Asked by Regina
Assuming P≥0, suppose that a population develops according to the logistic equation
dP/dt=0.03P−0.00015P^2
where tt is measured in weeks. Answer the following questions.
1. What is the carrying capacity?
I tried solving the differential equation and got 600 but it says it's wrong what am I doing wrong??
Carrying Capacity:
1
2. What is the value of kk?
Answer: k=k=
1
3. For what values of PP is the population increasing?
Answer (in interval notation):
4. For what values of PP is the population decreasing?
Answer (in interval notation):
dP/dt=0.03P−0.00015P^2
where tt is measured in weeks. Answer the following questions.
1. What is the carrying capacity?
I tried solving the differential equation and got 600 but it says it's wrong what am I doing wrong??
Carrying Capacity:
1
2. What is the value of kk?
Answer: k=k=
1
3. For what values of PP is the population increasing?
Answer (in interval notation):
4. For what values of PP is the population decreasing?
Answer (in interval notation):
Answers
Answered by
Steve
dp/dt=0.03p−0.00015p^2
This is a Bernoulli equation, with solution
200 e^<sup><sup>0.03t</sub></sub> / (e^<sup><sup>c</sub></sub>+e^<sup><sup>0.03t</sub></sub>)
No idea what k is supposed to be, or the carrying capacity. You will need some more info to determine c.
Since dp/dt = 0.00015p(200-p)
its roots are at p=0 and p=200
So, p is growing until it reaches 200, then starts decreasing. The problem is, p never reaches 200. So, p is always increasing. Read up on logistic growth.
This is a Bernoulli equation, with solution
200 e^<sup><sup>0.03t</sub></sub> / (e^<sup><sup>c</sub></sub>+e^<sup><sup>0.03t</sub></sub>)
No idea what k is supposed to be, or the carrying capacity. You will need some more info to determine c.
Since dp/dt = 0.00015p(200-p)
its roots are at p=0 and p=200
So, p is growing until it reaches 200, then starts decreasing. The problem is, p never reaches 200. So, p is always increasing. Read up on logistic growth.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.