YOU TRY IT Calculate

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d [( ln x) ln x ] dx

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65 Exponential Growth and Decay

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Many processes of nature and many mathematical applications involve logarithmic and exponential functions For example, if we examine a population of bacteria, we notice that the rate at which the population grows is proportional to the number of bacteria present To see that this makes good sense, suppose that a bacterium reproduces itself every 4 hours If we begin with 5 thousand bacteria then after 4 hours after 8 hours there are there are 10 thousand bacteria 20 thousand bacteria

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6 T R A N S C E N D E N T A L F U N C T I O N S

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FIGURE 612

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after 12 hours there are after 16 hours there are etc The point is that each new generation of bacteria also reproduces, and the older generations reproduce as well A sketch (Figure 612) of the bacteria population against time shows that the growth is certainly not linear---indeed the shape of the curve appears to be of exponential form Notice that, when the number of bacteria is large, then different generations of bacteria will be reproducing at different times So, averaging out, it makes sense to hypothesize that the growth of the bacteria population varies continuously as in Figure 613 Here we are using a standard device of mathematical analysis: even though the number of bacteria is always an integer, we represent the graph of the population of bacteria by a smooth curve This enables us to apply the tools of calculus to the problem 40 thousand bacteria 80 thousand bacteria

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651 A Differential Equation

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If B(t) represents the number of bacteria present in a given population at time t, then the preceding discussion suggests that dB = K B(t), dt

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CALCULUS DeMYSTiFieD

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FIGURE 613

where K is a constant of proportionality This equation expresses quantitatively the assertion that the rate of change of B(t) (that is to say, the quantity dB/dt) is proportional to B(t) To solve this equation, we rewrite it as dB 1 = K B(t) dt We integrate both sides with respect to the variable t: 1 dB dt = B(t) dt The left side is ln |B(t)| + C and the right side is Kt + C, where C and C are constants of integration We thus obtain ln |B(t)| = Kt + D, where we have amalgamated the two constants into a single constant D Exponentiating both sides gives |B(t)| = e Kt+D K dt

6 T R A N S C E N D E N T A L F U N C T I O N S

or B(t) = e D e Kt = P e Kt ( )

Notice that we have omitted the absolute value signs since the number of bacteria is always positive Also we have renamed the constant e D with the simpler symbol P Equation ( ) will be our key to solving exponential growth and decay problems We motivated our calculation by discussing bacteria, but in fact the calculation applies to any function that grows at a rate proportional to the size of the function Next we turn to some examples

652 Bacterial Growth

EXAMPLE

A population of bacteria tends to double every four hours If there are 5000 bacteria at 9:00 am, then how many will there be at noon

SOLUTION

To answer this question, let B( t) be the number of bacteria at time t For convenience, let t = 0 correspond to 9:00 am and suppose that time is measured in hours Thus noon corresponds to t = 3 Equation ( ) guarantees that B( t) = P eK t for some undetermined constants P and K We also know that 5000 = B( 0) = P eK 0 = P We see that P = 5000 and B( t) = 5000 eK t We still need to solve for K Since the population tends to double in four hours, there will be 10, 000 bacteria at time t = 4; hence 10000 = B( 4) = 5000 eK 4