−kt





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Hyperthyroidism 5


dR



Since the half-life of 131I is 8 days, after 8 days there willare 18 mCi of 131I





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Hyperthyroidism 5


dR




Thus, R(8) = 18 = 36 e−8k, so

e8k = 2 or 8k = ln(2)





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Hyperthyroidism 5


dR




Thus, R(8) = 18 = 36 e−8k, so

e8k = 2 or 8k = ln(2)

Thus, k = ln(2)8 = 0.0866 day−1





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Hyperthyroidism 6

Solution (cont): Since

R(t) = 36 e−kt with k = 0.0866 day−1





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Hyperthyroidism 6


R(t) = 36 e−kt with k = 0.0866 day−1

At the time of the patient’s release t = 4 days, so in thethyroid

R(4) = 36 e−4k =36√

2= 25.46 mCi





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Hyperthyroidism 6


R(t) = 36 e−kt with k = 0.0866 day−1

At the time of the patient’s release t = 4 days, so in thethyroid

R(4) = 36 e−4k =36√

2= 25.46 mCi

After 30 days, we find in the thyroid

R(30) = 36 e−30k = 2.68 mCi





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Hyperthyroidism 7

Graph of R(t)

0 5 10 15 20 25 300

10

20

30

half−life

t (days)

R(t

) (m

Ci13

1 I)

R(t) Remaining in Thyroid





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Solution of Linear Growth and Decay Models

General Solution to Linear Growth and Decay Models:Consider

dy

dt= a y with y(t0) = y0





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Solution of Linear Growth and Decay Models

General Solution to Linear Growth and Decay Models:Consider

dy

dt= a y with y(t0) = y0

The solution isy(t) = y0e

a(t−t0)





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Example: Linear Decay Model

Example: Linear Decay Model: Consider

dy

dt= −0.3 y with y(4) = 12





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Example: Linear Decay Model

Example: Linear Decay Model: Consider

dy

dt= −0.3 y with y(4) = 12

The solution isy(t) = 12 e−0.3(t−4)





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Murder InvestigationCooling Tea

Newton’s Law of Cooling 1

Newton’s Law of Cooling:





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After a murder (or death by other causes), the forensicscientist takes the temperature of the body





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Later the temperature of the body is taken again to findthe rate at which the body is cooling





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Two (or more) data points are used to extrapolate back towhen the murder occurred





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Two (or more) data points are used to extrapolate back towhen the murder occurred

This property is known as Newton’s Law of Cooling





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Newton’s Law of Cooling states that the rate of change intemperature of a cooling body is proportional to the differencebetween the temperature of the body and the surroundingenvironmental temperature





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If T (t) is the temperature of the body, then it satisfies thedifferential equation

dT

dt= −k(T (t) − Te) with T (0) = T0





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dT

dt= −k(T (t) − Te) with T (0) = T0

The parameter k is dependent on the specific properties ofthe particular object (body in this case)





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dT

dt= −k(T (t) − Te) with T (0) = T0


Te is the environmental temperature





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dT

dt= −k(T (t) − Te) with T (0) = T0


Te is the environmental temperature

T0 is the initial temperature of the object





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Murder Example 1

Murder Example





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Murder Example 1

Murder Example

Suppose that a murder victim is found at 8:30 am





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Murder Example 1

Murder Example


The temperature of the body at that time is 30◦C





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Murder Example 1

Murder Example



Assume that the room in which the murder victim lay wasa constant 22◦C





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Murder Example 1

Murder Example




Suppose that an hour later the temperature of the body is28◦C





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Murder Example 1

Murder Example





Normal temperature of a human body when it is alive is37◦C





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Murder Example 1

Murder Example





Normal temperature of a human body when it is alive is37◦C

Use this information to determine the approximate timethat the murder occurred





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Murder Example 2

Solution: From the model for Newton’s Law of Cooling andthe information that is given, if we set t = 0 to be 8:30 am, thenwe solve the initial value problem

dT

dt= −k(T (t) − 22) with T (0) = 30





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Murder Example 2


dT

dt= −k(T (t) − 22) with T (0) = 30

Make a change of variables z(t) = T (t) − 22





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Murder Example 2


dT

dt= −k(T (t) − 22) with T (0) = 30

Make a change of variables z(t) = T (t) − 22Then z ′(t) = T ′(t), so the differential equation abovebecomes

