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.

c. How fast (in fish per year) is the population growing at t=0? At t=5?

Accepted Answer

Welcome back, everyone. In a newly established wildlife reserve, 100 rabbits are introduced into an area with an estimated carrying capacity of 10,000 rabbits. A logistic model of the rabbit population is given by R of T equals 1 million divided by 100 + 9900 E to the negative 0.3 T, where T is measured in years. How fast in rabbits per year is the population growing at T equals 0, approximate to the nearest whole number? ASS is 15 rabbits per year, B 30, C 60, and D 120. Now, how are we going to figure out how fast, OK, in rabbits per year, the population is growing at T equals 0. Well, to figure out how fast it's growing, we'll need the rate of change for the population with respect to time, and then we can substitute T equals 0 into that rate of change. Notice here that we are already given the function for the population, OK? The population as a function of time. So that means if we can differentiate our function. RFT with respect to T, then it will give us the rate of change for the population, and we can use that to figure out how fast our population is growing. So let's differentiate R of T. So far We know that our function are of T. Is equal to 1 million. Divided by 100. Plus 9900 E to the negative 0.3 T. So our function is a quotient. What do we know about differentiating quotients? Well, recall that we can find the derivative of a quotient using the quotient rule. It basically tells us that if we have a function that's a quotient, Y equals u divided by V, where U and V are differentiable functions. Then the derivative of Y will be equal to Vu minus UV divided by V2, where U and V are the derivatives of U and V respectively. So if we look at our function R of T, we can tell that our numerator is U, our denominator is V. So if we differentiate both of these with respect to T, then we can substitute them into the quotient rule to solve for the derivative of R of T. And once we do that, then we'll be able to find out how fast our population is growing at that point. So let's go ahead and do that. So we know that you is going to be equal to 1 million. And when we differentiate you with respect to T, the derivative is 0 because U is our constant. Next, we know that our denominator V is 100 plus 9900 E to the negative 0.3 T. and now when we differentiate this function E, basically we differentiate what's in the power, bring it down, and write back our original function. So we know the derivative of 100 is zero, but for our second term it's going to be negative 0.3T. Multiplied by. 9900 to 0.3 T. 0.3, rather the derivative of negative 0.3 T multiplied by 9900 E to the negative 0.3 T which. Is going to be equal. The negative 2970 E to the negative 0.3 T. So this is V. Know that we have the derivatives of U and V, we can apply the quotient rule. So no, this means that R of T, sorry, the derivative of R rather. With respect to tea. Is going to be equal to. ZU, OK, so that's gonna be 0 multiplied by 100 plus 9900 E to the negative 0.3 T. Minus UV that has 1 million. Multiplied by negative 2970 E to the negative 0.3 T, OK? I know all of this is being divided by V2d. 100 + 9900 E to the negative 0.3 T. Squared Now, in this case, OK, here, our first term will be zero, and when we multiply 1 million by 2970, we're left with 2.97 billion, OK? So 2 and then we're at 970, and we know we have 60s after that. OK. So 2.97 billion multiplied by E to the negative 0.3T in our numerator, and of course we'll write back our denominator. So now we have the, the, the rate of change of the population with respect to time. Thus we can figure out the rate at T equals 0 because now at T equals 0, then E raised to the negative 0.3 T will be E raises to 0.3 multiplied by 0, which equals E 0 and anything raised to the power of 0 equals 1. So that means then that our rate of change. At T equals 0 is going to be equal to 2.97 billion. Multiplied by one. Divided by 100. Plus 9900 multiplied by 1 are squared. In this case, that's 2.97 billion. Divided by this would be 10,000, so 10,000 squared or 100 million, OK. I know when we divide these, then we'll get our rate of change to be equal to 29.7 or approximately 30 rabbits per year to the nearest whole number. Thus, when we look back on our answer choices, that tells us then that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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