97–100. Logistic growth Scientists often use the logistic grow...

Question

97–100. Logistic growth Scientists often use the logistic growth function P(t) = P₀K / P₀+(K−P₀)e^−r₀t to model population growth, where P₀ is the initial population at time t=0, K is the carrying capacity, and r₀ 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.

{Use of Tech} Gone fishing When a reservoir is created by a new 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) = 400,000 / 50+7950e^−0.5t, where t is measured in years.

d. Graph P' and use the graph to estimate the year in which the population is growing fastest.

Accepted Answer

In the newly established wildlife reserve, 100 rabbits are introduced to an area with an estimated carrying capacity of 10,000 rabbits. A logistical model of the rabbit population as given by RFT equals 1 million divided by 100 plus 9900 e to the 0.3 T. Where T is measured in years, plot the graph of the derivative of R and determine the year when the population is growing fastest. Round the answer to two decimal places. Now to solve this, let's first find our derivative. We have RT Equals 1 million Divided by 100 plus 9900 E to the negative 0.3 T. We can use the quotient rule for this derivative. That tells us if we have a quotient. Such as G. Divided by H. Its derivative can be written. As G multiplied by H. Minus G multiplied by H. All divided by h squared. In our case, G is the 1 million. H Is the 100. Plus 9900 eats the negative 0.3 T. G then would be the derivative of the 1 million, which is 0. In H based on our exponent rule. would be 9900 multiplied. By 0.3, E to the negative 0.3 T. Now, we can play this into a quotient rule. We end up getting 0, multiplied by our H 100 plus 9900 E to the negative 0.3 T. Minus G 1 million. Multiplied by 9900. Multiplied by 0.3, E to the negative 0.3 T. We can actually simplify by multiplying our negative 0.3 and 9900. To get -2970, he eats the negative 0.3 T. All of this is divided. By 9900. E to the negative 0.3 T. The 100 All squared. We can now simplify this. This would give us 2970 multiplied by a million. E to the negative 0.3 T. Divided by 100 plus 9900 E to the negative 0.3 T squared. Now that we have our derivative equation, let's plug in some values to plot our graph. She can make a table. And let's say we use X equals 05, 1015. 20 25 and 30. Now, you plug in these values. Let's say for 0, we end up getting 29.7. For 5, we get 124.3. For 10, we get 420.7. For 15 We get 7:48. 3. For 20 We get 474.7, 25, 147.7. And for 30 We get pretty close to 29.7. Now, Let's find our value where we have a maximum. Now, we need to check the carrying capacity. Our carrying capacity If we look, we're starting at 10,000. A carrying capacity 10,000 divided by 2 gives us 5000 This would be when the population is halfway to its carrying capacity, which should be the maximum point on our graph. So then let's take our equation. RFT and instead it equals to 5000. We have 5000 Equals 1 million. Divided by 100 plus 9900 E to the 0.3 T. Now, let's solve for T. We can multiply our denominator to get 5000. Multiplied by 100 plus 9900 E to the negative 0.3 T. Equals 1 million We can expand to get 500,000. Plus 49 million 500,000 E to the negative 0.3 T equals 1 million. Now we seem to keep simplifying by subtracting 500,000 on both sides. We get 49,500,000. E to the negative 0.3 T equals 500,000. Now, if we divide both sides, we get E to the negative 0.3 T equals 500,000. Divided by 49,500,000. We can simplify Giving us E to the negative 0.3 T equals 5 divided by 495. And then we can simplify again by taking the natural log of both sides. We would get 0.3 T equals natural log of 5 divided by 495. Now, we just divide both sides by negative 0.3. To give us T equals natural log, 5 divided by 4 and 5, all divided by negative 0.3. If we approximate that with the calculator, we would see our value is about 15.32. And this is in years. Now, we go back to our derivative graph. You will have a maximum of 15.32. Let's go ahead and graph our points. We had 0, 29.7. We then had 5, 124.3. We had 10 for 20.7. 15 748.3. 20 or 74.7. 25, 147.7. And finally 30 and about 29.7. We also have our maximum point, 15.32. Which is just around our 15. We can then make a smooth curve between the points. We end up getting this graph right here. And so, we can then see the answer to our problem. It's 15.32 years. Modeled by our first derivative graph. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

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