dz

dt= −kz(t), with z(0) = T (0) − 22 = 8





Pollution in a Lake


Murder Example 2


dT

dt= −k(T (t) − 22) with T (0) = 30


dz

dt= −kz(t), with z(0) = T (0) − 22 = 8

This is the radioactive decay problem that we solved





Pollution in a Lake


Murder Example 2


dT

dt= −k(T (t) − 22) with T (0) = 30


dz

dt= −kz(t), with z(0) = T (0) − 22 = 8

This is the radioactive decay problem that we solvedThe solution is

z(t) = 8 e−kt





Pollution in a Lake


Murder Example 3

Solution (cont): From the solution z(t) = 8 e−kt, we have

z(t) = T (t) − 22, so T (t) = z(t) + 22

T (t) = 22 + 8 e−kt





Pollution in a Lake


Murder Example 3


z(t) = T (t) − 22, so T (t) = z(t) + 22

T (t) = 22 + 8 e−kt

One hour later the body temperature is 28◦C

T (1) = 28 = 22 + 8 e−k





Pollution in a Lake


Murder Example 3


z(t) = T (t) − 22, so T (t) = z(t) + 22

T (t) = 22 + 8 e−kt


T (1) = 28 = 22 + 8 e−k

Solving6 = 8 e−k or ek = 4

3





Pollution in a Lake


Murder Example 3


z(t) = T (t) − 22, so T (t) = z(t) + 22

T (t) = 22 + 8 e−kt


T (1) = 28 = 22 + 8 e−k

Solving6 = 8 e−k or ek = 4

3

Thus, k = ln(

43

)

= 0.2877





Pollution in a Lake


Murder Example 4

Solution (cont): It only remains to find out when the murderoccurred





Pollution in a Lake


Murder Example 4


At the time of death, td, the body temperature is 37◦C

T (td) = 37 = 22 + 8 e−k





Pollution in a Lake


Murder Example 4



T (td) = 37 = 22 + 8 e−k

Thus,

8 e−ktd = 37 − 22 = 15 or e−ktd = 158 = 1.875





Pollution in a Lake


Murder Example 4



T (td) = 37 = 22 + 8 e−k

Thus,

8 e−ktd = 37 − 22 = 15 or e−ktd = 158 = 1.875

This gives −ktd = ln(1.875) or

td = − ln(1.875)

k= −2.19





Pollution in a Lake


Murder Example 4



T (td) = 37 = 22 + 8 e−k

Thus,

8 e−ktd = 37 − 22 = 15 or e−ktd = 158 = 1.875

This gives −ktd = ln(1.875) or

td = − ln(1.875)

k= −2.19

The murder occurred about 2 hours 11 minutes before the bodywas found, which places the time of death around 6:19 am





Pollution in a Lake


Cooling Tea 1

Cooling Tea: We would like to determine whether a cup of teacools more rapidly by adding cold milk right after brewing thetea or if you wait 5 minutes to add the milk





Pollution in a Lake


Cooling Tea 1


Begin with 45 cup of boiling hot tea, T (0) = 100◦C





Pollution in a Lake


Cooling Tea 1



Assume the tea cools according to Newton’s law of cooling

dT

dt= −k(T (t) − Te) with Te = 20◦C





Pollution in a Lake


Cooling Tea 1




dT


k is the cooling constant based on the properties of the cupto be calculated





Pollution in a Lake


Cooling Tea 1




dT


k is the cooling constant based on the properties of the cupto be calculated

a. In the first scenario, you let the tea cool for 5 minutes,then add 1

5 cup of cold milk, 5◦C





Pollution in a Lake


Cooling Tea 2

Cooling Tea (cont):





Pollution in a Lake


Cooling Tea 2

Cooling Tea (cont):

Assume that after 2 minutes the tea has cooled to atemperature of 95◦C





Pollution in a Lake


Cooling Tea 2

Cooling Tea (cont):


Determine the value of k, which we assume stays the samein this problem





Pollution in a Lake


Cooling Tea 2

Cooling Tea (cont):



Mix in the milk, assuming that the temperature mixesperfectly in proportion to the volume of the two liquids





