ch07_01 - 566 CHAPTER 7 . First-Order Differential...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
566 CHAPTER 7 . . First-Order Differential Equations 7-2 we return to the topic of differential equations and present additional examples. However, a more thorough examination of this vast field will need to wait for a course focused on this topic. 100 200 300 400 4 3 2 1 Number of bacteria (millions per ml) Time (hours) FIGURE 7.1 Growth of bacteria Number of Time Bacteria (hours) (millions per ml) 0 1.2 0.5 2.5 1 5.1 1.5 11.0 2 23.0 2.5 45.0 3 91.0 3.5 180.0 4 350.0 7.1 MODELING WITH DIFFERENTIAL EQUATIONS Growth and Decay Problems In this age, we are all keenly aware of how infection by microorganisms such as Escherichia coli ( E. coli ) causes disease. Many organisms (such as E. coli ) produce a toxin that can cause sickness or even death. Some bacteria can reproduce in our bodies at a surprisingly fast rate, overwhelming our bodies’ natural defenses with the sheer volume of toxin they are producing. The table shown in the margin 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. We have plotted the number of bacteria per milliliter versus time in Figure 7.1. What would you say the graph most resembles? If you said, “an exponential,” you guessed right. Careful analysis of experimental data has shown that many populations grow at a rate proportional to their current level. This is quite easily observed in bacterial cultures, where the bacteria reproduce by binary fission (i.e., each cell reproduces by dividing into two cells). In this case, the rate at which the bacterial culture grows is directly proportional to the current population (until such time as resources become scarce or overcrowding becomes a limiting factor). If we 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 y ± ( t ) = ky ( t ) , (1.1) for some constant of proportionality k (the growth constant ). Since equation (1.1) involves the derivative of an unknown function, we call it a differential equation. Our aim is to solve the differential equation, that is, find the function y ( t ). Assuming that y ( t ) > 0 (this is a reasonable assumption, since y ( t ) represents a population), we have y ± ( t ) y ( t ) = k . (1.2) Integrating both sides of equation (1.2) with respect to t ,we obtain ± y ± ( t ) y ( t ) dt = ± kdt . (1.3) Substituting y = y ( t )in the integral on the left-hand side, we have dy = y ± ( t ) and so, (1.3) becomes ± 1 y = ± . Evaluating these integrals, we obtain ln | y |+ c 1 = kt + c 2 , where c 1 and c 2 are constants of integration. Subtracting c 1 from both sides yields ln | y |= + ( c 2 c 1 ) = + c ,
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
7-3 SECTION 7.1 . . Modeling with Differential Equations 567 for some constant c . Since y ( t ) > 0, we have ln y ( t ) = kt + c and taking exponentials of both sides, we get y ( t ) = e ln y ( t ) = e + c = e e c .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/03/2010 for the course CHE 2C CHE 2C taught by Professor Atsumi during the Spring '10 term at UC Davis.

Page1 / 12

ch07_01 - 566 CHAPTER 7 . First-Order Differential...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online