Question

A biologist observed that a certain bacterial colony obeys the population growth law and that the colony triples every 4 hours.

If the colony occupied 2 square centimeters initially, find:

(a) An expression for the size P(t) of the colony at any time t.

(b) The area occupied by the colony after 12 hours.

(c) The doubling time for the colony?

Answer 1

Answer:

a) $P(t) = 2e^{0.275t}$

b) 54.225 square centimeters.

c) 2.52 hours

Step-by-step explanation:

The population growth law is:

$P(t) = P_{0}e^{rt}$

In which P(t) is the population after t hours, $P_{0}$ is the initial population and r is the growth rate, as a decimal.

In this problem, we have that:

The colony occupied 2 square centimeters initially, so $P_{0} = 2$

The colony triples every 4 hours. So

$P(4) = 3P_{0} = 6$

(a) An expression for the size P(t) of the colony at any time t.

We have to find the value of r. We can do this by using the P(4) equation.

$P(t) = P_{0}e^{rt}$

$6 = 2e^{4r}$

$e^{4r} = 3$

Applying ln to both sides, we get:

$4r = 1.1$

$r = 0.275$

So

$P(t) = 2e^{0.275t}$

(b) The area occupied by the colony after 12 hours.

$P(t) = 2e^{0.275t}$

$P(12) = 2e^{0.275*12}$

$P(12) = 54.225$

(c) The doubling time for the colony?

t when $P(t) = 2P_{0} = 2*2 = 4$.

$P(t) = 2e^{0.275t}$

$4 = 2e^{0.275t}$

$e^{0.275t} = 2$

Applying ln to both sides

$0.275t = 0.6931$

$t = 2.52$