Solved: (a) If a constant number h of fish are harvested from a fishery per unit time

The number N(t) of supermarkets throughout the country that are using a computerized checkout system is described by the initial-value problem \(\frac{d N}{d t}=N(1-0.0005 N), \quad N(0)=1\). (a) Use the phase portrait concept of Section 2.1 to predict how many supermarkets are expected to adopt the new procedure over a long period of time. By hand, sketch a solution curve of the given initial-value problem. (b) Solve the initial-value problem and then use a graphing utility to verify the solution curve in part (a). How many companies are expected to adopt the new technology when t = 10?

The number N(t) of people in a community who are exposed to a particular advertisement is governed by the logistic equation. Initially N(0) = 500, and it is observed that N(1) = 1000. Solve for N(t) if it is predicted that the limiting number of people in the community who will see the advertisement is 50,000.

A model for the population P(t) in a suburb of a large city is given by the initial-value problem \(\frac{d P}{d t}=P\left(10^{-1}-10^{-7} P\right), \quad P(0)=5000\), where t is measured in months. What is the limiting value of the population? At what time will the population be equal to one-half of this limiting value?

(a) Census data for the United States between 1790 and 1950 is given in the following table. Construct a logistic population model using the data from 1790, 1850, and 1910. (b) Construct a table comparing actual census population with the population predicted by the model in part (a). Compute the error and the percentage error for each entry pair.

(a) If a constant number h of fish are harvested from a fishery per unit time, then a model for the population P(t) of the fishery at time t is given by \(\frac{d P}{d t}=P(a-b P)-h, \quad P(0)=P_{0}\), where a, b, h, and \(P_{0}\) are positive constants. Suppose \(a=5\), \(b=1\), and \(h=4\). Since the DE is autonomous, use the phase portrait concept of Section 2.1 to sketch representative solution curves corresponding to the cases \(P_{0}>4\), \(1<P_{0}<4\), and \(0<P_{0}<1\). Determine the long-term behavior of the population in each case. (b) Solve the IVP in part (a). Verify the results of your phase portrait in part (a) by using a graphing utility to plot the graph of P(t) with an initial condition taken from each of the three intervals given. (c) Use the information in parts (a) and (b) to determine whether the fishery population becomes extinct in finite time. If so, find that time.

Investigate the harvesting model in Problem 5 both qualitatively and analytically in the case \(a=5\), \(b=1\), \(h=\frac{25}{4}\). Determine whether the population becomes extinct in finite time. If so, find that time.

Repeat Problem 6 in the case \(a=5\), \(b=1\), \(h={7}\).

(a) Suppose \(a=b=1\) in the Gompertz differential equation (7). Since the DE is autonomous, use the phase portrait concept of Section 2.1 to sketch representative solution curves corresponding to the cases \(P_{0}>e\) and \(0<P_{0}<e\). (b) Suppose \(a=1\), \(b=-1\) in (7). Use a new phase portrait to sketch representative solution curves corresponding to the cases \(P_{0}>e^{-1}\) and \(0<P_{0}<e^{-1}\).

Find an explicit solution of equation (7) subject to \(P(0)=P_{0}\).

For an initial population \(P_{0}\), where \(P_{0}>K\) the logistic population model (3) predicts that population cannot sustain itself over time so it decreases but yet never falls below the carrying capacity K of the ecosystem. Moreover, for \(0<P_{0}<K\), , the same model predicts that, regardless of how small \(P_{0}\) is, the population increases over time and does not surpass the carrying capacity K. See Figure 2.8.2, where \(a / b=K\) But the American ecologist Warder Clyde Allee (1885–1955) showed that by depleting certain fisheries beyond a certain level, the fishery population never recovers. How would you modify the differential equation (3) to describe a population P that has these same two characteristics of (3) but additionally has a threshold level A, \(0<A<K\), below which the population cannot sustain itself and becomes extinct. [Hint: Construct a phase portrait of what you want and then form a DE.]

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. Initially there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A are used. It is observed that 10 grams of C are formed in 5 minutes. How much is formed in 20 minutes? What is the limiting amount of C after a long time? How much of chemicals A and B remains after a long time?

Solve Problem 11 if 100 grams of chemical A are present initially. At what time is chemical C half-formed?

A tank in the form of a right circular cylinder standing on end is leaking water through a circular hole in its bottom. As we saw in (10) of Section 1.3, when friction and contraction of water at the hole are ignored, the height h of water in the tank is described by \(\frac{d h}{d t}=-\frac{A_{h}}{A_{w}} \sqrt{2 g h}\), where \(A_{w}\) and \(A_{h}\) are the cross-sectional areas of the water and the hole, respectively. (a) Solve for h(t) if the initial height of the water is H. By hand, sketch the graph of h(t) and give its interval I of definition in terms of the symbols \(A_{w}\), \(A_{h}\) and H. Use \(g=32 \mathrm{ft} / \mathrm{s}^{2}\). (b) Suppose the tank is 10 ft high and has radius 2 ft and the circular hole has radius \(\frac{1}{2} \text { in }\). If the tank is initially full, how long will it take to empty?

When friction and contraction of the water at the hole are taken into account, the model in Problem 13 becomes \(\frac{d h}{d t}=-c \frac{A_{h}}{A_{w}} \sqrt{2 g h}\), where \(0<c<1\). How long will it take the tank in Problem 13(b) to empty if \(c=0.6\)? See Problem 13 in Exercises 1.3.

A tank in the form of a right-circular cone standing on end, vertex down, is leaking water through a circular hole in its bottom. (a) Suppose the tank is 20 feet high and has radius 8 feet and the circular hole has radius 2 inches. In Problem 14 in Exercises 1.3 you were asked to show that the differential equation governing the height h of water leaking from a tank is \(\frac{d h}{d t}=-\frac{5}{6 h^{3 / 2}}\). In this model, friction and contraction of the water at the hole were taken into account with \(c=0.6\), and g was taken to be \(32 \mathrm{ft} / s^{2}\). See Figure 1.3.13. If the tank is initially full, how long will it take the tank to empty? (b) Suppose the tank has a vertex angle of \(60^{\circ}\), and the circular hole has radius 2 inches. Determine the differential equation governing the height h of water. Use \(c=0.6\) and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\). If the height of the water is initially 9 feet, how long will it take the tank to empty?

Suppose that the conical tank in Problem 15(a) is inverted, as shown in FIGURE 2.8.5, and that water leaks out a circular hole of radius 2 inches in the center of the circular base. Is the time it takes to empty a full tank the same as for the tank with vertex down in Problem 15? Take the friction/contraction coefficient to be \(c=0.6\) and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\).

A differential equation governing the velocity v of a falling mass m subjected to air resistance proportional to the square of the instantaneous velocity is \(m \frac{d v}{d t}=m g-k v^{2}\), where \(k>0\) is the drag coefficient. The positive direction is downward. (a) Solve this equation subject to the initial condition \(v(0)=v_{0}\). (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 41 in Exercises 2.1. (c) If distance s, measured from the point where the mass was released above ground, is related to velocity v by \(d s / d t=v(t)\), find an explicit expression for s(t) if \(s(0)=0\).

Consider the 16-pound cannonball shot vertically upward in Problems 36 and 37 in Exercises 2.7 with an initial velocity \(v_{0}=300 \mathrm{ft} / \mathrm{s}\). Determine the maximum height attained by the cannonball if air resistance is assumed to be proportional to the square of the instantaneous velocity. Assume the positive direction is upward and take the drag coefficient to be \(k=0.0003\). [Hint: Slightly modify the DE in Problem 17.]

(a) Determine a differential equation for the velocity v(t) of a mass m sinking in water that imparts a resistance proportional to the square of the instantaneous velocity and also exerts an upward buoyant force whose magnitude is given by Archimedes’ principle. See Problem 18 in Exercises 1.3. Assume that the positive direction is downward. (b) Solve the differential equation in part (a). (c) Determine the limiting, or terminal, velocity of the sinking mass.

The differential equation \(\frac{d y}{d x}=\frac{-x+\sqrt{x^{2}+y^{2}}}{y}\) describes the shape of a plane curve C that will reflect all incoming light beams to the same point and could be a model for the mirror of a reflecting telescope, a satellite antenna, or a solar collector. See Problem 29 in Exercises 1.3. There are several ways of solving this DE. (a) Verify that the differential equation is homogeneous (see Section 2.5). Show that the substitution \(y=u x\) yields \(\frac{u d u}{\sqrt{1+u^{2}}\left(1-\sqrt{1+u^{2}}\right)}=\frac{d x}{x}\). Use a CAS (or another judicious substitution) to integrate the left-hand side of the equation. Show that the curve C must be a parabola with focus at the origin and is symmetric with respect to the x-axis. (b) Show that the first differential equation can also be solved by means of the substitution \(u=x^{2}+y^{2}\).

(a) A simple model for the shape of a tsunami is given by \(\frac{d W}{d x}=W \sqrt{4-2 W}\), where \(W(x)>0\) is the height of the wave expressed as a function of its position relative to a point offshore. By inspection, find all constant solutions of the DE. (b) Solve the differential equation in part (a). A CAS may be useful for integration. (c) Use a graphing utility to obtain the graphs of all solutions that satisfy the initial condition \(W(0)=2\).

An outdoor decorative pond in the shape of a hemispherical tank is to be filled with water pumped into the tank through an inlet in its bottom. Suppose that the radius of the tank is \(R=10 \mathrm{ft}\), that water is pumped in at a rate of \(\pi \mathrm{ft}^{3} / \mathrm{min}\), and that the tank is initially empty. See FIGURE 2.8.6. As the tank fills, it loses water through evaporation. Assume that the rate of evaporation is proportional to the area A of the surface of the water and that the constant of proportionality is \(k=0.01\). (a) The rate of change dV/dt of the volume of the water at time t is a net rate. Use this net rate to determine a differential equation for the height h of the water at time t. The volume of the water shown in the figure is \(V=\pi R b^{2}-\frac{1}{2} \pi b^{3}\), where \(R=10\). Express the area of the surface of the water \(A=\pi r^{2}\) in terms of h. (b) Solve the differential equation in part (a). Graph the solution. (c) If there were no evaporation, how long would it take the tank to fill? (d) With evaporation, what is the depth of the water at the time found in part (c)? Will the tank ever be filled? Prove your assertion.

A long uniform piece of wood (cross sections are the same) is cut through perpendicular to its length by a vertical saw blade. See FIGURE 2.8.7. If the friction between the sides of the saw blade and the wood through which the blade passes is ignored, then it can be assumed that the rate at which the saw blade moves through the piece of wood is inversely proportional to the width of the wood in contact with its cutting edge. As the blade advances through the wood (moving left to right), the width of a cross section changes as a nonnegative continuous function w. If a cross section of the wood is described as a region in the xy-plane defined over an interval fa, bg, then as shown in Figure 2.8.7(c) the position x of the saw blade is a function of time t and the vertical cut made by the blade can be represented by a vertical line segment. The length of this vertical line is the width w(x) of the wood at that point. Thus the position x(t) of the saw blade and the rate dx>dt at which it moves to the right are related to w(x) by \(w(x) \frac{d x}{d t}=k, \quad x(0)=a\). Here k represents the number of square units of the material removed by the saw blade per unit time. In the problems that follow, we assume that the saw can be programmed so that \(k=1\). Find an explicit solution x(t) of the initial-value problem (14) when a cross section of the piece of wood is triangular and is bounded by the graphs of \(y=x\), \(x=1,\) and \(y=0\). How long does it take the saw to cut through this piece of wood?

(a) Find an implicit solution of the initial-value problem (14) in Problem 23 when the piece of wood is a circular log. Assume a cross section is a circle of radius 2 centered at (0, 0). [Hint: To save time, see formula 33 in the table of integrals given on the right page inside the front cover.] (b) Solve the implicit solution obtained in part (b) for time t as a function of x. Graph the function t(x). With the aid of the graph, approximate the time that it takes the saw to cut through this piece of wood. Then find the exact value of this time.

Solve the initial-value problem (14) in Problem 23 when a cross section of a uniform piece of wood is the triangular region given in FIGURE 2.8.8. Assume again that k = 1. How long does it take to cut through this piece of wood?

