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A conical tank, vertex down, has a depth of 8 feet and a radiusof 4 feet. From the vertex at the bottom, the liquid is leaking outso the height of the liquid is fal...

Question

A conical tank, vertex down, has a depth of 8 feet and a radiusof 4 feet. From the vertex at the bottom, the liquid is leaking outso the height of the liquid is falling at the rate of 0.75 feet perhour. What is the rate, in cubic feet per hour, at which the wateris leaving the reservoir when the water is 6 ft deep? The volume ofa cone is given by the formula V = 1/3 pi r^2 h.

A conical tank, vertex down, has a depth of 8 feet and a radius of 4 feet. From the vertex at the bottom, the liquid is leaking out so the height of the liquid is falling at the rate of 0.75 feet per hour. What is the rate, in cubic feet per hour, at which the water is leaving the reservoir when the water is 6 ft deep? The volume of a cone is given by the formula V = 1/3 pi r^2 h.



Answers

Water is flowing into a conical reservoir 50 feet high with a top radius of 10 feet, at the rate of 16 cubic feet per second. At what rate is the depth of the water in the reservoir changing when the water is 10 feet deep? (The volume of a cone is $\frac{1}{3} \pi r^{2} h,$ where $r$ is the radius and $h$ its height.)

In problem 22. We once again have related rates once again dealing with a cone, and that means they're going to be dealing with a volume of a cone. Specifically, we're gonna be using that same equation from the last couple problems. This case for solving for the height as the time is changing, given all the rest of the information we need. But before he could go ahead and take the derivative of V. With respect to time, we need to get this down to one variable. You need to find the relationship between R and H. And since we're trying to find a HR, ready T means we want to eliminate the R value. Well, we're given some constraints. You We are told that for this entire cone that the base is 10 ft in the height is 12 ft and census nice triangle is always gonna have that same ratio. So if I split it in half for him, it's gonna be dealing with half of the base, which is gonna be the radius and at having 5 ft. And then I'm gonna have 12 ft, and that's gonna have the same ratio, as are H or whenever I'm dealing with eventually when my height is 8 ft and then I'm just gonna have, um some are at eight, but And first, what I want to do is I'm trying to solve for R in terms of H. And I could do that because this ratio is the same. So 5/12 is gonna be equal to our over h are is going to be equal to five wealth each. Now, what if I wanted to solve this? When H is equal to 8 ft? Well, our of eight going to be equal to 5. 12 times eight, which is going to be equal to then go ahead and I can cross it out. Make that two gonna become three, going to end up with 13th. But now I have all the relevant information in order to complete this problem. But I can plug in that our value into my original equation. So I have one third. Hi. I'm gonna have have 5/12 h squared. That's going to be times H that's going to be equal to. I'm gonna have 25 that's going to be over 1 44 times. That's going to be of 400 32 Gonna be pi times h Q So I can go ahead and take the derivative so D v d over DT gonna be equal to 25 over 432 pi It's just the power rule, So it's gonna be three times h squared. And remember, I'm taking this with respect. T so d h over deep. I'm just going to simplify this as little as I can. Basically, that three can just cancel out in the denominator here, and that's going to turn back into 144. I'm going to get that. This is going to be equal to 25 over 144 pi h squared th over the team. As you can imagine, this isn't gonna end up being a very pretty number. So TV over DT is 10 going to be equal to? I have 25 over 144. I'm spy. That's going to be times eight squared. That's going to be times D h over d team. But I can solve for d h over DT I just dividing you side by all those constants. So I mean, I have 64 times, 25 times pi divided by 144 and divide 10 by that number, which is going to give me that th over DT gonna be equal to 0.2. I'll do it to three significant figures Or 286 And for this, I want to make sure I keep my units And in this case, in the beginning, as dealing with feet cubed per minute and I went down to just for a minute.

So here I have one. The container. Okay, so this is a right circular cone with vertex step. It has a radius of four and a height of 16. Now, water is being poured into the container at a constant rate. So The question is how fast is the water level rising when the water is eight m deep? So meaning let's we're looking for the change and the height When the water is at this level, which is eight m. So let's write down all the information that we have. So we have a D. V. D. T. Which is the change in volume with the water. It's increasing at 16 m cubed for a minute. And what we're looking for is so let's call the height of the water each. We want the H. D. T. evaluated at H. is equal to eight. Okay, so let's write down the volume of a cone. So we have V. Is equal to one third. Hi, R squared H. Okay, so our would be this radius here. So as you can see both H and R are going to change with respect to time because if my water level was higher, I would have a bigger age and a bigger art. If my water level was lower, it will be a smaller age and the smaller art. So neither of those two are constant. Okay, so this means when I take the derivative with respect to T on both sides, I'm going to have to use a product will here because both are an age depend on T. Okay, so let's go ahead and do that. DVD T is equal to. So are constants here would be just pi over three. And then the product rule, we would get to our times the R D. T times H plus r squared times. Yeah. Okay. So now what we're looking for is we're looking for D H D. T. Okay so we're looking for this thing here. We have DVD et And we have a choice because we want to evaluate that when ages eight but we don't have our the radius and the change of radius with respect to time. Okay So that tells us we need to go ahead and find that and we can do that using similar triangles. So if you look here if we if we're talking about this whole container we have a radius of four and a high the six takes. Okay so this triangle here and this triangle here are similar trying. Okay so meaning if we do um If we do radius over height that would be 4/16. Well that must be equal to our over H cross Multiplying would give us four H. is equal for 16 art. Okay so using this we can say that When H. is eight Or would be equal to two. Okay so that means are evaluated at H. equals eight is equal. Thank you. Okay next we're going to take the derivative with respect to H. On both sides. So we get four. The HDTV is equal to 16 D. R. D. T. And by doing this we can now express D. R. D. T. In terms of D. H. D. T. So we get the R. D. T. It's actually one quarter D. H. D. T. Okay so that means we can go ahead and substitute. But it so we get this is equal to hi over three times to our times. Um Let's put the ancient front times and then we have D. R. D. T. Which is actually 1/4 P. H. D. T. Plus R. Squared the H. D. T. And this is equal to D. V. D. T. Okay so now we can go ahead and isolate D. H. D. T. Okay so we get so we can pull out A. D. H. D. T. So we get D. V. D. T. Is equal to D. H. DT. Heinz. Hi over three times maybe if we want we can pull out an R. And then we would have 1/2 h. Plus are left. Okay? So that tells us if we want to isolate D. H. D. T. We get D. H. D. T. Is equal to D. V. B. T. Over. Mhm. Hi over three. Are times one half H. Plus are okay now finally we can do our evaluation because we want D. H. D. T evaluated at H. Is equal to eight. Okay So that's going to be a D. V. D. T. That's 16 over. Hi over three. And our we've evaluated that so our when H. Is equal to ages too. So that's times two and then we have one half of H which has eight plus our which is to Okay so let's go ahead and simplify that. That's 16 over. That's two pi over three times. This is four plus two which is six. So cancel we get to so this is 16/4 pi or simplify to four over pipe. And our units here will be readers her minute.

