2.7 Linear Models

INTRODUCTION

In this section we solve some of the linear first-order models that were introduced in Section 1.3.

Growth and Decay

The initial-value problem

(1)

where k is the constant of proportionality, serves as a model for diverse phenomena involving either growth or decay. We have seen in Section 1.3 that in biology, over short periods of time, the rate of growth of certain populations (bacteria, small animals) is observed to be proportional to the population present at time t. If a population at some initial time t0 is known, then the solution of (1) can be used to predict the population in the future—that is, at times t > t0. The constant of proportionality k in (1) can be determined from the solution of the initial-value problem using a subsequent measurement of x at some time t1 > t0. In physics and chemistry, (1) is seen in the form of a first-order reaction, that is, a reaction whose rate or velocity dx/dt is directly proportional to the first power of the reactant concentration x at time t. The decomposition or decay of U-238 (uranium) by radioactivity into Th-234 (thorium) is a first-order reaction.

EXAMPLE 1 Bacterial Growth

A culture initially has P0 number of bacteria. At t = 1 h the number of bacteria is measured to be P0. If the rate of growth is proportional to the number of bacteria P(t) present at time t, determine the time necessary for the number of bacteria to triple.

SOLUTION

We first solve the differential equation in (1) with the symbol x replaced by P. With t0 = 0 the initial condition is P(0) = P0. We then use the empirical observation that P(1) = P0 to determine the constant of proportionality k.

Notice that the differential equation dP/dt = kP is both separable and linear. When it is put in the standard form of a linear first-order DE,

,

we can see by inspection that the integrating factor is e-kt. Multiplying both sides of the equation by this term immediately gives

Integrating both sides of the last equation yields e-ktP = c or P(t) = cekt. At t = 0 it follows that P0 = ce0 = c, and so P(t) = P0ekt. At t = 1 we have P0 = P0ek or ek = . From the last equation we get k = ln = 0.4055. Thus . To find the time at which the number of bacteria has tripled, we solve 3P0 = P0e0.4055t for t. It follows that 0.4055t = ln 3, and so

See FIGURE 2.7.1. ≡

A graph. The horizontal axis is labeled t and the vertical axis is labeled P. A curve labeled P of t equals P subscript zero times e to the 0.4055 t, begins at (0, P subscript zero) and goes up and to the right with increasing steepness. A point on the curve corresponding to 3 P subscript zero on the P axis has a corresponding point on the t axis at t = 2.71.

FIGURE 2.7.1 Time in which initial population triples in Example 1

Notice in Example 1 that the actual number P0 of bacteria present at time t = 0 played no part in determining the time required for the number in the culture to triple. The time necessary for an initial population of, say, 100 or 1,000,000 bacteria to triple is still approximately 2.71 hours.

As shown in FIGURE 2.7.2, the exponential function ekt increases as t increases for k > 0 and decreases as t increases for k < 0. Thus problems describing growth (whether of populations, bacteria, or even capital) are characterized by a positive value of k, whereas problems involving decay (as in radioactive disintegration) yield a negative k value. Accordingly, we say that k is either a growth constant (k > 0) or a decay constant (k < 0).

Two curves are graphed. The horizontal axis is labeled t and the vertical axis is labeled y. The curve labeled e to the k t, with k greater than 0 is the growth curve. It goes up and to the right with increasing steepness from the second quadrant into the first quadrant. The curve labeled e to the k t with k less than 0 is the decay curve. It goes down and to the right with decreasing steepness from the second quadrant into the first quadrant. The curves intersect on the vertical axis. The portion of the curves in the second quadrant consists of dashes.

FIGURE 2.7.2 Growth (k > 0) and decay (k < 0)

Half-Life

In physics the half-life is a measure of the stability of a radioactive substance. The half-life is simply the time it takes for one-half of the atoms in an initial amount A0 of a substance to transmute into the atoms of another element. The longer the half-life of a substance, the more stable it is. For example, the most commonly occurring uranium isotope, U-238, has a half-life of approximately 4,500,000,000 years. In about 4.5 billion years, one-half of a quantity of U-238 is transmuted into lead, Pb-206. In contrast, the isotope Ra-223 of radium has a half-life of 11.4 days. Of the 33 isotopes of radium, Ra-226 is the most stable, with a half-life of approximately 1600 years. The nucleus of an atom of Ra-226 is transmuted into the nucleus of a radon atom Rn-222. Radon is a heavy, odorless, colorless, and highly radioactive gas that usually originates in the ground. Because radon can penetrate a sealed concrete floor, it frequently accumulates in the basements of new and highly insulated homes. Some medical organizations claim that after cigarette smoking, exposure to radon gas is the second leading cause of lung cancer. Because of its short half-life, the isotope Ra-223 is frequently used in nuclear medicine to treat forms of cancer, especially metastatic bone cancer and advanced prostate cancer.

EXAMPLE 2 Half-Life of Plutonium

A breeder reactor converts relatively stable uranium-238 into the isotope plutonium-239. After 15 years it is determined that 0.043% of the initial amount A0 of the plutonium has disintegrated. Find the half-life of this isotope if the rate of disintegration is proportional to the amount remaining.

