Starting with an initial value of p(0)equals30, the population of a prairie dog community grows at a rate of pprime(t)equals30minusstartfraction t over 2 endfraction (in units of prairie dogs/month), for 0less than or equalstless than or equals60.

a. what is the population 9 months later?
b. find the population p(t) for 0less than or equalstless than or equals60.

a) The population of prairie dogs after nine months is 280.

b) P(t) = 30 + 30 · t - t²/4 for 0 ≤ t ≤ 60

Explanation:

Hi there!

We have the following information:

The initial population is P(0) = 30.

The rate of growth of the population is the following:

P´(t) = 30 - t/2 where

a) Let´s find the function of the population of prairie dogs P(t). For that, let´s integrate the P´(t) function between t = 0 and t and between P = 30 and P

P(t) = ∫P´(t)

P´(t) = dP/dt = 30 - t/2

Separating variables:

dP = (30 - t/2) dt

∫dP = ∫(30 - t/2) dt

P - 30 = 30 · t - t²/4

P(t) = 30 + 30 · t - t²/4

The population of prairie dogs at t = 9 months will be equal to P(9):

P(9) = 30 + 30(9) - (9)²/ 4

P(9) = 280 prairie dogs

The population of prairie dogs after nine months is 280.

b) P(t) = 30 + 30 · t - t²/4 (it was obtained in part a).

the inner core.

good luck: )

klicker

because if the tool that you are using is not balanced it there would be false mass.