Read the documentation for your CAS on scatter plots (or scatter diagrams) and least-squares linear fit. The straight line that best fits a set of data points is called a regression line or a least squares line. Your task is to construct a logistic model for the population of the United States, defining f (P) in (2) as an equation of a regression line based on the population data in the table in Problem 4. One way of doing this is to approximate the left-hand side \(\frac{1}{P} \frac{d P}{d t}\) of the first equation in (2) using the forward difference quotient in place of dP/dt: \(Q(t)=\frac{1}{P(t)} \frac{P(t+h)-P(t)}{h}\). (a) Make a table of the values t, P(t), and Q(t) using \(t=0\), 10, 20, . . . , 160, and \(h=10\). For example, the first line of the table should contain \(t=0\) P(0), and Q(0). With \(P(0)=3.929\) and \(P(10)=5.308\), \(Q(0)=\frac{1}{P(0)} \frac{P(10)-P(0)}{10}=0.035\) Note that Q(160) depends on the 1960 census population P(170). Look up this value. (b) Use a CAS to obtain a scatter plot of the data (P(t), Q(t)) computed in part (a). Also use a CAS to find an equation of the regression line and to superimpose its graph on the scatter plot. (c) Construct a logistic model dP/dt Pf (P), where f (P) is the equation of the regression line found in part (b). (d) Solve the model in part (c) using the initial condition P(0) 3.929. (e) Use a CAS to obtain another scatter plot, this time of the ordered pairs (t, P(t)) from your table in part (a). Use your CAS to superimpose the graph of the solution in part (d) on the scatter plot. (f ) Look up the U.S. census data for 1970, 1980, and 1990. What population does the logistic model in part (c) predict for these years? What does the model predict for the U.S. population P(t) as \(t \rightarrow \infty\)?

(a) In Examples 3 and 4 of Section 2.1, we saw that any solution P(t) of (4) possesses the asymptotic behavior \(P(t) \rightarrow a / b\) as \(t \rightarrow \infty\) for \(P_{0}>a / b\) and for \(0<P_{0}<a / b\); as a consequence, the equilibrium solution \(P=a / b\) is called an attractor. Use a root-finding application of a CAS (or a graphic calculator) to approximate the equilibrium solution of the immigration model \(\frac{d P}{d t}=P(1-P)+0.3 e^{-P}\). (b) Use a graphing utility to graph the function \(F(P)=\) \(P(1-P)+0.3 e^{-P}\). Explain how this graph can be used to determine whether the number found in part (a) is an attractor. (c) Use a numerical solver to compare the solution curves for the IVPs \(\frac{d P}{d t}=P(1-P), \quad P(0)=P_{0}\) for \(P_{0}=0.2\) and \(P_{0}=1.2\). with the solution curves for the IVPs \(\frac{d P}{d t}=P(1-P)+0.3 e^{-P}, \quad P(0)=P_{0}\) for \(P_{0}=0.2\) and \(P_{0}=1.2\). Superimpose all curves on the same coordinate axes but, if possible, use a different color for the curves of the second initial-value problem. Over a long period of time, what percentage increase does the immigration model predict in the population compared to the logistic model?

In Problem 18 let \(t_{a}\) be the time it takes the cannonball to attain its maximum height and let td be the time it takes the cannonball to fall from the maximum height to the ground. Compare the value of \(t_{a}\) with the value of \(t_{a}\) and compare the magnitude of the impact velocity \(v_{i}\) with the initial velocity \(v_{0}\). See Problem 51 in Exercises 2.7. A root-finding application of a CAS may be useful here. [Hint: Use the model in Problem 17 when the cannonball is falling.]

A skydiver is equipped with a stopwatch and an altimeter. She opens her parachute 25 seconds after exiting a plane flying at an altitude of 20,000 ft and observes that her altitude is 14,800 ft. Assume that air resistance is proportional to the square of the instantaneous velocity, her initial velocity upon leaving the plane is zero, and \(g=32 \mathrm{ft} / \mathrm{s}^{2}\). (a) Find the distance s(t), measured from the plane, that the skydiver has traveled during free fall in time t. [Hint: The constant of proportionality k in the model given in Problem 17 is not specified. Use the expression for terminal velocity \(V_{t}\) obtained in part (b) of Problem 17 to eliminate k from the IVP. Then eventually solve for \(V_{t}\).] (b) How far does the skydiver fall and what is her velocity at \(t=15 \mathrm{~s}\)?

A helicopter hovers 500 feet above a large open tank full of liquid (not water). A dense compact object weighing 160 pounds is dropped (released from rest) from the helicopter into the liquid. Assume that air resistance is proportional to instantaneous velocity v while the object is in the air and that viscous damping is proportional to \(v^{2}\) after the object has entered the liquid. For air, take \(k=\frac{1}{4}\), and for the liquid, \(k=0.1\). Assume that the positive direction is downward. If the tank is 75 feet high, determine the time and the impact velocity when the object hits the bottom of the tank. [Hint: Think in terms of two distinct IVPs. If you use (13), be careful in removing the absolute value sign. You might compare the velocity when the object hits the liquid—the initial velocity for the second problem—with the terminal velocity vt of the object falling through the liquid.]