SOLUTION

Let A(t) denote the amount of plutonium remaining at any time. As in Example 1, the solution of the initial-value problem

, (2)

is A(t) = A0ekt. If 0.043% of the atoms of A0 have disintegrated, then 99.957% of the substance remains. To find the decay constant k, we use 0.99957A0 = A(15); that is, 0.99957A0 = A0e15k. Solving for k then gives k = ln 0.99957 = −0.00002867. Hence . Now the half-life is the corresponding value of time at which A(t) = A0. Solving for t gives A0 = A0e–0.00002867t or = e–0.00002867t. The last equation yields

≡

Carbon Dating

About 1950, a team of scientists at the University of Chicago led by the American physical chemist Willard Frank Libby (1908–1980) devised a method using a radioactive isotope of carbon as a means of determining the approximate ages of carbonaceous fossilized matter. The theory of carbon dating is based on the fact that the radioisotope carbon-14 is produced in the atmosphere by the action of cosmic radiation on nitrogen-14. The ratio of the amount of C-14 to the stable C-12 in the atmosphere appears to be a constant, and as a consequence the proportionate amount of the isotope present in all living organisms is the same as that in the atmosphere. When a living organism dies, the absorption of C-14, by breathing, eating, or photosynthesis, ceases. Thus by comparing the proportionate amount of C-14, say, in a fossil with the constant amount ratio found in the atmosphere, it is possible to obtain a reasonable estimation of its age. The method is based on the knowledge of the half-life of C-14. Libby’s calculated value for the half-life of C-14 was approximately 5600 years and is called the Libby half-life. Today the commonly accepted value for the half-life of C-14 is the Cambridge half-life that is close to 5730 years. For his work, Libby was awarded the Nobel Prize for chemistry in 1960. Libby’s method has been used to date wooden furniture in Egyptian tombs, the woven flax wrappings of the Dead Sea Scrolls, and the cloth of the enigmatic Shroud of Turin.

A photo represents Willard Frank Libby.

Willard Frank Libby

© Jack Fields/Science Source/Getty Images

EXAMPLE 3 Age of a Fossil

A fossilized bone is found to contain 0.1% of its original amount of C-14. Determine the age of the fossil.

SOLUTION

As in Example 2 the starting point is A(t) = A0ekt. To determine the value of the decay constant k we use the fact that A0 = A(5730) or A0 = A0e5730k. The last equation implies 5730k = ln = −ln 2 and so we get k = −(ln 2)/5730 = −0.00012097. Therefore . With A(t) = 0.001A0 we have 0.001A0 = A0e– 0.00012097t and −0.00012097t = ln (0.001) = −ln 1000. Thus

≡

The date found in Example 3 is really at the border of accuracy of this method. The usual carbon-14 technique is limited to about 10 half-lives of the isotope, or roughly 60,000 years. One fundamental reason for this limitation is the relatively short half-life of C-14. There are other problems as well; the chemical analysis needed to obtain an accurate measurement of the remaining C-14 becomes somewhat formidable around the point 0.001A0. Moreover, this analysis requires the destruction of a rather large sample of the specimen. If this measurement is accomplished indirectly, based on the actual radioactivity of the specimen, then it is very difficult to distinguish between the radiation from the specimen and the normal background radiation. But recently the use of a particle accelerator has enabled scientists to separate the C-14 from the stable C-12 directly. When the precise value of the ratio of C-14 to C-12 is computed, the accuracy can be extended to 70,000–100,000 years. For these reasons and the fact that the C-14 dating is restricted to organic materials this method is used mainly by archaeologists. On the other hand, geologists who are interested in questions about the age of rocks or the age of the Earth use radiometric dating techniques. Radiometric dating, invented by the New Zealand–born physicist and Nobel laureate Ernest Rutherford (1871–1937) around 1905, is based on the radioactive decay of a naturally occurring radioactive isotope with a very long half-life and a comparison between a measured quantity of this decaying isotope and one of its decay products using known decay rates. Radiometric methods using potassium-argon, rubidium-strontium, or uranium-lead can give ages of certain kinds of rocks of several billion years. The potassium-argon method of dating is also used to date terrestrial materials, such as minerals and lava, and extra-terrestrial materials, such as meteorites and lunar rocks. The age of a fossil, such as a dinosaur bone, cannot be dated by the C-14 method since it has very little organic material left in it. Its age can be estimated by determining the age of the rock strata in which it was found. See Problems 5 and 6 in Exercises 2.9 for a discussion of the potassium-argon method.

A photo represents Ernest Rutherford.

Ernest Rutherford

© Universal History Archive/UIG/Shutterstock

The size and location of the sample caused major difficulties when a team of scientists were invited to use carbon-14 dating on the Shroud of Turin in 1988. See Problem 14 in Exercises 2.7.

Newton’s Law of Cooling/Warming

In equation (3) of Section 1.3 we saw that the mathematical formulation of Newton’s empirical law of cooling of an object is given by the linear first-order differential equation

, (3)

where k is a constant of proportionality, T(t) is the temperature of the object for t > 0, and Tm is the ambient temperature—that is, the temperature of the medium around the object. In Example 4 we assume that Tm is constant.

EXAMPLE 4 Cooling of a Cake

When a cake is removed from an oven, its temperature is measured at 300°F. Three minutes later its temperature is 200°F. How long will it take for the cake to cool off to a room temperature of 70°F?