Pollution in a Lake


Cooling Tea 2

Cooling Tea (cont):




b. In the second case, add 15 cup of cold milk, 5◦C,

immediately and mix it thoroughly





Pollution in a Lake


Cooling Tea 2

Cooling Tea (cont):




b. In the second case, add 15 cup of cold milk, 5◦C,

immediately and mix it thoroughly

Find how long until each cup of tea reaches atemperature of 70◦C





Pollution in a Lake


Cooling Tea 3

Solution of Cooling Tea: Find the rate constant k for

dT

dt= −k(T (t) − 20), T (0) = 100 and T (2) = 95





Pollution in a Lake


Cooling Tea 3


dT

dt= −k(T (t) − 20), T (0) = 100 and T (2) = 95

Let z(t) = T (t) − 20, so z(0) − T (0) − 20 = 80





Pollution in a Lake


Cooling Tea 3


dT

dt= −k(T (t) − 20), T (0) = 100 and T (2) = 95

Let z(t) = T (t) − 20, so z(0) − T (0) − 20 = 80Since z ′(t) = T ′(t), the initial value problem becomes

dz

dt= −k z(t), z(0) = 80





Pollution in a Lake


Cooling Tea 3


dT

dt= −k(T (t) − 20), T (0) = 100 and T (2) = 95


dz

dt= −k z(t), z(0) = 80

The solution is

z(t) = 80 e−kt = T (t) − 20





Pollution in a Lake


Cooling Tea 3


dT

dt= −k(T (t) − 20), T (0) = 100 and T (2) = 95


dz

dt= −k z(t), z(0) = 80

The solution is

z(t) = 80 e−kt = T (t) − 20

Thus,T (t) = 80 e−kt + 20





Pollution in a Lake


Cooling Tea 4

Solution (cont): The solution is

T (t) = 80 e−kt + 20





Pollution in a Lake


Cooling Tea 4


T (t) = 80 e−kt + 20

Since T (2) = 95,

95 = 80e−2k + 20 or e2k = 8075





Pollution in a Lake


Cooling Tea 4


T (t) = 80 e−kt + 20

Since T (2) = 95,

95 = 80e−2k + 20 or e2k = 8075

k =ln( 80

75)2 = 0.03227





Pollution in a Lake


Cooling Tea 4


T (t) = 80 e−kt + 20

Since T (2) = 95,

95 = 80e−2k + 20 or e2k = 8075

k =ln( 80

75)2 = 0.03227

Find the temperature at 5 min

T (5) = 80e−5k + 20 = 88.1◦C





Pollution in a Lake


Cooling Tea 4


T (t) = 80 e−kt + 20

Since T (2) = 95,

95 = 80e−2k + 20 or e2k = 8075

k =ln( 80

75)2 = 0.03227

Find the temperature at 5 min

T (5) = 80e−5k + 20 = 88.1◦C

Now mix the 45 cup of tea at 88.1◦C with the 1

5 cup of milkat 5◦C, so

T+(5) = 88.1(

45

)

+(

515

)

= 71.5◦C





Pollution in a Lake


Cooling Tea 5

Solution (cont): For the first scenario, the temperature afteradding the milk after 5 min satisfies

T+(5) = 71.5◦C





Pollution in a Lake


Cooling Tea 5


T+(5) = 71.5◦C

The new initial value problem is

dT

dt= −k(T (t) − 20), T (5) = 71.5◦C





Pollution in a Lake


Cooling Tea 5


T+(5) = 71.5◦C


dT

dt= −k(T (t) − 20), T (5) = 71.5◦C

With the same substitution, z(t) = T (t) − 20,

dz

dt= −kz, z(5) = 51.5





Pollution in a Lake


Cooling Tea 5


T+(5) = 71.5◦C


dT

dt= −k(T (t) − 20), T (5) = 71.5◦C

With the same substitution, z(t) = T (t) − 20,

dz

dt= −kz, z(5) = 51.5

This has the solution

z(t) = 51.5e−k(t−5) = T (t) − 20





Pollution in a Lake


Cooling Tea 6

Solution (cont): For the first scenario, the temperaturesatisfies

T (t) = 51.5e−k(t−5) + 20





Pollution in a Lake


Cooling Tea 6


T (t) = 51.5e−k(t−5) + 20

To find when the tea is 70◦C, solve

70 = 51.5e−k(t−5) + 20





Pollution in a Lake


Cooling Tea 6


T (t) = 51.5e−k(t−5) + 20


70 = 51.5e−k(t−5) + 20

Thus,ek(t−5) = 51.5

50





Pollution in a Lake


Cooling Tea 6


T (t) = 51.5e−k(t−5) + 20


70 = 51.5e−k(t−5) + 20

Thus,ek(t−5) = 51.5

50

It follows that k(t − 5) = ln(51.5/50), so

t = 5 +ln(51.5/50)

k= 5.92 min





Pollution in a Lake


Cooling Tea 7

Solution (cont): For the second scenario, we mix the tea andmilk, so

T (0) = 100(

45

)

