Calculus Of One Real Variable – By Pheng Kim Ving Chapter 7: The Exponential And Logarithmic Functions – Section 7.5: Growth And Decay 7.5 Growth And Decay

 1. Rate Of Change Of Quantity Proportional To Quantity

In this section that discusses the processes of growth and decay, the quantities are functions of time and their rates of
change are with respect to time.

Suppose at one time there are 1,000 bacteria in a container. Suppose 1 hour later, the number of bacteria increases to 1,100.
So the rate of increase of the number of bacteria is 100 bacteria per hour (rate of change is +100). Suppose at a later time
there are 2,000 bacteria. Supppose 1 hour later, the number of bacteria increases to 2,200. So now the rate of increase of the
number of bacteria is 200 bacteria per hour. As the number of bacteria doubles, its rate of increase doubles also. In general, if
the number of bacteria is multiplied by a real number
k > 1, then its rate of increase is multiplied by k also. The rate of
increase of the number of bacteria is proportional to the number of bacteria. This is a reasonable situation. As the number of
bacteria increases, there are more bacteria to re-produce per unit of time.

Suppose at one time there are 1,000 grams of a radioactive material in a container. Suppose 1 year later, the mass of the
material decreases to 900 grams. So the rate of decrease of the mass of the material is 100 grams per year (rate of change is
–100). Suppose at a later time there are 500 grams of the material. Supppose 1 year later, the mass of the material
decreases to 450 grams. So now the rate of decrease of the mass of the material is 50 grams per year. As the mass of the
material is halved, its rate of decrease is halved also. In general, if the mass of the material is divided by a real number
k > 1,
then its rate of decrease is divided by
k also. The rate of decrease of the mass of the material is proportional to the mass of
the material. This is a reasonable situation. As the mass of the material decreases, there's less material to disappear per unit
of time.

Rate Of Change Of Amount Of Money # Compounded Annually

Suppose the interest rate is compounded annually: once per year, at the end of the year. The two 1-year periods are the
t-intervals [0, 1] and [1, 2], as shown in Fig. 1.1. The unit of the t-axis is the year. At the ends of the first and second
periods the amount grows to
A(1) and A(2) respectively where: Note that the average rate of change of the amount over any period is equal to the same constant i/100 multiplied by the
amount at the start of that period. The average rate of change of the amount is proportional to the amount, with the
constant of proportionality of
i/100. ## Fig. 1.1

Two 1-year periods. ## Fig. 1.2

Four half-year periods.

# Compounded Semi-Annually

Suppose the interest rate is compounded semi-annually instead: once per each half-year period, at the end of the period;
thus twice per year. There are four periods. They're the
t-intervals [0, 1/2], [1/2, 1], [1, 3/2], and [3/2, 2], as shown in
Fig. 1.2. The semi-annual interest rate is (
i/2)%. The amounts A(1/2), A(1), A(3/2), and A(2) at the ends of the first,
second, third, and fourth periods respectively are: The average rates of change of the amount over the four periods [0, 1/2], [1/2, 1], [1, 3/2], and [3/2, 2] respectively
are: Again the average rate of change of the amount is proportional to the amount, with the same constant of proportionality
of
i/100. Remark that t is a real number, not just an integer.

Compounded n Times Per Year

The amount A(t) in the case where the interest is compounded semi-annually can be written explicitly as a function of t
as follows: In the target expression for A(t), the t is the exponent over the outside pair of parentheses. We also re-write i/(100 x 2)
as (
i/100)/2. The number of times the interest is compounded per year, namely 2, appears as the denominator inside
and as the exponent over the inside pair of parentheses.

The amounts in the case where the interest is compounded annually can be written in the same form as follows: Observe that in this formula, i is the annual  (yearly, or per-year) interest rate and t is measured in year. The per-unit of i
and the unit of
t must be the same. For example, if i is the daily interest rate instead, then t must be measured in day.

As the number of yearly compoundings is n, each 1-year interval is divided into n equal sub-intervals, each of which is of
length 1/
n, and the compounding occurs at the end of each sub-interval. Refer to Fig. 1.2 for the case of n = 2. There are
2
n sub-intervals in the 2-year period. The average rate of change of the amount over any sub-interval is proportional to
the amount at the start of that sub-interval, with the same constant of proportionality of
i/100, ie: where t = t1 and h = 1/n.

