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The constant is required because there are no limits on the integrals In this problem, as with every problem in growth or decay, there is an initial amount of material In this case there is an initial number of bacteria at the start of the experiment Call this initial amount N o In the language of mathematics, at t N =N o Substitute these values into In N = kt + C
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InN,=k(O)+C
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With the constant evaluated in terms of the initial amount of material the basic relation is
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If you had any trouble manipulating the logarithms in the previous line, go back and review the manipulative rules for logarithms At this point switch to an exponential format
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This last statement correctly describes the model The number of bacteria at any time starts out at No ( eo = 1) and increases with time in an exponential manner
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This N = Noekr is the general growth law for something with growth proportional to the number present Some text authors begin the discussion of growth and decay with this equation This approach is simple but neglects the development of a mathematical model of a simple statement that "the growth of is proportional to the number of present at any time" A little reflection will convince you that this model fits many different phenomena
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Suppose in this bacteria growth problem that 100 bacteria are introduced into a growth environment (water, nutrients, etc) and that 2 hours later the bacteria are separated fiom the environment or otherwise identified and that their number has increased to 130 Can this information be used to determine the growth law With these two numbers, N and N o , and the time interval the constant k can be evaluated The calculation is a little logarithm and exponent intense but follow along wt your calculator Substitute as follows: ih 130 = 100e2k or 130 = e 2k
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To solve for k switch 130 = e2k to a logarithmic equation (Buzz Word Bee says "A logarithm is an exponent") The logarithm of something is an exponent so
ln130=2k
1 k=-h130=013 2
The specific law governing the growth of these bacteria in this environment is
With this law it is possible to predict how many of these bacteria would be present, say, after 12 hours and starting with 50 bacteria Put in the 50 for N o and the 12 hours for t and we get N - 50~013'12 50(476) = 238 = This model that starts with the statement that the growth rate is proportional to the amount present can, with a modest amount of calculus and initial information, be used to predict hture growth There is a standard pattern to growth and decay problems that always works The general procedure for these problems is outlined below
1) Any problem where the number of events is proportional to the number of participants present can be written as dN/dt equals a constant (+k for growth and dN - k for decay) times the number: -= +W
Pattern
dN dN 2) Rearrange to -= f k d t and integrate -= f k l dt to get In N = fkt + A Take N N aninitialnumber No,at t=O,toevaluate A=InN,,andwrite InN=+_kr+InN, N 3) Rearrange the equation to In---=M N O N -=eefkt or N = N o e f k t No and switch to an exponential format
4) One data point, a certain N at a specific time, allows calculation of k (For example, a
20% increase in N o in one hour means 1 2N0 = Noelk or 12 = e l k Switch to a logarithmic equation and k = In 12 = 069 and finally write N = N0e069')
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5 ) With the calculation of k, the specific growth or decay equation is written for the
same conditions that produced the initial data With this specific growth or decay equation N a any time can be predicted t
Refer to this procedure in subsequent problems It is a very logical procedure for growth and decay problems and it works Growth and decay problems are favorite test problems Know how to work them and especially know how to switch fiom exponential equations to logarithmic equations and vice versa and know how to take logs and perform exponentiation on your hand calculator
If "a fool and his money are soon parted," the rate at which it leaves is probably proportional to the amount remaining If a certain fool starting with $20,000 starts gambling his money away and after 2 hours has lost $2000, how long will it take for him to loose 90% of the original amount
9- 14
Solution: The basic assumption in this problem is that the fool will loose in proportion to the amount he has at any time Humans are a little harder to predict than bacteria, but this is a good assumption Follow the procedural steps as written previously and be aware of the logic in the problem
Stepl: The statement "the rate at which the fool looses money is proportional to the amount present" means that
Step 2: Rearrange, integrate, and evaluate the constant of integration with the initial data -=-kdt,
-= -kjdZ,
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