Question

The University of Arizona is monitoring the spread of a bacteria by placing 100 spores of the bacteria at the start of the project. The number, P, of the bacteria is expected to grow according to the differential equation dP over dt equals the product of 0.05 times P and the quantity 1 minus P over 2000, where P t is measured in days. When does the number of bacteria reach 1000?

Answer 1

Answer:

In 59 days, the number of bacteria reaches 1000.

Step-by-step explanation:

 Given the data in the question;

by placing 100 spores of the bacteria at the start of the project,

p(0) = 100

dP over dt equals the product of 0.05 times P and the quantity 1 minus P over 2000

dp/dt = 0.05p( 1 - p/200 )

2000dp / p( 2000 - p ) = 0.05dt

[1 / p + 1 / 200-p ]dp = 0.05 dt

Now, we integrate

∫[1 / p + 1 / 200-p ]dp = ∫ 0.05 dt

ln p - ln [2000 - p ]  = 0.05 + c'

ln[ p / 2000 - p ] = 0.05 + c'

p / 2000 - p  = $e^{0.05t + c'$

p / 2000 - p  = k$e^{0.05t$

Now, since p(0) = 100, we substitute

100 / (2000 - 100)  = k$e^{0.05(0)$

100 / 1900  = k × $e^{0$

100 / 1900  = k × 1

k = 1 / 19

Therefore,

p / 2000 - p  = k$e^{0.05t$

p / 2000 - p  = (1/19)$e^{0.05t$

when p( t ) = 1000, then t will be;

1000 / (2000 - 1000)  = (1/19)$e^{0.05t$

1000 / 1000  = (1/19)$e^{0.05t$

1  = (1/19 ) × $e^{0.05t$

1/(1/19 )  = $e^{0.05t$

(1 × 19)/1 = $e^{0.05t$

$e^{0.05t$ = 19

0.05t = ln( 19 )

0.05t = 2.9444

t = 2.9444 / 0.05

t = 58.888 ≈ 59 days

Therefore, In 59 days, the number of bacteria reaches 1000.