SOLUTION

In (3) we make the identification Tm = 70. We must then solve the initial-value problem

(4)

and determine the value of k so that T(3) = 200.

Equation (4) is both linear and separable. Separating variables,

yields ln |T − 70| = kt + c1, and so T = 70 + c2ekt. When t = 0, T = 300, so that 300 = 70 + c2 gives c2 = 230, and, therefore, T = 70 + 230ekt. Finally, the measurement T(3) = 200 leads to e3k = or k = ln = −0.19018. Thus

T(t) = 70 + 230e– 0.19018t. (5)

We note that (5) furnishes no solution to T(t) = 70 since limt → T(t) = 70. Yet intuitively we expect the cake to reach the room temperature after a reasonably long period of time. How long is “long”? Of course, we should not be disturbed by the fact that the model (4) does not quite live up to our physical intuition. Parts (a) and (b) of FIGURE 2.7.3 clearly show that the cake will be approximately at room temperature in about one-half hour. ≡

The first illustration is a graph with an open curve. The horizontal axis is labeled t and the vertical axis is labeled uppercase T. The curve begins at (0, 300) and goes down and to the right with decreasing steepness toward uppercase T equals 70. The second illustration represents a table. The table consists of six rows and two columns. The column headers are temperature and time in minutes. The row entries are as follows. Row 1. Temperature: 75 degrees, time: 20.1. Row 2. Temperature: 74 degrees, time: 21.3. Row 3. Temperature: 73 degrees, time: 22.8. Row 4. Temperature: 72 degrees, time: 24.9. Row 5. Temperature: 71 degrees, time: 28.6. Row 6. Temperature: 70.5 degrees, time: 32.3.

FIGURE 2.7.3 Temperature of cooling cake in Example 4

Mixtures

The mixing of two fluids sometimes gives rise to a linear first-order differential equation. When we discussed the mixing of two brine solutions in Section 1.3, we assumed that the rate x′(t) at which the amount of salt in the mixing tank changes was a net rate:

. (6)

In Example 5 we solve equation (8) of Section 1.3.

EXAMPLE 5 Mixture of Two Salt Solutions

Recall that the large tank considered in Section 1.3 held 300 gallons of a brine solution. Salt was entering and leaving the tank; a brine solution was being pumped into the tank at the rate of 3 gal/min, mixed with the solution there, and then the mixture was pumped out at the rate of 3 gal/min. The concentration of the salt in the inflow, or solution entering, was 2 lb/gal, and so salt was entering the tank at the rate Rin = (2 lb/gal) · (3 gal/min) = 6 lb/min and leaving the tank at the rate Rout = (x/300 lb/gal) · (3 gal/min) = x/100 lb/min. From this data and (6) we get equation (8) of Section 1.3. Let us pose the question: If there were 50 lb of salt dissolved initially in the 300 gallons, how much salt is in the tank after a long time?

SOLUTION

To find the amount of salt x(t) in the tank at time t, we solve the initial-value problem

Note here that the side condition is the initial amount of salt, x(0) = 50 in the tank, and not the initial amount of liquid in the tank. Now since the integrating factor of the linear differential equation is et/100, we can write the equation as

Integrating the last equation and solving for x gives the general solution x(t) = 600 + ce-t/100. When t = 0, x = 50, so we find that c = −550. Thus the amount of salt in the tank at any time t is given by

x(t) = 600 − 550e-t/100. (7)

The solution (7) was used to construct the table in FIGURE 2.7.4 (b). Also, it can be seen from (7) and Figure 2.7.4(a) that . Of course, this is what we would expect in this case; over a long time the number of pounds of salt in the solution must be (300 gal)(2 lb/gal) = 600 lb. ≡

The first illustration is a graph with an open curve. The horizontal axis is labeled t and the vertical axis is labeled x. A curve begins at approximately (0, 55) and goes up and to the right with decreasing steepness through the approximate points (52.7, 266.7), (100.0, 390.0), (151.8, 474.5), (205.9, 517.7), (307.2, 563.2), and (399.5, 580.5). The second illustration represents a table. The table consists of six rows and two columns. The column headers are time and pounds. The row entries are as follows. Row 1. Time: 50, pounds: 266.41. Row 2. Time: 100, pounds: 397.67. Row 3. Time: 150, pounds: 477.27. Row 4. Time: 200, pounds: 525.57. Row 5. Time: 300, pounds: 572.62. Row 6. Time: 400, pounds: 589.93.

FIGURE 2.7.4 Pounds of salt in tank as a function of time in Example 5

In Example 5 we assumed that the rate at which the solution was pumped in was the same as the rate at which the solution was pumped out. However, this need not be the situation; the mixed brine solution could be pumped out at a rate rout faster or slower than the rate rin at which the other brine solution was pumped in.

EXAMPLE 6 Example 5 Revisited

If the well-stirred solution in Example 5 is pumped out at the slower rate of rout = 2 gallons per minute, then liquid accumulates in the tank at a rate of

gal/min = 1 gal/min.

