11.6: Exercises

11.1. Checking solutions of differential equations. A differential equation is an equation in which some function is related to its own derivative(s).

For each of the following functions, calculate the appropriate derivative, and show that the function satisfies the indicated differential equation

(a) \(f(x)=2 e^{-3 x}, \quad f^{\prime}(x)=-3 f(x)\)

(b) \(f(t)=C e^{k t}, \quad f^{\prime}(t)=k f(t)\)

(c) \(f(t)=1-e^{-t}, \quad f^{\prime}(t)=1-f(t)\)

11.2. Linear differential equations. Consider the function \(y=f(t)=C e^{k t}\) where \(C\) and \(k\) are constants. For what value(s) of these constants does this function satisfy the equation (a) \(\frac{d y}{d t}=-5 y\), (b) \(\frac{d y}{d t}=3 y\).

Note: an equation which involves a function and its derivative is called a differential equation.

11.3. Checking initial value solution to a differential equation. Check that the function (11.6) satisfies the differential equation (11.2) and the initial condition \(N(0)=N_{0}\).

11.4. Solving linear differential equations. Find a function that satisfies each of the following differential equations.

Note: all your answers should be exponential functions, but they may have different dependent and independent variables.

(a) \(\frac{d y}{d t}=-y\)

(b) \(\frac{d c}{d x}=-0.1 c\) and \(c(0)=20\)

(c) \(\frac{d z}{d t}=3 z\) and \(z(0)=5\).

11.5. Andromeda strain, revisited. In Chapter 10 we discussed the growth of bacteria, starting from a single cell. The doubling time of the bacteria was given as 20 min.

Find the appropriate differential equation that describes this growth, the appropriate initial condition, and the exponential function (with base \(e\) ) that is the solution to that differential equation. Use units of hours for time \(t\).

11.6. Population growth in developed and developing countries. In Canada, women have only about 2 children during their 40 years of fertility, and people live to age 80. In underdeveloped countries, people on average live to age 60 and women have a child roughly every 4 years between ages 13 and 45 .

Compare the per capita birth and mortality rates and the predicted population growth or decay in each of these scenarios, using arguments analogous to those of Section 11.2.

Find the growth rate \(k\) in percent per year and the doubling time for the growing population.

11.7. Population growth and doubling. A population of animals has a per-capita birth rate of \(b=0.08\) per year and a per-capita death rate of \(m=0.01\) per year. The population density, \(P(t)\) is found to satisfy the differential equation

\[\frac{d P(t)}{d t}=b P(t)-m P(t) \nonumber \]

(a) If the population is initially \(P(0)=1000\), find how big the population is in 5 years.

(b) When does the population double?

11.8. Rodent population. The per capita birthrate of one species of rodent is \(0.05\) newborns per day. This means that, on average, each member of the population results in 5 newborn rodents every 100 days. Suppose that over the period of 1000 days there are no deaths, and that the initial population of rodents is 250 .

(a) Write a differential equation for the population size \(N(t)\) at time \(t\) (in days).

(b) Write down the initial condition that \(N\) satisfies.

(c) Find the solution, i.e. express \(N\) as some function of time \(t\) that satisfies your differential equation and initial condition.

(d) How many rodents are there after 1 year?

11.9. Growth and extinction of microorganisms.

(a) The population \(y(t)\) of a certain microorganism grows continuously and follows an exponential behavior over time. Its doubling time is found to be \(0.27\) hours. What differential equation would you use to describe its growth ?

Note: you must find the value of the rate constant, \(k\), using the doubling time.

(b) With exposure to ultra-violet radiation, the population ceases to grow, and the microorganisms continuously die off. It is found that the half-life is then \(0.1\) hours. What differential equation would now describe the population?

11.10. A bacterial population. A bacterial population grows at a rate proportional to the population size at time \(t\). Let \(y(t)\) be the population size at time \(t\). By experiment it is determined that the population at \(t=10 \mathrm{~min}\) is 15,000 and at \(t=30 \mathrm{~min}\) it is 20,000 .

(a) What was the initial population?

(b) What is the population at time \(t=60 \mathrm{~min}\) ?

11.11. Antibiotic treatment. A colony of bacteria is treated with a mild antibiotic agent so that the bacteria start to die. It is observed that the density of bacteria as a function of time follows the approximate relationship \(b(t)=85 e^{-0.5 t}\) where \(t\) is time in hours.

Determine the time it takes for half of the bacteria to disappear; this is called the half-life.

Find how long it takes for \(99 \%\) of the bacteria to die.

11.12. Two populations. Two populations are studied. Population \(\mathbf{1}\) is found to obey the differential equation

\[d y_{1} / d t=0.2 y_{1} \nonumber \]

and population 2 obeys

\[d y_{2} / d t=-0.3 y_{2} \nonumber \]

where \(t\) is time in years.

