Practice Word Problems

Slides:



Advertisements
Similar presentations
ORDINARY DIFFERENTIAL EQUATIONS (ODE)
Advertisements

Bellwork 1) C2) A3) B. A few things to discuss… Increasing vs. Decreasing Increasing vs. Decreasing Linear vs. Exponential Linear vs. Exponential Asymptotes.
6.2 Growth and Decay Law of Exponential Growth and Decay C = initial value k = constant of proportionality if k > 0, exponential growth occurs if k < 0,
Differential Equations
Differential Equations Separable Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Math 3C Practice Final Problems Solutions Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Systems of Linear Equations Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Math 3C Practice Word Problems Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Math 3C Euler’s Method Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Differential Equations Solving First-Order Linear DEs Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
9.3 Separable Equations.
Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1 4 Inverse, Exponential, and Logarithmic Functions Copyright © 2013, 2009, 2005 Pearson Education,
Chapter 3 – Differentiation Rules
Exponential and Logarithmic Functions Chapter 11.
Differential Equations Separable Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
2.5 Mixing Problems.  In these problems we will start with a substance that is dissolved in a liquid. Liquid will be entering and leaving a holding tank.
MTH 253 Calculus (Other Topics)
Any population of living creatures increases at a rate that is proportional to the number present (at least for a while). Other things that increase or.
Differential Equations Second-Order Linear DEs Variation of Parameters Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
Differential Equations Solving First-Order Linear DEs Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB.
9/27/2016Calculus - Santowski1 Lesson 56 – Separable Differential Equations Calculus - Santowski.
Advanced Higher Notes. Inverse Trigonometric Functions Integration By Partial Fractions 1 Integration By Partial Fractions 2 Integration By Partial.
Oh Sheep! 11.5 Exponential Growth and Decay
Application of Logarithms.
12.3 Geometric Series.
Differential Equations
Math 4B Practice Midterm Problems
Differential Equations
Differential Equations
Differential Equations
Differential Equations
Differential Equations
6.4 Growth and Decay.
Week 8 Second-order ODEs Second-order linear homogeneous ODEs
Systems of Linear Equations
Differential Equations
Differential Equations
Prepared by Vince Zaccone
Differential Equations
3.1 Growth and Decay.
4.7 Growth and Decay.
Chapter 9.3: Modeling with First-Order Differential Equations
Mixture Problems MAT 275.
6.4 Exponential Growth and Decay, p. 350
LSP 120 Exponential Modeling.
Setting up and Solving Differential Equations
6.2 Exponential Growth and Decay
7.4 Exponential Growth and Decay
6.4 day 2 Exponential Growth and Decay
Lesson 58 - Applications of DE
Systems of Differential Equations Nonhomogeneous Systems
Integration 2 and Differential equations
Differential Equations
Differential Equations
7.5b - Applications of Logarithms
Differential Equations
Section 10.1 Separable Equations II
Copyright © Cengage Learning. All rights reserved.
9.1/9.3 – Differential Equations
7.4 Exponential Growth and Decay Glacier National Park, Montana
Section 4.8: Exponential Growth & Decay
Differential Equations
Section 4.8: Exponential Growth & Decay
Chapter 5 APPLICATIONS OF ODE.
4.7 Growth and Decay.
Mathematical Explorations
5.2 Growth and Decay Law of Exponential Growth and Decay
Week 8 Second-order ODEs Second-order linear homogeneous ODEs
Differential Equations
Presentation transcript:

Practice Word Problems Math 4B 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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? 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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