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 y ( t )
 y
0
 e kt
Where y(t)=concentration after t hours, and y(0)=1500
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 y ( t )
 y
0
 e kt
Where y(t)=concentration after t hours, and y(0)=1500
The half-life is given, so we can find the value for k (it will be negative) y ( 7 )
 1
2 y
0
7 k
 ln(
1
2
)
 y
0
 e k ( 7 )  1
2 y
0 k
 ln(
1
2
)
7

 ln( 2 )
7
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 y ( t )
 y
0
 e kt
Where y(t)=concentration after t hours, and y(0)=1500
The half-life is given, so we can find the value for k (it will be negative) y ( 7 )
 1
2 y
0
7 k
 ln(
1
2
)
 y
0
 e k ( 7 )  1
2 y
0 k
 ln(
1
2
)
7

 ln( 2 )
7
If a half-life is given, the value of k is always
If a doubling time is given, k is always k
 ln( 1
2
) t half k
 ln( 2 ) t double
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 y ( t )
 y
0
 e kt
Where y(t)=concentration after t hours, and y(0)=1500
The half-life is given, so we can find the value for k (it will be negative) y ( 7 )
 1
2 y
0
7 k
 ln(
1
2
)
 y
0
 e k ( 7 )  1
2 y
0 k
 ln(
1
2
)
7

 ln( 2 )
7
So our formula is: y ( t )

1500
 e
 ln( 2 )  t
7
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 y ( t )
 y
0
 e kt
Where y(t)=concentration after t hours, and y(0)=1500
The half-life is given, so we can find the value for k (it will be negative) y ( 7 )
 1
2 y
0
7 k
 ln(
1
2
)
 y
0
 e k ( 7 )  1
2 y
0 k
 ln(
1
2
)
7

 ln( 2 )
7
So our formula is: y ( t )

1500
 e
 ln( 2 )  t
7
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 y ( t )
 y
0
 e kt
Where y(t)=concentration after t hours, and y(0)=1500
The half-life is given, so we can find the value for k (it will be negative) y ( 7 )
 1
2 y
0
7 k
 ln(
1
2
)
 y
0
 e k ( 7 )  1
2 y
0 k
 ln(
1
2
)
7

 ln( 2 )
7
So our formula is: y ( t )

1500
 e
 ln( 2 )  t
7
When is the concentration 100? Set y=100 and solve for t.
100

1500
 e
 ln( 2 )  t
7
1
15
7

 e
 ln(
7
2 )  t ln( 15 )
 t ln( 2 )

 t ln(
1
15
)

27
  ln( 2 ) hours
7
 t
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 y
 
0 .
04 y
 d with initial value y ( 0 )

10 , 000 Here d is the annual deposit amount.
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 y
 
0 .
04 y
 d with initial value y ( 0 )

10 , 000 Here d is the annual deposit amount.
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 y
 
0 .
04 y
 d with initial value y ( 0 )

10 , 000 Here d is the annual deposit amount.
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: y
 
0 .
04 y
 d
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 y
 
0 .
04 y
 d with initial value y ( 0 )

10 , 000 Here d is the annual deposit amount.
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: y
 
0 .
04 y
 d
  e
  .
04 dt  e

.
04 t
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 y
 
0 .
04 y
 d with initial value y ( 0 )

10 , 000 Here d is the annual deposit amount.
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: y
 
0 .
04 y
 d
  e
  .
04 dt  e

.
04 t
Multiply through by µ, then integrate and solve for y: e d dt


.
04 e
 t 
.
04 y
 t 
 y

0 .
04
 e

.
 e

.
04
04 t  d t e

.
04 t  y
  e

.
04 t  d
 dt
 y
 e

.
04 t  d e

.
04 t  y
 d

.
04 e

.
04 t 
C y
 d

.
04

C
 e
.
04 t
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 y
 
0 .
04 y
 d with initial value y ( 0 )

10 , 000 Here d is the annual deposit amount.
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: y
 
