Question

The rate of change in the number of bacteria in a culture is proportional to the number present. In a certain laboratory experiment, a culture has 10,000 bacterial initially, 20,000 bacteria at time t 1 minutes, and 100,000 bacteria at (t 1+10 ) minutes.a. In terms of t only, find the number of bacteria in the culture at any time t minutes, t ≥ 0b. How many bacteria were there after 20 minutes?c. How many minutes had elapsed when the 20,000 bacteria were observed?

Answer 1

Answer:

(a) $N(t)=10000e^{(\frac{ln5}{10})t }$

(b) 25,000

(c) 4.3068 min.

Step-by-step explanation:

Rate of change in the number of bacteria is proportional to the number present.

Let N is the population of bacteria.

$\frac{dN}{dt}$ ∝ N ⇒ $\frac{dN}{dt}=kt$ { k = proportionality constant}

initial population No. = 10,000

                          $N(t_{1} )$ = 20,000

         and $N(t_{1}+10 )=100,000$

(a) For population growth

$N(t)=N_{0}e^{kt}=10000e^{kt}$

$N(t_1)=10,000e^{kt_1}=20,000$

$e^{kt}=2$

$ln(e^{kt_1})=ln(2)$

$kt_1=ln(2)$

$t_{1}=\frac{ln2}{k}$ ----------(1)

$N(t_1+t_{10})=100,000$

$100,000=10,000e^{k(t_1+10)}$

$10=e^{k(t_1+10)}$

$ln10=ln[e^{k(t_1+10)}]$

$k(t_1+10)=ln10$

$k(t_1)=ln10-10k$

$t_1=\frac{ln10-10k}{k}$ ----------(2)

from equation (1) and (2)

$\frac{ln_2}{k}=\frac{ln10-10k}{k}$

$ln10-ln2=10k$

$k=\frac{ln5}{10}$

so expression will be

$N(t)=10000e^{(\frac{ln5}{10})t }$

(b) for t = 20

    $N_{(20)}=10,000e\frac{ln5}{10}\times 20$

             =  $10,000\times e^{2ln5}$

            = 10,000 × 25

            = 25,000

(c) Since $t_1=\frac{ln2}{k}$   [from equation (1)]

                  $=\frac{ln2}{\frac{ln5}{10} }$

                  $=\frac{ln2}{ln5}\times 10$

                  = 4.3068

                 = 4.3068 min.