After t minutes there are 300 + t gallons of brine in the tank and so the concentration of the outflow is c(t) = x/(300 + t). The output rate of salt is then Rout = c(t) · rout or

·

Hence equation (6) becomes

    or    

Multiplying the last equation by the integrating factor

yields

By integrating and applying the initial condition x(0) = 50 we obtain the solution See the discussion following (8) of Section 1.3, Problem 12 in Exercises 1.3, and Problems 27–30 in Exercises 2.7. ≡

Series Circuits

For a series circuit containing only a resistor and an inductor, Kirchhoff’s second law states that the sum of the voltage drop across the inductor (L(di/dt)) and the voltage drop across the resistor (iR) is the same as the impressed voltage (E(t)) on the circuit. See FIGURE 2.7.5.

An L R-series circuit diagram consists of a power source E, an inductor L, and a resistor R.

FIGURE 2.7.5 LR-series circuit

Thus we obtain the linear differential equation for the current i(t),

, (8)

where L and R are known as the inductance and the resistance, respectively. The current i(t) is also called the response of the system.

The voltage drop across a capacitor with capacitance C is given by q(t)/C, where q is the charge on the capacitor. Hence, for the series circuit shown in FIGURE 2.7.6, Kirchhoff’s second law gives

(9)

An L R-series circuit diagram consists of a power source E, a capacitor C, and a resistor R.

FIGURE 2.7.6 RC-series circuit

But current i and charge q are related by i = dq/dt, so (9) becomes the linear differential equation

. (10)

EXAMPLE 7 LR-Series Circuit

A 12-volt battery is connected to an LR-series circuit in which the inductance is henry and the resistance is 10 ohms. Determine the current i if the initial current is zero.

SOLUTION

From (8) we see that we must solve the equation

subject to i(0) = 0. First, we multiply the differential equation by 2 and read off the integrating factor e20t. We then obtain

Integrating each side of the last equation and solving for i gives i(t) = + ce–20t. Now i(0) = 0 implies 0 = + c or c = −. Therefore the response is . ≡

From (4) of Section 2.3 we can write a general solution of (8):

(11)

In particular, when E(t) = E0 is a constant, (11) becomes

(12)

Note that as t → , the second term in (12) approaches zero. Such a term is usually called a transient term; any remaining terms are called the steady-state part of the solution. In this case E0/R is also called the steady-state current; for large values of time it then appears that the current in the circuit is simply governed by Ohm’s law (E = iR). Georg Simon Ohm (1789–1854) was a physicist and mathematician born in Erlangen, Principality of Bayreuth, located then in the Holy Roman Empire.

REMARKS

The solution P(t) = P0e0.4055t of the initial-value problem in Example 1 described the population of a colony of bacteria at any time t ≥ 0. Of course, P(t) is a continuous function that takes on all real numbers in the interval defined by P0 ≤ P < . But since we are talking about a population, common sense dictates that P can take on only positive integer values. Moreover, we would not expect the population to grow continuously—that is, every second, every microsecond, and so on—as predicted by our solution; there may be intervals of time [t1, t2] over which there is no growth at all. Perhaps, then, the graph shown in FIGURE 2.7.7(a) is a more realistic description of P than is the graph of an exponential function. Using a continuous function to describe a discrete phenomenon is often more a matter of convenience than of accuracy. However, for some purposes we may be satisfied if our model describes the system fairly closely when viewed macroscopically in time, as in Figures 2.7.7(b) and 2.7.7(c), rather than microscopically, as in Figure 2.7.7(a). Keep firmly in mind, a mathematical model is not reality.

Three graphs. In each graph, the horizontal axis is labeled t and the vertical axis is labeled P. Graph a is a broken-line graph where five horizontal line segments begin above the end of the previous segment. The interval 0 to 1 on the t axis is divided into smaller unequal parts. The graph begins at (0, P superscript zero) and goes right to (t superscript one, P superscript zero). It then begins above (t superscript one, P superscript zero) and goes to the right up to t superscript two. It continues for three more intervals and ends at t equals 1. Graph b is a broken-line graph. The horizontal axis marks points 1 to 3. The graph is a curve consisting of small horizontal intervals. Each interval begins above the end of the previous one. The graph begins at (0, P superscript zero) and goes up and to the right with increasing steepness. Graph c is a smooth curve that goes up and to the right with increasing steepness from (0, P superscript zero).

FIGURE 2.7.7 Population growth is a discrete process

2.7 Exercises Answers to selected odd-numbered problems begin on page ANS-3.

Growth and Decay

  1. The population of a community is known to increase at a rate proportional to the number of people present at time t. If an initial population P0 has doubled in 5 years, how long will it take to triple? To quadruple?
  2. Suppose it is known that the population of the community in Problem 1 is 10,000 after 3 years. What was the initial population P0? What will the population be in 10 years? How fast is the population growing at t = 10?
  3. The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increases by 15% in 10 years. What will the population be in 30 years? How fast is the population growing at t = 30?
  4. The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at time t. After 3 hours it is observed that 400 bacteria are present. After 10 hours 2000 bacteria are present. What was the initial number of bacteria?
  5. The radioactive isotope of lead, Pb-209, decays at a rate proportional to the amount present at time t and has a half-life of 3.3 hours. If 1 gram of this isotope is present initially, how long will it take for 90% of the lead to decay?
  6. Initially, 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of decay is proportional to the amount of the substance present at time t, find the amount remaining after 24 hours.
  7. Determine the half-life of the radioactive substance described in Problem 6.

