Answer to Question #264526 in Differential Equations for joanne

Question

1. In 8:00 AM, the population of a bacteria is 1000. At 10:30 AM, the number of bacteria triples. At what time will the population become 100 times the initial population of bacteria? What will be the population at 2:00 PM?  Ans. 6:29 PM; 13,967
2.  If the present population of a certain country is 40 million and in 10 years, the population is 50 million, what will be its population 20 years from now? Ans. 62.5 million

Answer 1

1) Bacteria at initial state=1000,8:00am

Bacteria at final state=100000

Tripled after 2.5hrs

The exponential form is;

$100000=1000*3^{(\frac{t}{150})}$

$100=3^{(\frac{t}{150})}$

$ln\ 100=\frac{t}{150}ln\ 3$

$\frac{t}{150}=(\frac{ln\ 100}{ln\ 3})$

${t}=(\frac{ln\ 100}{ln\ 3})*150$

$t=628.77mins \approx629mins$

$t=10hrs \ 29 mins$

Time =8.00 am+10hr 29 mins

=6.29pm

Population at 2.00pm

$P=1000*3^{(\frac{t}{150})}$

t=2.00pm -8.00am=6hrs=360mins

$P=1000*3^{(\frac{360}{150})}$

=41899.83

2)$\frac{dN}{dt}=kn$

$\int \frac{dN}{N}=\int kdt$

$ln\ N=kt+C$

Given for t=0, N=40million

$\implies ln(40)=C$

And for t=10, N=50million

ln(50)=k*10+ln(40)

$k=\frac{1}{10}(ln\ 50-ln\ 40)=\frac{1}{10}ln\ (1.25)$

Hence, $ln\ (N)=\frac{1}{10}ln\ (1.25)t+ln\ (40)$

For t=20

$ln\ (N)=\frac{1}{10}ln\ (1.25)*20+ln\ (40)$

=$ln\ (1.25)^2+ln\ (40)=ln\ (40*1.25^2)$

N=40*1.25²

=62.5million