The population of endangered species is changing at a rate of P′(t) = 30 − 20t.

Question

The population of endangered species is changing at a rate of P′(t) = 30 − 20t. The known population at the time is a community of 200. Time is represented as the variable t.

a) What would the population of the species be after 3 years?

b) How long will it be till the population becomes extinct?

c) When will the population of the species be half of what it initially started with?

Answer 1

Po = original Population = 200

P'(t) = 30-20t= rate of change of the population, where t = number of years

Integrate

P(t) = Integral of 30-20t = 30t -10t^2 + C where C= an unknown constant, solve as follows for C with t=0

P(0) = 200 = 30(0) -10(0^2)+ C

C= 200

P(t)= -10t^2 +30t +200

P(3) = population after 3 years = 30(3)-10(3^2) +200 = 90-90=200

it goes extinct when P=0

P(t)= 30t -10t^2 +200 =0

divide by 10

3t -t^2 +20 =0

divide by -1

t^2 -3t -20=0

use the quadratic formula if you can't factor

t=3/2 +/-.5sqr(9+80)

t = 1.5 +.5sqr89, if ignore the negative square root

= about 1.5 +/-4.7

= about 6.22 years to go extinct

it reaches 100 when

P(t) = 100 = 30t-10t^2 +200

10t^2-30t-100 =0

t^2 -3t -10 =0

t= 3/2 +/-.5sqr(9+40)

=1.5+/-3.5

= -2 or 5 years

pop t

200 0

220 1

222.5 1.5 maximum population in 1 1/2 years

220 2

200 3

160 4

100 5 half the original population in 5 years

20 6

0 about 6.22 years when it goes extinct

population follows a parabolic curve, rising intially, then falling after 1 1/2 years

Answer 2

a)

Find P(t) by integrating P'(t) if P(0)=200

P(t)= -10t²+30t+200

b) Solve P(t) for t when P(t) = 0

c) Solve P(t) for t when P(t) = 100