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Notice that when the number of bacteria is large, then different generations of
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 · 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
· = K.
B(t) dt
We integrate both sides with respect to the variable t:
1 dB
· 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 · eKt = P · 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 · eKt
for some undetermined constants P and K. We also know that
5000 = B(0) = P · eK·0 = P .
We see that P = 5000 and B(t) = 5000 · 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 · eK·4 .
We divide by 5000 to obtain
2 = eK·4 .
175
CHAPTER 6 Transcendental Functions

Taking the natural logarithm of both sides yields
ln 2 = ln(eK·4 ) = 4K.
We conclude that K = [ln 2]/4. As a result,
B(t) = 5000 · 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 · 2t/4 .
The number of bacteria at noon (time t = 3) is then given by
B(3) = 5000 · 23/4 ≈ 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 · 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 = ’2. The initial data then tell us that
6000 = P · eK·0 (—)
10000 = P · eK·2 . (——)
From equation (—) we may immediately conclude that P = 6000. Substituting
this into (——) gives
10000 = 6000 · (eK )2 .
We conclude that

5
=√ .
eK
3
176 CHAPTER 6 Transcendental Functions

As a result,
√ t
5
B(t) = 6000 · √ .
3
At time t = ’2 (7:00 p.m.) the number of bacteria was therefore
√ ’2
5 3
B(’2) = 6000 · √ = · 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 ¬ts 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 “we lose two grams every hundred years
so ¦.” 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 · 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 ¬rst piece of information we learn that
5 = P · eK·0 = P .
Hence P = 5 and
R(t) = 5 · eKt = 5 · (eK )t .
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CHAPTER 6 Transcendental Functions

The second piece of information yields
3 = R(100) = 5 · (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 · .
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 ·
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 = · ln(3/5).
5 100
We conclude that there is 1 gram of radioactive material remaining when
ln(1/5)
t = 100 · ≈ 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 · M 2.
dt
178 CHAPTER 6 Transcendental Functions

Here C is a (negative) constant of proportionality. We apply the method of
“separation of variables” described earlier in the section. Thus
dM/dt
=C
M2
so that
dM/dt
dt = C dt.
M2
Evaluating the integrals, we ¬nd that
1
’ = Ct + D.
M
We have combined the constants from the two integrations. In summary,
1
M(t) = ’ .
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 · 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 · 1+ (—)
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 · 1+ .
100n
Let us set
n · 100
k=
p
and rewrite (—) as
k p/100
kp/100
1 1
P · 1+ =P · 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 in¬nity.
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 · ep/100 .
After t years of continuous compounding of interest the total money is
P · 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 · e5t/100 .
After four and one half years the size of the account is therefore
M(9/2) = 6000 · e5·(9/2)/100 ≈ $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-¬rst birthday.The endowment is set up on the day of the nephew™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
nephew™s birth. Using our compound interest equation ( ), we have
100000 = P · e11·21/100 ,
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 · e’0.11·21 = 100000 · e’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 ¬xed 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 2π -periodic (that is, the function
repeats itself every 2π units: f (x + 2π ) = 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 [’π/2, π/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 (’π/2, π/2). At the endpoints of the interval, and only there, the
function Sin x takes the values ’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 [’1, 1]. Thus Sin : [’π/2, π/2] ’ [’1, 1] is one-to-one and
onto; therefore f (x) = Sin x is an invertible function.
We can obtain the graph of Sin’1 x by the principle of re¬‚ection in the line
y = x (Fig. 6.16). The function Sin’1 : [’1, 1] ’ [’π/2, π/2] is increasing,
one-to-one, and onto.
y

F/2
_1
y = Sin x



x

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