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Unformatted text preview: 5E10(pp 622631) 1/18/06 9:18 AM Page 622 CHAPTER 10
By analyzing pairs of differential equations we gain
insight into population
cycles of predators and
prey, such as the Canada
lynx and snowshoe hare. W
150 100 R
3000 W R
W 50 120
2000
80 0 1000 2000 3000 R
1000
40 0 t¡ t™ t£ D ifferential Equations t 5E10(pp 622631) 1/18/06 9:18 AM Page 623 Perhaps the most important of all the applications of calculus is to differential equations. When physical scientists
or social scientists use calculus, more often than not it is to
analyze a differential equation that has arisen in the process of modeling some phenomenon that they are studying.
Although it is often impossible to ﬁnd an explicit formula for the solution of a differential equation, we will see that graphical and numerical approaches provide the
needed information.  10.1 Modeling with Differential Equations  Now is a good time to read (or reread) the
discussion of mathematical modeling on page 25. In describing the process of modeling in Section 1.2, we talked about formulating a mathematical model of a realworld problem either through intuitive reasoning about the phenomenon or from a physical law based on evidence from experiments. The mathematical
model often takes the form of a differential equation, that is, an equation that contains an
unknown function and some of its derivatives. This is not surprising because in a realworld problem we often notice that changes occur and we want to predict future behavior
on the basis of how current values change. Let’s begin by examining several examples of
how differential equations arise when we model physical phenomena. Models of Population Growth
One model for the growth of a population is based on the assumption that the population
grows at a rate proportional to the size of the population. That is a reasonable assumption
for a population of bacteria or animals under ideal conditions (unlimited environment, adequate nutrition, absence of predators, immunity from disease).
Let’s identify and name the variables in this model:
t
P time the independent variable the number of individuals in the population the dependent variable The rate of growth of the population is the derivative dP dt. So our assumption that the
rate of growth of the population is proportional to the population size is written as the
equation
1 dP
dt kP where k is the proportionality constant. Equation 1 is our ﬁrst model for population
growth; it is a differential equation because it contains an unknown function P and its
derivative dP dt.
Having formulated a model, let’s look at its consequences. If we rule out a population
of 0, then P t
0 for all t. So, if k 0, then Equation 1 shows that P t
0 for all t.
This means that the population is always increasing. In fact, as P t increases, Equation 1
shows that dP dt becomes larger. In other words, the growth rate increases as the population increases. 623 5E10(pp 622631) 624 ❙❙❙❙ 1/18/06 9:18 AM Page 624 CHAPTER 10 DIFFERENTIAL EQUATIONS P Let’s try to think of a solution of Equation 1. This equation asks us to ﬁnd a function
whose derivative is a constant multiple of itself. We know that exponential functions have
that property. In fact, if we let P t
Ce kt, then
Pt
t k Ce kt kP t Thus, any exponential function of the form P t
Ce kt is a solution of Equation 1. When
we study this equation in detail in Section 10.4, we will see that there is no other solution.
Allowing C to vary through all the real numbers, we get the family of solutions
Pt
Ce kt whose graphs are shown in Figure 1. But populations have only positive
values and so we are interested only in the solutions with C 0. And we are probably concerned only with values of t greater than the initial time t 0. Figure 2 shows the physically meaningful solutions. Putting t 0, we get P 0
Ce k 0
C, so the constant C
turns out to be the initial population, P 0 .
Equation 1 is appropriate for modeling population growth under ideal conditions, but
we have to recognize that a more realistic model must reﬂect the fact that a given environment has limited resources. Many populations start by increasing in an exponential
manner, but the population levels off when it approaches its carrying capacity K (or
decreases toward K if it ever exceeds K ). For a model to take into account both trends, we
make two assumptions: FIGURE 1 The family of solutions of d P/dt=kP
P 0 C ke kt t ■ dP
dt kP if P is small (Initially, the growth rate is proportional to P.) ■ dP
dt 0 if P FIGURE 2 The family of solutions P (t)=Ce kt
with C >0 and t ˘0 A simple expression that incorporates both assumptions is given by the equation 2 P P =K equilibrium
solutions
P =0
0 FIGURE 3 Solutions of the logistic equation K (P decreases if it ever exceeds K.) t dP
dt kP 1 P
K Notice that if P is small compared with K, then P K is close to 0 and so dP dt kP. If
P K , then 1 P K is negative and so dP dt 0.
Equation 2 is called the logistic differential equation and was proposed by the Dutch
mathematical biologist PierreFrançois Verhulst in the 1840s as a model for world population growth. We will develop techniques that enable us to ﬁnd explicit solutions of the
logistic equation in Section 10.5, but for now we can deduce qualitative characteristics of
the solutions directly from Equation 2. We ﬁrst observe that the constant functions
Pt
0 and P t
K are solutions because, in either case, one of the factors on the right
side of Equation 2 is zero. (This certainly makes physical sense: If the population is ever
either 0 or at the carrying capacity, it stays that way.) These two constant solutions are
called equilibrium solutions.
If the initial population P 0 lies between 0 and K, then the right side of Equation 2 is
positive, so dP dt 0 and the population increases. But if the population exceeds the carrying capacity P K , then 1 P K is negative, so dP dt 0 and the population
decreases. Notice that, in either case, if the population approaches the carrying capacity
P l K , then dP dt l 0, which means the population levels off. So we expect that the
solutions of the logistic differential equation have graphs that look something like the ones
in Figure 3. Notice that the graphs move away from the equilibrium solution P 0 and
move toward the equilibrium solution P K . 5E10(pp 622631) 1/18/06 9:18 AM Page 625 S ECTION 10.1 MODELING WITH DIFFERENTIAL EQUATIONS ❙❙❙❙ 625 A Model for the Motion of a Spring
Let’s now look at an example of a model from the physical sciences. We consider the
motion of an object with mass m at the end of a vertical spring (as in Figure 4). In Section 6.4 we discussed Hooke’s Law, which says that if the spring is stretched (or compressed) x units from its natural length, then it exerts a force that is proportional to x :
m equilibrium
position 0 x restoring force m where k is a positive constant (called the spring constant). If we ignore any external resisting forces (due to air resistance or friction) then, by Newton’s Second Law (force equals
mass times acceleration), we have x FIGURE 4 kx m 3 d 2x
dt 2 kx This is an example of what is called a secondorder differential equation because it
involves second derivatives. Let’s see what we can guess about the form of the solution
directly from the equation. We can rewrite Equation 3 in the form
d 2x
dt 2 k
x
m which says that the second derivative of x is proportional to x but has the opposite sign. We
know two functions with this property, the sine and cosine functions. In fact, it turns
out that all solutions of Equation 3 can be written as combinations of certain sine
and cosine functions (see Exercise 3). This is not surprising; we expect the spring to oscillate about its equilibrium position and so it is natural to think that trigonometric functions
are involved. General Differential Equations
In general, a differential equation is an equation that contains an unknown function and
one or more of its derivatives. The order of a differential equation is the order of the highest derivative that occurs in the equation. Thus, Equations 1 and 2 are ﬁrstorder equations
and Equation 3 is a secondorder equation. In all three of those equations the independent
variable is called t and represents time, but in general the independent variable doesn’t
have to represent time. For example, when we consider the differential equation
4 y xy it is understood that y is an unknown function of x.
A function f is called a solution of a differential equation if the equation is satisﬁed
when y f x and its derivatives are substituted into the equation. Thus, f is a solution of
Equation 4 if
fx xf x for all values of x in some interval.
When we are asked to solve a differential equation we are expected to ﬁnd all possible
solutions of the equation. We have already solved some particularly simple differential 5E10(pp 622631) 626 ❙❙❙❙ 1/18/06 9:18 AM Page 626 CHAPTER 10 DIFFERENTIAL EQUATIONS equations, namely, those of the form
y fx For instance, we know that the general solution of the differential equation
x3 y
is given by
x4
4 y C where C is an arbitrary constant.
But, in general, solving a differential equation is not an easy matter. There is no systematic technique that enables us to solve all differential equations. In Section 10.2, however, we will see how to draw rough graphs of solutions even when we have no explicit
formula. We will also learn how to ﬁnd numerical approximations to solutions.
EXAMPLE 1 Show that every member of the family of functions ce t
ce t 1
1 y 1
2 is a solution of the differential equation y y2 1. SOLUTION We use the Quotient Rule to differentiate the expression for y : 5 ce t ce t ce t
1 5 c 2e 2t ce t
1 ce t 2 y2 1 1
2 1
1 ce t
ce t 1
4ce t
2 1 ce t
_5 FIGURE 5 1
ce t ce t ce t 2 c 2e 2t 2ce t
1 ce t 2 The right side of the differential equation becomes
1
2 _5 1 y  Figure 5 shows graphs of seven members of
the family in Example 1. The differential equation
shows that if y
1, then y
0. That is
borne out by the ﬂatness of the graphs near
y 1 and y
1. 2 2 1
2ce t
1 ce t 1
2 1 ce t 2
1 ce t
1 ce t 2 2 2 Therefore, for every value of c, the given function is a solution of the differential
equation.
When applying differential equations, we are usually not as interested in ﬁnding a family of solutions (the general solution) as we are in ﬁnding a solution that satisﬁes some
additional requirement. In many physical problems we need to ﬁnd the particular solution
that satisﬁes a condition of the form y t0
y0 . This is called an initial condition, and the
problem of ﬁnding a solution of the differential equation that satisﬁes the initial condition
is called an initialvalue problem.
Geometrically, when we impose an initial condition, we look at the family of solution
curves and pick the one that passes through the point t0 , y0 . Physically, this corresponds
to measuring the state of a system at time t0 and using the solution of the initialvalue problem to predict the future behavior of the system. 5E10(pp 622631) 1/18/06 9:18 AM Page 627 S ECTION 10.1 MODELING WITH DIFFERENTIAL EQUATIONS EXAMPLE 2 Find a solution of the differential equation y initial condition y 0 1
2 y2 ❙❙❙❙ 627 1 that satisﬁes the 2. SOLUTION Substituting the values t 0 and y 2 into the formula
ce t
ce t y 1
1 1
1 ce 0
ce 0 from Example 1, we get
2
Solving this equation for c, we get 2
of the initialvalue problem is
y  10.1
y x x 1 1
3
1
3 1
3 c, which gives c 1
et
et 3
3 . So the solution et
et (d) Find a solution of the differential equation y
isﬁes the initial condition y 1
2. is a solution of the differential equation 2 x. 2. Verify that y sin x cos x y cos2 x tan x y
2 on the interval x y0 y
y 2 just by looking at the differential equation?
(b) Verify that all members of the family y 1 x C are
solutions of the equation in part (a).
(c) Can you think of a solution of the differential equation
y
y 2 that is not a member of the family in part (b)?
(d) Find a solution of the initialvalue problem 1 2. 3. (a) For what nonzero values of k does the function y sin k t
satisfy the differential equation y
9y 0?
(b) For those values of k, verify that every member of the family of functions
y A sin k t differential equation y y 6y 0? 5. Which of the following functions are solutions of the differen tial equation y
(a) y e t
(c) y te t 2y y 0?
(b) y
(d) y et
t 2e 0.5 ; tion y
x y 3 when x is close to 0? What if x is large?
(b) Verify that all members of the family y
c x 2 1 2 are
solutions of the differential equation y
x y 3.
(c) Graph several members of the family of solutions on a
common screen. Do the graphs conﬁrm what you predicted
in part (a)?
(d) Find a solution of the initialvalue problem
y xy 3 y0 2 t 6. (a) Show that every member of the family of functions
2 y0 8. (a) What can you say about the graph of a solution of the equa is also a solution.
e rt satisfy the y2 y B cos k t 4. For what values of r does the function y x y that sat 7. (a) What can you say about a solution of the equation cos x is a solution of the initial value problem ; 1
1 c
c Exercises 1. Show that y xy 2c 1
1 y Ce x 2 is a solution of the differential equation y
x y.
(b) Illustrate part (a) by graphing several members of the family of solutions on a common screen.
(c) Find a solution of the differential equation y
x y that satisﬁes the initial condition y 0
5. 9. A population is modeled by the differential equation dP
dt 1.2P 1 P
4200 (a) For what values of P is the population increasing?
(b) For what values of P is the population decreasing?
(c) What are the equilibrium solutions? 5E10(pp 622631) 628 ❙❙❙❙ 1/18/06 9:18 AM Page 628 CHAPTER 10 DIFFERENTIAL EQUATIONS 10. A function y t satisﬁes the differential equation dy
dt y4 6y 3 13. Psychologists interested in learning theory study learning 5y 2 (a) What are the constant solutions of the equation?
(b) For what values of y is y increasing?
(c) For what values of y is y decreasing?
11. Explain why the functions with the given graphs can’t be solu tions of the differential equation
dy
dt et y (a) y 1 2 dP
dt (b) y 1 t t 1 12. The function with the given graph is a solution of one of the following differential equations. Decide which is the correct
equation and justify your answer.
y 0 1  10.2 xy B. y x 2 xy kM k a positive constant P is a reasonable model for learning.
(c) Make a rough sketch of a possible solution of this differential equation. 1 1 A. y curves. A learning curve is the graph of a function P t , the
performance of someone learning a skill as a function of the
training time t. The derivative dP dt represents the rate at
which performance improves.
(a) When do you think P increases most rapidly? What
happens to dP dt as t increases? Explain.
(b) If M is the maximum level of performance of which the
learner is capable, explain why the differential equation C. y 1 2 xy 14. Suppose you have just poured a cup of freshly brewed coffee with temperature 95 C in a room where the temperature
is 20 C.
(a) When do you think the coffee cools most quickly? What
happens to the rate of cooling as time goes by? Explain.
(b) Newton’s Law of Cooling states that the rate of cooling of
an object is proportional to the temperature difference
between the object and its surroundings, provided that this
difference is not too large. Write a differential equation that
expresses Newton’s Law of Cooling for this particular situation. What is the initial condition? In view of your answer
to part (a), do you think this differential equation is an
appropriate model for cooling?
(c) Make a rough sketch of the graph of the solution of the
initialvalue problem in part (b). Direction Fields and Euler ’s Method
Unfortunately, it’s impossible to solve most differential equations in the sense of obtaining an explicit formula for the solution. In this section we show that, despite the absence
of an explicit solution, we can still learn a lot about the solution through a graphical
approach (direction ﬁelds) or a numerical approach (Euler’s method). Direction Fields
Suppose we are asked to sketch the graph of the solution of the initialvalue problem
y x y y0 1 We don’t know a formula for the solution, so how can we possibly sketch its graph? Let’s
think about what the differential equation means. The equation y
x y tells us that the
slope at any point x, y on the graph (called the solution curve) is equal to the sum of the 5E10(pp 622631) 1/18/06 9:18 AM Page 629 S ECTION 10.2 DIRECTION FIELDS AND EULER’S METHOD ❙❙❙❙ 629 x and ycoordinates of the point (see Figure 1). In particular, because the curve passes
through the point 0, 1 , its slope there must be 0 1 1. So a small portion of the solution curve near the point 0, 1 looks like a short line segment through 0, 1 with slope 1
(see Figure 2).
y y Slope at
(¤, ﬁ) is
¤+ﬁ. Slope at
(⁄, ›) is
⁄+›. (0, 1) 0 x Slope at (0, 1)
is 0+1=1. 0 x FIGURE 1 FIGURE 2 A solution of y ª=x+y Beginning of the solution curve through (0, 1) As a guide to sketching the rest of the curve, let’s draw short line segments at a number of points x, y with slope x y. The result is called a direction ﬁeld and is shown in
Figure 3. For instance, the line segment at the point 1, 2 has slope 1 2 3. The direction ﬁeld allows us to visualize the general shape of the solution curves by indicating the
direction in which the curves proceed at each point.
y y (0, 1)
0 1 2 x 0 1 2 FIGURE 3 FIGURE 4 Direction field for y ª=x+y x The solution curve through (0, 1) Now we can sketch the solution curve through the point 0, 1 by following the direction ﬁeld as in Figure 4. Notice that we have drawn the curve so that it is parallel to nearby
line segments.
In general, suppose we have a ﬁrstorder differential equation of the form
y F x, y where F x, y is some expression in x and y. The differential equation says that the slope
of a solution curve at a point x, y on the curve is F x, y . If we draw short line segments
with slope F x, y at several points x, y , the result is called a direction ﬁeld (or slope
ﬁeld). These line segments indicate the direction in which a solution curve is heading, so
the direction ﬁeld helps us visualize the general shape of these curves. 5E10(pp 622631) 630 ❙❙❙❙ 1/18/06 9:18 AM Page 630 CHAPTER 10 DIFFERENTIAL EQUATIONS y EXAMPLE 1 2 (a) Sketch the direction ﬁeld for the differential equation y
x 2 y 2 1.
(b) Use part (a) to sketch the solution curve that passes through the origin. 1 SOLUTION
_2 _1 0 1 2 (a) We start by computing the slope at several points in the following chart: x 1 x y FIGURE 5 2 1 0 1 2 y 2 1 0 1 2 2 1 0 1 2 ... 0 0 0 0 1 1 1 1 1 ... 3 0 1 0 3 4 1 0 1 4 ... Now we draw short line segments with these slopes at these points. The result is the
direction ﬁeld shown in Figure 5.
(b) We start at the origin and move to the right in the direction of the line segment
(which has slope 1 ). We continue to draw the solution curve so that it moves parallel
to the nearby line segments. The resulting solution curve is shown in Figure 6. Returning
to the origin, we draw the solution curve to the left as well. y _1 x 2 1 0 y _2 _2 2 x The more line segments we draw in a direction ﬁeld, the clearer the picture becomes.
Of course, it’s tedious to compute slopes and draw line segments for a huge number of
points by hand, but computers are well suited for this task. Figure 7 shows a more detailed,
computerdrawn direction ﬁeld for the differential equation in Example 1. It enables us to
draw, with reasonable accuracy, the solution curves shown in Figure 8 with yintercepts
2, 1, 0, 1, and 2. 1 _2 FIGURE 6 3 3 Module 10.2A shows direction ﬁelds and
solution curves for a variety of differential equations. _3 3 _3 3 _3 _3 F IGURE 7
R E L switch
FIGURE 9 FIGURE 8 Now let’s see how direction ﬁelds give insight into physical situations. The simple electric circuit shown in Figure 9 contains an electromotive force (usually a battery or generator) that produces a voltage of E t volts (V) and a current of I t amperes (A) at time t.
