# Models for Population Growth

1. Jul 23, 2017

### FritoTaco

1. The problem statement, all variables and given/known data
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.)

2. Relevant equations

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

3. The attempt at a solution

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

At time 0, biomass is $2\times 10^7 kg$ $\implies$$y(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?

2. Jul 23, 2017

### Staff: Mentor

The two sides here are not equal. The denominator is larger than 1, the fraction cannot be larger than the numerator.

3. Jul 23, 2017

### Ray Vickson

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. Jul 23, 2017

### FritoTaco

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

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

5. Jul 24, 2017

### Ray Vickson

6. Jul 24, 2017

### FritoTaco

Sorry for my bad English. For (a) I got $3.26 \times 10^7$