+ 5(

15

)

= 81◦C





Pollution in a Lake


Cooling Tea 7


T (0) = 100(

45

)

+ 5(

15

)

= 81◦C


dT

dt= −k(T (t) − 20), T (0) = 81◦C





Pollution in a Lake


Cooling Tea 7


T (0) = 100(

45

)

+ 5(

15

)

= 81◦C


dT

dt= −k(T (t) − 20), T (0) = 81◦C

With z(t) = T (t) − 20,

dz

dt= −k z(t), z(0) = 61





Pollution in a Lake


Cooling Tea 7


T (0) = 100(

45

)

+ 5(

15

)

= 81◦C


dT

dt= −k(T (t) − 20), T (0) = 81◦C

With z(t) = T (t) − 20,

dz

dt= −k z(t), z(0) = 61


z(t) = 61e−kt = T (t) − 20





Pollution in a Lake


Cooling Tea 8

Solution (cont): For the second scenario, the solution is

T (t) = 61 e−kt + 20





Pollution in a Lake


Cooling Tea 8


T (t) = 61 e−kt + 20


70 = 61e−kt + 20





Pollution in a Lake


Cooling Tea 8


T (t) = 61 e−kt + 20


70 = 61e−kt + 20

Thus,ekt = 61

50





Pollution in a Lake


Cooling Tea 8


T (t) = 61 e−kt + 20


70 = 61e−kt + 20

Thus,ekt = 61

50

Since kt = ln(

6150

)

,

t =ln(61/50

k= 6.16 min





Pollution in a Lake


Cooling Tea 8


T (t) = 61 e−kt + 20


70 = 61e−kt + 20

Thus,ekt = 61

50

Since kt = ln(

6150

)

,

t =ln(61/50

k= 6.16 min

Waiting to pour in the milk for 5 minutes, saves about 15seconds in cooling time





Pollution in a Lake



Graph of Cooling Tea

0 1 2 3 4 5 6 760

70

80

90

100

t (min)

Tem

pera

ture

(o C)

Cooling Tea

Part bPart a





Pollution in a Lake

Solution of General Linear Model 1

Solution of General Linear Model Consider the LinearModel

dy

dt= a y + b with y(t0) = y0





Pollution in a Lake



dy


Rewrite equation as

dy

dt= a

(

y +b

a

)





Pollution in a Lake



dy


Rewrite equation as

dy

dt= a

(

y +b

a

)

Make the substitution z(t) = y(t) + ba , so dz

dt = dydt and

z(t0) = y0 + ba

dz

dt= a z with z(t0) = y0 + b

a





Pollution in a Lake


Solution of General Linear Model The shifted model is

dz


a





Pollution in a Lake



dz


a

The solution to this problem is

z(t) =

(

y0 +b

a

)

ea(t−t0) = y(t) +b

a





Pollution in a Lake



dz


a


z(t) =

(

y0 +b

a

)

ea(t−t0) = y(t) +b

a

The solution is

y(t) =

(

y0 +b

a

)

ea(t−t0) − b

a





Pollution in a Lake

Example of Linear Model 1

Example of Linear Model Consider the Linear Model

dy

dt= 5 − 0.2 y with y(3) = 7





Pollution in a Lake



dy

dt= 5 − 0.2 y with y(3) = 7

Rewrite equation as

dy

dt= −0.2(y − 25)





Pollution in a Lake



dy

dt= 5 − 0.2 y with y(3) = 7

Rewrite equation as

dy

dt= −0.2(y − 25)

Make the substitution z(t) = y(t) − 25, so dzdt = dy

dt andz(3) = −18

dz

dt= −0.2 z with z(3) = −18





Pollution in a Lake


Example of Linear Model The substituted model is

dz

dt= −0.2 z with z(3) = −18





Pollution in a Lake



dz

dt= −0.2 z with z(3) = −18

Thus,z(t) = −18 e−0.2(t−3) = y(t) − 25





Pollution in a Lake



dz

dt= −0.2 z with z(3) = −18

Thus,z(t) = −18 e−0.2(t−3) = y(t) − 25

The solution isy(t) = 25 − 18 e−0.2(t−3)





Pollution in a Lake

Example of Pollution with Evaporation

Pollution in a Lake 1

Pollution in a Lake: Introduction





Pollution in a Lake




One of the most urgent problems in modern society is howto reduce the pollution and toxicity of our water sources





