# Mathematics

➤ The restriction on p in property 3 will be lifted shortly.

QUICK CHECK 2 Simplify eln 2x, ln 1e2x2, e2 ln x, and ln 12ex2. Derivatives and Integrals By Theorem 3.23 (derivatives of inverse functions), the

derivative of the exponential function exists for all x. To compute d dx 1ex2, we observe that

ln 1ex2 = x and then differentiate both sides with respect to x: d dx

1ln ex2 = d dx

1×2 1

1 ex

d dx

1ex2 = 1 d dx

1ln u1x22 = u′1×2 u1x2 (Chain Rule)

d dx

1ex2 = ex. Solve for d dx

1ex2.

e c

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6.8 Logarithmic and Exponential Functions Revisited 477

Once again, we obtain the remarkable result that the exponential function is its own deriva- tive. It follows that ex is its own antiderivative up to a constant; that is,

Lex dx = ex + C. Extending these results using the Chain Rule, we have the following theorem.

EXAMPLE 2 Integrals with ex Evaluate L ex

1 + ex dx.

SOLUTION The change of variables u = 1 + ex implies du = ex dx:

L 1

1 + ex e x dx = L

1 u

du u = 1 + ex, du = ex dx

u du

= ln ! u ! + C Antiderivative of u-1

= ln 11 + ex2 + C. Replace u with 1 + ex. Note that the absolute value may be removed from ln ! u ! because 1 + ex 7 0, for all x.

Related Exercises 21–26

Step 3: General Logarithmic and Exponential Functions We now turn to exponential and logarithmic functions with a general positive base b. The first step is to define the exponential function bx for positive bases with b ≠ 1 and for all real numbers x. We use property 3 of Theorem 6.5 and the fact that b = eln b. If x is a rational number, then

bx = 1eln b2x = ex ln b; b

this important relationship expresses bx in terms of ex. Because ex is defined for all real x, we use this relationship to define bx for all real x.

DEFINITION Exponential Functions with General Bases Let b be a positive real number with b ≠ 1. Then for all real x,

bx = ex ln b.

QUICK CHECK 3 What is the slope of the curve y = ex at x = ln 2? What is the area of the region bounded by the graph of y = ex and the x-axis between x = 0 and x = ln 2?

THEOREM 6.6 Derivative and Integral of the Exponential Function For real numbers x,

d dx

1eu1x22 = eu1x2u′1×2 and Lex dx = ex + C.

ee

c

This definition fills the gap in property 4 of Theorem 6.4 1ln xp = p ln x2. We use the definition of bx to write

xp = ep ln x, for x 7 0 and p real.

Taking the natural logarithm of both sides and using the inverse relationship between ex and ln x, we find that

ln xp = ln ep ln x = p ln x, for x 7 0 and p real.

In this way, we extend property 4 of Theorem 6.4 to real powers. Just as bx is defined in terms of ex, logarithms with base b 7 1 and b ≠ 1 may be

expressed in terms of ln x. All that is needed is the change of base formula (Section 1.3)

logb x = ln x ln b

.

➤ Knowing that ln xp = p ln x for real p, we can also extend property 3 of Theorem 6.5 to real numbers. For real x and y, we take the natural logarithm of both sides of z = 1ex2y, which gives ln z = y ln ex = xy, or z = exy. Therefore, 1ex2y = exy.

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6.8 Logarithmic and Exponential Functions Revisited 477

Once again, we obtain the remarkable result that the exponential function is its own deriva- tive. It follows that ex is its own antiderivative up to a constant; that is,

Lex dx = ex + C. Extending these results using the Chain Rule, we have the following theorem.

EXAMPLE 2 Integrals with ex Evaluate L ex

1 + ex dx.

SOLUTION The change of variables u = 1 + ex implies du = ex dx:

L 1

1 + ex e x dx = L

1 u

du u = 1 + ex, du = ex dx

u du

= ln ! u ! + C Antiderivative of u-1

= ln 11 + ex2 + C. Replace u with 1 + ex. Note that the absolute value may be removed from ln ! u ! because 1 + ex 7 0, for all x.

Related Exercises 21–26

Step 3: General Logarithmic and Exponential Functions We now turn to exponential and logarithmic functions with a general positive base b. The first step is to define the exponential function bx for positive bases with b ≠ 1 and for all real numbers x. We use property 3 of Theorem 6.5 and the fact that b = eln b. If x is a rational number, then

bx = 1eln b2x = ex ln b; b

this important relationship expresses bx in terms of ex. Because ex is defined for all real x, we use this relationship to define bx for all real x.

DEFINITION Exponential Functions with General Bases Let b be a positive real number with b ≠ 1. Then for all real x,

bx = ex ln b.

QUICK CHECK 3 What is the slope of the curve y = ex at x = ln 2? What is the area of the region bounded by the graph of y = ex and the x-axis between x = 0 and x = ln 2?

THEOREM 6.6 Derivative and Integral of the Exponential Function For real numbers x,

d dx

1eu1x22 = eu1x2u′1×2 and Lex dx = ex + C.

ee

c

This definition fills the gap in property 4 of Theorem 6.4 1ln xp = p ln x2. We use the definition of bx to write

xp = ep ln x, for x 7 0 and p real.

Taking the natural logarithm of both sides and using the inverse relationship between ex and ln x, we find that

ln xp = ln ep ln x = p ln x, for x 7 0 and p real.

In this way, we extend property 4 of Theorem 6.4 to real powers. Just as bx is defined in terms of ex, logarithms with base b 7 1 and b ≠ 1 may be

expressed in terms of ln x. All that is needed is the change of base formula (Section 1.3)

logb x = ln x ln b

.

➤ Knowing that ln xp = p ln x for real p, we can also extend property 3 of Theorem 6.5 to real numbers. For real x and y, we take the natural logarithm of both sides of z = 1ex2y, which gives ln z = y ln ex = xy, or z = exy. Therefore, 1ex2y = exy.