The circuit also contains a resistor with a resistance of R ohms ( ) and an inductor with
an inductance of L henries (H).
Ohm’s Law gives the drop in voltage due to the resistor as RI. The voltage drop due to
the inductor is L dI dt . One of Kirchhoff’s laws says that the sum of the voltage drops is
equal to the supplied voltage E t . Thus, we have
1 L dI
dt RI Et which is a ﬁrstorder differential equation that models the current I at time t . 5E10(pp 622631) 1/18/06 9:18 AM Page 631 S ECTION 10.2 DIRECTION FIELDS AND EULER’S METHOD ❙❙❙❙ 631 E XAMPLE 2 Suppose that in the simple circuit of Figure 9 the resistance is 12 , the
inductance is 4 H, and a battery gives a constant voltage of 60 V.
(a) Draw a direction ﬁeld for Equation 1 with these values.
(b) What can you say about the limiting value of the current?
(c) Identify any equilibrium solutions.
0, use the direc(d) If the switch is closed when t 0 so the current starts with I 0
tion ﬁeld to sketch the solution curve.
SOLUTION (a) If we put L 4, R
4 12, and E t
dI
dt 12 I 60 in Equation 1, we get
60 dI
dt or 15 3I The direction ﬁeld for this differential equation is shown in Figure 10.
I
6 4 2 0 1 2 3 t FIGURE 10 (b) It appears from the direction ﬁeld that all solutions approach the value 5 A, that is,
lim I t tl dI
dt 15 5 (c) It appears that the constant function I t
5 is an equilibrium solution. Indeed, we
can verify this directly from the differential equation. If I t
5, then the left side is
dI dt 0 and the right side is 15 3 5
0.
(d) We use the direction ﬁeld to sketch the solution curve that passes through 0, 0 , as
shown in red in Figure 11. 3I I
6 4 2 0 FIGURE 11 1 2 3 t 5E10(pp 632641) 632 ❙❙❙❙ 1/18/06 9:20 AM Page 632 CHAPTER 10 DIFFERENTIAL EQUATIONS Notice from Figure 10 that the line segments along any horizontal line are parallel. That
is because the independent variable t does not occur on the right side of the equation
I
15 3I . In general, a differential equation of the form
y fy in which the independent variable is missing from the right side, is called autonomous.
For such an equation, the slopes corresponding to two different points with the same
ycoordinate must be equal. This means that if we know one solution to an autonomous
differential equation, then we can obtain inﬁnitely many others just by shifting the graph
of the known solution to the right or left. In Figure 11 we have shown the solutions
that result from shifting the solution curve of Example 2 one and two time units (namely,
seconds) to the right. They correspond to closing the switch when t 1 or t 2. E uler ’s Method
The basic idea behind direction ﬁelds can be used to ﬁnd numerical approximations to
solutions of differential equations. We illustrate the method on the initialvalue problem
that we used to introduce direction ﬁelds:
y
y solution curve 1 y=L(x) 0 FIGURE 12 First Euler approximation 1 x x y y0 1 The differential equation tells us that y 0
0 1 1, so the solution curve has slope
1 at the point 0, 1 . As a ﬁrst approximation to the solution we could use the linear
approximation L x
x 1. In other words, we could use the tangent line at 0, 1 as a
rough approximation to the solution curve (see Figure 12).
Euler’s idea was to improve on this approximation by proceeding only a short distance
along this tangent line and then making a midcourse correction by changing direction as
indicated by the direction ﬁeld. Figure 13 shows what happens if we start out along the
tangent line but stop when x 0.5. (This horizontal distance traveled is called the step
size.) Since L 0.5
1.5, we have y 0.5
1.5 and we take 0.5, 1.5 as the starting point
for a new line segment. The differential equation tells us that y 0.5
0.5 1.5 2, so
we use the linear function
y 1.5 2x 0.5 2x 0.5 as an approximation to the solution for x 0.5 (the orange segment in Figure 13). If we
decrease the step size from 0.5 to 0.25, we get the better Euler approximation shown in
Figure 14.
y 1
0 y 1 1.5
0.5 1 x 0 0.25 1 x FIGURE 13 FIGURE 14 Euler approximation with step size 0.5 Euler approximation with step size 0.25 In general, Euler’s method says to start at the point given by the initial value and proceed in the direction indicated by the direction ﬁeld. Stop after a short time, look at the
slope at the new location, and proceed in that direction. Keep stopping and changing direc 5E10(pp 632641) 1/18/06 9:20 AM Page 633 S ECTION 10.2 DIRECTION FIELDS AND EULER’S METHOD y slope=F(x¸, y¸)
(⁄, ›) hF(x¸, y¸)
h
y¸ 0 ⁄ x FIGURE 15 y2 y1 In general, yn yn h F x 1, y1
h F x n 1, yn 1 1 EXAMPLE 3 Use Euler’s method with step size 0.1 to construct a table of approximate values for the solution of the initialvalue problem
y
SOLUTION We are given that h x y 0.1, x 0 y0 0, y0 y1 y0 h F x 0 , y0 1 y2 y1 h F x 1, y1 1.1 y3
Module 10.2B shows how Euler’s method
works numerically and visually for a
variety of differential equations and
step sizes. 633 tion according to the direction ﬁeld. Euler’s method does not produce the exact solution to
an initialvalue problem—it gives approximations. But by decreasing the step size (and
therefore increasing the number of midcourse corrections), we obtain successively better
approximations to the exact solution. (Compare Figures 12, 13, and 14.)
For the general ﬁrstorder initialvalue problem y
F x, y , y x 0
y0 , our aim is to
ﬁnd approximate values for the solution at equally spaced numbers x 0 , x 1 x 0 h,
x 2 x 1 h, . . . , where h is the step size. The differential equation tells us that the slope
at x 0 , y0 is y
F x 0 , y0 , so Figure 15 shows that the approximate value of the solution
when x x 1 is
y1 y0 h F x 0 , y0
Similarly, x¸ ❙❙❙❙ y2 h F x 2 , y2 1.22 1 1, and F x, y 0.1 0 1 0.1 0.1 x y. So we have 1.1
1.1 0.1 0.2 1.22 1.22 1.362 This means that if y x is the exact solution, then y 0.3
1.362.
Proceeding with similar calculations, we get the values in the table:
n xn yn n xn yn 1
2
3
4
5 0.1
0.2
0.3
0.4
0.5 1.100000
1.220000
1.362000
1.528200
1.721020 6
7
8
9
10 0.6
0.7
0.8
0.9
1.0 1.943122
2.197434
2.487178
2.815895
3.187485 For a more accurate table of values in Example 3 we could decrease the step size. But
for a large number of small steps the amount of computation is considerable and so we
need to program a calculator or computer to carry out these calculations. The following
table shows the results of applying Euler’s method with decreasing step size to the initialvalue problem of Example 3.
Step size
 Computer software packages that produce
numerical approximations to solutions of
differential equations use methods that are
reﬁnements of Euler’s method. Although Euler’s
method is simple and not as accurate, it is the
basic idea on which the more accurate methods
are based. Euler estimate of y 0.5 Euler estimate of y 1 0.500
0.250
0.100
0.050
0.020
0.010
0.005
0.001 1.500000
1.625000
1.721020
1.757789
1.781212
1.789264
1.793337
1.796619 2.500000
2.882813
3.187485
3.306595
3.383176
3.409628
3.423034
3.433848 5E10(pp 632641) 634 ❙❙❙❙ 1/18/06 9:20 AM Page 634 CHAPTER 10 DIFFERENTIAL EQUATIONS Notice that the Euler estimates in the table seem to be approaching limits, namely, the
true values of y 0.5 and y 1 . Figure 16 shows graphs of the Euler approximations with
step sizes 0.5, 0.25, 0.1, 0.05, 0.02, 0.01, and 0.005. They are approaching the exact solution curve as the step size h approaches 0.
y 1 FIGURE 16
0 Euler approximations
approaching the exact solution 0.5 1 x EXAMPLE 4 In Example 2 we discussed a simple electric circuit with resistance
12 , inductance 4 H, and a battery with voltage 60 V. If the switch is closed when
t 0, we modeled the current I at time t by the initialvalue problem dI
dt 15 3I I0 0 Estimate the current in the circuit half a second after the switch is closed.
SOLUTION We use Euler’s method with F t, I h 15 3I, t0 0, I0 0.1 second:
I1 0 0.1 15 I2 1.5 I3 2.55 I4 3.285 I5 3.7995 30 0.1 15 1.5 3 1.5 0.1 15
0.1 15 2.55 3 2.55 3.285 3 3.285 0.1 15 3 3.7995 I 0.5 4.16 A So the current after 0.5 s is 3.7995
4.15965 0, and step size 5E10(pp 632641) 1/18/06 9:20 AM Page 635 SECTION 10.2 DIRECTION FIELDS AND EULER’S METHOD  10.2 ❙❙❙❙ 635 Exercises
y (1 1. A direction ﬁeld for the differential equation y 1
4 y 2) I (iii) y 0 3 (iv) y 0 II y y 2 is shown.
(a) Sketch the graphs of the solutions that satisfy the given
initial conditions.
(i) y 0
(ii) y 0
1
1
3 2 2x _2 2x _2 (b) Find all the equilibrium solutions.
y _2 _2 3 III IV y y 2 2 2 1 _3 _2 _1 0 1 2 3 x 2x _2 2x _2 _1 _2 _2 _3 ■ ■ _2 ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 7. Use the direction ﬁeld labeled I (for Exercises 3–6) to
2. A direction ﬁeld for the differential equation y x sin y is sketch the graphs of the solutions that satisfy the given
initial conditions.
(a) y 0
(b) y 0
(c) y 0
1
0
1 shown.
(a) Sketch the graphs of the solutions that satisfy the given
initial conditions.
(i) y 0
(ii) y 0
(iii) y 0
1
2
(iv) y 0 4 (v) y 0 8. Repeat Exercise 7 for the direction ﬁeld labeled III.
9–10  Sketch a direction ﬁeld for the differential equation. Then
use it to sketch three solution curves. 5 (b) Find all the equilibrium solutions. 9. y
■ y 1
■ x2 10. y y
■ ■ ■ ■ ■ ■ y2
■ ■ ■ ■ 11–14  Sketch the direction ﬁeld of the differential equation.
Then use it to sketch a solution curve that passes through the
given point. 5 4 11. y
3 13. y
■ 2 x, y
y
■ x y,
■ 12. y 1, 0 14. y 0, 1
■ 1 ■ ■ ■ x
■ x y, 0, 0 x y, 1, 0 ■ ■ ■ ■ 2 C AS
1 _3 _2 _1 0 1 2 3 x 15–16  Use a computer algebra system to draw a direction ﬁeld
for the given differential equation. Get a printout and sketch on it
the solution curve that passes through 0, 1 . Then use the CAS to
draw the solution curve and compare it with your sketch. 15. y
■ 3–6  Match the differential equation with its direction ﬁeld
(labeled I–IV). Give reasons for your answer. 3. y
5. y y
y 4. y 1
2 x 2 6. y y
y x
3 x3 CAS 16. y y sin 2 x
■ ■ ■ ■ ■ ■ sin x
■ ■ y
■ ■ ■ 17. Use a computer algebra system to draw a direction ﬁeld for the differential equation y
y 3 4y. Get a printout and sketch
on it solutions that satisfy the initial condition y 0
c for
various values of c. For what values of c does lim t l y t
exist? What are the possible values for this limit? 5E10(pp 632641) 636 ❙❙❙❙ 1/18/06 9:20 AM Page 636 CHAPTER 10 DIFFERENTIAL EQUATIONS 22. Use Euler’s method with step size 0.2 to estimate y 1 , where 18. Make a rough sketch of a direction ﬁeld for the autonomous y x is the solution of the initialvalue problem y
y0
0. differential equation y
f y , where the graph of f is as
shown. How does the limiting behavior of solutions depend
on the value of y 0 ? 1 x y, 23. Use Euler’s method with step size 0.1 to estimate y 0.5 , where y x is the solution of the initialvalue problem y
y0
1. f(y) y x y, 24. (a) Use Euler’s method with step size 0.2 to estimate y 1.4 , _2 _1 0 1 2 where y x is the solution of the initialvalue problem
y
x x y, y 1
0.
(b) Repeat part (a) with step size 0.1. y ; 25. (a) Program a calculator or computer to use Euler’s method to
compute y 1 , where y x is the solution of the initialvalue
problem
19. (a) Use Euler’s method with each of the following step sizes to estimate the value of y 0.4 , where y is the solution of the
initialvalue problem y
y, y 0
1.
(i) h 0.4
(ii) h 0.2
(iii) h 0.1
(b) We know that the exact solution of the initialvalue
problem in part (a) is y e x. Draw, as accurately as you
can, the graph of y e x, 0 x 0.4, together with the
Euler approximations using the step sizes in part (a).
(Your sketches should resemble Figures 12, 13, and 14.)
Use your sketches to decide whether your estimates in
part (a) are underestimates or overestimates.
(c) The error in Euler’s method is the difference between
the exact value and the approximate value. Find the errors
made in part (a) in using Euler’s method to estimate the
true value of y 0.4 , namely e 0.4. What happens to the error
each time the step size is halved?
20. A direction ﬁeld for a differential equation is shown. Draw, with a ruler, the graphs of the Euler approximations to the
solution curve that passes through the origin. Use step sizes
h 1 and h 0.5. Will the Euler estimates be underestimates
or overestimates? Explain.
y dy
dx
(i) h
(iii) h 3x 2 y 6x 2 (ii) h
(iv) h 1
0.01 y0 3 0.1
0.001 3 (b) Verify that y 2 e x is the exact solution of the differential equation.
(c) Find the errors in using Euler’s method to compute y 1
with the step sizes in part (a). What happens to the error
when the step size is divided by 10?
C AS 26. (a) Program your computer algebra system, using Euler’s method with step size 0.01, to calculate y 2 , where y
is the solution of the initialvalue problem
x3 y y3 y0 1 (b) Check your work by using the CAS to draw the solution
curve.
27. The ﬁgure shows a circuit containing an electromotive force, a capacitor with a capacitance of C farads (F), and a resistor with
a resistance of R ohms ( ). The voltage drop across the capacitor is Q C, where Q is the charge (in coulombs), so in this case
Kirchhoff’s Law gives 2 RI
But I Q
C Et dQ dt, so we have
R 1 dQ
dt 1
Q
C Et Suppose the resistance is 5 , the capacitance is 0.05 F, and a
battery gives a constant voltage of 60 V.
(a) Draw a direction ﬁeld for this differential equation.
(b) What is the limiting value of the charge?
C
0 1 2x 21. Use Euler’s method with step size 0.5 to compute the approxi mate yvalues y1, y2 , y3 , and y4 of the solution of the initialvalue problem y
y 2 x, y 1
0. E R 5E10(pp 632641) 1/18/06 9:20 AM Page 637 S ECTION 10.3 SEPARABLE EQUATIONS (c) Is there an equilibrium solution?
(d) If the initial charge is Q 0
0 C, use the direction ﬁeld to
sketch the solution curve.
(e) If the initial charge is Q 0
0 C, use Euler’s method with
step size 0.1 to estimate the charge after half a second. 637 at a rate of 1 C per minute when its temperature is 70 C.
(a) What does the differential equation become in this case?
(b) Sketch a direction ﬁeld and use it to sketch the solution
curve for the initialvalue problem. What is the limiting
value of the temperature?
(c) Use Euler’s method with step size h 2 minutes to
estimate the temperature of the coffee after 10 minutes. 28. In Exercise 14 in Section 10.1 we considered a 95 C cup of cof fee in a 20 C room. Suppose it is known that the coffee cools  10.3 ❙❙❙❙ Separable Equations
We have looked at ﬁrstorder differential equations from a geometric point of view (direction ﬁelds) and from a numerical point of view (Euler’s method). What about the symbolic
point of view? It would be nice to have an explicit formula for a solution of a differential
equation. Unfortunately, that is not always possible. But in this section we examine a certain type of differential equation that can be solved explicitly.
A separable equation is a ﬁrstorder differential equation in which the expression for
d y d x can be factored as a function of x times a function of y. In other words, it can be
written in the form
dy
txf y
dx
The name separable comes from the fact that the expression on the right side can be “separated” into a function of x and a function of y. Equivalently, if f y
0, we could write
dy
dx 1 where h y 1 f y . To solve this equation we rewrite it in the differential form
h y dy  The technique for solving separable differential equations was ﬁrst used by James Bernoulli
(in 1690) in solving a problem about pendulums
and by Leibniz (in a letter to Huygens in 1691).
John Bernoulli explained the general method in a
paper published in 1694. tx
hy t x dx so that all y’s are on one side of the equation and all x’s are on the other side. Then we integrate both sides of the equation: yh y 2 dy ytx dx Equation 2 deﬁnes y implicitly as a function of x. In some cases we may be able to solve
for y in terms of x.
We use the Chain Rule to justify this procedure: If h and t satisfy (2), then
d
dx
so d
dy yh y yh y and
Thus, Equation 1 is satisﬁed. dy d
dx dy dy
dx tx hy dy
dx tx ytx dx 5E10(pp 632641) 638 ❙❙❙❙ 1/18/06 9:20 AM Page 638 CHAPTER 10 DIFFERENTIAL EQUATIONS EXAMPLE 1 dy
x2
.
dx
y2
(b) Find the solution of this equation that satisﬁes the initial condition y 0
(a) Solve the differential equation  Figure 1 shows graphs of several members
of the family of solutions of the differential
equation in Example 1. The solution of the initialvalue problem in part (b) is shown in red. 2. SOLUTION (a) We write the equation in terms of differentials and integrate both sides:
y 2 dy x 2 dx 3 _3 3 2 dy yx 1
3 yy y3 1
3 2 dx x3 C where C is an arbitrary constant. (We could have used a constant C1 on the left side and
another constant C 2 on the right side. But then we could combine these constants by
writing C C 2 C1.)
Solving for y, we get _3 3
sx 3 y FIGURE 1 3C We could leave the solution like this or we could write it in the form
3
sx 3 y K where K 3C. (Since C is an arbitrary constant, so is K .)
