Ask your own question, for FREE!
Mathematics 36 Online
OpenStudy (anonymous):

A population grows according to the logistic law with a limiting population of 5×109 individuals. The initial population of 109 begins growing by doubling every hour. What will the population be after 4 hours?

OpenStudy (anonymous):

109 should be \[10^9\]

OpenStudy (anonymous):

The equation I used was y' = r(1 - (y/L))y The solution is y(t) = L / [1 + [(L/y0) - 1] ]e^(-rt) where y0 is y naught and represents the initial population. I initially solved for the rate using the DE. 2*(10^9) = r(1-(10^9/5*10^9))10^9 I found r and then plugged and chugged with the solution. My answer was incorrect. Any ideas? How do I find the rate? I feel as if that is incorrect.

OpenStudy (anonymous):

Haha you're making it much more difficult than it has to be. Try just doubling the initial value 4 times?

OpenStudy (anonymous):

There is a limiting factor of 5*10^9 individuals, so that needs to be accounted for in the calculation. I have to use the equations above to calculate amount at t =4. I took the logistics law equation y' = r(1 - (y/L))y and used that to find the rate. (r= rate, y= population, L = limiting factor) 2*(10^9) = r(1-(10^9/5*10^9))10^9 I multiplied by 2 because the population initially doubled.

OpenStudy (anonymous):

What do you get when you double 4x?

OpenStudy (anonymous):

8*10^9 :/

OpenStudy (anonymous):

Then wouldn't it make sense that the population stops at 5e9? That's the property of the limit.

OpenStudy (anonymous):

It's carrying capacity. After a certain point, an ecosystem can't hold more individuals. At that point deaths or offspring decrease and growth stabilizes.

OpenStudy (anonymous):

Meaning a plateau, right?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

the logistics equation needs to be used

OpenStudy (anonymous):

Alright, well then by definition the highest a population can get is the limiting factor. So, take the limit of the equation, and you'd get the 5e9, so that would be one way to use the equation to show that that is the highest it can get.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!