Population problems! fun!

Hello everyone, im' having major problems doing these 2 population problems, because they don't seem to fit any of the population models we have went over. I'll try to give as much detail as to what i've been taught and to what the problem is, i'm not sure how i'm going to transform the equations given into the right population model.

Here are the 2 problems:
#1. As the salt KNO3 dissolves in methanol, the number of x(t) of grams of the salt in a solution after t seconds satisfies the differiental equation dx/dt = .8x - .004x^2

(a) what is the maximum amount of the salet that will ever dissolve in the methanol.

(b) If x = 50 when t = 0, how long will it take for an addtional 50 g of salt to dissolve?


answers:
A. lim x(t) = 200 grams
t->infinity

B. 5/4ln|3| == 1.37 seconds


#2. Suppose that the number x(t) (with t in months) of alligators in a swamp satisfies the differential equation:
dP/dt = .0001x^2 - .01x

(a). If initially there are 25 alligators in the swamp, solve this differential equation to determine what happens to the alligator population in the long run.

(b). Repeat part (a), except with 150 alligators initialy.


Answers:
A. the alligators eeventually die out.
B. Doomsday occurs after about 9 years 2 months.


Bounded populaitons and the logistics equation:
dP/dt = kP(M-P)
where k = b, and M = a/b are constants

population model, which is limiting population and carrying capacity
P(t) = MPo/[Po + (M-Po)e^(-kMt)]

I also saw a doomsday verus extiniction but we never went over it...it says:

dP/dt = kP^2 - (delta)P = kP(P-M);

Where M = (delta)/k > 0) as a mathematical model of population.


Alright, if anyone can help me that would be great!!

thanks!!



:afrog:

Quote:

---

Originally posted by voidflux
#1. As the salt KNO3 dissolves in methanol, the number of x(t) of grams of the salt in a solution after t seconds satisfies the differiental equation dx/dt = .8x - .004x^2

(a) what is the maximum amount of the salet that will ever dissolve in the methanol.

(b) If x = 50 when t = 0, how long will it take for an addtional 50 g of salt to dissolve?


answers:
A. lim x(t) = 200 grams
t->infinity

B. 5/4ln|3| == 1.37 seconds

---

If .8x = .004x^2 is the derivative, find when it is 0, then calculate the value of it... underive.

f(x) = x^n
f'(x) = nx^(n-1)

or, to underive

f'(x) = x^n
f(x) = x^(n+1)/(n+1)

Hey thanks for the responce, When i took the derivative of it, and I set it equal to 0 and solved for x, it came out to be x = 100 grams. And the answer in the back of the book is
lim x as t ->infinity = 200grams;


Any idea's what went wrong?
Thanks!:thumb:

How long does it take an alligator to dissolve in methanol?

I figured both these out, they where quite whorish, and it would probably take around two months.