Question

A bacteria culture grows with a constant relative growth rate. After 2 hours there are 400 bacteria and after 8 hours the count is 50,000.
(a) Find the initial population. P(0) = 80 )ãbacteria
(b) Find an expression for the population after t hours. r(t) = 180( 125(6 Plt) =180(125(2))-
(c) Find the number of cells after 7 hours. (Round your answer to the nearest integer.) P(7)=72.358- bacteria
(d) Find the rate of growth after 7 hours. (Round your answer to the nearest integer.) P(7) 2x bacteria/hour
(e) When will the population reach 200,000? (Round your answer to one decimal place.) hours

Answer 1

Answer:

a) P(0) = 80

b) $P(t) = 80(2.2361)^t$

c) 22,363 cells.

d) The rate of growth after 7 hours is of 18,000 bacteria per hour.

e) 9.7 hours.

Step-by-step explanation:

A bacteria culture grows with a constant relative growth rate.

This means that the population is given by:

$P(t) = P(0)(1+r)^t$

In which P(0) is the initial population and r is the growth rate, as a decimal.

After 2 hours there are 400 bacteria and after 8 hours the count is 50,000.

This means that in 6 hours, the population went from 400 bacteria to 50,000 bacteria. We use this to find r. So

$50000 = 400(1+r)^6$

$(1+r)^6 = \frac{50000}{400}$

$(1+r)^6 = 125$

$\sqrt[6]{(1+r)^6} = \sqrt[6]{125}$

$1 + r = 125^{\frac{1}{6}}$

$1 + r = 2.2361$

So

$P(t) = P(0)(2.2361)^t$

(a) Find the initial population. P(0)

We have that P(2) = 400. We use this to find P(0). So

$P(t) = P(0)(2.2361)^t$

$400 = P(0)(2.2361)^2$

$P(0) = \frac{400}{(2.2361)^2}$

$P(0) = 80$

So

$P(t) = 80(2.2361)^t$

(b) Find an expression for the population after t hours.

$P(t) = 80(2.2361)^t$

(c) Find the number of cells after 7 hours.

This is P(7). So

$P(7) = 80(2.2361)^7 = 22363$

22,363 cells.

(d) Find the rate of growth after 7 hours.

We have to find the derivative when t = 7. So

$P(t) = 80(2.2361)^t$

$P^{\prime}(t) = 80\ln{2.2361}(2.2361)^t$

$P^{\prime}(7) = 80\ln{2.2361}(2.2361)^7 = 18000$

The rate of growth after 7 hours is of 18,000 bacteria per hour.

(e) When will the population reach 200,000?

This is t for which $P(t) = 200000$. So

$P(t) = 80(2.2361)^t$

$200000 = 80(2.2361)^t$

$(2.2361)^t = \frac{200000}{80}$

$(2.2361)^t = 2500$

$\log{(2.2361)^t} = \log{2500}$

$t\log{2.2361} = \log{2500}$

$t = \frac{\log{2500}}{\log{2.2361}}$

$t = 9.7$

So 9.7 hours.