First Oder Equations-Applications

7/29/2019 First Oder Equations-Applications

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Copyright Ren Barrientos Page 1

APPLICTIONS OF FIRST ORDER EQUATIONS

This section presents examples of first order models whose equations are linear or separable (or in some

instances both).

Population Models (The Malthusian Model)

This model, named after the English Scholar Thomas Malthus (1766-1834), assumes that the per-capita rate of

change in population is constant.Let denote the population (humans, bacteria, or any other organism that increases in numbers) at time .This model assumes that where k is a constant. The expression on the left-hand side of this equation is the per-capita rate of change in

population and hence the models name. If the initial population is 0 , then we can write this model asthe initial value problem ; which is a separable equation. Separating variables,

; 0 Integrating, l n | | ; 0 Applying the initial condition, ln . Thus,l n | | l n ln| |ln

ln

Example 1 Assume a population obeys the constant per-capita rate law and suppose it starts with 100,200individuals. One year later, the population has grown by 0.6%. Determine the population at the five-year mark

and the percent increase in population over that time period.

SolutionThis population obeys the law With 100,200. Thus,

100,200

All we need to find is the parameter kin order to fully determine this population.From the information given, we have 1 100,2001.006

or

100,200 100,200 1.006Solving for k:

Working with this Model: This model involves two parameters, the initial population and the constantk.Enough information will be provided so that both can be obtained and the population formula determined.

Pop. Increased by 0.6%in one year


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ln1.006.Therefore: ,.We can now predict the population at any time in the future:

5 100,200 . 103,242The percent change in population over the five-years period is5 00 100%3.03%

Example 2 Assume a population obeys the Malthusian model. Suppose that there are 500,000 individuals atthe end of year 3 and 545,000 at the end of year 5. What was the initial population?Solution

Our model is We seek, the initial population. There are two quantities we need to determine: kand .The information we are given is:

3 500,0005 545,000Thus, 500,000

and

545,000These equations can be solved for the two unknown quantities and k: Dividing the equations (toeliminate ): 545,000500,000

1.09 .

Having obtained the value ofk, we can use either of the original equations to find , for example, 500,000 500,000 Thus,

500,000. ,Example 3 Assume a bacterial culture grows at a rate which is proportional to the bacterial population at time. Two hours after starting with a population a count shows a population of90 bacteria per square cm in a80-cm2 dish. Four hours later, a new count shows 140 bacteria per cm2. (a) How many bacteria were presentoriginally? (b) How long before the culture doubles in size relative to its original value?

Solution

The growth formula is given by Let denote the population per square centimeter so that 80 . The reason why thepopulation is measured on a per cm2 basis is because it is very difficult to accurately count the entirepopulation under a microscope. Hence, one concentrates on a small area of the dish and counts the

bacteria there at different times [we assume that the bacterial distribution is homogeneous and what

happens in one square centimeter more or less happens in all others].

(a) We seek, the original population in the dish. The information we are given is:


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2 90 bacteriacm 4 140 bacteriacm

Thus,

90

140Solve for k: 14090 1.5556Therefore, .. Solving for ,. 90 58So the initial population is approximately , bacteria.(b) we want to determine the time

at which

2.Thus, we solve the equation

. 2for . Cancelling, . 2 10.2209 ln23.14The time to double is approximately 3.14 hours.

Another Characterization of the Malthusian Model

The Malthusian model tell us that . Thus, the population in year 1 is 1 This means that 1 . In other words, populations that grow this way are also said to grow in ageometricprogression: the population in year

1is a constant multiple of the population in year

, that

multiple being . We write this relation as where . Malthus proposed that humanity would run out of food sooner rather than later because whereasthe human population was increasing in numbers in a geometric progression, food supply could be increased

only as anarithmetic progression:

"The power of population is so superior to the power of the earth to produce subsistence for man, that premature

death must in some shape or other visit the human race. The vices of mankind are active and able ministers of

depopulation. They are the precursors in the great army of destruction, and often finish the dreadful work

themselves. But should they fail in this war of extermination, sickly seasons, epidemics, pestilence, and plague

advance in terrific array, and sweep off their thousands and tens of thousands. Should success be still

incomplete, gigantic inevitable famine stalks in the rear, and with one mighty blow levels the population with

the food of the world".

Malthus T.R. 1798.An essay on the principle of population. Chapter VII, p61.

A Radioactive Decay Model

Radioactive decay is the process by which atomic nuclei of unstable atoms decay spontaneously into other,

more stable, nuclei. This decay process is accompanied by the release of energy which manifests itself in the

form of alpha, beta, and gamma particles.


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Let be the original amount of radioactive substance and be the amount present at time . This is adiscrete number, however, since it is so large (of the order of 10), we can treat is as if it were a continuousvariable.

