Question

consider the population model
dp/dt=0.3(1-p/200)(p/50-1)p
where p(t) is the population at time t
For what values of p is the population in equilibrium?
can someone help me with this question...=.="

Answer 1

derivative is zero ?

0 = 0.3(1-p/200)(p/50-1)p

Answer 2

p=0,p=50,p=200

Answer 3

when dp/dt=0 it will be a max or a min.

Answer 4

i think that i should write the equation more properly since it may give different meaning~~
the exact equation is
dp/dt=[0.3][1-(p/200)][(p/50)-1][p]
sorry for the inconvenience~~

Answer 5

i think my answer holds ..

Answer 6

-.3p^3/1000+1.5p^2/200-.3p=dp/dt

Answer 7

when the derivative is zero - the population isnt changing..
0 = 0.3(1-p/200)(p/50-1)p
p=0,p=50,p=200

Answer 8

why ?
ana_anne wrote : dp/dt=[0.3][1-(p/200)][(p/50)-1][p]

Answer 9

-.3p^4/4000+p^3/400-.15p^2+c=p(t)

Answer 10

nevermind read it wrong. 1-1 duh.

Answer 11

so they dont need the time, just the equilibrium. 0, 50, 200 are the points of inflection.
0 and 200 are not maxima or minima though. 50 is the equilibrium

Answer 12

dear coolsector~~
how can i help u~~
i was just trying to make the equation easier to be understand~~=)

Answer 13

i gave you the solution already..

Answer 14

the equilibrium occurs at p=50.

Answer 15

thankz tutordiffeq n coolsector~~
=)

Answer 16

what about 0,200 ?

Answer 17

they are points of inflection, not maxima or minima. graph it if you wanna see. i put the integral up

Answer 18

i dont really understand what's wrong with it .. the derivative is in fact zero there
so no change occures

Answer 19

dear tutordiffeq, what do u mean by u integral it up~~
i cant understand it~~~ =.="