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478 Chapter 6 Applications of Integration

Theorems 3.18 and 3.20 give us the derivative results for exponential and logarithmic functions with a general base b 7 0. Extending those results with the Chain Rule, we have the following derivatives and integrals.

SUMMARY Derivatives and Integrals with Other Bases Let b 7 0 and b ≠ 1. Then

d dx

1logb 0 u1x2 0 2 = u′1x2u1x2 ln b, for u1x2 ≠ 0 and ddx 1bu1x22 = 1ln b2bu1x2u′1×2. For b 7 0 and b ≠ 1, Lbx dx =

1 ln b

bx + C.

QUICK CHECK 4 Verify that the derivative and integral results for a general base b reduce to the expected results when b = e.

EXAMPLE 3 Integrals involving exponentials with other bases Evaluate the follow- ing integrals.

a. Lx 3x 2 dx b. L

4

1

6-1x1x dx SOLUTION

a. Lx 3x 2 dx =

1 2 L3u du u = x2, du = 2x dx

= 1 2

1

ln 3 3u + C Integrate.

= 1

2 ln 3 3x

2 + C Substitute u = x2.

b. L 4

1

6-1x1x dx = -2L -2-1 6u du u = – 1x, du = – 121x dx = – 2

ln 6 6u `

-1

-2 Fundamental Theorem

= 5

18 ln 6 Simplify.

Related Exercises 27–32

Step 4: General Power Rule With the identity xp = ep ln x, we can state and prove the final version of the Power Rule. In Chapter 3, we showed that

d dx

1xp2 = pxp – 1 when p is a rational number. This result is extended to all real values of p by differentiat- ing xp = ep ln x:

d dx

1xp2 = d dx

1ep ln x2 xp = ep ln x = ep ln x

p x

Chain Rule xp

= xp p x

ep ln x = xp

= pxp – 1. Simplify.

c

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6.8 Logarithmic and Exponential Functions Revisited 479

EXAMPLE 4 Derivative of a tower function Evaluate the derivative of f 1×2 = x2x. SOLUTION We use the inverse relationship eln x = x to write x2x = eln 1x2x2 = e2x ln x. It follows that

d dx

1x2x2 = d dx

1e2x ln x2 = e2x ln x

d dx

12x ln x2 d dx

1eu1x22 = eu1x2u′1×2 x2x

= x2x a2 ln x + 2x # 1 x b Product Rule

= 2x2x11 + ln x2. Simplify. Related Exercises 33–40

Computing e We have shown that the number e serves as a base for both ln x and ex, but how do we approximate its value? Recall that the derivative of ln x at x = 1 is 1. By the definition of the derivative, it follows that

1 = d dx

1ln x2 ` x = 1

= lim hS0

ln 11 + h2 – ln 1

h Derivative of ln x at x = 1

= lim hS0

ln 11 + h2

h ln 1 = 0

= lim hS0

ln 11 + h21>h. p ln x = ln xp The natural logarithm is continuous for x 7 0, so it is permissible to interchange the order of lim

hS0 and the evaluation of ln 11 + h21>h. The result is that

ln 1 lim hS0

11 + h21>h2 = 1. e

Observe that the limit within the brackets is e because ln e = 1 and only one number satisfies this equation. Therefore, we have isolated e as a limit:

e = lim hS0

11 + h21>h. It is evident from Table 6.2 that 11 + h21>h S 2.718282c as h S 0. The value of this limit is e, and it has been computed to millions of digits. A better approximation,

e ≈ 2.718281828459045,

is obtained by methods introduced in Chapter 9.

➤ Because d dx

1ln x2 = 1 x

,

d dx

1ln x2 ` x = 1

= 1 1

= 1.

➤ We rely on Theorem 2.12 of Section 2.6 here. If f is continuous at g1a2 and lim

xSa g1x2 exists, then

lim xSa

f 1g1x22 = f 1 lim xSa

g1x22.

THEOREM 6.7 General Power Rule For any real number p,

d dx

1xp2 = pxp – 1 and d dx

1u1x2p2 = pu1x2p – 1u′1×2.

Table 6.2

h 11 + h 21,h h 11 + h 21,h 10-1 2.593742 -10-1 2.867972 10-2 2.704814 -10-2 2.731999 10-3 2.716924 -10-3 2.719642 10-4 2.718146 -10-4 2.718418 10-5 2.718268 -10-5 2.718295 10-6 2.718280 -10-6 2.718283 10-7 2.718282 -10-7 2.718282

v

c

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482 Chapter 6 Applications of Integration

6.9 Exponential Models The uses of exponential functions are wide-ranging. In this section, you will see them ap- plied to problems in finance, medicine, ecology, biology, economics, pharmacokinetics, anthropology, and physics.

Exponential Growth Exponential growth functions have the form y1t2 = Cekt, where C is a constant and the rate constant k is positive (Figure 6.84). If we start with this function and take its derivative, we find that

y′1t2 = d dt

1Cekt2 = C # kekt = k1Cekt2; y

that is, y′1t2 = ky. Here is the first insight about exponential functions: Their rate of change is proportional to their value. If y represents a population, then y′1t2 is the growth rate with units such as people/month or cells/hr. We see that the larger the population, the faster its growth.

Another way to talk about growth rates is to use the relative growth rate, which is the growth rate divided by the current value of that quantity, or y′1t2>y1t2. For example, if y is a population, the relative growth rate is the fraction or percentage by which the population grows each unit of time. Examples of relative growth rates are 5% per year or a factor of 1.2 per month. Therefore, when the equation y′1t2 = ky is written in the form y′1t2>y = k, it has another interpretation. It says a quantity that grows exponentially has a constant relative growth rate. Constant relative or percentage change is the hallmark of exponential growth.

