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Example 2 Bacteria reproduce at a rate proportional to the number present.

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?

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Example 2 Bacteria reproduce at a rate proportional to the number present.

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  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= 100ektand 2700 = y(3) = 100e3k. Then e3k = 27, 3k = ln 27 and 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

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