A population of protozoa develops with a constant relative growth rate of 0.6671 per member per day. On day zero the population consists of 4 members. Find the population size after 7 days. Since the relative growth rate is 0.6671, then the differential equation that models this growth is

Answer :

Answer:

The population size after 7 days is about 427.

Step-by-step explanation:

If [tex]y(t)[/tex] is the value of a quantity [tex]y[/tex] at time [tex]t[/tex] and the if the rate of change of [tex]y[/tex] with respect to [tex]t[/tex] is proportional to its size [tex]y(t)[/tex] at any time, then

                                                  [tex]\frac{dy}{dt} =ky[/tex]

where [tex]k[/tex] is a constant.

This equation is sometimes called the law of natural growth (if [tex]k>0[/tex]).

The only solutions of the differential equation [tex]\frac{dy}{dt} =ky[/tex] are the exponential functions

                                                  [tex]y(t)=y(0)e^{kt}[/tex]

Let [tex]P[/tex] be the population size and let [tex]t[/tex] be the time variable, measured in hours. Since the relative growth rate is 0.6671, then the differential equation that models this growth is

                                                 [tex]\frac{dP}{dt} =0.6671\cdot P[/tex]

According with the above information the solution to this differential equation is

                                                [tex]P(t)=P(0)e^{0.6671t}[/tex]

On day zero the population consists of 4 members [tex]P(0)=4[/tex].

Therefore, the population size after 7 days is

[tex]P(7)=4e^{0.6671\cdot 7}\\\\P(7)=4e^{4.6697}=4\cdot \:106.66573=426.66295[/tex]

about 427.

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