Question 1

The radioactive isotope Bismuth-210 has a half-life of 5 days. How many days does it take for 87.5% of a given amount to decay?
A)15 days
B)8 days
C)10 days
D)13 days
E)11 days
F)9 days
G)12 days
H)14 days

Accepted Answer

After 15 days, or three half-lives, 87.5% of the Bismuth-210 would have decayed.

Question 2

When a child was born, her grandparents deposited $1000 in a saving account at 5% interest compounded continuously. The amount of money after t years is:
A) $1000 \left( 2 ^ { t } \right)$ 
B) $1000 \left( e ^ { t } \right)$ 
C) $500 \left( e ^ { t } \right)$ 
D) $500 \left( 3 ^ { t } \right)$ 
E) $1000 \left( e ^ { - 0.05 t } \right)$ 
F) $1000 \left( e ^ { 0.05 t } \right)$ 
G) $500 \left( e ^ { 0.1 t } \right)$ 
H) $1000 \left( e ^ { 2 t } \right)$

Accepted Answer

The formula for continuous compounding is $A = P e^{rt}$, where $A$ is the amount of money after time $t$, $P$ is the principal amount (initial investment), $r$ is the annual interest rate (in decimal), and $e$ is the base of the natural logarithm. Given a principal of $1000 and an interest rate of 5% (or 0.05 in decimal form), the correct formula is $1000 \left( e ^ { 0.05 t } \right)$.

Question 3

Suppose that a population grows according to a logistic model.(a) Write the differential equation for this situation with k = 0.01 and carrying capacity of 60 thousand.(b) Solve the differential equation in part (a) with the initial condition t = 0 (hours) and population P = 1 thousand.(c) Find the population for t = 10 hours, t = 100 hours, and t = 1000 hours.(d) After how many hours does the population reach 2 thousand? 30 thousand? 55 thousand?
(e) As the time t increases without bound, what happens to the population?
(f) Sketch the graph of the solution of the differential equation.

Accepted Answer

The answer of Suppose that a population grows according to...

Question 4

A bacteria culture starts with 200 bacteria and triples in size every half hour. The population of the bacteria after $t$ hours is:
A) $200 \left( 9 ^ { - t } \right)$ 
B) $200 \left( 9 ^ { t } \right)$ 
C) $200 \left( 3 ^ { - t } \right)$ 
D) $200 \left( 3 ^ { t } \right)$ 
E) $200 \left( e ^ { - t } \right)$ 
F) $200 \left( e ^ { t } \right)$ 
G) $200 \left( e ^ { - 3 t } \right)$ 
H) $200 \left( e ^ { 3 t } \right)$

Accepted Answer

The population triples every half hour, so in one hour (t=1), it will triple twice. This is represented by the formula $200 \times 3^{2t}$, which simplifies to $200 \times 9^t$, matching option B.

Question 5

In a model of epidemics, the number of infected individuals in a population at a time is a solution of the logistic differential equation $\frac { d y } { d t } = 0.6 y - 0.0002 y ^ { 2 }$ , where y is the number of infected individuals in the community and t is the time in days.(a) Describe the population for this situation.(b) Assume that 10 people were infected at the initial time t = 0. Find the solution for the differential equation.(c) How many days will it take for half of the population to be infected?

Accepted Answer

The answer of In a model of epidemics, the number...

Question 6

When a child was born, her grandparents placed $1000 in a savings account at 10% interest compounded continuously, to be withdrawn at age 20 to help pay for college. How much money is in the account at the time of withdrawal?
A) $1000 e$ 
B) $500 e$ 
C) $500 e ^ { 2 }$ 
D) $2000 e ^ { 2 }$ 
E) $4000 e$ 
F) $2000 e$ 
G) $1000 e ^ { 2 }$ 
H) $4000 e ^ { 2 }$

Accepted Answer

The formula for continuous compounding is $A = P e^{rt}$, where $A$ is the amount of money accumulated after n years, including interest, $P$ is the principal amount (the initial amount of money), $r$ is the annual interest rate (decimal), and $t$ is the time the money is deposited or borrowed for, in years. Here, $P = 1000$, $r = 0.10$, and $t = 20$. Plugging these values into the formula gives $A = 1000 e^{(0.10)(20)} = 1000 e^{2}$.

