Calculus Of One Real Variable –
By Pheng Kim Ving 
7.5 
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 reproduce 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
Suppose the
interest rate is compounded annually: once per year, at the end of the year. The
two 1year periods are the
tintervals [0, 1] and [1, 2], as shown in Fig. 1.1. The unit of the taxis 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 1year periods. 
Fig. 1.2
Four halfyear periods. 
Suppose the
interest rate is compounded semiannually instead: once per each halfyear
period, at the end of the period;
thus twice per year. There are four periods. They're the tintervals [0, 1/2], [1/2, 1], [1,
3/2], and [3/2, 2], as shown in
Fig. 1.2. The semiannual 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 semiannually 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 rewrite 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 peryear) interest rate and t is measured in year. The
perunit 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
1year interval is divided into n equal
subintervals, each of which is of
length 1/n, and the compounding occurs at the
end of each subinterval. Refer to Fig. 1.2 for the case of n = 2. There are
2n subintervals in the 2year period.
The average rate of change of the amount over any subinterval is proportional
to
the amount at the start of that subinterval, with the same constant of
proportionality of i/100, ie:
where t = t_{1} and h = 1/n.
see Section
7.2 Eq. [8.1]. 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) = (e^{i}^{/100})^{t}A(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.

Size of growth changes slightly as number of 
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} = 15000e^{t}^{/20}.
After 10
years the amount is:
EOS
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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 = t^{2} grows at a rate not proportional to the quantity: dy/dt = 2t, and there's no constant k such
that 2t = kt^{2}, because if 2t = kt^{2}, then k = 2/t, which is a nonconstant.
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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 = Ce^{kt} 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)e^{kt}.
As g(0) is a number, g(t) is of the form g(t) = Ce^{kt} 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) = Ce^{kt}. Now we see that the function y = y(t) =
Ce^{kt}, 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) = Ce^{0} = 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)e^{kt}. [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) = Ce^{kt} we have C = y(0). 
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.
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.
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4.
The Constant Of Proportionality 
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)e^{kt}, where P(0) = 30,000,000 =
3 x 10^{7}. As the population grows by 2% annually we
have:
P(0)e^{k}^{(}^{t}^{+1)} = P(t + 1) = P(t) + (2/100)P(t) = (1.02)P(t) = (1.02)P(0)e^{kt},
P(0)e^{kt}e^{k} = (1.02)P(0)e^{kt},
e^{k} = 1.02,
k = ln 1.02.
Consequently:
P(t) = (3 x 10^{7})e^{(}^{ln}^{
1.02)}^{t}.
Now 150,000
km^{2} = 150,000 x (1,000 m)^{2} = 150,000,000,000 m^{2} = 1.5 x
10^{11} m^{2}. 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 10^{11}, from which we get:
e^{(}^{ln}^{ 1..02)}^{t} = (1.5 x 10^{11})/(3 x
10^{7}) = 5 x 10^{3},
( ln 1.02)t = ln (5 x 10^{3}),
It follows
that there'll be one person for every square meter of land in the year 1985 +
431 = 2416.
EOS
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 interestrate 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 interestrate 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 interestrate case.
For continuous growth in interestrate 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.
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5.
HalfLife 
Fig. 5.1 Same amount of time, ln 2 units
of time, is required 
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.
y(t + T ) = y(0)e^{k}^{(}^{t}^{+}^{T}^{)} = y(0)e^{kt}^{+}^{kT} = y(0)e^{kt}e^{kT} = y(0)e^{kt}(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 halflife
of the substance.
Observe
that the halflife of a substance isn't half of the total life span of the
substance. A substance that now weighs
100 kg and whose halflife 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 halflife 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.
A
radioactive substance decays at a rate proportional to the quantity present.
Suppose 20% of such a substance decays in
35 years. Find the halflife 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)e^{kt}.
After 35 years, the quantity left is 80% of the initial quantity. So:
y(0)e^{35}^{k} = y(35) = (0.80)y(0),
e^{35}^{k} = 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)(e^{ln}^{ 0.8})^{t }^{/35} = y(0)(0.8)^{t}^{/35}.
Let T
be the halflife 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.
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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( y – a), where a is a constant. This situation is handled as follows. Let v = y – a, so that dv/dt = dy/dt =
k( y – a) = kv. Thus, v = v(t) = v(0)e^{kt}.
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 90^{o} C to 60^{o} 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.