(a) Which population is growing and which is declining?

(b) Find the doubling time (respectively half-life) associated with the given population.

(c) If the initial levels of the two populations were \(y_{1}(0)=100\) and \(y_{2}(0)=10,000\), how big would each population be at time \(t\) ?

(d) At what time would the two populations be exactly equal?

11.13. The human population. The human population on Earth doubles roughly every 50 years. In October 2000 there were \(6.1\) billion humans on earth.

(a) Determine what the human population would be 500 years later under the uncontrolled growth scenario.

(b) How many people would have to inhabit each square kilometer of the planet for this population to fit on earth? (Take the circumference of the earth to be \(40,000 \mathrm{~km}\) for the purpose of computing its surface area and assume that the oceans have dried up.)

11.14. Fish in two lakes. Two lakes have populations of fish, but the conditions are quite different in these lakes. In the first lake, the fish population is growing and satisfies the differential equation

\[\frac{d y}{d t}=0.2 y \nonumber \]

where \(t\) is time in years. At time \(t=0\) there were 500 fish in this lake. In the second lake, the population is dying due to pollution. Its population satisfies the differential equation

\[\frac{d y}{d t}=-0.1 y, \nonumber \]

and initially there were 4000 fish in this lake.

At what time are the fish populations in the two lakes identical?

11.15. First order chemical kinetics. When chemists say that a chemical reaction follows "first order kinetics", they mean that the concentration of the reactant at time \(t\), i.e. \(c(t)\), satisfies an equation of the form \(\frac{d c}{d t}=\) \(-r c\) where \(r\) is a rate constant, here assumed to be positive. Suppose the reaction mixture initially has concentration 1M ("1 molar") and that after 1 hour there is half this amount.

(a) Find the "half life" of the reactant.

(b) Find the value of the rate constant \(r\).

(c) Determine how much is left after 2 hours.

(d) When is only \(10 \%\) of the initial amount be left?

11.16. Chemical breakdown. In a chemical reaction, a substance \(S\) is broken down. The concentration of the substance is observed to change at a rate proportional to the current concentration. It was observed that 1 Mole/liter of \(S\) decreased to \(0.5\) Moles/liter in 10 minutes.

(a) How long does it take until only \(0.25\) Moles per liter remain?

(b) How long does it take until only \(1 \%\) of the original concentration remains?

11.17. Half-life. If \(10 \%\) of a radioactive substance remains after one year, find its half-life.

11.18. Carbon 14. Carbon 14 , or \({ }^{14} C\), has a half-life of 5730 years. This means that after 5730 years, a sample of Carbon 14, which is a radioactive isotope of carbon, has lost one half of its original radioactivity.

(a) Estimate how long it takes for the sample to fall to roughly \(0.001\) of its original level of radioactivity.

(b) Each gram of \({ }^{14} \mathrm{C}\) has an activity given here in units of 12 decays per minute. After some time, the amount of radioactivity decreases. For example, a sample 5730 years old has only one half the original activity level, i.e. 6 decays per minute. If a \(1 \mathrm{gm}\) sample of material is found to have 45 decays per hour, approximately how old is it?

Note: \({ }^{14} \mathrm{C}\) is used in radiocarbon dating, a process by which the age of materials containing carbon can be estimated. W. Libby received the Nobel prize in chemistry in 1960 for developing this technique.

11.19. Strontium-90. Strontium-90 is a radioactive isotope with a half-life of 29 years. If you begin with a sample of 800 units, how long does it take for the amount of radioactivity of the strontium sample to be reduced to (a) 400 units (b) 200 units (c) 1 unit

11.20. More radioactivity. The half-life of a radioactive material is 1620 years.

(a) What percentage of the radioactivity remains after 500 years?

(b) Cobalt 60 is a radioactive substance with half life \(5.3\) years. It is used in medical application (radiology). How long does it take for \(80 \%\) of a sample of this substance to decay?

11.21. Salt in a barrel. A barrel initially contains \(2 \mathrm{~kg}\) of salt dissolved in \(20 L\) of water. If water flows in the rate of \(0.4 L\) per minute and the well-mixed salt water solution flows out at the same rate, how much salt is present after 8 minutes?

11.22. Atmospheric pressure. Assume the atmospheric pressure \(y\) at a height \(x\) meters above the sea level satisfies the relation

\[\frac{d y}{d x}=k y \text {. } \nonumber \]

If one day at a certain location the atmospheric pressures are 760 and 675 torr (unit for pressure) at sea level and at 1000 meters above sea level, respectively, find the value of the atmospheric pressure at 600 meters above sea level.