Population growth Consider the following population functions....

Question

Population growth Consider the following population functions.

b. What is the instantaneous growth rate at t=5?

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 multiplied by the quantity of 2 T2 + 5 in quantity, divided by the quantity of 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 T equal to 8. We give them four possible choices as our answers. For choice A, we have 28.14, for choice B, we have 12.28. For choice C, we have 14.28, and for choice D, we have 10.28. Now we need to calculate the instantaneous rate of change of the chemicals concentration at T equal to 8. So we need to determine the instantaneous rate of change of the chemical's concentration as a function of time. And to do this, we'll need to take a derivative of our function CFT with respect to T. So we will be calculating C of T? And this is going to be equal to the derivative with respect to T. Of the quantity of 500 multiplied by the quantity of 2 T 2 plus 5 in quantity, divided by the quantity of T2 plus 7 in quantity. They're taking this derivative, this is going to be equal to 500 is a constant, so it'll move through the derivative, so we'll have 500. And then I'll be multiplying this by the derivative of a fraction. And so to take this derivative, we're going to need to use our quotient rule. So, we'll have 500 multiplied by the quantity, and here we call your quotient rule. So, for the quotient rule, we're gonna have our denominator, which is the quantity of T2 + 7. 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 new murder, 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 plus 7 in quantity, and all of that is divided by our denominator squared. Which is the quantity of T2 + 7 in quantity squared. So taking in the derivatives of the numerator, this is going to be equal to 500, multiplying the quantity of the quantity of T2 + 7 in quantity, multiplied by 4 T. When you take a derivative specta T of 2 T 2 + 5, we get 4 T. Then we'll have minus the quantity of 2 T 2 + 5. In quantity, multiplied by the quantity of 2 T. So here taking the derivative expected T of the quantity of T2 + 7 gives us 2 T. And all of that is still divided by the quantity of E2 + 7 in quantity squared. So now we just need to simplify this, and when we do so, this is going to be equal to 9000 T. Divided by the quantity of T2 + 7 in quantity squared. So now we need to actually calculate the of 8. And so doing that, we'll have D of 8. is equal to 9000. Multiplied by 8 Divided by the quantity of 8 squad. Plus 7 in quantity squared. And we can insert these values into our calculator, and in doing so, this is going to be approximately equal to. 14.28. So here we've just round it to two decimal places. So this is the answer that we're looking for, and we compare our answer that we've calculated to our answer choices. The correct answer choice corresponds to answer choice C. 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.
b. What is the instantaneous growth rate at t=5?
p(t) = 600 (t²+3/t²+9)

Key Concepts

Instantaneous Growth Rate

Derivatives

Population Functions

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