Source:Science.
a.Standard concert pitch for an A is 440 cycles per second.
Find the associated value of p.
b.An A one octave higher than standard concert pitch is 880 cycles per second. Find the associated value of p.
p569112 log2 1f/4402.
H1
pH5 2log3H14, E0
R1E2 52 3log aE
E0b,
I0.
YOUR TURN ANSWERS
1. 2.
3. 4. 3.561
5. 2 6.
7. 1.0253x
1ln 22/ln13/22 <1.7095.
2 loga x 23 loga y
24 log511/252 5 22
This question, which will be answered in Example 7, is one of many situations that occur in biology, economics, and the social sciences, in which a quantity changes at a rate propor- tional to the amount of the quantity present.
In cases such as continuous compounding described above, the amount present at time tis a function of t, called theexponential growth and decay function.(The derivation of this equation is presented in a later section on Differential Equations.)
Applications: Growth and Decay;
Mathematics of Finance
What interest rate will cause $5000 to grow to $7250 in 6 years if money is compounded continuously?
2.6
APPLY IT
Exponential Growth and Decay Function
Let be the amount or number of some quantity present at time The quantity is said to grow or decay exponentially if for some constant k, the amount present at time t is given by
y5y0ekt.
t50.
y0
If then kis called the growth constant; if then kis called the decay con- stant. A common example is the growth of bacteria in a culture. The more bacteria present, the faster the population increases.
Yeast Production
Yeast in a sugar solution is growing at a rate such that 1 g becomes 1.5 g after 20 hours.
Find the growth function, assuming exponential growth.
SOLUTION The values of and kin the exponential growth function must be found. Since is the amount present at time To find k, substitute
and into the equation.
Now take natural logarithms on both sides and use the power rule for logarithms and the fact that
Take of both sides.
Divide both sides by .
k0.02 (to the nearest hundredth)
The exponential growth function is ye0.02t, where yis the number of grams of yeast pre-
sent after thours. TRY YOUR TURN 1
The decline of a population or decay of a substance may also be described by the expo- nential growth function. In this case the decay constant kis negative, since an increase in time leads to a decrease in the quantity present. Radioactive substances provide a good example of exponential decay. By definition, the half-lifeof a radioactive substance is the time it takes for exactly half of the initial quantity to decay.
ln 1.5 20 20 5k
ln ex5x ln 1.5520k
ln 1.55ln e20k ln 1.55e20k ln e51.
1.551ek1202 y5y0ekt y051
t520, y0 t50,y051. y51.5,
y5y0ekt y0
k,0, k.0,
EXAMPLE 1
YOUR TURN 1 Find the
growth function if 5 g grows expo- nentially to 18 g after 16 hours.
2.6 Applications: Growth and Decay; Mathematics of Finance 103 Carbon Dating
Carbon-14 is a radioactive form of carbon that is found in all living plants and animals.
After a plant or animal dies, the carbon-14 disintegrates. Scientists determine the age of the remains by comparing its carbon-14 with the amount found in living plants and animals.
The amount of carbon-14 present after tyears is given by the exponential equation
with
(a) Find the half-life of carbon-14.
SOLUTION Let and
Divide by .
Take of both sides.
Multiply by .
The half-life is 5600 years.
(b)Charcoal from an ancient fire pit on Java had the amount of carbon-14 found in a living sample of wood of the same size. Estimate the age of the charcoal.
SOLUTION Let and
The charcoal is about 11,200 years old. TRY YOUR TURN 2
t511,200 25600
ln 2 ln 1 45t ln 1
45 2ln 2 5600t ln 1
45ln e231ln 22/56004t 1
45e231ln 22/56004t 1
4A05A0e231ln 22/56004t k5 23 1ln 22/56004.
A1t2 5 11/42A0
1/4
56005t
ln 150 2 5600
ln 212ln 22 5t
ln xy5ln x2ln y 25600
ln 21ln 12ln 22 5t
25600 2 5600 ln 2
ln 2 ln 1 25t
ln ex5x ln 1
252 ln 2 5600t ln 1 ln
25ln e231ln 22/56004t
A0 1
25e231ln 22/56004t 1
2 A05A0e231ln 22/56004t
k5 23 1ln 22/56004.
A1t2 5 11/22A0
k5 23 1ln 22/56004.
A1t2 5A0ekt,
EXAMPLE 2
YOUR TURN 2 Estimate the age of a sample with 1/10 the amount of carbon-14 as a live sample.