EXAMPLE 1 Linear versus exponential growth Suppose the population of the town of Pine is given by P1t2 = 1500 + 125t, while the population of the town of Spruce is given by S1t2 = 1500e0.1t, where t Ú 0 is measured in years. Find the growth rate and the relative growth rate of each town.

SOLUTION Note that Pine grows according to a linear function, while Spruce

grows exponentially (Figure 6.85). The growth rate of Pine is dP dt

= 125 people/year,

which is constant for all times. The growth rate of Spruce is

dS dt

= 0.111500e0.1t2 = 0.1S1t2, S1t2 showing that the growth rate is proportional to the population. The relative growth rate

of Pine is 1 P

dP dt

= 125

1500 + 125t, which decreases in time. The relative growth rate of

Spruce is

1 S

dS dt

= 0.1 # 1500e0.1t

1500e0.1t = 0.1,

which is constant for all times. In summary, the linear population function has a constant absolute growth rate and the exponential population function has a constant relative growth rate.

Related Exercises 9–10

QUICK CHECK 1 Population A increases at a constant rate of 4%>yr. Population B increases at a constant rate of 500 people/yr. Which population exhibits exponential growth? What kind of growth is exhibited by the other population?

r

➤ The derivative dy

dt is the absolute growth

rate but is usually simply called the growth rate.

➤ A consumer price index that increases at a constant rate of 4% per year increases exponentially. A currency that is devalued at a constant rate of 3% per month decreases exponentially. By contrast, linear growth is characterized by constant absolute growth rates, such as 500 people per year or $400 per month.

Figure 6.84

3

2

!1 1

y

t

y ” 2e2t

y ” e2t

y ” 0.1e10t

y ” et

0

4000

3000

2000

1000

y

t1284

Spruce: S(t) ! 1500e0.1t

exponential growth

Pine: P(t) ! 1500 ” 125t linear growth

0

Figure 6.85

t

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6.9 Exponential Models 483

The rate constant k in y1t2 = Cekt determines the growth rate of the exponential function. We adopt the convention that k 7 0; then it is clear that y1t2 = Cekt describes exponential growth and y1t2 = Ce-kt describes exponential decay, to be discussed shortly. For problems that involve time, the units of k are time-1; for example, if t is measured in months, the units of k are month-1. In this way, the exponent kt is dimen- sionless (without units).

Unless there is good reason to do otherwise, it is customary to take t = 0 as the refer- ence point for time. Notice that with y1t2 = Cekt, we have y102 = C. Therefore, C has a simple meaning: It is the initial value of the quantity of interest, which we denote y0. In the examples that follow, two pieces of information are typically given: the initial value and clues for determining the rate constant k. The initial value and the rate constant deter- mine an exponential growth function completely.

➤ The unit time-1 is read per unit time. For example, month-1 is read per month.

Exponential Growth Functions Exponential growth is described by functions of the form y1t2 = y0ekt. The initial value of y at t = 0 is y102 = y0, and the rate constant k 7 0 determines the rate of growth. Exponential growth is characterized by a constant relative growth rate.

Because exponential growth is characterized by a constant relative growth rate, the time required for a quantity to double (a 100% increase) is constant. Therefore, one way to describe an exponentially growing quantity is to give its doubling time. To compute the time it takes the function y1t2 = y0ekt to double in value, say from y0 to 2y0, we find the value of t that satisfies

y1t2 = 2y0 or y0ekt = 2y0. Canceling y0 from the equation y0e

kt = 2y0 leaves the equation ekt = 2. Taking logarithms

of both sides, we have ln ekt = ln 2, or kt = ln 2, which has the solution t = ln 2

k . We

denote this doubling time T2 so that T2 = ln 2

k . If y increases exponentially, the time it takes

to double from 100 to 200 is the same as the time it takes to double from 1000 to 2000.

➤ Note that the initial value y0 appears on both sides of this equation. It may be canceled, meaning that the doubling time is independent of the initial condition: The doubling time is constant for all t.

QUICK CHECK 2 Verify that the time needed for y1t2 = y0ekt to double from y0 to 2y0 is the same as the time needed to double from 2y0 to 4y0.

EXAMPLE 2 World population Human population growth rates vary geographically and fluctuate over time. The overall growth rate for world population peaked at an annual rate of 2.1% per year in the 1960s. Assume a world population of 6.0 billion in 1999 1t = 02 and 6.9 billion in 2009 1t = 102. a. Find an exponential growth function for the world population that fits the two data

points.

b. Find the doubling time for the world population using the model in part (a). c. Find the (absolute) growth rate y′1t2 and graph it, for 0 … t … 50. d. How fast was the population growing in 2014 1t = 152?

DEFINITION Doubling Time The quantity described by the function y1t2 = y0ekt, for k 7 0, has a constant dou- bling time of T2 =

ln 2 k

, with the same units as t.

➤ World population

1804 1 billion 1927 2 billion 1960 3 billion 1974 4 billion 1987 5 billion 1999 6 billion 2011 7 billion 2050 9 billion (proj.)

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484 Chapter 6 Applications of Integration

SOLUTION

a. Let y1t2 be world population measured in billions of people t years after 1999. We use the growth function y1t2 = y0ekt, where y0 and k must be determined. The initial value is y0 = 6 (billion). To determine the rate constant k, we use the fact that y1102 = 6.9. Substituting t = 10 into the growth function with y0 = 6 implies

y1102 = 6e10k = 6.9. Solving for k yields the rate constant k =

ln 16.9>62 10

≈ 0.013976 ≈ 0.014 year-1.