Pollution in a Lake





These are very complex issues that require amultidisciplinary approach and are often politically veryintractable because of the key role that water plays inhuman society and the many competing interests





Pollution in a Lake






Here we examine a very simplistic model for pollution of alake





Pollution in a Lake






Here we examine a very simplistic model for pollution of alake

The model illustrates some basic elements from which morecomplicated models can be built and analyzed





Pollution in a Lake



Pollution in a Lake: Problem set up





Pollution in a Lake




Consider the scenario of a new pesticide that is applied tofields upstream from a clean lake with volume V





Pollution in a Lake





Assume that a river receives a constant amount of this newpesticide into its water, and that it flows into the lake at aconstant rate, f





Pollution in a Lake






This assumption implies that the river has a constantconcentration of the new pesticide, p





Pollution in a Lake






This assumption implies that the river has a constantconcentration of the new pesticide, p

Assume that the lake is well-mixed and maintains aconstant volume by having a river exiting the lake with thesame flow rate, f , of the inflowing river





Pollution in a Lake



Diagram for Lake Problem Design a model using a linearfirst order differential equation for the concentration of thepesticide in the lake, c(t)

f : flow rate

f

p: pollutant

V: Volume

c t( ): concentration ofpollutant in the lake





Pollution in a Lake



Differential Equation for Pollution in a Lake





Pollution in a Lake




Set up a differential equation that describes the massbalance of the pollutant





Pollution in a Lake




Set up a differential equation that describes the massbalance of the pollutantThe change in amount of pollutant =

Amount entering - Amount leaving





Pollution in a Lake





Amount entering - Amount leavingThe amount entering is simply the concentration of thepollutant, p, in the river times the flow rate of the river, f





Pollution in a Lake





Amount entering - Amount leavingThe amount entering is simply the concentration of thepollutant, p, in the river times the flow rate of the river, fThe amount leaving has the same flow rate, f





Pollution in a Lake





Amount entering - Amount leavingThe amount entering is simply the concentration of thepollutant, p, in the river times the flow rate of the river, fThe amount leaving has the same flow rate, fSince the lake is assumed to be well-mixed, theconcentration in the outflowing river will be equal to theconcentration of the pollutant in the lake, c(t)





Pollution in a Lake





Amount entering - Amount leavingThe amount entering is simply the concentration of thepollutant, p, in the river times the flow rate of the river, fThe amount leaving has the same flow rate, fSince the lake is assumed to be well-mixed, theconcentration in the outflowing river will be equal to theconcentration of the pollutant in the lake, c(t)The product f c(t) gives the amount of pollutant leavingthe lake per unit time





Pollution in a Lake



Differential Equations for Amount and Concentrationof Pollutant





Pollution in a Lake




The change in amount of pollutant satisfies the model

da(t)

dt= f p − f c(t)





Pollution in a Lake





da(t)

dt= f p − f c(t)

Since the lake maintains a constant volume V , thenc(t) = a(t)/V , which also implies that c ′(t ) = a ′(t)/V





Pollution in a Lake





da(t)

dt= f p − f c(t)

Since the lake maintains a constant volume V , thenc(t) = a(t)/V , which also implies that c ′(t ) = a ′(t)/VDividing the above differential equation by the volume V ,

dc(t)

dt=

f

V(p − c(t))





Pollution in a Lake





da(t)

dt= f p − f c(t)

Since the lake maintains a constant volume V , thenc(t) = a(t)/V , which also implies that c ′(t ) = a ′(t)/VDividing the above differential equation by the volume V ,

dc(t)

dt=

f

V(p − c(t))

If the lake is initially clean, then c(0) = 0Joseph M. Mahaffy, 〈[email protected]〉





Pollution in a Lake



Soluton of the Differential Equation: Rewrite thedifferential equation for the concentration of pollutant as

dc(t)

dt= − f

V(c(t) − p) with c(0) = 0





Pollution in a Lake




dc(t)

dt= − f

V(c(t) − p) with c(0) = 0

This DE should remind you of Newton’s Law of Coolingwith f/V acting like k and p acting like Te





Pollution in a Lake




dc(t)

dt= − f

V(c(t) − p) with c(0) = 0


Make the substitution, z(t) = c(t) − p, so z ′(t) = c ′(t)





Pollution in a Lake




dc(t)

dt= − f

V(c(t) − p) with c(0) = 0


Make the substitution, z(t) = c(t) − p, so z ′(t) = c ′(t)The initial condition becomes z(0) = c(0) − p = −p





