Exponential Growth and Decay - Welcome to the Department of 指数的增长和衰变教研室.pptVIP

Exponential Growth and Decay Section 6.7 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. 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). 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 For part e, we know . We want to know when will . We’ll do that next! 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. Half life formula Worksheet “Logarithmic and exponential applications: Exponential decay and growth” will provide some practice problems. 6.7 homework: 1acdef, 3bcde, 5, 9, 11, 21 * … 12 9 6 3 0 N(t) (# of bacteria) t (seconds) The graph has the shloopy shape of an exponential function. Since one times a number is that number, we can ignore the 1 in front and just write . Why does the graph not go to the left of the y-axis? 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 means 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, decay if k 0, growt