Therefore, the growth function is

y1t2 = 6e0.014t. b. The doubling time of the population is

T2 = ln 2

k ≈

ln 2 0.014

≈ 50 years.

c. Working with the growth function y1t2 = 6e0.014t, we find that y′1t2 = 6 10.0142e0.014t = 0.084e0.014t,

which has units of billions of people/year. As shown in Figure 6.86 the growth rate itself increases exponentially.

d. In 2014 1t = 152, the growth rate was y′1152 = 0.084e10.01421152 ≈ 0.104 billion people>year,

or roughly 104 million people/year. Related Exercises 11–20

QUICK CHECK 3 Assume y1t2 = 100e0.05t. By (exactly) what percentage does y increase when t increases by 1 unit?

A Financial Model Exponential functions are used in many financial applications, sev- eral of which are explored in the exercises. For now, consider a simple savings account in which an initial deposit earns interest that is reinvested in the account. Interest payments are made on a regular basis (for example, annually, monthly, daily), or interest may be compounded continuously. In all cases, the balance in the account increases exponentially at a rate that can be determined from the advertised annual percentage yield (or APY) of the account. Assuming that no additional deposits are made, the balance in the account is given by the exponential growth function y1t2 = y0ekt, where y0 is the initial deposit, t is measured in years, and k is determined by the annual percentage yield.

EXAMPLE 3 Compounding The APY of a savings account is the percentage increase in the balance over the course of a year. Suppose you deposit $500 in a savings account that has an APY of 6.18% per year. Assume that the interest rate remains constant and that no additional deposits or withdrawals are made. How long will it take the balance to reach $2500?

SOLUTION Because the balance grows by a fixed percentage every year, it grows expo- nentially. Letting y1t2 be the balance t years after the initial deposit of y0 = $500, we have y1t2 = y0ekt, where the rate constant k must be determined. Note that if the initial balance is y0, one year later the balance is 6.18% more, or

y112 = 1.0618 y0 = y0ek. Solving for k, we find that the rate constant is

k = ln 1.0618 ≈ 0.060 yr-1.

➤ It is a common mistake to assume that if the annual growth rate is 1.4% per year, then k = 1.4% = 0.014 year-1. The rate constant k must be calculated, as it is in Example 2, to give k = 0.013976. For larger growth rates, the difference between k and the actual growth rate is greater.

Figure 6.86

0.1

0.2

0

y!

t4020

y!(t) ” 0.084e0.014t

Years (after 1999)

G ro

w th

ra te

(b ill

io ns

o f p

eo pl

e/ ye

ar )

➤ Converted to a daily rate (dividing by 365), the world population in 2014 increased at a rate of roughly 284,000 people per day.

➤ The concept of continuous compounding was introduced in Exercise 109 of Section 4.7.

➤ If the balance increases by 6.18% in one year, it increases by a factor of 1.0618 in one year.

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6.9 Exponential Models 485

Therefore, the balance at any time t Ú 0 is y1t2 = 500e0.060t. To determine the time required for the balance to reach $2500, we solve the equation

y1t2 = 500e0.060t = 2500. Dividing by 500 and taking the natural logarithm of both sides yields

0.060t = ln 5.

The balance reaches $2500 in t = 1ln 52>0.060 ≈ 26.8 yr. Related Exercises 11–20

Resource Consumption Among the many resources that people use, energy is cer- tainly one of the most important. The basic unit of energy is the joule (J), roughly the energy needed to lift a 0.1-kg object (say an orange) 1 m. The rate at which energy is con- sumed is called power. The basic unit of power is the watt (W), where 1 W = 1 J>s. If you turn on a 100-W lightbulb for 1 min, the bulb consumes energy at a rate of 100 J>s, and it uses a total of 100 J>s # 60 s = 6000 J of energy.

A more useful measure of energy for large quantities is the kilowatt-hour (kWh). A kilowatt is 1000 W or 1000 J>s. So if you consume energy at the rate of 1 kW for 1 hr (3600 s), you use a total of 1000 J>s # 3600 s = 3.6 * 106 J, which is 1 kWh. A person running for one hour consumes roughly 1 kWh of energy. A typical house uses on the order of 1000 kWh of energy in a month.

Assume that the total energy used (by a person, machine, or city) is given by the function E1t2. Because the power P1t2 is the rate at which energy is used, we have P1t2 = E′1t2. Using the ideas of Section 6.1, the total amount of energy used between the times t = a and t = b is

total energy used = L b

a E′1t2 dt = Lba P1t2 dt.

We see that energy is the area under the power curve. With this background, we can inves- tigate a situation in which the rate of energy consumption increases exponentially.

EXAMPLE 4 Energy consumption At the beginning of 2010, the rate of energy consumption for the city of Denver was 7000 megawatts (MW), where 1 MW = 106 W. That rate is expected to increase at an annual growth rate of 2% per year.

a. Find the function that gives the power or rate of energy consumption for all times after the beginning of 2010.

b. Find the total amount of energy used during 2014. c. Find the function that gives the total (cumulative) amount of energy used by the city

between 2010 and any time t Ú 0.

SOLUTION

a. Let t Ú 0 be the number of years after the beginning of 2010 and let P1t2 be the power function that gives the rate of energy consumption at time t. Because P increases at a constant rate of 2% per year, it increases exponentially. Therefore, P1t2 = P0ekt, where P0 = 7000 MW. We determine k as before by setting t = 1; after one year the power is

P112 = P0ek = 1.02P0. Canceling P0 and solving for k, we find that k = ln 1.02 ≈ 0.0198. Therefore, the power function (Figure 6.87) is

P1t2 = 7000e0.0198t, for t Ú 0.

➤ In one year, the power function increases by 2% or by a factor of 1.02.

Figure 6.87

8000

4000

0

y

t302010

Power ! rate of energy consumption

Years after 2010

Po w

er (M

W )

P(t) ! 7000e0.0198t

M06_BRIG7345_02_SE_C06.9.indd 485 05/11/13 7:10 AM

486 Chapter 6 Applications of Integration

b. The entire year 2014 corresponds to the interval 4 … t … 5. Substituting P1t2 = 7000e0.0198t, the total energy used in 2014 was

L 5

4 P1t2 dt = L54 7000e0.0198t dt Substitute for P1t2.

= 7000

0.0198 e0.0198t `

4

5

Fundamental Theorem

≈ 7652. Evaluate.

