A bacteria culture initially contains 500 cells and grows at a rate proportional

Question

A bacteria culture initially contains 500 cells and grows at a rate proportional to its size. After 5 hours the population has increased to 570. (a) Find an expression for the number of bacteria after t hours. P(t)= (b) Find the number of bacteria after 7 hours. (c) Find the rate of growth after 7 hours. (d) When will the population reach 5000?

Explanation / Answer

The unit used to measure time seems to be hours. So we know that: t=0=>x(0)=500 t=1=>x(1)=570 and that dx/dt=ax, where x=x(t), for some constant a. dx/x=adt ln(x)=?dx/x=?adt=a?dt=at+c where c is another constant. So the required function is: ln(x)=at+c or, better: x(t)=exp(at+c)=Cexp(at) [C=exp(c)] Now, use initial data: t=0=>x(0)=500 t=1=>x(1)=570 to determine constants a and c: 500=x(0)=Cexp(a0)=C so x(t)=500exp(at) 570=x(1)=100exp(a) exp(a)=570/500=1.14 a=ln(1.14)= 0.0569 Finally we get the function x(t)=500exp(0.569t) (a) P(t)=500exp(0.569t) (b) P(7)=24050 cells (c) P'(7)=0.569*2401=13684.75 cells/hour (d) P(t)=5000 The equation to solve is 500exp(0.569t)=5000 exp(0.569t)=10 t=ln(10)/0.569=1.75hours

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