Let y = y(t) be the temperature of the water t minutes after the water was at 90^{o} 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)e^{kt},
where v(0) = y(0) – 20 = 90 – 20 = 70. Now y(15) = 60; consequently:
70e^{15}^{k} = v(0)e^{15}^{k} = v(15) = y(15) – 20 = 60 – 20 = 40,
e^{15}^{k} = 40/70 = 4/7,
15k = ln (4/7),
k = ( ln (4/7))/15.
The water
will be at 50^{o} C at time t where y(t) = 50, and hence:
v(0)e^{kt} = v(t) = y(t) – 20 = 50 – 20 = 30,
e^{kt} = 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 50^{o} C.
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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)e^{kt}.
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/sec^{2}, but its
magnitude decreases from 5 m/sec^{2} to 3 m/sec^{2}.
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.
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)e^{kt}, where v(0) = 100; hence v = v(t) = 100e^{kt}. We have:
20 = v(3) = 100e^{3}^{k},
e^{3}^{k} = 20/100 = 1/5,
3k = ln 1/5 = – ln 5,
k = – ( ln 5)/3.
It follows that:
v = v(t) = 100e^{(– }^{ln}^{ 5)}^{t}^{/3} = 100(e^{ln}^{ 5})^{–}^{t}^{/3} = 100(5^{–}^{t}^{/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:
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 1.5hour 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)e^{kt} = 1,200e^{kt}. At t = 1 we have 2 x 1,200 = 2y(0) = y(1) =
1,200e^{k}^{(1)} =1,200e^{k}, thus e^{k} = 2, yielding k = ln 2. Consequently:
y(t) = 1,200e^{(}^{ln}^{ 2)}^{t} = 1,200(e^{ln}^{ 2})^{t} = 1,200(2^{t}).
3.
Radioactive carbon14 in fossils decays at a rate proportional to its amount
present and has a halflife of
approximately 5,700 years. Estimate
the age of a fossil in which 80% of its original carbon14 has decayed.
Solution
Let y = y(t) be the amount of carbon14 at t years
after the initial time t = 0 when it started to decay. As dy/dt = ky we
have y(t) = y(0)e^{kt}. Now:
(1/2)y(0) = y(5,700) = y(0)e^{5,700}^{k},
e^{5,700}^{k} = 1/2,
5,700k = ln (1/2) = – ln 2,
k = (– ln 2)/5,700.
So:
y(t) = y(0)e^{((–}^{ln}^{ 2)/5,700)}^{t} = y(0)(e^{ln}^{ 2})^{–}^{t}^{/5,700} = y(0)2^{–}^{t}^{/5,700}.
Let A be the age
of the fossil. Thus, A is the time when 80% of carbon14
has decayed. As 20% remains, we get:
(0.2)y(0) = y(A) = y(0)2^{–}^{A}^{/5,700},
0.2 = 2^{–}^{A}^{/5,700},
ln 0.2 = ln 2^{–}^{A}^{/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 – 5^{o} C. The
object cools from 45^{o} C to 20^{o} 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)e^{kt}. Now, v(0) = y(0) + 5 = 45 + 5 = 50. Consequently, v = v(t) = 50e^{kt}.
Since the
temperature is 20^{o} C after 40 minutes we get:
20 = y(40) = v(40) – 5 = 50e^{40}^{k} – 5,
e^{40}^{k} = (20 + 5)/50 = 1/2,
40k = ln (1/2) = – ln 2,
k = (– ln 2)/40.
Hence:
v = v(t) = 50e^{((–}^{ln}^{ 2)/40)}^{t} = 50(e^{ln}^{ 2})^{– }^{t}^{/40} = 50(2^{–}^{t}^{/40}).
The object
cools to 0^{o} C at time t where:
0 = y(t) = v(t) – 5 = 50(2^{–}^{t}^{/40}) – 5,
2^{–}^{t}^{/40} = 5/50
= 1/10,
ln (2^{–}^{t}^{/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 0^{o} 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)e^{kt}. 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)e^{kt}, 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)e^{k}^{(}^{t}^{+1)} = (1.05)S(0)e^{kt},
e^{kt}e^{k} = (1.05)e^{kt},
e^{k} = 1.05,
k = ln 1.05.
Thus:
S(t) = S(0)e^{(}^{ln}^{ 1.05)}^{t} = 21,000(e^{ln}^{ 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.
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