By following the steps in Example 2, we get the general equation giving the half-life T in terms of the decay constant kas
For example, the decay constant for potassium-40, where tis in billions of years, is approx- imately 0.5545 so its half-life is
We can rewrite the growth and decay function as
where This is sometimes a helpful way to look at an exponential growth or decay function.
Radioactive Decay
Rewrite the function for radioactive decay of carbon-14 in the form SOLUTION From the previous example, we have
This last expression shows clearly that every time tincreases by 5600 years, the amount of carbon-14 decreases by a factor of
Effective Rate We could use a calculator to see that $1 at 8% interest (per year) compounded semiannually is or $1.0816. The actual increase of
$0.0816 is 8.16% rather than the 8% that would be earned with interest compounded annu- ally. To distinguish between these two amounts, 8% (the annual interest rate) is called the nominalor statedinterest rate, and 8.16% is called the effectiveinterest rate. We will con- tinue to use rto designate the stated rate and we will use rEfor the effective rate.
111.042251.0816 1/2.
5A022t/56005A0 12212t/56005A0a1 2bt/5600. 5A01eln 222t/5600
A1t2 5A0ekt5A0e231ln 22/56004t
A1t2 5A0af1t2. a5ek.
y5y0ekt5y01ek2t5y0at,
<1.25 billion years.
T5 2 ln 2 120.55452 T5 2ln 2
k .
EXAMPLE 3
Effective Rate for Compound Interest
If ris the annual stated rate of interest and mis the number of compounding periods per year, the effective rate of interest is
. rE5 a11 r
mbm21 Effective rate is sometimes called annual yield.
With continuous compounding, $1 at 8% for 1 year becomes
The increase is 8.33% rather than 8%, so a stated interest rate of 8% produces an effective rate of 8.33%.
e0.08<1.0833.
112e110.0825
Effective Rate for Continuous Compounding
If interest is compounded continuously at an annual stated rate of r, the effective rate of interest is
. rE5er21
2.6 Applications: Growth and Decay; Mathematics of Finance 105 Effective Rate
Find the effective rate corresponding to each stated rate.
(a) 6% compounded quarterly
SOLUTION Using the formula, we get
The effective rate is 6.14%.
(b)6% compounded continuously
SOLUTION The formula for continuous compounding gives
so the effective rate is 6.18%. TRY YOUR TURN 3
The formula for interest compounded m times a year, has five variables: A, P, r, m, and t. If the values of any four are known, then the value of the fifth can be found.
Interest
Meghan Moreau has received a bonus of $25,000. She invests it in an account earning 7.2%
compounded quarterly. Find how long it will take for her $25,000 investment to grow to
$40,000.
SOLUTION Here P $25,000, r 0.072, and m 4. We also know the amount she wishes to end up with, A$40,000. Substitute these values into the compound interest formula and solve for time, t.
Divide both sides by 25,000. Take of both sides.
Divide both sides by 4 1.018.
Note that the interest is calculated quarterly and is added only at the endof each quarter.
Therefore, we need to round up to the nearest quarter. She will have $40,000 in 6.75
years. TRY YOUR TURN 4
When calculating the time it takes for an investment to grow, take into account that interest is added only at the endof each compounding period. In Example 5, interest is added quarterly. At the end of the second quarter of the sixth year (t 6.5), she will have only $39,754.13, but at the end of the third quarter of that year (t 6.75), she will have $40,469.70.
If A, the amount of money we wish to end up with, is given as well as r, m, and t, then Pcan be found using the formula for compounded interest. Here Pis the amount that should be deposited today to produce Adollars in tyears. The amount Pis called the present valueof Adollars.
t5 ln 1.6 ln
4 ln 1.018 <6.586 ln1.654t. ln1.018 ln1.65ln11.01824t ln
1.651.0184t
40,000525,00011.01824t 40,000525,000 a11 0.072
4 b4t
A5P111r/m2tm,
e0.0621<0.0618,
a110.06
4 b4215 11.0152421<0.0614.
YOUR TURN 3 Find the effec- tive rate corresponding to each stated rate. (a)4.25% compounded monthly (b)3.75% compounded continuously.
EXAMPLE 4
EXAMPLE 5
YOUR TURN 4 Find the time needed for $30,000 to grow to
$50,000 when invested in an account that pays 3.15% com- pounded quarterly.
CAUTION
Present Value
Tom Shaffer has a balloon payment of $100,000 due in 3 years. What is the present value of that amount if the money earns interest at 4% annually?