0 .
04 y
 d
  e
  .
04 dt  e

.
04 t
Multiply through by µ, then integrate and solve for y:
Use the initial value to find C:
10 , 000
 d

.
04

C

10 , 000
 d
.
04
C
 e
.
04 e d dt


.
04 e
 t 
.
04 y
 t 
 y

0 .
04
 e

.
 e

.
04
04 t  d t e

.
04 t  y
  e

.
04 t  d
 dt
 y
 e

.
04 t  d e

.
04 t  y
 d

.
04 e

.
04 t 
C y
 d

.
04

C
 e
.
04 t
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 y
 
0 .
04 y
 d with initial value y ( 0 )

10 , 000 Here d is the annual deposit amount.
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: y
 
0 .
04 y
 d
  e
  .
04 dt  e

.
04 t
Multiply through by µ, then integrate and solve for y:
Use the initial value to find C:
10 , 000
 d

.
04

C
 e
.
04
C

10 , 000
 d
.
04
So our solution becomes: y
 d

.
04

10 , 000
 d
.
04
 e
.
04 t e d dt


.
04 e
 t 
.
04 y
 t 
 y

0 .
04
 e

.
 e

.
04
04 t  d t e

.
04 t  y
  e

.
04 t  d
 dt
 y
 e

.
04 t  d e

.
04 t  y
 d

.
04 e

.
04 t 
C y
 d

.
04

C
 e
.
04 t
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 y
 
0 .
04 y
 d with initial value y ( 0 )

10 , 000 Here d is the annual deposit amount.
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: y
 
0 .
04 y
 d
  e
  .
04 dt  e

.
04 t
Multiply through by µ, then integrate and solve for y:
Use the initial value to find C:
10 , 000
 d

.
04

C
 e
.
04
C

10 , 000
 d
.
04
So our solution becomes: y
 d

.
04

10 , 000
 d
.
04
 e
.
04 t e d dt


.
04 e
 t 
.
04 y
 t 
 y

0 .
04
 e

.
 e

.
04
04 t  d t e

.
04 t  y
  e

.
04 t  d
 dt
 y
 e

.
04 t  d e

.
04 t  y
 d

.
04 e

.
04 t 
C y
 d

.
04

C
 e
.
04 t
Now we need to find the d that will give $50,000 at t=5.
50 , 000
 d

.
04

10 , 000
 d
.
04
 e
.
04
 a lg ebra  d

.
04 ( 50 , 000

10000
 e
.
2
) e
.
2 
1

$ 6 , 827 / year
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.
T(t)=temperature at time t
For heating or cooling the DE is T
  k ( M

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 T
  k ( M

T )
T(t)=temperature at time t
M=ambient (constant) temperature
In 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 T
  k ( M

T )
T(t)=temperature at time t
M=ambient (constant) temperature
In 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 T
  k ( M

T )
T(t)=temperature at time t
M=ambient (constant) temperature
In 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.
dT dt
 k ( 20

T )
 dT

20
 ln 20
T


T
 kdt
 kt

C
20

T
T


Ce
 kt
20

Ce
 kt
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 T
  k ( M

T )
T(t)=temperature at time t
M=ambient (constant) temperature
In 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.
dT dt
 k ( 20

T )
 dT

20
 ln 20
T


T
 kdt
 kt

C
20

T
T


Ce
 kt
20

Ce
 kt
We can use the initial value to find C.
T ( 0 )

90

20

Ce
 k ( 0 ) 
C

70
Now our solution is T ( t )

20

70 e
 kt
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 T
  k ( M

T )
T(t)=temperature at time t
M=ambient (constant) temperature
In 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.
dT dt
 k ( 20

T )
 dT

20
 ln 20
T


T
 kdt
 kt

C
20

T
T


Ce
 kt
20

Ce
 kt
Using T(15)=70 we can find k
T (
50
15

)

70
70 e


15 k
20

70 e
 k ( 15 )
 ln(
5
7
)
 
