math 3c practice word problems prepared by vince zaccone for campus learning assistance services at...
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Math 3C
Practice Word Problems
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1) A dangerous substance known as Chemical X is lethal if its concentration in the air is 100 parts per million by volume (ppmv). The half-life of chemical X is known to be 7 hours. If I accidentally release a quantity of chemical X in my secret underground lab such that its initial concentration is 1500 ppmv, how long do I have to wait before I can enter the lab again?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1) A dangerous substance known as Chemical X is lethal if its concentration in the air is 100 parts per million by volume (ppmv). The half-life of chemical X is known to be 7 hours. If I accidentally release a quantity of chemical X in my secret underground lab such that its initial concentration is 1500 ppmv, how long do I have to wait before I can enter the lab again?
We are given the half-life, so we can assume exponential decay.
We know the general formula will be something like
Where y(t)=concentration after t hours, and y(0)=1500
kt0 ey)t(y
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1) A dangerous substance known as Chemical X is lethal if its concentration in the air is 100 parts per million by volume (ppmv). The half-life of chemical X is known to be 7 hours. If I accidentally release a quantity of chemical X in my secret underground lab such that its initial concentration is 1500 ppmv, how long do I have to wait before I can enter the lab again?
We are given the half-life, so we can assume exponential decay.
We know the general formula will be something like
Where y(t)=concentration after t hours, and y(0)=1500
kt0 ey)t(y
The half-life is given, so we can find the value for k (it will be negative)
7
)2ln(
7
)ln(k
)ln(k7
yeyy)7(y
2121
021)7(k
0021
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1) A dangerous substance known as Chemical X is lethal if its concentration in the air is 100 parts per million by volume (ppmv). The half-life of chemical X is known to be 7 hours. If I accidentally release a quantity of chemical X in my secret underground lab such that its initial concentration is 1500 ppmv, how long do I have to wait before I can enter the lab again?
We are given the half-life, so we can assume exponential decay.
We know the general formula will be something like
Where y(t)=concentration after t hours, and y(0)=1500
kt0 ey)t(y
The half-life is given, so we can find the value for k (it will be negative)
7
)2ln(
7
)ln(k
)ln(k7
yeyy)7(y
2121
021)7(k
0021
If a half-life is given, the value of k is always
If a doubling time is given, k is always
half
21
t
)ln(k
doublet
)2ln(k
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1) A dangerous substance known as Chemical X is lethal if its concentration in the air is 100 parts per million by volume (ppmv). The half-life of chemical X is known to be 7 hours. If I accidentally release a quantity of chemical X in my secret underground lab such that its initial concentration is 1500 ppmv, how long do I have to wait before I can enter the lab again?
We are given the half-life, so we can assume exponential decay.
We know the general formula will be something like
Where y(t)=concentration after t hours, and y(0)=1500
kt0 ey)t(y
The half-life is given, so we can find the value for k (it will be negative)
7
)2ln(
7
)ln(k
)ln(k7
yeyy)7(y
2121
021)7(k
0021
So our formula is:t
7)2ln(
e1500)t(y
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1) A dangerous substance known as Chemical X is lethal if its concentration in the air is 100 parts per million by volume (ppmv). The half-life of chemical X is known to be 7 hours. If I accidentally release a quantity of chemical X in my secret underground lab such that its initial concentration is 1500 ppmv, how long do I have to wait before I can enter the lab again?
We are given the half-life, so we can assume exponential decay.
We know the general formula will be something like
Where y(t)=concentration after t hours, and y(0)=1500
kt0 ey)t(y
The half-life is given, so we can find the value for k (it will be negative)
7
)2ln(
7
)ln(k
)ln(k7
yeyy)7(y
2121
021)7(k
0021
So our formula is:t
7)2ln(
e1500)t(y
When is the concentration 100? Set y=100 and solve for t.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
1) A dangerous substance known as Chemical X is lethal if its concentration in the air is 100 parts per million by volume (ppmv). The half-life of chemical X is known to be 7 hours. If I accidentally release a quantity of chemical X in my secret underground lab such that its initial concentration is 1500 ppmv, how long do I have to wait before I can enter the lab again?
We are given the half-life, so we can assume exponential decay.