Pollution in a Lake




dc(t)

dt= − f

V(c(t) − p) with c(0) = 0


Make the substitution, z(t) = c(t) − p, so z ′(t) = c ′(t)The initial condition becomes z(0) = c(0) − p = −pThe initial value problem in z(t) becomes,

dz(t)

dt= − f

Vz(t), with z(0) = −p





Pollution in a Lake



Soluton of the Differential Equation (cont): Since

dz(t)

dt= − f






Pollution in a Lake




dz(t)

dt= − f



z(t) = −p e−ft

V = c(t) − p





Pollution in a Lake




dz(t)

dt= − f



z(t) = −p e−ft

V = c(t) − p

c(t) = p(

1 − e−ft

V

)





Pollution in a Lake




dz(t)

dt= − f



z(t) = −p e−ft

V = c(t) − p

c(t) = p(

1 − e−ft

V

)

The exponential decay in this solution shows

limt→∞

c(t) = p





Pollution in a Lake




dz(t)

dt= − f



z(t) = −p e−ft

V = c(t) − p

c(t) = p(

1 − e−ft

V

)

The exponential decay in this solution shows

limt→∞

c(t) = p

This is exactly what you would expect, as the enteringriver has a concentration of p





Pollution in a Lake


Example: Pollution in a Lake 1

Example: Pollution in a Lake Part 1





Pollution in a Lake




Suppose that you begin with a 10,000 m3 clean lake





Pollution in a Lake





Assume the river entering has a flow of 100 m3/day andthe concentration of some pesticide in the river is measuredto have a concentration of 5 ppm (parts per million)





Pollution in a Lake






Form the differential equation describing the concentrationof pollutant in the lake at any time t and solve it





Pollution in a Lake






Form the differential equation describing the concentrationof pollutant in the lake at any time t and solve it

Find out how long it takes for this lake to have aconcentration of 2 ppm





Pollution in a Lake



Solution: This example follows the model derived above, sothe differential equation for the concentration of pollutant is

dc(t)

dt= − f

V(c(t) − p) with c(0) = 0





Pollution in a Lake




dc(t)

dt= − f

V(c(t) − p) with c(0) = 0

Since V = 10, 000, f = 100, and p = 5,

dc(t)

dt= − 100

10000(c(t) − 5) with c(0) = 0





Pollution in a Lake




dc(t)

dt= − f

V(c(t) − p) with c(0) = 0

Since V = 10, 000, f = 100, and p = 5,

dc(t)

dt= − 100

10000(c(t) − 5) with c(0) = 0

Let z(t) = c(t) − 5, then the differential equation becomes,

dz

dt= −0.01z(t), with z(0) = −5





Pollution in a Lake




dc(t)

dt= − f

V(c(t) − p) with c(0) = 0

Since V = 10, 000, f = 100, and p = 5,

dc(t)

dt= − 100

10000(c(t) − 5) with c(0) = 0

Let z(t) = c(t) − 5, then the differential equation becomes,

dz

dt= −0.01z(t), with z(0) = −5

This has a solution

z(t) = −5e−0.01t = c(t) − 5Joseph M. Mahaffy, 〈[email protected]〉





Pollution in a Lake



Solution (cont): The concentration of pollutant in the lake is

c(t) = 5(

1 − e−0.01t)





Pollution in a Lake




c(t) = 5(

1 − e−0.01t)

To find how long it takes for the concentration to reach2 ppm, solve the equation

2 = 5 − 5e−0.01t)





Pollution in a Lake




c(t) = 5(

1 − e−0.01t)


2 = 5 − 5e−0.01t)

Thus,e−0.01t) = 3

5 or e0.01t) = 53





Pollution in a Lake




c(t) = 5(

1 − e−0.01t)


2 = 5 − 5e−0.01t)

Thus,e−0.01t) = 3

5 or e0.01t) = 53

Solving this for t, we obtain

t = 100 ln(

53

)

= 51.1 days





Pollution in a Lake








Pollution in a Lake




Suppose that when the concentration reaches 4 ppm, thepesticide is banned





Pollution in a Lake





For simplicity, assume that the concentration of pesticidedrops immediately to zero in the river





Pollution in a Lake






Assume that the pesticide is not degraded or lost by anymeans other than dilution





Pollution in a Lake






Assume that the pesticide is not degraded or lost by anymeans other than dilution

