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Tts) Teluc - Wlen the pond urgnic nalural 4st level, uxyg dumpee cutenle inlu pOnd Suppaee Honutt , [XIN Rven that thc tirne oxi aliuu mKE Colllent naluit #ill prOC...

Question

Tts) Teluc - Wlen the pond urgnic nalural 4st level, uxyg dumpee cutenle inlu pOnd Suppaee Honutt , [XIN Rven that thc tirne oxi aliuu mKE Colllent naluit #ill prOClS days alter Tutoto in pce UGAnkc exygen cnient wistc huis 1+7I dumetd Finto prcent f() = I( + I of its nornal after organic Wlrt ' level. Find the Iu brn nertake dumped inlu colltent of OxygCT Hlnt: thie pou] Slur Chat (er tha first 10 days+l0t 100 - + 20t IO0 } =[- 1+710 (+ 10)7

Tts) Teluc - Wlen the pond urgnic nalural 4st level, uxyg dumpee cutenle inlu pOnd Suppaee Honutt , [XIN Rven that thc tirne oxi aliuu mKE Colllent naluit #ill prOClS days alter Tutoto in pce UGAnkc exygen cnient wistc huis 1+7I dumetd Finto prcent f() = I( + I of its nornal after organic Wlrt ' level. Find the Iu brn nertake dumped inlu colltent of OxygCT Hlnt: thie pou] Slur Chat (er tha first 10 days +l0t 100 - + 20t IO0 } =[- 1+710 (+ 10)7



Answers

The oxygen content $t$ days after organic waste has been dumped into a pond is given by $$f(t)=100\left(\frac{t^{2}+10 t+100}{t^{2}+20 t+100}\right)$$ percent of its normal level. a. Show that $f(0)=100$ and $f(10)=75$ b. Use the Intermediate Value Theorem to conclude that the oxygen content of the pond must have been at a level of $80 \%$ at some time. c. At what time(s) is the oxygen content at the $80 \%$ level?

Yeah. For the given problem, um we know that the oxygen content two days after organic waste was dumped into a pond is going to be given by F. F. T. Is equal to 100. I'm t squared Plus 10 T Plus 100. Okay. Can um over t squared plus 20 P plus 100. Mhm. That's our function that's given to us. Um And we want to show that f of zero is 100 f of 10 is 75. So f of 0 100 ff 10 At 75. So what that tells us is that from zero days to 10 days, Um we want to show that the oxygen content of the pond must have been 80% at some point. We know that since this is a continuous function, it has to be. Since its oxygen content We know that it must have crossed over 80% at one point time because it went from 100 to 75.

For this problem here, 84 we have the oxygen content of a pond and the function given to us is F F. T, which will call um Half of acts in this case Will be 100 times X squared plus 10 X crossed 100. Try to buy X squared Best 20 x plus 100. And then we want to show that f of zero Equals 100, which it does, and then after 10 equals 75, which does so by the intermediate value theorem, we see that the oxygen content of the pond Must have been 80% at some point between zero and 10. Um We want another time at which the oxygen content is going to be 80. And we see that that occurs at 26.18, where T is equal to 26.18 as well as T equals 3.82.

Uh we're told that when organic waste is dumped into a pond, the oxidation process that takes place reduces the ponds oxygen content. However, given time nature where we started the african contents to its natural level in the phonograph uh p uh gives the I guess I I call it why here, why why it gives the actual content as the percent of its normal level. Two days after organic police has been dumped in the fund explained the significance of the inflection point. So we the organic goes waste goes in and so it starts using up some oxygen to decay um as it acted ISES. And then we get to a minimum point, the level of the uh the oxygen levels stops going down. Um And that's probably when the bacteria basically, you know, there's a there's uh basically because it wants to be so much oxygen absorbed in the water. There is constantly being some more then. But at some point around here the amount that's being drawn in from the surface overcomes. That's what's being used by the organic waste. And so we turn around and then it starts building up again. And now at at this point basically we get this inflection point I think basically says the radar change has an extreme at this. So this is where kind of the rate of change um is the maximum um As you know, as far as it's it's increasing. But so here the rate of change is decreasing and then here the rate of change is increasing. No no the rate of change goes from increasing here to decrease in here. So it's negative here, negative negative then turns around positive, but then it starts to decrease. Um And what what that is is basically, you know, it's a saturation effect, right? Um It's probably something where the more oxygen that's in there, the less that that is absorbed it and you know the rate of absorption. So that's where basically we have this um The rate now has a has a change where we now are the rate of change of why starts to decrease. And again, there's a maximum here where this is probably maybe this point here is where the oh the organic waste has been uh has been oxidized. Now, it's just a matter of you know, absorbing oxygen through the surface of the water.

Uh huh. We want to find the horizontal and vertical ascent. Oats of the function F F T is equal to plus the quantity five over c minus two squared. Since we're not giving a graph, we're going to have to find these horizontal and vertical athletes analytically as such. Let's make let's define the equations that we can find this more easily. So for a vertical ascent, Oh you have to the limit as X approaches A of F of X equals infinity infinity. But that means that for F of X equals P of X over Q of X. You can simply find the essentials for Q of X equals zero or the denominator. In this case it's t minus two squared for horizontal asuntos We simply take the limit of F as X goes to infinity, negative infinity evidence equals some constant be. Then we have a horizontal aceto. So we can go ahead and solve now for the vertical axis we have t minus two squared equals zero. This gives us into it equals two for the horizontal axis coats The limit, as X goes to infinity of two plus 5/10 minutes to squared and negative infinity as well. It's just too because the term 5/10 minutes to square dose zero, so we have a sentence vertically and Y equals two and horizontally at T equals two. Or more appropriately, we have assumed totes at vertically at T equals two, and horizontally at Y equals two. Or rather F equals two.


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