Population growth Consider the following population functions....

Question

Population growth Consider the following population functions.

c. Estimate the time when the instantaneous growth rate is greatest.

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. TFT 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. Estimate the time when the instantaneous rate of change of the chemical's concentration is highest. We're given four possible choices as our answers. For choice A, we have T is equal to 1.53 hours. For choice B, we have T is equal to 3.06 hours, for choice C, we have T is equal to 5.53 hours, and for choice D, we have T is equal to 15.3 hours. Now we need to estimate the time when the instantaneous rate of change of the chemicals concentration is the highest. So we're going to first need to determine the instantaneous rate of change of the chemicals concentration as a function of time. That means taking a derivative of our function CFT with respect to T. So we're going to be calculating C. Of tea, and this is going to be equal to the derivative with respect to tea. 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. So this is going to be equal to the 500 will move through the derivative, so we'll have 500. And then we'll be multiplying that by the derivative of this fraction. So we're going to need to use our quotient rule. 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 of our numerator. When we take the derivative of 2 T 2 + 5, we get 4 T, so we'll multiply this by 4 T. Then we'll have minus our numerator, which is the quantity of 2 T 2 + 5. In quantity, and multiply this by the derivative of our denominator, which when you take a derivative of T2 + 7, we get 2 T. We'll multiply that by 2 T. And this whole quantity is divided by our denominator squared, which is the quantity of T2 + 7 in quantity squared. So multiplying out our numerator, this is going to be equal to 500. Multiplied by the quantity of 40 cubed. Plus 28 T. -40 cubed. Minus 10 T. and that quantity is divided by the quantity of T2 + 7 in quantity squared. So, simplifying this expression, this is going to be equal to 9000 T. Divided by the quantity of T2 + 7 in quantity squared. So now that we have our C of T calculated, we can now calculate its maximum by taking another derivative and setting that derivative equal to 0. So, we're now going to need to calculate C of T. And this is going to be equal to the derivative with respect to T of T T. So this is going to be the quantity of 9000 multiplied by T divided by the quantity of T2 + 7 in quantity squared. So again, we're going to need to apply our quotient rule. So this is going to be equal to our our denominator, which is the quantity of T2 plus 7 in quantity squared multiplied by the derivative of our numerator, which we take a derivative of 9000 T, we just get 9000. We'll multiply that by 9000, and we'll have minus our numerator, which is 9000 T. Multiplied by the derivative of our denominator. So when we take a derivative of our denominator, which is the quantity of T2 + 7 in quantity squad, we'll get 2 multiplied by the quantity of T2 + 7. In quantity, multiplied by the quantity of 2 T. In quanti And all this is going to be divided by The quantity of T2 + 7 in quantity raised to the 4th power since I need to square my denominator in our quotient rule. Now we can simplify this expression by factoring out a quantity of T2 + 7 out of the numerator, and that will cancel out one of the factors of T2 + 7 in the denominator. So doing that, this is going to be equal to the quantity of T2 + 7 in quantity multiplied by 9000. Minus 9000 Multiplied by T Multiplied by The quantity of or tea. And all of this is going to be divided by the quantity of T2 + 7 in quantity viewed. Now, to find the maximum of the rate of change for our concentration of our chemical, we'll need to set this second derivative equal to 0 and solve for T. Now, since the quantity of T2 + 7 is going to be positive, I only need to find when the numerator is equal to 0. So we're gonna have the quantity of T2 + 7 in quantity multiplied by 9000. Minus the quantity of 9000. T In quantity, multiplied by 4 T is equal to 0. So, multiplying out everything on the left hand side of this question will have 9000 Multiplied by Esquad. Plus 63,000. Then we'll have minus. 36,000 multiplied by He squared. And this is going to be equal to 0. So compiling like terms, we're going to have 63,000. Minus 27,000 Multiplied by T2 is equal to 0. So now just solving for tea, we're going to see that. 63,000. is equal to, and we'll move the 27,000 T2 to the other side by adding 27,000 T2 to both sides. We'll have that equal to 27,000 T squad. Divide both sides by 27,000. We're going to get 63 divided by 27 is equal to T2, and then taking the square root of both sides, we'll get T is equal to the square root, and here we can reduce that fraction from 63 divided by 27 to 7 divided by 3. And so when we insert the square root of 7 divided by 3 into our calculator, this is going to be approximately equal to 1.53, and that will have units of hours, since our time is given in hours. So that is the answer that we're looking for, and when we come back up to our answer choices and compare our answers that we calculated to the 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.
c. Estimate the time when the instantaneous growth rate is greatest.
p(t) = 600 (t²+3/t²+9)

Key Concepts

Instantaneous Growth Rate

Derivatives

Critical Points

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