Answer:
a) the number of bacteria in the dish after an additional hour has passed = 8
b) the carrying capacity of the dish based on the developed logistic equation. = 9
Step-by-step explanation:
Given that ;
The Logistic growth model is :
[tex]\frac{dB}{dt}= B (a-bB)[/tex]
Solving the above equation ; we have a MODEL EQUATION;
[tex]B_t = \frac{aB_o}{bB_o+(a-bB_o)e^{-at}}[/tex]
in which:
[tex]B_o[/tex] represents the initial population of the bateria.
We are given that the number of the bacteria in the Petri dish was initially determined to be 1 , then ;
[tex]B_t = \frac{a*1}{b*1+(a-b*1)e^{-at}}[/tex]
[tex]B_t = \frac{a}{b+(a-b)e^{-at}}[/tex]
After an hour ; we were told that the number of the bacteria increased to 3 ; So:
[tex]3 = \frac{a}{b+(a-b)e^{-a(1)}}[/tex]
[tex]3b+3(a-b)e^{-a} =a ---- equation (1)[/tex]
Similarly after (2) hours; the number of the bacteria increased to 6; then
[tex]6 = \frac{a}{b+(a-b)e^{-a(2)}}[/tex]
[tex]6b+6(a-b)e^{-2a} =a ---- equation (2)[/tex]
So;
[tex]3b+3(a-b)e^{-a} =a ---- equation (1)[/tex]
[tex]6b+6(a-b)e^{-2a} =a ---- equation (2)[/tex]
Solving for a and b from the above two eqautions: Then,
a = 1.386
b = 0.154
Substituting the value of a and b into our MODEL EQUATION; we have
[tex]B_t = \frac{1.386}{0.154+(1.386-0.154*1)e^{-1.386t}}[/tex]
[tex]B_t = \frac{1.386}{0.154+1.232e^{-1.386t}}[/tex]
a) Determine the number of bacteria in the dish after an additional hour has passed.
i.e at t = 3
[tex]B = \frac{1.386}{0.154+1.232e^{-1.386(3)}}[/tex]
[tex]B = \frac{1.386}{0.154+1.232 * 0.01563}[/tex]
[tex]B= \frac{1.386}{0.173267}[/tex]
[tex]B = 7.9992[/tex]
[tex]B= 8[/tex]
b) Determine the carrying capacity of the dish based on the developed logistic equation.
The carrying capacity B can be expressed as :
[tex]B = \frac{a}{b}[/tex]
[tex]B = \frac{1.386}{0.154}[/tex]
B =9
