Logistic growth Scientists often use the logistic growth...

Logistic growth Scientists often use the logistic growth function $P(t)=\frac{P_{0} K}{P_{0}+\left(K-P_{0}\right) e^{-r_{0} t}}$ to model population growth where $P_{0}$ is the initial population at time $t=0, K$ is the carrying capacity, and $r_{0}$ is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model. (FIGURE CAN'T COPY) When a reservoir is created by a new $\operatorname{dam}, 50$ fish are introduced into the reservoir, which has an estimated carrying capacity of 8000 fish. A logistic model of the fish population is $P(t)=\frac{400,000}{50+7950 e^{-0.5 t}},$ where $t$ is measured in years. a. Graph $P$ using a graphing utility. Experiment with different windows until you produce an $S$ -shaped curve characteristic of the logistic model. What window works well for this function? b. How long does it take for the population to reach 5000 fish? How long does it take for the population to reach $90 \%$ of the carrying capacity? c. How fast (in fish per year) is the population growing at $t=0 ?$ At $t=5 ?$ d. Graph $P^{\prime}$ and use the graph to estimate the year in which the population is growing fastest.

Logistic growth Scientists often use the logistic growth function $P(t)=\frac{P_{0} K}{P_{0}+\left(K-P_{0}\right) e^{-r_{0} t}}$ to model population growth
where $P_{0}$ is the initial population at time $t=0, K$ is the carrying capacity, and $r_{0}$ is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model.
(FIGURE CAN'T COPY)

When a reservoir is created by a new $\operatorname{dam}, 50$ fish are introduced into the reservoir, which has an estimated carrying capacity of 8000 fish. A logistic model of the fish population is $P(t)=\frac{400,000}{50+7950 e^{-0.5 t}},$ where $t$ is measured in years.
a. Graph $P$ using a graphing utility. Experiment with different windows until you produce an $S$ -shaped curve characteristic of the logistic model. What window works well for this function?
b. How long does it take for the population to reach 5000 fish? How long does it take for the population to reach $90 \%$ of the carrying capacity?
c. How fast (in fish per year) is the population growing at $t=0 ?$ At $t=5 ?$
d. Graph $P^{\prime}$ and use the graph to estimate the year in which the population is growing fastest.

Key Concepts

Logistic Growth Function

The logistic growth function is used to model populations that grow exponentially when small but level off as they approach a maximum limit. This mathematical model reflects constraints in the environment by incorporating a carrying capacity, meaning that the population cannot grow indefinitely and will instead level off in an S-shaped or sigmoid curve.

Carrying Capacity

Carrying capacity is the maximum population size that an environment can sustain indefinitely. In logistic growth models, it acts as the upper limit and represents the resources available in the environment. As the population approaches the carrying capacity, the growth rate decreases due to resource limitations.

Sigmoidal (S-shaped) Curve

A sigmoidal, or S-shaped, curve is characteristic of logistic growth. It shows slow initial growth, a phase of rapid increase, and finally a plateau as the population nears its carrying capacity. This curve visually represents the transition from exponential growth to a stable equilibrium.

Rate of Change and its Derivative

In population dynamics, the derivative of the logistic growth function represents the instantaneous rate of change of the population. Analyzing this derivative helps in understanding at what time the growth is fastest, which is crucial for studies involving population management and resource allocation.

Parameter Interpretation in Logistic Models

Parameters in the logistic function have specific interpretations; for example, the initial population represents the starting point at time zero, the growth rate determines how quickly the population increases, and the carrying capacity indicates the environmental limit. Understanding these parameters is essential for modeling real-life scenarios and interpreting the outcomes of the logistic growth model.

Transcript

00:01 To start with the problem of logistic growth, we have a equation here for the population growth of fish.

00:15 And the equation is written as such, and the first part of our problem is asking us to graph it and find a good window for that.

00:25 So i went ahead and did that.

00:29 I just bring the window into here.

00:33 And we have the equation as you can see, and i'm just going to hide it to get a better view of what we're looking at.