#### Compounded Continuously It follows that the amount A(t) at t years after the initial amount A(0) is deposited at an
annual interest rate of
i% compounded continuously is A(t) = (ei/100)tA(0). Therefore: Note that even though the interest is compounded an infinite number of times per year, the resulting interest and thus the
amount are finite, not  infinite. Thus:

 If the annual interest rate of i% is compounded continuously, then the rate of change of the amount at any instant is proportional to the amount at that instant, with the constant of proportionality of i/100.

This fact can be verified by differentiating Eq. [1.3] as follows:

dA/dt = A(0)(i/100)e(i/100)t = (i/100)(A(0)e(i/100)t) = (i/100)A(t),

as desired.

#### Comparison  Fig. 1.3   Size of growth changes slightly as number of compoundings per year changes.

Example 1.1

An amount of \$15,000 is deposited in a savings account at an annual interest rate of 5% compounded continuously.
Determine the amount in the account after 10 years.

Solution
Let A(t) be the amount after t years. So:

A(t) = A(0)e(5/100)t = 15000et/20.

After 10 years the amount is: EOS

 2. General Case

The amount in a savings account with an annual interest rate of i% compounded continuously grows at a rate that's at
any instant proportional to the amount itself at that instant:
dA/dt = (i/100)A(t), with the constant of proportionality of
i/100. Compounding the interest continuously means that the amount grows continuously. We've got this behavior of the
amount when the interest is compounded continuously: it grows continuously and its rate of growth with respect to time
at any instant is proportional to it at that instant.

In general, in many natural processes, a quantity grows (increases) or decays (decreases) at a rate with respect to time that's
proportional to the quantity itself. For example, under ideal conditions with adequate food supply and with no disease,
overcrowding, or predators, the rate of growth of a population (people, elephants, or bacteria) is proportional to the size
of the population. The larger the population, the larger its rate of growth, ie the faster it grows. As another example, the
rate of decay of a radioactive substance is proportional to the mass of the substance present. The smaller the mass, the
smaller its rate of decay, ie, the slower it decays.

If a quantity Q(t) grows or decays at a rate with respect to time proportional to the quantity itself, then dQ/dt = kQ(t),
where
k is the constant of proportionality. Remark that, since Q(t) > 0, we have k > 0 if Q(t) grows and k < 0 if Q(t)
decays; that's a consequence of Section 5.3 Theorem 2.1.

The quantity y = e(1/10)t grows at a rate with respect to time t proportional to the quantity itself: dy/dt = e(1/10)t(1/10) =
(1/10)
y. The quantity y = t2 grows at a rate not  proportional to the quantity: dy/dt = 2t, and there's no constant k such
that 2
t = kt2, because if 2t =  kt2, then k = 2/t, which is a non-constant.

 3. The Equations where k is the constant of proportionality. Growth and decay processes of quantities with rates proportional to the
quantities are governed by this differential equation. For an annual interest rate of
i% compounded continuously, we have
k = i/100.

Let's find the solution y = y(t) of the differential equation [3.1]. In Part 1, we saw that the amount function A(t) in the
case where the interest is compounded continuously is given by
A(t) = A(0)e(i/100)t as seen in Eq. [1.3] and satisfies the
differential equation
dA/dt = (i/100)A(t) as seen in Eq. [1.4]. This leads us to suspect that a solution of Eq. [3.1] is given
by
y = Cekt for some constant C. Let's prove it: Indeed our suspicion is correct. Is there any other solution? Let's find out. Suppose g(t) is also a solution, so that g'(t) =
kg(t). Then: g(t) = g(0)ekt.

As g(0) is a number, g(t) is of the form g(t) = Cekt for some constant C. Hence the answer is no: there's no other
solution. If a function
f(t) is a solution, then it must be of the form f(t) = Cekt. Now we see that the function y = y(t) =
Cekt, where C is an arbitrary constant, is the general  solution of Eq. [3.1].

