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bacteria will be reproducing at different times. So, averaging out, it makes sense

to hypothesize that the growth of the bacteria population varies continuously as

in Fig. 6.13. Here we are using a standard device of mathematical analysis: even

though the number of bacteria is always an integer, we represent the graph of the

population of bacteria by a smooth curve. This enables us to apply the tools of

calculus to the problem.

Fig. 6.13

6.5.1 A DIFFERENTIAL EQUATION

If B(t) represents the number of bacteria present in a given population at time t,

then the preceding discussion suggests that

dB

= K В· B(t),

dt

where K is a constant of proportionality. This equation expresses quantitatively

the assertion that the rate of change of B(t) (that is to say, the quantity dB/dt) is

proportional to B(t). To solve this equation, we rewrite it as

1 dB

В· = K.

B(t) dt

We integrate both sides with respect to the variable t:

1 dB

В· dt = K dt.

B(t) dt

The left side is

ln |B(t)| + C

and the right side is

Kt + C,

174 CHAPTER 6 Transcendental Functions

where C and C are constants of integration. We thus obtain

ln |B(t)| = Kt + D,

where we have amalgamated the two constants into a single constant D. Exponen-

tiating both sides gives

|B(t)| = eKt+D

or

B(t) = eD В· eKt = P В· eKt . ()

Notice that we have omitted the absolute value signs since the number of bacteria is

always positive. Also we have renamed the constant eD with the simpler symbol P .

Equation ( ) will be our key to solving exponential growth and decay problems.

We motivated our calculation by discussing bacteria, but in fact the calculation

applies to any function which grows at a rate proportional to the size of the function.

Next we turn to some examples.

6.5.2 BACTERIAL GROWTH

EXAMPLE 6.30

A population of bacteria tends to double every four hours. If there are 5000

bacteria at 9:00 a.m., then how many will there be at noon?

SOLUTION

To answer this question, let B(t) be the number of bacteria at time t. For con-

venience, let t = 0 correspond to 9:00 a.m. and suppose that time is measured

in hours. Thus noon corresponds to t = 3.

Equation ( ) guarantees that

B(t) = P В· eKt

for some undetermined constants P and K. We also know that

5000 = B(0) = P В· eKВ·0 = P .

We see that P = 5000 and B(t) = 5000 В· eKt . We still need to solve for K.

Since the population tends to double in four hours, there will be 10,000

bacteria at time t = 4; hence

10000 = B(4) = 5000 В· eKВ·4 .

We divide by 5000 to obtain

2 = eKВ·4 .

175

CHAPTER 6 Transcendental Functions

Taking the natural logarithm of both sides yields

ln 2 = ln(eKВ·4 ) = 4K.

We conclude that K = [ln 2]/4. As a result,

B(t) = 5000 В· e([ln 2]/4)t .

We simplify this equation by noting that

e([ln 2]/4)t = (eln 2 )t/4 = 2t/4 .

In conclusion,

B(t) = 5000 В· 2t/4 .

The number of bacteria at noon (time t = 3) is then given by

B(3) = 5000 В· 23/4 в‰€ 8409.

It is important to realize that population growth problems cannot be described

using just arithmetic. Exponential growth is nonlinear, and advanced analytical

ideas (such as calculus) must be used to understand it.

EXAMPLE 6.31

Suppose that a certain petri dish contains 6000 bacteria at 9:00 p.m. and

10000 bacteria at 11:00 p.m. How many of the bacteria were there at

7:00 p.m.?

SOLUTION

We know that

B(t) = P В· eKt .

The algebra is always simpler if we take one of the times in the initial data to

correspond to t = 0. So let us say that 9:00 p.m. is t = 0. Then 11:00 p.m. is

t = 2 and 7:00 p.m. is t = в€’2. The initial data then tell us that

6000 = P В· eKВ·0 (в€—)

10000 = P В· eKВ·2 . (в€—в€—)

From equation (в€—) we may immediately conclude that P = 6000. Substituting

this into (в€—в€—) gives

10000 = 6000 В· (eK )2 .

We conclude that

в€љ

5

=в€љ .

eK

3

176 CHAPTER 6 Transcendental Functions

As a result,

в€љ t

5

B(t) = 6000 В· в€љ .

3

At time t = в€’2 (7:00 p.m.) the number of bacteria was therefore

в€љ в€’2

5 3

B(в€’2) = 6000 В· в€љ = В· 6000 = 3600.

