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

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Homework Help Overview

The discussion revolves around modeling the growth of the Pacific halibut biomass using the logistic equation, represented by a differential equation. Participants are tasked with calculating the biomass after one year and determining the time required for the biomass to reach a specific value.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to apply the logistic growth formula to find the biomass after one year and question whether they are using the correct values for variables. There is also discussion about the relationship between the variables P and y in the context of the equation.

Discussion Status

Some participants have identified errors in their calculations and have corrected them, while others are still clarifying their understanding of the variables involved. There is an acknowledgment of a mistake related to parentheses in calculations, and one participant expresses confidence in their answer for part (a).

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. There is also a language barrier noted by one participant, which may affect their contributions.

FritoTaco
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Homework Statement


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

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

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 M = 7\times 10^7 kg, and k=0.78 per year.

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

Homework Equations



\displaystyle P= \dfrac{K}{1+Ce^{-kt}}

The Attempt at a Solution



K = carrying capacity \implies 7\times 10^7 kg
k = 0.78 per year

At time 0, biomass is 2\times 10^7 kg \impliesy(0)= 2\times 10^7 kg

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

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

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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FritoTaco said:
\dfrac{7 \times 10^7}{1+\dfrac{5}{2}e^{-0.78\cdot1}}= 70000001.15 kg
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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FritoTaco said:

Homework Statement


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

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

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 M = 7\times 10^7 kg, and k=0.78 per year.

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

Homework Equations



\displaystyle P= \dfrac{K}{1+Ce^{-kt}}

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

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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Thanks, guys, I got the right answer now. Major parentheses mistake for the calculator.

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

Why the "but"? Your answer for (a) is correct.
 
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Sorry for my bad English. For (a) I got 3.26 \times 10^7
 

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