216
CHAPTER 26
26.10
26.11
26.12
26.13
26.14
26.15
26.16
26.17
26.18
26.19
By Problem 26.8, y = y 0 (l+ r/100n)*"
if the money is compounded n times per year. If we
let n approach infinity, we get
(Here we have used Problem 24.74.) Thus, the money grows exponentially, with growth
*,(«"
IOO
)* = */0
"*.
constant K = 0.0lr.
If an amount of money earning 8 percent per year is compounded quarterly, what is the equivalent yearly rate of
return?
By Problem 26.8, the amount present after 1 year will be
Thus, the
equivalent yearly rate is 8.24 percent.
If an amount of money earning 8 percent per year is compounded 10 times per year, what is the equivalent yearly
rate of return?
By Problem 26.8, the amount after 1 year will be .y 0 (H-Tggg)
10 = 3> 0 (1.008)
IO = 1.0829>>. Thus, the
equivalent yearly rate is 8.29 percent.
If an amount of money receiving interest of 8 percent per year is compounded continuously, what is the equivalent
yearly rate of return?
By Problem 26.9, the amount after 1 year will be y a e° °
8
, which, by a table of values of e*, is approximately
l.OS33y 0 . Hence, the equivalent yearly rate is about 8.33 percent.
A sum of money, compounded continuously, is multiplied by 5 in 8 years. If it amounts to $10,000 after 24 years,
what was the initial sum of money?
Hence, the initial quantity y 0 was
80 dollars.
If a quantity of money, earning interest compounded continuously, is worth 55 times the initial amount after 100
years, what was the yearly rate of interest?
By Problem 26.9,
is the percentage rate of interest, y a is the initial amount, and k is the
number of years. Then,
and, by a table of logarithms,
Hence,
Thus, the rate is approximately 4 percent per year.
In
Assume that a quantity y decays exponentially, with a decay constant K. The half-life Tis defined to be the time
interval after which half of the original quantity remains. Find the relationship between K and T.
y = y 0 e
K> . By definition,
The half-life of radium is 1690 years. If 10 percent of an original quantity of radium remains, how long ago was
the radium created?
Let y be the number of grams of radium t years after the radium was created. Then y = y 0 e
Kl , where
1690K = -In 2, by Problem 26.15. If at the present time
then
Thus, the radium was
created about 5614 years ago.
Hence, -(In2/1690)f = -In 10, t= 1690 In 10/ln 2 = 5614.477.
If radioactive carbon-14 has a half-life of 5750 years, what will remain of 1 gram after 3000 years?
We know that
Since
and
Thus, about 0.7 gram will remain.
(from a table for e *).
The amount of carbon is
If 20 percent of a radioactive element disappears in 1 year, compute its half-life.
Let y n be the original amount, and let T be the half-life. Then 0.8y 0 remains when t = 1. Thus,
0.&y 0 = y 0 e, 0.8 = e
K , K = In 0.8 « -0.2231 (from a table of logarithms). But KT = -In 2= -0.6931.
So -0.2231 T= -0.6931, 7=3.1067 years.
Fruit flies are being bred in an enclosure that can hold a maximum of 640 flies. If the flies multiply exponentially,
with a growth constant K = 0.05 and where time is measured in days, how long will it take an initial population
of 20 to fill the enclosure?
So
By the formula of Problem 26.9, y = y 0 e
0 '
airk , where r is the rate of interest and k is the number of
years. Hence, 5y 0 = y 0 e°'"*', 5 = e°
08
'. After 24 years, 10,000 = y 0 e°
024r = ^ 0 (e°08r )
3 = >- 0 (5)
3 = 125>- 0 .
CHAPTER 26
26.10
26.11
26.12
26.13
26.14
26.15
26.16
26.17
26.18
26.19
By Problem 26.8, y = y 0 (l+ r/100n)*"
if the money is compounded n times per year. If we
let n approach infinity, we get
(Here we have used Problem 24.74.) Thus, the money grows exponentially, with growth
*,(«"
IOO
)* = */0
"*.
constant K = 0.0lr.
If an amount of money earning 8 percent per year is compounded quarterly, what is the equivalent yearly rate of
return?
By Problem 26.8, the amount present after 1 year will be
Thus, the
equivalent yearly rate is 8.24 percent.
If an amount of money earning 8 percent per year is compounded 10 times per year, what is the equivalent yearly
rate of return?
By Problem 26.8, the amount after 1 year will be .y 0 (H-Tggg)
10 = 3> 0 (1.008)
IO = 1.0829>>. Thus, the
equivalent yearly rate is 8.29 percent.
If an amount of money receiving interest of 8 percent per year is compounded continuously, what is the equivalent
yearly rate of return?
By Problem 26.9, the amount after 1 year will be y a e° °
8
, which, by a table of values of e*, is approximately
l.OS33y 0 . Hence, the equivalent yearly rate is about 8.33 percent.
A sum of money, compounded continuously, is multiplied by 5 in 8 years. If it amounts to $10,000 after 24 years,
what was the initial sum of money?
Hence, the initial quantity y 0 was
80 dollars.
If a quantity of money, earning interest compounded continuously, is worth 55 times the initial amount after 100
years, what was the yearly rate of interest?
By Problem 26.9,
is the percentage rate of interest, y a is the initial amount, and k is the
number of years. Then,
and, by a table of logarithms,
Hence,
Thus, the rate is approximately 4 percent per year.
In
Assume that a quantity y decays exponentially, with a decay constant K. The half-life Tis defined to be the time
interval after which half of the original quantity remains. Find the relationship between K and T.
y = y 0 e
K> . By definition,
The half-life of radium is 1690 years. If 10 percent of an original quantity of radium remains, how long ago was
the radium created?
Let y be the number of grams of radium t years after the radium was created. Then y = y 0 e
Kl , where
1690K = -In 2, by Problem 26.15. If at the present time
then
Thus, the radium was
created about 5614 years ago.
Hence, -(In2/1690)f = -In 10, t= 1690 In 10/ln 2 = 5614.477.
If radioactive carbon-14 has a half-life of 5750 years, what will remain of 1 gram after 3000 years?
We know that
Since
and
Thus, about 0.7 gram will remain.
(from a table for e *).
The amount of carbon is
If 20 percent of a radioactive element disappears in 1 year, compute its half-life.
Let y n be the original amount, and let T be the half-life. Then 0.8y 0 remains when t = 1. Thus,
0.&y 0 = y 0 e, 0.8 = e
K , K = In 0.8 « -0.2231 (from a table of logarithms). But KT = -In 2= -0.6931.
So -0.2231 T= -0.6931, 7=3.1067 years.
Fruit flies are being bred in an enclosure that can hold a maximum of 640 flies. If the flies multiply exponentially,
with a growth constant K = 0.05 and where time is measured in days, how long will it take an initial population
of 20 to fill the enclosure?
So
By the formula of Problem 26.9, y = y 0 e
0 '
airk , where r is the rate of interest and k is the number of
years. Hence, 5y 0 = y 0 e°'"*', 5 = e°
08
'. After 24 years, 10,000 = y 0 e°
024r = ^ 0 (e°08r )
3 = >- 0 (5)
3 = 125>- 0 .