Answer to Question #264526 in Differential Equations for joanne

Question #264526
  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
1
Expert's answer
2021-11-14T16:22:48-0500

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

Bacteria at final state=100000

Tripled after 2.5hrs

The exponential form is;

100000=10003(t150)100000=1000*3^{(\frac{t}{150})}

100=3(t150)100=3^{(\frac{t}{150})}

ln 100=t150ln 3ln\ 100=\frac{t}{150}ln\ 3

t150=(ln 100ln 3)\frac{t}{150}=(\frac{ln\ 100}{ln\ 3})

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

t=628.77mins629minst=628.77mins \approx629mins

t=10hrs 29minst=10hrs \ 29 mins

Time =8.00 am+10hr 29 mins

=6.29pm


Population at 2.00pm

P=10003(t150)P=1000*3^{(\frac{t}{150})}

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

P=10003(360150)P=1000*3^{(\frac{360}{150})}

=41899.83


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

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

ln N=kt+Cln\ N=kt+C

Given for t=0, N=40million

    ln(40)=C\implies ln(40)=C

And for t=10, N=50million

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

k=110(ln 50ln 40)=110ln (1.25)k=\frac{1}{10}(ln\ 50-ln\ 40)=\frac{1}{10}ln\ (1.25)

Hence, ln (N)=110ln (1.25)t+ln (40)ln\ (N)=\frac{1}{10}ln\ (1.25)t+ln\ (40)

For t=20

ln (N)=110ln (1.25)20+ln (40)ln\ (N)=\frac{1}{10}ln\ (1.25)*20+ln\ (40)

=ln (1.25)2+ln (40)=ln (401.252)ln\ (1.25)^2+ln\ (40)=ln\ (40*1.25^2)

N=40*1.252

=62.5million


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