  8. (a) Consider the initial-value problem dA/dt = kA, A(0) = A0, as the model for the decay of a radioactive substance. Show that, in general, the half-life T of the substance is T = −(ln 2)/k.

    (b) Show that the solution of the initial-value problem in part (a) can be written A(t) = A02-t/T.

    (c) If a radioactive substance has a half-life T given in part (a), how long will it take an initial amount A0 of the substance to decay to A0?

  9. The amount A(t) of a radioactive substance remaining at any time can be determined directly from the initial amount A0 and the amount A1 of the substance present at a time t1 > 0 Show that

  10. Using the result in Problem 9, show that the half-life T of a radioactive substance is

    Use this result to verify the half-life in Problem 7.

  11. When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional to I(t), where t represents the thickness of the medium (in feet). In clear seawater, the intensity 3 feet below the surface is 25% of the initial intensity I0 of the incident beam. What is the intensity of the beam 15 feet below the surface?
  12. When interest is compounded continuously, the amount of money increases at a rate proportional to the amount S present at time t, that is, dS/dt = rS, where r is the annual rate of interest.

    (a) Find the amount of money accrued at the end of 5 years when $5000 is deposited in a savings account drawing 5 % annual interest compounded continuously.

    (b) In how many years will the initial sum deposited have doubled?

    (c) Use a calculator to compare the amount obtained in part (a) with the amount S = 5000(1 + (0.0575))5(4) that is accrued when interest is compounded quarterly.

Carbon Dating

  1. Archaeologists used pieces of burned wood, or charcoal, found at the site to date prehistoric paintings and drawings on walls and ceilings in a cave in Lascaux, France. See the photo below. Use the information on pages 78 and 79 to determine the approximate age of a piece of burned wood, if it was found that 85.5% of the C-14 found in living trees of the same type had decayed.
    A photo of a cave wall painting.

    Cave wall painting in Problem 13

    © thipjang/Shutterstock

  2. The Shroud of Turin, which shows the negative image of the body of a man who appears to have been crucified, is believed by many to be the burial shroud of Jesus of Nazareth. See the photo on the next page. In 1988 the Vatican granted permission to have the shroud carbon dated. Three independent scientific laboratories analyzed the cloth and concluded that the shroud was approximately 660 years old,* an age consistent with its historical appearance. Using this age, determine what percentage of the original amount of C-14 remained in the cloth as of 1988.
    A photo of a shroud.

    Shroud image in Problem 14

    © imageBROKER/Shutterstock

Newton’s Law of Cooling/Warming

  1. A thermometer is removed from a room where the temperature is 70°F and is taken outside, where the air temperature is 10°F. After one-half minute the thermometer reads 50°F. What is the reading of the thermometer at t = 1 min? How long will it take for the thermometer to reach 15°F?
  2. A thermometer is taken from an inside room to the outside, where the air temperature is 5°F. After 1 minute the thermometer reads 55°F, and after 5 minutes it reads 30°F. What is the initial temperature of the inside room?
  3. A small metal bar, whose initial temperature was 20°C, is dropped into a large container of boiling water. How long will it take the bar to reach 90°C if it is known that its temperature increased 2° in 1 second? How long will it take the bar to reach 98°C?
  4. Two large containers A and B of the same size are filled with different fluids. The fluids in containers A and B are maintained at 0°C and 100°C, respectively. A small metal bar, whose initial temperature is 100°C, is lowered into container A. After 1 minute the temperature of the bar is 90°C. After 2 minutes the bar is removed and instantly transferred to the other container. After 1 minute in container B the temperature of the bar rises 10°. How long, measured from the start of the entire process, will it take the bar to reach 99.9°C?
  5. A thermometer reading 70°F is placed in an oven preheated to a constant temperature. Through a glass window in the oven door, an observer records that the thermometer read 110°F after minute and 145°F after 1 minute. How hot is the oven?
  6. At t = 0 a sealed test tube containing a chemical is immersed in a liquid bath. The initial temperature of the chemical in the test tube is 80°F. The liquid bath has a controlled temperature (measured in degrees Fahrenheit) given by Tm(t) = 100 − 40e– 0.1t, t ≥ 0, where t is measured in minutes.

    (a) Assume that k = −0.1 in (2). Before solving the IVP, describe in words what you expect the temperature T(t) of the chemical to be like in the short term. In the long term.

    (b) Solve the initial-value problem. Use a graphing utility to plot the graph of T(t) on time intervals of various lengths. Do the graphs agree with your predictions in part (a)?

  7. A dead body was found within a closed room of a house where the temperature was a constant 70°F. At the time of discovery, the core temperature of the body was determined to be 85°F. One hour later a second measurement showed that the core temperature of the body was 80°F. Assume that the time of death corresponds to t = 0 and that the core temperature at that time was 98.6°F. Determine how many hours elapsed before the body was found.
  8. Repeat Problem 21 if evidence indicated that the dead person was running a fever of 102°F at the time of death.

