Question

A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 290. (a) Find an expression for the number of bacteria after t hours. P(t) = (b) Find the number of bacteria after 2 hours. (Round your answer to the nearest whole number.) P(2) = bacteria (c) Find the rate of growth after 2 hours. (Round your answer to the nearest whole number.) P'(2) = bacteria per hour (d) When will the population reach 10,000? (Round your answer to one decimal place.) t = hr

Answer 1

Answer:

a) $P(t) = 100e^{1.0647t}$

b) P(2) = 841 bacteria

c) P'(2) = 895 bacteria per hour

d) t = 4.3hr

Step-by-step explanation:

The equation of growth for the number of bacteria can be modeled by the following differential equation:

$\frac{dP}{dt} = r$

With solution.

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

In which P(t) is the population after t hours, P(0) is the initial population and r is the growth rate.

(a) Find an expression for the number of bacteria after t hours.

Initially 100 cells, so P(0) = 100.

After 1 hour, 290 cells, so P(1) = 290.

We apply this to the equation to find the value of r.

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

$290 = 100e^{r}$

$e^{r} = 2.9$

Applying ln to both sides of the equality.

$\ln{e^{r}} = \ln{2.9}$

$r = 1.0647$

So

$P(t) = 100e^{1.0647t}$

(b) Find the number of bacteria after 2 hours.

This is P(2).

$P(t) = 100e^{1.0647t}$

$P(2) = 100e^{1.0647*2}$

$P(2) = 841$

841 bacteria after 2 hours.

(c) Find the rate of growth after 2 hours.

$P(t) = 100e^{1.0647t}$

$P'(t) = 100*1.0647e^{1.0647t}$

$P'(t) = 106.47e^{1.0647t}$

$P'(2) = 106.47e^{1.0647*2}$

$P'(2) = 895$

(d) When will the population reach 10,000?

This is t when P(t) = 10,000. So

$P(t) = 100e^{1.0647t}$

$10000 = 100e^{1.0647t}$

$e^{1.0647t} = \frac{10000}{100}$

$e^{1.0647t} = 100$

Applying ln to both sides

$\ln{e^{1.0647t}} = \ln{100}$

$1.0647t = \ln{100}$

$t = \frac{\ln{100}}{1.0647}$

$t = 4.3$