We know the general formula will be something like
Where y(t)=concentration after t hours, and y(0)=1500
kt0 ey)t(y
The half-life is given, so we can find the value for k (it will be negative)
7
)2ln(
7
)ln(k
)ln(k7
yeyy)7(y
2121
021)7(k
0021
So our formula is:t
7)2ln(
e1500)t(y
When is the concentration 100? Set y=100 and solve for t.
hours27tt)2ln(
)15ln(7
t)ln(e
e1500100
7)2ln(
151t
151
t
7)2ln(
7)2ln(
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2) Sally is saving for the down payment on a house. She already has $10,000 in her savings account, and figures that she will need $50,000 for the down payment. Assuming she earns a 4% annual interest rate (compounded continuously), how much should she deposit in her account each year (again, continuously) to end up with enough money for the down payment after 5 years?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2) Sally is saving for the down payment on a house. She already has $10,000 in her savings account, and figures that she will need $50,000 for the down payment. Assuming she earns a 4% annual interest rate (compounded continuously), how much should she deposit in her account each year (again, continuously) to end up with enough money for the down payment after 5 years?
We can set up a DE for the account balance. If y(t)=$ in the account after t years, then
Here d is the annual deposit amount.dy04.0y 000,10)0(y with initial value
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2) Sally is saving for the down payment on a house. She already has $10,000 in her savings account, and figures that she will need $50,000 for the down payment. Assuming she earns a 4% annual interest rate (compounded continuously), how much should she deposit in her account each year (again, continuously) to end up with enough money for the down payment after 5 years?
We can set up a DE for the account balance. If y(t)=$ in the account after t years, then
Here d is the annual deposit amount.dy04.0y 000,10)0(y with initial value
This DE is first-order, linear, and separable. So we have lots of options for solving it.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2) Sally is saving for the down payment on a house. She already has $10,000 in her savings account, and figures that she will need $50,000 for the down payment. Assuming she earns a 4% annual interest rate (compounded continuously), how much should she deposit in her account each year (again, continuously) to end up with enough money for the down payment after 5 years?
We can set up a DE for the account balance. If y(t)=$ in the account after t years, then
Here d is the annual deposit amount.dy04.0y 000,10)0(y with initial value
This DE is first-order, linear, and separable. So we have lots of options for solving it.Let’s use an integrating factor. We need to rewrite the equation in standard form: dy04.0y
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2) Sally is saving for the down payment on a house. She already has $10,000 in her savings account, and figures that she will need $50,000 for the down payment. Assuming she earns a 4% annual interest rate (compounded continuously), how much should she deposit in her account each year (again, continuously) to end up with enough money for the down payment after 5 years?
We can set up a DE for the account balance. If y(t)=$ in the account after t years, then
Here d is the annual deposit amount.dy04.0y 000,10)0(y with initial value
This DE is first-order, linear, and separable. So we have lots of options for solving it.Let’s use an integrating factor. We need to rewrite the equation in standard form:
t04.dt04. ee
dy04.0y
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2) Sally is saving for the down payment on a house. She already has $10,000 in her savings account, and figures that she will need $50,000 for the down payment. Assuming she earns a 4% annual interest rate (compounded continuously), how much should she deposit in her account each year (again, continuously) to end up with enough money for the down payment after 5 years?
We can set up a DE for the account balance. If y(t)=$ in the account after t years, then
Here d is the annual deposit amount.dy04.0y 000,10)0(y with initial value
This DE is first-order, linear, and separable. So we have lots of options for solving it.Let’s use an integrating factor. We need to rewrite the equation in standard form:
t04.dt04. ee
dy04.0y
Multiply through by µ, then integrate and solve for y:
t04.04.
t04.04.
t04.
t04.t04.
t04.t04.dtd
t04.t04.t04.
eCd
y
Ced
ye
dtdeye
deye
deye04.0ye
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2) Sally is saving for the down payment on a house. She already has $10,000 in her savings account, and figures that she will need $50,000 for the down payment. Assuming she earns a 4% annual interest rate (compounded continuously), how much should she deposit in her account each year (again, continuously) to end up with enough money for the down payment after 5 years?
We can set up a DE for the account balance. If y(t)=$ in the account after t years, then
Here d is the annual deposit amount.dy04.0y 000,10)0(y with initial value
This DE is first-order, linear, and separable. So we have lots of options for solving it.Let’s use an integrating factor. We need to rewrite the equation in standard form:
t04.dt04. ee
dy04.0y
Multiply through by µ, then integrate and solve for y:
t04.04.
t04.04.
t04.
t04.t04.
t04.t04.dtd
t04.t04.t04.
eCd
y
Ced
ye
dtdeye
deye
deye04.0ye
Use the initial value to find C:
04.
004.04.
d000,10C
eCd
000,10
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2) Sally is saving for the down payment on a house. She already has $10,000 in her savings account, and figures that she will need $50,000 for the down payment. Assuming she earns a 4% annual interest rate (compounded continuously), how much should she deposit in her account each year (again, continuously) to end up with enough money for the down payment after 5 years?
We can set up a DE for the account balance. If y(t)=$ in the account after t years, then
Here d is the annual deposit amount.dy04.0y 000,10)0(y with initial value
This DE is first-order, linear, and separable. So we have lots of options for solving it.Let’s use an integrating factor. We need to rewrite the equation in standard form:
t04.dt04. ee
dy04.0y
Multiply through by µ, then integrate and solve for y:
t04.04.
t04.04.
t04.
t04.t04.
t04.t04.dtd
t04.t04.t04.
eCd
y
Ced
ye
dtdeye
deye
deye04.0ye
Use the initial value to find C:
04.
004.04.
d000,10C
eCd
000,10
t04.04.04.
ed
000,10d
y
So our solution becomes:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
2) Sally is saving for the down payment on a house. She already has $10,000 in her savings account, and figures that she will need $50,000 for the down payment. Assuming she earns a 4% annual interest rate (compounded continuously), how much should she deposit in her account each year (again, continuously) to end up with enough money for the down payment after 5 years?
We can set up a DE for the account balance. If y(t)=$ in the account after t years, then
Here d is the annual deposit amount.dy04.0y 000,10)0(y with initial value
This DE is first-order, linear, and separable. So we have lots of options for solving it.Let’s use an integrating factor. We need to rewrite the equation in standard form:
t04.dt04. ee
dy04.0y
Multiply through by µ, then integrate and solve for y:
t04.04.
t04.04.
t04.
t04.t04.
t04.t04.dtd
t04.t04.t04.
eCd
y
Ced
ye
dtdeye
deye
deye04.0ye
Use the initial value to find C:
04.
004.04.
d000,10C
eCd
000,10
t04.04.04.
ed
000,10d
y
So our solution becomes:
Now we need to find the d that will give $50,000 at t=5.
year/827,6$1e
)e10000000,50(04.d
ebralga
ed
000,10d
000,50
2.
2.
504.04.04.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
3) My cup of coffee this morning was initially 90 degrees celsius, and after 15 minutes it was 70 degrees Celsius. If the room temperature is 20 degrees Celsius, how long will I have to wait until my coffee is at my ideal drinking temperature of 60 degrees Celsius? Assume that the coffee cools according to Newton’s law of cooling.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
3) My cup of coffee this morning was initially 90 degrees celsius, and after 15 minutes it was 70 degrees Celsius. If the room temperature is 20 degrees Celsius, how long will I have to wait until my coffee is at my ideal drinking temperature of 60 degrees Celsius? Assume that the coffee cools according to Newton’s law of cooling.
For heating or cooling the DE is
)TM(kT T(t)=temperature at time t
M=ambient (constant) temperature
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
3) My cup of coffee this morning was initially 90 degrees celsius, and after 15 minutes it was 70 degrees Celsius. If the room temperature is 20 degrees Celsius, how long will I have to wait until my coffee is at my ideal drinking temperature of 60 degrees Celsius? Assume that the coffee cools according to Newton’s law of cooling.
For heating or cooling the DE is
)TM(kT T(t)=temperature at time t
M=ambient (constant) temperatureIn our problem M=20 and we also know two values for T: T(0)=90 and T(15)=70
Time in minutes
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
3) My cup of coffee this morning was initially 90 degrees celsius, and after 15 minutes it was 70 degrees Celsius. If the room temperature is 20 degrees Celsius, how long will I have to wait until my coffee is at my ideal drinking temperature of 60 degrees Celsius? Assume that the coffee cools according to Newton’s law of cooling.