15 k
 k
 ln( 5
7

15
)
We can use the initial value to find C.
T ( 0 )

90

20

Ce
 k ( 0 ) 
C

70
Now our solution is T ( t )

20

70 e
 kt
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 T
  k ( M

T )
T(t)=temperature at time t
M=ambient (constant) temperature
In 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.
dT dt
 k ( 20

T )
 dT

20
 ln 20
T


T
 kdt
 kt

C
20

T
T


Ce
 kt
20

Ce
 kt
Using T(15)=70 we can find k
T (
50
15

)

70
70 e


15 k
20

70 e
 k ( 15 )
 ln(
5
7
)
 
Our final formula is thus
15 k

T ( t )
 k

20 ln( 5 )
7


15
70 e

 ln(
5
7
)
15

 t
We can use the initial value to find C.
T ( 0 )

90

20

Ce
 k ( 0 ) 
C

70
Now our solution is T ( t )

20

70 e
 kt
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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 T
  k ( M

T )
T(t)=temperature at time t
M=ambient (constant) temperature
In 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.
dT dt
 k ( 20

T )
 dT

20
 ln 20
T


T
 kdt
 kt

C
20

T
T


Ce
 kt
20

Ce
 kt
Using T(15)=70 we can find k
T (
50
15

)

70
70 e


15 k
20

70 e
 k ( 15 )
 ln(
5
7
)
 
Our final formula is thus
15 k

T ( t )
 k

20 ln( 5 )
7


15
70 e

 ln(
5
7
)
15

 t
We can use the initial value to find C.
T ( 0 )

90

20

Ce
 k ( 0 ) 
C

70
Now our solution is T ( t )

20

70 e
 kt
Finally we can answer our question – set T=60 and solve
60

20

70 e

 ln(
15
5
7
)
 t
 ln( 4
7
)
 ln(
15
5
7
)
 t t

15
 ln( ln( 5
7
)
4
7
)

25 min
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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?
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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
Mixing problems will always have the same basic form for their DE: Total Rate = (Rate In) – (Rate Out) saltwater output
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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
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) saltwater output
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
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) saltwater output
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
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)
The input is straightforward: saltwater output
Input
In this case we are keeping track of grams of salt, so concentration should have units of grams/liter.

 
L


10 L min


60 g min
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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?
60 g min
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)
The input is straightforward: saltwater output
Input
In this case we are keeping track of grams of salt, so concentration should have units of grams/liter.

 
L


10 L min


60 g min
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
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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?
60 g min
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)
The input is straightforward: saltwater output
Input
In this case we are keeping track of grams of salt, so concentration should have units of grams/liter.

 
L


10 L min


60 g min
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
Now we can write down the output rate:
Output
 x
100

5 t L g 
  min
 x
20
 t g min 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
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?
60 g min
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)
The input is straightforward:
20 x
 t
Input g min
In this case we are keeping track of grams of salt, so concentration should have units of grams/liter.

 
L


10 L min


60 g min
Now we can write down the output rate:
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
Output
 x
100

5 t L g 
  min
 x
20
 t g min
Now we have the DE: need to define our variables.
Define x(t)=grams of salt in tank after t minutes x
 
60

20 x
 t
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
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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?
x
 
60

20 x
 t
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
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For Campus Learning
Assistance Services at UCSB

x
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?
x
 
60

20 x
 t
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
 
1
20
 t
 x

0 Start with the homogeneous equation.
dx dt
 
 20
1
 t
 x
 dx x ln x


 
 ln t
20

1

20 t

 dt

C x h
 t

C
20
Skipped a couple steps here – let me know if you want more details
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For Campus Learning
Assistance Services at UCSB

x
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?
x
 
60

20 x
 t
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
 
1
20
 t
 x

0 Start with the homogeneous equation.
dx dt
 
 20
1
 t
 x
 dx x ln x


 
 ln t
20

1

20 t

 dt

C x h
 t

C
20
Skipped a couple steps here – let me know if you want more details
Now we get the particular solution by modifying the homogeneous solution x p
 t v ( t )

