The size of a certain bacterial colony inceases at a rateproportional to the siz
ID: 2938312 • Letter: T
Question
The size of a certain bacterial colony inceases at a rateproportional to the size of the colony. Suppose the colonoccupied an area of 0.25 square centimeters initially, and after 8hours it occupied an area of 0.35 square centimeters. (a) Estimate the size of the colony t hours afterthe initial measurement. (b) What is the expected size of the colony after 12hours? (c) Find the doubling time of the colony. Your help is greatly appreciated. The size of a certain bacterial colony inceases at a rateproportional to the size of the colony. Suppose the colonoccupied an area of 0.25 square centimeters initially, and after 8hours it occupied an area of 0.35 square centimeters. (a) Estimate the size of the colony t hours afterthe initial measurement. (b) What is the expected size of the colony after 12hours? (c) Find the doubling time of the colony. Your help is greatly appreciated.Explanation / Answer
The size of a certain bacterial colony inceases at a rateproportional to the size of the colony. Suppose the colonoccupied an area of 0.25 square centimeters initially, and after 8hours it occupied an area of 0.35 square centimeters. (a) Estimate the size of the colony t hours afterthe initial measurement. p(t) = p0 . exp(k.t) 0.25 square centimeters initially ==> p(0) = 0.25 p(t) = 0.25 . exp(k.t) after 8 hours it occupied an area of 0.35 squarecentimeters==> p(8) = 0.35 0.35= 0.25 exp(8k) exp(8k) = 0.35 / 0.25 =1.4 8 k = ln(1.4) = 0.336472 k= 0.336472 / 8 k= 0.042059 HENCE p(t) = 0.25 . exp( 0.042059 t) (b) What is the expected size of the colony after 12hours? p(t) = 0.25 . exp( 0.042059 t) ==> p(12) = 0.25 .exp( 0.042059 (12)) = 0.414125 square centimeters. (c) Find the doubling time of the colony. p(t) = 0.25 . exp( 0.042059 t) 0.50 = 0.25 .exp( 0.042059 t) 2 = exp( 0.042059 t) 0.042059 t = ln(2) t = ln(2 ) / 0.042059 t= 16.4804 hoursRelated Questions
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