Let's determine what the constant C is. We have y(0) = Ce0 = C . 1 = C. It follows that C is the initial value y(0) of the
quantity
y(t) at the initial time t = 0. See Fig. 3.1. A particular solution is obtained by specifying a value for C. So a
particular solution is obtained by specifying its initial value
y(0). Thus the solution of Eq. [3.1] is:

 y = y(0)ekt.                                                                                                                                                           [3.2]

Remark that Eq. [1.3] is in the form of Eq. [3.2]. If a quantity grows or decays with a rate proportional to the quantity
itself, then it must be of the form of Eq. [3.2]. That's why it's also referred to as exponential growth or decay. # Fig. 3.1

For y(t) = Cekt we have C = y(0).

### Sign Ofk

Clearly k > 0 iff y increases as time increases, and k < 0 iff y decreases as time increases. This can be seen from Eq.
[3.1] or Eq. [3.2]. Remember that
y > 0 and y(0) > 0 as we deal only with positive quantities.

#### Speed Of Change

In growth and decay problems, the quantities are functions of time and their rates of change are with respect to time.
Therefore we can think of the rate of change of a quantity as the speed of change of the quantity.

 4. The Constant Of Proportionality

### Example 4.1

Suppose the population of a country in 1985 was 30,000,000. The population has been growing by 2% annually. Assume
the population was growing and continues to grow continuously at a rate proportional to the population. If the country's
land area is 150,000 km
2, find the year when there'll be one person for every square meter of its land area.

Solution

Let P = P(t) be the population t years after 1985. As dP/dt = kP we have P(t) = P(0)ekt, where P(0) = 30,000,000 =
3 x 10
7. As the population grows by 2% annually we have:

P(0)ek(t+1) = P(t + 1) = P(t) + (2/100)P(t) = (1.02)P(t) = (1.02)P(0)ekt,
P(0)ektek = (1.02)P(0)ekt,
ek = 1.02,
k = ln 1.02.

Consequently:

P(t) = (3 x 107)e(ln 1.02)t.

Now 150,000 km2 = 150,000 x (1,000 m)2 = 150,000,000,000 m2 = 1.5 x 1011 m2. Hence, when there'll be one person for
every square meter, the population will be 1.5 x 10
11. The year when this will happen will be 1985 + t, where
(3 x 10
7)e(ln 1..02)t = P(t) = 1.5 x 1011, from which we get:

e(ln 1..02)t = (1.5 x 1011)/(3 x 107) = 5 x 103,
(
ln 1.02)t = ln (5 x 103), It follows that there'll be one person for every square meter of land in the year 1985 + 431 = 2416.
EOS

#### The Constant Of Proportionality

Suppose dy/dt = ky(t), where k is the constant of proportionality. The instantaneous percentage rate of change is the
instantaneous rate of change as a percentage of the quantity. We have: Let's take a look back at the case of the amount A(t) with an annual interest rate of i%. If the interest is compounded
annually, then the amount after
t + 1 years is A(t + 1) = A(t) + (i/100)A(t). The annual rate of growth is: The amount grows by i% annually. If the interest is compounded continuously then A(t) grows continuously with a rate
proportional to
A(t) itself, ie dA/dt = (i/100)A(t), the constant of proportionality, which is the instantaneous percentage
rate of growth, being
i% = i/100, the same as the annual percentage rate of growth; see Eq. [1.4]. which is larger than the 2% as given by the statement of the problem, and thus is incorrect.

So why do we use the annual percentage rate of growth i/100 as the instantaneous percentage rate of growth (constant
of proportionality) in the interest-rate case
? Let's take a look at Fig. 1.3, where i = 5. The amount A(1) resulting from
continuous compounding is greater than that resulting from annual (
n = 1) compounding. This is because we use the
annual percentage rate of growth,
i/100, as also the instantaneous percentage rate of growth: dA/dt = (i/100)A(t).

In order for a continuous growth to yield the same amount as the annual compounding, the instantaneous percentage
rate of growth
k in dA/dt = kA(t) would have to be smaller than i/100; it can be found by the same method as in
Example 4.1. But that continuous growth isn't the one resulting from continuous compounding, because the one resulting
from continuous compounding has the instantaneous percentage rate of growth of
i/100. Now, in interest-rate problems,
the continuous growth considered is the one resulting from continuous compounding. That's why we use the annual
percentage rate of growth
i/100 as the instantaneous percentage rate of growth in the interest-rate case.

