Blank B.E., Krantz S.G. Calculus: Single Variable

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62:428 πð20

2 y

Þydy 5 62:428π

ð400y 2 y

Þdy

5 62:428π



200y





y520

y50

5 62:428π



200  20



;

which comes to about 7.84493 3 10

foot-pounds. ¥

QUICK QUIZ

1. To overcome friction, a force of 12 2 3x

N is used to push an object from x 5 0

to x 5 2 m. How much work is done?

2. If the amount of work in stretching a spring 0.02 meter beyond its equilibrium

position is 8 J, then what force is necessary to maintain the spring at that

position?

3. A cube of side length 1 m is ﬁlled with a ﬂuid that weighs 1 newton per cubic

meter. What work is done in pumping the ﬂuid to the surface?

Answers

1. 16 J 2. 800 N 3.

EXERCISES

Problems for Practice

1. A steam shovel lifts a 500 pound load of gravel from the

ground to a point 80 feet above the ground. The gravel is

ﬁne, however, and it leaks from the shovel at the rate of

1 pound per second. If it takes the steam shovel one

minute to lift its load at a constant rate, then how much

work is performed?

2. On the surface of the earth, a rocket weighs 10,000

newtons. How much work is performed lifting the rocket

to a height 100 kilometers above the surface of the earth,

assuming that the direction of the force, as well as the

motion, is straight up?

3. An object that weighs W

at the surface of the earth

weighs

15697444 

ð3962 1 yÞ

when it is y miles above the surface of the earth. How

much work is done lifting an object from the surface of

the earth, where it weighs 100 pounds, straight up to a

height 100 miles above Earth?

4. When a mass M measured in slugs is y miles above the

surface of the earth, its weight in pounds is

502318208 

ð3962 1 yÞ

How much work is done lifting a satellite from the sur-

face of the earth, where it weighs 800 pounds, to an orbit

200 miles above Earth?

5. To lift an object straight up from the surface of the earth

to a height 25 km above the surface of the earth requires

58610091 J of work. What is the mass of the object?

6. A 100 kg mass is lifted to a height at which its weight is

98% of its weight at the surface of Earth. How much

work is done?

7. A rocket climbs straight up. Its initial total weight,

including fuel, is 7000 pounds. Fuel is consumed at the

constant rate of 30 pounds per mile. Taking into account

the decrease in weight due to fuel consumption but dis-

regarding the decrease in weight due to increasing ele-

vation, approximate the work done in lifting the rocket

the ﬁrst 20 miles into space.

8. A man stands at the top of a tall building and pulls a

chain up the side of the building. The chain is 50 feet long

and weighs 3 pounds per linear foot. How much work

does the man do in pulling the chain to the top?

9. With regard to the preceding exercise, how much work

did the man do in pulling up the ﬁrst 30 feet of the chain

to the top of the building?

10. A heavy uniform cable is used to lift a 300 kg load from

ground level to the top of a building that is 40 m high. If

the cable weighs 220 newtons per linear meter, then how

much work is done?

11. With regard to the preceding exercise, how much work is

done in lifting the load from ground level to a height

24 meters above ground level?

12. With regard to the cable and load of Exercises 10 and 11,

how much work was done in lifting the load from a height

7.5 Work 585

24 meters above ground level to a height of 30 meters

above ground level?

13. A heavy uniform cable is used to lift a 300 pound load

from ground level to the top of a 100 foot high building. If

the cable weighs 20 pounds per linear foot, then how

much work is done?

14. With regard to the preceding exercise, how much work is

done in lifting the load from ground level to a height 30

feet above ground level?

15. With regard to the cable and load of Exercises 13 and 14,

how much work is done in lifting the load from a height

30 feet above ground level to a height 80 feet above

ground level?

16. A crane lifts a large boat out of its berth to place it in dry

dock. The vessel, including its contents, weighs 80 tons.

But as the boat is lifted, it releases water from its holds at

the constant rate of 50 cubic feet per minute. The crane

raises the boat a distance of 50 feet at the constant rate of

5 feet per minute. How much work is performed in lifting

the boat?

17. If a spring with spring constant 8 pounds per inch is

stretched 7 inches beyond its equilibrium position, then

how much work is done?