SOLUTION Here P in the compound interest formula is unknown, with
and Substitute the known values into the formula to get Solve forP, using a calculator to find
The present value of $100,000 in 3 years at 4% per year is $88,889.64.
In general, to find the present value for an interest rate rcompounded mtimes per year for tyears, solve the equation
for the variable P. To find the present value for an interest rate rcompounded continuously for tyears, solve the equation
for the variable P.
Continuous Compound Interest
Find the interest rate that will cause $5000 to grow to $7250 in 6 years if the money is com- pounded continuously.
SOLUTION Use the formula for continuous compounding, with and Solve first for then forr.
Divide by . Take of both sides.
The required interest rate is 6.19%. TRY YOUR TURN 5
Limited Growth Functions The exponential growth functions discussed so far all continued to grow without bound. More realistically, many populations grow exponen- tially for a while, but then the growth is slowed by some external constraint that eventually limits the growth. For example, an animal population may grow to the point where its habi- tat can no longer support the population and the growth rate begins to dwindle until a stable population size is reached. Models that reflect this pattern are called limited growth func- tions.The next example discusses a function of this type that occurs in industry.
Employee Turnover
Assembly-line operations tend to have a high turnover of employees, forcing companies to spend much time and effort in training new workers. It has been found that a worker new to a task on the line will produce items according to the function defined by
P1x2 525225e20.3x, r<0.0619
r5ln 1.45 6
ln ex5x ln 1.4556r
ln 1.455ln e6r ln 1.455e6r 5000 725055000e6r
A5Pert ert, t56.
P55000, A5Pert, A57250,
A5Pert A5Pa11 r
mbtm P5 100,000
11.0423 <88,889.64 11.0423. 100,0005tP511.043, 23. m51.
r50.04,
A5100,000,
EXAMPLE 6
YOUR TURN 5 Find the inter- est rate that will cause $3200 to grow to $4500 in 7 years if the money is compounded continuously.
EXAMPLE 7
EXAMPLE 8 APPLY IT
2.6 Applications: Growth and Decay; Mathematics of Finance 107 where items are produced by the worker on day x.
(a) What happens to the number of items a worker can produce as xgets larger and larger?
SOLUTION As xgets larger, becomes closer to 0, so approaches 25. This represents the limit on the number of items a worker can produce in a day. Note that this limit represents a horizontal asymptote on the graph of P, shown in Figure 58.
(b)How many days will it take for a new worker to produce at least 20 items in a day?
SOLUTION Let and solve for x.
Now take natural logarithms of both sides and use properties of logarithms.
This means that 5 days are not quite enough; on the fifth day, a new worker produces items. It takes 6 days, and on the sixth day, a new
worker produces items.
Graphs such as the one in Figure 58 are called learning curves. According to such a graph, a new worker tends to learn quickly at first; then learning tapers off and approaches some upper limit. This is characteristic of the learning of certain types of skills involving the repetitive performance of the same task.
P162 525225e20.3152 <20.9 P152 525225e20.3152<19.4
x5 ln 0.2 20.3 <5.4
ln eu5u ln 0.25 20.3x
ln 0.25ln e20.3x 0.25e20.3x 255 225e20.3x
20525225e20.3x P1x2 525225e20.3x P1x2 520
P1x2 e20.3x
P1x2
P x
x
P x e x
FIGURE 58
2.6 EXERCISES
1. What is the difference between stated interest rate and effec- tive interest rate?
2. In the exponential growth or decay function what does represent? What does krepresent?
3. In the exponential growth or decay function, explain the cir- cumstances that cause kto be positive or negative.
4. What is meant by the half-life of a quantity?
5. Show that if a radioactive substance has a half-life of T, then the corresponding constant kin the exponential decay function is given by
6. Show that if a radioactive substance has a half-life of T, then the corresponding exponential decay function can be written as y5y011/221t/T2.
k5 21ln 22/T.
y0
y5y0ekt,
APPLICATIONS
B u s i n e s s a n d E c o n o m i c s
Effective Rate Find the effective rate corresponding to each nominal rate of interest.
7.4% compounded quarterly 8.6% compounded monthly 9.8% compounded continuously 10.5% compounded continuously
Present Value Find the present value of each amount.
11.$10,000 if interest is 6% compounded quarterly for 8 years 12.$45,678.93 if interest is 7.2% compounded monthly for
11 months