Transcript Slide 1

Example 2 Bacteria reproduce at a rate proportional to the number present.
100 bacteria are placed in a dish of agar. After 3 hours there are 2700 bacteria in
this dish.
(a) How many bacteria will be in this dish after 4 hours?
(b) How long will it take until there are 10,000 bacteria in this dish?
Solution Let y(t) denote the number of bacteria in the dish after t hours. The
rate of reproduction y / of these bacteria is k times the amount y present.
Therefore this situation satisfies the differential equation y / = ky with solution
y= y(0)ekt.
We are given that y(0) = 100 and y(3) = 2700. In other words,
y= 100ekt and 2700 = y(3) = 100e3k.
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Then e3k = 27, 3k = ln 27 and k  ln 27  ln 27 1 / 3  ln 3. Thus
y = 100 e(ln 3)t = 100(eln 3) t = 100(3 t ).
(a) After 4 hours, y(4) = 100(34) = 100 (81) = 8100 bacteria.
(b) There will be 10,000 bacteria in the dish when
10,000  y (t )  100(3t ), 3t  100, t  log3 100  4 hours, 12 minutes