Question 7

An object cools at a rate (measured in ${ } ^ { \circ } \mathrm { C } / \mathrm { min }$ ) equal to $k$ times the difference between its temperature and that of the surrounding air. Suppose the object takes 10 minutes to cool from 60 $^\circ$ C to 40 $^\circ$ C in a room kept at 20 $^\circ$ C. Find the value of $k$ .
A) $e ^ { - 20 }$ 
B) $\ln 2$ 
C) $10 e ^ { - 20 }$ 
D) $40 \ln 10$
E) $\frac { 1 } { 2 }$ 
F) $e ^ { - 1 / 20 }$ 
G) $\frac { 1 } { 10 } \ln \frac { 1 } { 2 }$ 
H) $60 \ln \frac { 1 } { 2 }$

Accepted Answer

The cooling process is described by Newton's Law of Cooling, which states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature. The formula for the temperature $T(t)$ of an object at time $t$ in this context is given by $T(t) = T_{\text{ambient}} + (T_{\text{initial}} - T_{\text{ambient}})e^{-kt}$, where $T_{\text{ambient}}$ is the ambient temperature, $T_{\text{initial}}$ is the initial temperature of the object, and $k$ is the constant of proportionality.Given that the object cools from 60°C to 40°C in 10 minutes in a room at 20°C, we can set up the equation $40 = 20 + (60 - 20)e^{-10k}$. Simplifying, we get $20 = 40e^{-10k}$, or $1/2 = e^{-10k}$. Taking the natural logarithm of both sides gives $\ln(1/2) = -10k$, so $k = \frac{1}{10}\ln(1/2)$, which corresponds to choice G.

Question 8

Radium has a half-life of 1600 years. How many years does it take for 90% of a given amount of radium to decay?
A) $\frac { 1600 } { \ln 5 }$ 
B) $1600 \ln 2$ 
C) $\frac { 1600 \ln 10 } { \ln 2 }$ 
D) $1600 \ln 5$ 
E) $1600 \ln 10$ 
F) $\frac { 1600 } { \ln 2 }$ 
G) $1500 \ln 6$ 
H) $\frac { 1600 \ln 2 } { \ln 10 }$

Accepted Answer

The formula to calculate the time it takes for a certain percentage of a substance to decay is derived from the half-life formula: $t = \frac{\ln(\frac{1}{remaining\ fraction})}{\ln(2)} \times half\ life$. For 90% decay, the remaining fraction is 0.1 (100% - 90% = 10% remaining). Substituting the values: $t = \frac{\ln(10)}{\ln(2)} \times 1600$, which matches option C.

Question 9

The radioactive isotope Bismuth-210 has a half-life of 5 days. Suppose we have an initial amount of 100 mg. The amount of Bismuth-210 remaining after $t$ days is
A) $100 \left( 2 ^ { 0.2 t } \right)$ 
B) $5 \left( 2 ^ { 0.2 t } \right)$ 
C) $50 \left( 2 ^ { - 0.2 t } \right)$ 
D) $100 \left( 2 ^ { - 0.2 t } \right)$ 
E) $100 \left( e ^ { 0.2 t } \right)$ 
F) $5 \left( e ^ { 0.2 t } \right)$ 
G) $50 \left( e ^ { - 0.2 t } \right)$ 
H) $100 \left( e ^ { - 0.2 t } \right)$

Accepted Answer

The formula for the decay of a radioactive substance is $A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T}}$, where $A$ is the amount remaining, $A_0$ is the initial amount, $t$ is the time elapsed, and $T$ is the half-life. Substituting $A_0 = 100$, $T = 5$, and converting the base from $2$ to $e$ is unnecessary here. The correct formula simplifies to $100 \left( 2 ^ { - 0.2 t } \right)$, matching option D.

Question 10

A bacteria culture starts with 200 bacteria and in 1 hour contains 400 bacteria. How many hours does it take to reach 2000 bacteria?
A) $\ln 400$ 
B) $\ln 10$ 
C) $10$ 
D) $\ln 1600$ 
E) $\ln 2000$ 
F) $\ln 200$ 
G) $5$ 
H) $\frac { \ln 10 } { \ln 2 }$

Accepted Answer

The bacteria culture doubles every hour. To find the number of hours it takes to grow from 200 to 2000 bacteria, we can use the formula for exponential growth: $N = N_0 \cdot 2^t$, where $N$ is the final amount, $N_0$ is the initial amount, and $t$ is the time in hours. Rearranging to solve for $t$, we get $t = \log_2 \frac{N}{N_0}$. Substituting $N = 2000$ and $N_0 = 200$, we get $t = \log_2 \frac{2000}{200} = \log_2 10$. Since $\log_2 10$ is equivalent to $\frac{\ln 10}{\ln 2}$, the correct answer is H.

Question 11

Assume that a population grows at a rate summarized by the equation $\frac { d P } { d t } = \frac { b k P e ^ { - k t } } { 1 - b e ^ { - k t } }$ , where b and k are positive constants (b > 1), and P is the population at time t. Show that $P = \frac { P _ { o } } { 1 - b } \left( 1 - b e ^ { - k t } \right)$ is the general solution for the differential equation (where $P _ { o }$ is the initial population). [Note: This is known as the monomolecular growth curve.]

Accepted Answer

The answer of Assume that a population grows at a...

Question 12

Carbon 14, with a half-life of 5700 years, is used to estimate the age of organic materials. What fraction of the original amount of carbon 14 would an object have if it were 2000 years old?
A) $e ^ { - ( 57 / 20 ) \ln 2 }$ 
B) $\frac { 57 } { 20 } \ln 2$ 
C) $e ^ { - ( 20 / 57 ) \ln 2 }$ 
D) $\frac { 20 } { 57 } \ln 2$ 
E) $e ^ { ( 57 / 20 ) \ln 2 }$ 
F) $\frac { 1 } { 57 } \ln 20$ 
G) $e ^ { ( 20 / 57 ) \ln 2 }$ 
H) $\frac { 1 } { 20 } \ln 57$

Accepted Answer

The fraction of the original amount of carbon-14 remaining after time $ t $ is given by $ e^{-kt} $, where $ k = \frac{\ln 2}{\text{half-life}} $. For 2000 years, $ k = \frac{\ln 2}{5700} $, so the fraction is $ e^{-(2000/5700)\ln 2} = e^{-(20/57)\ln 2} $.

Question 13

Suppose that a population, P, grows at a rate given by the equation $\frac { d P } { d t } = b P e ^ { - k t }$ , where P is the population (in thousands) at time t (in hours), and b and k are positive constants.(a) Find the solution to the differential equation when b = 0.04, k = 0.01 and P (0) = 1.(b) Find P (10), P (100), and P (1000).(c) After how many hours does the population reach 2 thousand? 30 thousand? 54 thousand?
(d) As time t increases without bound, what happens to the population?
(e) Sketch the graph of the solution of the differential equation.

Accepted Answer

The answer of Suppose that a population, P, grows at...

Question 14

(a) Solve the differential equation $\frac { d P } { d t } = \frac { b k P e ^ { - k t } } { 1 - b e ^ { - k t } }$ , with b = 2 and k = 0.1, and $P _ { o }$ = 1.(b) Sketch a graph of the solution you produced for part (a) and discuss the major characteristics of this monomolecular growth curve.

Accepted Answer

The answer of (a) Solve the differential equation \(\frac {...

Question 15

Suppose that a certain population grows according to an exponential model.(a) Write the differential equation for this situation with a relative growth rate of k = 0.01. Produce a solution for the initial condition t = 0 (in hours) and population P = 1 (in thousands).(b) Find the population when t = 10 hours, t = 100 hours, and t = 1000 hours.(c) After how many hours does the population reach 2 thousand? 30 thousand? 55 thousand?
(d) As the time t increases without bound, what happens to the population?
(e) Sketch the graph of the solution of the differential equation.

Accepted Answer

The answer of Suppose that a certain population grows according...

Question 16

An outbreak of a previously unknown influenza occurred on the campus of the University of Northern South Dakota at Roscoe during the first semester. Due to the contagious nature of the disease, the campus was quarantined and the disease was allowed to run its course. The table below shows the total number P of infected students for the first four weeks of the outbreak on this campus of 2,500 students. $$\begin{array} { | l | c | c | c | c | c | c | c | c | c | c | c | c | c | c | c | } 
\hline t \text { (days) } & 0 & 2 & 4 & 6 & 8 & 10 & 12 & 14 & 16 & 18 & 20 & 22 & 24 & 26 & 28 \
\hline P & 100 & 121 & 146 & 176 & 212 & 254 & 304 & 361 & 428 & 503 & 589 & 683 & 787 & 899 & 1017 \
\hline \text { Logistic } & & & & & & & & & & & & & & & \
\hline \text { Exponential } & & & & & & & & & & & & & & & \
\hline
\end{array}$$ (a) Find a logistic model for the data. Complete the table with predicted values using this model.(b) Find an exponential model for these data. Complete the table with predicted values using this model.(c) Compare your findings in parts (a) and (b) above. For what values would you consider both models to be a good fit for the data? Which model provides the best fit for the data? Justify your choice.

Accepted Answer

The answer of An outbreak of a previously unknown influenza...

