Respuesta :
Answer:
We are given that the rate of change is proportional to its size S
So, [tex]\frac{dS}{dt} \propto S[/tex]
[tex]\frac{dS}{dt} = kS[/tex]
[tex]\frac{dS}{S} = kdt[/tex]
Integrating both sides
[tex]\log(S)= kt + log c[/tex]
[tex]\frac{S}{S_0}=e^{kt}[/tex]
[tex]S=S_0 e^{kt}[/tex]
S is the population after t hours
[tex]S_0[/tex] is the initial population
Now we are given that After 6 hours there will be 1200 bacteria
[tex]1200=200 e^{6k}[/tex]
[tex]6=e^{6k}[/tex]
[tex]6^{\frac{1}{6}=e^{k}[/tex]
So, [tex]S=200 \times 6^{\frac{t}{6}[/tex]
a)Now the population after t hours as a function of t; [tex]S=200 \times 6^{\frac{t}{6}[/tex]
b) What will be the population after 7 hours?
Substitute t = 7 hours
A bacteria culture starts with 200 bacteria
[tex]S=200 \times 6^{\frac{7}{6}}[/tex]
[tex]S=1617.607[/tex]
c) How long will it take for the population to reach 1750 ?
[tex]1750=200 \times 6^{\frac{t}{6}[/tex]
[tex]\frac{1750}{200} =6^{\frac{t}{6}[/tex]
[tex]8.75 =6^{\frac{t}{6}[/tex]
[tex]t=7.26[/tex]
So, it will take 7.2 hours for the population to reach 1750