For heating or cooling the DE is
)TM(kT T(t)=temperature at time t
M=ambient (constant) temperatureIn our problem M=20 and we also know two values for T: T(0)=90 and T(15)=70
Time in minutes
Note that our DE is first-order, linear and separable, so we have lots of options. Let’s try separation this time.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
3) My cup of coffee this morning was initially 90 degrees celsius, and after 15 minutes it was 70 degrees Celsius. If the room temperature is 20 degrees Celsius, how long will I have to wait until my coffee is at my ideal drinking temperature of 60 degrees Celsius? Assume that the coffee cools according to Newton’s law of cooling.
For heating or cooling the DE is
)TM(kT T(t)=temperature at time t
M=ambient (constant) temperatureIn our problem M=20 and we also know two values for T: T(0)=90 and T(15)=70
Time in minutes
Note that our DE is first-order, linear and separable, so we have lots of options. Let’s try separation this time.
kt
kt
dtdT
Ce20T
CeT20
CktT20ln
kdtT20
dT
)T20(k
Note that the constant C in the last line will not be the same number as in the previous lines, but since it is arbitrary, we can still just call it C.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
3) My cup of coffee this morning was initially 90 degrees celsius, and after 15 minutes it was 70 degrees Celsius. If the room temperature is 20 degrees Celsius, how long will I have to wait until my coffee is at my ideal drinking temperature of 60 degrees Celsius? Assume that the coffee cools according to Newton’s law of cooling.
For heating or cooling the DE is
)TM(kT T(t)=temperature at time t
M=ambient (constant) temperatureIn our problem M=20 and we also know two values for T: T(0)=90 and T(15)=70
Time in minutes
Note that our DE is first-order, linear and separable, so we have lots of options. Let’s try separation this time.
kt
kt
dtdT
Ce20T
CeT20
CktT20ln
kdtT20
dT
)T20(k
We can use the initial value to find C.
70CCe2090)0(T )0(k
Now our solution is kte7020)t(T
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
3) My cup of coffee this morning was initially 90 degrees celsius, and after 15 minutes it was 70 degrees Celsius. If the room temperature is 20 degrees Celsius, how long will I have to wait until my coffee is at my ideal drinking temperature of 60 degrees Celsius? Assume that the coffee cools according to Newton’s law of cooling.
For heating or cooling the DE is
)TM(kT T(t)=temperature at time t
M=ambient (constant) temperatureIn our problem M=20 and we also know two values for T: T(0)=90 and T(15)=70
Time in minutes
Note that our DE is first-order, linear and separable, so we have lots of options. Let’s try separation this time.
kt
kt
dtdT
Ce20T
CeT20
CktT20ln
kdtT20
dT
)T20(k
We can use the initial value to find C.
70CCe2090)0(T )0(k
Now our solution is kte7020)t(T
Using T(15)=70 we can find k
15
)ln(kk15)ln(e7050
e702070)15(T
75
75k15
)15(k
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
3) My cup of coffee this morning was initially 90 degrees celsius, and after 15 minutes it was 70 degrees Celsius. If the room temperature is 20 degrees Celsius, how long will I have to wait until my coffee is at my ideal drinking temperature of 60 degrees Celsius? Assume that the coffee cools according to Newton’s law of cooling.
For heating or cooling the DE is
)TM(kT T(t)=temperature at time t
M=ambient (constant) temperatureIn our problem M=20 and we also know two values for T: T(0)=90 and T(15)=70
Time in minutes
Note that our DE is first-order, linear and separable, so we have lots of options. Let’s try separation this time.
kt
kt
dtdT
Ce20T
CeT20
CktT20ln
kdtT20
dT
)T20(k
We can use the initial value to find C.
70CCe2090)0(T )0(k
Now our solution is kte7020)t(T
Using T(15)=70 we can find k
15
)ln(kk15)ln(e7050
e702070)15(T
75
75k15
)15(k
Our final formula is thus
t15
)ln(75
e7020)t(T
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
3) My cup of coffee this morning was initially 90 degrees celsius, and after 15 minutes it was 70 degrees Celsius. If the room temperature is 20 degrees Celsius, how long will I have to wait until my coffee is at my ideal drinking temperature of 60 degrees Celsius? Assume that the coffee cools according to Newton’s law of cooling.
For heating or cooling the DE is
)TM(kT T(t)=temperature at time t
M=ambient (constant) temperatureIn our problem M=20 and we also know two values for T: T(0)=90 and T(15)=70
Time in minutes
Note that our DE is first-order, linear and separable, so we have lots of options. Let’s try separation this time.
kt
kt
dtdT
Ce20T
CeT20
CktT20ln
kdtT20
dT
)T20(k
We can use the initial value to find C.
70CCe2090)0(T )0(k
Now our solution is kte7020)t(T
Using T(15)=70 we can find k
15
)ln(kk15)ln(e7050
e702070)15(T
75
75k15
)15(k
Our final formula is thus
t15
)ln(75
e7020)t(T
Finally we can answer our question – set T=60 and solve
min25)ln(
)ln(15t
t15
)ln()ln(e702060
75
74
75
74
t15
)ln(75
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
saltwater input
saltwater output
Mixing problems will always have the same basic form for their DE: Total Rate = (Rate In) – (Rate Out)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
saltwater input
saltwater output
Mixing problems will always have the same basic form for their DE: Total Rate = (Rate In) – (Rate Out)
The input and output rates will typically be calculated as (concentration) x (flow rate)
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
saltwater input
saltwater output
Mixing problems will always have the same basic form for their DE: Total Rate = (Rate In) – (Rate Out)
The input and output rates will typically be calculated as (concentration) x (flow rate)
In this case we are keeping track of grams of salt, so concentration should have units of grams/liter.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
saltwater input
saltwater output
Mixing problems will always have the same basic form for their DE: Total Rate = (Rate In) – (Rate Out)
The input and output rates will typically be calculated as (concentration) x (flow rate)
In this case we are keeping track of grams of salt, so concentration should have units of grams/liter.
The input is straightforward: min
gminL
Lg 60106Input
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
saltwater output
Mixing problems will always have the same basic form for their DE: Total Rate = (Rate In) – (Rate Out)
The input and output rates will typically be calculated as (concentration) x (flow rate)
In this case we are keeping track of grams of salt, so concentration should have units of grams/liter.
The input is straightforward: min
gminL
Lg 60106Input
Output is trickier, because the concentration is not constant. As more saltwater is poured in, the concentration increases. Also, the volume of water in the tank is not constant. Before we can find the output rate we need to define our variables.
Define x(t)=grams of salt in tank after t minutes
The water in the tank starts at 100 liters, but then increases by 5 liters each minute (10L in, 5L out)
So the amount of water in the tank is 100+5t
ming60
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
saltwater output
Mixing problems will always have the same basic form for their DE: Total Rate = (Rate In) – (Rate Out)
The input and output rates will typically be calculated as (concentration) x (flow rate)
In this case we are keeping track of grams of salt, so concentration should have units of grams/liter.
The input is straightforward: min
gminL
Lg 60106Input
Output is trickier, because the concentration is not constant. As more saltwater is poured in, the concentration increases. Also, the volume of water in the tank is not constant. Before we can find the output rate we need to define our variables.
Define x(t)=grams of salt in tank after t minutes
The water in the tank starts at 100 liters, but then increases by 5 liters each minute (10L in, 5L out)
So the amount of water in the tank is 100+5t
Now we can write down the output rate:
min
gminL
Lg
t20
x5
t5100
xOutput
ming60
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
Mixing problems will always have the same basic form for their DE: Total Rate = (Rate In) – (Rate Out)
The input and output rates will typically be calculated as (concentration) x (flow rate)
In this case we are keeping track of grams of salt, so concentration should have units of grams/liter.
The input is straightforward: min
gminL
Lg 60106Input
ming60
Output is trickier, because the concentration is not constant. As more saltwater is poured in, the concentration increases. Also, the volume of water in the tank is not constant. Before we can find the output rate we need to define our variables.