(b) If we put x 0 in the general solution in part (a), we get y 0
3
2 and so K 8.
initial condition y 0
2, we must have sK
Thus, the solution of the initialvalue problem is
3
sx 3 y  Some computer algebra systems can plot
curves deﬁned by implicit equations. Figure 2
shows the graphs of several members of the
family of solutions of the differential equation
in Example 2. As we look at the curves from left
to right, the values of C are 3, 2, 1, 0, 1, 2,
and 3. EXAMPLE 2 Solve the differential equation 3 _4 FIGURE 2 6x 2
.
2y cos y dy
dx 2y y 2 8 SOLUTION Writing the equation in differential form and integrating both sides, we have 4 _2 3
sK . To satisfy the cos y d y 6x 2 dx 2y cos y d y y 6x y2 sin y 2x 3 2 dx
C where C is a constant. Equation 3 gives the general solution implicitly. In this case it’s
impossible to solve the equation to express y explicitly as a function of x.
EXAMPLE 3 Solve the equation y x 2 y. SOLUTION First we rewrite the equation using Leibniz notation: dy
dx x2y 5E10(pp 632641) 1/18/06 9:20 AM Page 639 S ECTION 10.3 SEPARABLE EQUATIONS  If a solution y is a function that satisﬁes
yx
0 for some x, it follows from a
uniqueness theorem for solutions of differential
equations that y x
0 for all x. If y ❙❙❙❙ 639 0, we can rewrite it in differential notation and integrate:
dy
y x 2 dx dy
y yx ln y x3
3 y 2 y 0 dx
C This equation deﬁnes y implicitly as a function of x. But in this case we can solve
explicitly for y as follows:
e ln y y so e x3 3 C e Ce x y 3 e Ce x 3 3 3 We can easily verify that the function y 0 is also a solution of the given differential
equation. So we can write the general solution in the form
y Ae x e C, or A where A is an arbitrary constant ( A 3 3 e C, or A 0). y
6
4  Figure 3 shows a direction ﬁeld for the differential equation in Example 3. Compare it with
Figure 4, in which we use the equation
3
y A e x / 3 to graph solutions for several values
of A. If you use the direction ﬁeld to sketch
solution curves with yintercepts 5, 2, 1, 1,
and 2, they will resemble the curves in
Figure 4. 6
2 _2 _1 0 1 2 x
_2 _2 2 _4
_6 _6 FIGURE 4 FIGURE 3
R E L switch
FIGURE 5 EXAMPLE 4 In Section 10.2 we modeled the current I t in the electric circuit shown in
Figure 5 by the differential equation L dI
dt RI Et Find an expression for the current in a circuit where the resistance is 12 , the inductance is 4 H, a battery gives a constant voltage of 60 V, and the switch is turned on when
t 0. What is the limiting value of the current?
SOLUTION With L 4, R 12, and E t
4 60, the equation becomes
dI
dt 12I 60 5E10(pp 632641) 640 ❙❙❙❙ 1/18/06 9:20 AM Page 640 CHAPTER 10 DIFFERENTIAL EQUATIONS dI
dt or 15 3I and the initialvalue problem is
dI
dt 15 3I I0 0 We recognize this equation as being separable, and we solve it as follows: y 15
1
3 dI y dt 3I ln 15 3I t 15 3I e 0 3t C 15 3I 3C e I 1
3 5 6 Since I 0 3I C  Figure 6 shows how the solution in
Example 4 (the current) approaches its limiting
value. Comparison with Figure 11 in Section 10.2
shows that we were able to draw a fairly accurate solution curve from the direction ﬁeld. y=5 15 0, we have 5 1
3 A 0, so A
It e 3t Ae Ae 3t 3t 15 and the solution is
5 5e 3t The limiting current, in amperes, is
0 lim I t tl FIGURE 6 lim 5 5e 5 2.5 3t 0 5 5 lim e 3t 5 tl tl Orthogonal Trajectories
An orthogonal trajectory of a family of curves is a curve that intersects each curve of the
family orthogonally, that is, at right angles (see Figure 7). For instance, each member of
the family y m x of straight lines through the origin is an orthogonal trajectory of the
family x 2 y 2 r 2 of concentric circles with center the origin (see Figure 8). We say that
the two families are orthogonal trajectories of each other.
y x orthogonal
trajectory
FIGURE 7 FIGURE 8 EXAMPLE 5 Find the orthogonal trajectories of the family of curves x k y 2, where k is an arbitrary constant.
k y 2 form a family of parabolas whose axis of symmetry is
the xaxis. The ﬁrst step is to ﬁnd a single differential equation that is satisﬁed by all
SOLUTION The curves x 5E10(pp 632641) 1/18/06 9:20 AM Page 641 S ECTION 10.3 SEPARABLE EQUATIONS 2 ky dy
dx 641 k y 2, we get members of the family. If we differentiate x
1 ❙❙❙❙ dy
dx or 1
2 ky This differential equation depends on k, but we need an equation that is valid for all
values of k simultaneously. To eliminate k we note that, from the equation of the given
general parabola x k y 2, we have k x y 2 and so the differential equation can be
written as
dy
dx
dy
dx or 1
2ky 1
x
2 2y
y y
2x This means that the slope of the tangent line at any point x, y on one of the parabolas is
y
y 2 x . On an orthogonal trajectory the slope of the tangent line must be the negative reciprocal of this slope. Therefore, the orthogonal trajectories must satisfy the differential equation
dy
dx y 2x
y This differential equation is separable, and we solve it as follows: y y dy
y2
2 x x2 4 FIGURE 9 y 2 x dx y2
2 x2 C C where C is an arbitrary positive constant. Thus, the orthogonal trajectories are the family
of ellipses given by Equation 4 and sketched in Figure 9.
Orthogonal trajectories occur in various branches of physics. For example, in an electrostatic ﬁeld the lines of force are orthogonal to the lines of constant potential. Also, the
streamlines in aerodynamics are orthogonal trajectories of the velocityequipotential
curves. Mixing Problems
A typical mixing problem involves a tank of ﬁxed capacity ﬁlled with a thoroughly mixed
solution of some substance, such as salt. A solution of a given concentration enters the tank
at a ﬁxed rate and the mixture, thoroughly stirred, leaves at a ﬁxed rate, which may differ
from the entering rate. If y t denotes the amount of substance in the tank at time t, then
y t is the rate at which the substance is being added minus the rate at which it is being
removed. The mathematical description of this situation often leads to a ﬁrstorder separable differential equation. We can use the same type of reasoning to model a variety of
phenomena: chemical reactions, discharge of pollutants into a lake, injection of a drug into
the bloodstream. 5E10(pp 642651) 642 ❙❙❙❙ 1/18/06 9:21 AM Page 642 CHAPTER 10 DIFFERENTIAL EQUATIONS EXAMPLE 6 A tank contains 20 kg of salt dissolved in 5000 L of water. Brine that contains 0.03 kg of salt per liter of water enters the tank at a rate of 25 L min. The solution
is kept thoroughly mixed and drains from the tank at the same rate. How much salt
remains in the tank after half an hour?
SOLUTION Let y t be the amount of salt (in kilograms) after t minutes. We are given that y0
20 and we want to ﬁnd y 30 . We do this by ﬁnding a differential equation satisﬁed by y t . Note that dy dt is the rate of change of the amount of salt, so
dy
dt 5 rate in rate out where (rate in) is the rate at which salt enters the tank and (rate out) is the rate at which
salt leaves the tank. We have
rate in kg
L 0.03 25 L
min 0.75 kg
min The tank always contains 5000 L of liquid, so the concentration at time t is y t 5000
(measured in kilograms per liter). Since the brine ﬂows out at a rate of 25 L min, we
have
y t kg
5000 L rate out 25 L
min y t kg
200 min Thus, from Equation 5 we get
dy
dt yt
200 0.75 150 y t
200 Solving this separable differential equation, we obtain
dy y 150
ln 150  Figure 10 shows the graph of the function
y t of Example 6. Notice that, as time goes by,
the amount of salt approaches 150 kg. Since y 0 20, we have y
y ln 130 y dt
200 t
200 C C, so ln 150 y t
200 150 y 130e y ln 130
t 200 150 Therefore 100 Since y t is continuous and y 0
20 and the right side is never 0, we deduce that
150 y t is always positive. Thus, 150 y
150 y and so 50 yt
0 FIGURE 10 200 400 t 150 130e t 200 The amount of salt after 30 min is
y 30 150 130e 30 200 38.1 kg 5E10(pp 642651) 1/18/06 9:21 AM Page 643 S ECTION 10.3 SEPARABLE EQUATIONS  10.3
1–10 1.  3. x 2
5. 1 y
x
1y
tan y y x dy
dt
du
dt 2 1 ■ 2u
■ 1
1 ■ dz
dt 10. tu
■ ■ ■ sr
su et
■ z 0
■ ■ ■ ■  Find the solution of the differential equation that satisﬁes
the given initial condition. dy
dx dy
12.
dx y2 dP
dt 15. du
dt 16. dy
dt sPt, ■ ■ 0 y0 sec 2t
,
2u
y1 ■ ■ ■ ■ ■ ■ ■ ■  Find the orthogonal trajectories of the family of curves.
Use a graphing device to draw several members of each family on
a common screen. kx 2 28. x 2 x
■ k
■ 1 y2 30. y
■ ■ ■ ke ■ ■ k
x
■ ■ ■ ■ tion that models the temperature of a 95 C cup of coffee in a
20 C room. Solve the differential equation to ﬁnd an expression
for the temperature of the coffee at time t. 5 0 33. In Exercise 13 in Section 10.1 we formulated a model for 3 y1 a, 0 x learning in the form of the differential equation 2 1
■ ■ ■ ■ ■ ■ ■ ■ 4 x 3 y and whose yintercept is 7.
20. Find an equation of the curve that passes through the point 1, 1 and whose slope at x, y is y 2 x 3.
21. (a) Solve the differential equation y (b) Solve the initialvalue problem y
and graph the solution.
(c) Does the initialvalue problem y
have a solution? Explain. 2 x s1
2 x s1 y 2.
y 2, y 0 0, 2 x s1 y 2, y 0 2, y
; 22. Solve the equation e y cos x 0 and graph several members of the family of solutions. How does the solution curve
change as the constant C varies? CAS ■ 31. Solve the initialvalue problem in Exercise 27 in Section 10.2 19. Find an equation of the curve that satisﬁes dy d x ; ■ 32. In Exercise 28 in Section 10.2 we discussed a differential equa y, y y 2, y ■ to ﬁnd an expression for the charge at time t. Find the limiting
value of the charge. u0 te y, 0 2 2t y0 ■ ; 27–30 29. y 1 P1 ■ x2 y 26. y 1y 27. y e 3y y , a ■ ■ 2y 17. y tan x
18. x y y1 y cos x
,
1
y2 13. x cos x
14. 1,  25. y 11–18 11. 25–26 (a) Use a computer algebra system to draw a direction ﬁeld
for the differential equation. Get a printout and use it to sketch
some solution curves without solving the differential equation.
(b) Solve the differential equation.
(c) Use the CAS to draw several members of the family of solutions
obtained in part (b). Compare with the curves from part (a). xy
2 ln y 8. y t CAS y 2 sin x du
6.
dr te t
y s1 y 2 9. 2 24. Solve the equation y e 2x
4y 3 4. y xy x s x 2 1 ye y and graph several
members of the family of solutions (if your CAS does implicit
plots). How does the solution curve change as the constant C
varies? C AS dy
dx 2. 7. ■ 643 Exercises Solve the differential equation. dy
dx ❙❙❙❙ 23. Solve the initialvalue problem y sin x sin y, y 0
and graph the solution (if your CAS does implicit plots). 2, dP
dt kM P where P t measures the performance of someone learning a
skill after a training time t, M is the maximum level of performance, and k is a positive constant. Solve this differential
equation to ﬁnd an expression for P t . What is the limit of this
expression?
34. In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C:
A B l C. The law of mass action states that the rate
of reaction is proportional to the product of the concentrations
of A and B:
dC
dt kA B (See Example 4 in Section 3.4.) Thus, if the initial concentrations are A
a moles L and B
b moles L and we write 5E10(pp 642651) ❙❙❙❙ 644 x 1/18/06 9:21 AM CHAPTER 10 DIFFERENTIAL EQUATIONS C , then we have
dx
dt CAS Page 644 ka xb x (a) Assuming that a b, ﬁnd x as a function of t. Use the fact
that the initial concentration of C is 0.
(b) Find x t assuming that a b. How does this expression
for x t simplify if it is known that C
a 2 after
20 seconds?
35. In contrast to the situation of Exercise 34, experiments show that the reaction H 2 Br 2 l 2HBr satisﬁes the rate law d HBr
dt k H 2 Br 2 12 and so for this reaction the differential equation becomes
dx
dt ka xb x 12 where x
HBr and a and b are the initial concentrations of
hydrogen and bromine.
(a) Find x as a function of t in the case where a b. Use the
fact that x 0
0.
(b) If a b, ﬁnd t as a function of x. [Hint: In performing the
integration, make the substitution u sb x .]
36. A sphere with radius 1 m has temperature 15 C. It lies inside a concentric sphere with radius 2 m and temperature 25 C. The
temperature T r at a distance r from the common center of the
spheres satisﬁes the differential equation
d 2T
dr 2 2 dT
r dr 0 If we let S d T d r, then S satisﬁes a ﬁrstorder differential
equation. Solve it to ﬁnd an expression for the temperature T r
between the spheres.
37. A glucose solution is administered intravenously into the bloodstream at a constant rate r. As the glucose is added, it is
converted into other substances and removed from the bloodstream at a rate that is proportional to the concentration at that
time. Thus, a model for the concentration C C t of the glucose solution in the bloodstream is
dC
dt r kC where k is a positive constant.
(a) Suppose that the concentration at time t 0 is C0.
Determine the concentration at any time t by solving the
differential equation.
(b) Assuming that C0 r k, ﬁnd lim t l C t and interpret
your answer.
38. A certain small country has $10 billion in paper currency in circulation, and each day $50 million comes into the country’s
banks. The government decides to introduce new currency by
having the banks replace old bills with new ones whenever old currency comes into the banks. Let x x t denote the amount
of new currency in circulation at time t, with x 0
0.
(a) Formulate a mathematical model in the form of an initialvalue problem that represents the “ﬂow” of the new
currency into circulation.
(b) Solve the initialvalue problem found in part (a).
(c) How long will it take for the new bills to account for 90%
of the currency in circulation?
39. A tank contains 1000 L of brine with 15 kg of dissolved salt. Pure water enters the tank at a rate of 10 L min. The solution
is kept thoroughly mixed and drains from the tank at the same
rate. How much salt is in the tank (a) after t minutes and
(b) after 20 minutes?
40. A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of
5 L min. Brine that contains 0.04 kg of salt per liter of water
enters the tank at a rate of 10 L min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L min.
How much salt is in the tank (a) after t minutes and (b) after
one hour?
41. When a raindrop falls, it increases in size and so its mass at time t is a function of t, m t . The rate of growth of the mass is
km t for some positive constant k. When we apply Newton’s
Law of Motion to the raindrop, we get mv
t m, where v is
the velocity of the raindrop (directed downward) and t is the
acceleration due to gravity. The terminal velocity of the
raindrop is lim t l v t . Find an expression for the terminal
velocity in terms of t and k.
42. An object of mass m is moving horizontally through a medium which resists the motion with a force that is a function of the
velocity; that is,
m d 2s
dt 2 m dv
dt fv where v v t and s s t represent the velocity and position
of the object at time t, respectively. For example, think of a
boat moving through the water.
(a) Suppose that the resisting force is proportional to the
velocity, that is, f v
k v, k a positive constant. (This
model is appropriate for small values of v.) Let v 0
v0
and s 0
s0 be the initial values of v and s. Determine v
and s at any time t . What is the total distance that the
object travels from time t 0?
(b) For larger values of v a better model is obtained by supposing that the resisting force is proportional to the square
of the velocity, that is, f v
k v 2, k 0. (This model
was ﬁrst proposed by Newton.) Let v0 and s0 be the initial
values of v and s. Determine v and s at any time t . What is
the total distance that the object travels in this case?
43. Let A t be the area of a tissue culture at time t and let M be the ﬁnal area of the tissue when growth is complete. Most cell
divisions occur on the periphery of the tissue and the number
of cells on the periphery is proportional to sA t . So a reason 5E10(pp 642651) 1/18/06 9:21 AM Page 645 A PPLIED PROJECT HOW FAST DOES A TANK DRAIN? C AS able model for the growth of tissue is obtained by assuming
that the rate of growth of the area is jointly proportional to
sA t and M A t .
(a) Formulate a differential equation and use it to show that
the tissue grows fastest when A t
M 3.
(b) Solve the differential equation to ﬁnd an expression
for A t . Use a computer algebra system to perform the
integration. ❙❙❙❙ 645 and so
m dv
dt mtR 2
xR 2 (a) Suppose a rocket is ﬁred vertically upward with an initial
velocity v0. Let h be the maximum height above the surface
reached by the object. Show that 44. According to Newton’s Law of Universal Gravitation, the gravitational force on an object of mass m that has been
projected vertically upward from Earth’s surface is
F mtR 2
xR 2 where x x t is the object’s distance above the surface at
time t , R is Earth’s radius, and t is the acceleration due to
gravity. Also, by Newton’s Second Law, F ma m dv dt v0 2 tRh
Rh [Hint: By the Chain Rule, m dv dt
mv dv d x .]
(b) Calculate ve lim h l v0. This limit is called the escape
velocity for Earth.
(c) Use R 3960 mi and t 32 ft s2 to calculate ve in feet
per second and in miles per second. A PPLIED PROJECT
H ow Fast Does a Tank Drain?
If water (or other liquid) drains from a tank, we expect that the ﬂow will be greatest at ﬁrst
(when the water depth is greatest) and will gradually decrease as the water level decreases. But
we need a more precise mathematical description of how the ﬂow decreases in order to answer
the kinds of questions that engineers ask: How long does it take for a tank to drain completely?
How much water should a tank hold in order to guarantee a certain minimum water pressure for
a sprinkler system?
Let h t and V t be the height and volume of water in a tank at time t. If water leaks through
a hole with area a at the bottom of the tank, then Torricelli’s Law says that
dV
dt 1 a s2t h where t is the acceleration due to gravity. So the rate at which water ﬂows from the tank is proportional to the square root of the water height.
