Problem

The family of bell-shaped curves $$ y = \frac{1}…

07:30

Problem 71 Hard Difficulty

A drug response curve describes the level of medication in the bloodstream after a drug is administered. A surge function $ S(t) = At^p e^{-kt} $ is often used to model the response curve, reflecting an initial surge in the drug level and then a more gradual decline. If, for a particular drug, $ A = 0.01 $, $ p = 4 $, $ k = 0.07 $, and $ t $ is measured in minutes, estimate the times corresponding to the inflection points and explain their significance. If you have graphing device, use it to graph the drug response curve.

Answer

$S(t)=A t^{p} e^{-k t}$ with $A=0.01, p=4,$ and $k=0.07 .$ We will find the
zeros of $f^{\prime \prime}$ for $f(t)=t^{p} e^{-k t}$
\[
\begin{aligned}
f^{\prime}(t) &=t^{p}\left(-k e^{-k t}\right)+e^{-k t}\left(p t^{p-1}\right)=e^{-k t}\left(-k t^{p}+p t^{p-1}\right) \\
f^{\prime \prime}(t) &=e^{-k t}\left(-k p t^{p-1}+p(p-1) t^{p-2}\right)+\left(-k t^{p}+p t^{p-1}\right)\left(-k e^{-k t}\right) \\
&=t^{p-2} e^{-k t}\left[-k p t+p(p-1)+k^{2} t^{2}-k p t\right] \\
&=t^{p-2} e^{-k t}\left(k^{2} t^{2}-2 k p t+p^{2}-p\right)
\end{aligned}
\]

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Video Transcript

called the search function and it's equal to uh some number eight times T. To the p power times E. To the negative K. Times T. Power S. Is the level of medication and bloodstream T. Is the number of minutes that have elapsed for this particular Search function A. is .01 times T. To the P. Power where P. Is given to be four. And then that multiplies E. To the negative K. T. K. Is given to be point oh seven. So that's E. To the negative zero. Actually .07. Each of the negative .07. Each of negative Katie each of negative .07 R. T. So we are going to graft dysfunction using dez mose. And then we're going to look for the inflection points and kind of describe what's going on with this search function at those uh inflection points. Okay so using Dez most I graphed the search function. Dez most uh prefers to work with the variable X. So instead of the variable T. Which was representing minutes, we're just using the variable X. And since the number of minutes that have elapsed since the injection uh should be greater than or equal to zero. Uh graphing dysfunction for uh time less than zero really doesn't mean anything. So to uh make the graph, you know look a little less confusing. I restricted uh dysfunction to being graft off when the number of minutes X. Is greater than or equal to zero. Uh The goal here is to estimate the time which would be the horizontal X. Axis, estimate the time at which inflection points occur. Now, an inflection point occurs when the second derivative of the function is zero. Uh Basically it's signals an inflection point signals when the first derivative uh will change from increasing to decreasing if the second derivative is zero, that means the first derivative reaches a maximum point meaning the first derivative or the steepness to slope. If you will was increasing, reached a maximum point now will start decreasing or vice versa. Uh An inflection point being where the second derivative is zero. Inflection point could also be where the first derivative reaches a minimum, the first derivative is decreasing, reaches a minimum and then starts increasing. Now if we're looking for the first derivative to be at a maximum point or minimum point. That means, for example, if we say the first derivative is reaching a maximum, that means to first derivative was increasing, reaches a maximum point will start decreasing. Well, the first derivative lets you know, you know just how steep the function itself is increasing or decreasing. So if we're looking for where the first derivative is going to change from increasing or decreasing, we want to look for a change in the steepness of the curve. Let me go back to our uh whiteboard here and give you an idea of what um what an inflection point would look like. Just so you have an idea of what we're looking for. If I have a function it's going up steeply. Okay, But then at this point it's still gonna be increasing but at a at a slower rate. Okay, this would be an inflection point. The function is still increasing, but the rate at which it's increasing uh as indicated by the first derivative um is changing here to function is increasing at a quicker and quicker rate. So to slope or the first derivative is increasing. Then at this inflection point, uh the second derivative zero. The first derivative reaches a maximum. Uh The function itself still increases but now at a slower rate because F prime of X now is starting to decrease. So this would be the graph of the function F and uh f prime of X would change from the slope. Would at this inflection point changes from uh increasing and increasing to where the slope or the derivative is now uh decreasing. Okay, so at prime of X changes from positive two negative. The second derivative at this inflection point would be zero. So this is what happens at an inflection point. At an inflection point. Uh function will still be increasing. And of course this can also be shown for a function that is decreasing through an inflection point. Let me give you no idea what that looks looks like. Um If I uh three very set. If I have a function that's decreasing reaches an inflection point still decreases. But now at a slower rate. This would also be an inflection point Um at an inflection point the second derivative is zero and the first derivative changes signs. So here the first derivative is changing signs at this point here the first derivative is positive. Uh Here the first derivative will be negative at this inflection point, The second derivative will be zero here, the first derivative is negative and then at the inflection point, first derivative becomes positive even though the function is still decreasing through that inflection point. So we're looking for a bend or you change in the bend of the curve. That's where these, that's where these inflection points are. C Okay, we're going up but then we change how steeply you can think of it as common cavity kind of concave duck. And then we start con caving down here, we're concave down and then we start uh at the inflection point becoming concave uh So we can estimate um Here you can see that the function is increasing. And then approximately around here there might be an inflection point right around here at approximately uh t equals 25 Maybe 30. Um so when T is 30 minutes, uh the function is still increasing fastest inflection point. Um but not as deeply. Okay, here it's climbing up steeply reaches an inflection point and then starts, you know kind of rounding off. Think of this as concave up Until time is 30 minutes and then concave down. So it's a little hard to see here if we zoomed in and be a little bit easier but we have a point of inflection here and then we have another point of inflection here here the function is coming down. it's the inflection point. The function keeps decreasing but now at a slower rate. Okay here is kind of concave down if I exaggerate it. And then at this inflection point here at approximately 100 minutes uh to function changes come cavity and starts becoming concave, let's go back to our white board and write down those two times. So the two values of T. Of course on our graft the variables X 22 values of T. That are inflection points would be, It looks like when T. 35 minutes We have an inflection point and then we have another inflection point uh when X in the graph or are variable t. 100 minutes. So let's go back to our graph and uh kind of describe what's going on with our search function at these inflection points. So uh when uh t or in this case ex uh number the number of minutes. The time is 30 minutes. We have an inflection point. So for 30 minutes we are at this point on the graph. So the surge is increasing uh rapidly. Um until we reach a time of 30 minutes and then the surge continues increasing but at a slower rate that's what happens. Uh the rate at which it's increasing uh changes. Okay, the surge is increasing and increasing the rate at which the surge, the rate at which the function is increasing is itself increasing until the inflection point. Then the function the surge keeps increasing after 30 minutes. But at a slower rate. Now, when the time was 100 minutes, we estimated there was another inflection point here, at 100 minutes. Uh here the surge is decreasing rapidly until we reach 100 minutes. That's another point of inflection and then the search is still decreasing. But at a slower rate, that's what's happening at those two inflection points.

Leon D.

Temple University

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Problem 71 Hard Difficulty

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