Mixtures

  1. A tank contains 200 liters of fluid in which 30 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 4 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.
  2. Solve Problem 23 assuming that pure water is pumped into the tank.
  3. A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into the tank at a rate of 5 gal/min. The well-mixed solution is pumped out at the same rate. Find the number A(t) of pounds of salt in the tank at time t.
  4. In Problem 25, what is the concentration c(t) of the salt in the tank at time t? At t = 5 min? What is the concentration of the salt in the tank after a long time; that is, as t → ? At what time is the concentration of the salt in the tank equal to one-half this limiting value?
  5. Solve Problem 25 under the assumption that the solution is pumped out at a faster rate of 10 gal/min. When is the tank empty?
  6. Determine the amount of salt in the tank at time t in Example 5 if the concentration of salt in the inflow is variable and given by cin(t) = 2 + sin(t/4) lb/gal. Without actually graphing, conjecture what the solution curve of the IVP should look like. Then use a graphing utility to plot the graph of the solution on the interval [0, 300]. Repeat for the interval [0, 600] and compare your graph with that in Figure 2.7.4(a).
  7. A large tank is partially filled with 100 gallons of fluid in which 10 pounds of salt is dissolved. Brine containing pound of salt per gallon is pumped into the tank at a rate of 6 gal/min. The well-mixed solution is then pumped out at a slower rate of 4 gal/min. Find the number of pounds of salt in the tank after 30 minutes.
  8. In Example 5 the size of the tank containing the salt mixture was not given. Suppose, as in the discussion following Example 5, that the rate at which brine is pumped into the tank is 3 gal/min but that the well-stirred solution is pumped out at a rate of 2 gal/min. It stands to reason that since brine is accumulating in the tank at the rate of 1 gal/min, any finite tank must eventually overflow. Now suppose that the tank has an open top and has a total capacity of 400 gallons.

    (a) When will the tank overflow?

    (b) What will be the number of pounds of salt in the tank at the instant it overflows?

    (c) Assume that, although the tank is overflowing, the brine solution continues to be pumped in at a rate of 3 gal/min and the well-stirred solution continues to be pumped out at a rate of 2 gal/min. Devise a method for determining the number of pounds of salt in the tank at t = 150 min.

    (d) Determine the number of pounds of salt in the tank as t → . Does your answer agree with your intuition?

    (e) Use a graphing utility to plot the graph A(t) on the interval [0, 500).

Series Circuits

  1. A 30-volt electromotive force is applied to an LR-series circuit in which the inductance is 0.1 henry and the resistance is 50 ohms. Find the current i(t) if i(0) = 0. Determine the current as t → .
  2. Solve equation (8) under the assumption that E(t) = E0 sin ωt and i(0) = i0.
  3. A 100-volt electromotive force is applied to an RC-series circuit in which the resistance is 200 ohms and the capacitance is 10– 4 farad. Find the charge q(t) on the capacitor if q(0) = 0. Find the current i(t).
  4. A 200-volt electromotive force is applied to an RC-series circuit in which the resistance is 1000 ohms and the capacitance is 5 × 10– 6 farad. Find the charge q(t) on the capacitor if i(0) = 0.4. Determine the charge and current at t = 0.005 s. Determine the charge as t → .
  5. An electromotive force

    is applied to an LR-series circuit in which the inductance is 20 henries and the resistance is 2 ohms. Find the current i(t) if i(0) = 0.

  6. An LR-series circuit has a variable inductor with the inductance defined by

    Find the current i(t) if the resistance is 0.2 ohm, the impressed voltage is E(t) = 4, and i(0) = 0 Graph i(t).

Additional Linear Models

  1. Air Resistance     In (14) of Section 1.3 we saw that a differential equation describing the velocity v of a falling mass subject to air resistance proportional to the instantaneous velocity is

    where k > 0 is a constant of proportionality called the drag coefficient. The positive direction is downward.

    (a) Solve the equation subject to the initial condition v(0) = v0.

    (b) Use the solution in part (a) to determine the limiting, or terminal, velocity of the mass. We saw how to determine the terminal velocity without solving the DE in Problem 40 in Exercises 2.1.

    (c) If the distance s, measured from the point where the mass was released above ground, is related to velocity v by ds/dt = v, find an explicit expression for s(t) if s(0) = 0.

  2. How High?—No Air Resistance    Suppose a small cannonball weighing 16 pounds is shot vertically upward with an initial velocity v0 = 300 ft/s. The answer to the question, “How high does the cannonball go?” depends on whether we take air resistance into account.
    1. Suppose air resistance is ignored. If the positive direction is upward, then a model for the state of the cannonball is given by d2s/dt2 = −g (equation (12) of Section 1.3). Since ds/dt = v(t) the last differential equation is the same as dv/dt = −g, where we take g = 32 ft/s2. Find the velocity v(t) of the cannonball at time t.
    2. Use the result obtained in part (a) to determine the height s(t) of the cannonball measured from ground level. Find the maximum height attained by the cannonball.
  3. How High?—Linear Air Resistance     Repeat Problem 38, but this time assume that air resistance is proportional to instantaneous velocity. It stands to reason that the maximum height attained by the cannonball must be less than that in part (b) of Problem 38. Show this by supposing that the drag coefficient is k = 0.0025. [Hint: Slightly modify the DE in Problem 37.]
  4. Skydiving     A skydiver weighs 125 pounds, and her parachute and equipment combined weigh another 35 pounds. After exiting from a plane at an altitude of 15,000 feet, she waits 15 seconds and opens her parachute. See FIGURE 2.7.8. Assume the constant of proportionality in the model in Problem 37 has the value k = 0.5 during free fall and k = 10 after the parachute is opened. Assume that her initial velocity on leaving the plane is zero. What is her velocity and how far has she traveled 20 seconds after leaving the plane? How does her velocity at 20 seconds compare with her terminal velocity? How long does it take her to reach the ground? [Hint: Think in terms of two distinct IVPs.]
    A skydiver is in free fall. Air resistance is indicated by a small arrow pointing up. After a distance, the parachute is opened. Air resistance is indicated by a long arrow pointing up.