We consider the one-decay process whereby nuclide A decays into a more stable nuclide B. The number ofdecay events in a time interval of length is a random variable. However, we will assume that this number isjointly proportional to the number of nuclide present at the beginning of the time interval and its duration

:

where is a positive constant1 called the decay constant and its value changes from one species of radionuclide to another [the negative sign is necessary in order to ensure that 0].Dividing by and letting 0, ; Observe that this is the same model as the population model we studied earlier except for the negative sign. Its

solution is Example 4 Suppose we start with 80 grams of a radioactive substance which exhibits exponential decay.Suppose also that

1hour later only

77.5grams are left. When will the substance decay to

20% of its original

amount?

SolutionThe model is We need, which we obtain by applying the additional information stated in the problem:

1 77.5Thus, 80 77.5

77.580 Solving for

:

ln 77.580 .Therefore, this substance is governed by

.When will be 20% of what we started with? That is, when will 16? The answer isobtained by solving the equation 80. 16Solving for

1

0.03175 l n 16

80

. Example 5 (Carbon Dating) Carbon occurs naturally in the earths atmosphere in the form of two stable isotopes(and ) and the radioactive isotope whose half-life is 5730 years2. This radioactive isotope is1Physical constants are taken to be positive numbers.2The half-lifeof a substance is thetime it takes for half of the substance to decay. Thus, if one starts with 80 grams ofcarbon-14 today, there will be 40 grams of it left in 5730.


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produced when cosmic rays interact with the Nitrogen in the earths upper atmosphere and the assumption is

that this rate of creation is constant. It is also assumed that the ratio of carbon-14 to carbon is constant, butthere are studies that indicate the contrary and for that reason one must be careful when using carbon dating.

Nevertheless, here is the idea: when an organism is alive, the concentration of present in it is the same asthat of the surroundings. When it dies, it ceases to consume the isotope which then starts to decay. A

measurement of the amount present can be used to estimate how long the organism has been dead and therefore

date it.The method is fairly accurate up to about 50,000 years. For longer periods, other radioactive materials withlonger half-lives can be used, but these too are susceptible to errors.

Suppose a sample of material from an ancient burial case is found to have 60% of carbon-14 found on a sampleof similar material (wood in this situation). Approximately how old is the case?

SolutionCarbon-14, like any other radioactive substance, decays more or less according to

We begin by finding because we know the half-life of this isotope. Since this value is 5730,

12

Thus,

15730 ln 12Using the properties of logarithms, we can also write

ln 25730 1.209710Back to our problem. Now we know that

.If the burial case contained carbon atoms (or grams of the isotope) originally, once it was placedunderground it ceased to absorb the isotope and what was originally present started to decay without

being replenished. We know that what remains is 60% of the normal value. Therefore,

0.6 .Solving for t: 0 . 6 . 1.209710ln0.6Thus,

ln0.61.209710 ,. The half-life formula is worth remembering:

Using this relationship, we can also write

2//

/

Half-Life Formula

If a radioactive substance has a half-life of

/time units, then the decay constant is given

by


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Heat Conduction (Newtons Law of Cooling)

This model describes something that is very familiar to us: if a hot object is placed in contact with a cold object,

the hot object cools down and the cold object heats up until an equilibrium temperature is attained. The process

by which this happens is calledheat conduction.

Imagine an insulated container containing a fluid whose initial temperature is and suppose that we place in ita hot object whose initial temperature is

:

We know from experience that the temperature, , of the object will decrease and that of the fluid, ,will increase until both the fluid and object are at the same equilibrium temperature , at which time no moreheat exchange takes place (we say that equilibrium has been reached). The qualitative behavior of the

functions

and

is shown below:

Newtons model states thatthe rate of change in temperature of the object is proportional to the difference in

temperature between the substance and the object:

; 0 Clearly when , 0. Furthermore, notice that if the object is hotter than its surroundings then 0. Thus 0 as expected.We now make a simplifying assumption: The substance is of such extent and characteristic that its temperatureremains essentially constant, that is, for all . Then In addition, we assume that 0 .This model makes sense: as the object cools, it does so at a slower rate as it approaches the equilibrium

temperature .

EQ0

t

Q(t)

ET0

t

T(t)

The magnitude of this slope is proportionalto , that is, .

T0

time

T(t)

t


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Exercise: Suppose instead that the fluid in the contained is hotter than the object. Show that equation (4) is still a validmodel for.Thus, we have the first order linear initial value problem ; Let us solve this equation in general terms. By separation of variables, ; Integrating,

Note: is a constant. The solution is ln| | ln| | Simplifying,

ln Solving for

,

Thus, Observe that and must always have the same sign (why?). Therefore, their quotient must alwaysbe positive. Thus, Hence,

Exercise: Solve the model for the case and show that for all 0.

Example 6 Suppose that a small object initially at temperature 20 C is placed in a bath whosetemperature is kept constant at 40 C. Two minutes later the objects temperature is 24.5o C. a) find theobjects temperature at 1 0 minutes. b) how long will it take for the temperature to attain 99% of that of thesurroundings?

Solution

The initial value problem that we need to solve is: 4 0; 0 20Solving the equation 4 0 Thus, ln| 4 0| Solving for :

Working With This Model There is one parameter that need to be determined: the constantk. The temperatureof the surroundings will be given or will be implied by additional information and an additional temperature readingat some time 0 allows us to find .