1. (a) Suppose the tank is cylindrical with height 6 ft and radius 2 ft and the hole is circular with radius 1 in. If we take t 32 ft s2, show that y satisﬁes the differential equation
dh
dt 1
sh
72 (b) Solve this equation to ﬁnd the height of the water at time t, assuming the tank is full at
time t 0.
(c) How long will it take for the water to drain completely?
2. Because of the rotation and viscosity of the liquid, the theoretical model given by Equation 1 isn’t quite accurate. Instead, the model
2 dh
dt k sh 5E10(pp 642651) 646 ❙❙❙❙ 1/18/06 9:21 AM Page 646 CHAPTER 10 DIFFERENTIAL EQUATIONS  This part of the project is best done as a
classroom demonstration or as a group project
with three students in each group: a timekeeper
to call out seconds, a bottle keeper to estimate
the height every 10 seconds, and a record keeper
to record these values. is often used and the constant k (which depends on the physical properties of the liquid) is
determined from data concerning the draining of the tank.
(a) Suppose that a hole is drilled in the side of a cylindrical bottle and the height h of the
water (above the hole) decreases from 10 cm to 3 cm in 68 seconds. Use Equation 2 to
ﬁnd an expression for h t . Evaluate h t for t 10, 20, 30, 40, 50, 60.
(b) Drill a 4mm hole near the bottom of the cylindrical part of a twoliter plastic softdrink
bottle. Attach a strip of masking tape marked in centimeters from 0 to 10, with 0 corresponding to the top of the hole. With one ﬁnger over the hole, ﬁll the bottle with water
to the 10cm mark. Then take your ﬁnger off the hole and record the values of h t for
t 10, 20, 30, 40, 50, 60 seconds. (You will probably ﬁnd that it takes 68 seconds for
the level to decrease to h 3 cm.) Compare your data with the values of h t from
part (a). How well did the model predict the actual values?
3. In many parts of the world, the water for sprinkler systems in large hotels and hospitals is supplied by gravity from cylindrical tanks on or near the roofs of the buildings. Suppose
such a tank has radius 10 ft and the diameter of the outlet is 2.5 inches. An engineer has to
guarantee that the water pressure will be at least 2160 lb ft 2 for a period of 10 minutes.
(When a ﬁre happens, the electrical system might fail and it could take up to 10 minutes for
the emergency generator and ﬁre pump to be activated.) What height should the engineer
specify for the tank in order to make such a guarantee? (Use the fact that the water pressure
at a depth of d feet is P 62.5 d. See Section 9.3.)
4. Not all water tanks are shaped like cylinders. Suppose a tank has crosssectional area A h at height h. Then the volume of water up to height h is V
Theorem of Calculus gives dV dh A h . It follows that
dV
dt dV dh
dh dt Ah x0h A u du and so the Fundamental dh
dt and so Torricelli’s Law becomes
Ah dh
dt a s2 th (a) Suppose the tank has the shape of a sphere with radius 2 m and is initially half full of
water. If the radius of the circular hole is 1 cm and we take t 10 m s2, show that h
satisﬁes the differential equation
4h h2 dh
dt 0.0001 s20h (b) How long will it take for the water to drain completely? APPLIED PROJECT
Which Is Faster, Going Up or Coming Down?
Suppose you throw a ball into the air. Do you think it takes longer to reach its maximum
height or to fall back to Earth from its maximum height? We will solve the problem in this project but, before getting started, think about that situation and make a guess based on your physical intuition.
1. A ball with mass m is projected vertically upward from Earth’s surface with a positive initial
velocity v0. We assume the forces acting on the ball are the force of gravity and a retarding force of air resistance with direction opposite to the direction of motion and with magnitude
p v t , where p is a positive constant and v t is the velocity of the ball at time t. In both 5E10(pp 642651) 1/18/06 9:21 AM Page 647 S ECTION 10.4 EXPONENTIAL GROWTH AND DECAY  In modeling force due to air resistance,
various functions have been used, depending
on the physical characteristics and speed of
the ball. Here we use a linear model, pv,
but a quadratic model ( pv 2 on the way up
and pv 2 on the way down) is another possibility for higher speeds (see Exercise 42 in
Section 10.3). For a golf ball, experiments
have shown that a good model is pv 1.3
going up and p v 1.3 coming down. But no
matter which force function f v is used
0
[where f v
0 for v 0 and f v
for v 0], the answer to the question
remains the same. See F. Brauer, “What
Goes Up Must Come Down, Eventually,”
Amer. Math. Monthly 108 (2001),
pp. 437–440. ❙❙❙❙ 647 the ascent and the descent, the total force acting on the ball is pv m t. [During ascent,
v t is positive and the resistance acts downward; during descent, v t is negative and the
resistance acts upward.] So, by Newton’s Second Law, the equation of motion is
mv pv mt Solve this differential equation to show that the velocity is
vt mt
e
p v0 mt
p pt m 2. Show that the height of the ball, until it hits the ground, is yt v0 mt
p m
1
p e pt m mtt
p 3. Let t1 be the time that the ball takes to reach its maximum height. Show that t1 m
mt pv0
ln
p
mt Find this time for a ball with mass 1 kg and initial velocity 20 m s. Assume the air
1
resistance is 10 of the speed. ; 4. Let t2 be the time at which the ball falls back to Earth. For the particular ball in Problem 3,
estimate t2 by using a graph of the height function y t . Which is faster, going up or coming
down?
5. In general, it’s not easy to ﬁnd t2 because it’s impossible to solve the equation y t 0
explicitly. We can, however, use an indirect method to determine whether ascent or descent
is faster; we determine whether y 2 t1 is positive or negative. Show that
m 2t
p2 y 2t1
where x e pt1 m. Then show that x
fx 1
x x 2 ln x 1 and the function
x 1
x 2 ln x is increasing for x 1. Use this result to decide whether y 2 t1 is positive or negative. What
can you conclude? Is ascent or descent faster?  10.4 Exponential Growth and Decay
One of the models for population growth that we considered in Section 10.1 was based
on the assumption that the population grows at a rate proportional to the size of the
population:
dP
kP
dt
Is that a reasonable assumption? Suppose we have a population (of bacteria, for instance) 5E10(pp 642651) 648 ❙❙❙❙ 1/18/06 9:21 AM Page 648 CHAPTER 10 DIFFERENTIAL EQUATIONS 300 bacteria per
with size P 1000 and at a certain time it is growing at a rate of P
hour. Now let’s take another 1000 bacteria of the same type and put them with the ﬁrst population. Each half of the new population was growing at a rate of 300 bacteria per hour.
We would expect the total population of 2000 to increase at a rate of 600 bacteria per
hour initially (provided there’s enough room and nutrition). So if we double the size, we
double the growth rate. In general, it seems reasonable that the growth rate should be proportional to the size.
The same assumption applies in other situations as well. In nuclear physics, the mass
of a radioactive substance decays at a rate proportional to the mass. In chemistry, the rate
of a unimolecular ﬁrstorder reaction is proportional to the concentration of the substance.
In ﬁnance, the value of a savings account with continuously compounded interest increases
at a rate proportional to that value.
In general, if y t is the value of a quantity y at time t and if the rate of change of y with
respect to t is proportional to its size y t at any time, then
dy
dt 1 ky where k is a constant. Equation 1 is sometimes called the law of natural growth (if k 0)
or the law of natural decay (if k 0). Because it is a separable differential equation we
can solve it by the methods of Section 10.3: y dy
y y k dt ln y kt y e kt y C
C e Ce kt Ae kt e C or 0) is an arbitrary constant. To see the signiﬁcance of the constant A,
where A (
we observe that
y0
Ae k 0 A
Therefore, A is the initial value of the function.
Because Equation 1 occurs so frequently in nature, we summarize what we have just
proved for future use.
2 The solution of the initialvalue problem
dy
dt ky y0
y0 e kt yt is y0 Population Growth
What is the signiﬁcance of the proportionality constant k ? In the context of population
growth, we can write
3 dP
dt kP or 1 dP
P dt k 5E10(pp 642651) 1/18/06 9:21 AM Page 649 SECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 649 The quantity
1 dP
P dt
is the growth rate divided by the population size; it is called the relative growth rate.
According to (3), instead of saying “the growth rate is proportional to population size” we
could say “the relative growth rate is constant.” Then (2) says that a population with constant relative growth rate must grow exponentially. Notice that the relative growth rate k
appears as the coefﬁcient of t in the exponential function y0 e kt. For instance, if
dP
dt 0.02P and t is measured in years, then the relative growth rate is k
0.02 and the population
grows at a rate of 2% per year. If the population at time 0 is P0 , then the expression for the
population is
Pt
TABLE 1 Year Population
(millions) 1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000 1650
1750
1860
2070
2300
2560
3040
3710
4450
5280
6080 P0 e 0.02 t EXAMPLE 1 Assuming that the growth rate is proportional to population size, use the data
in Table 1 to model the population of the world in the 20th century. What is the relative
growth rate? How well does the model ﬁt the data?
SOLUTION We measure the time t in years and let t
0 in the year 1900. We measure the
population P t in millions of people. Then the initial condition is P 0
1650. We are
assuming that the growth rate is proportional to population size, so the initialvalue problem is dP
dt kP P0 1650 From (2) we know that the solution is
Pt 1650e kt One way to estimate the relative growth rate k is to use the fact that the population in
1910 was 1750 million. Therefore
1650e k 10 P 10 1750 We solve this equation for k :
e 10k k 1750
1650
1
1750
ln
10
1650 0.005884 Thus, the relative growth rate is about 0.6% per year and the model becomes
Pt 1650e 0.005884t 5E10(pp 642651) 650 ❙❙❙❙ 1/18/06 9:21 AM Page 650 CHAPTER 10 DIFFERENTIAL EQUATIONS Table 2 and Figure 1 allow us to compare the predictions of this model with the actual
data. You can see that the predictions become quite inaccurate after about 30 years and
they underestimate by a factor of more than 2 in 2000.
TABLE 2 P Year Model Population 1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000 1650
1750
1856
1969
2088
2214
2349
2491
2642
2802
2972 1650
1750
1860
2070
2300
2560
3040
3710
4450
5280
6080 6000 Population
(in millions)
P=1650e 0.005884t 20 40 60 80 100 t Years since 1900
FIGURE 1 A possible model for world population growth Another possibility for estimating k would be to use the given population for 1950,
for instance, instead of 1910. Then
P 50  In Sections 7.2 and 7.4* we modeled the
same data with an exponential function, but
there we used the method of least squares. k 1650e 50k 2560 1
2560
ln
50
1650 0.0087846 The estimate for the relative growth rate is now 0.88% per year and the model is
Pt 1650e 0.0087846t The predictions with this second model are shown in Table 3 and Figure 2. This exponential model is more accurate over a longer period of time, but it too lags behind reality
in recent years.
TABLE 3 P Year Model Population 1900
1910
1920
1930
1940
1950
1960
1970
1980
1990
2000 1650
1802
1967
2148
2345
2560
2795
3052
3332
3638
3972 1650
1750
1860
2070
2300
2560
3040
3710
4450
5280
6080 6000 Population
(in millions) P=1650e 0.0087846t 20 40 60 80 Years since 1900
FIGURE 2 Another model for world population growth 100 t 5E10(pp 642651) 1/18/06 9:21 AM Page 651 S ECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 651 EXAMPLE 2 Use the data in Table 1 to model the population of the world in the second
half of the 20th century. Use the model to estimate the population in 1993 and to predict
the population in the year 2010.
SOLUTION Here we let t 0 in the year 1950. Then the initialvalue problem is
dP
dt kP P0 2560 Pt 2560e kt and the solution is Let’s estimate k by using the population in 1960:
P 10
k 2560e 10k 3040 1
3040
ln
10
2560 0.017185 The relative growth rate is about 1.7% per year and the model is
Pt 2560e 0.017185t We estimate that the world population in 1993 was
P 43 2560e 0.017185 43 5360 million The model predicts that the population in 2010 will be
P 60 2560e 0.017185 60 7179 million The graph in Figure 3 shows that the model is fairly accurate to date, so the estimate for
1993 is quite reliable. But the prediction for 2010 is riskier.
P
6000 P=2560e 0.017185t Population
(in millions) FIGURE 3 A model for world population growth
in the second half of the 20th century 20 Years since 1950 40 t 5E10(pp 652661) 652 ❙❙❙❙ 1/18/06 9:23 AM Page 652 CHAPTER 10 DIFFERENTIAL EQUATIONS Radioactive Decay
Radioactive substances decay by spontaneously emitting radiation. If m t is the mass
remaining from an initial mass m0 of the substance after time t, then the relative decay rate
1 dm
m dt
has been found experimentally to be constant. (Since dm dt is negative, the relative decay
rate is positive.) It follows that
dm
dt km where k is a negative constant. In other words, radioactive substances decay at a rate proportional to the remaining mass. This means that we can use (2) to show that the mass
decays exponentially:
m0 e kt mt Physicists express the rate of decay in terms of halflife, the time required for half of
any given quantity to decay.
EXAMPLE 3 The halflife of radium226 ( .226 Ra) is 1590 years.
88 (a) A sample of radium226 has a mass of 100 mg. Find a formula for the mass of .226 Ra
88
that remains after t years.
(b) Find the mass after 1000 years correct to the nearest milligram.
(c) When will the mass be reduced to 30 mg?
SOLUTION (a) Let m t be the mass of radium226 (in milligrams) that remains after t years. Then
dm dt km and y 0
100, so (2) gives
m 0 e kt mt 100e kt In order to determine the value of k, we use the fact that y 1590
100e 1590k 50 e 1590k so 1
2 100 . Thus 1
2 and
ln 1
2 1590k ln 2 ln 2
1590 k
mt We could use the fact that e ln 2
form 2 to write the expression for m t in the alternative
mt 100e ln 2 1590 t Therefore 100 2 t 1590 (b) The mass after 1000 years is
m 1000 100e ln 2 1590 1000 65 mg 5E10(pp 652661) 1/18/06 9:23 AM Page 653 S ECTION 10.4 EXPONENTIAL GROWTH AND DECAY (c) We want to ﬁnd the value of t such that m t
100e ln 2 1590 t 30 ❙❙❙❙ 653 30, that is, or e ln 2 1590 t 0.3 We solve this equation for t by taking the natural logarithm of both sides:
ln 2
t
1590 150 ln 0.3 m=100e _( ln 2)t/1590 Thus
m=30
0 FIGURE 4 4000 t 1590 ln 0.3
ln 2 2762 years As a check on our work in Example 3, we use a graphing device to draw the graph of
m t in Figure 4 together with the horizontal line m 30. These curves intersect when
t 2800, and this agrees with the answer to part (c). Newton ’s Law of Cooling
Newton’s Law of Cooling states that the rate of cooling of an object is proportional to
the temperature difference between the object and its surroundings, provided that this
difference is not too large. (This law also applies to warming.) If we let T t be the temperature of the object at time t and Ts be the temperature of the surroundings, then we
can formulate Newton’s Law of Cooling as a differential equation:
dT
dt kT Ts where k is a constant. We could solve this equation as a separable differential equation
by the method of Section 10.3, but an easier method is to make the change of variable
yt
Tt
Ts . Because Ts is constant, we have y t
T t and so the equation
becomes
dy
ky
dt
We can then use (2) to find an expression for y, from which we can find T .
EXAMPLE 4 A bottle of soda pop at room temperature (72 F) is placed in a refrigerator
where the temperature is 44 F. After half an hour the soda pop has cooled to 61 F.
(a) What is the temperature of the soda pop after another half hour?
(b) How long does it take for the soda pop to cool to 50 F?
SOLUTION (a) Let T t be the temperature of the soda after t minutes. The surrounding temperature
is Ts 44 F, so Newton’s Law of Cooling states that
dT
dt dy
dt 44) T0 If we let y T 44, then y 0
initialvalue problem kT
44 72 ky y0 44
28 28, so y is a solution of the 5E10(pp 652661) 654 ❙❙❙❙ 1/18/06 9:23 AM Page 654 CHAPTER 10 DIFFERENTIAL EQUATIONS and by (2) we have
y 0 e kt yt
We are given that T 30 61, so y 30 61 28 e 30 k 28 e kt
44 17 and
17
28 e 30 k 17 Taking logarithms, we have
ln ( 17 )
28
30 k 0.01663 Thus
0.01663 t yt 28 e Tt 44 28e 0.01663 t T 60 44 28 e 0.01663 60 54.3 So after another half hour the pop has cooled to about 54 F.
(b) We have T t
50 when T
72 28 e 0.01663t 50 e 44 0.01663 t 6
28 6
ln ( 28 )
0.01663 t 44 92.6 The pop cools to 50 F after about 1 hour 33 minutes.
Notice that in Example 4, we have 0 FIGURE 5 30 60 90 t lim T t tl lim 44 tl 28 e 0.01663 t 44 28 0 44 which is to be expected. The graph of the temperature function is shown in Figure 5. Continuously Compounded Interest
EXAMPLE 5 If $1000 is invested at 6% interest, compounded annually, then after 1 year
the investment is worth $1000 1.06
$1060, after 2 years it’s worth
$ 1000 1.06 1.06 $1123.60, and after t years it’s worth $1000 1.06 t. In general, if
an amount A0 is invested at an interest rate r r 0.06 in this example), then after
t years it’s worth A0 1 r t. Usually, however, interest is compounded more frequently,
say, n times a year. Then in each compounding period the interest rate is r n and there
are nt compounding periods in t years, so the value of the investment is A0 1 r
n nt 5E10(pp 652661) 1/18/06 9:23 AM Page 655 S ECTION 10.4 EXPONENTIAL GROWTH AND DECAY ❙❙❙❙ 655 For instance, after 3 years at 6% interest a $1000 investment will be worth
$1000 1.06 3 $1191.02 with annual compounding $1000 1.03 6 $1194.05 with semiannual compounding $1000 1.015 12 $1195.62 with quarterly compounding $1000 1.005 36 $1196.68 with monthly compounding $1197.20 with daily compounding $1000 1 0.06
365 365 3 You can see that the interest paid increases as the number of compounding periods n
increases. If we let n l , then we will be compounding the interest continuously and
the value of the investment will be At nt r
n lim A0 1 nl lim A0 nl A0 lim 1 r
n A0 lim 1 1
m nl ml 1 nr nr rt rt m r
n rt (where m n r) But the limit in this expression is equal to the number e (see Equation 7.4.9 or 7.4*.9).
So with continuous compounding of interest at interest rate r, the amount after t years is
At A0 e rt If we differentiate this equation, we get
dA
dt rA0 e rt rA t which says that, with continuous compounding of interest, the rate of increase of an
investment is proportional to its size.
Returning to the example of $1000 invested for 3 years at 6% interest, we see that
with continuous compounding of interest the value of the investment will be
A3 $1000e 0.06 3
$1000e 0.18 $1197.22 Notice how close this is to the amount we calculated for daily compounding, $1197.20.
But the amount is easier to compute if we use continuous compounding. 5E10(pp 652661) 656 ❙❙❙❙ 1/18/06 9:23 AM Page 656 CHAPTER 10 DIFFERENTIAL EQUATIONS  10.4 Exercises 1. A population of protozoa develops with a constant relative Year 2. A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrientbroth
medium divides into two cells every 20 minutes. The initial
population of a culture is 60 cells.
(a) Find the relative growth rate.
(b) Find an expression for the number of cells after t hours.
(c) Find the number of cells after 8 hours.
(d) Find the rate of growth after 8 hours.
(e) When will the population reach 20,000 cells?
3. A bacteria culture starts with 500 bacteria and grows at a rate proportional to its size. After 3 hours there are 8000 bacteria.
(a) Find an expression for the number of bacteria after t hours.
(b) Find the number of bacteria after 4 hours.
(c) Find the rate of growth after 4 hours.
(d) When will the population reach 30,000?
4. A bacteria culture grows with constant relative growth rate. After 2 hours there are 600 bacteria and after 8 hours the count
is 75,000.
(a) Find the initial population.
(b) Find an expression for the population after t hours.
(c) Find the number of cells after 5 hours.
(d) Find the rate of growth after 5 hours.
(e) When will the population reach 200,000?
5. The table gives estimates of the world population, in millions, from 1750 to 2000:
Year Population Year 790
980
1260 1900
1950
2000 1650
2560
6080 Year Population 76
92
106
123
131
150 1960
1970
1980
1990
2000 179
203
227
250
275 (a) Use the exponential model and the census ﬁgures for 1900
and 1910 to predict the population in 2000. Compare with
the actual ﬁgure and try to explain the discrepancy.
(b) Use the exponential model and the census ﬁgures for 1980
and 1990 to predict the population in 2000. Compare with
the actual population. Then use this model to predict the
population in the years 2010 and 2020.
(c) Graph both of the exponential functions in parts (a) and (b)
together with a plot of the actual population. Are these
models reasonable ones?
7. Experiments show that if the chemical reaction N2O5 l 2NO 2 1
2 O2 takes place at 45 C, the rate of reaction of dinitrogen pentoxide is proportional to its concentration as follows:
d N2O5
dt 0.0005 N2O5 (See Example 4 in Section 3.4.)
(a) Find an expression for the concentration N2O5 after
t seconds if the initial concentration is C.
(b) How long will the reaction take to reduce the concentration
of N2O5 to 90% of its original value? Population 1750
1800
1850 ; Population 1900
1910
1920
1930
1940
1950 growth rate of 0.7944 per member per day. On day zero the
population consists of two members. Find the population size
after six days. (a) Use the exponential model and the population ﬁgures for
1750 and 1800 to predict the world population in 1900 and
1950. Compare with the actual ﬁgures.
(b) Use the exponential model and the population ﬁgures for
1850 and 1900 to predict the world population in 1950.
Compare with the actual population.
(c) Use the exponential model and the population ﬁgures for
1900 and 1950 to predict the world population in 2000.
Compare with the actual population and try to explain the
discrepancy.
6. The table gives the population of the United States, in millions, for the years 1900–2000. 8. Bismuth210 has a halflife of 5.0 days. (a) A sample originally has a mass of 800 mg. Find a formula
for the mass remaining after t days.
(b) Find the mass remaining after 30 days.
(c) When is the mass reduced to 1 mg?
(d) Sketch the graph of the mass function.
9. The halflife of cesium137 is 30 years. Suppose we have a 100mg sample.
(a) Find the mass that remains after t years.
(b) How much of the sample remains after 100 years?
(c) After how long will only 1 mg remain?
10. After 3 days a sample of radon222 decayed to 58% of its orig inal amount.
(a) What is the halflife of radon222?
(b) How long would it take the sample to decay to 10% of its
original amount? 5E10(pp 652661) 1/18/06 9:23 AM Page 657 S ECTION 10.4 EXPONENTIAL GROWTH AND DECAY 11. Scientists can determine the age of ancient objects by a method called radiocarbon dating. The bombardment of the upper
atmosphere by cosmic rays converts nitrogen to a radioactive
isotope of carbon, 14 C, with a halflife of about 5730 years.
Vegetation absorbs carbon dioxide through the atmosphere and
animal life assimilates 14 C through food chains. When a plant
or animal dies, it stops replacing its carbon and the amount of
14
C begins to decrease through radioactive decay. Therefore,
the level of radioactivity must also decay exponentially. A
parchment fragment was discovered that had about 74% as
much 14 C radioactivity as does plant material on Earth today.
Estimate the age of the parchment.
12. A curve passes through the point 0, 5 and has the property that the slope of the curve at every point P is twice the
ycoordinate of P. What is the equation of the curve?
13. A roast turkey is taken from an oven when its temperature has reached 185 F and is placed on a table in a room where the
temperature is 75 F.
(a) If the temperature of the turkey is 150 F after half an hour,
what is the temperature after 45 min?
(b) When will the turkey have cooled to 100 F?
14. A thermometer is taken from a room where the temperature is 20 C to the outdoors, where the temperature is 5 C. After one
minute the thermometer reads 12 C.
(a) What will the reading on the thermometer be after one
more minute?
(b) When will the thermometer read 6 C?
15. When a cold drink is taken from a refrigerator, its temperature is 5 C. After 25 minutes in a 20 C room its temperature has
increased to 10 C.
(a) What is the temperature of the drink after 50 minutes?
(b) When will its temperature be 15 C?
16. A freshly brewed cup of coffee has temperature 95 C in a 20 C room. When its temperature is 70 C, it is cooling at a rate of
1 C per minute. When does this occur?
17. The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is
constant. At 15 C the pressure is 101.3 kPa at sea level and
87.14 kPa at h 1000 m.
(a) What is the pressure at an altitude of 3000 m?
(b) What is the pressure at the top of Mount McKinley, at an
altitude of 6187 m?
18. (a) If $500 is borrowed at 14% interest, ﬁnd the amounts due at the end of 2 years if the interest is compounded
(i) annually, (ii) quarterly, (iii) monthly, (iv) daily,
(v) hourly, and (vi) continuously. ; ❙❙❙❙ 657 (b) Suppose $500 is borrowed and the interest is compounded
continuously. If A t is the amount due after t years, where
0 t 2, graph A t for each of the interest rates 14%,
10%, and 6% on a common screen.
19. (a) If $3000 is invested at 5% interest, ﬁnd the value of the investment at the end of 5 years if the interest is compounded (i) annually, (ii) semiannually, (iii) monthly,
(iv) weekly, (v) daily, and (vi) continuously.
(b) If A t is the amount of the investment at time t for the case
of continuous compounding, write a differential equation
and an initial condition satisﬁed by A t .
20. (a) How long will it take an investment to double in value if the interest rate is 6% compounded continuously?
(b) What is the equivalent annual interest rate?
21. Consider a population P P t with constant relative birth and
death rates and , respectively, and a constant emigration
rate m, where , , and m are positive constants. Assume that
. Then the rate of change of the population at time t is
modeled by the differential equation
dP
dt kP m where k (a) Find the solution of this equation that satisﬁes the initial
condition P 0
P0.
(b) What condition on m will lead to an exponential expansion
of the population?
(c) What condition on m will result in a constant population?
A population decline?
(d) In 1847, the population of Ireland was about 8 million and
the difference between the relative birth and death rates was
1.6% of the population. Because of the potato famine in the
1840s and 1850s, about 210,000 inhabitants per year emigrated from Ireland. Was the population expanding or
declining at that time?
22. Let c be a positive number. A differential equation of the form dy
dt ky 1 c where k is a positive constant, is called a doomsday equation
because the exponent in the expression ky 1 c is larger than that
for natural growth (that is, ky).
(a) Determine the solution that satisﬁes the initial condition
y0
y0.
(b) Show that there is a ﬁnite time t T (doomsday) such that
.
lim t l T y t
(c) An especially proliﬁc breed of rabbits has the growth term
ky 1.01. If 2 such rabbits breed initially and the warren has
16 rabbits after three months, then when is doomsday? 5E10(pp 652661) 658 ❙❙❙❙ 1/18/06 9:23 AM Page 658 CHAPTER 10 DIFFERENTIAL EQUATIONS APPLIED PROJECT
Calculus and Baseball
In this project we explore three of the many applications of calculus to baseball. The physical
interactions of the game, especially the collision of ball and bat, are quite complex and their
models are discussed in detail in a book by Robert Adair, The Physics of Baseball, 3d ed.
(New York: HarperPerennial, 2002).
1. It may surprise you to learn that the collision of baseball and bat lasts only about a thou sandth of a second. Here we calculate the average force on the bat during this collision by
ﬁrst computing the change in the ball’s momentum.
The momentum p of an object is the product of its mass m and its velocity v, that is,
p mv. Suppose an object, moving along a straight line, is acted on by a force F F t
that is a continuous function of time.
(a) Show that the change in momentum over a time interval t0 , t1 is equal to the integral of
F from t0 to t1; that is, show that
p t1 p t0 y Batter’s box An overhead view of the position of a
baseball bat, shown every ﬁftieth of
a second during a typical swing.
(Adapted from The Physics of Baseball ) t1 t0 F t dt This integral is called the impulse of the force over the time interval.
(b) A pitcher throws a 90mi h fastball to a batter, who hits a line drive directly back
to the pitcher. The ball is in contact with the bat for 0.001 s and leaves the bat with
velocity 110 mi h. A baseball weighs 5 oz and, in U.S. Customary units, its mass is
measured in slugs: m w t where t 32 ft s 2.
(i) Find the change in the ball’s momentum.
(ii) Find the average force on the bat.
2. In this problem we calculate the work required for a pitcher to throw a 90mi h fastball by ﬁrst considering kinetic energy.
The kinetic energy K of an object of mass m and velocity v is given by K
pose an object of mass m, moving in a straight line, is acted on by a force F
depends on its position s. According to Newton’s Second Law
Fs ma m 1
2 mv 2. SupF s that dv
dt where a and v denote the acceleration and velocity of the object.
(a) Show that the work done in moving the object from a position s0 to a position s1 is equal
to the change in the object’s kinetic energy; that is, show that
W y s1 s0 F s ds 1
2 2
mv1 1
2 2
mv 0 where v0 v s0 and v1 v s1 are the velocities of the object at the positions s0 and s1.
Hint: By the Chain Rule,
dv
dv ds
dv
m
m
mv
dt
ds dt
ds
(b) How many footpounds of work does it take to throw a baseball at a speed of
90 mi h?
3. (a) An outﬁelder ﬁelds a baseball 280 ft away from home plate and throws it directly to the
catcher with an initial velocity of 100 ft s. Assume that the velocity v t of the ball after
t seconds satisﬁes the differential equation dv dt
v 10 because of air resistance. How long does it take for the ball to reach home plate? (Ignore any vertical motion of
the ball.)
(b) The manager of the team wonders whether the ball will reach home plate sooner if it
is relayed by an inﬁelder. The shortstop can position himself directly between the outﬁelder and home plate, catch the ball thrown by the outﬁelder, turn, and throw the ball to 5E10(pp 652661) 1/18/06 9:23 AM Page 659 S ECTION 10.5 THE LOGISTIC EQUATION ;  10.5 ❙❙❙❙ 659 the catcher with an initial velocity of 105 ft s. The manager clocks the relay time of the
shortstop (catching, turning, throwing) at half a second. How far from home plate should
the shortstop position himself to minimize the total time for the ball to reach the plate?
Should the manager encourage a direct throw or a relayed throw? What if the shortstop
can throw at 115 ft s?
(c) For what throwing velocity of the shortstop does a relayed throw take the same time as a
direct throw? The Logistic Equation
In this section we discuss in detail a model for population growth, the logistic model, that
is more sophisticated than exponential growth. In doing so we use all the tools at our disposal—direction ﬁelds and Euler’s method from Section 10.2 and the explicit solution of
separable differential equations from Section 10.3. In the exercises we investigate other
possible models for population growth, some of which take into account harvesting and
seasonal growth. The Logistic Model
As we discussed in Section 10.1, a population often increases exponentially in its early
stages but levels off eventually and approaches its carrying capacity because of limited
resources. If P t is the size of the population at time t, we assume that
dP
dt kP if P is small This says that the growth rate is initially close to being proportional to size. In other words,
the relative growth rate is almost constant when the population is small. But we also want
to reﬂect the fact that the relative growth rate decreases as the population P increases and
becomes negative if P ever exceeds its carrying capacity K, the maximum population that
the environment is capable of sustaining in the long run. The simplest expression for the
relative growth rate that incorporates these assumptions is
1 dP
P dt k1 P
K Multiplying by P, we obtain the model for population growth known as the logistic differential equation: 1 dP
dt kP 1 P
K Notice from Equation 1 that if P is small compared with K, then P K is close to 0 and so
dP dt kP. However, if P l K (the population approaches its carrying capacity), then
P K l 1, so dP dt l 0. We can deduce information about whether solutions increase or
decrease directly from Equation 1. If the population P lies between 0 and K, then the right
side of the equation is positive, so dP dt 0 and the population increases. But if the population exceeds the carrying capacity P K , then 1 P K is negative, so dP dt 0
and the population decreases. 5E10(pp 652661) 660 ❙❙❙❙ 1/18/06 9:23 AM Page 660 CHAPTER 10 DIFFERENTIAL EQUATIONS Direction Fields
Let’s start our more detailed analysis of the logistic differential equation by looking at a
direction ﬁeld.
EXAMPLE 1 Draw a direction ﬁeld for the logistic equation with k capacity K 0.08 and carrying 1000. What can you deduce about the solutions? SOLUTION In this case the logistic differential equation is dP
dt 0.08P 1 P
1000 A direction ﬁeld for this equation is shown in Figure 1. We show only the ﬁrst quadrant
because negative populations aren’t meaningful and we are interested only in what happens after t 0.
P
1400
1200
1000
800
600
400
200 FIGURE 1 Direction field for the logistic
equation in Example 1 0 20 40 60 80 t The logistic equation is autonomous (dP dt depends only on P, not on t ), so the
slopes are the same along any horizontal line. As expected, the slopes are positive for
0 P 1000 and negative for P 1000.
The slopes are small when P is close to 0 or 1000 (the carrying capacity). Notice that
the solutions move away from the equilibrium solution P 0 and move toward the
equilibrium solution P 1000.
In Figure 2 we use the direction ﬁeld to sketch solution curves with initial populations
P0
100, P 0
400, and P 0
1300. Notice that solution curves that start below
P 1000 are increasing and those that start above P 1000 are decreasing. The slopes
are greatest when P 500 and, therefore, the solution curves that start below P 1000
have inﬂection points when P 500. In fact we can prove that all solution curves that
start below P 500 have an inﬂection point when P is exactly 500 (see Exercise 9).
P
1400
1200
1000
800
600
400
200 FIGURE 2 Solution curves for the logistic
equation in Example 1 0 20 40 60 80 t 5E10(pp 652661) 1/18/06 9:23 AM Page 661 S ECTION 10.5 THE LOGISTIC EQUATION ❙❙❙❙ 661 E uler ’s Method
Next let’s use Euler’s method to obtain numerical estimates for solutions of the logistic differential equation at speciﬁc times.
EXAMPLE 2 Use Euler’s method with step sizes 20, 10, 5, 1, and 0.1 to estimate the population sizes P 40 and P 80 , where P is the solution of the initialvalue problem dP
dt
SOLUTION With step size h P
1000 0.08P 1
20, t0
F t, P 0, P0 P0 100 100, and
P
1000 0.08P 1 we get, using the notation of Section 10.2,
P1 100 20F 0, 100 244
P2 244 20F 20, 244 P3 539.14 20F 40, 539.14 936.69 P4 936.69 20F 60, 936.69 1031.57 539.14 Thus, our estimates for the population sizes at times t
P 40 539 P 80 40 and t 80 are 1032 For smaller step sizes we need to program a calculator or computer. The table gives
the results.
Step size Euler estimate of P 40 Euler estimate of P 80 20
10
5
1
0.1 539
647
695
725
731 1032
997
991
986
985 Figure 3 shows a graph of the Euler approximations with step sizes h 10 and h 1.
We see that the Euler approximation with h 1 looks very much like the lower solution
curve that we drew using a direction ﬁeld in Figure 2.
P
1000 h=1
h=10 FIGURE 3 Euler approximations of the
solution curve in Example 2 0 20 40 60 80 t 5E10(pp 662671) 662 ❙❙❙❙ 1/18/06 9:25 AM Page 662 CHAPTER 10 DIFFERENTIAL EQUATIONS The Analytic Solution
The logistic equation (1) is separable and so we can solve it explicitly using the method of
Section 10.3. Since
dP
dt P
K kP 1 we have
dP
PK y P1 2 y k dt To evaluate the integral on the left side, we write
1
P1 K
PK PK P Using partial fractions (see Section 8.4), we get
K
PK 1
P P 1
K P This enables us to rewrite Equation 2: y 1
P ln P 1
K dP ln K
ln K y k dt P P kt P P
P K 3 where A kt P
K P
P C
C kt C e Ae e Ce kt kt e C. Solving Equation 3 for P, we get
K
P 1 Ae kt P so
We ﬁnd the value of A by putting t
population), so P
K ? 1 K
Ae 1 P0
P0 Ae 0 kt kt 0 in Equation 3. If t K 1
Ae A 0, then P P0 (the initial 5E10(pp 662671) 1/18/06 9:25 AM Page 663 S ECTION 10.5 THE LOGISTIC EQUATION ❙❙❙❙ 663 Thus, the solution to the logistic equation is
Pt 4 1 K
Ae K where A kt P0
P0 Using the expression for P t in Equation 4, we see that
lim P t K tl which is to be expected.
EXAMPLE 3 Write the solution of the initialvalue problem dP
dt P
1000 0.08P 1 P0 100 and use it to ﬁnd the population sizes P 40 and P 80 . At what time does the population
reach 900?
SOLUTION The differential equation is a logistic equation with k
0.08, carrying capacity K 1000, and initial population P0 100. So Equation 4 gives the population at
time t as Pt 1 1000
Ae 0.08t Thus Pt So the population sizes when t  Compare these values with the Euler
estimates from Example 2:
P 40
731
P 80
985 P 40 1000
1 9e 3.2 1000 100
100 where A 1 9 1000
9e 0.08t 40 and 80 are
731.6 1000
1 9e P 80 6.4 985.3 The population reaches 900 when
1000
1 9e 0.08t 900 Solving this equation for t, we get
 Compare the solution curve in Figure 4 with
the lowest solution curve we drew from the
direction ﬁeld in Figure 2. 9e 0.08t 10
9 e 1 0.08t 1
81 1000 0.08t P=900 t
P=
0 FIGURE 4 1000
1+9e _0.08t
80 1
ln 81 ln 81 ln 81
0.08 54.9 So the population reaches 900 when t is approximately 55. As a check on our work, we
graph the population curve in Figure 4 and observe where it intersects the line P 900.
The cursor indicates that t 55. 5E10(pp 662671) 664 ❙❙❙❙ 1/18/06 9:25 AM Page 664 CHAPTER 10 DIFFERENTIAL EQUATIONS Comparison of the Natural Growth and Logistic Models
In the 1930s the biologist G. F. Gause conducted an experiment with the protozoan Paramecium and used a logistic equation to model his data. The table gives his daily count
of the population of protozoa. He estimated the initial relative growth rate to be 0.7944 and
the carrying capacity to be 64.
t (days) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 P (observed) 2 3 22 16 39 52 54 47 50 76 69 51 57 70 53 59 57 EXAMPLE 4 Find the exponential and logistic models for Gause’s data. Compare the
predicted values with the observed values and comment on the ﬁt.
SOLUTION Given the relative growth rate k
exponential model is 0.7944 and the initial population P0 P0 e kt Pt 2, the 2e 0.7944 t Gause used the same value of k for his logistic model. [This is reasonable because
P0 2 is small compared with the carrying capacity (K 64). The equation
1 dP
P0 dt 2
64 k1
t0 k shows that the value of k for the logistic model is very close to the value for the exponential model.]
Then the solution of the logistic equation in Equation 4 gives
Pt 1 A where K
Ae So K kt P0 64 P0
Pt 1 64
Ae
2 31 2
64
31e 1 0.7944 t 0.7944 t We use these equations to calculate the predicted values (rounded to the nearest integer)
and compare them in the table.
t (days) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 P (observed) 2 3 22 16 39 52 54 47 50 76 69 51 57 70 53 59 57 P (logistic model) 2 4 9 17 28 40 51 57 61 62 63 64 64 64 64 64 64 P (exponential model) 2 4 10 22 48 106 ... We notice from the table and from the graph in Figure 5 that for the ﬁrst three or four
days the exponential model gives results comparable to those of the more sophisticated
logistic model. For t 5, however, the exponential model is hopelessly inaccurate, but
the logistic model ﬁts the observations reasonably well. 5E10(pp 662671) 1/18/06 9:25 AM Page 665 S ECTION 10.5 THE LOGISTIC EQUATION ❙❙❙❙ 665 P P=2e0.7944t
60
40 P=
20 64
1+31e _0.7944t FIGURE 5
0 The exponential and logistic
models for the Paramecium data 4 8 16 t 12 Other Models for Population Growth
The Law of Natural Growth and the logistic differential equation are not the only equations that have been proposed to model population growth. In Exercise 14 we look at
the Gompertz growth function and in Exercises 15 and 16 we investigate seasonalgrowth models.
Two of the other models are modiﬁcations of the logistic model. The differential
equation
dP
P
kP 1
c
dt
K
has been used to model populations that are subject to “harvesting” of one sort or another.
(Think of a population of ﬁsh being caught at a constant rate). This equation is explored
in Exercises 11 and 12.
For some species there is a minimum population level m below which the species tends
to become extinct. (Adults may not be able to ﬁnd suitable mates.) Such populations have
been modeled by the differential equation
dP
dt
where the extra factor, 1
tion (see Exercise 13).  10.5 kP 1 P
K m
P 1 m P, takes into account the consequences of a sparse popula Exercises 1. Suppose that a population develops according to the logistic equation
dP
dt 0.05P 0.0005P 2 where t is measured in weeks.
(a) What is the carrying capacity? What is the value of k ?
(b) A direction ﬁeld for this equation is shown at the right.
Where are the slopes close to 0? Where are they largest?
Which solutions are increasing? Which solutions are
decreasing?
(c) Use the direction ﬁeld to sketch solutions for initial populations of 20, 40, 60, 80, 120, and 140. What do these P
150 100 50 0 20 40 60 t 5E10(pp 662671) 666 ❙❙❙❙ 1/18/06 9:25 AM Page 666 CHAPTER 10 DIFFERENTIAL EQUATIONS solutions have in common? How do they differ? Which
solutions have inﬂection points? At what population levels
do they occur?
(d) What are the equilibrium solutions? How are the other solutions related to these solutions? ; 2. Suppose that a population grows according to a logistic model
with carrying capacity 6000 and k 0.0015 per year.
(a) Write the logistic differential equation for these data.
(b) Draw a direction ﬁeld (either by hand or with a computer
algebra system). What does it tell you about the solution
curves?
(c) Use the direction ﬁeld to sketch the solution curves for
initial populations of 1000, 2000, 4000, and 8000. What
can you say about the concavity of these curves? What is
the signiﬁcance of the inﬂection points?
(d) Program a calculator or computer to use Euler’s method
with step size h 1 to estimate the population after
50 years if the initial population is 1000.
(e) If the initial population is 1000, write a formula for the
population after t years. Use it to ﬁnd the population after
50 years and compare with your estimate in part (d).
(f) Graph the solution in part (e) and compare with the
solution curve you sketched in part (c).
3. The Paciﬁc halibut ﬁshery has been modeled by the differential equation
dy
dt ky 1 y
K 5. The population of the world was about 5.3 billion in 1990. Birth rates in the 1990s ranged from 35 to 40 million per year
and death rates ranged from 15 to 20 million per year. Let’s
assume that the carrying capacity for world population is
100 billion.
(a) Write the logistic differential equation for these data.
(Because the initial population is small compared to the
carrying capacity, you can take k to be an estimate of the
initial relative growth rate.)
(b) Use the logistic model to estimate the world population in
the year 2000 and compare with the actual population of
6.1 billion.
(c) Use the logistic model to predict the world population in
the years 2100 and 2500.
(d) What are your predictions if the carrying capacity is
50 billion?
6. (a) Make a guess as to the carrying capacity for the U.S. population. Use it and the fact that the population was
250 million in 1990 to formulate a logistic model for the
U.S. population.
(b) Determine the value of k in your model by using the fact
that the population in 2000 was 275 million.
(c) Use your model to predict the U.S. population in the years
2100 and 2200.
(d) Use your model to predict the year in which the U.S. population will exceed 300 million.
7. One model for the spread of a rumor is that the rate of spread where y t is the biomass (the total mass of the members of
the population) in kilograms at time t (measured in years), the
carrying capacity is estimated to be K 8 10 7 kg, and
k 0.71 per year.
(a) If y 0
2 10 7 kg, ﬁnd the biomass a year later.
(b) How long will it take for the biomass to reach 4 10 7 kg?
4. The table gives the number of yeast cells in a new laboratory culture.
Time (hours) Yeast cells Time (hours) Yeast cells 0
2
4
6
8 18
39
80
171
336 10
12
14
16
18 509
597
640
664
672 (a) Plot the data and use the plot to estimate the carrying
capacity for the yeast population.
(b) Use the data to estimate the initial relative growth rate.
(c) Find both an exponential model and a logistic model for
these data.
(d) Compare the predicted values with the observed values,
both in a table and with graphs. Comment on how well
your models ﬁt the data.
(e) Use your logistic model to estimate the number of yeast
cells after 7 hours. is proportional to the product of the fraction y of the population
who have heard the rumor and the fraction who have not heard
the rumor.
(a) Write a differential equation that is satisﬁed by y.
(b) Solve the differential equation.
(c) A small town has 1000 inhabitants. At 8 A.M., 80 people
have heard a rumor. By noon half the town has heard it. At
what time will 90% of the population have heard the rumor?
8. Biologists stocked a lake with 400 ﬁsh and estimated the carrying capacity (the maximal population for the ﬁsh of that
species in that lake) to be 10,000. The number of ﬁsh tripled in
the ﬁrst year.
(a) Assuming that the size of the ﬁsh population satisﬁes the
logistic equation, ﬁnd an expression for the size of the population after t years.
(b) How long will it take for the population to increase
to 5000?
9. (a) Show that if P satisﬁes the logistic equation (1), then d 2P
dt 2 k 2P 1 P
K 1 2P
K (b) Deduce that a population grows fastest when it reaches half
its carrying capacity. ; 10. For a ﬁxed value of K (say K 10), the family of logistic
functions given by Equation 4 depends on the initial value P0
and the proportionality constant k. Graph several members of 5E10(pp 662671) 1/18/06 9:25 AM Page 667 S ECTION 10.5 THE LOGISTIC EQUATION this family. How does the graph change when P0 varies? How
does it change when k varies? 14. Another model for a growth function for a limited population is follows:
dP
dt CAS 667 (d) Use the solution in part (c) to show that if P0 m, then the
species will become extinct. [Hint: Show that the numerator in your expression for P t is 0 for some value of t.] 11. Let’s modify the logistic differential equation of Example 1 as CAS ❙❙❙❙ 0.08P 1 P
1000 given by the Gompertz function, which is a solution of the
differential equation 15 dP
dt (a) Suppose P t represents a ﬁsh population at time t, where t
is measured in weeks. Explain the meaning of the term 15.
(b) Draw a direction ﬁeld for this differential equation.
(c) What are the equilibrium solutions?
(d) Use the direction ﬁeld to sketch several solution curves.
Describe what happens to the ﬁsh population for various
initial populations.
(e) Solve this differential equation explicitly, either by using
partial fractions or with a computer algebra system. Use the
initial populations 200 and 300. Graph the solutions and
compare with your sketches in part (d). 15. In a seasonalgrowth model, a periodic function of time is P
1000 0.08P 1 c as a model for a ﬁsh population, where t is measured in weeks
and c is a constant.
(a) Use a CAS to draw direction ﬁelds for various values of c.
(b) From your direction ﬁelds in part (a), determine the values
of c for which there is at least one equilibrium solution. For
what values of c does the ﬁsh population always die out?
(c) Use the differential equation to prove what you discovered
graphically in part (b).
(d) What would you recommend for a limit to the weekly catch
of this ﬁsh population? introduced to account for seasonal variations in the rate of
growth. Such variations could, for example, be caused by
seasonal changes in the availability of food.
(a) Find the solution of the seasonalgrowth model
dP
dt ; P
K 1 m
P (a) Use the differential equation to show that any solution is
increasing if m P K and decreasing if 0 P m.
(b) For the case where k 0.08, K 1000, and m 200,
draw a direction ﬁeld and use it to sketch several solution
curves. Describe what happens to the population for various
initial populations. What are the equilibrium solutions?
(c) Solve the differential equation explicitly, either by using
partial fractions or with a computer algebra system. Use the
initial population P0 . P0 where k, r, and are positive constants.
(b) By graphing the solution for several values of k, r, and ,
explain how the values of k, r, and affect the solution.
What can you say about lim t l P t ?
follows: some species there is a minimum population m such that the
species will become extinct if the size of the population falls
below m. This condition can be incorporated into the logistic
equation by introducing the factor 1 m P . Thus, the modiﬁed logistic model is given by the differential equation
kP 1 P0 k P cos r t 16. Suppose we alter the differential equation in Exercise 15 as 13. There is considerable evidence to support the theory that for dP
dt K
P
P where c is a constant and K is the carrying capacity.
(a) Solve this differential equation.
(b) Compute lim t l P t .
(c) Graph the Gompertz growth function for K 1000,
P0 100, and c 0.05, and compare it with the logistic
function in Example 3. What are the similarities? What are
the differences?
(d) We know from Exercise 9 that the logistic function grows
fastest when P K 2. Use the Gompertz differential equation to show that the Gompertz function grows fastest
when P K e. 12. Consider the differential equation dP
dt c ln dP
dt ; kP cos 2 r t P0 P0 (a) Solve this differential equation with the help of a table of
integrals or a CAS.
(b) Graph the solution for several values of k, r, and . How do
the values of k, r, and affect the solution? What can you
say about lim t l P t in this case?
17. Graphs of logistic functions (Figures 2 and 4) look suspiciously similar to the graph of the hyperbolic tangent function
(Figure 3 in Section 7.6). Explain the similarity by showing
that the logistic function given by Equation 4 can be written as
Pt 1
2 [ K1 tanh ( 1 k t
2 c )] where c
ln A k. Thus, the logistic function is really just a
shifted hyperbolic tangent. 5E10(pp 662671) 668 ❙❙❙❙ 1/18/06 9:25 AM Page 668 CHAPTER 10 DIFFERENTIAL EQUATIONS  10.6 Linear Equations
A ﬁrstorder linear differential equation is one that can be put into the form
dy
dx 1 Pxy Qx where P and Q are continuous functions on a given interval. This type of equation occurs
frequently in various sciences, as we will see.
An example of a linear equation is xy
y 2 x because, for x 0, it can be written
in the form
1
y
x y 2 2 Notice that this differential equation is not separable because it’s impossible to factor the
expression for y as a function of x times a function of y. But we can still solve the equation by noticing, by the Product Rule, that
xy y xy and so we can rewrite the equation as
xy 2x If we now integrate both sides of this equation, we get
xy x2 or C y C
x x If we had been given the differential equation in the form of Equation 2, we would have
had to take the preliminary step of multiplying each side of the equation by x.
It turns out that every ﬁrstorder linear differential equation can be solved in a similar
fashion by multiplying both sides of Equation 1 by a suitable function I x called an
integrating factor. We try to ﬁnd I so that the left side of Equation 1, when multiplied by
I x , becomes the derivative of the product I x y:
Ix y 3 Pxy Ixy If we can ﬁnd such a function I , then Equation 1 becomes
Ixy IxQx Integrating both sides, we would have
Ixy yI xQx dx C so the solution would be
4 yx 1
Ix yI xQx dx C 5E10(pp 662671) 1/18/06 9:26 AM Page 669 SECTION 10.6 LINEAR EQUATIONS ❙❙❙❙ 669 To ﬁnd such an I, we expand Equation 3 and cancel terms:
Ixy IxPxy Ixy IxPx I xy Ixy Ix This is a separable differential equation for I , which we solve as follows:
dI
I yPx dx ln I yPx dx Ae x P x dx y I where A
e C. We are looking for a particular integrating factor, not the most general
one, so we take A 1 and use
ex P x Ix 5 dx Thus, a formula for the general solution to Equation 1 is provided by Equation 4, where I
is given by Equation 5. Instead of memorizing this formula, however, we just remember
the form of the integrating factor. To solve the linear differential equation y
P x y Q x , multiply both sides by
the integrating factor I x
e x P x d x and integrate both sides. EXAMPLE 1 Solve the differential equation dy
dx 3x 2 y 6 x 2. SOLUTION The given equation is linear since it has the form of Equation 1 with Px 3x 2 and Q x 6 x 2. An integrating factor is
Ix e x 3x 2 dx ex 3 3 Multiplying both sides of the differential equation by e x , we get
 Figure 1 shows the graphs of several members of the family of solutions in Example 1.
Notice that they all approach 2 as x l . ex 3 dy
dx 6
C=2 3 3x 2e x y
d x3
ey
dx or C=1
C=0 3 6x 2e x 3 Integrating both sides, we have C=_1
_1.5 1.8
C=_2 3 ex y 2 x3 y 6x e dx _3 FIGURE 1 6 x 2e x y 2 Ce x3 2e x 3 C 5E10(pp 662671) 670 ❙❙❙❙ 1/18/06 9:26 AM Page 670 CHAPTER 10 DIFFERENTIAL EQUATIONS EXAMPLE 2 Find the solution of the initialvalue problem x2y xy 1 x 0 y1 2 SOLUTION We must ﬁrst divide both sides by the coefﬁcient of y to put the differential
equation into standard form: 1
y
x y 6 1
x2 x 0 e ln x x The integrating factor is
ex Ix 1 x dx Multiplication of Equation 6 by x gives
y 1
x xy y xy Then
 The solution of the initialvalue problem in
Example 2 is shown in Figure 2. and so or
1
dx
x ln x ln x y 1
x xy C C
x 5 Since y 1 2, we have (1, 2)
0 ln 1 2 4 C C 1 Therefore, the solution to the initialvalue problem is
_5 ln x
x y FIGURE 2 EXAMPLE 3 Solve y 2 xy 2 1. SOLUTION The given equation is in the standard form for a linear equation. Multiplying by
the integrating factor
 Even though the solutions of the differential
equation in Example 3 are expressed in terms of
an integral, they can still be graphed by a computer algebra system (Figure 3). we get 2 2 xe x y 2 ex 2 (e x y) 2 2 ex y or 2.5 ex ex 2 e x 2x dx C=2 2 ex y Therefore
_2.5 ye x2 dx C 2.5
2 C=_2 Recall from Section 8.5 that x e x d x can’t be expressed in terms of elementary functions.
Nonetheless, it’s a perfectly good function and we can leave the answer as
_2.5 FIGURE 3 y e x2 ye x2 dx Ce x2 5E10(pp 662671) 1/18/06 9:26 AM Page 671 S ECTION 10.6 LINEAR EQUATIONS ❙❙❙❙ 671 Another way of writing the solution is
y x2 e y x 0 2 e t dt x2 Ce (Any number can be chosen for the lower limit of integration.) Application to Electric Circuits
R E L switch In Section 10.2 we considered the simple electric circuit shown in Figure 4: An electromotive force (usually a battery or generator) produces a voltage of E t volts (V) and a current of I t amperes (A) at time t . The circuit also contains a resistor with a resistance of
R ohms ( ) and an inductor with an inductance of L henries (H).