Define x(t)=grams of salt in tank after t minutes
The water in the tank starts at 100 liters, but then increases by 5 liters each minute (10L in, 5L out)
So the amount of water in the tank is 100+5t
Now we can write down the output rate:
min
gminL
Lg
t20
x5
t5100
xOutput
Now we have the DE:
t20
x60x
ming
t20
x
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
t20
x60x
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
t20
x60x
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
0xt20
1x
Start with the homogeneous equation.
20t
Cx
C20tlnxln
dtt20
1
x
dx
xt20
1
h
dtdx
Skipped a couple steps here – let me know if you want more details
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
t20
x60x
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
0xt20
1x
Start with the homogeneous equation.
20t
Cx
C20tlnxln
dtt20
1
x
dx
xt20
1
h
dtdx
Skipped a couple steps here – let me know if you want more details
Now we get the particular solution by modifying the homogeneous solution
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
20t
)t(vxp
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
t20
x60x
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
0xt20
1x
Start with the homogeneous equation.
20t
Cx
C20tlnxln
dtt20
1
x
dx
xt20
1
h
dtdx
Skipped a couple steps here – let me know if you want more details
Now we get the particular solution by modifying the homogeneous solution
20t
)t(vxp
quotient rule
2p
2p
)20t(
v
)20t(
vx
)20t(
)1(v)20t(vx
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
t20
x60x
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
0xt20
1x
Start with the homogeneous equation.
20t
Cx
C20tlnxln
dtt20
1
x
dx
xt20
1
h
dtdx
Skipped a couple steps here – let me know if you want more details
Now we get the particular solution by modifying the homogeneous solution
20t
)t(vxp
2p
2p
)20t(
v
)20t(
vx
)20t(
)1(v)20t(vx
quotient rule
Plug these in to the DE:
60x20t
1x
60)20t(
v
6020t
v
20t
1
)20t(
v
)20t(
vxx
2
Solve for v
20t
t1200t30x
)t202
t(60v)20t(60v
2
p
2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
t20
x60x
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
0xt20
1x
Start with the homogeneous equation.
20t
Cx
C20tlnxln
dtt20
1
x
dx
xt20
1
h
dtdx
Skipped a couple steps here – let me know if you want more details
Now we get the particular solution by modifying the homogeneous solution
20t
)t(vxp
2p
2p
)20t(
v
)20t(
vx
)20t(
)1(v)20t(vx
quotient rule
Plug these in to the DE:
60x20t
1x
60)20t(
v
6020t
v
20t
1
)20t(
v
)20t(
vxx
2
Solve for v
20t
t1200t30x
)t202
t(60v)20t(60v
2
p
2
Now we have our general solution:
20t
t1200t30
20t
C)t(x
2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
20t
t1200t30
20t
C)t(x
2
We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
20t
t1200t30
20t
C)t(x
2
We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water.
0C
200
01200030
200
C0
2
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
20t
t1200t30
20t
C)t(x
2
We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water.
0C
200
01200030
200
C0
2
20t
t1200t30)t(x
2
Here is our final formula for the amount of salt in the tank.
Now we can answer the question – all we need now is to figure out when the tank is full, then plug that in for t.
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
20t
t1200t30
20t
C)t(x
2
We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water.
0C
200
01200030
200
C0
2
20t
t1200t30)t(x
2
Here is our final formula for the amount of salt in the tank.
Now we can answer the question – all we need now is to figure out when the tank is full, then plug that in for t.
Recall our formula for the amount of water in the tank, and set it to 1000 liters:
min180tt51001000
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
4) A tank with capacity 1000 liters is initially filled with 100 liters of fresh water. Salt water with a concentration of 6 grams of salt per liter is poured into the tank at a rate of 10 liters per minute. The well stirred mixture flows out of the tank at a rate of 5 liters per minute. Note that the volume of water in the tank will be increasing until the tank is full. What will the concentration of salt be at the moment the tank is full?
20t
t1200t30
20t
C)t(x
2
We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water.
0C
200
01200030
200
C0
2
20t
t1200t30)t(x
2
Here is our final formula for the amount of salt in the tank.
Now we can answer the question – all we need now is to figure out when the tank is full, then plug that in for t.
Recall our formula for the amount of water in the tank, and set it to 1000 liters:
min180tt51001000
grams5940
20180
180120018030)180(x
2
To get the concentration, divide by 1000 liters:
L
g94.5
ionConcentrat
Final Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB