population question calculus

Question

A population of bacteria follows the continuous exponential growth model  P(t)=P0ekt​,

where t is in days. The relative​ (daily) growth rate is 4​%. The current population is 909.

​A) Find the growth model.​ (the function that represents the population after t​ days).

​P(t) = enter your response here

​B) Find the population exactly 1 weeks from now.  Round to the nearest bacterium.

The population in 1 weeks will be

enter your response here.

​C) Find the rate of change in the population exactly 1 weeks from now.Round to the nearest

unit.

The population will be increasing by about

enter your response here bacteria per day exactly 1 weeks from now.

​D) When will the popualtion reach 6727​? ROUND TO 2 DECIMAL

PLACES.The population will reach 6727

about enter your response here

days from now.

Answer 1

There are two different forms of an exponential function that will be of use here:

P(t) = P₀ · e^kt and P(t) = P₀ · (1 + r)^t

The first uses e, Euler's constant, an irrational number that is ∼ 2.71828, as the base of the exponential function.

The second uses (1 + r) as the base, where r is the % (daily) growth rate, expressed as a decimal.

The constant, P₀, is the same in both, and represents the initial population.

We will use the 2nd form here, which we could leave in that form to answer all of the questions, except the question directs us explicitly to find the function in the first form. So we will convert the 2nd form into the first as follows:

P(t) = 909·(1.04)^t

Now, we want to find k, which we can do by noting 1.04 = e^k. Taking the natural log of both sides of this equation gives us k = ln(1.04) ∼ .03922.

P(t) = 909 · e^.03922t

We calculate the answer to B by evaluating P(7).

We calculate the answer C by noting that the rate of change at t = 7 is given by k · P(7).

We get the answer to D by setting P(t) = 6,727 and solving for t using natural logs as above.