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I am not sure where I am going wrong, I will admit I do not fully understand all the equations but I do feel like I have a pretty good grasp on them.

This is from Stewart Calculus 7e pg 613 #3

"The pacific halibut fishery has been modeled by the differential $$\frac{dy}{dt} = ky \left(1 - \frac{y}{M}\right)$$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 $M = 8 * 10^7 kg$ and $k = .071$ per year.

a) if $y(0) = 2 \cdot 10^7 kg$ find the biomass a year later.

b) How long will it take for the biomass to reach $4 \cdot 10^7 kg$?"

First I start by putting in the given information into the logistic equation. $$\frac{dy}{dt} = .071y\left(1 - \frac{y}{8*10^7}\right)$$

Starting with $a$ I need to find $y$ when $t$ is $0$ so I use another equation from the book. I find that $A = 3$ $$A = \frac{M - y_0}{y_0}$$ Where I am assuming $y_0$ is the initial value. $y(0) = 2 \cdot 10^7$

$$A = \frac{(8 \cdot 10^7) - (2 \cdot 10^7)}{ 2 \cdot 10^7}$$

This comes out to 3 so I put it in the other equation that solves for $y(t)$

$$y(t) = \frac{M}{1+ Ae^{-kt}}$$

$$y(1) = \frac{(8 \cdot 10^7) }{1+ 3e^{-.071}}$$

This comes out to $21083781.89$ which I know is wrong, going through my work it all seems to be correct, I am not sure where I have gone wrong.

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  • $\begingroup$ Why do you know this is wrong? $\endgroup$
    – Did
    Jun 23, 2012 at 19:41
  • $\begingroup$ The book gets $3.23 \cdot 10^7$ $\endgroup$
    – user138246
    Jun 23, 2012 at 19:44
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    $\begingroup$ Then $k=0.71$ instead of $k=0.071$. $\endgroup$
    – Did
    Jun 23, 2012 at 19:46

1 Answer 1

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The solution is correct. If the value of $k=0.71$ is used instead of $k=0.071$, the answer comes out to $$y(1) = \frac{(8 \cdot 10^7) }{1+ 3e^{-0.71}} \approx 3.23\cdot 10^7$$ in accordance with the textbook answer. (Observed by Did)

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