Section 10.1 Separable Equations II MAT 1235 Calculus II Section 10.1 Separable Equations II http://myhome.spu.edu/lauw

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

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

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.

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.

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?

Example 1: Mixtures Step 1 Define the variable.

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

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

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

Example 1: Mixtures Changes over time…

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.

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

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.)

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

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

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

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

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

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

Law of Mass Action

Example 2 Second-Order Reaction

Example 2 Second-Order Reaction

Example 2 Second-Order Reaction

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

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

Law of Mass Action

Law of Mass Action

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

Example 2

Example 2

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

Example 2 (c) Interpret the solution as

Example 2 (c) Interpret the solution as