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A flask has the shape sketched at right The bottom part of the flask is in the shape of a truncated cone with bottom radius 10 cm; top radius 5 cm; and height 10 cm...

Question

A flask has the shape sketched at right The bottom part of the flask is in the shape of a truncated cone with bottom radius 10 cm; top radius 5 cm; and height 10 cm Sitting on top of the truncated cone is a cylinder whose radius is 5 cm_ Liquid is being poured into the flask at a rate of 2 cm'/s. Note that the volume of the cone whose bottom radius is and whose vertical height is h is given by V = Trh When the liquid is 15 cm high in the flask; how fast is the height of the liquid increasin

A flask has the shape sketched at right The bottom part of the flask is in the shape of a truncated cone with bottom radius 10 cm; top radius 5 cm; and height 10 cm Sitting on top of the truncated cone is a cylinder whose radius is 5 cm_ Liquid is being poured into the flask at a rate of 2 cm'/s. Note that the volume of the cone whose bottom radius is and whose vertical height is h is given by V = Trh When the liquid is 15 cm high in the flask; how fast is the height of the liquid increasing? If the flask is initially empty, how long does it take the liquid to reach a height of 10 cm?



Answers

A conical flask contains water to height $H=36.8 \mathrm{mm}$ where the flask diameter is $D=29.4 \mathrm{mm}$. Water drains out through a smoothly rounded hole of diameter $d=7.35 \mathrm{mm}$ at the apex of the cone. The flow speed at the exit is approximately $V=\sqrt{2 g y},$ where $y$ is the height of the liquid free surface above the hole. A stream of water flows into the top of the flask at constant volume flow rate, $Q=3.75 \times 10^{-7} \mathrm{m}^{3} / \mathrm{hr}$ Find the volume flow rate from the bottom of the flask. Evaluate the direction and rate of change of water surface level in the flask at this instant.

Number 187. Which related to the application of the deliberative question says a vessel is like inverted corn Hi 10 fit some vertical angle alpha 30° Philippine solution. And this solution is drained through our office at the bottom of cylindrical baker. Okay, two inches Per minute. This is rate six ft. Okay, so This is 187 number of questions. So basically we have to find out, okay, the rate at which height of the solution increases in the beaker. So it is like this, this cylindrical and phone is inverted like this. Okay, Yeah. And how it has been given as 10 ft height of the corn. It's 10 ft and semi vertical. And girl, this angle alpha is given as 30 degrees and we have radius after cylinder sound like a big radio 606 centimeter. So this release of the cylinder is under six centimetre. And okay, so let small art is the radius of the corn. It is of the corn and B B V. Is the volume of the solution in the Chronicle vessel and that is the height of the liquid in conical vessel. This is edge and V is the volume. Okay, No, if you look at in the conical flask, it will look like our by at equal to 10 30 degrees, 10 30°,, which means uh I will be equal to one by route three at Now. The is volume is equal to one x 3 by square. It So one x 3 by into our square means it's quite of this one x 3 at Square into it. So I ask you by nine, this is a volume now rate or rate of change of volume DV by detail will be going to buy a Squire by the into into the earth by duty. Okay, Uh if at 21 be the height and volume of solution in cylinder. This is excellent. Edwin And Volume v. one. Okay, This is 606 cm. So we have volume, we won Equal to buy our Square H. one. So five are 1 square. Uh Excellent. So Darwinism Road 6 65. Which one? Sorry? It will be TV run by, they take 4- 65. The S one by duty. Now that's equal to six Bs. Quality is given as minus two minus two due to the fact that the volume is decreasing and we have to convert it into a permanent. No problem. So the H one by the T will be equal to six by Okay. and ah one more thing to be given the flow out of the conical flask, flow into the bigger. So we have to just it created but with negative sign because one is rate is increasing and another volume is decreasing. So D. V. By DT it was decreasing because in conical flask water is turning out equal to tv even by DT. Which means minus by I just acquired by three D. H. By D. T. Equal to six by D. As one by ADT. Okay so When as equal to six given as equal to six centimetres Ds by DT will be equal to that's just given us six or six ft sorry not centimeter six ft 60 death but it is -2 inches one minute. But we have to convert it in six, divide by 12 -1 x six it Per minute. So let's find the value of the H. one x 80. The S one by beating people to Get one by the day where it is. Okay. Yeah four inches five minutes From here. The sun by date. If we could find out as 4" per minute. This is dancing. Thank you so much.

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.

We have some cylinder that is leaking. It has a height of two meters in a race of two meters. We want to find the rate at which the water is leaking. Alvis older if the rate at which the height is decreasing is 10 centimeters per minute when the high is one meter. All right, So we are wanting to do with how much water is leaking. And that's a volume question. So we should probably find the volume of a cylinder. So the volume is going to be high, are square h and we want to find Devi by DTs those Go ahead and take the derivative of this with respect to time and doing that, we'd get TV by d. T is equal to hi r squared times d h by d. T. Now the reason why we don't have to do anything with our here is because noticed that the radius of this, despite the height, never changes. So our radius is always going to be constant in this with respect to talk. So that's why we don't have anything, um, any derivative for that. So now we have our equation to tell us, Devi by DT So let's just figure out what things we need to plug it. So we know the height is one meter and then we know the height is changing, so D h by DT would be changing by negative 10 centimeters her minute. Now all of our units are going to be in meters. So we should probably cover are right here two meters as well. So we're gonna want to divide Bye. So we want one meter up top and then there's 100 centimeters interviewed her smaller units for seven years comes out and then dividing those, we'd get negative 1/10 meters per minute. So now we could just go ahead and plug everything in. So we have DVD by DT would be so high. So our radius is just constantly too so squared and the d h by d t. We had negative 1 10 negative 1 10 meters per a minute. And our radius over here was also in meters. It's going to put that so simplifying this down. We should end up with negative or hi over 10 meters cube her a minute and then we can simplify that too negative to pie ifs cubic meters per minute so the ball you is decreasing at a rate of two pi fifths cubic meters per minute.

Let's write out what we know. We know that B is three over four age before even solving are doing any calculations. This is what's been given in the problem right in the South and mathematical form. This indicates that V equals 3/8 h squared times l okay, expressing the volume of the water in terms of the level of water Devi over D t equals three l h d h over d t gives us d h over D t is four over three out H devi over dt. So we drive in respected time. We know that this means that d h over dt plugging in the information we know four over three times 10 times one and then we have toe multiply this by five multiplying by Devi over dt Such five This gives us 2/3 permanent So this means that the height of the water is increasing at 2/3 meters per minute. So 2/3 meters per minute is the rate of change


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