Question

A bacteria population starts with 400 bacteria and grows at a rate of r(t) = (450.268)e^{1.12567t} bacteria per hour. How many bacteria will there be after three hours

Answer 1

Answer

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Explanation:

Answer 2

Initial population is 400, rate of growth of the population is

$r(t)=450.268(e^(1.12567t)$

Which means the antiderivative of r(t) is the population size at time t.

Lets call it p(t)

p(t)=$\int\limits}(450.268e^1.12567t) \, dt$

which means the antiderivative of r(t) is the population size at time t.

Lets call it p(t)

$p(t)=\int\limits {450.268(e^1.12567t)} \, dt\\ u=1.12567t,\\du=1.12567dt\\=450.268\int\limits e^u.(1/1.12567du)$

$450.268/1.12567(e^u) +c\\=400e^(1.12567t)+c$

At time t=0, there are 400 bacteria so,

p(0)=400=$400e^(1.12567)(0)+c\\=400+C\\c=0$

So, the population function is

$p(t)=400e^(1.12567t)$

and at t=3 hours.

p(t)=$400e^(1.12567)(3)=11713$

so after three hours, there will be 11713bacteria.