Population models The population of a species is given by the ...

Question

Population models The population of a species is given by the function P(t) = Kt²/(t² + b) , where t ≥ 0 is measured in years and K and b are positive real numbers.

c. For arbitrary positive values of K and b, when does the maximum growth rate occur (in terms of K and b)?

Accepted Answer

The concentration of a drug in the bloodstream is given by C of T equals 120 T2 divided by T2 + 30, where T is the time in hours after the drug is administered, and C of T is the concentration in micrograms per liter. For arbitrary positive values, when does the maximum rate of change in drug concentration occur? And there are 4 possible answers, square of 1020 square of 20, and 10 all in hours. Now, to solve this, we want the maximum rate of change, which means we are going to find the 2nd derivative. Set equals to 0. But to solve this, we first need to find the first derivative. So, we're looking for C of T first. Now, we have C of T, which is equal to some function G of T divided by H of T. We can use the quotient rule for the derivative. Which states that our derivative will be G T. HOT Minus G of T. H T All divided By HFT squared. We can then see GFT Will be our 120 T squad. H of T will be our T2 + 30. G Would then be based on the parable, 240T. And H5 of T. Based on the power rule again, will be 2 T. We can then write our quotient rule. G 240T. Multiplied by H of T T 2 + 30. Minus GFT, which is 120 T squared. Multiplied by H2 T. All divided By H of T2, which is T2 + 30 squared. We can now simplify this. We have 240 Tea cuts Plus 7200 tee. -240 TQ. Divided by T2 + 30 squad. We noticed 240 T cube cancels, making our first derivative, 7,200T. Divided by T2 + 30, all squared. Now, we need to find our second derivative. So we will continue from the C T and find the next derivative. This GFT is 7,200T. The age of Is T2 + 30. Squared G prime then, will just be 7200. And now H T. is given by a chain rule. This will be 2 multiplied by T2 + 30. Multiplied by the inside derivative to T. Which is just 4 T multiplied by T2 plus 30. Now, we can plug this into our quotient rule. We have G, which is 7200. Multiplied by H D 2 + 30 squad. Minus 7200T. Multiplied By H, which is 4 T. T2 + 30. All divided by H2, which will be T2 + 30, raised to the 4th. We can now simplify again. We combine all of our life terms. We end up getting 7200. Multiplied By T2 + 30. Which is all multiplied. By T squared. Plus 30 Minus 4 T squared. This is all divided. By T2 + 30. Raised to the 4th. We can even further simplify this, to get 7200. Multiplied by T2 + 30. Multiplied by -3 T2 plus 30. All divided by T2 + 30, raised to the 4th. Now, We will set our 2nd derivative equal to 0 to find our point. Now, the denominator will cancel out if we set the equal to 0. So, we will just check the numerator, 7200. Multiplied by T2 + 30, multiplied by -3 T2 + 30. Equals 0. Now, 7200 is going to go away. We would set T2 + 30 equal to 0 and -3 T2 + 30 equal to 0. Now, we noticed the T2 + 30 term. It never equals 0. Because of the plus 30. So this, we don't use. We can solve our other equation for T. By subtracting 31st. To get -3 T 2 equals -30. T2 equals 10. Which then leads us to T equals the square root of 10. Finally, if we were to check our value, By plugging in values to the left and to the right of the square root of 10. We would then confirm That when T is less than the square of 10, our second derivative. Is positive. If it's greater than the square root of 10, our second derivative is negative. This confirms that we have a maximum rate of change. We can then say our answer. T equals square of 10. Corresponds to answer A square of 10 in hours. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Population models The population of a species is given by the function P(t) = Kt²/(t² + b) , where t ≥ 0 is measured in years and K and b are positive real numbers.c. For arbitrary positive values of K and b, when does the maximum growth rate occur (in terms of K and b)?

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Population models The population of a species is given by the function P(t) = Kt²/(t² + b) , where t ≥ 0 is measured in years and K and b are positive real numbers.

c. For arbitrary positive values of K and b, when does the maximum growth rate occur (in terms of K and b)?