Deck 3: Modeling With First-Order Differential Equations

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

A) A(t)=200+200.3et/40A ( t ) = - 200 + 200.3 e ^ { t / 40 }
B) A(t)=200199.7et/40A ( t ) = 200 - 199.7 e ^ { - t / 40 }
C) A(t)=8079.7et/40A ( t ) = 80 - 79.7 e ^ { - t / 40 }
D) A(t)=80+80.3et/25A ( t ) = - 80 + 80.3 e ^ { t / 25 }
E) A(t)=200+100et/40A ( t ) = 200 + 100 e ^ { - t / 40 }
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Question
The solution of the equation dPdt=P(123P)\frac { d P } { d t } = P ( 12 - 3 P ) with initial condition P(0)=3P ( 0 ) = 3 is

A) P=12/(3+e12t)P = 12 / \left( 3 + e ^ { - 12 t } \right)
B) P=4/(3+e12t)P = 4 / \left( 3 + e ^ { - 12 t } \right)
C) P=4/(3e12t)P = 4 / \left( 3 - e ^ { - 12 t } \right)
D) P=3/(12+e12t)P = 3 / \left( 12 + e ^ { - 12 t } \right)
E) P=3/(4+e12t)P = 3 / \left( 4 + e ^ { - 12 t } \right)
Question
In the logistic model for population growth, dPdt=P(123P)\frac { d P } { d t } = P ( 12 - 3 P ) , what is the carrying capacity of the population P(t)P ( t ) ?

A) 4
B) 1/41 / 4
C) 12
D) 3
Question
Radioactive element X decays to element Y with decay constant 0.5- 0.5 . Y, in turn, decays to stable element Z with decay constant 0.1- 0.1 . What is the system of differential equations for the amounts, x(t),y(t),z(t)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)=0x ( 0 ) = 10 , y ( 0 ) = 0 , z ( 0 ) = 0 .

A) dxdt=0.5x,dydt=0.1x0.5y,dzdt=0.2y\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) dxdt=0.5x,dydt=0.5x0.1y,dzdt=0.5y\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) dxdt=0.5x,dydt=0.5x0.1y,dzdt=0.1y\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) dxdt=0.1x,dydt=0.1x0.5y,dzdt=0.1y\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) dxdt=0.1y,dydt=0.5x0.1z,dzdt=0.5y\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
Question
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
Question
A chicken is taken out of the freezer (0C)\left( 0 ^ { \circ } \mathrm { C } \right) and placed on a table in a 23C23 ^ { \circ } \mathrm { C } room. Forty-five minutes later the temperature is 10C10 ^ { \circ } \mathrm { C } . It warms according to Newton's Law. How long does it take before the temperature reaches 20C20 ^ { \circ } \mathrm { C } ?

A) 147 minutes
B) 153 minutes
C) 157 minutes
D) 161 minutes
E) 165 minutes
Question
In the Lotka-Volterra predator-prey model dxdt=ax+bxy,dydt=eycxy\frac { d x } { d t } = - a x + b x y , \frac { d y } { d t } = e y - c x y , where x(t)x ( t ) is the predator population and y(t)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
Question
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
Question
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
Question
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=5x = 0 , y = 5 , z = 5
B) x=5,y=5,z=0x = 5 , y = 5 , z = 0
C) x=5,y=0,z=5x = 5 , y = 0 , z = 5
D) x=0,y=0,z=10x = 0 , y = 0 , z = 10
E) none of the above
Question
An object is taken out of a 21C21 ^ { \circ } \mathrm { C } room and placed outside where the temperature is 4C4 ^ { \circ } \mathrm { C } room. Twenty-five minutes later the temperature is 17C17 ^ { \circ } \mathrm { C } . It cools according to Newton's Law. The temperature of the object after one hour is

A) 12.2C12.2 ^ { \circ } \mathrm { C }
B) 12.9C12.9 ^ { \circ } \mathrm { C }
C) 13.6C13.6 ^ { \circ } \mathrm { C }
D) 14.3C14.3 ^ { \circ } \mathrm { C }
E) 15.0C15.0 ^ { \circ } \mathrm { C }
Question
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)x ( t ) and y(t)y ( t ) , of salt in tanks A and B, respectively, at time t are

A) dxdt=32x/25+y/5,dydt=x/25y/15,x(0)=2,y(0)=3\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) dxdt=3x/25+y/15,dydt=2x/252y/15,x(0)=2,y(0)=3\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) dxdt=2.42x/25+y/30,dydt=2x/252y/15,x(0)=2,y(0)=3\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) dxdt=2.4x/50+y/30,dydt=x/40y/3,x(0)=2,y(0)=3\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) dxdt=2.4x/25+y/15,dydt=x/50y/30,x(0)=2,y(0)=3\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
Question
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
Question
In Newton's Law of cooling, dTdt=k(TTm),Tm\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
Question
In the previous problem, the amount of chemical C,X(t)C , X ( t ) , produced by time t is

A) x=2000(1e25kt/3)/(1615e25kt/3), where k=3ln(5/4)/250x = 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(1e125kt)/(4e125kt), where k=ln(19/16)/1250x = 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(1e25kt/3)/(3e25kt/3), where k=3ln(3)/250x = 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(1e125kt)/(3e125kt), where k=ln(3)/1250x = 400 \left( 1 - e ^ { - 125 k t } \right) / \left( 3 - e ^ { - 125 k t } \right) , \text { where } k = \ln ( 3 ) / 1250
E) x=800(1e25kt/3)/(4e25kt/3), where k=3ln(7/4)/250x = 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
Question
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)A ( t ) represent the amount of salt in the tank at time t. The correct initial value problem for A(t)A ( t ) is

