CHAPTER 26
218
26.30
26.31
26.32
26.33
26.34
26.35
26.36
26.37
A radioactive substance decreases from 8 grams to 7 grams in 1 hour. Find its half-life.
At noon, t = 6 and y =
grams.
y = 2OOe
Kl , where f = 0 at 6a.m. Then, 500 = 200e
3
*,
y = 8e
Kl .
Then
7 = 8e*,
e" = 0.875,
K = In 0.875 =-0.1335.
The half-life
T=-ln2/K~
0.6931/0.1335 = 5.1918 hours.
A doomsday equation is an equation of the form
with
Solve this equation and show
that
for some
Setting t = 0, we find that
Now, P°-°
l = ~lOO/(Kt+C). As t^-C/K from below, />-»+°°.
Cobalt-60, with a half-life of 5.3 years, is extensively used in medical radiology. How long does it take for 90
percent of a given quantity to decay?
y=y 0 e
Kl . Since T = 5.3 and ,KT=-ln2, K = -In2/5.3=-0.1308. When 90 percent of y 0 has
decayed, y = 0.1>> 0 = y 0 e , 0.1 = e,
years.
In a chemical reaction, a compound decomposes exponentially. If it is found by experiment that 8 grams
diminishes to 4 grams in 2 hours, when will 1 gram be left?
The half-life is 2 hours.
n = 3. Thus 1 gram remains after three half-lifes, or 6 hours.
A tank initially contains 400 gallons of brine in which 100 pounds of salt are dissolved. Pure water is running into
the tank at the rate of 20 gallons per minute, and the mixture (which is kept uniform by stirring) is drained off at
the same rate. How many pounds of salt remain in the tank after 30 minutes?
Let y be the number of pounds of salt in the mixture at time t. Since the concentration of salt at any given
time is y/400 pounds per gallon, and 20 gallons flow out per minute, the rate at which y is diminishing is
20-y/400 = 0.05y pounds per minute. Hence, D t y = — 0.05y, and, thus, y is decaying exponentially with a
decay constant of-0.05. Hence, y = 100e~° °
5
'. So, after 30 minutes, y = WOe'
15 = 100(0.2231) = 22.31
pounds.
Solve Problem 26.34 with the modification that instead of pure water, brine containing ^ pound per gallon is run
into the tank at 20 gallons per minute, the mixture being drained off at the same rate.
As before, the tank is losing salt at the rate of O.OSy pounds per minute. However, it is also gaining
salt at the rate of
pounds per minute.
Hence,
ways > 40, since y(0) = 100 and y = 40 is impossible, fin |2 - 0.05(40)1 = In 0 is undefined.! Hence,
0.05>» 0.05(40) = 2,
and
|2-0.05y| = 0.05y - 2. Thus,
(O.OSy - 2)/3 = e °
5
'.
When
t = 30,
(0.05;y-2)/3 = e~
15 ==0.2231, O.OSy = 2.6693, y «53.386 pounds.
-201n|2-0.05y| = r+C. When t = 0, -20 In |2- 5| = C, C=-201n3. Hence, -20 In |2 -0.05y| =
Note that y is alA country has 5 billion dollars of paper money in circulation. Each day 30 million dollars is brought into the
banks for deposit and the same amount is paid out. The government decides to issue new paper money;
whenever the old money comes into the banks, it is destroyed and replaced by the new money. How long will it
take for the paper money in circulation to become 90 percent new?
Let y be the number of millions of dollars in old money. Each day, (y/5000)- 30 = 0.006}' millions of
dollars of old money is turned in at the banks. Hence, D,y = —0.006y, and, thus, y is decreasing
exponentially, with a decay constant of —0.006. Hence, y = 5000e~°'
006
'. When 90 percent of the money is
new, y = 500, 500 = 5000e~° °
06
', 0.1 = e"
0006 ', -0.006f = In & = -In 10, 0.006f = In 10 = 2.3026, r»
383.77 days.
The number of bacteria in a culture doubles every hour. How long does it take for a thousand bacteria to
produce a billion?