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4 0

[where ]. Applying the initial condition,0 4 0 20

Thus,

and

Notice that is conforms with the formula derived earlier. Since 2 24.5C: 4020 24.5Solving for k: 4024.520

or.

Hence, the temperature function for this object is

.To answer a), 10 4 0 2 0 ..

To answer b), we seek the time at which 0.9940, that is,4020. 39.6

Solving for , . 0.02 10.12745 ln0.02

. An application of Newtons Law of Cooling can be used to describe the temperature inside a building.Suppose we have a building inside of which there are objects that might generate heat (lighting, people,

machines, etc.). We also have a devise which acts as asource orsinkof heat a heater or air-conditioning unit.

Finally, we have the environment outside, whose temperature is not significantly affected by what goes on

inside the building, but which will have an effect in its interior since there is no such thing as perfect insulation.

From basic science, the of heat required to change the temperature of an object by an amount is given by

Thus,

If heat is added 0 to a system its temperature increases and if heat is removed 0 , then it decreases.Here where and is a constant called the objectsspecific heat. The mass in question is the massof air inside the building.Let 0

: 0 if heat generated 0 if heat is removed


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Let be the temperature inside the building and be that of the surroundings. In time the buildingstemperature will change due to the heat produced by the objects, exchange with the surroundings, and operation

of the devises. Then,

or

where and are positive constants [note that the last term corresponds to Newtons Law of Cooling].Dividing by and taking l i m 0: We may absorb the constant into the functions and and rename them and . Then this equationbecomes the initial value problem:

; Problems Involving Motion and Newtons Second Law

Free Fall with Air Resistance

Motion problems can be solved by applying Newtons SecondLaw, which states that an object that is actedupon by a force will accelerate and this acceleration is proportional to the applied force:

where is its linear momentum defined as the product of its mass and velocity. When several forces acton an object of fixed mass Newtons Second Law states

The units we use when we work with motion are

Quantity SI British

Distance Meter (m) Foot (ft)

Time Second (sec) Second (sec)

Mass Kilogram (Kg) Slug

Example 7 (Free Fall Motion with Air Resistance) An object is released from some height and allowed to dropunder the influence of the force of gravity. As it falls, it experiencesdrag which is assumed to be in magnitude

proportional to the objects speed but in the opposite direction of its motion. Determine the velocity vector.

Does the object attain a steady state of motion under these two forces?

Solution

The two force that act on the object are the force of gravity and the drag force . The selection ofa frame of reference is crucial in problems like this because the wrong choice can make themathematics unnecessarily complicated. Let us choose our coordinate system with the positive y-axis

pointing down, in the direction of the force of gravity. This might be at first confusing since we

usually select up as the positive direction, but as you will see this simplifies the mathematics without

affecting the physics.

Let denote the unit vector in the positive y- direction (see figure) and let denote the objectsvertical component of velocity at time so that


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The force of gravity in this coordinate system is given by

where has a value of9.8 m/s2 in the SI system and 32 ft/s2 in English units.We also assume that the drag force is of the form

where bis a positive constant. This is a typical assumption made in cases of low speeds in a normalmedium (e.g. not too viscous) and we will refer to these media as Type I media.

Applying Newtons Second Law,

or

Hence,

vecor equationEquating components,

scalar equationThe corresponding initial value problem is

; 0 0 Equation (7) is an autonomous equation which we can write as ; 0 0and it has one critical point:

This critical value is called the objects terminal velocity, denoted by,and it represents the constantspeed which it will eventually attain.

D

Objectmovingdown


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Let us solve the equation using Separation of Variables:

is a separable equation:

or Integrating,

Let . Then and

or

ln| | Since 0 when 0, we have ln||Therefore, ln| | ln||Multiplying by /:

ln| | l n||

ln We can solve for :

Since 0, | |

What is the sign of ? it must be positive for the case under consideration because of thechoice of coordinates as well as the initial velocity assumption.

Exercise: show that in this problem,

0

Solving for , 1 Once again, lim /Furthermore, observe that / at all times.


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When does the object actually attain its terminal velocity? Considering the equation for we derived theanswer is never. The terminal velocity is a limiting value because never becomes identically 0.Therefore, never attains its limiting value Example 8 A 120 lb parachutist jumps from an altitude of 10,000 ft and free falls for 3 minutes. Suppose 0 . 7 and that we are in a Type I environment. a) what is her terminal velocity? b) does she attain it beforethe three-minute free fall period? c) at what time does she attain 99% of her free fall velocity?

SolutionUsing the equation for : 1200.7 171.43 /

Remarks: 1) weight is not mass. When we are given the weight of something, we are actually given its value. Thus, the mass of a person whose weight is 120 lb is 3.75 . 2)171.43 ft/sec isapproximately 116.88mph.To answer b) we need do nothing since we know, from the formula we derived, that the terminal

velocity is never attained.