For continuous growth in interest-rate problems where the interest rate (percentage rate of growth) is i% and where the
interest is compounded continuously, the constant of proportionality is
k = i/100. For continuous change (growth or decay)
in other situations, the constant of proportionality
k has to be determined if it's not given but it's needed, as done in
Example 4.1.

 5. Half-Life

#### A Particular Case  # Fig. 5.1

Same amount of time, ln 2 units of time, is required
for y to decrease by half from any value. This shows that the function has decreased by half at time t + ln 2, ie ln 2 units of time later. Now we see that the
function takes the same amount of time,
ln 2 units of time, to decrease by half from any value.

### The General Case y(t + T ) = y(0)ek(t+T) = y(0)ekt+kT = y(0)ektekT = y(0)ekt(1/2) = (1/2)y(t).

Thus, y requires the same amount of time, T, to decay by half from any  size. This amount of time is called the half-life
of the substance.

Observe that the half-life of a substance isn't half of the total life span of the substance. A substance that now weighs
100 kg and whose half-life is 1,000 years will weigh 50 kg at 1,000 years from now, and will weigh 25 kg, not 0 kg (it
hasn't totally disappeared), at 2,000 years from now. The half-life is actually the (fixed) halving time, ie the (same)
amount of time required by the substance to decay by half from any size. Also recall that that's in accordance with the
property that the smaller the quantity, the slower it decays.

### Example 5.1

A radioactive substance decays at a rate proportional to the quantity present. Suppose 20% of such a substance decays in
35 years. Find the half-life of the substance.

Solution
Let
y = y(t) be the quantity of the substance t years after the initial time t = 0. Since dy/dt = ky we get  y(t) = y(0)ekt.
After 35 years, the quantity left is 80% of the initial quantity. So:

y(0)e35k = y(35) = (0.80)y(0),
e35k = 0.80,
35
k = ln 0.8,
k = ( ln 0.8)/35.

Thus y(t) = y(0)e((ln 0.8)/35)t = y(0)(eln 0.8)t /35 = y(0)(0.8)t/35.

Let T  be the half-life of the substance. Then:

y(0)(0.8)T/35 = y(T ) = (1/2)y(0),
(0.8)
T/35 = 1/2 = 0.5,
ln (0.8)T/35 = ln 0.5,
(T/35) ln 0.8 = ln 0.5,
T/35 = ( ln 0.5)/( ln 0.8), EOS

Note that k < 0 because ln 0.8 < ln 1 = 0.

 6.  Rate Of Change Proportional To A Difference

Suppose a quantity y = y(t) changes with a rate that's proportional to the difference between the quantity and a constant,
ie
dy/dt = k( ya), where a is a constant. This situation is handled as follows. Let v = ya, so that dv/dt = dy/dt =
k( ya) = kv. Thus, v = v(t) = v(0)ekt.

### Example 6.1

Newton's law of cooling states that a hot object brought into a cooler environment will cool at a rate that's proportional to
the excess of its temperature above that of the environment. A glass of hot water is placed on a table in a room whose
temperature is kept at 20
o C. The water cools from 90o C to 60o C in 15 minutes. How many more minutes will it take to
cool to 50
o C?

Note.  Newton's law means that the object's temperature decreases at a rate proportional to the mentioned excess.

# Solution

Let y = y(t) be the temperature of the water t minutes after the water was at 90o C. We have dy/dt = k( y – 20). Let v =
v(t) = y – 20. So dv/dt = dy/dt = k( y – 20) = kv. Thus:

v = v(t) = v(0)ekt,

where v(0) = y(0) – 20 = 90 – 20 = 70.  Now y(15) = 60; consequently:

70e15k = v(0)e15k = v(15) = y(15) – 20 = 60 – 20 = 40,
e15k = 40/70 = 4/7,
15
k = ln (4/7),
k = ( ln (4/7))/15.

The water will be at 50o C at time t where y(t) = 50, and hence:

v(0)ekt = v(t) = y(t) – 20 = 50 – 20 = 30,
ekt = 30/v(0),
kt = ln (30/v(0)), It follows that the water will take approximately 22.71 – 15 = 7.71 more minutes to cool to 50o C.