Because the units of P are MW and t is measured in years, the units of energy are MW-yr. To convert to MWh, we multiply by 8760 hr>yr to get the total energy of about 6.7 * 107 MWh (or 6.7 * 1010 kWh).

c. The total energy used between t = 0 and any future time t is given by the future value formula (Section 6.1):

E1t2 = E102 + L t0 E′1s2 ds = E102 + L t0 P1s2 ds. Assuming t = 0 corresponds to the beginning of 2010, we take E102 = 0. Substitut- ing again for the power function P, the total energy (in MW-yr) at time t is

E1t2 = E102 + L t0 P1s2 ds = 0 + L

t

0 7000e0.0198s ds Substitute for P1s2 and E102.

= 7000

0.0198 e0.0198s `

0

t

Fundamental Theorem

≈ 353,5351e0.0198t – 12. Evaluate. As shown in Figure 6.88, when the rate of energy consumption increases exponentially, the total amount of energy consumed also increases exponentially.

Related Exercises 11–20

Exponential Decay Everything you have learned about exponential growth carries over directly to exponen- tial decay. A function that decreases exponentially has the form y1t2 = y0e-kt, where y0 = y102 is the initial value and k 7 0 is the rate constant.

Exponential decay is characterized by a constant relative decay rate and by a constant half-life. For example, radioactive plutonium has a half-life of 24,000 years. An initial sample of 1 mg decays to 0.5 mg after 24,000 years and to 0.25 mg after 48,000 years. To compute the half-life, we determine the time required for the quantity y1t2 = y0e-kt to reach one-half of its current value; that is, we solve y0e

-kt = y0>2 for t. Canceling y0 and taking logarithms of both sides, we find that

e-kt = 1 2 1 -kt = ln 1

2 = – ln 2 1 t = ln 2

k .

The half-life is given by the same formula as the doubling time.

Figure 6.88

0 t302010

200,000

100,000

y

Years after 2010

Energy consumed t years after 2010

E ne

rg y

(M W

-y r)

y ! E(t)

QUICK CHECK 4 If a quantity decreases by a factor of 8 every 30 years, what is its half-life?

Exponential Decay Functions Exponential decay is described by functions of the form y1t2 = y0e-kt. The initial value of y is y102 = y0, and the rate constant k 7 0 determines the rate of decay. Exponential decay is characterized by a constant relative decay rate. The constant

half-life is T1>2 = ln 2k , with the same units as t.

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6.9 Exponential Models 487

Radiometric Dating A powerful method for estimating the age of ancient objects (for example, fossils, bones, meteorites, and cave paintings) relies on the radioactive decay of certain elements. A common version of radiometric dating uses the carbon isotope C@14, which is present in all living matter. When a living organism dies, it ceases to replace C@14, and the C@14 that is present decays with a half-life of about T1>2 = 5730 years. Comparing the C@14 in a living organism to the amount in a dead sample provides an estimate of its age.

EXAMPLE 5 Radiometric dating Researchers determine that a fossilized bone has 30% of the C@14 of a live bone. Estimate the age of the bone. Assume a half-life for C@14 of 5730 years.

SOLUTION The exponential decay function y1t2 = y0e-kt represents the amount of C-14 in the bone t years after its owner died. By the half-life formula, T1>2 = 1ln 22>k. Substituting T1>2 = 5730 yr, the rate constant is

k = ln 2 T1>2 = ln 25730 yr ≈ 0.000121 yr-1.

Assume that the amount of C@14 in a living bone is y0. Over t years, the amount of C@14 in the fossilized bone decays to 30% of its initial value, or 0.3y0. Using the decay function, we have

0.3y0 = y0e-0.000121t.

Solving for t, the age of the bone (in years) is

t = ln 0.3

-0.000121 ≈ 9950.

Related Exercises 21–26

Pharmacokinetics Pharmacokinetics describes the processes by which drugs are assimilated by the body. The elimination of most drugs from the body may be modeled by an exponential decay function with a known half-life (alcohol is a notable exception). The simplest models assume that an entire drug dose is immediately absorbed into the blood. This assumption is a bit of an idealization; more refined mathematical models account for the absorption process.

EXAMPLE 6 Pharmacokinetics An exponential decay function y1t2 = y0e-kt models the amount of drug in the blood t hr after an initial dose of y0 = 100 mg is administered. Assume the half-life of the drug is 16 hours.

a. Find the exponential decay function that governs the amount of drug in the blood. b. How much time is required for the drug to reach 1% of the initial dose (1 mg)? c. If a second 100-mg dose is given 12 hr after the first dose, how much time is required

for the drug level to reach 1 mg?

SOLUTION

a. Knowing that the half-life is 16 hr, the rate constant is

k = ln 2 T1>2 = ln 216 hr ≈ 0.0433 hr-1.

Therefore, the decay function is y1t2 = 100e-0.0433t. b. The time required for the drug to reach 1 mg is the solution of

100e-0.0433t = 1.

Solving for t, we have

t = ln 0.01

-0.0433 hr-1 ≈ 106 hr.

It takes more than 4 days for the drug to be reduced to 1% of the initial dose.

➤ Half-lives of common drugs

Penicillin 1 hr Amoxicillin 1 hr Nicotine 2 hr Morphine 3 hr Tetracycline 9 hr Digitalis 33 hr Phenobarbitol 2–6 days

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6.10 Hyperbolic Functions 491

6.10 Hyperbolic Functions In this section, we introduce a new family of functions called the hyperbolic functions, which are closely related to both trigonometric functions and exponential functions. Hyperbolic functions find widespread use in applied problems in fluid dynamics, projectile motion, architecture, and electrical engineering, to name just a few areas. Hyperbolic functions are also important in the development of many theoretical results in mathematics.