To answer c) we want to know when will

0.99171.43170 /. Thus, we must solve

the equation

170 1 for . Substituting the given values,

170 1200.7 1 ..Solving for the exponential part,

.. 1 1700.7120 0.00833Hence,

3.750.7 ln0.00833

.So make a note next time you jump out of a plane you will have attained your terminal velocity in

about 30 seconds after you are out. Safe landing!

Example 9 Obtain a more general expression for the position function of an object that is free to movevertically in aType I medium assuming it has an initial velocity and initial position0 .Solution

Unlike the previous discussion example, we no longer know in which direction the object is initially

moving because we are not given the sign of . Therefore, we choose a coordinate system that isfamiliar to us, as shown below.

Objectmovingdown

D

ObjectmovingupD


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At some instant the object is acted upon by the forces and where . The function is now arbitrary and its sign will depend in which direction the object ismoving. Newton tells us that

Thus,

Equating components, ; 0

we will solve this problem using an integrating factor. Writing the equation in standard form,

; 0 An integrating factor is /. Therefore,

/ /

Integrating from

0to

,

0 / or

/ / 1Solving for :

Observe that 0 as . This must be so because we selected a coordinate system in whichdown is negative; eventually (provided that is not too large) the object will move in thedownward direction and will only stop when it makes impact with the gorund.

We integrate to obtain the position function:

Applying the initial condition 0 :

Therefore,

The position function is Simplifying,

1 Let us graph a typical position function with the following assumptions: 50 Kg, 1.00,


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1000 m, and 1 m/s and compare it to the position function for an object which free-fallswithout any resistance. The position function for such free-falling body is given by

1000490501490 1 1000490245501

Since 0, the object must initially move up. Here is the graph in a different time scale:

Exercise: Compare these results to the ones obtained when no resistive forces are present, in which case

12 Example 10 (Two Regimes Problem) A 5Kg object is released from rest and free falls under the influence ofgravity from a height of 30 m above a body of water. When it strikes the water, it sinks and its motion isgoverned by three forces: the force of gravity, the buoyant, and a drag force. Assume that the buoyant force is

0.5mg. Also assume that air resistance is proportional to velocity with proportionality constant 10 and water

offers a resistive force proportional to velocity with proportionality constant 60. Find the objects equation ofmotion and its velocity one minute after it is released. Use 1 0 to simplify the calculations.Solution

For falling objects, it is better to place the origin of coordinates at the starting point of the motion and

align the axis so that positive is in the direction of motion:

6 8 10

600

700

800

900

1000

0.1 0.2 0.3 0.4 0.5

999.4

999.6

999.8

1000.0

1 0 ; 0 0

6 0

12 ; 0 ,

Part 1: During free fall:

Part 2: After entering the water:

Where , is the speed with which the object hits thewater and will be obtained from part 1. is the time asmeasured by a clock at the surface of the water. It startsthe moment the object strikes the water.

O

30


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Let us solve each part:

Part 1:

5 5 1 0 1 0 ; 0 0a separable equation

5 2; 0 0Integrating, 5

2

ln 55 2

Solving for : | 5 | 5 So that 5 5 We select the function that satisfies the initial condition and that would be

5 5 Exercise: show that both5 5 and 5 5 satisfy the differential equation.Thus, the solution for part 1 is 51 To determine the objects velocity one minute after it is released, we need to know whether it is still

free falling, or it is already in the water, so let us first calculate how long it takes for the object to hit

the water.

The objects position function before impact is found by integration:

0 51 [note: our origin was located at the point where the object started its motion, therefore, 0 0.Furthermore, is an increasing function oftbecause we picked down as the positive direction].

51

The object hits the water when 30. Thus, we need to solve the equation52 2 1 30There is no simple way to solve this equation. Numerical techniques may be used and of course a

computer can help a great deal. Let T the time of impact. Then it can be shown3 that

. .

Thus, the oneminute mark takes place once the object is in the water and we need to solve thevelocity function for this portion.First, we need to find the velocity just as the object hits the water because this is the initial condition

for Part 2. This velocity is 6.5 51 .. We denoted this vectors magnitude by ,:, 5 m/sec3The value6.5was found using the NSolve command inMathematica.


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We are ready forPart 2. The objects motion is now governed by the equation45 5 1 0 6 0 25; 0 5

Simplifying:

1 0 1 2 ; 0 5

Note: is measured by a clock at the water level and it starts ticking the moment the object enters thewater. At that time, the towers clock already reads 6.5 sec. Thus, the connection between these twoclocks is . .Separating variables, 5 6 2; 0 5

Integrating,

16 65 6

2

ln|56| 12

l n | 2 5 | l n|5 6 | 12Solving for , 56 1 5 In order to find the velocity at the oneminute mark from the moment the object was released, we usethe relation 6 . 5. The one-minute mark corresponds to 60 6.5 53.5 sec. Thus,

53.5 512 1 . . /or about 42 cm/ sec.