Ohm’s Law gives the drop in voltage due to the resistor as RI . The voltage drop due to
the inductor is L dI dt . One of Kirchhoff’s laws says that the sum of the voltage drops is
equal to the supplied voltage E t . Thus, we have FIGURE 4 dI
dt L 7 RI Et which is a ﬁrstorder linear differential equation. The solution gives the current I at time t .
EXAMPLE 4 Suppose that in the simple circuit of Figure 4 the resistance is 12 and the
inductance is 4 H. If a battery gives a constant voltage of 60 V and the switch is closed
when t 0 so the current starts with I 0
0, ﬁnd (a) I t , (b) the current after 1 s, and
(c) the limiting value of the current.
SOLUTION
 The differential equation in Example 4 is
both linear and separable, so an alternative
method is to solve it as a separable equation
(Example 4 in Section 10.3). If we replace the
battery by a generator, however, we get an equation that is linear but not separable (Example 5). (a) If we put L
problem 4, R 12, and E t 4 dI
dt
dI
dt or 60 in Equation 7, we obtain the initialvalue 12 I 60 I0 0 3I 15 I0 0 Multiplying by the integrating factor e x 3 dt
e 3t dI
dt 3e 3 tI
d 3t
eI
dt e 3 t, we get
15e 3 t
15e 3 t e 3 tI
It
Since I 0 0, we have 5 C y 15e
5 0, so C
It 3t dt Ce 3t 5 and
51 e 3t 5e 3 t C 5E10(pp 672681) ❙❙❙❙ 672 1/18/06 5:16 PM Page 672 CHAPTER 10 DIFFERENTIAL EQUATIONS  Figure 5 shows how the current in Example 4
approaches its limiting value. (b) After 1 second the current is
I1 51 e 3 4.75 A 6 2.5 3t 5 lim e 3t 5
0 e 5 (c) y=5 0 lim I t lim 5 1 tl tl tl 5 EXAMPLE 5 Suppose that the resistance and inductance remain as in Example 4
but, instead of the battery, we use a generator that produces a variable voltage of
Et
60 sin 30t volts. Find I t . FIGURE 5 SOLUTION This time the differential equation becomes 4 dI
dt 12 I 60 sin 30t dI
dt or 3I 15 sin 30t The same integrating factor e 3 t gives
d 3t
eI
dt  Figure 6 shows the graph of the current
when the battery is replaced by a generator. e 3t dI
dt 3e 3 tI Using Formula 98 in the Table of Integrals, we have 2 e 3 tI
0 15e 3 t sin 30t I 2.5 Since I 0 y 15e
5
101 3t sin 30t dt sin 30t e 3t
3 sin 30t
909 15 10 cos 30t Ce 30 cos 30t C 3t 0, we get
50
101 C 0 _2 so FIGURE 6  10.6
1–4 e xy 1. y
3. xy
■ 5–14  5. y 2y 9. x y y dy
dx ■ ■ ■ ■ x 3y sin x 4. y 0 12. xy
2 xy 2e x ■ x2
sx 6. y
8. x 2 y dy
dx 50
101 10 cos 30t ■ ■ ■ t e 3t dr
dt ■ x x2 u 2 t, t 1 x 0 ■ te t r
■ ■ ■ ■ ■ ■ 5y
2 xy cos 2 x 2 ■ 15–20  15. y
16. t xy du
dt 13. 1 tan x y tan x, x sin 2 x 14. t ln t cos y Solve the differential equation. 2y 11. 2. y x 2y ■ 7. xy 10. 1 x 2y 2 ln x ■ sin 30t Exercises Determine whether the differential equation is linear.  5
101 It 17. Solve the initialvalue problem.
x dy
dt dv
dt y,
2y 2tv y0
t 3, 2
t 0, y1 3t 2e t , v 0 5 2 0 ■ ■ ■ ■ 5E10(pp 672681) 1/18/06 5:17 PM Page 673 S ECTION 10.6 LINEAR EQUATIONS 18. 2 xy y 19. xy
20. x
■ x y y x, 1 ■ ■ y4 y x sin x,
x ■ 0, 2 dy
dx ; 21–22 6 x, 20 ■ 0, ■ x 0 ■ ■ ■ ■ ■ 22. y cos x y ■ ■ x cos x,
■ x But I dQ dt (see Example 3 in Section 3.4), so we have
R 1
Q
C dQ
dt Et 0 cos x
■ Et ■ Solve the differential equation and use a graphing calculator or computer to graph several members of the family of
solutions. How does the solution curve change as C varies?
y Q
C RI  21. xy 673 capacitor is Q C, where Q is the charge (in coulombs), so in
this case Kirchhoff’s Law gives 0
y1 ❙❙❙❙ ■ ■ ■ ■ ■ ■ ■ ■ Suppose the resistance is 5 , the capacitance is 0.05 F, a
battery gives a constant voltage of 60 V, and the initial charge
is Q 0
0 C. Find the charge and the current at time t. 23. A Bernoulli differential equation (named after James
30. In the circuit of Exercise 29, R Bernoulli) is of the form
dy
dx and E t
Q x yn Pxy 31. Let P t be the performance level of someone learning a skill Observe that, if n 0 or 1, the Bernoulli equation is linear.
For other values of n, show that the substitution u y 1 n transforms the Bernoulli equation into the linear equation
du
dx
24–26  1 nPxu 1 Use the method of Exercise 23 to solve the differential 26. y
■ xy 2 y ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ 27. In the circuit shown in Figure 4, a battery supplies a constant voltage of 40 V, the inductance is 2 H, the resistance is 10
and I 0
0.
(a) Find I t .
(b) Find the current after 0.1 s. , 28. In the circuit shown in Figure 4, a generator supplies a voltage ; Pt 32. Two new workers were hired for an assembly line. Jim xy 3 y kM as a reasonable model for learning, where k is a positive
constant. Solve it as a linear differential equation and use your
solution to graph the learning curve. y3
x2 2
y
x 25. y as a function of the training time t. The graph of P is called a
learning curve. In Exercise 13 in Section 10.1 we proposed the
differential equation
dP
dt nQx equation.
24. xy 2 , C 0.01 F, Q 0
0,
10 sin 60t. Find the charge and the current at time t. of E t
40 sin 60t volts, the inductance is 1 H, the resistance
is 20 , and I 0
1 A.
(a) Find I t .
(b) Find the current after 0.1 s.
(c) Use a graphing device to draw the graph of the current
function.
29. The ﬁgure shows a circuit containing an electromotive force, a capacitor with a capacitance of C farads (F), and a resistor
with a resistance of R ohms ( ). The voltage drop across the processed 25 units during the ﬁrst hour and 45 units during the
second hour. Mark processed 35 units during the ﬁrst hour and
50 units the second hour. Using the model of Exercise 31 and
assuming that P 0
0, estimate the maximum number of
units per hour that each worker is capable of processing.
33. In Section 10.3 we looked at mixing problems in which the volume of ﬂuid remained constant and saw that such problems
give rise to separable equations. (See Example 6 in that
section.) If the rates of ﬂow into and out of the system are different, then the volume is not constant and the resulting differential equation is linear but not separable.
A tank contains 100 L of water. A solution with a salt concentration of 0.4 kg L is added at a rate of 5 L min. The
solution is kept mixed and is drained from the tank at a rate
of 3 L min. If y t is the amount of salt (in kilograms) after
t minutes, show that y satisﬁes the differential equation
dy
dt 2 3y
100 2t C Solve this equation and ﬁnd the concentration after 20 minutes.
E R 34. A tank with a capacity of 400 L is full of a mixture of water and chlorine with a concentration of 0.05 g of chlorine per liter.
In order to reduce the concentration of chlorine, fresh water is
pumped into the tank at a rate of 4 L s. The mixture is kept 5E10(pp 672681) 674 ❙❙❙❙ 1/18/06 5:17 PM Page 674 CHAPTER 10 DIFFERENTIAL EQUATIONS stirred and is pumped out at a rate of 10 L s. Find the amount
of chlorine in the tank as a function of time. (a) Solve this as a linear equation to show that
mt
v
1 e ct m
c 35. An object with mass m is dropped from rest and we assume that the air resistance is proportional to the speed of the object.
If s t is the distance dropped after t seconds, then the speed is
v
s t and the acceleration is a v t . If t is the acceleration
due to gravity, then the downward force on the object is
m t cv, where c is a positive constant, and Newton’s Second
Law gives
m  10.7 dv
dt mt (b) What is the limiting velocity?
(c) Find the distance the object has fallen after t seconds.
36. If we ignore air resistance, we can conclude that heavier objects fall no faster than lighter objects. But if we take air
resistance into account, our conclusion changes. Use the
expression for the velocity of a falling object in Exercise 35(a)
to ﬁnd dv dm and show that heavier objects do fall faster than
lighter ones. cv PredatorPrey Systems
We have looked at a variety of models for the growth of a single species that lives alone in
an environment. In this section we consider more realistic models that take into account
the interaction of two species in the same habitat. We will see that these models take the
form of a pair of linked differential equations.
We ﬁrst consider the situation in which one species, called the prey, has an ample food
supply and the second species, called the predators, feeds on the prey. Examples of prey
and predators include rabbits and wolves in an isolated forest, food ﬁsh and sharks, aphids
and ladybugs, and bacteria and amoebas. Our model will have two dependent variables and
both are functions of time. We let R t be the number of prey (using R for rabbits) and W t
be the number of predators (with W for wolves) at time t.
In the absence of predators, the ample food supply would support exponential growth
of the prey, that is,
dR
dt kR where k is a positive constant In the absence of prey, we assume that the predator population would decline at a rate proportional to itself, that is,
dW
dt rW where r is a positive constant With both species present, however, we assume that the principal cause of death among the
prey is being eaten by a predator, and the birth and survival rates of the predators depend
on their available food supply, namely, the prey. We also assume that the two species
encounter each other at a rate that is proportional to both populations and is therefore proportional to the product RW. (The more there are of either population, the more encounters there are likely to be.) A system of two differential equations that incorporates these
assumptions is as follows:
W represents the predator.
R represents the prey. 1 dR
dt kR aRW dW
dt rW bRW where k, r, a, and b are positive constants. Notice that the term aRW decreases the natural growth rate of the prey and the term bRW increases the natural growth rate of the
predators. 5E10(pp 672681) 1/18/06 5:17 PM Page 675 S ECTION 10.7 PREDATORPREY SYSTEMS  The LotkaVolterra equations were proposed
as a model to explain the variations in the shark
and foodﬁsh populations in the Adriatic Sea
by the Italian mathematician Vito Volterra
(1860–1940). ❙❙❙❙ 675 The equations in (1) are known as the predatorprey equations, or the LotkaVolterra
equations. A solution of this system of equations is a pair of functions R t and W t that
describe the populations of prey and predator as functions of time. Because the system is
coupled (R and W occur in both equations), we can’t solve one equation and then the other;
we have to solve them simultaneously. Unfortunately, it is usually impossible to ﬁnd
explicit formulas for R and W as functions of t. We can, however, use graphical methods
to analyze the equations.
E XAMPLE 1 Suppose that populations of rabbits and wolves are described by the LotkaVolterra equations (1) with k 0.08, a 0.001, r 0.02, and b 0.00002. The time t
is measured in months.
(a) Find the constant solutions (called the equilibrium solutions) and interpret
the answer.
(b) Use the system of differential equations to ﬁnd an expression for dW dR.
(c) Draw a direction ﬁeld for the resulting differential equation in the RWplane. Then
use that direction ﬁeld to sketch some solution curves.
(d) Suppose that, at some point in time, there are 1000 rabbits and 40 wolves. Draw the
corresponding solution curve and use it to describe the changes in both population levels.
(e) Use part (d) to make sketches of R and W as functions of t.
SOLUTION (a) With the given values of k, a, r, and b, the LotkaVolterra equations become
dR
dt 0.08R dW
dt 0.001RW 0.02W 0.00002RW Both R and W will be constant if both derivatives are 0, that is,
R R 0.08 W W 0.001W 0.02 0 0.00002R 0 One solution is given by R 0 and W 0. (This makes sense: If there are no rabbits or
wolves, the populations are certainly not going to increase.) The other constant solution is
W 0.08
0.001 80 R 0.02
0.00002 1000 So the equilibrium populations consist of 80 wolves and 1000 rabbits. This means that
1000 rabbits are just enough to support a constant wolf population of 80. There are neither too many wolves (which would result in fewer rabbits) nor too few wolves (which
would result in more rabbits).
(b) We use the Chain Rule to eliminate t :
dW
dt so dW
dR dW
dt
dR
dt dW dR
dR dt
0.02W
0.08R 0.00002RW
0.001RW 5E10(pp 672681) 676 ❙❙❙❙ 1/18/06 5:17 PM Page 676 CHAPTER 10 DIFFERENTIAL EQUATIONS (c) If we think of W as a function of R, we have the differential equation
dW
dR 0.02W
0.08R 0.00002RW
0.001RW We draw the direction ﬁeld for this differential equation in Figure 1 and we use it to
sketch several solution curves in Figure 2. If we move along a solution curve, we
observe how the relationship between R and W changes as time passes. Notice that the
curves appear to be closed in the sense that if we travel along a curve, we always return
to the same point. Notice also that the point (1000, 80) is inside all the solution curves.
That point is called an equilibrium point because it corresponds to the equilibrium solution R 1000, W 80.
W W 150 150 100 100 50 50 0 FIGURE 1 1000 2000 0 3000 R FIGURE 2 Direction ﬁeld for the predatorprey system 3000 R 2000 1000 Phase portrait of the system When we represent solutions of a system of differential equations as in Figure 2, we
refer to the RWplane as the phase plane, and we call the solution curves phase trajectories. So a phase trajectory is a path traced out by solutions R, W as time goes by. A
phase portrait consists of equilibrium points and typical phase trajectories, as shown in
Figure 2.
(d) Starting with 1000 rabbits and 40 wolves corresponds to drawing the solution curve
through the point P0(1000, 40). Figure 3 shows this phase trajectory with the direction
ﬁeld removed. Starting at the point P0 at time t 0 and letting t increase, do we move
clockwise or counterclockwise around the phase trajectory? If we put R 1000 and
W P™
140
120
100
80 P£ P¡ 60
40 P¸ (1000, 40) 20 F IGURE 3 Phase trajectory through (1000, 40) 0 500 1000 1500 2000 2500 3000 R 5E10(pp 672681) 1/18/06 5:18 PM Page 677 SECTION 10.7 PREDATORPREY SYSTEMS W ❙❙❙❙ 677 40 in the ﬁrst differential equation, we get
dR
dt 0.08 1000 0.001 1000 40 80 40 40 Since dR dt 0, we conclude that R is increasing at P0 and so we move counterclockwise around the phase trajectory.
We see that at P0 there aren’t enough wolves to maintain a balance between the populations, so the rabbit population increases. That results in more wolves and eventually
there are so many wolves that the rabbits have a hard time avoiding them. So the number
of rabbits begins to decline (at P1 , where we estimate that R reaches its maximum population of about 2800). This means that at some later time the wolf population starts to
fall (at P2 , where R 1000 and W 140). But this beneﬁts the rabbits, so their population later starts to increase (at P3 , where W 80 and R 210). As a consequence, the wolf population eventually starts to increase as well. This happens when the
populations return to their initial values of R 1000 and W 40, and the entire cycle
begins again.
(e) From the description in part (d) of how the rabbit and wolf populations rise and fall,
we can sketch the graphs of R t and W t . Suppose the points P1 , P2 , and P3 in Figure 3
are reached at times t1 , t2 , and t3 . Then we can sketch graphs of R and W as in Figure 4.
R W
140 2500 120
2000 100 1500 80
60 1000 40
500
0 20
t¡ t™ 0 t t£ t¡ t™ t t£ FIGURE 4 Graphs of the rabbit and wolf
populations as functions of time To make the graphs easier to compare, we draw the graphs on the same axes but with
different scales for R and W, as in Figure 5. Notice that the rabbits reach their maximum
populations about a quarter of a cycle before the wolves.
R
3000 W R
W 120
Number 2000
of
rabbits 80 Number
of
wolves 1000
40 FIGURE 5 Comparison of the rabbit
and wolf populations 0 t¡ t™ t£ t 5E10(pp 672681) 678 ❙❙❙❙ 1/18/06 5:18 PM Page 678 CHAPTER 10 DIFFERENTIAL EQUATIONS An important part of the modeling process, as we discussed in Section 1.2, is to interpret our mathematical conclusions as realworld predictions and to test the predictions
against real data. The Hudson’s Bay Company, which started trading in animal furs in
Canada in 1670, has kept records that date back to the 1840s. Figure 6 shows graphs of the
number of pelts of the snowshoe hare and its predator, the Canada lynx, traded by the company over a 90year period. You can see that the coupled oscillations in the hare and lynx
populations predicted by the LotkaVolterra model do actually occur and the period of
these cycles is roughly 10 years.
160 hare
120 9 lynx
Thousands 80
of hares 6 Thousands
of lynx 40 3 FIGURE 6 Relative abundance of hare and lynx
from Hudson’s Bay Company records 0 1850 1875 1900 1925 Although the relatively simple LotkaVolterra model has had some success in explaining and predicting coupled populations, more sophisticated models have also been proposed. One way to modify the LotkaVolterra equations is to assume that, in the absence
of predators, the prey grow according to a logistic model with carrying capacity K. Then
the LotkaVolterra equations (1) are replaced by the system of differential equations
dR
dt kR 1 R
K aRW dW
dt rW bRW This model is investigated in Exercises 9 and 10.