20
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Assistance Services at UCSB

x
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?
x
 
60

20 x
 t
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
 
1
20
 t
 x

0 Start with the homogeneous equation.
dx dt
 
 20
1
 t
 x
 dx x ln x


 
 ln t
20

1

20 t

 dt

C x h
 t

C
20
Skipped a couple steps here – let me know if you want more details
Now we get the particular solution by modifying the homogeneous solution x p
 t v ( t )

20 x
 p x
 p

 v
 
( t
( t

20 )
 v

( 1 )
 v

( t

20 )
20

( t
)
2
 v
20 )
2 quotient rule
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For Campus Learning
Assistance Services at UCSB

x
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?
x
 
60

20 x
 t
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
 
1
20
 t
 x

0 Start with the homogeneous equation.
Plug these in to the DE: x
  t
1

20 dx dt
 
 20
1
 t
 x
 dx x ln x


 
 ln t
20

1

20 t

 dt

C
 x

60 x h
 t

C
20
Skipped a couple steps here – let me know if you want more details
Now we get the particular solution by modifying the homogeneous solution


    

   
( t
 v

20 )

( t
 v
20 )
2



( t v


 v

20 )
60 ( t


60
20 )
 v t
Solve for v

60 ( t
2
2

1
20

 t
  
 v
20
20 t )

60 x p

30 t
2 t

1200 t

20 x p
 t v ( t )

20 x
 p x
 p

 v
 
( t
( t

20 )
 v

( 1 )
 v

( t

20 )
20

( t
)
2
 v
20 )
2 quotient rule
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For Campus Learning
Assistance Services at UCSB

x
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?
x
 
60

20 x
 t
This is first-order and linear, so we have a couple of options for solving.
Let’s try variation of parameters this time:
 
1
20
 t
 x

0 Start with the homogeneous equation.
Plug these in to the DE: x
  t
1

20 dx dt
 
 20
1
 t
 x
 dx x ln x


 
 ln t
20

1

20 t

 dt

C
 x

60 x h
 t

C
20
Skipped a couple steps here – let me know if you want more details
Now we get the particular solution by modifying the homogeneous solution


    

   
( t
 v

20 )

( t
 v
20 )
2



( t v


 v

20 )
60 ( t


60
20 )
 v t
Solve for v

60 ( t
2
2

1
20

 t
  
 v
20
20 t )

60 x p

30 t
2 t

1200 t

20 x p
 t v ( t )

20 x
 p x
 p

 v
 
( t
( t

20 )
 v

( 1 )
 v

( t

20 )
20

( t
)
2
 v
20 )
2 quotient rule Now we have our general solution: x ( t )
 t
C

20

30 t
2 t

1200 t

20
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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?
x ( t )
 t
C

20

30 t
2 t

1200 t

20
We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water.
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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?
x ( t )
 t
C

20

30 t
2 t

1200 t

20
We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water.
0

0

C
20

30
 
2
0

1200
 

20

C

0
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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?
x ( t )
 t
C

20

30 t
2 t

1200 t

20
We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water.
0

0

C
20

30
 
2
0

1200
 

20

C

0 x ( t )

30 t
2 t

1200 t

20
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.
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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?
x ( t )
 t
C

20

30 t
2 t

1200 t

20
We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water.
0

0

C
20

30
 
2
0

1200
 

20

C

0 x ( t )

30 t
2 t

1200 t

20
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:
1000

100

5 t
 t

180 min
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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?
x ( t )
 t
C

20

30 t
2 t

1200 t

20
We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water.
0

0

C
20

30
 
2
0

1200
 

20

C

0 x ( t )

30 t
2 t

1200 t

20
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:
1000

100

5 t
 t

180 min x ( 180 )

30
 
2
180


1200
 
20

5940 grams
To get the concentration, divide by 1000 liters:
Final
Concentrat ion
 g
5 .
94
L
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB