Question

a bacteria population starts with 200 bacteriaa 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:

11,513

Explanation:

Data provided in the question:

Initial bacteria = 200

Growth rate r(t) = $(450.268)e^{1.12567t}$

Now,

Total growth after 3 hours = $\int\limits^3_0 {(450.268)e^{1.12567t}} \, dt$

or

Total growth after 3 hours = $450.268\int\limits^3_0 {e^{1.12567t}} \, dt$

or

Total growth after 3 hours =$450.268[\frac{e^{1.12567t}}{1.12567}]^3_0$

[ ∵ $\int{d(e^x)}{dt}=\frac{e^x}{\frac{dx}{dt}}$]

Thus,

Total growth after 3 hours = 400 × $[{e^{1.12567t}]^3_0$

or

Total growth after 3 hours = 400 × $[{e^{1.12567(3)}-e^{1.12567(0)}]$

or

Total growth after 3 hours ≈ 11313

Hence,

Total bacteria after 3 hours = Initial bacteria + Total growth after 3 hours

= 200 + 11313

= 11,513