Exponential Growth and Decay Help
Introduction to Exponential Growth and Decay
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 there are 10 thousand bacteria
after 8 hours there are 20 thousand bacteria
after 12 hours there are 40 thousand bacteria
after 16 hours there are 80 thousand bacteria ...
The point is that each new generation of bacteria also reproduces , and the older generations reproduce as well. A sketch (Fig. 6.12) 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 Fig. 6.13. 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.
A Differential Equation
If B ( t ) represents the number of bacteria present in a given population at time t , then the preceding discussion suggests that
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
We integrate both sides with respect to the variable t :
The left side is
ln | B ( t )| + C
and the right side is
where C and 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
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 which grows at a rate proportional to the size of the function.
Next we turn to some examples.
Bacterial Growth Examples
A population of bacteria tends to double every four hours. If there are 5000 bacteria at 9:00 a.m., then how many will there be at noon?
To answer this question, let B ( t ) be the number of bacteria at time t . For convenience, let t = 0 correspond to 9:00 a.m. and suppose that time is measured in hours. Thus noon corresponds to t = 3.
Equation (*) guarantees that
B ( t ) = P · e Kt
for some undetermined constants P and K . We also know that
5000 = B (0) = P · e K · 0 = P .
We see that P = 5000 and B ( t ) = 5000 · e Kt . 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 · e K ·4 .
We divide by 5000 to obtain
2 = e K ·4 .
Taking the natural logarithm of both sides yields
ln 2 = ln( e K ·4 ) = 4 K .
We conclude that K = [ln 2]/4. As a result,
B ( t ) = 5000 · ( e ([ln 2]/4) t ).
We simplify this equation by noting that
e ([ln 2]/4) t = ( e ln 2 ) t /4 = 2 t /4 .
B ( t ) = 5000 · 2 t /4 .
The number of bacteria at noon (time t = 3) is then given by
B (3) = 5000 · 2 3/4 ≈ 8409.
It is important to realize that population growth problems cannot be described using just arithmetic. Exponential growth is nonlinear, and advanced analytical ideas (such as calculus) must be used to understand it.
Suppose that a certain petri dish contains 6000 bacteria at 9:00 p.m. and 10000 bacteria at 11:00 p.m. How many of the bacteria were there at 7:00 p.m.?
We know that
B ( t ) = P · e Kt .
The algebra is always simpler if we take one of the times in the initial data to correspond to t = 0. So let us say that 9:00 p.m. is t = 0. Then 11:00 p.m. is t = 2 and 7:00 p.m. is t = −2. The initial data then tell us that
From equation (*) we may immediately conclude that P = 6000. Substituting this into (**) gives
10000 = 6000 · ( e K ) 2 .
We conclude that
As a result,
At time t = −2 (7:00 p.m.) the number of bacteria was therefore
You Try It : A petri dish has 5000 bacteria at 1:00 p.m. on a certain day and 8000 bacteria at 5:00 p.m. that same day. How many bacteria were there at noon?
Find practice problems and solutions for these concepts at: Transcendental Functions Practice Test.
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