Models have also been proposed to describe and predict population levels of two
species that compete for the same resources or cooperate for mutual beneﬁt. Such models
are explored in Exercise 2.  10.7 Exercises 1. For each predatorprey system, determine which of the vari ables, x or y, represents the prey population and which
represents the predator population. Is the growth of the prey
restricted just by the predators or by other factors as well? Do
the predators feed only on the prey or do they have additional
food sources? Explain.
dx
(a)
0.05x 0.0001xy
dt
dy
0.1y 0.005xy
dt
(b) dx
dt
dy
dt 0.2 x 0.0002 x 2 0.015y 0.006xy 0.00008xy 2. Each system of differential equations is a model for two species that either compete for the same resources or cooperate
for mutual beneﬁt (ﬂowering plants and insect pollinators, for
instance). Decide whether each system describes competition
or cooperation and explain why it is a reasonable model. (Ask
yourself what effect an increase in one species has on the
growth rate of the other.)
dx
(a)
0.12 x 0.0006x 2 0.00001xy
dt
dy
0.08x 0.00004xy
dt
dx
(b)
0.15x 0.0002 x 2 0.0006xy
dt
dy
0.2y 0.00008y 2 0.0002 xy
dt 5E10(pp 672681) 1/18/06 5:18 PM Page 679 ❙❙❙❙ S ECTION 10.7 PREDATORPREY SYSTEMS y 6. 3–4  A phase trajectory is shown for populations of rabbits R
and foxes F .
(a) Describe how each population changes as time goes by.
(b) Use your description to make a rough sketch of the graphs of R
and F as functions of time. species 1 1200
1000
800 F 600 300 3. 679 400 species 2 200
0 200
■ ■ ■ ■ ■ ■ ■ 800 400 1200 1600 R 2000 0.02W
0.08R ■ ■ t
■ ■ 0.00002RW
0.001RW By solving this separable differential equation, show that
R 0.02W 0.08
e
e C 0.00002R 0.001W 4. 15 tions satisfy the differential equation
dW
dR 0 10 7. In Example 1(b) we showed that the rabbit and wolf popula t=0 100 ■ 5 F where C is a constant.
It is impossible to solve this equation for W as an explicit
function of R (or vice versa). If you have a computer algebra
system that graphs implicitly deﬁned curves, use this equation
and your CAS to draw the solution curve that passes through
the point 1000, 40 and compare with Figure 3. t=0 160 120 8. Populations of aphids and ladybugs are modeled by the 80 equations
dA
dt 40 0 ■ ■ 400 ■ ■ 800 ■ ■ 1200 ■ ■ R 1600 ■ ■ ■ ■ 5–6  Graphs of populations of two species are shown. Use them
to sketch the corresponding phase trajectory. 5. y dL
dt 2A 0.01AL 0.5L 0.0001AL (a) Find the equilibrium solutions and explain their
signiﬁcance.
(b) Find an expression for dL dA.
(c) The direction ﬁeld for the differential equation in part (b) is
shown. Use it to sketch a phase portrait. What do the phase
trajectories have in common?
L species 1 200 400 species 2 300
100 200
100 0 1 t 0 5000 10000 15000 A 5E10(pp 672681) ❙❙❙❙ 680 1/18/06 5:18 PM Page 680 CHAPTER 10 DIFFERENTIAL EQUATIONS (d) Suppose that at time t 0 there are 1000 aphids and
200 ladybugs. Draw the corresponding phase trajectory and
use it to describe how both populations change.
(e) Use part (d) to make rough sketches of the aphid and ladybug populations as functions of t. How are the graphs
related to each other? (b) Find all the equilibrium solutions and explain their
signiﬁcance.
(c) The ﬁgure shows the phase trajectory that starts at the point
1000, 40 . Describe what eventually happens to the rabbit
and wolf populations.
(d) Sketch graphs of the rabbit and wolf populations as functions of time. 9. In Example 1 we used LotkaVolterra equations to model popu lations of rabbits and wolves. Let’s modify those equations as
follows:
dR
0.08R 1 0.0002R
0.001RW
dt
dW
dt 0.02W CAS 10. In Exercise 8 we modeled populations of aphids and ladybugs with a LotkaVolterra system. Suppose we modify those equations as follows:
dA
dt 0.00002RW (a) According to these equations, what happens to the rabbit
population in the absence of wolves? dL
dt W 60 50 40  800 1000 1200 1400 1600 10 Review R ■ CONCEPT CHECK 1. (a) What is a differential equation? (b) What is the order of a differential equation?
(c) What is an initial condition?
2. What can you say about the solutions of the equation y x2 y 2 just by looking at the differential equation? 3. What is a direction ﬁeld for the differential equation y F x, y ? 4. Explain how Euler’s method works.
5. What is a separable differential equation? How do you solve it?
6. What is a ﬁrstorder linear differential equation? How do you solve it? 0.5L 0.0001A 0.01AL 0.0001AL (a) In the absence of ladybugs, what does the model predict
about the aphids?
(b) Find the equilibrium solutions.
(c) Find an expression for dL dA.
(d) Use a computer algebra system to draw a direction ﬁeld for
the differential equation in part (c). Then use the direction
ﬁeld to sketch a phase portrait. What do the phase trajectories have in common?
(e) Suppose that at time t 0 there are 1000 aphids and
200 ladybugs. Draw the corresponding phase trajectory and
use it to describe how both populations change.
(f) Use part (e) to make rough sketches of the aphid and
ladybug populations as functions of t. How are the graphs
related to each other? 70 30
600 2A 1 ■ 7. (a) Write a differential equation that expresses the law of natural growth. What does it say in terms of relative
growth rate?
(b) Under what circumstances is this an appropriate model for
population growth?
(c) What are the solutions of this equation?
8. (a) Write the logistic equation. (b) Under what circumstances is this an appropriate model for
population growth?
9. (a) Write LotkaVolterra equations to model populations of food ﬁsh F and sharks S .
(b) What do these equations say about each population in the
absence of the other? 5E10(pp 672681) 1/18/06 5:19 PM Page 681 C HAPTER 10 REVIEW ■ TRUEFALSE QUIZ 1. All solutions of the differential equation y equation x 2 y ln x x is a solution of the differential
1. xy 3. The equation y x 4. The equation y e y is linear. xy 7. If y is the solution of the initialvalue problem decreasing functions.
2. The function f x y is linear. 6. The equation y y 4 are 1 681 ■ 5. The equation e x y Determine whether the statement is true or false. If it is true, explain why.
If it is false, explain why or give an example that disproves the statement. ❙❙❙❙ 3y dy
dt y is separable.
2x 6 xy 1 is separable. then lim t l y ■ EXERCISES y
5 2y 1 y0 1 5. ■ y
3 1. (a) A direction ﬁeld for the differential equation y
y y 2 y 4 is shown. Sketch the graphs of the
solutions that satisfy the given initial conditions.
0.3
1
(i) y 0
(ii) y 0
3
4.3
(iii) y 0
(iv) y 0
(b) If the initial condition is y 0
c, for what values of c is
lim t l y t ﬁnite? What are the equilibrium solutions? 2
1 _3 y
6 _2 0 _1 1 3x 2 _1
_2 4 _3 2 0 1 2 (b) Use Euler’s method with step size 0.1 to estimate y 0.3
where y x is the solution of the initialvalue problem in
part (a). Compare with your estimate from part (a).
(c) On what lines are the centers of the horizontal line
segments of the direction ﬁeld in part (a) located? What
happens when a solution curve crosses these lines? x 4. (a) Use Euler’s method with step size 0.2 to estimate y 0.4 , 2. (a) Sketch a direction ﬁeld for the differential equation y
x y. Then use it to sketch the four solutions that
satisfy the initial conditions y 0
1, y 0
1, y 2
and y 2
1.
(b) Check your work in part (a) by solving the differential
equation explicitly. What type of curve is each solution
curve? where y x is the solution of the initialvalue problem x2 y2
is shown. Sketch the solution of the initialvalue problem 3. (a) A direction ﬁeld for the differential equation y y x y 2 y0 y0 5–8 Solve the differential equation.  sin x xe 6. y cos x 1
7. 3y 2 Use your graph to estimate the value of y 0.3 . 1 (b) Repeat part (a) with step size 0.1.
(c) Find the exact solution of the differential equation and
compare the value at 0.4 with the approximations in
parts (a) and (b). 5. y
2 2 xy 2 y 1, ■ ■ 2y y
■ 8. x 2 y x cos x
■ ■ dx
dt ■ ■ ■ 1
y
■ t x
2 x 3e
■ tx
1x
■ ■ 5E10(pp 682683) 1/18/06 9:30 AM Page 682 682 ❙❙❙❙ CHAPTER 10 DIFFERENTIAL EQUATIONS 9–11  Solve the initialvalue problem. 9. xyy
10. 1 x 11. y
■ ln x, 2 xyy , 2
x s x e x, y
■ y1 ■ 19. The von Bertalanffy growth model is used to predict the length ■ 0, y1 y0
■ 2 3
■ ■ ■ ■ ■ xe x, y 0 ; 12. Solve the initialvalue problem 2yy ■ ■ 1, and graph the solution.
13–14  13. kx 2
■ ■ 20. A tank contains 100 L of pure water. Brine that contains Find the orthogonal trajectories of the family of curves.
y2
■ ■ ■ ■ ■ 0.1 kg of salt per liter enters the tank at a rate of 10 L min.
The solution is kept thoroughly mixed and drains from the tank
at the same rate. How much salt is in the tank after 6 minutes? k 14. y 1 1
■ x2
■ ■ ■ L t of a ﬁsh over a period of time. If L is the largest length
for a species, then the hypothesis is that the rate of growth in
length is proportional to L
L , the length yet to be achieved.
(a) Formulate and solve a differential equation to ﬁnd an
expression for L t .
(b) For the North Sea haddock it has been determined that
53 cm, L 0
L
10 cm, and the constant of proportionality is 0.2. What does the expression for L t become
with these data? ■ 15. A bacteria culture starts with 1000 bacteria and the growth rate is proportional to the number of bacteria. After 2 hours the
population is 9000.
(a) Find an expression for the number of bacteria after t hours.
(b) Find the number of bacteria after 3 h.
(c) Find the rate of growth after 3 h.
(d) How long does it take for the number of bacteria to double?
16. An isotope of strontium, 90Sr, has a halflife of 25 years. (a) Find the mass of 90Sr that remains from a sample of 18 mg
after t years.
(b) How long would it take for the mass to decay to 2 mg?
17. Let C t be the concentration of a drug in the bloodstream. As the body eliminates the drug, C t decreases at a rate that is
proportional to the amount of the drug that is present at the
time. Thus, C t
kC t , where k is a positive number
called the elimination constant of the drug.
(a) If C0 is the concentration at time t 0, ﬁnd the concentration at time t.
(b) If the body eliminates half the drug in 30 h, how long does
it take to eliminate 90% of the drug?
18. (a) The population of the world was 5.28 billion in 1990 and 6.07 billion in 2000. Find an exponential model for these
data and use the model to predict the world population in
the year 2020.
(b) According to the model in part (a), when will the world
population exceed 10 billion?
(c) Use the data in part (a) to ﬁnd a logistic model for the population. Assume a carrying capacity of 100 billion. Then
use the logistic model to predict the population in 2020.
Compare with your prediction from the exponential model.
(d) According to the logistic model, when will the world population exceed 10 billion? Compare with your prediction in
part (b). 21. One model for the spread of an epidemic is that the rate of spread is jointly proportional to the number of infected people
and the number of uninfected people. In an isolated town of
5000 inhabitants, 160 people have a disease at the beginning of
the week and 1200 have it at the end of the week. How long
does it take for 80% of the population to become infected?
22. The BrentanoStevens Law in psychology models the way that a subject reacts to a stimulus. It states that if R represents the
reaction to an amount S of stimulus, then the relative rates of
increase are proportional:
1 dR
R dt k dS
S dt where k is a positive constant. Find R as a function of S.
23. The transport of a substance across a capillary wall in lung physiology has been modeled by the differential equation
R
V dh
dt h
k h where h is the hormone concentration in the bloodstream, t is
time, R is the maximum transport rate, V is the volume of the
capillary, and k is a positive constant that measures the afﬁnity
between the hormones and the enzymes that assist the process.
Solve this differential equation to ﬁnd a relationship between h
and t.
24. Populations of birds and insects are modeled by the equations dx
dt
dy
dt 0.4 x
0.2y 0.002 xy
0.000008 xy (a) Which of the variables, x or y, represents the bird population and which represents the insect population? Explain. 5E10(pp 682683) 1/18/06 9:30 AM Page 683 C HAPTER 10 R EVIEW (b) Find the equilibrium solutions and explain their
signiﬁcance.
(c) Find an expression for dy dx.
(d) The direction ﬁeld for the differential equation in part (c) is
shown. Use it to sketch the phase trajectory corresponding
to initial populations of 100 birds and 40,000 insects. Then
use the phase trajectory to describe how both populations
change. ❙❙❙❙ 683 y
260
240
220
200
180
160
140
120
100 y
400 10000 15000 35000 45000 x (d) Sketch graphs of the bird and insect populations as
functions of time. 300 26. Barbara weighs 60 kg and is on a diet of 1600 calories per day, 200
100 0 25000 20000 40000 60000 x of which 850 are used automatically by basal metabolism. She
spends about 15 cal kg day times her weight doing exercise. If
1 kg of fat contains 10,000 cal and we assume that the storage
of calories in the form of fat is 100% efﬁcient, formulate a differential equation and solve it to ﬁnd her weight as a function
of time. Does her weight ultimately approach an equilibrium
weight?
27. When a ﬂexible cable of uniform density is suspended between (e) Use part (d) to make rough sketches of the bird and insect
populations as functions of time. How are these graphs
related to each other?
25. Suppose the model of Exercise 24 is replaced by the equations dx
dt 0.4 x 1 dy
dt 0.2y 0.000005x 0.002 xy 0.000008 xy two ﬁxed points and hangs of its own weight, the shape
y f x of the cable must satisfy a differential equation of the
form
dy 2
d 2y
k1
2
dx
dx
where k is a positive constant. Consider the cable shown in the
ﬁgure.
(a) Let z d y d x in the differential equation. Solve the resulting ﬁrstorder differential equation (in z ), and then integrate
to ﬁnd y.
(b) Determine the length of the cable.
y (a) According to these equations, what happens to the insect
population in the absence of birds?
(b) Find the equilibrium solutions and explain their
signiﬁcance.
(c) The ﬁgure shows the phase trajectory that starts with
100 birds and 40,000 insects. Describe what eventually
happens to the bird and insect populations. (b, h) (_b, h) (0, a)
_b 0 b x 5E10(pp 684685) 1/18/06 9:30 AM Page 684 PROBLEMS
PLUS 1. Find all functions f such that f is continuous and [ f x ]2 y 100 x 0 [ f t ]2 [ f t ] 2 dt for all real x 2. A student forgot the Product Rule for differentiation and made the mistake of thinking that f t
f t . However, he was lucky and got the correct answer. The function f that he
2
used was f x
e x and the domain of his problem was the interval ( 1 , ). What was the
2
function t ?
3. Let f be a function with the property that f 0 1, f 0
1, and f a b
f x for all x and deduce that f x all real numbers a and b. Show that f x f a f b for
e x. 4. Find all functions f that satisfy the equation yf x dx y 1
dx
fx 1 5. A peach pie is taken out of the oven at 5:00 P.M. At that time it is piping hot: 100 C. At 5:10 P.M. its temperature is 80 C; at 5:20 P.M. it is 65 C. What is the temperature of the
room?
6. Snow began to fall during the morning of February 2 and continued steadily into the after noon. At noon a snowplow began removing snow from a road at a constant rate. The plow
traveled 6 km from noon to 1 P.M. but only 3 km from 1 P.M. to 2 P.M. When did the snow
begin to fall? [Hints: To get started, let t be the time measured in hours after noon; let x t
be the distance traveled by the plow at time t; then the speed of the plow is dx dt. Let b be
the number of hours before noon that it began to snow. Find an expression for the height
of the snow at time t. Then use the given information that the rate of removal R (in m3 h)
is constant.]
y 7. A dog sees a rabbit running in a straight line across an open ﬁeld and gives chase. In a rectan gular coordinate system (as shown in the ﬁgure), assume:
(i) The rabbit is at the origin and the dog is at the point L, 0 at the instant the dog ﬁrst
sees the rabbit.
(ii) The rabbit runs up the yaxis and the dog always runs straight for the rabbit.
(iii) The dog runs at the same speed as the rabbit.
(a) Show that the dog’s path is the graph of the function y
ential equation (x, y) x
0 FIGURE FOR PROBLEM 7 (L, 0) x d 2y
dx 2 1 dy
dx f x , where y satisﬁes the differ 2 (b) Determine the solution of the equation in part (a) that satisﬁes the initial conditions
yy
0 when x L. [Hint: Let z d y d x in the differential equation and solve the
resulting ﬁrstorder equation to ﬁnd z; then integrate z to ﬁnd y.]
(c) Does the dog ever catch the rabbit?
8. (a) Suppose that the dog in Problem 7 runs twice as fast as the rabbit. Find a differential equation for the path of the dog. Then solve it to ﬁnd the point where the dog catches the
rabbit.
(b) Suppose the dog runs half as fast as the rabbit. How close does the dog get to the rabbit?
What are their positions when they are closest? 684 5E10(pp 684685) 1/18/06 9:30 AM Page 685 9. A planning engineer for a new alum plant must present some estimates to his company regard ing the capacity of a silo designed to contain bauxite ore until it is processed into alum. The
ore resembles pink talcum powder and is poured from a conveyor at the top of the silo. The
silo is a cylinder 100 ft high with a radius of 200 ft. The conveyor carries 60,000 ft 3 h and
the ore maintains a conical shape whose radius is 1.5 times its height.
(a) If, at a certain time t, the pile is 60 ft high, how long will it take for the pile to reach the
top of the silo?
(b) Management wants to know how much room will be left in the ﬂoor area of the silo when
the pile is 60 ft high. How fast is the ﬂoor area of the pile growing at that height?
(c) Suppose a loader starts removing the ore at the rate of 20,000 ft 3 h when the height of
the pile reaches 90 ft. Suppose, also, that the pile continues to maintain its shape. How
long will it take for the pile to reach the top of the silo under these conditions?
10. Find the curve that passes through the point 3, 2 and has the property that if the tangent line is drawn at any point P on the curve, then the part of the tangent line that lies in the ﬁrst
quadrant is bisected at P.
11. Recall that the normal line to a curve at a point P on the curve is the line that passes through P and is perpendicular to the tangent line at P. Find the curve that passes through the point
3, 2 and has the property that if the normal line is drawn at any point on the curve, then the
yintercept of the normal line is always 6.
12. Find all curves with the property that if the normal line is drawn at any point P on the curve, then the part of the normal line between P and the xaxis is bisected by the yaxis. 685 ...
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