    FIGURE 2.7.8 Skydiver in Problem 40

  5. Evaporating Raindrop     As a raindrop falls, it evaporates while retaining its spherical shape. If we make the further assumptions that the rate at which the raindrop evaporates is proportional to its surface area and that air resistance is negligible, then a model for the velocity v(t) of the raindrop is

    Here ρ is the density of water, r0 is the radius of the raindrop at t = 0, k < 0 is the constant of proportionality, and the downward direction is taken to be the positive direction.

    (a) Solve for v(t) if the raindrop falls from rest.

    (b) Reread Problem 36 of Exercises 1.3 and then show that the radius of the raindrop at time t is r(t) = (k/ρ)t + r0.

    (c) If r0 = 0.01 ft and r = 0.007 ft 10 seconds after the raindrop falls from a cloud, determine the time at which the raindrop has evaporated completely.

  6. Fluctuating Population     The differential equation dP/dt = (k cos t)P, where k is a positive constant, is a mathematical model for a population P(t) that undergoes yearly seasonal fluctuations. Solve the equation subject to P(0) = P0. Use a graphing utility to obtain the graph of the solution for different choices of P0.
  7. Population Model     In one model of the changing population P(t) of a community, it is assumed that

    where dB/dt and dD/dt are the birth and death rates, respectively.

    (a) Solve for P(t) if dB/dt = k1P and dD/dt = k2P.

    (b) Analyze the cases k1 > k2, k1 = k2, and k1 < k2.

  8. Memorization     When forgetfulness is taken into account, the rate of memorization of a subject is given by

    where k1 > 0, k2 > 0, A(t) is the amount to be memorized in time t, M is the total amount to be memorized, and M − A is the amount remaining to be memorized. See Problems 25 and 26 in Exercises 1.3.

    (a) Since the DE is autonomous, use the phase portrait concept of Section 2.1 to find the limiting value of A(t) as t → . Interpret the result.

    (b) Solve for A(t) subject to A(0) = 0. Sketch the graph of A(t) and verify your prediction in part (a).

  9. Drug Dissemination     A mathematical model for the rate at which a drug disseminates into the bloodstream is given by dx/dt = r − kx, where r and k are positive constants. The function x(t) describes the concentration of the drug in the bloodstream at time t.

    (a) Since the DE is autonomous, use the phase portrait concept of Section 2.1 to find the limiting value of x(t) as t → .

    (b) Solve the DE subject to x(0) = 0. Sketch the graph of x(t) and verify your prediction in part (a). At what time is the concentration one-half this limiting value?

  10. Rocket Motion     Suppose a small single-stage rocket of total mass m(t) is launched vertically and that the rocket consumes its fuel at a constant rate. If the positive direction is upward and if we take air resistance to be linear, then a differential equation for its velocity v(t) is given by

    where k is the drag coefficient, λ is the rate at which fuel is consumed, R is the thrust of the rocket, m0 is the total mass of the rocket at t = 0, and g is the acceleration due to gravity. See Problem 21 in Exercises 1.3.

    (a) Find the velocity v(t) of the rocket if m0 = 200 kg, R = 2000 N, λ = 1 kg/s, g = 9.8 m/s2, k = 3 kg/s, and v(0) = 0.

    (b) Use ds/dt = v and the result in part (a) to find the height s(t) of the rocket at time t.

  11. Rocket Motion—Continued     In Problem 46, suppose that of the rocket’s initial mass m0, 50 kg is the mass of the fuel.

    (a) What is the burnout time tb, or the time at which all the fuel is consumed? See Problem 22 in Exercises 1.3.

    (b) What is the velocity of the rocket at burnout?

    (c) What is the height of the rocket at burnout?

    (d) Why would you expect the rocket to attain an altitude higher than the number in part (b)?

    (e) After burnout what is a mathematical model for the velocity of the rocket?

Discussion Problems

  1. Cooling and Warming     A small metal bar is removed from an oven whose temperature is a constant 300°F into a room whose temperature is a constant 70°F. Simultaneously, an identical metal bar is removed from the room and placed into the oven. Assume that time t is measured in minutes. Discuss: Why is there a future value of time, call it t* > 0, at which the temperature of each bar is the same?
  2. Heart Pacemaker A heart pacemaker, shown in FIGURE 2.7.9, consists of a switch, a battery, a capacitor, and the heart as a resistor. When the switch S is at P, the capacitor charges; when S is at Q, the capacitor discharges, sending an electrical stimulus to the heart. In Problem 57 in Exercises 2.3, we saw that during the time the electrical stimulus is being applied to the heart, the voltage E across the heart satisfies the linear DE

    (a) Let us assume that over the time interval of length t1, (0, t1), the switch S is at position P shown in Figure 2.7.9 and the capacitor is being charged. When the switch is moved to position Q at time t1 the capacitor discharges, sending an impulse to the heart over the time interval of length t2: [t1, t1 + t2). Thus, over the initial charging/discharging interval (0, t1 + t2) the voltage to the heart is actually modeled by the piecewise-linear differential equation

    By moving S between P and Q, the charging and discharging over time intervals of lengths t1 and t2 is repeated indefinitely. Suppose t1 = 4 s, t2 = 2 s, E0 = 12 V, and E(0) = 0, E(4) = 12, E(6) = 0, E(10) = 12, E(12) = 0, and so on. Solve for E(t) for 0 ≤ t ≤ 24.

