Presentation is loading. Please wait.

Presentation is loading. Please wait.

Practice Word Problems

Similar presentations


Presentation on theme: "Practice Word Problems"— Presentation transcript:

1 Practice Word Problems
Math 4B Practice Word Problems Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

2 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

3 We are given the half-life, so we can assume exponential decay.
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 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4 We are given the half-life, so we can assume exponential decay.
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 The half-life is given, so we can find the value for k (it will be negative) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

5 We are given the half-life, so we can assume exponential decay.
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 The half-life is given, so we can find the value for k (it will be negative) If a half-life is given, the value of k is always If a doubling time is given, k is always Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

6 We are given the half-life, so we can assume exponential decay.
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 The half-life is given, so we can find the value for k (it will be negative) So our formula is: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

7 We are given the half-life, so we can assume exponential decay.
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 The half-life is given, so we can find the value for k (it will be negative) So our formula is: When is the concentration 100? Set y=100 and solve for t. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

8 We are given the half-life, so we can assume exponential decay.
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 The half-life is given, so we can find the value for k (it will be negative) So our formula is: When is the concentration 100? Set y=100 and solve for t. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

9 2) Sally is saving for the down payment on a house
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

10 Here d is the annual deposit amount.
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 with initial value Here d is the annual deposit amount. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

11 Here d is the annual deposit amount.
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 with initial value 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

12 Here d is the annual deposit amount.
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 with initial value 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: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

13 Here d is the annual deposit amount.
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 with initial value 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: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

14 Here d is the annual deposit amount.
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 with initial value 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: Multiply through by µ, then integrate and solve for y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

15 Here d is the annual deposit amount.
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 with initial value 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: Use the initial value to find C: Multiply through by µ, then integrate and solve for y: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

16 Here d is the annual deposit amount.
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 with initial value 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: Use the initial value to find C: Multiply through by µ, then integrate and solve for y: So our solution becomes: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

17 Here d is the annual deposit amount.
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 with initial value 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: Use the initial value to find C: Multiply through by µ, then integrate and solve for y: So our solution becomes: Now we need to find the d that will give $50,000 at t=5. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

18 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

19 T(t)=temperature at time t M=ambient (constant) temperature
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 M=ambient (constant) temperature For heating or cooling the DE is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

20 T(t)=temperature at time t M=ambient (constant) temperature
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 M=ambient (constant) temperature For heating or cooling the DE is 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

21 T(t)=temperature at time t M=ambient (constant) temperature
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 M=ambient (constant) temperature For heating or cooling the DE is 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

22 T(t)=temperature at time t M=ambient (constant) temperature
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 M=ambient (constant) temperature For heating or cooling the DE is 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. 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

23 T(t)=temperature at time t M=ambient (constant) temperature
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 M=ambient (constant) temperature For heating or cooling the DE is 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. We can use the initial value to find C. Now our solution is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

24 T(t)=temperature at time t M=ambient (constant) temperature
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 M=ambient (constant) temperature For heating or cooling the DE is 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. Using T(15)=70 we can find k We can use the initial value to find C. Now our solution is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

25 T(t)=temperature at time t M=ambient (constant) temperature
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 M=ambient (constant) temperature For heating or cooling the DE is 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. Using T(15)=70 we can find k Our final formula is thus We can use the initial value to find C. Now our solution is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

26 T(t)=temperature at time t M=ambient (constant) temperature
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 M=ambient (constant) temperature For heating or cooling the DE is 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. Using T(15)=70 we can find k Our final formula is thus We can use the initial value to find C. Finally we can answer our question – set T=60 and solve Now our solution is Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

27 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

28 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) saltwater input saltwater output Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

29 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) saltwater input 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

30 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) saltwater input 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. saltwater output Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

31 The input is straightforward:
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) saltwater input 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. saltwater output The input is straightforward: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

32 The input is straightforward:
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. saltwater output The input is straightforward: 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 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

33 The input is straightforward: Now we can write down the output rate:
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. saltwater output The input is straightforward: 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 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

34 The input is straightforward: Now we can write down the output rate:
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: 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 need to define our variables. Define x(t)=grams of salt in tank after t minutes Now we have the DE: 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

35 Let’s try variation of parameters this time:
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? 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

36 Let’s try variation of parameters this time:
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? This is first-order and linear, so we have a couple of options for solving. Let’s try variation of parameters this time: Start with the homogeneous equation. Skipped a couple steps here – let me know if you want more details Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

37 Let’s try variation of parameters this time:
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? This is first-order and linear, so we have a couple of options for solving. Let’s try variation of parameters this time: Start with the homogeneous equation. 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

38 Let’s try variation of parameters this time:
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? This is first-order and linear, so we have a couple of options for solving. Let’s try variation of parameters this time: Start with the homogeneous equation. Skipped a couple steps here – let me know if you want more details Now we get the particular solution by modifying the homogeneous solution quotient rule Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

39 Let’s try variation of parameters this time:
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? This is first-order and linear, so we have a couple of options for solving. Let’s try variation of parameters this time: Start with the homogeneous equation. Plug these in to the DE: Skipped a couple steps here – let me know if you want more details Solve for v Now we get the particular solution by modifying the homogeneous solution quotient rule Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

40 Let’s try variation of parameters this time:
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? This is first-order and linear, so we have a couple of options for solving. Let’s try variation of parameters this time: Start with the homogeneous equation. Plug these in to the DE: Skipped a couple steps here – let me know if you want more details Solve for v Now we get the particular solution by modifying the homogeneous solution quotient rule Now we have our general solution: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

41 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? 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

42 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? 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

43 Here is our final formula for the amount of salt in the tank.
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? We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water. 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

44 Here is our final formula for the amount of salt in the tank.
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? We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water. 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: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

45 Here is our final formula for the amount of salt in the tank.
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? We need to find a value for C. Use the initial value – x(0)=0 because we start with fresh water. 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: To get the concentration, divide by 1000 liters: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB


Download ppt "Practice Word Problems"

Similar presentations


Ads by Google