A) dAdt=2+A/40,A(0)=0.3\frac { d A } { d t } = 2 + A / 40 , A ( 0 ) = 0.3
B) dAdt=2A/40,A(0)=0.3\frac { d A } { d t } = 2 - A / 40 , A ( 0 ) = 0.3
C) dAdt=5+A/40,A(0)=300\frac { d A } { d t } = 5 + A / 40 , A ( 0 ) = 300
D) dAdt=5A/40,A(0)=300\frac { d A } { d t } = 5 - A / 40 , A ( 0 ) = 300
E) dAdt=0.4A/40,A(0)=300\frac { d A } { d t } = 0.4 - A / 40 , A ( 0 ) = 300
Question
The differential equation dPdt=(kcost)P\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=ecksintP = e ^ { c k \sin t }
B) P=cekcostP = c e ^ { k \cos t }
C) P=cekcostP = c e ^ { - k \cos t }
D) P=ceksintP = c e ^ { - k \sin t }
E) P=ceksintP = c e ^ { k \sin t }
Question
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
Question
The solution of the system of differential equations in the two previous problems is

A) x=10e0.1t,y=12.5(e0.1te0.5t),z=1012.5e0.1t+2.5e0.5tx = 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=10e0.1t,y=12.5(e0.5te0.1t),z=1012.5e0.5t+2.5e0.1tx = 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=10e0.5t,y=12.5(e0.5te0.1t),z=1012.5e0.2t+2.5e0.3tx = 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=10e0.5t,y=12.5(e0.1te0.5t),z=1012.5e0.5t+2.5e0.1tx = 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=10e0.5t,y=12.5(e0.1te0.5t),z=1012.5e0.1t+2.5e0.5tx = 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 }
Question
In the competition model dxdt=axbxy,dydt=cydxy\frac { d x } { d t } = a x - b x y , \frac { d y } { d t } = c y - d x y where x(t)x ( t ) and y(t)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
Question
A chicken is taken out of the freezer (0C)\left( 0 ^ { \circ } \mathrm { C } \right) and placed on a table in a 20C20 ^ { \circ } \mathrm { C } room. Ten minutes later the temperature is 2C2 ^ { \circ } \mathrm { C } . It warms according to Newton's Law. How long does it take before the temperature reaches 15C15 ^ { \circ } \mathrm { C } ?

A) 122 minutes
B) 127 minutes
C) 132 minutes
D) 137 minutes
E) 142 minutes
Question
An object is taken out of a 65F65 ^ { \circ } \mathrm { F } room and placed outside where the temperature is 35F35 ^ { \circ } \mathrm { F } room. Five minutes later the temperature is 63F63 ^ { \circ } \mathrm { F } . It cools according to Newton's Law. The temperature of the object after one hour is

A) 50.5F50.5 ^ { \circ } \mathrm { F }
B) 49.9F49.9 ^ { \circ } \mathrm { F }
C) 49.3F49.3 ^ { \circ } \mathrm { F }
D) 48.7F48.7 ^ { \circ } \mathrm { F }
E) 48.1F48.1 ^ { \circ } \mathrm { F }
Question
The solution of the logistic equation dPdt=P(82P)\frac { d P } { d t } = P ( 8 - 2 P ) with initial condition P(0)=2P ( 0 ) = 2 is

A) P=4/(2e8t)P = 4 / \left( 2 - e ^ { - 8 t } \right)
B) P=2/(8+e8t)P = 2 / \left( 8 + e ^ { - 8 t } \right)
C) P=8/(2+e8t)P = 8 / \left( 2 + e ^ { - 8 t } \right)
D) P=8/(2e8t)P = 8 / \left( 2 - e ^ { - 8 t } \right)
E) P=4/(1+e8t)P = 4 / \left( 1 + e ^ { - 8 t } \right)
Question
A tank contains 50 gallons of water in which 2 pounds of salt is dissolved. A brine solution containing 1.5 pounds of salt per gallon of water is pumped into the tank at the rate of 4 gallons per minute, and the well-stirred mixture is pumped out at the same rate. Let A(t)A ( t ) represent the amount of salt in the tank at time t. The correct initial value problem for A(t)A ( t ) is

A) dAdt=62A/25,A(0)=2\frac { d A } { d t } = 6 - 2 A / 25 , A ( 0 ) = 2
B) dAdt=6+2A/25,A(0)=2\frac { d A } { d t } = 6 + 2 A / 25 , A ( 0 ) = 2
C) dAdt=4+2A/25,A(0)=2\frac { d A } { d t } = 4 + 2 A / 25 , A ( 0 ) = 2
D) dAdt=1.52A/25,A(0)=0\frac { d A } { d t } = 1.5 - 2 A / 25 , A ( 0 ) = 0
E) dAdt=42A/25,A(0)=0\frac { d A } { d t } = 4 - 2 A / 25 , A ( 0 ) = 0
Question
Tank A contains 80 gallons of water in which 20 pounds of salt has been dissolved. Tank B contains 30 gallons of water in which 5 pounds of salt has been dissolved. A brine mixture with a concentration of 0.5 pounds of salt per gallon of water is pumped into tank A at the rate of 4 gallons per minute. The well-mixed solution is then pumped from tank A to tank B at the rate of 6 gallons per minute. The solution from tank B is also pumped through another pipe into tank A at the rate of 2 gallons per minute, and the solution from tank B is also pumped out of the system at the rate of 4 gallons per minute. The correct differential equations with initial conditions for the amounts, x(t)x ( t ) and y(t)y ( t ) , of salt in tanks A and B, respectively, at time t are

A) dxdt=2x/40+y/5,dydt=x/40y/3,x(0)=20,y(0)=5\frac { d x } { d t } = 2 - x / 40 + y / 5 , \frac { d y } { d t } = x / 40 - y / 3 , x ( 0 ) = 20 , y ( 0 ) = 5
B) dxdt=23x/40+y/15,dydt=3x/40y/5,x(0)=20,y(0)=5\frac { d x } { d t } = 2 - 3 x / 40 + y / 15 , \frac { d y } { d t } = 3 x / 40 - y / 5 , x ( 0 ) = 20 , y ( 0 ) = 5
C) dxdt=43x/40+y/15,dydt=3x/40y/5,x(0)=20,y(0)=5\frac { d x } { d t } = 4 - 3 x / 40 + y / 15 , \frac { d y } { d t } = 3 x / 40 - y / 5 , x ( 0 ) = 20 , y ( 0 ) = 5
D) dxdt=4x/40+y/5,dydt=x/40y/3,x(0)=20,y(0)=5\frac { d x } { d t } = 4 - x / 40 + y / 5 , \frac { d y } { d t } = x / 40 - y / 3 , x ( 0 ) = 20 , y ( 0 ) = 5
E) dxdt=23x/40+y/5,dydt=x/40y/5,x(0)=20,y(0)=5\frac { d x } { d t } = 2 - 3 x / 40 + y / 5 , \frac { d y } { d t } = x / 40 - y / 5 , x ( 0 ) = 20 , y ( 0 ) = 5
Question
In the logistic model for population growth, dPdt=P(82P)\frac { d P } { d t } = P ( 8 - 2 P ) , the carrying capacity of the population P(t)P ( t ) is

A) 8
B) 2
C) 4
D) 1/41 / 4
E) 16
Question
Radioactive element X decays to element Y with decay constant -0.3. Y decays to stable element Z with decay constant 0.2- 0.2 . What is the system of differential equations for the amounts, x(t),y(t),z(t)x ( t ) , y ( t ) , z ( t ) of the elements X, Y, Z, respectively, at time t.

A) dxdt=0.2x,dydt=0.2x0.3y,dzdt=0.2y\frac { d x } { d t } = - 0.2 x , \frac { d y } { d t } = 0.2 x - 0.3 y , \frac { d z } { d t } = 0.2 y
B) dxdt=0.2x,dydt=0.2x0.3y,dzdt=0.3y\frac { d x } { d t } = - 0.2 x , \frac { d y } { d t } = 0.2 x - 0.3 y , \frac { d z } { d t } = 0.3 y
C) dxdt=0.3x,dydt=0.3x0.2y,dzdt=0.2y\frac { d x } { d t } = - 0.3 x , \frac { d y } { d t } = 0.3 x - 0.2 y , \frac { d z } { d t } = 0.2 y
D) dxdt=0.3x,dydt=0.3x0.2y,dzdt=0.3y\frac { d x } { d t } = - 0.3 x , \frac { d y } { d t } = 0.3 x - 0.2 y , \frac { d z } { d t } = 0.3 y
E) dxdt=0.3y,dydt=0.3x0.2y,dzdt=0.2y\frac { d x } { d t } = - 0.3 y , \frac { d y } { d t } = 0.3 x - 0.2 y , \frac { d z } { d t } = 0.2 y
Question
In the previous problem, how much salt will there be in tanks A and B after a long period of time?

A) 5 pounds in A, 20 pounds in B
B) 20 pounds in A, 5 pounds in B
C) 5 pounds in A, 40 pounds in B
D) 40 pounds in A, 15 pounds in B
E) none of the above
Question
In the previous problem, the amount of chemical C,X(t)C , X ( t ) , produced by time t is

A) x=800(1e400kt)/(2e400kt), where k=ln(29/26)/4000x = 800 \left( 1 - e ^ { - 400 k t } \right) / \left( 2 - e ^ { - 400 k t } \right) , \text { where } k = \ln ( 29 / 26 ) / 4000
B) x=800(1e800kt)/(4e800kt), where k=ln(29/20)/8000x = 800 \left( 1 - e ^ { - 800 k t } \right) / \left( 4 - e ^ { - 800 k t } \right) , \text { where } k = \ln ( 29 / 20 ) / 8000
C) x=400(1e800kt)/(4e800kt), where k=ln(29/20)/8000x = 400 \left( 1 - e ^ { - 800 k t } \right) / \left( 4 - e ^ { - 800 k t } \right) , \text { where } k = \ln ( 29 / 20 ) / 8000
D) x=400(1e400kt)/(4e400kt), where k=ln(29/26)/4000x = 400 \left( 1 - e ^ { - 400 k t } \right) / \left( 4 - e ^ { - 400 k t } \right) , \text { where } k = \ln ( 29 / 26 ) / 4000
E) x=600(1e800kt)/(4e800kt), where k=ln(29/20)/8000x = 600 \left( 1 - e ^ { - 800 k t } \right) / \left( 4 - e ^ { - 800 k t } \right) , \text { where } k = \ln ( 29 / 20 ) / 8000
Question
In Newton's Law of cooling, dTdt=k(TTm)\frac { d T } { d t } = k \left( T - T _ { m } \right) the constant k is

A) a constant of integration evaluated from an initial condition
B) a constant of integration evaluated from another condition
C) a proportionality constant evaluated from an initial condition
D) a proportionality constant evaluated from another condition
Question
In the Lotka-Volterra predator-prey model dxdt=ax+bxy,dydt=eycxy\frac { d x } { d t } = - a x + b x y , \frac { d y } { d t } = e y - c x y , where x(t)x ( t ) is the predator population and y(t)y ( t ) is the prey population, the coefficient c 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
Question
The population of a certain town doubles in 14 years. How long will it take for the population to triple? Assume that the rate of increase of the population is proportional to the population.

A) 18.2 years
B) 20.2 years
C) 22.2 years
D) 23.2 years
E) 24.2 years
Question
In the previous problem, how much salt will there be in the tank after a long period of time?

A) 2 pounds
B) 50 pounds
C) 75 pounds
D) 200 pounds
E) none of the above
Question
In the competition model dxdt=ax+bxy,dydt=eycxy\frac { d x } { d t } = - a x + b x y , \frac { d y } { d t } = e y - c x y where x(t)x ( t ) and y(t)y ( t ) are the populations of the competing species, moose and deer, respectively, the coefficient c 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
Question
The half-life of radium is 1700 years. Assume that the decay rate is proportional to the amount. An initial amount of 5 grams of radium deca to 3 grams in

A) 850 years
B) 1050 years
C) 1150 years
D) 1250 years
E) 1350 years
Question
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 200 grams of A and 300 grams of B, and, during the reaction, for each gram of A used up in the conversion, there are three grams of B used up. An experiments shows that 75 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) 200 grams of A, 0 grams of B, 300 grams of C
B) 0 grams of A, 0 grams of B, 500 grams of C
C) 100 grams of A, 0 grams of B, 400 grams of C
D) 0 grams of A, 100 grams of B, 400 grams of C
E) 0 grams of A, 200 grams of B, 300 grams of C
Question
In the previous two problems, the amount of salt in the tank at time t is

A) A(t)=75+77e2t/25A ( t ) = - 75 + 77 e ^ { 2 t / 25 }
B) A(t)=7573e2t/25A ( t ) = 75 - 73 e ^ { - 2 t / 25 }
C) A(t)=5048e2t/25A ( t ) = 50 - 48 e ^ { - 2 t / 25 }
D) A(t)=50+52e2t/25A ( t ) = - 50 + 52 e ^ { 2 t / 25 }
E) A(t)=7573e2t/25A ( t ) = 75 - 73 e ^ { 2 t / 25 }
Question
In the previous problem, the solution of the initial value problem is

A) x=16t2+40t+200x = 16 t ^ { 2 } + 40 t + 200
B) x=16t2+200t+40x = - 16 t ^ { 2 } + 200 t + 40
C) x=16t2+40t+200x = - 16 t ^ { 2 } + 40 t + 200
D) x=32t2+40t+200x = 32 t ^ { 2 } + 40 t + 200
E) x=32t2+40t+200x = - 32 t ^ { 2 } + 40 t + 200
Question
A ball is thrown upward from the top of a 200 foot tall building with a velocity of 40 feet per second. Take the positive direction upward and the origin of the coordinate system at ground level. What is the initial value problem for the position, x(t)x ( t ) , of the ball at time t?

A) d2xdt2=40,x(0)=200,dxdt(0)=40\frac { d ^ { 2 } x } { d t ^ { 2 } } = 40 , x ( 0 ) = 200 , \frac { d x } { d t } ( 0 ) = 40
B) d2xdt2=40,x(0)=200,dxdt(0)=40\frac { d ^ { 2 } x } { d t ^ { 2 } } = - 40 , x ( 0 ) = 200 , \frac { d x } { d t } ( 0 ) = 40
C) d2xdt2=32,x(0)=200,dxdt(0)=40\frac { d ^ { 2 } x } { d t ^ { 2 } } = 32 , x ( 0 ) = 200 , \frac { d x } { d t } ( 0 ) = 40
D) d2xdt2=32,x(0)=200,xAdt(0)=40\frac { d ^ { 2 } x } { d t ^ { 2 } } = - 32 , x ( 0 ) = 200 , \frac { x A } { d t } ( 0 ) = 40
E) d2xdt2=200,x(0)=32,dxdt(0)=40\frac { d ^ { 2 } x } { d t ^ { 2 } } = 200 , x ( 0 ) = 32 , \frac { d x } { d t } ( 0 ) = 40
Question
The solution of the system of differential equations in the previous problem is

A) x=c1e0.3t,y=3c1e03t+c2e02t,z=c33c1e0.3tc2e0.2tx = c _ { 1 } e ^ { - 0.3 t } , y = - 3 c _ { 1 } e ^ { - 03 t } + c _ { 2 } e ^ { - 02 t } , z = c _ { 3 } - 3 c _ { 1 } e ^ { - 0.3 t } - c _ { 2 } e ^ { - 0.2 t } .

B) x=c1e02t,y=2c1e0.2t+c2e0.3t,z=c3+2c1e0.3tc2e0.2tx = c _ { 1 } e ^ { - 02 t } , y = - 2 c _ { 1 } e ^ { - 0.2 t } + c _ { 2 } e ^ { - 0.3 t } , z = c _ { 3 } + 2 c _ { 1 } e ^ { - 0.3 t } - c _ { 2 } e ^ { - 0.2 t } .

C) x=c1e0.3t,y=3c1e03t+c2e02t,z=c3+2c1e0.3t+c2e0.2tx = c _ { 1 } e ^ { - 0.3 t } , y = - 3 c _ { 1 } e ^ { - 03 t } + c _ { 2 } e ^ { - 02 t } , z = c _ { 3 } + 2 c _ { 1 } e ^ { - 0.3 t } + c _ { 2 } e ^ { - 0.2 t }

D) x=c1e0.3t,y=3c1e03t+c2e02t,z=c3+2c1e0.3tc2e0.2tx = c _ { 1 } e ^ { - 0.3 t } , y = - 3 c _ { 1 } e ^ { - 03 t } + c _ { 2 } e ^ { - 02 t } , z = c _ { 3 } + 2 c _ { 1 } e ^ { - 0.3 t } - c _ { 2 } e ^ { - 0.2 t } .

E) x=c1e02t,y=2c1e0.2t+c2e0.3t,z=c3+2c1e0.3tc2e0.2tx = c _ { 1 } e ^ { - 02 t } , y = - 2 c _ { 1 } e ^ { - 0.2 t } + c _ { 2 } e ^ { - 0.3 t } , z = c _ { 3 } + 2 c _ { 1 } e ^ { - 0.3 t } - c _ { 2 } e ^ { - 0.2 t }
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Deck 3: Modeling With First-Order Differential Equations
1
The amount of salt in the tank at time t in the previous two problems is

A) A(t)=200+200.3et/40A ( t ) = - 200 + 200.3 e ^ { t / 40 }
B) A(t)=200199.7et/40A ( t ) = 200 - 199.7 e ^ { - t / 40 }
C) A(t)=8079.7et/40A ( t ) = 80 - 79.7 e ^ { - t / 40 }
D) A(t)=80+80.3et/25A ( t ) = - 80 + 80.3 e ^ { t / 25 }
E) A(t)=200+100et/40A ( t ) = 200 + 100 e ^ { - t / 40 }
A(t)=8079.7et/40A ( t ) = 80 - 79.7 e ^ { - t / 40 }
2
The solution of the equation dPdt=P(123P)\frac { d P } { d t } = P ( 12 - 3 P ) with initial condition P(0)=3P ( 0 ) = 3 is

A) P=12/(3+e12t)P = 12 / \left( 3 + e ^ { - 12 t } \right)
B) P=4/(3+e12t)P = 4 / \left( 3 + e ^ { - 12 t } \right)
C) P=4/(3e12t)P = 4 / \left( 3 - e ^ { - 12 t } \right)
D) P=3/(12+e12t)P = 3 / \left( 12 + e ^ { - 12 t } \right)
E) P=3/(4+e12t)P = 3 / \left( 4 + e ^ { - 12 t } \right)
P=12/(3+e12t)P = 12 / \left( 3 + e ^ { - 12 t } \right)
3
In the logistic model for population growth, dPdt=P(123P)\frac { d P } { d t } = P ( 12 - 3 P ) , what is the carrying capacity of the population P(t)P ( t ) ?

A) 4
B) 1/41 / 4
C) 12
D) 3
4
4
Radioactive element X decays to element Y with decay constant 0.5- 0.5 . Y, in turn, decays to stable element Z with decay constant 0.1- 0.1 . What is the system of differential equations for the amounts, x(t),y(t),z(t)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)=0x ( 0 ) = 10 , y ( 0 ) = 0 , z ( 0 ) = 0 .

A) dxdt=0.5x,dydt=0.1x0.5y,dzdt=0.2y\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) dxdt=0.5x,dydt=0.5x0.1y,dzdt=0.5y\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) dxdt=0.5x,dydt=0.5x0.1y,dzdt=0.1y\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) dxdt=0.1x,dydt=0.1x0.5y,dzdt=0.1y\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) dxdt=0.1y,dydt=0.5x0.1z,dzdt=0.5y\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
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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
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6
A chicken is taken out of the freezer (0C)\left( 0 ^ { \circ } \mathrm { C } \right) and placed on a table in a 23C23 ^ { \circ } \mathrm { C } room. Forty-five minutes later the temperature is 10C10 ^ { \circ } \mathrm { C } . It warms according to Newton's Law. How long does it take before the temperature reaches 20C20 ^ { \circ } \mathrm { C } ?

A) 147 minutes
B) 153 minutes
C) 157 minutes
D) 161 minutes
E) 165 minutes
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7
In the Lotka-Volterra predator-prey model dxdt=ax+bxy,dydt=eycxy\frac { d x } { d t } = - a x + b x y , \frac { d y } { d t } = e y - c x y , where x(t)x ( t ) is the predator population and y(t)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
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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
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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
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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=5x = 0 , y = 5 , z = 5
B) x=5,y=5,z=0x = 5 , y = 5 , z = 0
C) x=5,y=0,z=5x = 5 , y = 0 , z = 5
D) x=0,y=0,z=10x = 0 , y = 0 , z = 10
E) none of the above
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11
An object is taken out of a 21C21 ^ { \circ } \mathrm { C } room and placed outside where the temperature is 4C4 ^ { \circ } \mathrm { C } room. Twenty-five minutes later the temperature is 17C17 ^ { \circ } \mathrm { C } . It cools according to Newton's Law. The temperature of the object after one hour is

A) 12.2C12.2 ^ { \circ } \mathrm { C }
B) 12.9C12.9 ^ { \circ } \mathrm { C }
C) 13.6C13.6 ^ { \circ } \mathrm { C }
D) 14.3C14.3 ^ { \circ } \mathrm { C }
E) 15.0C15.0 ^ { \circ } \mathrm { C }
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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)x ( t ) and y(t)y ( t ) , of salt in tanks A and B, respectively, at time t are

A) dxdt=32x/25+y/5,dydt=x/25y/15,x(0)=2,y(0)=3\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) dxdt=3x/25+y/15,dydt=2x/252y/15,x(0)=2,y(0)=3\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) dxdt=2.42x/25+y/30,dydt=2x/252y/15,x(0)=2,y(0)=3\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) dxdt=2.4x/50+y/30,dydt=x/40y/3,x(0)=2,y(0)=3\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) dxdt=2.4x/25+y/15,dydt=x/50y/30,x(0)=2,y(0)=3\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
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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
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14
In Newton's Law of cooling, dTdt=k(TTm),Tm\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
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15
In the previous problem, the amount of chemical C,X(t)C , X ( t ) , produced by time t is

A) x=2000(1e25kt/3)/(1615e25kt/3), where k=3ln(5/4)/250x = 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(1e125kt)/(4e125kt), where k=ln(19/16)/1250x = 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(1e25kt/3)/(3e25kt/3), where k=3ln(3)/250x = 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(1e125kt)/(3e125kt), where k=ln(3)/1250x = 400 \left( 1 - e ^ { - 125 k t } \right) / \left( 3 - e ^ { - 125 k t } \right) , \text { where } k = \ln ( 3 ) / 1250
E) x=800(1e25kt/3)/(4e25kt/3), where k=3ln(7/4)/250x = 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
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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)A ( t ) represent the amount of salt in the tank at time t. The correct initial value problem for A(t)A ( t ) is

A) dAdt=2+A/40,A(0)=0.3\frac { d A } { d t } = 2 + A / 40 , A ( 0 ) = 0.3
B) dAdt=2A/40,A(0)=0.3\frac { d A } { d t } = 2 - A / 40 , A ( 0 ) = 0.3
C) dAdt=5+A/40,A(0)=300\frac { d A } { d t } = 5 + A / 40 , A ( 0 ) = 300
D) dAdt=5A/40,A(0)=300\frac { d A } { d t } = 5 - A / 40 , A ( 0 ) = 300
E) dAdt=0.4A/40,A(0)=300\frac { d A } { d t } = 0.4 - A / 40 , A ( 0 ) = 300
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17
The differential equation dPdt=(kcost)P\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=ecksintP = e ^ { c k \sin t }
B) P=cekcostP = c e ^ { k \cos t }
C) P=cekcostP = c e ^ { - k \cos t }
D) P=ceksintP = c e ^ { - k \sin t }
E) P=ceksintP = c e ^ { k \sin t }
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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
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19
The solution of the system of differential equations in the two previous problems is

A) x=10e0.1t,y=12.5(e0.1te0.5t),z=1012.5e0.1t+2.5e0.5tx = 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=10e0.1t,y=12.5(e0.5te0.1t),z=1012.5e0.5t+2.5e0.1tx = 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=10e0.5t,y=12.5(e0.5te0.1t),z=1012.5e0.2t+2.5e0.3tx = 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=10e0.5t,y=12.5(e0.1te0.5t),z=1012.5e0.5t+2.5e0.1tx = 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=10e0.5t,y=12.5(e0.1te0.5t),z=1012.5e0.1t+2.5e0.5tx = 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 }
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20
In the competition model dxdt=axbxy,dydt=cydxy\frac { d x } { d t } = a x - b x y , \frac { d y } { d t } = c y - d x y where x(t)x ( t ) and y(t)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
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21
A chicken is taken out of the freezer (0C)\left( 0 ^ { \circ } \mathrm { C } \right) and placed on a table in a 20C20 ^ { \circ } \mathrm { C } room. Ten minutes later the temperature is 2C2 ^ { \circ } \mathrm { C } . It warms according to Newton's Law. How long does it take before the temperature reaches 15C15 ^ { \circ } \mathrm { C } ?

A) 122 minutes
B) 127 minutes
C) 132 minutes
D) 137 minutes
E) 142 minutes
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22
An object is taken out of a 65F65 ^ { \circ } \mathrm { F } room and placed outside where the temperature is 35F35 ^ { \circ } \mathrm { F } room. Five minutes later the temperature is 63F63 ^ { \circ } \mathrm { F } . It cools according to Newton's Law. The temperature of the object after one hour is

A) 50.5F50.5 ^ { \circ } \mathrm { F }
B) 49.9F49.9 ^ { \circ } \mathrm { F }
C) 49.3F49.3 ^ { \circ } \mathrm { F }
D) 48.7F48.7 ^ { \circ } \mathrm { F }
E) 48.1F48.1 ^ { \circ } \mathrm { F }
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23
The solution of the logistic equation dPdt=P(82P)\frac { d P } { d t } = P ( 8 - 2 P ) with initial condition P(0)=2P ( 0 ) = 2 is

A) P=4/(2e8t)P = 4 / \left( 2 - e ^ { - 8 t } \right)
B) P=2/(8+e8t)P = 2 / \left( 8 + e ^ { - 8 t } \right)
C) P=8/(2+e8t)P = 8 / \left( 2 + e ^ { - 8 t } \right)
D) P=8/(2e8t)P = 8 / \left( 2 - e ^ { - 8 t } \right)
E) P=4/(1+e8t)P = 4 / \left( 1 + e ^ { - 8 t } \right)
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24
A tank contains 50 gallons of water in which 2 pounds of salt is dissolved. A brine solution containing 1.5 pounds of salt per gallon of water is pumped into the tank at the rate of 4 gallons per minute, and the well-stirred mixture is pumped out at the same rate. Let A(t)A ( t ) represent the amount of salt in the tank at time t. The correct initial value problem for A(t)A ( t ) is

A) dAdt=62A/25,A(0)=2\frac { d A } { d t } = 6 - 2 A / 25 , A ( 0 ) = 2
B) dAdt=6+2A/25,A(0)=2\frac { d A } { d t } = 6 + 2 A / 25 , A ( 0 ) = 2
C) dAdt=4+2A/25,A(0)=2\frac { d A } { d t } = 4 + 2 A / 25 , A ( 0 ) = 2
D) dAdt=1.52A/25,A(0)=0\frac { d A } { d t } = 1.5 - 2 A / 25 , A ( 0 ) = 0
E) dAdt=42A/25,A(0)=0\frac { d A } { d t } = 4 - 2 A / 25 , A ( 0 ) = 0
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25
Tank A contains 80 gallons of water in which 20 pounds of salt has been dissolved. Tank B contains 30 gallons of water in which 5 pounds of salt has been dissolved. A brine mixture with a concentration of 0.5 pounds of salt per gallon of water is pumped into tank A at the rate of 4 gallons per minute. The well-mixed solution is then pumped from tank A to tank B at the rate of 6 gallons per minute. The solution from tank B is also pumped through another pipe into tank A at the rate of 2 gallons per minute, and the solution from tank B is also pumped out of the system at the rate of 4 gallons per minute. The correct differential equations with initial conditions for the amounts, x(t)x ( t ) and y(t)y ( t ) , of salt in tanks A and B, respectively, at time t are

A) dxdt=2x/40+y/5,dydt=x/40y/3,x(0)=20,y(0)=5\frac { d x } { d t } = 2 - x / 40 + y / 5 , \frac { d y } { d t } = x / 40 - y / 3 , x ( 0 ) = 20 , y ( 0 ) = 5
B) dxdt=23x/40+y/15,dydt=3x/40y/5,x(0)=20,y(0)=5\frac { d x } { d t } = 2 - 3 x / 40 + y / 15 , \frac { d y } { d t } = 3 x / 40 - y / 5 , x ( 0 ) = 20 , y ( 0 ) = 5
C) dxdt=43x/40+y/15,dydt=3x/40y/5,x(0)=20,y(0)=5\frac { d x } { d t } = 4 - 3 x / 40 + y / 15 , \frac { d y } { d t } = 3 x / 40 - y / 5 , x ( 0 ) = 20 , y ( 0 ) = 5
D) dxdt=4x/40+y/5,dydt=x/40y/3,x(0)=20,y(0)=5\frac { d x } { d t } = 4 - x / 40 + y / 5 , \frac { d y } { d t } = x / 40 - y / 3 , x ( 0 ) = 20 , y ( 0 ) = 5
E) dxdt=23x/40+y/5,dydt=x/40y/5,x(0)=20,y(0)=5\frac { d x } { d t } = 2 - 3 x / 40 + y / 5 , \frac { d y } { d t } = x / 40 - y / 5 , x ( 0 ) = 20 , y ( 0 ) = 5
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26
In the logistic model for population growth, dPdt=P(82P)\frac { d P } { d t } = P ( 8 - 2 P ) , the carrying capacity of the population P(t)P ( t ) is

A) 8
B) 2
C) 4
D) 1/41 / 4
E) 16
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27
Radioactive element X decays to element Y with decay constant -0.3. Y decays to stable element Z with decay constant 0.2- 0.2 . What is the system of differential equations for the amounts, x(t),y(t),z(t)x ( t ) , y ( t ) , z ( t ) of the elements X, Y, Z, respectively, at time t.

A) dxdt=0.2x,dydt=0.2x0.3y,dzdt=0.2y\frac { d x } { d t } = - 0.2 x , \frac { d y } { d t } = 0.2 x - 0.3 y , \frac { d z } { d t } = 0.2 y
B) dxdt=0.2x,dydt=0.2x0.3y,dzdt=0.3y\frac { d x } { d t } = - 0.2 x , \frac { d y } { d t } = 0.2 x - 0.3 y , \frac { d z } { d t } = 0.3 y
C) dxdt=0.3x,dydt=0.3x0.2y,dzdt=0.2y\frac { d x } { d t } = - 0.3 x , \frac { d y } { d t } = 0.3 x - 0.2 y , \frac { d z } { d t } = 0.2 y
D) dxdt=0.3x,dydt=0.3x0.2y,dzdt=0.3y\frac { d x } { d t } = - 0.3 x , \frac { d y } { d t } = 0.3 x - 0.2 y , \frac { d z } { d t } = 0.3 y
E) dxdt=0.3y,dydt=0.3x0.2y,dzdt=0.2y\frac { d x } { d t } = - 0.3 y , \frac { d y } { d t } = 0.3 x - 0.2 y , \frac { d z } { d t } = 0.2 y
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28
In the previous problem, how much salt will there be in tanks A and B after a long period of time?

A) 5 pounds in A, 20 pounds in B
B) 20 pounds in A, 5 pounds in B
C) 5 pounds in A, 40 pounds in B
D) 40 pounds in A, 15 pounds in B
E) none of the above
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29
In the previous problem, the amount of chemical C,X(t)C , X ( t ) , produced by time t is

A) x=800(1e400kt)/(2e400kt), where k=ln(29/26)/4000x = 800 \left( 1 - e ^ { - 400 k t } \right) / \left( 2 - e ^ { - 400 k t } \right) , \text { where } k = \ln ( 29 / 26 ) / 4000
B) x=800(1e800kt)/(4e800kt), where k=ln(29/20)/8000x = 800 \left( 1 - e ^ { - 800 k t } \right) / \left( 4 - e ^ { - 800 k t } \right) , \text { where } k = \ln ( 29 / 20 ) / 8000
C) x=400(1e800kt)/(4e800kt), where k=ln(29/20)/8000x = 400 \left( 1 - e ^ { - 800 k t } \right) / \left( 4 - e ^ { - 800 k t } \right) , \text { where } k = \ln ( 29 / 20 ) / 8000
D) x=400(1e400kt)/(4e400kt), where k=ln(29/26)/4000x = 400 \left( 1 - e ^ { - 400 k t } \right) / \left( 4 - e ^ { - 400 k t } \right) , \text { where } k = \ln ( 29 / 26 ) / 4000
E) x=600(1e800kt)/(4e800kt), where k=ln(29/20)/8000x = 600 \left( 1 - e ^ { - 800 k t } \right) / \left( 4 - e ^ { - 800 k t } \right) , \text { where } k = \ln ( 29 / 20 ) / 8000
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30
In Newton's Law of cooling, dTdt=k(TTm)\frac { d T } { d t } = k \left( T - T _ { m } \right) the constant k is

A) a constant of integration evaluated from an initial condition
B) a constant of integration evaluated from another condition
C) a proportionality constant evaluated from an initial condition
D) a proportionality constant evaluated from another condition
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31
In the Lotka-Volterra predator-prey model dxdt=ax+bxy,dydt=eycxy\frac { d x } { d t } = - a x + b x y , \frac { d y } { d t } = e y - c x y , where x(t)x ( t ) is the predator population and y(t)y ( t ) is the prey population, the coefficient c 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
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32
The population of a certain town doubles in 14 years. How long will it take for the population to triple? Assume that the rate of increase of the population is proportional to the population.

A) 18.2 years
B) 20.2 years
C) 22.2 years
D) 23.2 years
E) 24.2 years
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33
In the previous problem, how much salt will there be in the tank after a long period of time?

A) 2 pounds
B) 50 pounds
C) 75 pounds
D) 200 pounds
E) none of the above
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34
In the competition model dxdt=ax+bxy,dydt=eycxy\frac { d x } { d t } = - a x + b x y , \frac { d y } { d t } = e y - c x y where x(t)x ( t ) and y(t)y ( t ) are the populations of the competing species, moose and deer, respectively, the coefficient c 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
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35
The half-life of radium is 1700 years. Assume that the decay rate is proportional to the amount. An initial amount of 5 grams of radium deca to 3 grams in

A) 850 years
B) 1050 years
C) 1150 years
D) 1250 years
E) 1350 years
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36
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 200 grams of A and 300 grams of B, and, during the reaction, for each gram of A used up in the conversion, there are three grams of B used up. An experiments shows that 75 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) 200 grams of A, 0 grams of B, 300 grams of C
B) 0 grams of A, 0 grams of B, 500 grams of C
C) 100 grams of A, 0 grams of B, 400 grams of C
D) 0 grams of A, 100 grams of B, 400 grams of C
E) 0 grams of A, 200 grams of B, 300 grams of C
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37
In the previous two problems, the amount of salt in the tank at time t is

A) A(t)=75+77e2t/25A ( t ) = - 75 + 77 e ^ { 2 t / 25 }
B) A(t)=7573e2t/25A ( t ) = 75 - 73 e ^ { - 2 t / 25 }
C) A(t)=5048e2t/25A ( t ) = 50 - 48 e ^ { - 2 t / 25 }
D) A(t)=50+52e2t/25A ( t ) = - 50 + 52 e ^ { 2 t / 25 }
E) A(t)=7573e2t/25A ( t ) = 75 - 73 e ^ { 2 t / 25 }
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38
In the previous problem, the solution of the initial value problem is

A) x=16t2+40t+200x = 16 t ^ { 2 } + 40 t + 200
B) x=16t2+200t+40x = - 16 t ^ { 2 } + 200 t + 40
C) x=16t2+40t+200x = - 16 t ^ { 2 } + 40 t + 200
D) x=32t2+40t+200x = 32 t ^ { 2 } + 40 t + 200
E) x=32t2+40t+200x = - 32 t ^ { 2 } + 40 t + 200
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39
A ball is thrown upward from the top of a 200 foot tall building with a velocity of 40 feet per second. Take the positive direction upward and the origin of the coordinate system at ground level. What is the initial value problem for the position, x(t)x ( t ) , of the ball at time t?

A) d2xdt2=40,x(0)=200,dxdt(0)=40\frac { d ^ { 2 } x } { d t ^ { 2 } } = 40 , x ( 0 ) = 200 , \frac { d x } { d t } ( 0 ) = 40
B) d2xdt2=40,x(0)=200,dxdt(0)=40\frac { d ^ { 2 } x } { d t ^ { 2 } } = - 40 , x ( 0 ) = 200 , \frac { d x } { d t } ( 0 ) = 40
C) d2xdt2=32,x(0)=200,dxdt(0)=40\frac { d ^ { 2 } x } { d t ^ { 2 } } = 32 , x ( 0 ) = 200 , \frac { d x } { d t } ( 0 ) = 40
D) d2xdt2=32,x(0)=200,xAdt(0)=40\frac { d ^ { 2 } x } { d t ^ { 2 } } = - 32 , x ( 0 ) = 200 , \frac { x A } { d t } ( 0 ) = 40
E) d2xdt2=200,x(0)=32,dxdt(0)=40\frac { d ^ { 2 } x } { d t ^ { 2 } } = 200 , x ( 0 ) = 32 , \frac { d x } { d t } ( 0 ) = 40
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40
The solution of the system of differential equations in the previous problem is

A) x=c1e0.3t,y=3c1e03t+c2e02t,z=c33c1e0.3tc2e0.2tx = c _ { 1 } e ^ { - 0.3 t } , y = - 3 c _ { 1 } e ^ { - 03 t } + c _ { 2 } e ^ { - 02 t } , z = c _ { 3 } - 3 c _ { 1 } e ^ { - 0.3 t } - c _ { 2 } e ^ { - 0.2 t } .

B) x=c1e02t,y=2c1e0.2t+c2e0.3t,z=c3+2c1e0.3tc2e0.2tx = c _ { 1 } e ^ { - 02 t } , y = - 2 c _ { 1 } e ^ { - 0.2 t } + c _ { 2 } e ^ { - 0.3 t } , z = c _ { 3 } + 2 c _ { 1 } e ^ { - 0.3 t } - c _ { 2 } e ^ { - 0.2 t } .

C) x=c1e0.3t,y=3c1e03t+c2e02t,z=c3+2c1e0.3t+c2e0.2tx = c _ { 1 } e ^ { - 0.3 t } , y = - 3 c _ { 1 } e ^ { - 03 t } + c _ { 2 } e ^ { - 02 t } , z = c _ { 3 } + 2 c _ { 1 } e ^ { - 0.3 t } + c _ { 2 } e ^ { - 0.2 t }

D) x=c1e0.3t,y=3c1e03t+c2e02t,z=c3+2c1e0.3tc2e0.2tx = c _ { 1 } e ^ { - 0.3 t } , y = - 3 c _ { 1 } e ^ { - 03 t } + c _ { 2 } e ^ { - 02 t } , z = c _ { 3 } + 2 c _ { 1 } e ^ { - 0.3 t } - c _ { 2 } e ^ { - 0.2 t } .

E) x=c1e02t,y=2c1e0.2t+c2e0.3t,z=c3+2c1e0.3tc2e0.2tx = c _ { 1 } e ^ { - 02 t } , y = - 2 c _ { 1 } e ^ { - 0.2 t } + c _ { 2 } e ^ { - 0.3 t } , z = c _ { 3 } + 2 c _ { 1 } e ^ { - 0.3 t } - c _ { 2 } e ^ { - 0.2 t }
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