1. Section 10.1 Separable Equations II
MAT 1235 Calculus II
Section 10.1
Separable Equations II

2. HW
WebAssign 10.1 Part II (A) Due Friday WebAssign 10.1 Part II (B) Due Monday
Quiz: 10.1

3. Preview (Version 2, No second order chemical reaction)
We are going to look into some applications of Separable Equations.

4. Example 1: Mixtures
A tank contains 20 kg of salt dissolved in 5000 L of water.
Brine that contains 0.03 kg of salt per liter of water enters the tank at a rate of 25 L/min.

5. Example 1: Mixtures
A tank contains 20 kg of salt dissolved in 5000 L of water.
Brine that contains 0.03 kg of salt per liter of water enters the tank at a rate of 25 L/min.

6. Example 1: Mixtures
The solution is kept thoroughly mixed and drains from the tank at the same rate.
How much salt remains in the tank after half an hour?

7. Example 1: Mixtures
Step 1
Define the variable.

8. Example 1: Mixtures
Step 2
Find rate in and rate out.

9. Example 1: Mixtures Step 3
Set up the DE and solve it with the initial condition.
(Do not erase this part!)

10. Example 1: Mixtures Step 4 Compute the amount of salt after 30 min.
Conclusion:

11. Example 1: Mixtures
Changes over time…

12. Example 2: Mixtures
The air in a room with volume 180 m3 contains 0.15% carbon dioxide initially.
Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 2 m3/min and the mixed air flows out at the same rate.

13. Example 2: Mixtures
Find the percentage of carbon dioxide in the room as a function of time.
What happens in the long run?

14. Example 2: Mixtures Step 1 Define the variable.
(Note that percentage is a ratio, it does not make sense to talk about rate in, rate out.)

15. Example 2: Mixtures
Step 2
Find rate in and rate out.

16. Example 2: Mixtures Step 3
Set up the DE and solve it with the initial condition.

17. Example 2: Mixtures Step 3
Set up the DE and solve it with the initial condition.

18. Example 2: Mixtures Step 4
Compute the percentage of carbon dioxide in the long run?
Conclusion:

19. Just for Fun,….
You can take a look at the PPT for the second order chemical reaction below.

20. Example 2 Second-Order Reaction
X(t) = amount of C
The rate of formation of C is given by

21. Law of Mass Action

22. Example 2 Second-Order Reaction

23. Example 2 Second-Order Reaction

24. Example 2 Second-Order Reaction

25. Example 2 Second-Order Reaction
Proportion of mass from A
Proportion of mass from B

26. Example 2 Second-Order Reaction
Proportion of mass from A
Proportion of mass from B

27. Law of Mass Action

28. Law of Mass Action

29. Example 2
A compound C is formed when two chemicals A and B are combined
The resulting reaction between the two chemicals is such that for each gram of A, 4 grams of B is used
It is observed that 30 grams of the compound C is formed in 10 minutes
Initially there are 50 grams of A and 32 grams of B

30. Example 2

31. Example 2

32. Example 2 (b)
How much of the compound C is present at 15 minutes?

33. Example 2 (c)
Interpret the solution as

34. Example 2 (c)
Interpret the solution as