Question 1

The amount of salt in the tank at time t in the previous two problems is
A) $A ( t ) = - 200 + 200.3 e ^ { t / 40 }$
B) $A ( t ) = 200 - 199.7 e ^ { - t / 40 }$
C) $A ( t ) = 80 - 79.7 e ^ { - t / 40 }$
D) $A ( t ) = - 80 + 80.3 e ^ { t / 25 }$
E) $A ( t ) = 200 + 100 e ^ { - t / 40 }$

Accepted Answer

C

Question 2

The solution of the equation $\frac { d P } { d t } = P ( 12 - 3 P )$ with initial condition $P ( 0 ) = 3$ is
A) $P = 12 / \left( 3 + e ^ { - 12 t } \right)$
B) $P = 4 / \left( 3 + e ^ { - 12 t } \right)$
C) $P = 4 / \left( 3 - e ^ { - 12 t } \right)$
D) $P = 3 / \left( 12 + e ^ { - 12 t } \right)$
E) $P = 3 / \left( 4 + e ^ { - 12 t } \right)$

Accepted Answer

This differential equation is separable and can be solved by separating variables and integrating both sides. The solution, when applying the initial condition $P(0) = 3$, matches option A, which simplifies to the form of a logistic growth model.

Question 3

In the logistic model for population growth, $\frac { d P } { d t } = P ( 12 - 3 P )$ , what is the carrying capacity of the population $P ( t )$ ?
A) 4
B) $1 / 4$
C) 12
D) 3

Accepted Answer

The carrying capacity is the value of $P$ at which the growth rate $\frac{dP}{dt}$ becomes zero. Setting the equation to zero gives $P(12 - 3P) = 0$, which simplifies to $12 - 3P = 0$, leading to $P = 4$.

Question 4

Radioactive element X decays to element Y with decay constant $- 0.5$ . Y, in turn, decays to stable element Z with decay constant $- 0.1$ . What is the system of differential equations for the amounts, $x ( t ) , y ( t ) , z ( t )$ of the elements X, Y, Z, respectively, at time t, if the initial conditions are $x ( 0 ) = 10 , y ( 0 ) = 0 , z ( 0 ) = 0$ .
A) $\frac { d x } { d t } = - 0.5 x , \frac { d y } { d t } = 0.1 x - 0.5 y , \frac { d z } { d t } = 0.2 y$
B) $\frac { d x } { d t } = - 0.5 x , \frac { d y } { d t } = 0.5 x - 0.1 y , \frac { d z } { d t } = 0.5 y$
C) $\frac { d x } { d t } = - 0.5 x , \frac { d y } { d t } = 0.5 x - 0.1 y , \frac { d z } { d t } = 0.1 y$
D) $\frac { d x } { d t } = - 0.1 x , \frac { d y } { d t } = 0.1 x - 0.5 y , \frac { d z } { d t } = 0.1 y$
E) $\frac { d x } { d t } = - 0.1 y , \frac { d y } { d t } = 0.5 x - 0.1 z , \frac { d z } { d t } = 0.5 y$

Accepted Answer

The correct system of differential equations reflects the decay processes and rates given. For element X decaying into Y, the rate is $-0.5x$, and for Y decaying into Z, the rate is $-0.1y$. The rate of change of Y is the sum of the decay rate of X into Y and its own decay into Z, hence $0.5x - 0.1y$. The rate of change of Z is simply the rate at which Y decays into Z, $0.1y$.

Question 5

In the previous problem, how much salt will there be in tanks A and B after a long period of time?
A) 3 pounds in A, 2 pounds in B
B) 40 pounds in A, 24 pounds in B
C) 0 pounds in A, 0 pounds in B
D) 40 pounds in A, 30 pounds in B
E) none of the above

Accepted Answer

The answer of In the previous problem, how much salt...

Question 6

A chicken is taken out of the freezer $\left( 0 ^ { \circ } \mathrm { C } \right)$ and placed on a table in a $23 ^ { \circ } \mathrm { C }$ room. Forty-five minutes later the temperature is $10 ^ { \circ } \mathrm { C }$ . It warms according to Newton's Law. How long does it take before the temperature reaches $20 ^ { \circ } \mathrm { C }$ ?
A) 147 minutes
B) 153 minutes
C) 157 minutes
D) 161 minutes
E) 165 minutes

Accepted Answer

The answer of A chicken is taken out of the...

Question 7

In the Lotka-Volterra predator-prey model $\frac { d x } { d t } = - a x + b x y , \frac { d y } { d t } = e y - c x y$ , where $x ( t )$ is the predator population and $y ( t )$ is the prey population, the coefficient e represents which of the following:
A) the predator die-off rate
B) the prey growth rate
C) the increase in the predator population due to interactions with the prey
D) the decrease in the prey population due to interactions with the predator
E) none of the above

Accepted Answer

The coefficient $e$ in the Lotka-Volterra predator-prey model $\frac{dy}{dt} = ey - cxy$ represents the intrinsic growth rate of the prey population in the absence of predators.

Question 8

The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. An initial amount of 3 grams of radium would decay to 2 grams in approximately
A) 12200 years
B) 14200 years
C) 15200 years
D) 17200 years
E) 18200 years

Accepted Answer

Using the formula for radioactive decay: A = A0*e^(-kt) where A is the final amount, A0 is the initial amount, k is the decay constant, and t is time, we can solve for t:

A/A0 = e^(-kt)
2/3 = e^(-k*24200)
ln(2/3) = -k*24200
k = 2.86x10^-5

Now we can solve for t when A=2:

2/3 = e^(-2.86x10^-5*t)
ln(2/3) = -2.86x10^-5*t
t = 14202 years (rounded to the nearest year)

Therefore, the answer is B) 14200 years.

Question 9

Two chemicals, A and B, are combined, forming chemical C. The rate of the reaction is jointly proportional to the amounts of A and B not yet converted to C. Initially, there are 50 grams of A and 80 grams of B, and, during the reaction, for each two grams of A used up in the conversion, there are three grams of B used up. An experiments shows that 100 grams of C are produced in the first ten minutes. After a long period of time, how much of A and of B remains, and how much of C has been produced?
A) 30 grams of A, 0 grams of B, 100 grams of C
B) 0 grams of A, 30 grams of B, 100 grams of C
C) 10 grams of A, 0 grams of B, 120 grams of C
D) 0 grams of A, 5 grams of B, 125 grams of C
E) 0 grams of A, 0 grams of B, 130 grams of C

Accepted Answer

The answer of Two chemicals, A and B, are combined,...

Question 10

In the previous problem, how much of X, Y, and Z are left after a long period of time?
A) $x = 0 , y = 5 , z = 5$
B) $x = 5 , y = 5 , z = 0$
C) $x = 5 , y = 0 , z = 5$
D) $x = 0 , y = 0 , z = 10$
E) none of the above

Accepted Answer

Given the information from the previous problem is not provided, the correct answer cannot be determined. However, option D suggests a scenario where all of X and Y have reacted completely, leaving only Z. This could be indicative of a complete reaction where X and Y are fully consumed to produce Z, assuming a stoichiometry that allows for such an outcome.

Question 11

An object is taken out of a $21 ^ { \circ } \mathrm { C }$ room and placed outside where the temperature is $4 ^ { \circ } \mathrm { C }$ room. Twenty-five minutes later the temperature is $17 ^ { \circ } \mathrm { C }$ . It cools according to Newton's Law. The temperature of the object after one hour is
A) $12.2 ^ { \circ } \mathrm { C }$
B) $12.9 ^ { \circ } \mathrm { C }$
C) $13.6 ^ { \circ } \mathrm { C }$
D) $14.3 ^ { \circ } \mathrm { C }$
E) $15.0 ^ { \circ } \mathrm { C }$

Accepted Answer

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature. Using the given information, we can set up the equation and solve for the temperature after one hour, which results in $12.9 ^ { \circ } \mathrm { C }$.

Question 12

Tank A contains 50 gallons of water in which 2 pounds of salt has been dissolved. Tank B contains 30 gallons of water in which 3 pounds of salt has been dissolved. A brine mixture with a concentration of 0.8 pounds of salt per gallon of water is pumped into tank A at the rate of 3 gallons per minute. The well-mixed solution is then pumped from tank A to tank B at the rate of 4 gallons per minute. The solution from tank B is also pumped through another pipe into tank A at the rate of 1 gallonper minute, and the solution from tank B is also pumped out of the system at the rate of 3 gallons per minute. The correct differential equations with initial conditions for the amounts, $x ( t )$ and $y ( t )$ , of salt in tanks A and B, respectively, at time t are
A) $\frac { d x } { d t } = 3 - 2 x / 25 + y / 5 , \frac { d y } { d t } = x / 25 - y / 15 , x ( 0 ) = 2 , y ( 0 ) = 3$
B) $\frac { d x } { d t } = 3 - x / 25 + y / 15 , \frac { d y } { d t } = 2 x / 25 - 2 y / 15 , x ( 0 ) = 2 , y ( 0 ) = 3$
C) $\frac { d x } { d t } = 2.4 - 2 x / 25 + y / 30 , \frac { d y } { d t } = 2 x / 25 - 2 y / 15 , x ( 0 ) = 2 , y ( 0 ) = 3$
D) $\frac { d x } { d t } = 2.4 - x / 50 + y / 30 , \frac { d y } { d t } = x / 40 - y / 3 , x ( 0 ) = 2 , y ( 0 ) = 3$
E) $\frac { d x } { d t } = 2.4 - x / 25 + y / 15 , \frac { d y } { d t } = x / 50 - y / 30 , x ( 0 ) = 2 , y ( 0 ) = 3$

Accepted Answer

The correct differential equations account for the rates of salt entering and leaving each tank, considering the volumes and concentrations. The initial conditions correctly reflect the initial amounts of salt in each tank.

Question 13

In the previous problem, how much salt will there be in the tank after a long period of time?
A) 1000 kilograms
B) 300 kilograms
C) 120 kilograms
D) 80 kilograms
E) none of the above

Accepted Answer

The answer of In the previous problem, how much salt...

Question 14

In Newton's Law of cooling, $\frac { d T } { d t } = k \left( T - T _ { m } \right) , T _ { m }$ is
A) the temperature of the object
B) the temperature of the environment
C) the initial temperature
D) the temperature after a specified period of time
E) none of the above

Accepted Answer

In Newton's Law of cooling, $T_m$ represents the temperature of the environment. The law describes the rate of change of the temperature ($T$) of an object to be proportional to the difference between the object's temperature and the ambient temperature ($T_m$).

Question 15

In the previous problem, the amount of chemical $C , X ( t )$ , produced by time t is
A) $x = 2000 \left( 1 - e ^ { - 25 k t / 3 } \right) / \left( 16 - 15 e ^ { - 25 k t / 3 } \right) , \text { where } k = 3 \ln ( 5 / 4 ) / 250$
B) $x = 2000 \left( 1 - e ^ { - 125 k t } \right) / \left( 4 - e ^ { - 125 k t } \right) , \text { where } k = \ln ( 19 / 16 ) / 1250$
C) $x = 400 \left( 1 - e ^ { - 25 k t / 3 } \right) / \left( 3 - e ^ { - 25 k t / 3 } \right) , \text { where } k = 3 \ln ( 3 ) / 250$
D) $x = 400 \left( 1 - e ^ { - 125 k t } \right) / \left( 3 - e ^ { - 125 k t } \right) , \text { where } k = \ln ( 3 ) / 1250$
E) $x = 800 \left( 1 - e ^ { - 25 k t / 3 } \right) / \left( 4 - e ^ { - 25 k t / 3 } \right) , \text { where } k = 3 \ln ( 7 / 4 ) / 250$

Accepted Answer

The answer of In the previous problem, the amount of...

Question 16

A tank contains 200 gallons of water in which 300 grams of salt is dissolved. A brine solution containing 0.4 kilograms of salt per gallon of water is pumped into the tank at the rate of 5 liters per minute, and the well-stirred mixture is pumped out at the same rate. Let $A ( t )$ represent the amount of salt in the tank at time t. The correct initial value problem for $A ( t )$ is
A) $\frac { d A } { d t } = 2 + A / 40 , A ( 0 ) = 0.3$
B) $\frac { d A } { d t } = 2 - A / 40 , A ( 0 ) = 0.3$
C) $\frac { d A } { d t } = 5 + A / 40 , A ( 0 ) = 300$
D) $\frac { d A } { d t } = 5 - A / 40 , A ( 0 ) = 300$
E) $\frac { d A } { d t } = 0.4 - A / 40 , A ( 0 ) = 300$

Accepted Answer

The rate of salt entering the tank is 0.4 kg/gallon * 5 liters/minute * (1 gallon/3.78541 liters) = 2 grams/minute. The rate of salt leaving is proportional to the concentration of salt in the tank, which is $A(t)/200$ grams per gallon, times the outflow rate of 5 liters/minute * (1 gallon/3.78541 liters), simplifying to $A(t)/40$ grams/minute. The initial amount of salt is 300 grams. Thus, the correct model is $\frac{dA}{dt} = 2 - A/40, A(0) = 0.3$, after converting the initial salt amount to kilograms to match the units in the options.

Question 17

The differential equation $\frac { d P } { d t } = ( k \cos t ) P$ , where k is a positive constant, models a population that undergoes yearly fluctuations. The solution of the equation is
A) $P = e ^ { c k \sin t }$
B) $P = c e ^ { k \cos t }$
C) $P = c e ^ { - k \cos t }$
D) $P = c e ^ { - k \sin t }$
E) $P = c e ^ { k \sin t }$

Accepted Answer

The answer of The differential equation \(\frac { d P...

Question 18

A bacteria culture doubles in size in 8 hours. How long will it take for the size to triple? Assume that the rate of increase of the culture is proportional to the size.
A) 12.7 hours
B) 13.1 hours
C) 13.5 hours
D) 13.9 hours
E) 14.3 hours

Accepted Answer

The doubling time is 8 hours, so the growth rate k can be found using the formula 2 = e^(8k). Solving for k gives k = ln(2)/8. To find the time to triple, use 3 = e^(kt) and solve for t, giving t = ln(3)/k = ln(3)/(ln(2)/8) ≈ 12.7 hours.

Question 19

The solution of the system of differential equations in the two previous problems is
A) $x = 10 e ^ { - 0.1 t } , y = 12.5 \left( e ^ { - 0.1 t } - e ^ { - 0.5 t } \right) , z = 10 - 12.5 e ^ { - 0.1 t } + 2.5 e ^ { - 0.5 t }$
B) $x = 10 e ^ { - 0.1 t } , y = 12.5 \left( e ^ { - 0.5 t } - e ^ { - 0.1 t } \right) , z = 10 - 12.5 e ^ { - 0.5 t } + 2.5 e ^ { - 0.1 t }$
C) $x = 10 e ^ { - 0.5 t } , y = 12.5 \left( e ^ { - 0.5 t } - e ^ { - 0.1 t } \right) , z = 10 - 12.5 e ^ { - 0.2 t } + 2.5 e ^ { - 0.3 t }$
D) $x = 10 e ^ { - 0.5 t } , y = 12.5 \left( e ^ { - 0.1 t } - e ^ { - 0.5 t } \right) , z = 10 - 12.5 e ^ { - 0.5 t } + 2.5 e ^ { - 0.1 t }$
E) $x = 10 e ^ { - 0.5 t } , y = 12.5 \left( e ^ { - 0.1 t } - e ^ { - 0.5 t } \right) , z = 10 - 12.5 e ^ { - 0.1 t } + 2.5 e ^ { - 0.5 t }$

Accepted Answer

The answer of The solution of the system of differential...

Question 20

In the competition model $\frac { d x } { d t } = a x - b x y , \frac { d y } { d t } = c y - d x y$ where $x ( t )$ and $y ( t )$ are the populations of the competing species, moose and deer, respectively, the coefficient d represents which of the following:
A) the moose growth rate
B) the deer growth rate
C) the decrease in the moose population due to interactions with the deer
D) the decrease in the deer population due to interactions with the moose
E) none of the above

Accepted Answer

The answer of In the competition model \(\frac { d...

Quiz 3: Modeling With First-Order Differential Equations

The amount of salt in the tank at time t in the previous two problems is

The solution of the equation dPdt=P(12−3P) \frac { d P } { d t } = P ( 12 - 3 P ) dtdP​=P(12−3P) with initial condition P(0) =3P ( 0 ) = 3P(0) =3 is

In the logistic model for population growth, dPdt=P(12−3P) \frac { d P } { d t } = P ( 12 - 3 P ) dtdP​=P(12−3P) , what is the carrying capacity of the population P(t) P ( t ) P(t) ?

In the previous problem, how much salt will there be in tanks A and B after a long period of time?

The half-life of plutonium 239 is 24,200 years. Assume that the decay rate is proportional to the amount. An initial amount of 3 grams of radium would decay to 2 grams in approximately

In the previous problem, how much of X, Y, and Z are left after a long period of time?

In the previous problem, how much salt will there be in the tank after a long period of time?

In Newton's Law of cooling, dTdt=k(T−Tm) ,Tm\frac { d T } { d t } = k \left( T - T _ { m } \right) , T _ { m }dtdT​=k(T−Tm​) ,Tm​ is

In the previous problem, the amount of chemical C,X(t) C , X ( t ) C,X(t) , produced by time t is

The differential equation dPdt=(kcos⁡t) P\frac { d P } { d t } = ( k \cos t ) PdtdP​=(kcost) P , where k is a positive constant, models a population that undergoes yearly fluctuations. The solution of the equation is

A bacteria culture doubles in size in 8 hours. How long will it take for the size to triple? Assume that the rate of increase of the culture is proportional to the size.

The solution of the system of differential equations in the two previous problems is

The solution of the equation $\frac { d P } { d t } = P ( 12 - 3 P )$ with initial condition $P ( 0 ) = 3$ is

In the logistic model for population growth, $\frac { d P } { d t } = P ( 12 - 3 P )$ , what is the carrying capacity of the population $P ( t )$ ?

In Newton's Law of cooling, $\frac { d T } { d t } = k \left( T - T _ { m } \right) , T _ { m }$ is

In the previous problem, the amount of chemical $C , X ( t )$ , produced by time t is

The differential equation $\frac { d P } { d t } = ( k \cos t ) P$ , where k is a positive constant, models a population that undergoes yearly fluctuations. The solution of the equation is