18. If a spring with spring constant 20 newtons per cen-

timeter is stretched 12 cm beyond its equilibrium posi-

tion, then how much work is done?

19. A spring is stretched 10 cm beyond its equilibrium posi-

tion. If the force required to maintain it in its stretched

position is 240 N, then how much work was done?

20. A spring is stretched 2 inches beyond its equilibrium

position. If the force required to maintain it in its stret-

ched position is 60 pounds, then how much work was

done?

21. A force of 280 pounds compresses a spring 4 inches from

its natural length. How much work is done compressing it

a further 4 inches?

22. A force of 10

newtons compresses a spring 10 cm from

its natural length. How much work is done compressing it

a further 10 cm?

23. If 40 joules of work are done in stretching a spring 8 cm

beyond its natural length, then how much work is done

stretching a further 4 cm?

24. If 30 foot-pounds of work are done in stretching a spring

3 inches beyond its natural length, then how much work is

done stretching it 1 inch more?

25. The work done in extending a spring at rest 2 inches

beyond its equilibrium position is equal to 2/3 ft-lb. How

much force is required to maintain the spring in that

position?

26. The work done in extending a spring at rest 4 cm beyond

its equilibrium position is 1.2 N. What force is required to

maintain the spring in that position?

27. A swimming pool full of water has square base of side 15

feet. It is 10 feet deep and is being pumped dry. The

pump ﬂoats on the surface of the water and pumps the

water to the top of the pool, at which point the water runs

off. How much work is done in emptying the pool?

28. A tank in the shape of a box has square base of side 5 m.

It is 4 m deep and is being pumped dry. The pump ﬂoats

on the surface of the water and pumps the water to the

top of the tank, at which point the water runs off. How

much work is done in emptying the tank?

29. A swimming pool has rectangular base with side lengths

of 6 m and 10 m. The pool is 3 m deep but not completely

ﬁlled: the depth of the water it contains is 2 m. A pump

ﬂoats on the surface of the water and pumps the water to

the top of the pool, at which point the water runs off.

How much work is done emptying the pool?

30. A swimming pool has rectangular base with side lengths

of 16 and 30 feet. The pool is 10 feet deep but not com-

pletely ﬁlled: the depth of the water it contains is 8 feet.

A pump ﬂoats on the surface of the water and pumps the

water to the top of the pool, at which point the water runs

off. How much work does the pump perform in emptying

the pool?

31. Referring to the swimming pool of the preceding pro-

blem, if power is interrupted when half the water has

been pumped, then how much work has been done?

32. A cube-shaped tank has side length 4 m. The depth

of water it holds is 3 m. A pump ﬂoats on the surface of

the water and pumps the water to the top of the tank, at

which point the water runs off. How much work is per-

formed removing 32 m

of water?

33. A semi-cylindrical tank ﬁlled with water is 16 feet long. It

has rectangular horizontal cross-sections and vertical

cross-sections that are semicircular of radius 6 feet (see

Figure 8). A pump ﬂoats on the surface of the tank water

and pumps water to the top, at which point the water runs

off. How much work is done in emptying the tank?

34. A semi-cylindrical tank ﬁlled with water is shaped as in

Figure 8 but with different dimensions: It is 5 m long, has

rectangular horizontal cross-sections, and has vertical

cross-sections that are semicircular of radius 2 m. A pump

Semi-c

lindrical tank

m Figure 8

586 Chapter

7 Applications of the Integral

ﬂoats on the surface of the water and pumps water to the

top, at which point the water runs off. How much work is

done in emptying the tank?

35. A ﬁlled reservoir is in the shape of a frustum of an

inverted cone as shown in Figure 9. The radius of the

circular base of the cone is 100 feet. The radius of the

circular top of the frustum and the depth at the center

are both 50 feet. A pump that ﬂoats on the surface of

the water pumps water to the top of the reservoir, at

which point the water runs off. How much work is done

emptying the reservoir?

36. A tank ﬁlled with water is in the shape of a frustum of

pyramid with square base. The base of the pyramid

measures 8 m on a side, the side length of the square top

is 4 m, and the height of the frustum is 6 m (see Figure

10). A pump ﬂoats on the surface of the water and pumps

water to the upper edge. If the depth of water in the tank

is reduced to 2 m, then what amount of work has been

performed?

Further Theory and Practice

37. When a mass M measured in slugs is y feet above the

surface of the earth, its weight in pounds is

1:4077 3 10



ð2:0917 3 10

1 yÞ

(One pound of force accelerates one slug one foot per

second squared.) How much work is done lifting a satel-

lite from the surface of the earth, where it weighs 800

pounds, to an orbit 200 miles above Earth?

38. The work done lifting a 1 kg mass straight up from the

surface of the earth is 1202260 J. What height above

the surface of the earth does the mass reach?

39. A spring is stretched a certain length beyond its natural

length. What percentage of the total amount of work is

expended by the ﬁrst half of the stretch?

40. A stonemason carries a 50-pound sack of mortar 120 feet

straight up a ladder at the rate of 20 feet per minute. The

sack leaks at the rate of 2 pounds per minute. When

the mason is 60 feet off the ground, he shifts the sack and

accidentally enlarges the hole. Now the sack leaks at the

rate of 3 pounds per minute, but he continues on up at

the same rate. If the mason himself weighs 180 pounds,

how much work does he perform climbing the ladder with

the sack?

41. Suppose a cylinder with cross-sectional area A in

has one

ﬁxed end (the cylinder head). Suppose that a movable

piston closes the other end a variable distance x inches

away from the cylinder head. The pressure (force per area)

p of gas conﬁned in the cylinder is a continuous function of

volume: p 5 p(v). If 0 , x

, x

, show that the work done

by the piston in compressing the gas from a volume Ax

W 5 A

pðAxÞdx:

42. Suppose that a cylinder with cross-sectional area 2 square

inches contains 20 cubic inches of a gas under 50 pounds

per square inch pressure. If the expression pv

1.4

is con-

stant as the piston moves, then how much work does a

piston do in compressing the gas to 4 cubic inches? (Refer

to Exercise 4.1.)

c A tank has the shape of a paraboloid of revolution that

results

from rotating about the y-axis the region that is

bounded above by the horizontal line y 5 18, on the left by

the y-axis, and on the right by the graph of y 5 2x

(x and y

measured in feet). In Exercises 43246,

a pump ﬂoats on the

surface of the water and pumps the water to the top of the

tank. Calculate the work that is done in performing the task

that is described. b

43. The

tank, initially ﬁlled, is pumped until the remaining

water is 3 feet deep at the center.

44. The tank, initially ﬁlled, is pumped until the remaining

water has half the original volume.

45. The tank, initially ﬁlled to a depth of 4 feet at the center,

is pumped until it is empty.

Frustum of an inverted cone

m Figure 9

Frustum of a pyramid

m Figure 10

7.5 Work 587

46. The tank, initially ﬁlled to a depth of 4 feet at the center,

is pumped until the remaining water is 2 feet deep at the

center.

47. A force F(x) . 0 causes a mass m, initially at rest at x 5 0,

to move along the x-axis with velocity v(x). Let W(b)be

the work done in moving the body from x 5 0tox 5 b.

Show that W(b) 5 mv (b)

/2. In other words, the work

done is equal to the gain in kinetic energy. Hint: Start

from Newton’s Law, which involves the derivative of

velocity with respect to time. Use the Chain Rule to

calculate the derivative of v with respect to x.

48. The gas tank of a tractor is in the shape of a cylinder lying on

its side. The tank is positioned on the tractor so that its

central axis is 5 feet above the ground. The radius of the

tank is 1.5 feet, and its length is 5 feet. For winter storage of

the tractor, gas is pumped straight up out of the tank, from

the surface of the gasoline into a holding tank with opening

15 feet from ground level. If gasoline weighs 35 pounds per

cubic foot, then how much work is done in emptying a full

tank of gas on the tractor into the holding tank?

49. The magnitude of a force that stretches a spring is given

by F(x) 5 kx when the spring is extended a distance x

beyond its equilibrium position. Let W(x) denote the

work done in stretching the spring a distance x from its

equilibrium position. Describe the curve that results from

plotting the parametric equations F 5 F(x), W 5 W(x)in

the FW-plane.

50. An object sits on a ﬂat surface that presents an irregular

frictional force. The object is pushed forward. In pushing it

x units from its rest position, the work done is x 1 arctan

(x). Describe the pushing force as a function of x.

Calculator/Computer Exercises

c Exercises 51254 are concerned with a tank ﬁlled to the top

with water. Its shape is obtained by rotating about the y-axis

the region that is bounded above by the horizontal line y 5 10

(121/e

), on the left by the y-axis, and on the right by the

graph of y 5 10 (1 2 exp(2x

/2)). Both x and y are measured

in feet. A pump ﬂoats on the surface of the water and pumps

water to the top, at which point the water runs off. b

51. How

much work is done pumping out a volume of water

that leaves the remaining water 4 feet deep at the center?

52. How much work is done in pumping out half the water in

the tank?

53. How much work is done in pumping out all the water in

the tank?

54. In pumping out all the water in the tank, how deep is the

remaining water at the center when half the work is done?

7.6 First Order Differential Equations—Separable

Equations

A differential equation is an equation that involves one or more derivatives of an

unknown function. For example, the rate at which the mass m of a radioactive

substance decreases at time t is proportional to the value of m at t. That is, m

satisﬁes the differential equation

52λ m where λ is a positive constant. This

differential equation is a ﬁrst order differential equation because the ﬁrst derivative

is the highest order derivative of the unknown function that appears in the equa-

tion. In this section, we will study ﬁrst order differential equations that have the

form

5 Fðx; yÞ: ð7:6:1Þ

Here F(x, y) is an expression that involves both x and y (in general). For example,

F(x, y) could be 2 2 xy

. It is also possible for the expression F(x, y) to depend on

only one of the variables. Thus if F(x, y) 5 1/x, then equation (7.6.1) becomes

. Similarly, if F(x, y) 5 2y (10 2 y), then equation (7.6.1) becomes

5 2yð10 2 yÞ.

588 Chapter 7 Applications of the Integral

Solutions of Differential

Equations

Solving an algebraic equation generally consists of ﬁnding the ﬁnite collection of

numbers that satisfy the equation. By contra st, to solve a differential equation, we

must ﬁnd the collection of functions that satisfy the equation.

DEFINITION

We say that a differentiable function y is a solution of differential

equation (7.6.1) if y

(x) 5 F(x, y(x)) for every x in some open interval. The graph

of a solution is said to be a solution curve of the differential equation.

Our ﬁrst example shows that there can be inﬁni tely many different funct ions

that satisfy a given differential equation.

⁄ EX

AMPLE 1 Let C denote an arbitrary constant. Verify that the function

y(x) 5 x 1 Ce

2 1 is a solution of the differential equation

5 x 2 y.

Solution We

calculate the left side of the differential equation,

ðx 1 Ce

2 1Þ5 1 2 Ce

;

and the right side,

x 2 y 5 x 2 ðx 1 Ce

2 1Þ5 1 2 Ce

Because these expressions are equal, it follows that the function y(x) 5

x 1 Ce

2 1 satisﬁes the given differential equation. Notice that this veriﬁcation

does not show how the solution y(x) 5 x 1 Ce

2 1isfound. The technique for

discovering this solution is the subject of Section 7.7.

Slope Fields We can represent equation (7.6.1) graphically by interpreting dy/dx as the slope of

a tangent line to a solution curve. To implement this observation, we select a grid of

points in the plane, and, for every point (x, y ) in the grid, we draw a short line

segment with slope F(x, y) and initial point (x, y). The lengths of the line segments

have no signiﬁcance. They are usually drawn so that they have the same length,

which is chosen to be short enough relative to the chosen grid so that different line

segments do not intersect. The collection of line segments drawn in this way is

called a slope ﬁeld.

⁄ EX

AMPLE 2 Sketch a slope ﬁeld for the differential equation,

5 x 2 y,

of Example 1.

Solution F

or this differential equation, the right side of equation (7.6.1) is F(x, y) 5

x 2 y. For points (x, y)onthex-axis, we have y 5 0andF(x,0)5 x. This tells us that

the line segments with initial points along the x-axis have slopes that increase as x

increases, from negative values when x , 0 to positive values when x . 0. Along the

line y 5 x,wehaveF(x, x) 5 0 for every x. As a result, the line segments with initial

points on the line y 5 x are all horizontal. In Figure 1, we start our sketch of the slope

ﬁeld by drawing the two families of line segments that we have described. The other

line segments in the slope ﬁeld are determined in a similar manner. For example,

the line segments with initial points on the line y 5 x 2 1 all have slope 1 because

F(x, x 2 1) 5 x 2 (x 2 1) 5 1. After we ﬁll in the slopes for the rest of the grid, we

obtain our ﬁnal sketch in Figure 2.

m Figure 1

7.6 First Order Differential Equations—Separable Equations 589

If a solution curve y(x ) to equation (7.6.1) is drawn in a slope ﬁeld and passes

through the point (x

, y

), then y(x

) 5 y

and y

) 5 F(x

, y

). In other words, the

line segment with initial point (x

, y

) is tangent to the solution curve. With a little

practice, we can use the slope ﬁeld to predict the shape of the solution curves to a

differential equation without actually ﬁnding an analytic solution to the equation.

⁄ EX

AMPLE 3 Superimpose the plots of several solution curves of the dif-

ferential equation,

5 x 2 y, from Example 1 on a slope ﬁeld plot for this

differential equation.

Solution We

veriﬁed in Example 1 that the equation y 5 x 1 Ce

2 1 is a solution

curve for any constant C. The graphs of these equations for C 525, 29/2, 24, ...,

9/2, 5 have been superimposed on the slope ﬁeld of Figure 2, resulting in Figure 3.

INSIGHT

An experienced eye can often use a slope ﬁeld to detect the behavior of a

solution to a differential equation (even when an explicit formula for the solution has not

been found). For example, the “ﬂow” toward y 5 x of the line segments in Figure 2

suggests that lim

x-N

yðxÞ

5 1 for every solution y(x)of

5 x 2 y. The solution curves

in Figure 3 provide supporting evidence for this inference. A rigorous demonstration can

be obtained from the explicit formula for the solution curves:

lim

x-N

yðxÞ

5 lim

x-N

x 1 Ce

2 1

5 1 1 lim

x-N

2 lim

x-N

5 1 1 0 1 0 5 1:

Initial Value Problems When differential equations arise in practice, we often know the value y

of the

unknown function y(x) at one particular value of x

. For example, if x represents

time, and y represents the depth of water in a tank, then a leak may develop that

leads to the differential equation

ﬃﬃﬃ

. If the tank has height y

and is

initially ﬁlled, and if the leak begins at time x 5 x

, then we know that y(x

) 5 y

Such an equation is called an initial condition.

DEFINITION

If x

and y

are speciﬁed values, then the pair of equations

5 Fðx; y Þðdiffe rential equationÞ ; yðx

Þ5 y

ðinitial conditionÞð7:6:2Þ

is said to be an initial value problem (IVP). We say that a differentiable function

y is a solution of initial value problem (7.6.2) if y(x

) 5 y

and y

(x) 5 F(x, y(x))

for all x in some open interval containing x

It can be shown that, unde r mild restrictions on F (satisﬁed by all examples in

this book), the initial value problem (7.6.2) has a unique solution.

⁄ EX

AMPLE 4 Use the computation of Example 1 to solve the initial value

problem

5 x 2 y ; yð0Þ5 2.

m Figure 2

m Figure 3

590 Chapter

7 Applications of the Integral

Solution In Example 1, we veriﬁed that y(x) 5 x 1 Ce

2 1 satisﬁes the given

differential equation for every constant C. Notice that y(0) 5 0 1 Ce

2 1 5 C 2 1.

Therefore to satisfy the initial condition y(0) 5 2, we set C 2 1 5 2 and ﬁnd that

C 5 3. Thus y(x) 5 x 1 3e

2 1 solves the given initial value problem. ¥

Separable Equations There is no single technique by which equation (7.6.1) can be solved for every

F(x, y). Sev eral methods have been developed to handle special cases according to

the form of the expression F(x, y). In this section, we will study the case in which

F(x, y) 5 g(x) h( y). The expressions

Fðx; yÞ5 4 cosðxÞ5 4 cosðxÞ

|ﬄﬄﬄﬄ{zﬄﬄﬄﬄ}

gðxÞ

 1

|{z}

hðyÞ

; Fðx; yÞ5 7y

5 7

|{z}

gðxÞ

 y

|{z}

hðyÞ

;

Fðx; yÞ5

2 1 x

1 1 y

5 ð2 1 xÞ

|ﬄﬄﬄ{zﬄﬄﬄ}

gðxÞ



1 1 y



|ﬄﬄﬄﬄﬄﬄﬄ{zﬄﬄﬄﬄﬄﬄﬄ}

hðyÞ

are all of this type. When F(x, y) factors as g(x) h( y ), the differential equati on

5 Fðx; yÞ5 gðxÞ hðyÞð7:6:3Þ

is said to be separable because the expressions involving x may be placed on one side

of the equation and the terms involving y may be moved to the other side of the

equation. To carry out this separation of variables, we rewrite equation (7.6.3) as

hðyÞ

dy 5 gðxÞdx:

This equation is only suggestive, but it does hint at the answer,

hðyÞ

dy 5

gðxÞdx:

Let us see how we can obtain this result more carefully. Assuming that h( y) 6¼ 0, we

rewrite equation (7.6.3) as

hðyÞ

5 gðxÞ

and integrate with respect to x to obtain

hðyÞ

dx 5

gðxÞdx: ð7:6:4Þ

Let H(u) be an antiderivative of 1/h(u), and let G(x) be an antiderivative of g(x).

Using the Chain Rule, we have

HðyÞ5 H

ðyÞ

hðyÞ

Therefore equation (7.6.4) can be stated as

HðyÞ5 GðxÞ1 C: ð7:6:5Þ

(There is no need to add a constant of integration to each side; if that were done,

then the two constants could be combined into one.) This procedure for solving a

differential equation is called the method of separation of variables. When we use this

method, we will generally simplify the notation and write

7.6 First Order Differential Equations—Separable Equations 591

hðyÞ

dy for

hðyÞ

dx:

⁄ EX

AMPLE 5 Solve the initial value problem

1 1

; yð0Þ5 3:

Solution The

given differential equation is separable because it may be written as

5 gðxÞhðyÞ with g(x) 5 x and hðyÞ5

ðy

1 1Þ

. Following the method of

solution that we have just discussed, we separate variables:

ðy

1 1Þ

|ﬄﬄ{zﬄﬄ}

treat as dy

xdx;

1 y 5

1 C:

If this equation is to satisfy the given initial condition y(0) 5 3, then C must satisfy

ð3Þ

1 ð3Þ5

ð0Þ

1 C

or C 5 12. Thus

1 y 5

1 12 is the solution of the given initial value problem.

The solution curve is shown in Figure 4.

INSIGHT

As can be seen in Example 5, the method of separation of variables does

not generally yield a solution y(x) of the differential equation dy=dx 5 gðxÞhðyÞ in explicit

form. That is, the method of separation of variables does not ordinarily lead to an

equation of the form y(x) 5 (expression in x). Instead, it usually results in an equation of

the form

ðexpression in yÞ5 ðexpression in xÞ:

Such an equation deﬁnes y implicitly as a function of x. It is possible to show that the

explicit solution to the initial value problem of Example 5 is y(x) 5 u(x) 2 1/u(x) where

uðxÞ5



144 1 6x

1 2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

5200 1 432x

1 9x



1=3

The explicit formula for y(x) is so complicated that it is a chore to even verify that it

is correct. By contrast, it is easy to implicitly differentiate the equation

1 y 5

1 12

and verify that the implicitly deﬁned function y does solve the given differential equation.

Equations of the Form

dy/dx 5 g(x)

The differential equation dy/dx 5 g(x) is nothing more than the statement “y(x)is

an antiderivative of g(x).” Theorem 1 provides the solution of the corresponding

initial value problem.

3.8

55

3.6

3.4

3.2

3.0

m Figure 4 The graph of

1 y 5

1 12

592 Chapter 7 Applications of the Integral

THEOREM 1

If g is a continuous function on an open interval containing a,

then the initial value problem

5 gðxÞ; yðaÞ5 b

has a unique solution:

yðxÞ5 b 1

gðtÞ dt: ð7:6:6Þ

Proof. Acco

rding to the Fundamental Theorem of Calculus,

gðtÞ dt is an

antiderivative of g. All other antiderivatives of g have the form

gðtÞ dt 1 C.Thus

yðxÞ5

gðtÞ dt 1 C for some constant C.WeusetheinitialconditiontosolveforC,

b 5 yðaÞ5

gðtÞ dt 1 C 5 0 1 C:

We obtain equation (7.6.6) by substituting C 5 b in our formula for y(x). ’

Many useful functions in science and engineering are deﬁned by equation

(7.6.6)

for particular choices of g: the Fresnel functions

FresnelCðxÞ5

cos





dt and FresnelSðxÞ5

sin





in optics, the Debye function

x /



2 1



in thermodynamics, and so on. Although the function y deﬁned by equation (7.6.6)

might seem complicated or abstract at ﬁrst, remember that the way it is deﬁned

allows us to instantly calculate its derivative. By now we have accumulated a good

deal of experience using the derivative y

(x) to deduce properties of y(x). It is

through those properties that the function y becomes familiar to us. If you review

Section 5.5 with Theorem 1 in mind, you will see that all the properties of the

logarithm and exponential functions were developed by deﬁning y(x) 5 ln(x)tobe

the solution of the initial value problem

; yð1Þ5 0.

Examples from the

Physical Sciences

In the physical sciences, often the most natural way to describe the relationship

between two variables is by means of a differential equation. The differential

equation is then solved to determine explicitly the relationship between the vari-

ables. The next three examples illustrate these ideas.

⁄ EX

AMPLE 6 According to Torricelli’s Law, the rate at which water ﬂows

out of a tank from a small hole in the bottom is proportional to the area A of the

hole and to the square root of the height y( t) of the water in the tank (Figure 5).

That is, there is a positive constant k such that

VðtÞ52k  A 

ﬃﬃﬃﬃﬃﬃﬃﬃ

yðtÞ

;

y(t)

V(t)

2.5

0.4

m Figure 5

522:6  A 

ﬃﬃﬃ

7.6 First Order Differential Equations—Separable Equations 593

where V(t) is the volume of water in the tank at time t. Consider a cylindrical tank

of height 2.5 m and radius 0.4 m that has a hole 2 cm in diameter in its bottom.

Suppose that the tank is full at time t 5 0. If k 5 2.6 m

1/2

/s, ﬁnd the height y(t)of

water as a function of time. How long will it take for the tank to empty?

Solution The

volume V (t) of water in the tank at time t is related to the height

y(t) of the water by the equati on V (t) 5 π(0.4)

y(t). Because A 5 π(0.01)

Torricelli’s Law tells us that

πð0:4Þ

522:6  πð0:01Þ



ﬃﬃﬃ

;

522:6

πð0:01Þ

πð0:4Þ

ﬃﬃﬃ

520:001625

ﬃﬃﬃ

On separating variables, we have

21=2

dy 5

ð20:001625Þdt 1 C;

2yðtÞ

1=2

520:001625t 1 C:

When we substitute t 5 0 into this equation, we obtain C 5 2

ﬃﬃﬃﬃﬃﬃﬃﬃﬃ

yð0Þ

. But y(0) 5 2.5

because the tank is initially ﬁlled to its height of 2.5 meters. Therefore

C 5 2

ﬃﬃﬃﬃﬃﬃﬃ

2:5

and 2yðtÞ

1=2

520:001625t 1 2

ﬃﬃﬃﬃﬃﬃﬃ

2:5

; or

yðtÞ5



ﬃﬃﬃﬃﬃﬃﬃ

2:5

0:001625



This is the height of the water in the tank at time t. The equation y(t) 5 0has

solution t 5 2

ﬃﬃﬃﬃﬃﬃﬃ

2:5

=0:001625  1946 seconds. Thus the tank will be empty in about

1946/60 32.4 minutes.

⁄ EXAMPLE 7 In chemical kinetics, the rate at which the concentration [S]

of a substance S changes is often described by means of a differential equation. For

example, the dimerization of butadiene

has reaction rate

d½C



52k½C



;

where k is a positive constant. If [C

] 5 1/60 moles/L when t 5 0 seconds, and

] 5 1/140 moles/L when t 5 6000 seconds, then determine [C

] as a function

of time. When is [C

] 5 1/200?

Solution Let y 5 [C

]. We solve as follows

52ky

dy 52

kdt. 2

yðtÞ

52kt 1 C;

594 Chapter 7 Applications of the Integral