Respuesta :
Answer:
a) [tex] P(t) = 660 e^{\frac{ln(6)}{6} t}[/tex]
b) [tex] P(t=3) = 660 e^{\frac{ln(6)}{6}*3} = 1616.663 \approx 1617[/tex]
Step-by-step explanation:
For this case we have the following info:
[tex] P(0) = 660[/tex] represent the initial amount of bacteria
[tex] P(6) = 3960[/tex] represent the population of bacteria after 6 hours
Part a
For this case we use the proportional model given by:
[tex] \frac{dP}{dt}= kP[/tex]
Where P represent the population at time t and k is a constant of proportionality.
We can rewrite the model like this:
[tex] \frac{dP}{P} = k dt[/tex]
And if we integrate both sides we got:
[tex] ln |P|= kt + C[/tex]
Now if we apply exponential we got:
[tex] P = e^{kt +c}= e^{kt} e^C[/tex]
[tex] P(t)= P_o e^{kt}[/tex]
For this case using the initial condition:
[tex] 660 = P_o e^{0k}, P_o = 660[/tex]
We have the following model:
[tex] P(t) = 660 e^{kt}[/tex]
Using the second condition [tex] P(6) = 3960[/tex] we have this:
[tex] 3960 = 660e^{6k}[/tex]
We can divide both sides by 660 and we got:
[tex] 6= e^{6k}[/tex]
Now we can apply natural log on both sides and we got:
[tex] ln (6)= 6k[/tex]
[tex] k = \frac{ln(6)}{6}=0.2986265782[/tex]
So then the model would be given by:
[tex] P(t) = 660 e^{\frac{ln(6)}{6} t}[/tex]
Part b
For this case we just need to replace t=3 into the model and we got:
[tex] P(t=3) = 660 e^{\frac{ln(6)}{6}*3} = 1616.663 \approx 1617[/tex]