So we have water being put down at a rate of two cubic feet per minute. So that means that we're actually given um tv with respect to T. is equals to two. Right? So we know that The volume of a cone is actually equals two pi over three R squared times age. Okay. But now the problem is we have more variables that we can deal with. So we have to eliminate one of them. And since the question is actually looking for um the level of the water rising so that the rate at which the water is rising. So this means since I have indicated therein raid the day which is the level of the water rising in age, what are looking for sth with respect to t. Okay so what we're going to eliminate here is this are And how are actually going to do that is we are given that across the corners eight ft but we know that um this forms a circle and the radius is actually equals to four half of eight. So that's four and then um the Which is H. is equal to six so if I can do um like a ratio so it's going to be are over age And this is going to be four over six. Yeah. Right so this is our mm So that means that our is equals to two or 3 H. So now I can substitute it into this formula so we have this is equals two. Sophie is equal to pi over three Times to over three H. old squid times age. And this After you know simplifying you will find that it's equals two four pie Age to the poor three over 27. So now we can find the derivative respected time. So tv T. T. Is equals to take out the constant for part of the 27. Now we find distributor of H. To the power three. That's going to be three times H. Squared. The age over D. T. And this will be equals two four hi over nine times age squared. But however we are told that when the container is half full. So that means if we're looking at our container here when it is half full they're talking about the container not the water. So maybe the water is here. And we already told that this height is equal to six so h must be equal to three half full. Okay so That means HS three. Let's go ahead and substitute that. So this is going to be um three squared D. H. Tiki. Okay we already know that Germans of the volume with respect to T. Is supposed to. We actually said this in the beginning right here. So yeah now we can solve for the HD T. Which is what we're looking for. So um if you make th t t the subject of the formula you're going to end up with 1/2 pi feet for a minute. Uh huh. Yeah

And this problem. We have a conical water tank that's being filled at a rate DC DT of 20 feet cubed for a minute. And we're interested in finding out how fast the depth or the height is changing at the moment when the height of water is 16 ft. And so from here, we know that this is an unknown currently, and so we could start off with the volume of a conical well of a cone is one third high R squared times h, um, but because we have three variables and we want and we only know one of the rates, we need to convert both of these variables what we need to convert the radius to a height. And we could do so with similar triangles because we have the triangular region of a height of 24 and a radius of 10 when the whole thing is full. And then that shrinks down to a generic triangle. Similar triangle of radius H and height. Uh, H and we know that those two triangles, no matter what size, that they're going to remain similar. So we could go ahead and set up the portion here and that would get us, um, what we could just say like the hot. The ratio of the hype to the radius is equal to H over our and then we want to solve for R. So we'll go ahead and buy it over to the other side and then multiply both sides by the inverse of that fraction. So we get the radius is equal to 10 24th of the height. Um, but we could simplify that and see dividing each of those by two. That would just be 5/12. And now we have a substitution for the radius of top here. So let's go ahead and get rid of this because this now just turned into five 12th h That was the radius, or that is the radius. And that's squared. Bye bye. Another factor of H. So before we differentiate will just clean this up once more and more more Step is, of course, is going to give us each square. I was gonna give us 25 up top and then 1 44 in the denominator, so we'll go ahead. And now differentiate this, considering that he's all combined to N h. Q. So we've moved that power of three out front and that would cancel with one third. So from there we'd have 25 high over 1 44 times h squared multiplied by D H duty. And that's what we're interested in solving for is that d h d t value. Now we know that the height in this instance is 16 because that's the moment that we're interested in. So we could change that to 16 sway. Sorry, that should have been a d v D. T over on the left side, which we know that value as well. That is 20. And so now we know everything so we could go ahead and solve or d h d t and solving that will end up getting if we clean this up just a little bit here, one of getting nine 28 months. Hi. And this is how fast the height is changing up the instant when the height is 16 feet and this is an feet, Her minutes. Yeah, and that is our final value.


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