    (b) Suppose for the sake of illustration that R = C = 1. Use a graphing utility to graph the solution for the IVP in part (a) for 0 ≤ t ≤ 24.

    The circuit consists of a power source E subscript zero, a heart as resistor R, a switch S with nodes P and Q, and a capacitor C. When the switch is at P, the closed circuit consists of the power source and the capacitor. When the switch is at Q, the closed circuit consists of the capacitor and the heart.

    FIGURE 2.7.9 Model of a pacemaker in Problem 49

  3. Sliding Box     (a) A box of mass m slides down an inclined plane that makes an angle θ with the horizontal as shown in FIGURE 2.7.10. Find a differential equation for the velocity v(t) of the box at time t in each of the following three cases:

    (i) No sliding friction and no air resistance

    (ii) With sliding friction and no air resistance

    (iii) With sliding friction and air resistance

    In cases (ii) and (iii), use the fact that the force of friction opposing the motion of the box is µN, where µ is the coefficient of sliding friction and N is the normal component of the weight of the box. In case (iii) assume that air resistance is proportional to the instantaneous velocity.

    (b) In part (a), suppose that the box weighs 96 lb, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is µ = /4, and that the additional retarding force due to air resistance is numerically equal to v. Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground.

    An inclined plane makes an angle theta with the horizontal. The height of the inclined plane is 50 feet. Three forces act on a block sliding down the plane: the motion acts downward along the plane, the friction acts in the opposite direction along the plane, and the weight W equals m g acts vertically down.

    FIGURE 2.7.10 Box sliding down inclined plane in Problem 50

Contributed Problem

Pierre Gharghouri, Professor Emeritus
Jean-Paul Pascal, Associate Professor
Department of Mathematics, Ryerson University, Toronto, Canada

  1. Air Exchange     A large room has a volume of 2000 m3. The air in this room contains 0.25% by volume of carbon dioxide (CO2). Starting at 9:00 A.M. fresh air containing 0.04% by volume of CO2 is circulated into the room at the rate of 400 m3/min. Assume that the stale air leaves the room at the same rate as the incoming fresh air and that the stale air and fresh air mix immediately in the room. See FIGURE 2.7.11.

    (a) If v(t) denotes the volume of CO2 in the room at time t, what is v(0)? Find v(t) for t > 0. What is the percentage of CO2 in the air of the room at 9:05 A.M?

    (b) When does the air in the room contain 0.06% by volume of CO2?

    (c) What should be the flow rate of the incoming fresh air if it is required to reduce the level of CO2 in the room to 0.08% in 4 minutes?

    A room measures 2000 cubic meters. Fresh air enters through the top and stale air exits from the sides.

    FIGURE 2.7.11 Air exchange in Problem 51

Computer Lab Assignments

  1. Sliding Box—Continued     (a) In Problem 50, let s(t) be the distance measured down the inclined plane from the highest point. Use ds/dt = v(t) and the solution for each of the three cases in part (b) of Problem 50 to find the time that it takes the box to slide completely down the inclined plane. A root-finding application of a CAS may be useful here.

    (b) In the case in which there is friction (µ ≠ 0) but no air resistance, explain why the box will not slide down the plane starting from rest from the highest point above ground when the inclination angle θ satisfies tan θ ≤ µ.

    (c) The box will slide downward on the plane when tan θ ≤ µ if it is given an initial velocity v(0) = v0 > 0. Suppose that µ = /4 and θ = 23°. Verify that tan θ ≤ µ. How far will the box slide down the plane if v0 = 1 ft/s?

    (d) Using the values µ = /4 and θ = 23°, approximate the smallest initial velocity v0 that can be given to the box so that, starting at the highest point 50 ft above ground, it will slide completely down the inclined plane. Then find the corresponding time it takes to slide down the plane.

  2. What Goes Up     (a) It is well known that the model in which air resistance is ignored, part (a) of Problem 38, predicts that the time ta it takes the cannonball to attain its maximum height is the same as the time td it takes the cannonball to fall from the maximum height to the ground. Moreover, the magnitude of the impact velocity vi will be the same as the initial velocity v0 of the cannonball. Verify both of these results.

    (b) Then, using the model in Problem 39 that takes linear air resistance into account, compare the value of ta with td and the value of the magnitude of vi with v0. A root-finding application of a CAS (or graphic calculator) may be useful here.

 

*Some scholars have disagreed with the finding. For more information on this fascinating mystery, see the Shroud of Turin website home page at www.shroud.com.