Find how long until the concentration reaches 1 ppm





Pollution in a Lake



Solution: The new initial value problem becomes

dc

dt= −0.01(c(t) − 0) = −0.01c(t) with c(0) = 4





Pollution in a Lake




dc

dt= −0.01(c(t) − 0) = −0.01c(t) with c(0) = 4

This problem is in the form of a radioactive decay problem





Pollution in a Lake




dc

dt= −0.01(c(t) − 0) = −0.01c(t) with c(0) = 4

This problem is in the form of a radioactive decay problemThis has the solution

c(t) = 4e−0.01t





Pollution in a Lake




dc

dt= −0.01(c(t) − 0) = −0.01c(t) with c(0) = 4


c(t) = 4e−0.01t

To find how long it takes for the concentration to return to1 ppm, solve the equation

1 = 4e−0.01t or e0.01t = 4





Pollution in a Lake




dc

dt= −0.01(c(t) − 0) = −0.01c(t) with c(0) = 4


c(t) = 4e−0.01t

To find how long it takes for the concentration to return to1 ppm, solve the equation

1 = 4e−0.01t or e0.01t = 4

Solving this for t

t = 100 ln(4) = 138.6 days





Pollution in a Lake


Pollution in a Lake: Complications

Pollution in a Lake: Complications The above discussion forpollution in a lake fails to account for many significant complications





Pollution in a Lake




There are considerations of degradation of the pesticide,stratefication in the lake, and uptake and reentering of thepesticide through interaction with the organisms living in thelake





Pollution in a Lake





The river will vary in its flow rate, and the leeching of thepesticide into river is highly dependent on rainfall, ground watermovement, and rate of pesticide application





Pollution in a Lake






Obviously, there are many other complications that wouldincrease the difficulty of analyzing this model





Pollution in a Lake






Obviously, there are many other complications that wouldincrease the difficulty of analyzing this model

The next section shows numerical methods to handle morecomplicated models





Pollution in a Lake


Example: Lake Pollution with Evaporation 1

Example: Lake Pollution with Evaporation





Pollution in a Lake




Suppose that a new industry starts up river from a lake at t = 0days, and this industry starts dumping a toxic pollutant, P (t),into the river at a rate of 7 g/day, which flows directly into thelake





Pollution in a Lake





The flow of the river is 1000 m3/day, which goes into the lakethat maintains a constant volume of 400,000 m3





Pollution in a Lake






The lake is situated in a hot area and loses 50 m3/day of water toevaporation (pure water with no pollutant), while the remainderof the water exits at a rate of 950 m3/day through a river





Pollution in a Lake






The lake is situated in a hot area and loses 50 m3/day of water toevaporation (pure water with no pollutant), while the remainderof the water exits at a rate of 950 m3/day through a river

Assume that all quantities are well-mixed and that there are notime delays for the pollutant reaching the lake from the river





Pollution in a Lake



Example: Lake Pollution with Evaporation (cont) Part a





Pollution in a Lake




Write a differential equation that describes theconcentration, c(t), of the pollutant in the lake, using unitsof mg/m3





Pollution in a Lake





Solve the differential equation





Pollution in a Lake






If a concentration of only 2 mg/m3 is toxic to the fishpopulation, then find how long until this level is reached





Pollution in a Lake






If a concentration of only 2 mg/m3 is toxic to the fishpopulation, then find how long until this level is reached

If unchecked by regulations, then find what the eventualconcentration of the pollutant is in the lake, assumingconstant output by the new industry





Pollution in a Lake



Solution: Let P (t) be the amount of pollutantThe change in amount of pollutant =






Pollution in a Lake





The change in amount is dPdt





Pollution in a Lake






The concentration is given by c(t) = P (t)/V andc ′(t) = P ′(t)/V





Pollution in a Lake







The amount entering is the constant rate of pollutantdumped into the river, which is given by k = 7000 mg/day





Pollution in a Lake







The amount entering is the constant rate of pollutantdumped into the river, which is given by k = 7000 mg/day

The amount leaving is given by the concentration of thepollutant in the lake, c(t) (in mg/m3), times the flow ofwater out of the lake, f = 950 m3/day





Pollution in a Lake



Solution (cont): The conservation of amount of pollutant isgiven by the equation:

dP

dt= k − f c(t) = 7000 − 950 c(t)





Pollution in a Lake




dP

dt= k − f c(t) = 7000 − 950 c(t)

Evaporation concentrates the pollutant by allowing waterto leave without the pollutant





Pollution in a Lake




dP

dt= k − f c(t) = 7000 − 950 c(t)

Evaporation concentrates the pollutant by allowing waterto leave without the pollutantDivide the equation above by the volume, V = 400, 000 m3