00:44 So we can see that in the course of time from maybe zero to 25 years after the fish are placed into the environment, the population will increase at first exponentially and then, it will keep increasing, but at a decreasing rate until it nears its capacity at 8 ,000.

01:19 So this is the graph, the graph settings, if it will display it.

01:26 The x -axis i currently have is about 20 to 40, and the y -axis, negative 250 to, i originally set it to 8500, so it's 82505.

01:44 So that is the, that is part one.

01:52 So we can take that away.

01:55 Or just bring back up the whiteboard here.

02:00 And the next part is asking us, when does the population reach 5 ,000? and when does reach 90 % capacity? so we need to solve, what we could do is solve this twice, but what i'm going to do is just rewrite the equation for x.

02:22 So we get 50p plus 7950p e to the negative point 5t is equal to 400 ,000.

02:47 So we multiplied it by the bottom half of the bottom part of the fraction, the denominator.

02:53 So then we can subtract 50p from both sides.

03:04 So minus 50p, minus 50p, horrible p, so this will give us 7950 p to the negative 0 .5 t is equal to the negative 0 .5 t is equal to to 400 ,000 to 3 minus 50p then we can divide both sides over 795 p so we get e to the negative point 5t equal to 400 ,000 minus 50p divided by 7950p divided by 7950 then we can take the natural log of both sides.

04:23 We can move down here to say that negative 0 .5 t is equal to the natural log of 400 ,000 minus 50p divided by 79 50p then we can multiply both sides by 2 or negative 2.

05:04 So we get negative 2 times and negative 2 times.

05:13 So on this side, the 1 1 half and the 2 will cancel.

05:20 So we are left with t is equal to negative 2 natural log of 400 ,000 minus 50p over 7 ,950p.

05:34 Now so they want to know when the population will be 5.

05:38 Thousand.

05:40 So we can plug this in.

05:43 One to pause this real quick so i can plug these in and get both of my answers and not waste your time.

05:51 All right, i'm going to resume here.

05:53 So when we plug in 5 ,000, so i'm going to say t 5 ,000 equal to, you can plug this into your calculators to figure this out.

06:19 9, 4, 5 approximately.

06:26 And then they say to check for 90 % of the maximum population.

06:32 So just to double check, our maximum population is 8 ,000.

06:45 Multiply that by 0 .9.

06:48 We get 7200.

06:51 And if you plug that in, we get t7200.

07:00 7 is more like a 2.

07:03 So we get 7, 200 is equal to 14 .5, 3, 2, 6.

07:21 So that is the, those are the solutions for the second part.

07:31 Just circle those.

07:36 Now it is asking us, how fast is it growing? at t is equal to zero so to find the growth rate that is what the derivative is it's the rate of change of the original equation so we need to calculate p prime of t and to quickly copy that up down so you get so now we can get that p of t equal to, again 400 ,000 divided by 50 plus 7950 e to the negative 0 .5 t.

08:54 So we need for our derivative here, we can recognize that the denominator is one large function, so we need to take the derivative of this function.

09:10 So we can call that g of t, g of t equal to 50 plus 7950, e to the negative point 5t, raise this g of t, negative 0, negative 0 ,0, negative 0 .5t, all to the negative 1.

09:50 So the derivative of g of t, g prime, t is equal to using the power rule of derivatives, we multiply by negative one as our current exponent, and then we subtract one from our exponent, which will give us negative 2.

10:15 So we get negative 1 times anything is just so i'm just going to have the parentheses over that.

10:24 50 plus 7, 9.

10:29 Horrible 7.

10:32 Fix that.

10:35 Still bad.

10:39 7 .9.

10:42 5 .0.

10:47 E to the negative 0 .5t.

10:58 All to the negative 2 power.

11:02 And now, since we have that, we can substitute that in for the denominator, but we still need to multiply it by the derivative of the inside function, which is just 7950 or 50 plus 7950, e to the negative 0 .5t.

11:30 So get the, you can rewrite, so we still have 400 ,000 divided by, since we have negative 1 ,000 we multiply them by negative 1.

11:53 50 plus 7 .9 .0.

12:00 E to the negative point 5t...

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