Example 11 In order to dispose of used up nuclear material form operation of nuclear reactors, waste is placedin drums and dropped in the ocean. As the drums sink, their speed

varies in response to the forces acting

on it.

Suppose a typical drum weighs 640 lb and has volume 8 ft3. If it is released into the ocean and it isassumed that the force of resistance in water is equal to 1 lb for each foot per second of the drums velocity,what is the maximum depth to which it can be dropped given that it is likely to burst upon impact at speeds

higher than 75 ft/sec?Solution

Assume a vertical axis in which positive displacement is measured downwards. Newtons Second

Law tells us that

where the weight of the drum, buoyant force, and force of water resistance.We have: 640 lb, the buoyant force which can be obtained using Archimedes Principle, and theresistive force which is assumed to be equal to 1 lb for each foot per second of the drums velocity.hence,

4We made the following assumption: as the object hits the water, its terminal velocity at the end of motion in the firstregime equals its initial speed in regime 2.


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Since the drums are dropped in the ocean, the weight of the water displaced is equal to the specific

weight of ocean water volume of the drum. Thus5, 6 2 . 5 8 ft 500 lb

The resistive force is given to us to be proportional to the velocity of the drum with proportionality

constant

1: 1The mass of a 640 lb drum is 20 slug. Therefore, by Newtons Second Law,20 6 4 0 5 0 0 ; 0 0or 120 7 ; 0 0

This is a first order initial value problem with integrating factor e x p . Applying theintegrating factor,

7

Integrating,

0 7 or

1 4 0 1Thus,

1401 We wish to find the depth to which a tank must sink when its speed just hits the 75 ft/sec mark.Therefore, we must first find the time it take to do that.

The equation

7 5 1 4 0 1 gives us . .From the expression for we can find the depth function. Let be that depth. Then Integrating,

0 Since we placed the origin at the surface of the ocean, 0 0.

1 4 0 1

14020

14020 1400 2 05We assume that the drums sink so quickly that we can ignore the variable buoyant force at the instant they hit the water. We also take

the specific weight of ocean water to be approximately62.5


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Simplifying,

14020 2800When 15.35 sec,

15.35 140 15.35 20. 2800

. The tanks should be dropped in waters no deeper than 648 feet.Inclines

Example 12 (The Incline Problem) a wood block of mass m is placed on an incline of length which makes anangle with the horizontal (see figure below). The coefficient of kinetic friction between the object and thesurface is so that the frictional force is given by where is the normal force on the object.Assume that the only retarding force is the frictional force. 1) Find the velocity and position relativeto the top of the incline. 2) Find the time it takes the object to reach the bottom of the incline.

SolutionThe figure illustrates the forces acting on the object as well as the components of these forces which

are needed in the calculations. The only forces responsible for the motion along the incline are the

component of the gravitational force sin and the frictional force whose magnitude iscos [ For all intensive purposes this object lies on a horizontal plane and is pulled by two opposing forcesof magnitudes

sinand

cos].

Let us introduce a coordinate system with its origin at the top of the incline where the blockbegins its

motion and whose positive x-axis runs along the length of the incline:

Newton tells us that

s i n c os ; 0 0or s i n c o s ; 0 0

cos

sin

O

sin


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Note that we assumed that there was an initial nudge to overcome friction because the coefficient of

static friction is larger than that of kinetic friction. We will assume that this is the case but that 0 isvery small.

This differential equation could not be easier; all the quantities on the right-hand side are constant.

Thus,

s i n c os

or s i n c os The position function is found by another integration:

0 Thus, 2 s i n c os Finally, the time the block takes to reach the bottom of the incline is calculated from the equation

2 s i n c os which gives us 2s i n c os

Exercise When an object falls under the influence of gravity only from a height its potential energy isconverted into kinetic energy and conservation of energy requires that at all times the total mechanical energy be a constant. Calculate the time it takes the block in example12to reach the bottom of the incline if there isno frictional losses.

Example 13 (an application to physics) the Work-Energy Theorem in physics states that if a force Facts on anobject, then the work done by that force as the object moves from point A to point B is equal to the objects

change in kinetic energy: 12 12 where is the component of the force along the objects motion, m is the objects mass, and its speed. Usethe method of separation of variables to establish this result.

SolutionLet us place the x-axis along the objects motion so that is a function of , denote it by [actually is the component of a vector fieldalong the objects motion]. We wish to show that

12 12

According to Newtons second law of motion6,

Since /,

But is a function of. Therefore, by the Chain Rule,6 Actually / where is the correct statement in this law.


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The condition that we are given is that the object starts at with speed : Separating variables:

Thus,

12

which gives us the important result

Tanks and Mixture Problems

Draining TanksWe are now interested in deriving and solving the differential equation that governs the height of the watercolumn in a tank of cross-section area which has a hole of area through which water can escape.

In time , the amount of fluid that comes out of the tank forms a small cylinder of volume Height or where is the speed of the fluid just as it leaves the tank. This is precisely the volume lost by the tank in thistime interval. Hence, the change in volume of fluid in the tank is

or

as 0 ,

The formula for is obtained from energy considerations: A particle of water that starts at the surface of the ofthe column of fluid and drops to the point of exit will give up itspotential energy tokinetic energy 12 2at the point of exit:

12 Therefore, 2

Areaa

h

Cross-section Height =


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Hence, the volume in the tank changes according to

2

where is the height of the water column in the tank at time . Finally, and are related by the equation

from which we obtain Solving for /, 2This formula represents the perfect fluid and it is an idealized situation. In reality there are energy loses due

to friction and other factors, so we make an adjustment by introducing a empirical constant with 0 1:

Note that 1 corresponds to the perfect case and 0 could represent a situation of an extremely viscousfluid for example.

Example 14 A water tower 5 meters height and rectangular cross-section 4 has a circular hole 12 cm indiameter. Using the model above with 0 . 7, a) how high is the water level two minutes after the waterstarts to drain? b) How long will it take to empty the tank?

Solution

We have 0.01131 , 4 and 0 5. Thus,

0.011314 0 . 7 2

0.0019792Separating variables,

Where we let 0.00197920.008762 to keep things simple. The integral solution is

2/ Since 5 when 0, 2 5/ 0 so 4.4721. The solution is therefore

2/ 0.008762 4.4721From which it follows that

0.0087624.47212

To answer a), 120 . Thus,120 0.0087621204.47212

. To answer b), the tank will be empty when 0. This will happen when

0.0087624.47210 .


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We should put this model to the test. Let us perform the experiment with at tin can whose inside dimensions are 7.33 , 9.3 (this is the mark at which the can is considered to be filled) and with ahole 0.6 cmin diameter at its bottom. Since all the dimensions are in centimeters we will convert the formula 2to these units by replacing

9.8 /with

980 /. Also, since the container is a cylinder and the hole was

carefully made to ensure that it is circular, we can rewrite the formula in terms of the measured variables, which

in this case are the diameters:

2 22 24

Thus,

2 2whose solution is

/ 12 2

The time to empty the can is

2 2

[verify the algebra]. Using our data,

20.7 0.67.33 9.31960 .

Let us calculate the perfect time with 1: 2

2 0

2.

Here are the actual data for the experiment:

Trial # Time (sec)

1 25.00

2 24.00

3 24.99

4 25.40

5 25.00

6 25.60

24.99What we learn from this experiment is the following: If we assume the model we derived is the correct one,

then our choice of, which was completely arbitrary, is wrong. As we discussed in class, I chose it practicallyat random so it is not surprising that our calculated value of29.37 sec is substantially off.Using the average value found on the table we can re-calculate the empirical value of for this particular can.Once we have it, we can reformulate our model using it and test it again. Here is the empirical value of:


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2

24.99 0.67.33 9.3

1960

.Phase two of this experiment consists of verifying the formula we derived by using this new value of but withvarying heights of the water column. The table below shows the results of trials performed with three different

water heights. Each trial was conducted three times in order to average out the time. The new formula used to

obtain the theoretical values of is 2. 0.67.33

1960Water level

(cm)

Theoretical Time

(sec)

Measured Time (sec) %

Error8.1 23.32 24 2.9

7.2 21.99 22 0.05

6.5 20.89 20 4.0

Considering the crudeness of the experiment these are very good values and so it looks like our model is

acceptable.

The Draining Tank - a General Case

Suppose now that the tank has a more general geometry with horizontal cross-sectional area as shownbelow and that the hole is of area (and of no particular shape). Notice also that we will be using the variable instead of.

Now we need to connect the variables and . Previously, because of the geometry of the tank, we were ableto express

as a function of

(previously denote by h). The relation between

and

is not simple any more

because the shape. However, consider the following figure:

2

In time

, the amount of fluid leaving the tank is

given by whereis the speed of thefluid as it leaves that tank. Therefore, the rate ofchange in volume of fluid in the tank is

Up to this point the derivation is as before. theequation that ensues from this argument is thesame:

Cross-section area


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The volume of the fluid obeys the equation

By the chain rule and the FTC, Thus,

2Solving for /:

2 ; 0

If we adjust for frictional losses, 2 ; 0 9where 0 1.Thus, if we know the geometry of the cross-sectional area, we can solve the problem.

Example 15 A gas tank has the shape shown in the figure below and the outlet valve is a circular hole 10 cm indiameter. Using the model above with 1, how long will it take to drain the tank? (Assume the top lid isrectangular). This problem will require technology for the integral.

Solution

We need to obtain a formula for the cross-sectional area:

The figure on the right allows us to write 1 1 since we have a circular cross-section. Thus, thearea of the slab is 2 8 1 61 1 Substituting in equation (9),

10cmSemicircle of radius 1 m

8 m1 m

y

1m

x

x

Lower portion of the circle 1 1

8 m

y

y


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0.05

161 1 2 ; 0 1 0.0022 1 1 ; 0 1

Thus,

1 1 0.0022 MATHEMATICA tells us that

1 1

23 23 2

We have 23 23 2 0.0022 The tank is empty when 0. Then

10.0022 23 23 2 0

.Problems Involving First Order Linear ModelsMixture Problems

Mixture problems involve the influx of a chemical into a mixture tank accompanied by an outflow of mixture

from of the tank. The physical units in problems involving mixtures are:

MKS Brit ish Variations

Volume Liter (L) Gallon (gal) Milliliter, cubic ft., etc

mass Kilogram (Kg) Pound (lb)7 Grams, milligrams, etc.

Time Second (sec) Second (sec) Minute, hour, day.

Suppose we have a tank containing a chemical mixture initially containing gallons of solution in which lbs of some chemical have been dissolved. Suppose also that additional solution containing the chemical with

concentration flows into the tank at a rate , and the well-stirred mixture is drained from the tank at a rate. What is the amount of chemical present in the tank at a later time ?The analysis of this problem can be summarized in the following diagram which will be used thought the

mixture examples:

7 This is not a measure of mass, it is a measure of weight. Nevertheless, it is used as such in most textbooks.

Solution of concentrationlb/gal goes in at a rateof gal/min

Mixture comes out at a

rate of gal/minInitially: gal, lb. ofchemical


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Here is the model for this situation:

The rate at which the amountof chemical changes must equal the rate at which chemical is added minus therate at which chemical is removed. In the language of mathematics:

How do we obtain these rates? The in rate is easy: if solution flows in at a rate of gal/min (the units arenot important as long as we are consistent) and each gallon contains lb/gal, then the rate at which chemicalis added is given by Similarly, the rate at which the chemical is removed is given by

The problem is: we dont know what is because it is changing all the time. However, concentration isgiven by In general these two quantities are changing in time, but we know how much solution there was initially in the

tank, the rate at which it is being added, and the rate at which it is being removed. Therefore, we can determinethe volume of solution in the tank at any time t. We also denoted the amount of chemical by .Therefore,

where . Thus, we can write our differential equation in:

This is a first order linear initial value problem which we can write in standard form as

; 0 where "chemical" . "chemical" . .

Example 16 A tank contains a 10-gallon mixture in which 4 lbs of salt are dissolved. A solution containing salt(called brine) of concentration 0.5 lb/gal enters the tank at a rate of 1 gal/min and is quickly mixed. Themixture leaves the tank at rate of2 gal/min. (a) How long does it take for the tank to be completely empty?(b) Find the salinity (measured by the amount of salt per gallon of mixture) of the tank at

3.

Solution

;

Solution of concentration0.5 lb/gal goes in at a rateof1 gal/min

Mixture comes out atarate of2gal/min

Initially: 10 gal,4 lb of salt


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The equation is 1 0.5 .

Observe that we obtain the rate at which salt goes in by multiplying the rate at which solution goes in

times the concentration of that solution. Similarly,

rate of salt out = rate of mixture coming out times the concentration of the mixture

The concentration of the solution coming out of the tank is changing in time and is given by

Thus, if we know the volume of solution in the tank present at any time t, we can obtain anexpression for the concentration of the mixture coming out. We can write a formula for the volume as

a function of timet: Where

is the initial volume of the mixture in the tank,

and

are the rates at which mixture

enters and exist the tank, respectively. In this example, a fluid (which could be pure water) goes in ata rate of 1 gal/min and the mixture goes out at a rate of 2 gal/min. Therefore themixtures volume is changing at a rate of1 /, that is, it is decreasing by one gallon everyminute. Thus, The concentration in the tank, and also that of the out-flowing solution, is given by

1 0 (a) from the volume equation, the tank will be empty when 0, that is, when .(b) To answer the second question, we solve the equation that governs this problem:

0.5 2 1 0 Since we originally had 4 lb of salt in the tank, 0 4 . Therefore, the initial value problem is:

0.5 2 1 0 ; 0 4Observe that this equation is not separable. However it is a linear first order equation and therefore we

can find an integrating factor:

Writing it instandard form:

21 0 0.5 ; 0 4 an integrating factor for this equation is

1 0


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Multiplying (a) by the I.F.

10 21 0 10 0.510Thus, 1 0 0.510 Integrating,

1 0 0.5 10 10 100 0.5 10

10 4100 12 1 0

110 125 12 11 0 110

10

1

210 1

25 1

2010 1210 1100Solving,

Graphically,

We can find the salinity at 3 from the graph: 3 3 . Also, 3 7. Therefore, the salinity at 3, 33 37 0.429

Example 17 A tank containing a 40 liters mixture in which 6 Kg of salt are dissolved. Pure water enters thetank at a rate of3 L/min and is quickly mixed. The mixture leaves the tank at the same rate of3 L/min. Findthe amount of salt

in the tank two minutes after pure water starts flowing into the tank.

Solution

2 4 6 8 10

1

2

3

4

t

A(t)

Pure water goes in at a rateof 3 L/min

Mixture comes out at arate of 3 L/min

Initially: 40 L,6 Kg of salt.