Relationship Between Trigonometric and Hyperbolic Functions The trigonometric functions defined in Chapter 1 are based on relationships involving a circle—for this reason, trigonometric functions are also known as circular functions. Specifically, cos t and sin t are equal to the x- and y-coordinates, respectively, of the point P1x, y2 on the unit circle that corresponds to an angle of t radians (Figure 6.90). We can also regard t as the length of the arc from 11, 02 to the point P1x, y2.

Figure 6.90

x2 ! y2 ” 1 1

y

x Area ” t/2

Three ways to interpret t : t = angle in radians t = arc length t = 2 · (area of sector)

t

P(x, y) ” (cos t, sin t)

Arc length ” t

1

There is yet another way to interpret the number t, and it is this third interpretation that links the trigonometric and hyperbolic functions. Observe that t is twice the area of the circular sector in Figure 6.90. The functions cos t and sin t are still defined as the x- and y-coordinates of the point P, but now we associate P with a sector whose area is one-half of t.

The hyperbolic cosine and hyperbolic sine are defined in an analogous fashion us- ing the hyperbola x2 – y2 = 1 instead of the circle x2 + y2 = 1. Consider the region bounded by the x-axis, the right branch of the unit hyperbola x2 – y2 = 1, and a line seg- ment from the origin to a point P1x, y2 on the hyperbola (Figure 6.91); let t equal twice the area of this region.

The hyperbolic cosine of t, denoted cosh t, is the x-coordinate of P and the hyperbolic sine of t, denoted sinh t, is the y-coordinate of P. Expressing x and y in terms of t leads to the standard definitions of the hyperbolic functions. We accomplish this task by writing t, which is an area, as an integral that depends on the coordinates of P. In Exercise 112, we ask you to carry out the calculations to show that

x = cosh t = et + e-t

2 and y = sinh t =

et – e-t

2 .

Everything that follows in this section is based on these two definitions.

Definitions, Identities, and Graphs of the Hyperbolic Functions Once the hyperbolic cosine and hyperbolic sine are defined, the four remaining hyperbolic functions follow in a manner analogous to the trigonometric functions.

➤ Recall that the area of a circular sector of radius r and angle u is A = 12 r2u. With r = 1 and u = t, we have A = 12t, which implies t = 2A.

Figure 6.91

1

x2 ! y2 ” 1

y

P(x, y) ” (cosh t, sinh t)

x

t ” twice the area of hyperbolic sector

M06_BRIG7345_02_SE_C06.10.indd 491 21/10/13 11:30 AM

492 Chapter 6 Applications of Integration

The hyperbolic functions satisfy many important identities. Let’s begin with the fun- damental identity for hyperbolic functions, which is analogous to the familiar trigonomet- ric identity cos2 x + sin2 x = 1:

cosh2 x – sinh2 x = 1.

This identity is verified by appealing to the definitions:

cosh2 x – sinh2 x = a ex + e-x 2

b2 – a ex – e-x 2

b2 Definition of cosh x and sinh x =

e2x + 2 + e-2x – 1e2x – 2 + e-2×2 4

Expand and combine fractions.

= 4 4

= 1. Simplify.

EXAMPLE 1 Deriving hyperbolic identities

a. Use the fundamental identity cosh2 x – sinh2 x = 1 to prove that 1 – tanh2 x = sech2 x. b. Derive the identity sinh 2x = 2 sinh x cosh x.

SOLUTION

a. Dividing both sides of the fundamental identity cosh2 x – sinh2 x = 1 by cosh2 x leads to the desired result:

cosh2 x – sinh2 x = 1 Fundamental identity

cosh2 x cosh2 x

– sinh 2 x

cosh2 x =

1 cosh2 x

Divide both sides by cosh2 x. (+)+* (+)+* tanh2 x sech2 x

1 – tanh2 x = sech2 x. Identify functions.

b. Using the definition of the hyperbolic sine, we have

sinh 2x = e2x – e-2x

2 Definition of sinh

= 1ex – e-x21ex + e-x2

2 Factor; difference of perfect squares

= 2 sinh x cosh x. Identify functions. Related Exercises 11–18

The identities in Example 1 are just two of many useful hyperbolic identities, some of which we list next.

➤ There is no universally accepted pronunciation of the names of the hyperbolic functions. In the United States, cohsh x (long oh sound) and sinch x are common choices for cosh x and sinh x. The pronunciations tanch x, cotanch x, seech x or sech x, and coseech x or cosech x are used for the other functions. International pronunciations vary as well.

➤ The fundamental identity for hyperbolic functions can also be understood in terms of the geometric definition of the hyperbolic functions. Because the point P1cosh t, sinh t2 is on the hyperbola x2 – y2 = 1, the coordinates of P satisfy the equation of the hyperbola, which leads immediately to

cosh2 t – sinh2 t = 1.

DEFINITION Hyperbolic Functions Hyperbolic cosine Hyperbolic sine

cosh x = ex + e-x

2 sinh x =

ex – e-x

2

Hyperbolic tangent Hyperbolic cotangent

tanh x = sinh x cosh x

= ex – e-x

ex + e-x coth x = cosh x sinh x

= ex + e-x

ex – e-x

Hyperbolic secant Hyperbolic cosecant

sech x = 1

cosh x =

2 ex + e-x csch x =

1 sinh x

= 2

ex – e-x

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6.10 Hyperbolic Functions 493

Graphs of the hyperbolic functions are relatively easy to produce be- cause they are based on the familiar graphs of ex and e-x. Recall that lim

xS ∞ e-x = 0 and that lim

xS- ∞ ex = 0. With these facts in mind, we see that

the graph of cosh x (Figure 6.92) approaches the graph of y = 1 2

ex as x S ∞

because cosh x = ex + e-x

2 ≈

ex

2 for large values of x. A similar argument

shows that as x S -∞ , cosh x approaches y = 1 2

e-x. Note also that cosh x is an even function:

cosh1-x2 = e-x + e-1-x2 2

= ex + e-x

2 = cosh x.

