How Does the Logistic Equation Model the Growth of the Pacific Halibut Biomass?

In summary: I calculated the value of P, I used 1 instead of t. That is why I said I can solve b, but not a.English is not my native language too. I do not understand what you are saying. You have got the correct answer for (a). Why do you say you cannot solve it?
  • #1
FritoTaco
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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?
 
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  • #2
FritoTaco said:
[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.
 
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  • #3
FritoTaco said:

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
$$ \frac{7 \times 10^7}{1 +\displaystyle \frac{5}{2} e^{-0.78 \times 1}} = 4 \times 10^7$$
 
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  • #4
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
FritoTaco said:
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.
 
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  • #6
Sorry for my bad English. For (a) I got [itex]3.26 \times 10^7[/itex]
 

Related to How Does the Logistic Equation Model the Growth of the Pacific Halibut Biomass?

1. What is a population growth model?

A population growth model is a mathematical representation of how a population changes over time, taking into account factors such as birth rate, death rate, and migration. It can help scientists predict future population trends and understand the dynamics of a population.

2. What are the different types of population growth models?

There are several types of population growth models, including exponential growth, logistic growth, and the Lotka-Volterra model. Each model makes different assumptions about population growth and can be used to study different types of populations.

3. How can population growth models be used to inform policy decisions?

Population growth models can be used to help policymakers make informed decisions about issues such as resource allocation, urban planning, and environmental management. By understanding how a population is likely to change in the future, policymakers can better plan for the needs of their communities.

4. What are the limitations of population growth models?

Population growth models are simplifications of complex real-world populations, so they may not perfectly reflect all factors that influence population growth. Additionally, they rely on data that may be incomplete or inaccurate, which can affect the accuracy of their predictions.

5. How can population growth models be validated?

Population growth models can be validated by comparing their predictions to real-world data. If the model accurately predicts population trends, it can be considered a valid representation of the population. Additionally, multiple models can be compared to see which one best fits the available data.

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