Models for Population Growth

  • Thread starter FritoTaco
  • Start date
  • #1
133
23

Homework Statement


The Pacific halibut fishery has been modeled by the differential equation.

[itex]\displaystyle\dfrac{dy}{dt}=ky\left(1-\dfrac{y}{M} \right)[/itex]

where y(t) is the biomass (the total mass of the members of the population) in kilograms at time t (measured in years), the carrying capacity is estimated to be [itex]M = 7\times 10^7 kg[/itex], and [itex]k=0.78[/itex] per year.


(a) If [itex]y(0)= 2\times 10^7 kg[/itex], find the biomass a year later. (Round your answer to two decimal places.)
(b) How long will it take for the biomass to reach [itex]4\times 10^7 kg[/itex]? (Round your answer to two decimal places.)


Homework Equations



[itex]\displaystyle P= \dfrac{K}{1+Ce^{-kt}}[/itex]

The Attempt at a Solution



K = carrying capacity [itex]\implies 7\times 10^7 kg[/itex]
k = [itex]0.78[/itex] per year

At time 0, biomass is [itex]2\times 10^7 kg[/itex] [itex]\implies [/itex][itex]y(0)= 2\times 10^7 kg[/itex]

C = the difference between the carrying capacity and the initial capacity subtracted by 1.

[itex]C= \dfrac{7\times 10^7}{2 \times 10^7}-1=\dfrac{5}{2}[/itex]



[itex]P=\dfrac{7 \times 10^7}{1+\dfrac{5}{2}e^{-0.78\cdot1}}= 70000001.15 kg[/itex]

I'm trying to solve for the biomass (P) after 1 year. This answer doesn't seem correct. Am I using the wrong number for variable t? Or I'm not solving for P right away?
 

Answers and Replies

  • #2
35,442
11,883
[itex]\dfrac{7 \times 10^7}{1+\dfrac{5}{2}e^{-0.78\cdot1}}= 70000001.15 kg[/itex]
The two sides here are not equal. The denominator is larger than 1, the fraction cannot be larger than the numerator.
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722

Homework Statement


The Pacific halibut fishery has been modeled by the differential equation.

[itex]\displaystyle\dfrac{dy}{dt}=ky\left(1-\dfrac{y}{M} \right)[/itex]

where y(t) is the biomass (the total mass of the members of the population) in kilograms at time t (measured in years), the carrying capacity is estimated to be [itex]M = 7\times 10^7 kg[/itex], and [itex]k=0.78[/itex] per year.

(a) If [itex]y(0)= 2\times 10^7 kg[/itex], find the biomass a year later. (Round your answer to two decimal places.)
(b) How long will it take for the biomass to reach [itex]4\times 10^7 kg[/itex]? (Round your answer to two decimal places.)


Homework Equations



[itex]\displaystyle P= \dfrac{K}{1+Ce^{-kt}}[/itex]

[itex]P=\dfrac{7 \times 10^7}{1+\dfrac{5}{2}e^{-0.78\cdot1}}= 70000001.15 kg[/itex]

I'm trying to solve for the biomass (P) after 1 year. This answer doesn't seem correct. Am I using the wrong number for variable t? Or I'm not solving for P right away?

Is ##P## the same as ##y##? I assume so.

Your final equation is incorrect; it should be
$$ \frac{7 \times 10^7}{1 +\displaystyle \frac{5}{2} e^{-0.78 \times 1}} = 4 \times 10^7$$
 
  • #4
133
23
Thanks, guys, I got the right answer now. Major parentheses mistake for the calculator.

I can solve b, but for a.) I got [itex]3.26\times10^7[/itex]
 
  • #5
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
Thanks, guys, I got the right answer now. Major parentheses mistake for the calculator.

I can solve b, but for a.) I got [itex]3.26\times10^7[/itex]

Why the "but"? Your answer for (a) is correct.
 
  • #6
133
23
Sorry for my bad English. For (a) I got [itex]3.26 \times 10^7[/itex]
 

Related Threads on Models for Population Growth

Replies
9
Views
14K
  • Last Post
Replies
2
Views
1K
  • Last Post
2
Replies
28
Views
5K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
12
Views
825
Replies
2
Views
3K
Replies
2
Views
2K
Top