Question

If the growth rate of bacteria at any time t is proportional to the number present at t and triples in 1 week

Answer 1

The population after 20 weeks will be 403.42$x_{0}$ in which $x_{0}$ is the initial population.

Given that the growth rate of bacteria at any time t is proportional to the number present at t and triples in 1 week.

We are required to find the number of bacteria present after 10 weeks.

let the number of bacteria present at t is x.

So,

dx/dt∝x

dx/dt=kx

1/x dx=k dt

Now integrate both sides.

$\int\limits {1/x} \, dx$=$\int\limits{k} \, dt$

log x=kt+log c----------1

Put t=0

log $x_{0}$=0 +log c    ($x_{0}$ shows the population in beginning)

Cancelling log from both sides.

c=$x_{0}$

So put c=$x_{0}$ in 1

log x=kt+log $x_{0}$

log x=log $e^{kt}$+log $x_{0}$

log x=log $e^{kt}x_{0}$

x=$e^{kt}x_{0}$

We have been given that the population triples in a week so we have to put the value of x=2$x_{0}$ and t=1 to get the value of k.

2$x_{0}$=$e^{k} x_{0}$

2=$e^{k}$

log 2=k

We have to now put the value of t=20 and k=log 2 ,to get the population after 20 weeks.

x=$e^{20log 2}$$x_{0}$

x=$e^{0.30*2}$$x_{0}$

x=$e^{6}x_{0}$

x=403.42$x_{0}$

Hence the population after 20 weeks will be 403.42$x_{0}$ in which $x_{0}$ is the initial population.

Learn more about growth rate at brainly.com/question/25849702

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The given question is incomplete as the question incudes the following:

Calculate the population after 20 weeks.