5

3

You Try It: A petri dish has 5000 bacteria at 1:00 p.m. on a certain day and 8000

bacteria at 5:00 p.m. that same day. How many bacteria were there at noon?

6.5.3 RADIOACTIVE DECAY

Another natural phenomenon which п¬Ѓts into our theoretical framework is radio-

active decay. Radioactive material, such as C14 (radioactive carbon), has a half life.

Saying that the half life of a material is h years means that if A grams of material is

present at time t then A/2 grams will be present at time t + h. In other words, half

of the material decays every h years. But this is another way of saying that the rate

at which the radioactive material vanishes is proportional to the amount present.

So equation ( ) will apply to problems about radioactive decay.

EXAMPLE 6.32

Five grams of a certain radioactive isotope decay to three grams in 100

years. After how many more years will there be just one gram?

SOLUTION

First note that the answer is not вЂњwe lose two grams every hundred years

so вЂ¦.вЂќ The rate of decay depends on the amount of material present. That is

the key.

Instead, we let R(t) denote the amount of radioactive material at time t.

Equation ( ) guarantees that R has the form

R(t) = P В· eKt .

Letting t = 0 denote the time at which there are 5 grams of isotope, and

measuring time in years, we have

R(0) = 5 R(100) = 3.

and

From the п¬Ѓrst piece of information we learn that

5 = P В· eKВ·0 = P .

Hence P = 5 and

R(t) = 5 В· eKt = 5 В· (eK )t .

177

CHAPTER 6 Transcendental Functions

The second piece of information yields

3 = R(100) = 5 В· (eK )100 .

We conclude that

3

(eK )100 =

5

or

1/100

3

=

K

e .

5

Thus the formula for the amount of isotope present at time t is

t/100

3

R(t) = 5 В· .

5

Thus we have complete information about the function R, and we can answer

the original question.

There will be 1 gram of material present when

t/100

3

1 = R(t) = 5 В·

5

or

t/100

1 3

= .

5 5

We solve for t by taking the natural logarithm of both sides:

t/100

3 t

ln(1/5) = ln = В· ln(3/5).

5 100

We conclude that there is 1 gram of radioactive material remaining when

ln(1/5)

t = 100 В· в‰€ 315.066.

ln(3/5)

So at time t = 315.066, or after 215.066 more years, there will be 1 gram of

the isotope remaining.

You Try It: Our analysis of exponential growth and decay is derived from the

hypothesis that the rate of growth is proportional to the amount of matter present.

Suppose instead that we are studying a system in which the rate of decay is propor-

tional to the square of the amount of matter. Let M(t) denote the amount of matter

at time t. Then our physical law is expressed as

dM

= C В· M 2.

dt

178 CHAPTER 6 Transcendental Functions

Here C is a (negative) constant of proportionality. We apply the method of

вЂњseparation of variablesвЂќ described earlier in the section. Thus

dM/dt

=C

M2

so that

dM/dt

dt = C dt.

M2

Evaluating the integrals, we п¬Ѓnd that

1

в€’ = Ct + D.

M

We have combined the constants from the two integrations. In summary,

1

M(t) = в€’ .

Ct + D

Y

For the problem to be realistic, we will require that C < 0 (so that M > 0 for

FL

large values of t) and we see that the population decays like the reciprocal of a

linear function when t becomes large.

AM

Re-calculate Example 6.32 using this new law of exponential decay.

6.5.4 COMPOUND INTEREST

TE

Yet a third illustration of exponential growth is in the compounding of interest. If

principal P is put in the bank at p percent simple interest per year then after one

year the account has

p

P В· 1+

100

dollars. [Here we assume, of course, that all interest is reinvested in the account.]

But if the interest is compounded n times during the year then the year is divided

into n equal pieces and at each time interval of length 1/n an interest payment of

percent p/n is added to the account. Each time this fraction of the interest is added

to the account, the money in the account is multiplied by

p/n

1+ .

100

Since this is done n times during the year, the result at the end of the year is that

the account holds

pn

P В· 1+ (в€—)

100n

179

CHAPTER 6 Transcendental Functions

dollars at the end of the year. Similarly, at the end of t years, the money accumulated

will be

p nt

P В· 1+ .

