Solving Diff. Equations: Population Growth & Changes

Click For Summary
SUMMARY

This discussion focuses on solving two differential equations related to population growth. The first equation, P=2000/(1+9e^.06t), requires finding the time when the population is growing the fastest by setting the second derivative d²P/dt² to zero. The second equation, dp/dt=0.08p(1-p/1000)(1-200/p), involves demonstrating that the population increases when 200 PREREQUISITES

  • Understanding of differential equations
  • Familiarity with population growth models
  • Knowledge of calculus, specifically derivatives and second derivatives
  • Ability to solve inequalities
NEXT STEPS
  • Learn how to find critical points in differential equations
  • Study the application of the second derivative test in population models
  • Explore the implications of population dynamics using the logistic growth model
  • Review techniques for solving inequalities in calculus
USEFUL FOR

Students and professionals in mathematics, biology, and environmental science who are interested in modeling population dynamics and analyzing growth patterns using differential equations.

gordda
Messages
20
Reaction score
0
Hello
I am having major difficulties with two Differential equation.
1. If I have the equation P=2000/(1+9e^.06t). where p=population and t=time. How do I work out when the population is growing the fastest? To work this question out do I just make t the subject and diff twice and let it equal to zero?

2. The diff equation dp/dt= .08p(1-p/1000)(1-200/p) has an inital population of Po how do I show that the population is increasing if 200<P<1000 and decreasing if 0<P<200.

Please help

Thanx:)
 
Physics news on Phys.org
gordda said:
Hello
I am having major difficulties with two Differential equation.
1. If I have the equation P=2000/(1+9e^.06t). where p=population and t=time. How do I work out when the population is growing the fastest? To work this question out do I just make t the subject and diff twice and let it equal to zero?

2. The diff equation dp/dt= .08p(1-p/1000)(1-200/p) has an inital population of Po how do I show that the population is increasing if 200<P<1000 and decreasing if 0<P<200.
1) set d2P/dt2 = 0 and solve for t is a good beginning.
2) show that (dp/dt)>0 when 200<P<1000, and (dp/dt)<0 when 0<P<200. it's not difficult.
 
ok i get 1 but in 2 how do i show that dp/dt is >0 or <0.
 
gordda said:
ok i get 1 but in 2 how do i show that dp/dt is >0 or <0.
you're thinking it's more difficult than it really is.
begin by evaluating (dp/dt) at p=100 and p=500.
which term changes sign? do any other terms change sign?
now generalize these observations over the 2 given ranges of p.
 
Last edited:
shouldn't i use values more closer to the range
 
gordda said:
shouldn't i use values more closer to the range
you can.
however, what are the solutions to the 2 inequalities below (for p > 0)?
(1-200/p) < 0
(1-200/p) > 0
and under what condition does the following inequality hold:
(1-p/1000) > 0
the above inequality solutions give the 2 ranges of p.
 
geosonel said:
you can.
however, what are the solutions to the 2 inequalities below (for p > 0)?
(1-200/p) < 0
(1-200/p) > 0
and under what condition does the following inequality hold:
(1-p/1000) > 0
the above inequality solutions give the 2 ranges of p.
if you need review of solving inequality equations, try here:
http://www.sosmath.com/algebra/inequalities/ineq01/ineq01.html
proceed page by page thru the various review topics to the extent needed.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
14
Views
3K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
13
Views
3K
  • · Replies 33 ·
2
Replies
33
Views
12K