(

1

V

)

dP (t)

dt=

k

V− f

Vc(t) = 7

400 − 950400000c(t)





Pollution in a Lake




dP

dt= k − f c(t) = 7000 − 950 c(t)

Evaporation concentrates the pollutant by allowing waterto leave without the pollutantDivide the equation above by the volume, V = 400, 000 m3

(

1

V

)

dP (t)

dt=

k

V− f

Vc(t) = 7

400 − 950400000c(t)

The concentration equation is

dc

dt= 7

400 − 950400000c(t) = − f

V

(

c(t) − kf

)





Pollution in a Lake



Solution (cont): The concentration equation is

dc

dt= − 95

40000

(

c(t) − 70095

)





Pollution in a Lake




dc

dt= − 95

40000

(

c(t) − 70095

)

Make the change of variables, z(t) = c(t) − 70095 , with

z(0) = −70095





Pollution in a Lake




dc

dt= − 95

40000

(

c(t) − 70095

)


z(0) = −70095


dz

dt= − 95

40000z(t) with z(0) = −70095





Pollution in a Lake




dc

dt= − 95

40000

(

c(t) − 70095

)


z(0) = −70095


dz

dt= − 95

40000z(t) with z(0) = −70095

The solution is

z(t) = −70095 e−95t/40000 = c(t) − 700

95





Pollution in a Lake




c(t) = 70095

(

1 − e−95t/40000)

≈ 7.368(

1 − e−0.002375t)





Pollution in a Lake




c(t) = 70095

(

1 − e−95t/40000)

≈ 7.368(

1 − e−0.002375t)

If a concentration of 2 mg/m3 is toxic to the fishpopulation, then find when c(t) = 2 mg/m3





Pollution in a Lake




c(t) = 70095

(

1 − e−95t/40000)

≈ 7.368(

1 − e−0.002375t)


Solve

2 = 7.368(

1 − e−0.002375t)

or e0.002375t ≈ 1.3726





Pollution in a Lake




c(t) = 70095

(

1 − e−95t/40000)

≈ 7.368(

1 − e−0.002375t)


Solve

2 = 7.368(

1 − e−0.002375t)

or e0.002375t ≈ 1.3726

Thus, t = ln(1.3726)0.002375 ≈ 133.3 days





Pollution in a Lake




c(t) = 70095

(

1 − e−95t/40000)

≈ 7.368(

1 − e−0.002375t)


Solve

2 = 7.368(

1 − e−0.002375t)

or e0.002375t ≈ 1.3726

Thus, t = ln(1.3726)0.002375 ≈ 133.3 days

The limiting concentration is

limt→∞

c(t) = 70095 ≈ 7.368





Pollution in a Lake



Example: Lake Pollution with Evaporation (cont) Part b





Pollution in a Lake




Suppose that the lake is at the limiting level of pollutantand a new environmental law is passed that shuts down theindustry at a new time t = 0 days





Pollution in a Lake





Write a new differential equation describing the situationfollowing the shutdown of the industry and solve thisequation





Pollution in a Lake





Write a new differential equation describing the situationfollowing the shutdown of the industry and solve thisequation

Calculate how long it takes for the lake to return to a levelthat allows fish to survive





Pollution in a Lake



Solution: Now k = 0, so the initial value problem becomes

dc

dt= − 95

40000c(t) = −0.002375 c(t) with c(0) = 70095





Pollution in a Lake




dc

dt= − 95

40000c(t) = −0.002375 c(t) with c(0) = 70095


c(t) = 70095 e−0.002375t ≈ 7.368 e−0.002375t





Pollution in a Lake




dc

dt= − 95

40000c(t) = −0.002375 c(t) with c(0) = 70095


c(t) = 70095 e−0.002375t ≈ 7.368 e−0.002375t

The concentration is reduced to 2 mg/m3 when

2 = 7.368 e−0.002375t or e0.002375t = 3.684





Pollution in a Lake




dc

dt= − 95

40000c(t) = −0.002375 c(t) with c(0) = 70095


c(t) = 70095 e−0.002375t ≈ 7.368 e−0.002375t

The concentration is reduced to 2 mg/m3 when

2 = 7.368 e−0.002375t or e0.002375t = 3.684

The lake is sufficiently clean for fish when

t = ln(3.684)0.002375 ≈ 549 days


Calculus for the Life Sciences II - Joseph M. Mahaffy

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Transcript of Calculus for the Life Sciences II - Joseph M. Mahaffy