Given:
The population was 100 after 10 mins.
The population was 1500 after 30 mins.
To fill the blanks:
Explanation:
According to the problem, we write,
[tex]\begin{gathered} P=P_0e^{kt} \\ 100=P_0e^{10k}.........(1) \\ 1500=P_0e^{30k}............(2) \end{gathered}[/tex]Dividing equation (2) by equation (1), we get
[tex]\begin{gathered} \frac{1500}{100}=\frac{P_0e^{30k}}{P_0e^{10k}} \\ 15=e^{20k} \\ ln15=20k \\ 2.708=20k \\ k=\frac{2.708}{20} \\ k=0.1354 \end{gathered}[/tex]So, the equation becomes,
[tex]P=P_0e^{0.1354t}....................(3)[/tex]a) To find: The initial population
When P = 100 and t = 10, then the initial population would be,
[tex]\begin{gathered} 100=P_0e^{0.1354(10)} \\ 100=P_0e^{1.354} \\ 100=P_0(3.873) \\ P_0=\frac{100}{3.873} \\ P_0\approx25.82 \end{gathered}[/tex]Therefore, the initial population is 25.82.
b) To find: The doubling time
Using the formula,
[tex]\begin{gathered} t=\frac{\ln2}{k} \\ t=\frac{\ln2}{0.1354} \\ t=5.1192 \\ t\approx5.12mins \end{gathered}[/tex]The doubling time is 5.12 mins.
c) To find: The population after 65 mins
Substituting t = 65 and the initial population is 25.82 in equation (3) we get,
[tex]\begin{gathered} P=25.82e^{0.1354(65)} \\ P\approx171467.56 \end{gathered}[/tex]Therefore, the population after 65 mins is 171467.56.
d) To find: The time taken for the population to reach 10000
Substituting P = 10000 and the initial population is25.82 in equation (3) we get,
[tex]\begin{gathered} 10000=25.82e^{0.1354t} \\ e^{0.1354t}=\frac{10000}{25.82} \\ e^{0.1354t}=387.297 \\ 0.1354t=\ln(387.297) \\ 0.1354t=5.959 \\ t=\frac{5.959}{0.1354} \\ t\approx44.01 \end{gathered}[/tex]Therefore, the time taken for the population to reach 10000 is 44.01 mins.
Final answer:
• The initial population is 25.82.
,• The doubling time is 5.12 mins.
,• The population after 65 mins is 171467.56.
,• The time taken for the population to reach 10000 is 44.01 mins.