A culture of the bacterium Salmonella enteritidis initially...

A culture of the bacterium Salmonella enteritidis initially contains 50 cells. When introduced into a nutrient broth, the culture grows at a rate proportional to its size. After 1.5 hours, the population has increased to 925. (a) Find an expression for the number of bacteria after t hours. (Round your numeric values to four decimal places.) P(t) = (b) Find the number of bacteria after 7 hours. (Round your answer to the nearest whole number.) P(7) = bacteria (c) Find the rate of growth (in bacteria per hour) after 7 hours. (Round your answer to the nearest whole number.) P'(7)= bacteria per hour (d) After how many hours will the population reach 250,000? (Round your answer to one decimal place.) t = hr

Transcript

00:01 Hi, today we are solving the question in which we are given that a culture of bacterium initially contains 50 cells.

00:24 So when introduced into a nutrient broad, the culture grows at a rate proportional to its size.

00:32 So after 1 .5 hours, so at 1 .5 hours, the population has increased to 925.

00:45 So we need to find an expression for the number of bacteria after tr's first.

00:53 So here, the exponential growth model is taken as p is equals to p .0 into e.

01:02 To the power r into t here p not is the initial growth p is the population r is the growth rate t is the time period so it is given that initially there are 50 cells and after 1 .5 hours so it led to 925 cells so we can substitute our values as here 920 25 is equals to 50 into e raised to the power r into 1 .5.

01:44 So from here we get it as on solving it further, we get it as 925 by 50 is equals to e raised to the power 1 .5 r.

02:03 Now solving further we get it as so here we get it as 18 .5 is equals to e raised to the power 1 .5 r so on solving it further we get it as so from here on solving it further we get it as loan of 18.

02:30 Point five is equals to loan e raised to the power 1 .5 r so loan 18 .5 is 2 .917 so that will be equal to 1 .5 r so from here solving for r we get 2 .917 by 1 .5 is equal to 1 .5 is equal to r so we get r value as e is equals to so we get r is equals to 1 .94 now from here we get the required expression as therefore pt is equals to 50 e raised to the power 1 .94 into t.

03:29 Now taking the next part so here we are given with t is equals to seven hours so we have to find the number of bacteria so as the we get the required expression in party we'll substitute t value so p seven is equal to 50 e raised to the power 1 .94 into seven on solving it we get it equals to we get it equals to 3 .950 so therefore population of bacteria after 7 will be equal to 3 .950.

04:08 Now taking the next one, so we are given that we have to take the rate of growth after 7 hours.

04:17 So t is given as 7 hours and we are, we know that pt is equals to 50 into e raised to the power 1 .94 into t.

04:33 Now differentiating pt is with respect to t here...

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