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An overview of exponential growth and decay, focusing on solving initial value problems and integrals. It covers the concepts of general and particular solutions, exponential growth and decay, and examples of chemical dilution. The document also includes notes and review materials for further study.
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We say that y = Ce kx is the general solution to y' = k y. We call it a general solution because we do not know the value of C. If we also determine the constant C , we call the resulting solution the particular solution. For this, we need additional information: must have an initial value problem.
y! = " 2 y !!!!!!!!!!!and!!!!!!!!!!! y ( 1 ) = 3
! 2 t
! 2 * 1
! 2
2
2
2
! 2 t
2! 2 t
Chemical Dilution Tank: 90 gallons water and 10 gallons chemical. Water flows in, mixture flows out at 20 gal/min. How many minutes to reduce concentration to 1%? Let y ( t ): concentration at t. Then What’s y'? Rate of chemical flowing out is (outflow rate)(concentration) = 20 y ( t ) Rate of change of concentration? Solve: y (0) = 0.1, y' = – 0.2 y.
Chemical Dilution Tank: 90 gallons water and 10 gallons chemical. Water flows in, mixture flows out at 20 gal/min. How many minutes to reduce concentration to 1%?
In Lab 1, you integrated the rate of change of concentration.
In Lab 1, you integrated the rate of change of concentration. If we integrate the rate of change of quantity Q, we get the total change in Q, over the interval a to b. So you got
0 24
Which means the change in concentration is about 0.000196298 mmol/L.
! 2 3 "
3
! 2 3 "
3
! 2 3 "
Problem 2: Smooth function, but fast growth!
x
! 7 7 "
Problem 2: Smooth function, but fast growth!
x
! 7 7 " n****! x^ Result 2 7 1921610 4 3.5 965037. 8 1.75 545318. 16 0.875 437759. 32 0.4375 424353. 64 0.2188 423292.