100n

Let us set

n В· 100

k=

p

and rewrite (в€—) as

k p/100

kp/100

1 1

P В· 1+ =P В· 1+ .

k k

It is useful to know the behavior of the account if the number of times the inter-

est is compounded per year becomes arbitrarily large (this is called continuous

compounding of interest). Continuous compounding corresponds to calculating the

limit of the last formula as k (or, equivalently, n), tends to inп¬Ѓnity.

We know from the discussion in Subsection 6.2.3 that the expression (1 + 1/k)k

tends to e. Therefore the size of the account after one year of continuous

compounding of interest is

P В· ep/100 .

After t years of continuous compounding of interest the total money is

P В· ept/100 . ()

EXAMPLE 6.33

If $6000 is placed in a savings account with 5% annual interest compounded

continuously, then how large is the account after four and one half years?

SOLUTION

If M(t) is the amount of money in the account at time t, then the preceding

discussion guarantees that

M(t) = 6000 В· e5t/100 .

After four and one half years the size of the account is therefore

M(9/2) = 6000 В· e5В·(9/2)/100 в‰€ $7513.94.

EXAMPLE 6.34

A wealthy woman wishes to set up an endowment for her nephew. She

wants the endowment to pay the young man $100,000 in cash on the day of

his twenty-п¬Ѓrst birthday.The endowment is set up on the day of the nephewвЂ™s

180 CHAPTER 6 Transcendental Functions

birth and is locked in at 11% interest compounded continuously. How much

principal should be put into the account to yield the desired payoff?

SOLUTION

Let P be the initial principal deposited in the account on the day of the

nephewвЂ™s birth. Using our compound interest equation ( ), we have

100000 = P В· e11В·21/100 ,

expressing the fact that after 21 years at 11% interest compounded continuously

we want the value of the account to be $100,000.

Solving for P gives

P = 100000 В· eв€’0.11В·21 = 100000 В· eв€’2.31 = 9926.13.

The aunt needs to endow the fund with an initial $9926.13.

You Try It: Suppose that we want a certain endowment to pay $50,000 in cash

ten years from now. The endowment will be set up today with $5,000 principal and

locked in at a п¬Ѓxed interest rate. What interest rate (compounded continuously) is

needed to guarantee the desired payoff?

6.6 Inverse Trigonometric Functions

6.6.1 INTRODUCTORY REMARKS

Figure 6.14 shows the graphs of each of the six trigonometric functions. Notice that

each graph has the property that some horizontal line intersects the graph at least

twice. Therefore none of these functions is invertible. Another way of seeing this

point is that each of the trigonometric functions is 2ПЂ -periodic (that is, the function

repeats itself every 2ПЂ units: f (x + 2ПЂ ) = f (x)), hence each of these functions is

not one-to-one.

If we want to discuss inverses for the trigonometric functions, then we must

restrict their domains (this concept was introduced in Subsection 1.8.5). In this

section we learn the standard methods for performing this restriction operation

with the trigonometric functions.

6.6.2 INVERSE SINE AND COSINE

Consider the sine function with domain restricted to the interval [в€’ПЂ/2, ПЂ/2]

(Fig. 6.15). We use the notation Sin x to denote this restricted function. Observe

that

d

Sin x = cos x > 0

dx

181

CHAPTER 6 Transcendental Functions

3

2

sec x

1

cos x sin x

_3 _2 _1 1 2 3

_1

csc x

_2

tan x cot x

_3

Fig. 6.14

y

y = Sin x

1

F/2 x

Fig. 6.15

182 CHAPTER 6 Transcendental Functions

on the interval (в€’ПЂ/2, ПЂ/2). At the endpoints of the interval, and only there, the

function Sin x takes the values в€’1 and +1. Therefore Sin x is increasing on its

entire domain. So it is one-to-one. Furthermore the Sine function assumes every

value in the interval [в€’1, 1]. Thus Sin : [в€’ПЂ/2, ПЂ/2] в†’ [в€’1, 1] is one-to-one and

onto; therefore f (x) = Sin x is an invertible function.

We can obtain the graph of Sinв€’1 x by the principle of reп¬‚ection in the line

y = x (Fig. 6.16). The function Sinв€’1 : [в€’1, 1] в†’ [в€’ПЂ/2, ПЂ/2] is increasing,

one-to-one, and onto.

y

F/2

_1

y = Sin x

x

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