# EOS

 7. Retarded Motion

When an object travels thru a medium like air or water, it's acted upon by a force F that retards its motion. Suppose that
this force is proportional to the velocity of the object thru the medium (the slower the object, the smaller the force), so
that
F = cv, where c is the constant of proportionality. Now F and v have opposite signs because they point in opposite
directions, so
c < 0. Let v = v(t) be the velocity of the object at time t. Assume that the retarding force F is the only
force acting upon the object. Then, according to Newton's second law of motion, we get
F = ma, where m > 0 and a are
the mass and acceleration respectively of the object. Now
a = dv/dt, thus cv = F = m(dv/dt), which yields: where k = c/m < 0. Thus:

v = v(0)ekt.

Note that we assume that the retarding force is proportional to the velocity and we arrive at the property that the velocity
decreases at a rate proportional to the velocity. The smaller the velocity, the slower it decreases.

As an example, suppose k = – 1/2 and v decreases from 10 m/sec to 6 m/sec over a period of time. Then dv/dt
increases from – 5 m/sec
2 to – 3 m/sec2, but its magnitude decreases from 5 m/sec2 to 3 m/sec2.

### Example 7.1

Suppose an object travelling thru a medium is acted upon only by a retarding force proportional to its velocity. At entry
into the medium it travels at 100 m/sec and 3 seconds later it travels at 20 m/sec. Find:
a.  Its velocity 9 more seconds later.
b.  The total distance it travels thru the medium.

# Solution

a. Let the initial time t = 0 be the time when the object enters the medium and v = v(t) its velocity at time t seconds.
The retarding force is
F = cv, where c is a constant. But F = m(dv/dt), where m is the mass of the object, by
Newton's second law of motion. So
cv = m(dv/dt) and thus dv/dt = kv, where k = c/m. Consequently v = v(t) =
v(0)ekt, where v(0) = 100; hence v = v(t) = 100ekt. We have:

20 = v(3) = 100e3k,
e3k = 20/100 = 1/5,
3
k = ln 1/5 = – ln 5,
k = – ( ln 5)/3.

It follows that:

v = v(t) = 100e(– ln 5)t/3 = 100(eln 5)t/3 = 100(5t/3).

At 9 more seconds later the time is t = 3 + 9 = 12 seconds, and the velocity is:

v(12) = 100(5–12/3) = 0.16 m/sec.

b. Let s = s(t) be the distance travelled thru the medium after t seconds. Then: # EOS

 Problems & Solutions

1.  An amount of \$1,000 invested at an annual interest rate compounded continuously grows to \$1,500 in 6 years.
Determine the time period required for it to double.

Solution

Let i be the annual interest rate in % and A(t) the amount in \$ after t years. So: Thus: The initial amount doubles in about 10.26 years. 2.  A culture of bacteria grows at a rate proportional to the amount present. If there are 1,200 bacteria present initially
and the amount doubles in 1 hour, how many will there be after a further 1.5
-hour period?

Solution

Let y = y(t)  be the number of bacteria t hours after the initial time t = 0 when the number is 1,200. We're given dy/dt =
ky, where k is the constant of proportionality. So y(t) = y(0)ekt = 1,200ekt. At t = 1 we have 2 x 1,200 = 2y(0) = y(1) =
1,200
ek(1) =1,200ek, thus ek = 2, yielding k = ln 2. Consequently:

y(t) = 1,200e(ln 2)t = 1,200(eln 2)t = 1,200(2t).  3.  Radioactive carbon-14 in fossils decays at a rate proportional to its amount present and has a half-life of
approximately 5,700 years. Estimate the age of a fossil in which 80% of its original carbon-14 has decayed.

Solution

Let y = y(t) be the amount of carbon-14 at t years after the initial time t = 0 when it started to decay. As dy/dt = ky we
have
y(t) = y(0)ekt. Now:

(1/2)y(0) = y(5,700) = y(0)e5,700k,
e5,700k = 1/2,
5,700
k = ln (1/2) = – ln 2,
k = (– ln 2)/5,700.

So:

y(t) = y(0)e((–ln 2)/5,700)t = y(0)(eln 2)t/5,700 = y(0)2t/5,700.

