Population growth Consider the following population functions....

Question

Population growth Consider the following population functions.

a. Find the instantaneous growth rate of the population, for t≥0.

p(t) = 600 (t²+3/t²+9)

Accepted Answer

Hi everyone, let's take a look at this practice problem dealing with derivatives. This problem says the concentration of a certain chemical in a reaction over time can be described by the function CFT is equal to 500, multiplying the quantity of 2 T2 + 5 in quantity divided by T2 + 7. Where T is the time and hour since the reaction started. Calculate the instantaneous rate of change of the chemical's concentration at any given time. T. We're given four possible choices as our answers. For choice A, we have 9000 T divided by the quantity of T2 + 7 in quantity squad. For choice B, we have 9000 divided by the quantity of T2 + 7 in quantity squared. For choice C, we have 9000 T divided by quantity of T2 + 7, and for choice D, we have 9000 divided by the quantity of T2 + 7. Now we're asked to calculate the instantaneous rate of change of the chemicals concentration. And so to do this, we need to take a derivative with respect to time of our function CFT since that is the chemicals concentration as a function of time. So, we're going to be calculating C of T. And that is going to be equal to the derivative with respect to T. Of the quantity of 500, multiplying the quantity of 2 T 2 plus 5 in quantity, divided by the quantity of T2 plus 7 in quantity. So taking this derivative, this is going to be equal to 500 is a constant, so I can pull it outside the derivative. So I'll have 500, multiplying the quantity, and here I'm gonna be taking a derivative of a fraction. So we're going to have to use our quotient rules. So we call for your quotient rule that we're going to have our denominator, which is the quantity of T2 + 7. In quantity, multiplied by the derivative with respect to T. Of our numerator, which is the quantity of 2 t 2 + 5 in quantity, then we'll have minus our numerator, which is the quantity of 2 T 2 + 5 in quantity multiplied by the derivative with respect to T of our denominator, which is the quantity of T2 + 7 in quantity. And all of that is divided by our denominator squared. This is this is going to be uh divided by the quantity of T2 + 7 in quantity squared. So now we need to take the derivatives that are in the numerator, so this is going to be equal to 500. Multiplying the quantity of the quantity of T2 + 7. In quantity, multiplied by the quantity of 4 T, since the derivative with respect to T of 2 T squad plus 5 is 4 T. Then we'll have minus the quantity of 2 T squared. Plus 5. In quantity, multiplied by the quantity of 2 T. Since the derivative of T2 + 7 with respect to T is 2T. And all of that is still being divided by the quantity of T2 + 7 in quantity squared. So now we need to perform the multiplication in the numerator. This is going to be equal to 500, multiplying the quantity. Of course he cut. Plus 28 T. -4 T cubed. Minus 10 T. And that quantity is divided by the quantity of T2 + 7 in quantity squared. So, in the numerator, I see that I have 40 cubed minus 4 cubes. So when I add those two together, I get 0. And then I will have 28T minus 10T, so it's gonna give me 18T. So this is going to be equal to 500 multiplied by 18T. Divided by the quantity of T2 + 7 in quantity squared. So this is going to be equal to 9000 T. Divided by the quantity of T2 + 7 in quantity squared. So this is the answer that we were actually looking for for this problem. And so when we compare our answer here to our answer choices, the correct answer choice corresponds to answer choice A. So I hope that this explanation has been useful, and I'll see you in the next video.

Population growth Consider the following population functions.
a. Find the instantaneous growth rate of the population, for t≥0.
p(t) = 600 (t²+3/t²+9)

Key Concepts

Instantaneous Growth Rate

Derivatives

Population Functions

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