Finally, cosh 0 = e0 + e0

2 = 1, so its y-intercept is 10, 12. The behavior of

sinh x, an odd function also shown in Figure 6.92, can be explained in much the same way.

Hyperbolic Identities cosh2 x – sinh2 x = 1 cosh1-x2 = cosh x 1 – tanh2 x = sech2 x sinh1-x2 = -sinh x coth2 x – 1 = csch2 x tanh1-x2 = – tanh x cosh1x + y2 = cosh x cosh y + sinh x sinh y sinh1x + y2 = sinh x cosh y + cosh x sinh y cosh 2x = cosh2 x + sinh2 x sinh 2x = 2 sinh x cosh x

cosh2 x = cosh 2x + 1

2 sinh2 x =

cosh 2x – 1 2

Figure 6.92

1 2

y

x!2 !1 1 2

2

!1

!2

y ” !ex 1 2

y ” !e!x

1 2

y ” !!e!x

y ” cosh x domain (!#, #) range [1, #)

(0, 1)

cosh x is an even function

y ” sinh x domain (!#, #) range (!#, #)

sinh x is an odd function

Figure 6.93

!2 !1

2

1

!1

!2

(0, 1)

x1 2

y ” tanh x domain (!#, #) range (!1, 1)

y ” coth x domain x ! 0 range |y| $ 1

!2 !1

y

2

1

!1

!2

y

x1 2

y ” sech x domain (!#, #) range (0, 1]

y ” csch x domain x ! 0 range y ! 0

QUICK CHECK 1 Use the definition of the hyperbolic sine to show that sinh x is an odd function.

The graphs of the other four hyperbolic functions are shown in Figure 6.93. As a con- sequence of their definitions, we see that the domain of cosh x, sinh x, tanh x, and sech x is 1- ∞ , ∞2, whereas the domain of coth x and csch x is the set of all real numbers excluding 0.

QUICK CHECK 2 Explain why the graph of tanh x has the horizontal asymptotes y = 1 and y = -1.

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494 Chapter 6 Applications of Integration

Derivatives and Integrals of Hyperbolic Functions Because the hyperbolic functions are defined in terms of ex and e-x, computing their derivatives is straightforward. The derivatives of the hyperbolic functions are given in Theorem 6.8—reversing these formulas produces corresponding integral formulas.

➤ The identities, derivative formulas, and integral formulas for the hyperbolic functions are similar to the corresponding formulas for the trigonometric functions, which makes them easy to remember. However, be aware of some subtle differences in the signs associated with these formulas. For instance,

d>dx1cos x2 = -sin x, whereas

d>dx1cosh x2 = sinh x.

THEOREM 6.8 Derivative and Integral Formulas

1. d dx 1cosh x2 = sinh x 1 Lsinh x dx = cosh x + C

2. d dx 1sinh x2 = cosh x 1 Lcosh x dx = sinh x + C

3. d dx 1tanh x2 = sech2 x 1 Lsech2 x dx = tanh x + C

4. d dx 1coth x2 = -csch2 x 1 Lcsch2 x dx = -coth x + C

5. d dx 1sech x2 = -sech x tanh x 1 Lsech x tanh x dx = -sech x + C

6. d dx 1csch x2 = -csch x coth x 1 Lcsch x coth x dx = -csch x + C

Proof: Using the definitions of cosh x and sinh x, we have

d dx

1cosh x2 = d dx

a ex + e-x 2

b = ex – e-x 2

= sinh x and

d dx

1sinh x2 = d dx

a ex – e-x 2

b = ex + e-x 2

= cosh x.

To prove formula (3), we begin with tanh x = sinh x>cosh x and then apply the Quotient Rule:

d dx

1tanh x2 = d dx

a sinh x cosh x

b Definition of tanh x =

cosh x1cosh x2 – sinh x1sinh x2 cosh2 x

Quotient Rule

= 1

cosh2 x cosh2 x – sinh2 x = 1

= sech2 x. sech x = 1>cosh x The proofs of the remaining derivative formulas are assigned in Exercises 19–21. The

integral formulas are a direct consequence of their corresponding derivative formulas.

EXAMPLE 2 Derivatives and integrals of hyperbolic functions Evaluate the follow- ing derivatives and integrals.

a. d dx

1sech 3×2 b. d2 dx2

1sech 3×2 c. L

csch2 1x1x dx d. L ln 30 sinh3 x cosh x dx

M06_BRIG7345_02_SE_C06.10.indd 494 21/10/13 11:30 AM

6.10 Hyperbolic Functions 495

SOLUTION

a. Combining formula (5) of Theorem 6.8 with the Chain Rule gives

d dx

1sech 3×2 = -3 sech 3x tanh 3x. b. Applying the Product Rule and Chain Rule to the result of part (a), we have

d2

dx2 1sech 3×2 = d

dx 1-3 sech 3x tanh 3×2

= d dx

1-3 sech 3×2 # tanh 3x + 1-3 sech 3×2 # d dx

1tanh 3×2 Product Rule v u

9 sech 3x tanh 3x 3 sech2 3x

= 9 sech 3x tanh2 3x – 9 sech3 3x Chain Rule = 9 sech 3x1tanh2 3x – sech2 3×2. Simplify.

c. The integrand suggests the substitution u = 1x: L

csch2 1x1x dx = 2Lcsch2 u du Let u = 1x ; du = 121x dx. = -2 coth u + C Formula (4), Theorem 6.8 = -2 coth 1x + C. u = 1x

d. The derivative formula d>dx1sinh x2 = cosh x suggests the substitution u = sinh x: L

ln 3

0 sinh3 x cosh x dx = L

4>3 0

u3 du. Let u = sinh x; du = cosh x dx.

