First Order Differential Equations

Transcription

1 First Order Differential Equations CHAPTER SEPARABLE DIFFERENTIAL 7.3 DIRECTION FIELDS AND EULER S METHOD 7.4 SYSTEMS OF FIRST ORDER DIFFERENTIAL Slide 1

2 Exponential Growth The table indicates the number of E. coli bacteria (in millions of bacteria per ml) in a laboratory culture measured at half hour intervals during the course of an experiment. Slide 2

3 Exponential Growth The plot appears to indicate that the bacterial culture is growing exponentially. Careful analysis of experimental data has shown that the rate at which the bacterial culture grows is directly proportional to the current population. Slide 3

4 Exponential Growth Let y(t) represent the number of bacteria in a culture at time t, then the rate of change of the population with respect to time is y (t). Thus, since y (t) is proportional to y(t), we have the following differential equation where k is the growth constant. We wish to solve this equation for y(t). We begin by rewriting and integrating. Slide 4

5 Exponential Growth We may combine c 1 and c 2 as a single constant we lablel c. Since y(t) represent a population, y(t) > 0. Solving for y(t) and defining A = e c we obtain the solution. We call this the general solutionofof the differential equation. For k > 0, it is called an exponential growth law and for k < 0, it is an exponential decay law. Slide 5

6 EXAMPLE 1.1 Exponential Growth in a Bacterial Colony A freshly inoculated bacterial culture of Streptococcus A contains 100 cells. When the culture is checked 60 minutes later, it is determined that there are 450 cells present. Assuming exponential growth, determine the number of cells present at any time t (measured in minutes) and find the doubling time. Slide 6

7 7.1 EXAMPLE 1.1 Exponential Growth in a Bacterial Colony Solution Exponential growth means where A and k are constants to be determined. We were given the initial condition. Substituting i this into the general solution gives us Slide 7

8 7.1 EXAMPLE 1.1 Exponential Growth in a Bacterial Colony Solution We can use the second observation to find k. We now have a formula representing the number of cells present at any time t: Slide 8

9 7.1 EXAMPLE 1.1 Exponential Growth in a Bacterial Colony Solution To find the doubling time, we wish to find t such that We substitute this into y(t) and solve for t. We see that the population doubles about every 28 minutes. Slide 9

10 Exponential Decay Experiments have shown that the rate at which a radioactive element decays is directly proportional to the amount present. Let y(t) represent the amount (mass) of a radioactive element present at time t, then the rate of decay satisfies We know the general solution to this differential equation. Slide 10

11 Exponential Decay It is common to discuss the decay rate of a radioactive element in terms of its half life, the time required for half of the initial quantity to decay into other elements. For instance, scientists have calculated that the half life of carbon 14 ( 14 C) is approximately 5730 years. That is, if you have 2 grams of 14 C today and you come back in 5730 years, you will have approximately 1 gram of 14 C remaining. Slide 11

12 EXAMPLE 1.2 Radioactive Decay If you have 50 grams of 14 C today, how much will be left in 100 years? Slide 12

13 7.1 EXAMPLE 1.2 Radioactive Decay Solution Let y(t) be the mass (in grams) of 14 C present at time t. Then, we have The initial condition y(0) = 50 gives Using the half life, we can find the decay constant k. Slide 13

14 7.1 EXAMPLE 1.2 Radioactive Decay Solution The graph shows the incredibly slow decay of 14 C. If we start with 50 grams, then the amount left after 100 years is Slide 14

15 Newton s Law of Cooling The rate at which the object cools (or warms) is not proportional to its temperature, but rather, to the difference in temperature between the object and its surroundings. Symbolically, if we let y(t) be the temperature of the object at time t and let T a be the temperature of the surroundings (the ambient temperature, which we assume to be constant), we have the differential equation Slide 15

16 Newton s Law of Cooling We wish ihto solve this equation for y(t). We begin by rewriting and integrating. To evaluate the integral, we will use the substitution u = y(t) T a and we assume y(t) T a > 0. Slide 16

17 Newton s Law of Cooling We now have which h can be written as. Solving for y(t) we obtain the general solution where we define A = e c. Slide 17

18 EXAMPLE 1.3 Newton s Law of Cooling for a Cup of Coffee A cup of fast food coffee is 180 o F when freshly poured. After 2 minutes in a room at 70 o F, the coffee has cooled ld to 165 o F. Find the temperature at any time t and find the time at which h the coffee has cooled to 120 o F. Slide 18

19 EXAMPLE 1.3 Newton s Law of Cooling for a Cup of Coffee Solution Let y(t) be the temperature of the coffee at time t, we have Using the initial condition y(0) = 180, gives us Using the second measured temperature, we have Slide 19

20 7.1 EXAMPLE 1.3 Newton s Law of Cooling for a Cup of Coffee Solution We now solve this equation for k. Ui Using this value for k, we can now solve Slide 20

21 Compound Interest If a bank agrees to pay you 8% (annual) interest on an investment of $10,000, then at the end of a year, you will have If a bank agrees to pay you interest twice a year at 8% annually, then you will receive (8/2)% interest twice per year, and at the end of a year you will have Slide 21

22 Compound Interest Compounding interest monthly would pay (8/12)% each month resulting in a balance of Compounding interest daily would result in a balance of Slide 22

23 Compound Interest Is there is a limit to how much interest can accrue on a given investment at a given interest rate? If n is the number of times per year that interest is compounded, we wish to calculate the annual percentage yield (APY) under continuous compounding, Slide 23

24 Compound Interest Recall that If we make the change of variables n = 0.08m, we have Continuous compounding would earn approximately 8.3%. Slide 24

25 Compound Interest Suppose that you invest $P at an annual interest rate r, compounded n times per year. Then the value of your investment after t years is Under continuous compounding (i.e., limit as n ) this becomes Slide 25

26 Compound Interest If y(t) is the value of your investment after t years, with continuous compounding, the rate of change of y(t) is proportional to y(t). For an initial i i linvestment of $P, we have Slide 26

27 7.1 EXAMPLE 1.5 Depreciation of Assets (a) Suppose that the value of a $10,000 asset decreases continuously at a constant rate of 24% per year. Find its worth after 10 years; after 20 years. (b) Compare these values to a $10,000 asset that is depreciated dto no value in 20 years using linear depreciation. Slide 27

28 7.1 EXAMPLE 1.5 Depreciation of Assets Solution The value v(t) satisfies v = rv, where r = Thus, we have Using the initial value, we have t =10: t =20: Slide 28

29 7.1 EXAMPLE 1.5 Depreciation of Assets Solution For linear depreciation we use the following information. Using this information we find t =10: v(10) = $5000 t =20: v(20) = $0 Slide 29