Fish Population A small lake is stocked with a certain species of fish. The fish popu

Fish Population A small lake is stocked with a certain...

$$ \begin{array}{l}{\text { Fish Population A small lake is stocked with a certain }} \\ {\text { species of fish. The fish population is modeled by the }} \\ {\text { function }}\end{array} $$ $$ P=\frac{10}{1+4 e^{-0.8 t}} $$ $$ \begin{array}{l}{\text { where } P \text { is the number of fish in thousands and } t \text { is }} \\ {\text { measured in years since the lake was stocked. }} \\ {\text { (a) Find the fish population after } 3 \text { years. }} \\ {\text { (b) After how many years will the fish population reach }} \\ {5000 \text { fish? }}\end{array} $$

$$
\begin{array}{l}{\text { Fish Population A small lake is stocked with a certain }} \\ {\text { species of fish. The fish population is modeled by the }} \\ {\text { function }}\end{array}
$$
$$
P=\frac{10}{1+4 e^{-0.8 t}}
$$
$$
\begin{array}{l}{\text { where } P \text { is the number of fish in thousands and } t \text { is }} \\ {\text { measured in years since the lake was stocked. }} \\ {\text { (a) Find the fish population after } 3 \text { years. }} \\ {\text { (b) After how many years will the fish population reach }} \\ {5000 \text { fish? }}\end{array}
$$

Key Concepts

Logistic Growth Model

This model represents how populations grow in a limited environment. It starts with an exponential growth phase and then slows down as it approaches a maximum capacity. The logistic function is widely used in biology and ecology to model populations with limited resources, capturing the initial rapid growth and the eventual stabilization.

Exponential Functions and Equations

Exponential functions appear in the growth processes, where the rate of change is proportional to the current size. In population models, exponential terms modulate the growth initially. Solving equations that contain exponential terms, such as finding the time when the population reaches a specific level, typically involves logarithmic manipulation to isolate the variable of interest.

Transcript

00:01 This question gives us a population model and asks us a couple questions about it.

00:06 The population model, which models fish in a lake, looks like this.

00:10 The population p is equal to 10 divided by 1 plus 4e, 4 times e to the negative 0 .8 t, where t is the number of years since the population was recorded.

00:29 So the first part of this question asks, how many fish will be in this lake three years later? well, all we have to do here is plug in three for t.

00:40 Doing this will get p equals 10 divided by 1 plus 4e to the negative 0 .8 times 3.

00:49 And this is all something that we can just plug into our calculator.

00:53 When we do that, we'll find that the population, remember p is in thousands of fishes, population p will be equal to 7 .3 ,4 ,000 fish.

01:11 That's a lot of fish.

01:13 Next, the question asks, how long will it take before the population is 5 ,000 fish? remember, we're measuring in thousands, so p is just equal to 5.

01:24 Here, this was going to take a little bit more work.

01:26 We can start, again, by plugging it in, but we're going to have to do some rearranging.

01:30 We'll have 5 equals 10, divided by 1 plus 4e to the negative 0 .8t, and we want to solve for t.

01:41 Our first step, we're gonna be getting t out of the denominator.

01:46 I'm gonna multiply both sides by 1 plus 4e to the negative 0 .8t and divide both sides by five...

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