The new limits of integration are determined by the calculations

x = 0 1 u = sinh 0 = 0 and

x = ln 3 1 u = sinh1ln 32 = eln 3 – e-ln 3 2

= 3 – 1>3

2 =

4 3

.

We now evaluate the integral in the variable u:

L 4>3

0 u3 du =

1 4

u4 ` 4>3 0

= 1 4

aa 4 3 b4 – 04b = 64

81 .

Related Exercises 19–40

QUICK CHECK 3 Find both the derivative and indefinite integral of f 1×2 = 4 cosh 2x. Theorem 6.9 presents integral formulas for the four hyperbolic functions not covered

in Theorem 6.8.

THEOREM 6.9 Integrals of Hyperbolic Functions

1. L tanh x dx = ln cosh x + C 2. Lcoth x dx = ln ! sinh x ! + C 3. Lsech x dx = tan-11sinh x2 + C 4. Lcsch x dx = ln ! tanh1x>22 ! + C

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496 Chapter 6 Applications of Integration

Proof: Formula (1) is derived by first writing tanh x in terms of sinh x and cosh x:

L tanh x dx = L sinh x cosh x

dx Definition of tanh x

= L 1 u

du Let u = cosh x; du = sinh x dx.

= ln ! u ! + C Evaluate integral. = ln cosh x + C. u = cosh x 7 0

Formula (2) is derived in a similar fashion (Exercise 44). The more challenging proofs of formulas (3) and (4) are considered in Exercises 107 and 108.

EXAMPLE 3 Integrals involving hyperbolic functions Determine the indefinite integral 1x coth 1×22 dx. SOLUTION The integrand suggests the substitution u = x2:

Lx coth x2 dx = 1 2 Lcoth u du. Let u = x2; du = 2x dx.

= 1 2

ln ! sinh u ! + C Evaluate integral; use Theorem 6.9.

= 1 2

ln1sinh x22 + C. u = x2; sinh x2 Ú 0 Related Exercises 41–44

QUICK CHECK 4 Determine the indefinite integral 1csch 2x dx. Inverse Hyperbolic Functions

At present, we don’t have the tools for evaluating an integral such as L dx2x2 + 4.

By studying inverse hyperbolic functions, we can discover new integration formulas. Inverse hyperbolic functions are also useful for solving equations involving hyperbolic functions.

Figures 6.92 and 6.93 show that the functions sinh x, tanh x, coth x, and csch x are all one-to-one on their respective domains. This observation implies that each of these func- tions has a well-defined inverse. However, the function y = cosh x is not one-to-one on 1- ∞ , ∞2, so its inverse, denoted y = cosh-1 x, exists only if we restrict the domain of cosh x. Specifically, when y = cosh x is restricted to the interval 30, ∞2, it is one-to-one, and its inverse is defined as follows:

y = cosh-1 x if and only if x = cosh y, for x Ú 1 and 0 … y 6 ∞ .

Figure 6.94a shows the graph of y = cosh-1 x, obtained by reflecting the graph of y = cosh x on 30, ∞2 over the line y = x. The definitions and graphs of the other five inverse hyperbolic functions are also shown in Figure 6.94. Notice that the domain of y = sech x (Figure 6.94d) must be restricted to 30, ∞2 to ensure the existence of its inverse.

Because hyperbolic functions are defined in terms of exponential functions, we can find explicit formulas for their inverses in terms of logarithms. For example, let’s start with the definition of the inverse hyperbolic sine. For all real x and y, we have

y = sinh-1 x 3 x = sinh y.

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6.10 Hyperbolic Functions 497

Following the procedure outlined in Section 1.3, we solve

x = sinh y = ey – e-y

2

for y to give a formula for sinh-1 x:

x = ey – e-y

2 1 ey – 2x – e-y = 0 Rearrange equation.

1 1ey22 – 2xey – 1 = 0. Multiply by ey. At this stage, we recognize a quadratic equation in ey and solve for ey using the quadratic formula, with a = 1, b = -2x, and c = -1:

ey = 2x { 24×2 + 4

2 = x { 2×2 + 1 = x + 2×2 + 1.

choose positive root

Because ey 7 0 and 2×2 + 1 7 x, the positive root must be chosen. We now solve for y by taking the natural logarithm of both sides:

ey = x + 2×2 + 1 1 y = ln1x + 2×2 + 12. Therefore, the formula we seek is sinh-1 x = ln1x + 2×2 + 12.

A similar procedure can be carried out for the other inverse hyperbolic functions (Exercise 110). Theorem 6.10 lists the results of these calculations.

y ! cosh”1 x

y ! cosh x

y ! cosh”1 x ⇔ x ! cosh y for x # 1 and 0 $ y % &

(a)

y

x

2

1

1 2

y ! sinh x

y ! sinh”1 x

y ! sinh”1 x ⇔ x ! sinh y for “# $ x $ # and “# $ y $ #

(b)

y

x21″1″2

“2

“1

1

2

y ! tanh”1 x ⇔ x ! tanh y for “1 # x # 1 and “$ # y # $

(c)

y ! tanh x

y ! tanh”1 x y

x321 “1

1

2

3

“2

“3

“1”2″3

Figure 6.94

y ! sech”1 x

y ! sech x

y ! sech”1 x ⇔ x ! sech y for 0 # x $ 1 and 0 $ y # %

(d)

y

x321

3

2

1

y ! csch”1 x

y ! csch x

y ! csch”1 x ⇔ x ! csch y for x ! 0 and y ! 0

(e)

y

x21″1″2

“2

“1

1

2

y ! coth”1 x ⇔ x ! coth y for |x| # 1 and y ! 0

( f )

y

x21″1″2

“2

“1

1

2

y ! coth”1 x

y ! coth x

v