Question 17

The half-life of Carbon 14 is 5700 years. A wooden table is measured with 80% of Carbon 14 compared with newly cut tree. Find the age of the table.
A)2, 933 years
B)1,000 years
C)500 years
D)2,000 years
E)13,235 years
F)4,200 years
G)1,835 years
H)3,000 years

Accepted Answer

Using the formula for exponential decay: 
(amount remaining) = (initial amount) x (1/2)^(time/half-life)
Let t be the age of the table, then we have:
0.8 = 1 x (1/2)^(t/5700)
Taking the logarithm of both sides:
log(0.8) = log(1/2)^(t/5700)
log(0.8) = (t/5700) x log(1/2)
t/5700 = log(0.8) / log(1/2)
t/5700 = 0.322
t = 5700 x 0.322
t = 1835.4 years
Therefore, the age of the table is approximately 1,835 years. The best choice is letter G.

Question 18

The following table contains population data for a Minnesota county for the decades from 1900 to 1980: $$\begin{array} { | l | c | c | c | } 
\hline \text { Year } & \text { Clay County Population } & \begin{array} { c } 
\text { Population Estimate } \
\text { (Exponential Model) }
\end{array} & \begin{array} { c } 
\text { Population Estimate } \
\text { (Logistic Model) }
\end{array} \
\hline 1900 & 17,942 & & \
\hline 1910 & 19,640 & & \
\hline 1920 & 21,780 & & \
\hline 1930 & 23,120 & & \
\hline 1940 & 25,337 & & \
\hline 1950 & 30,363 & & \
\hline 1960 & 39,080 & & \
\hline 1970 & 46,585 & & \
\hline 1980 & 49,327 & & \
\hline
\end{array}$$ (a) Produce a scatter plot for the data.(b) Find an exponential model using the data from 1900 through 1950.(c) Find a logistic model using the data from 1900 through 1950. (Assume the carrying capacity is 440,000.)
(d) Use your models to estimate the population for 1960, 1970, and 1980. Enter your data in the table provided above.

Accepted Answer

The answer of The following table contains population data for...

Question 19

A bacteria population grows at a rate proportional to its size. The initial count was 400 and 1600 after 1 hour. In how many minutes does the population double?
A)20
B)25
C)30
D)35
E)40
F)45
G)50
H)55

Accepted Answer

Let P(t) be the population after time t in minutes.

P(t) = 400e^kt   ------(1) (since the population is proportional to its size)

When t = 60, P(60) = 1600

1600 = 400e^(60k)

e^(60k) = 4

Taking logarithm both sides,

60k = ln(4)

k = ln(4)/60

Now we need to find t such that the population doubles i.e. P(t) = 800.

800 = 400e^(kt)

2 = e^(kt)

Taking logarithm both sides,

ln(2) = kt

t = ln(2)/k = ln(2)/(ln(4)/60) = 30 minutes (approx.)

Therefore, the answer is choice C, 30.

Question 20

A bacteria culture starts with 200 bacteria and triples in size every half hour. After 2 hours, how many bacteria are there?
A)17,800
B)16,200
C)23,500
D)24,000
E)19,300
F)14,800
G)15,700
H)21,000

Accepted Answer

The answer of A bacteria culture starts with 200 bacteria...

Quiz 7: Differential Equations

The radioactive isotope Bismuth-210 has a half-life of 5 days. How many days does it take for 87.5% of a given amount to decay?

When a child was born, her grandparents deposited $1000 in a saving account at 5% interest compounded continuously. The amount of money after t years is:

A bacteria culture starts with 200 bacteria and triples in size every half hour. The population of the bacteria after ttt hours is:

When a child was born, her grandparents placed $1000 in a savings account at 10% interest compounded continuously, to be withdrawn at age 20 to help pay for college. How much money is in the account at the time of withdrawal?

Radium has a half-life of 1600 years. How many years does it take for 90% of a given amount of radium to decay?

The radioactive isotope Bismuth-210 has a half-life of 5 days. Suppose we have an initial amount of 100 mg. The amount of Bismuth-210 remaining after ttt days is

A bacteria culture starts with 200 bacteria and in 1 hour contains 400 bacteria. How many hours does it take to reach 2000 bacteria?

Carbon 14, with a half-life of 5700 years, is used to estimate the age of organic materials. What fraction of the original amount of carbon 14 would an object have if it were 2000 years old?

The half-life of Carbon 14 is 5700 years. A wooden table is measured with 80% of Carbon 14 compared with newly cut tree. Find the age of the table.

A bacteria population grows at a rate proportional to its size. The initial count was 400 and 1600 after 1 hour. In how many minutes does the population double?

A bacteria culture starts with 200 bacteria and triples in size every half hour. After 2 hours, how many bacteria are there?

A bacteria culture starts with 200 bacteria and triples in size every half hour. The population of the bacteria after $t$ hours is:

The radioactive isotope Bismuth-210 has a half-life of 5 days. Suppose we have an initial amount of 100 mg. The amount of Bismuth-210 remaining after $t$ days is