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Notice that in this example there is no net gain or loss of solution in the tank since the rate at which it

is filled equal the rate at which it is evacuated: . Therefore, at all times.We can organize the given information in a table:

IN OUT

RATE OF SOLUTION 3 L/min 3 L/minCONCENTRATION

0 /RATE OF SALT 3 0 /The equation becomes, 0 3 All we need is . Since as much stuff goes in as it comes out, is constant and equal to 40L atall times, because this is what we had in the tank to begin with. Also, 0 6. Therefore, thisinitial value problem is

340 ; 0 6Written in standard form as a first order linear equation, 340 0 ; 0 6However, the problem is so straightforward for separation of variables that we will solve it using that

method instead:

Thus, the amount of salt 2 minutes later is

2 6

. It does not look like a whole lot of salt was removed. How can we be one hundred percent sure that

our answer is the correct one? first, the function we found certainly obeys the initial condition [verify].

Next, it satisfies the differential equation: 1840 340 6

340

340 l n l n 6 340 0

ln 6 340

Using the definite integral:

Integrating,

340

340 l n 340

Using the indefinite integral:

Integrating,

Applying the initial condition 0 6 gives usl n 6 . Thus,and the same answer follows.


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By the Existence and Uniqueness Theorem, we found the only solution. If our answer of5.16 isincorrect, it is only because we might have made operational mistakes along the way.

Example 18 A reservoir with a volume of8 billion cubic feet contains a pollutant whose initial concentration is

0.25%.The reservoir receives a daily inflow of

500million cubic feet of water which contains a concentration

of0.05% of the same pollutant, and leaves the reservoir at the same rate. Assuming that we can neglect the timeit takes for the pollutant to diffuse throughout the reservoir, in how many days will the pollutant concentrationin the reservoir be reduced to 0.10%?Solution

When dealing with different orders of magnitude it is useful to convert them to the same denomination.

Since we are dealing with millions and billions, 8 billion ft3 is the same as 8000 million ft3 so we willoperate in millions only [we are using billion as it is used in the United States].

The situation is shown in the following figure [note: all numbers are in millions]:

Let be the amount of pollutant in the tank (in million ft 3). Since the volume of the reservoirremains a constant 8000 (million cubic feet), we want to know the value oftfor which

0.001 8000 8 ftThe equation governing the amount of pollutant in the reservoir is

5000.0005500 8000 ; 0 20ft

The corresponding first order initial value problem is:

5000.0005500 8000 ; 0 20ftor

. ;

Using /: /0.25/Integrating from 0 to arbitrary 0:

2 0 0 . 2 5 / Therefore,

2 0 4 1

0.25% of 8000

Polluted water goes in at a rate 500 ft3/day and withpollutant concentration of 0.05%

Mixture comes out at arate of 500 ft3/dayInitially: 8000 ft3 pollutedwater with pollutant

concentration of 0.25%


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Solving for : 20 4 1

Simplifying,

Finally we solve 8 for :

16 4 8 0.25Solving fort: . days

Example 19 A tank containing a 100 liters mixture of10% sulfuric acid by volume is flushed by adding purewater at a rate of 5 L/min and simultaneously evacuating it at the same rate. The flushed solution entersanother 100L tank which initially is full of pure water and which is allowed to overflow into a containmenttank. Determine the time at which the acid concentration is the greatest in the second tank.

SolutionLet be the volume of acid in the mixture of tankI and that of tankII. As usual we will uselower case to denote the volume of solution in the tanks.

TANK I TANK II

IN OUT IN OUT

RATE OF SOLUTION 5L/min 5L/min 5 L/min 5 L/minCONCENTRATION 0 / / /

RATE OF ACID 5 0 / / /Notice that there is no net gain or loss of solution in either tank. Therefore, and .The differential equations are:

0 5 100 ; 0 10 5 100 5 100 ; 0 0 This is an example of a linear system. We are interested in finding the maximum value of

whichrepresents the concentration in tank II. We know from calculus that the critical points to consider are

ones at which

III

Pure water goes in at arate of 5 L/min

Mixture comes out of tankI at a rate of 5 L/min.Initially: 100 L,10L ofsulfuricacid

Overflow 5 L/min

100 L of water receivesthe solution from tank I.Acid enters this tank.


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100 0 0But the problem is that / depends of both and . Therefore, we must solve this system whichis fairly straight forward in this case because the first equation does not involve .The equation for tank I is separable:

5100 ; 0 10Integrating,

120

Thus,

ln 10 120 Since 0, 10/We may substitute this in the equation for tank II to obtain:

5 10/100 5 100 ; 0 0or 12 / 120 ; 0 0

This is also linear equation with integrating factor /. Thus, 12 ; 0 0Integrating,

12

0 12 Thus 100 1200 Differentiating and setting the derivative to 0:

100 1200 20 0gives us the critical point . The figure below corresponds to and shows that here iswhere

attains a max.

5 10 15 20 25 30

0.005

0.010

0.015

0.020

0.025

0.030

0.035

Documents

First Oder Equations-Applications