But,
-In 10 =-2.3026. So, f« -2.3026/(-0.1308) = 17.604
r-20 In 3, In |2 - 0.05y| = -O.OSr + In 3, In
218
26.30
26.31
26.32
26.33
26.34
26.35
26.36
26.37
A radioactive substance decreases from 8 grams to 7 grams in 1 hour. Find its half-life.
At noon, t = 6 and y =
grams.
y = 2OOe
Kl , where f = 0 at 6a.m. Then, 500 = 200e
3
*,
y = 8e
Kl .
Then
7 = 8e*,
e" = 0.875,
K = In 0.875 =-0.1335.
The half-life
T=-ln2/K~
0.6931/0.1335 = 5.1918 hours.
A doomsday equation is an equation of the form
with
Solve this equation and show
that
for some
Setting t = 0, we find that
Now, P°-°
l = ~lOO/(Kt+C). As t^-C/K from below, />-»+°°.
Cobalt-60, with a half-life of 5.3 years, is extensively used in medical radiology. How long does it take for 90
percent of a given quantity to decay?
y=y 0 e
Kl . Since T = 5.3 and ,KT=-ln2, K = -In2/5.3=-0.1308. When 90 percent of y 0 has
decayed, y = 0.1>> 0 = y 0 e , 0.1 = e,
years.
In a chemical reaction, a compound decomposes exponentially. If it is found by experiment that 8 grams
diminishes to 4 grams in 2 hours, when will 1 gram be left?
The half-life is 2 hours.
n = 3. Thus 1 gram remains after three half-lifes, or 6 hours.
A tank initially contains 400 gallons of brine in which 100 pounds of salt are dissolved. Pure water is running into
the tank at the rate of 20 gallons per minute, and the mixture (which is kept uniform by stirring) is drained off at
the same rate. How many pounds of salt remain in the tank after 30 minutes?
Let y be the number of pounds of salt in the mixture at time t. Since the concentration of salt at any given
time is y/400 pounds per gallon, and 20 gallons flow out per minute, the rate at which y is diminishing is
20-y/400 = 0.05y pounds per minute. Hence, D t y = — 0.05y, and, thus, y is decaying exponentially with a
decay constant of-0.05. Hence, y = 100e~° °
5
'. So, after 30 minutes, y = WOe'
15 = 100(0.2231) = 22.31
pounds.
Solve Problem 26.34 with the modification that instead of pure water, brine containing ^ pound per gallon is run
into the tank at 20 gallons per minute, the mixture being drained off at the same rate.
As before, the tank is losing salt at the rate of O.OSy pounds per minute. However, it is also gaining
salt at the rate of
pounds per minute.
Hence,
ways > 40, since y(0) = 100 and y = 40 is impossible, fin |2 - 0.05(40)1 = In 0 is undefined.! Hence,
0.05>» 0.05(40) = 2,
and
|2-0.05y| = 0.05y - 2. Thus,
(O.OSy - 2)/3 = e °
5
'.
When
t = 30,
(0.05;y-2)/3 = e~
15 ==0.2231, O.OSy = 2.6693, y «53.386 pounds.
-201n|2-0.05y| = r+C. When t = 0, -20 In |2- 5| = C, C=-201n3. Hence, -20 In |2 -0.05y| =
Note that y is alA country has 5 billion dollars of paper money in circulation. Each day 30 million dollars is brought into the
banks for deposit and the same amount is paid out. The government decides to issue new paper money;
whenever the old money comes into the banks, it is destroyed and replaced by the new money. How long will it
take for the paper money in circulation to become 90 percent new?
Let y be the number of millions of dollars in old money. Each day, (y/5000)- 30 = 0.006}' millions of
dollars of old money is turned in at the banks. Hence, D,y = —0.006y, and, thus, y is decreasing
exponentially, with a decay constant of —0.006. Hence, y = 5000e~°'
006
'. When 90 percent of the money is
new, y = 500, 500 = 5000e~° °
06
', 0.1 = e"
0006 ', -0.006f = In & = -In 10, 0.006f = In 10 = 2.3026, r»
383.77 days.
The number of bacteria in a culture doubles every hour. How long does it take for a thousand bacteria to
produce a billion?
But,
-In 10 =-2.3026. So, f« -2.3026/(-0.1308) = 17.604
r-20 In 3, In |2 - 0.05y| = -O.OSr + In 3, In