Let A be the age of the fossil. Thus, A is the time when 80% of carbon-14 has decayed. As 20% remains, we get:

(0.2)y(0) = y(A) = y(0)2A/5,700,
0.2 = 2
A/5,700,
ln 0.2 = ln 2A/5,700 = – (A/5,700) ln 2,
A/5,700 = ( ln 0.2)/ ln 2,  4.  An object is brought into a freezer that is maintained at a temperature of – 5o C.  The object cools from 45o C to 20o C
in 40 minutes. According to Newton's law of cooling, the object cools at a rate proportional to the excess of its present
temperature over that of the freezer. Determine how many more minutes the object will take to cool to 0
o C.

Solution

Let y = y(t) be the temperature of the object t minutes after the initial time t = 0 when it's brought into the freezer.
According to Newton's law of cooling, we have
dy/dt = k( y – (– 5)) = k( y + 5). Let v = v(t) = y(t) + 5. So dv/dt =
dy/dt = k( y + 5) = kv. Thus, v = v(t) = v(0)ekt. Now, v(0) = y(0) + 5 = 45 + 5 = 50. Consequently, v = v(t) = 50ekt.

Since the temperature is 20o C after 40 minutes we get:

20 = y(40) = v(40) – 5 = 50e40k – 5,
e40k = (20 + 5)/50 = 1/2,
40
k = ln (1/2) = – ln 2,
k = (– ln 2)/40.

Hence:

v = v(t) = 50e((­–ln 2)/40)t = 50(eln 2)t/40 = 50(2t/40).

The object cools to 0o C at time t where:

0 = y(t) = v(t) – 5 = 50(2t/40) – 5,
2
t/40 = 5/50 = 1/10,
ln (2t/40) = ln (1/10) = – ln 10,
– (
t/40) ln 2 = – ln 10, It follows that the object will take approximately 132.88 – 40 = 92.88 more minutes to cool to 0o C. 5.  An object enters a medium and travels thru it. The object's motion is retarded by a force proportional to its velocity.
Suppose that 5 seconds after entry the velocity is 70 m/sec and 7 seconds after entry it is 60 m/sec.

a.  Find the initial velocity of the object.
b.  Find the total distance the object travels thru the medium.

Solution

a.  Let F = F(t) be the force of retardation and v = v(t) the velocity of the object t seconds after entry. We're given F =

cv, where c is a constant of proportionality. By Newton's second law of motion, F = m(dv/dt), where m is the mass
of the object. So:  b.  Let s = s(t) be the distance the object has travelled thru the medium t seconds after entry. Then ds/dt = v(t) =

v(0)ekt. So:  6.  Suppose that the salary of the average worker in 1995 was \$21,000. Suppose that salaries increase by 5% annually.
Assume that the salary increases continuously with a rate proportional to the salary.

a.  Find the year when the salary of the average worker will be \$1,000,000.
b.  Find the year when the salary of the average worker will be \$1,000,000,000.
c.  Imagine a way to keep workers happy with raises but also to prevent working with such cumbersome large numbers.

Solution

a.  Let S = S(t) be the salary of the average worker t years after 1995. We're given dS/dt = kS(t), where k is the
constant of proportionality. So
S(t) = S(0)ekt, where S(0) = \$21,000. Also, S(t + 1) = S(t) + (5%)S(t) = S(t) +
(0.05)
S(t) = (1.05)S(t). That yields:

S(0)ek(t+1) = (1.05)S(0)ekt,

ektek = (1.05)ekt,

ek = 1.05,

k = ln 1.05.

Thus:

S(t) = S(0)e(ln 1.05)t = 21,000(eln 1.05)t = 21,000(1.05)t.

The salary will be \$1,000,000 at t years after 1995 where:

1,000,000 = S(t) = 21,000(1.05)t,
(1.05)
t = 1,000,000/21,000 = 1,000/21,

ln (1.05)t = ln (1,000/21),

t ln 1.05 = ln (1,000/21), Consequently the salary of the average worker will be \$1,000,000 in the year 1995 + 80 = 2075.

b.  Similarly the salary will be \$1,000,000,000 at t years after 1995 where:

1,000,000,000 = S(t) = 21,000(1.05)t, Hence the salary of the average worker will be \$1,000,000,000 in the year 1995 + 221 = 2216.

c.  Whenever the salary of the average worker reaches \$21,000 x 10 = \$210,000, create a new money, with 1 unit of the
new money being equivalent to 10 units of the old. Keep the name of the unit of the old money. For example, 1 new
dollar will be equivalent to 10 old dollars. Take the old money out of circulation.