{Use of Tech} Cell population The population of a culture of...

Question

{Use of Tech} Cell population The population of a culture of cells after t days is approximated by the function P(t)=1600 / 1 + 7e^−0.02t, for t≥0.

e. Graph the growth rate. When is it a maximum and what is the population at the time that the growth rate is a maximum?

Accepted Answer

Welcome back, everyone. In this problem, the number of fish in a small pond over time can be described by the function F of T equals 2000 divided by 1 + 4 E to the negative 0.05 T, where T is the time in days since observations began. Find them when the maximum growth rate occurs and the population of fish at this rate, running our answer to two decimal places. A says the maximum growth rate occurs at T equals 13.86 days, and the population of fish at the maximum growth rate is 2000. B says they are at 13.86 days and 1000 respectively. C 27.73 days and 1000, and D 27.73 days and 2000. Now, what are we trying to figure out here? We want to know when the maximum growth rate occurs and the population of fish at the maximum growth rate, where in our problem statement, we were given a function that describes how or that can give us, sorry, the number of fish over time. So how can we link that to our max growth rate to help us solve our problem? Well, if we know the number of fish over time from our function, then. The derivative of our function will give us the rate of change. In other words, giving F of T. OK, even FFT. Then the derivative of FF FFT prime represents OK. The growth rate. OK With, uh, that's the, it's derivative with respect to time. I know if F of T represents the growth rate, then at the maximum. Sorry, at the maximum growth rate. At the maximum growth rate. OK. Then we know the derivative of our expression. In other words, the derivative of the growth, growth rate, which is gonna be F, F T, which is the second derivative of F of T, OK, is going to be equal to 0. So if we can figure out first, our growth rate. Let me put that in in blue. If we can first figure out our growth rate. And then we can figure out our, our second derivative to our growth rate, the second derivative of, sorry, the derivative of our growth rate, which is the second derivative of FFT, and set it to zero, then we should be able to solve for T when the maximum growth rate occurs and substitute that value into FFT to find a population of fish at that time, OK? So, let's go ahead and first start by trying to find the derivative of FFT. Now we know that FFT. OK. It is going to be equal or it's given to be equal to 2000 divided by 1 + 4 E to the negative 0.05T. And now we can rewrite that as 2000 multiplied by 1 + 4 E to the negative 0.05T raised to the power of -1. And now to find a derivative, we can use the chain rule. Remember that the chain rule basically tells us that first we bring down the power. We rewrite or. Alter expression and subtract one from the power, OK. So in this case, instead of being raised to -1, it's going to be raised to -2. And then we multiply that by the derivative of the inner function. The derivative of 1 is 0 and the derivative of 4 E to the negative 0.05T is going to be equal to 4 E 0.05T multiplied by negative 0.05. We differentiate the poor and bring it down. I know when we simplify that. We get our growth rate to be. Equal to 400 multiplied by e to the negative 0.05 T. All divided by 1 + 4 E to the negative 0.05 T squared, OK? So this is our growth rate, F T. Next, now that we have our growth rate, let's differentiate it to get the second derivative of FFT, OK? And I'm gonna put that in red, yeah. So now solving for the second derivative of FFT. Notice that the first derivative is a quoent. So to differentiate it we can use the the the quotient rule and recall that the quotient rule tells us that if we have an X have a function Y equals u divided by V where U and V are functions of T. Then the derivative of Y is going to be equal to U minus V U divided by V2. So essentially If we set our numerator to be U from F T and the denominator to be V, we can find their derivatives and plug it into the equation to get the second derivative of F of T, OK? So now, let's let U be equal to 400 E to the negative 0.05T, OK? And V be equal to 1 + 4 E to the negative 0.05 T squared. Now, notice that we can differentiate you using what we know about the derivatives of terms with E, OK? And I'll allow you to solve that, but when you solve and simplify in the interest of time, it's gonna be -20 multiplied by E to the negative 0.05T. On the other hand, differentiating V using the chain rule, we're going to get it to eventually be equal to negative 0.4, E to the negative 0.05 T. Multiplied by 1 + 4 to the 1 + 4 E to the negative 0.05T. I know that we have the derivatives of U and V. We can use our caution rule, OK. Is that a caution. To get the derivative of F T to be equal to U 20 to 0.05T multiplied by V 1 + 4 E to the negative 0.05T all squared minus V 0.4 E to the negative 0.05T multiplied by 1 + 4 E to the negative 0.05T. OK. Multiplied by U 400 E to the negative 0.05T. I know all of this, OK, let's write that. Let's, let's just do that properly here. All of this is being divided by V2. 1 + 4 E to the negative 0.05 T squared squared. Now we can simplify our second derivative here, and when you do that, then you should get F T to be equal to E to the negative 0. Let me write that properly, negative 0.05T multiplied by -20 plus 80 E to the negative 0.05T. All divided by 1 + 4 E to the negative 0.05 T cubed. I know that we have an expression for the second derivative of FFT or the derivative of the growth rate, OK? At the max growth rate, like we said earlier, at the max growth rate. OK. Then the second derivative of FFT is going to be equal to 0. What does that mean? Well, basically, this entire expression, let's just copy and paste it here. That means this entire expression. Is going to be equal to 0. Yeah. And now we can simplify to help us solve for tea. Now notice here, if we divide through by E to the negative 0.05 T and multiply through the entire equation by 1 + 4 E to the negative 0.5 T or cubed, OK, then we're going to end up with negative 20 plus 80 E to the negative 0.05 T equal to 0. This makes it a bit easier to solve. No, we want to solve for T, so let's try to isolate the E term. We can divide our entire equation through by 80, OK. And the negative 20 divided by AT equals negative quarter. If we add a quarter to both sides, then we'll get E to the negative 0.05T because A0 will cancel in that term, OK? Being equal To a quarter Now, using what we know about logs, we could rewrite this as negative 0.05T being equal to the natural log of a quarter, OK. And no, this tells us then that our time, the time at the max growth rate, let's call that T max, is going to be equal to negative 20 multiplied by the natural log of 4. Which, if we were to evaluate that with a calculator and write it to two decimal places, will approximately be equal to 27.73 days, OK? So this is the this is when the maximum growth rate occurs. Now remember, we also need to figure out what the population of fish is at this rate, OK? So at T max. And the maximum growth rate, OK. And our population of fish, which is gonna be F of T max or F off, let's use here, sorry, this should have been 2020 multiplied by the natural log of 4. So let's use that. So F multiplied by 20. Multiplied by the natural log of 4 and. Or F of that expression is going to be equal to 2000, or we're going back to F of T divided by 1 + 4 E to the negative 0.05 multiplied by 20 LN4. And now that would simplify to 2000, OK, divided by 1 + 4 E to the negative and then 4. And when we simplify that even further, we get it to be 2000 divided by 2, which equals 1000. Thus, at our maximum time, at the time of the maximum growth rate, 27.73 days, there are 1000 fish in the pond. That means C. Is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

{Use of Tech} Cell population The ​population ​of a culture of cells after t days is approximated by the function P(t)=1600 / 1 + 7e^−0.02t, for t≥0.e. Graph the growth rate. When is it a maximum and what is the population at the time that the growth rate is a maximum?

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{Use of Tech} Cell population The population of a culture of cells after t days is approximated by the function P(t)=1600 / 1 + 7e^−0.02t, for t≥0.
e. Graph the growth rate. When is it a maximum and what is the population at the time that the growth rate is a maximum?