exponentialgrowthanddecay指数增长和衰变

1、Exponential Growth and Decay,Section 5.7,Problem: A single bacterium is in a Petri dish. Every 3 seconds the bacteria doubles. Find the relationship between t, the number of seconds, and N(t), the number of bacteria.,When will bacteria population reach 1000?,Variable in exponent,Take log (base 2) of both sides to undo the exponential function.,Change of base,The bacteria will increase its population from 1 to 1000 in 29.90 seconds.,We say the bacteria obey the law of uninhibited growth. This mea

2、ns the number of bacteria grows exponentially, the relationship between the number of bacteria and time is given by an exponential function.,Formula for uninhibited growth / decay,A0 = initial amount (at time 0) A(t) = amount after t years, days, etc t = time (years, days, etc) k = growth / decay constant You do not need this formula. You can derive a formula like we did using the table.,k is specific to substance if k 0, growth,To use the formula, you need to find k first. Then you can use the

3、formula to answer any questions, like when will the population reach 1000. Lets redo the problem this way.,We know A0, the initial amount, is 1. We also know the number of bacteria will be 2 at t = 3; meaning A(3) = 2.,Variable in exponent,Take log (base e) of both sides to undo the exponential function,Substituting this value of k and A0 = 1 into the equation , we get . This is the equation that relates time t to the number of bacteria present A(t).,When will the bacteria population reach 1000?

4、,Variable in exponent,Take log (base e) of both sides to undo the exponential function.,(in seconds),Half life of radioactive substances the amount of time it takes for one half of the substance present to decay expl: half life of radium is 1690 years,Recall the formula below.,A0 = initial amount (at time 0) A = amount after t years, days, etc t = time in years, days, etc k = growth / decay constant,expl: #4 Iodine 131 is a radioactive material that decays according to the function,where A0 is t

5、he initial amount present and A(t) is the amount present at time t (in days). What is the half life of iodine 131?,When will ?,Variable in exponent,Take log (base e) of both sides to undo the exponential function.,So the half-life of iodine 131 is 7.97 days.,expl: #14 A fossilized leaf contains 70% of its carbon 14. How old is it? (In other words, how long ago did it die?) Use 5600 years as the half life of carbon 14.,where A(t) = percent of original carbon 14 after t years,The question was how

6、many years have gone by if the percent of carbon 14 present, which is A(t), is 70%?,Percent present is 70,Variable in exponent,Take log (base ) of both sides to undo exponential function.,5.4 rule,Divide by 100.,The leaf died approximately 2882 years ago.,Change of Base,Half life formula,A0 = initial amount (at time 0) A(t) = amount after t years, days, etc t = time (years, days, etc) h = half life (years, days, etc),You can use the half life formula for half life problems if you want. I personally just develop the formula from a table of values like I demonstrated. 5.7 homework: 1, 3, 5, 7, 11, 13, 23,

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