Oxford Maths Admissions Interview Question with @blackpenredpen

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Steve from blackpenredpen answers a real Oxford maths admissions interview question set by University of Oxford Mathematician Dr Tom Crawford. This exact question was used by Tom in the 2018 Oxford maths admissions interviews. The question looks at surfaces and volumes of revolution via a famous shape known as Gabriel’s Horn, which has a volume of pi but an infinite surface area.

This is part 1 of the interview, with the second part on infinite series coming soon!

Check out Steve’s brilliant channel blackpenredpen here: https://www.youtube.com/user/blackpenredpen

Produced by Dr Tom Crawford at the University of Oxford. Tom is an Early-Career Teaching and Outreach Fellow at St Edmund Hall: https://www.seh.ox.ac.uk/people/tom-crawford

👍︎︎ 3 👤︎︎ u/tomrocksmaths 📅︎︎ Jul 03 2020 🗫︎ replies
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hello my fans I'm very excited to bring you another special guest to the channel this is Steve from black pen red pen Steve do you want to tell everyone a little bit about what you do over on your channel yes hello everyone my name is Steve I'm from black to repent now on to thanks to Tom for having me on his channel my channel as well I do a lot of like calculus videos and what make the map up on my topics are pretty random I'll say but I enjoy what I'm doing and I hope you guys find my videos interesting as well yeah really highly recommend everything Steve does it's super fun lots of marathons doing integrals and derivatives it's amazing some of the stuff that you end up getting up to so on the channel Steve has quite a few videos where he's trying to answer some Oxbridge interview questions and I thought this was quite fun being a Oxford professor myself and I actually do the interviews and ask these questions of candidates every year so we decided perhaps I wouldn't say reluctantly you were actually a very willing volunteer so I suggested that perhaps I could sort of interview Steve and give him some of the questions that I've actually used myself in previous interviews so these are real Oxford interview questions put forward by an actual Oxford professor to a potential student so Steve of course is just pretending here this is not a real admissions interview but hopefully we should have some fun with some of the questions and also do some really interesting maths yeah yeah I am actually nervous I make you feel like I am being interviewed right now so I don't know cuz I don't think I can ever get into Oxford so we'll see I hope I do okay okay okay awesome I can't thank you so much for being willing to give this a go so I have my my interview notes in front of me so me at the first problem that I thought we would look at is something called Gabriel's horn is the name of the problem so first of all have you heard of this yes I have okay awesome so so maybe you will quite enjoy this one bed so so just to start with we'll start with something hopefully reasonably straightforward so you just first draw the graph of 1 over X sure definitely so I'm going to put it down right here so let's see and I'll just try my best to make my picture clear for you guys so this is y equals 1 over X perfect and then if we were to also look at let's say 1 over x squared on the same set of axes how would that look you want to on the same one or a different one let's go with the same one because I think the interesting behavior is in comparing them ok so first of all I will put a 1 over x squared in red and of course 1 over 1 + 1 over 1 square it's the same so they will intersect each other at 1 comma 1 so let me see it would look like this you look like this 1 over x squared is going to decrease a lot faster than 1 over X so I'm trying too much yeah there we go 1 is 1 over x squared here we go these are only sketches so don't worry that's that's absolutely perfect ok great now we're going to start with 1 over X and what we're going to do now is actually create a volume of revolution for this function so we want to take the 1 over X curve and then rotate it a full 2pi around the x axis ok so if we want to take the kerfing black which is 1 over X I would rotate about the x axis which is like this so the picture is going to look like this and I usually I will try to do a mirror image like this and then at the end I'll just put it over and there maybe here well actually it does not it's just when X is equal to 0 so I will actually just do like this and here is still my y-axis ah yeah that's the picture no that's great that's great so we're actually going to think about it starting at 1 and going all the way to infinity so we can if you want you can close it up so this is the horn this is gay Gabriel's horn where the name of the problem comes from because you can imagine it's a very long almost like horn or trumpet like the blow down right so this right yeah I should technically keep on going forever yes perfect so it's an infinite infinite horn right okay okay yeah good um now do you know the the formula for the volume of a revolution when we've constructed it in this way yes could you possibly tell us the formula and also could you perhaps explain why that is actually equal to the volume okay so the formula is will we can kill just take a disk right so we can make a cut like so and the formula of the volume of a disk is pi times the radius square times the height but the hide right here is just a small piece our code is the thickness so the formula right here you can see that the thickness it is just DX and the radius is from here to here which is the Y so the volume for this little disc is PI Y squared DX and then if you want to get the whole volume we're going to add the up and the way we're going to add the app is the density of Y which is integration so the whole volume we are just going to write it down like this all your equals and we have to add them all so I just put on integral and yeah that's pretty much the idea and depending on where you want to start and where you're going to add we'll just have to put this down right here and one last touch is that we have to also pay attention to the Y because Y it's now allowed in the X world that's how I like to say it so we had to plug in one of X into the Y and which was tax to write that out Thanks very good so I'm glad you mentioned the limits because we're actually going to use we want to do this and calculate the volume between the point one to some large number a let's start with from one is it yes so starting at one we want to know the volume from one to some large number a all right so one to some a I'm sorry about a picture right here one to a alright okay you want a to be large so maybe you have to take the limit for that that's where we're going yes for now let's just calculate what would it be for a and then we'll take the limits at the end I think all right stop the volume this right here it's equal to if we take the limit and they a equal to infinity so it's right here that means this right here keeps on going forever and then the integral going from 1 to a and then we have pi well I'll just say this right here is my new volume because I'm doing the whole thing pi and then the Y is 1 over x squared and then we have the DX after that yeah we're telling me to work this out for yes yes yes so and I straight explained this along the way - yes please yeah it's I think it's some something we do encourage in all of the interviews is is for you know as you're working we want to hear your thought process so yes if you can explain what you're doing that would be perfect okay and I think in order to go into Oxford how to be x rigorous as possible so I will know okay I would write on the limits all right here we have to take the limit but we have to worked out the integral first so this is still the limit as a goes to infinity and this right here is just the interval going from 1 to a and we have PI and this is 1 over X square and we can write here as X to the negative 2 power the X and we can totally integrate this by adding the 1 and then that's going to be negative 1 divided by the new power so we see this is the limit as a goes to infinity and we have 1/2 a actually no we did the integration step already so this right here will give us negative pi and extra negative once already has one of X like this and I will have to plug in 1 and a like so and then I just have to worked out the limit as a goes to infinity parenthesis plucking a so we have negative PI over a and then - this is the first part - the second part is you plug in 1 so negative PI oops negative PI over 1 like this yeah if you put infinity here negative PI over infinity worked p0 and I think that step is pretty clear so yeah and then minus minus one in the a actually get PI for this yes yeah awesome yeah perfect so R so the volume of our Gabriel's horn is going to be pi yes yes okay fantastic now what we want to do next is try and think about the surface area of the same shape so to start with we are going to think about well what is the the formula for the surface area of a revolution I see for the surface area we cannot really look at a cylinder anymore because this part is slanted this part isn't it so you have to look here like a comb and you're cutting to like pieces you can do to a cross section and then you just unrolled the comb so it's going to look like this so if you have a comb and you cut it so it's actually going to look like this part and there's a formula to calculate the area of this which I think we will need to know how long this is and then also the diameter so just to illustrate I will actually just erase this part and I'll show you a new picture well imagine you just put down a part of the comb like this again this is not the whole cone yeah and we look at this area here and the formula for that I was just cut out to be the s yeah which is meaning the small change in the surface area this right here you will have to actually know the change in the are legs which are coated to be DL and then for the DL you are going to multiply that with the circumference so you have to know the yes and in the formula for the circumference is 2 PI R so that's the idea yeah yes so it's like this and to rotate it and that will give you the that will give you the surface area and again just like earlier we will have to use integral to add up all the surface area from the starting all the way to the end yeah so finally L say yes it's equal to the integral and we'll just put down 2 pi the radius is going to be our Y again so write down Y for now yeah and because we know DL actually is just by the Pythagorean theorem is DX square plus dy square no this is not a person I won't use mu the final result okay that was a correct step but yes you probably want to jump ahead so you know give you the final result from there I will have to just start out the DX square so in the end I actually get square root 1 plus the Y DX square and then the DX all the way yet so this right here represents the DL I will have to put it down right here those square root of 1 plus from the C is dy DX square Y X like this so this is the general formula and of course we were hard to plug in the function in here and do the derivative here in order to worked out the surface area perfect thank you now we're going to you may have guessed oh we're going we're going to now actually work this out for y equals 1 over X and we're going to take the full Gabriel's horn so we want to have our lower limit as 1 and our upper limit as infinity and it's probably I think it's easiest here to just plug in infinity and you know rigorously we are taking a limit exactly like you did but I think we can just plug in infinity for now and worry about the limit at the end okay yeah cuz nobody related to dread on the Li M exactly as long as we know you have to do it I think it's ok to leave it to the egg all right yes for the surface area we will go from one to infinity and we have two pi the Y is 1 over X I will just put that down and then we do the square root and here we the DX and let me actually get a new marker all right so here we have the 1 plus dy DX the derivative of 1 over X is just negative 1 over X square so I will just write that down so we have the negative 1 over X square so I'll put that down nefi 1 over X square like so and then we square that and again this is the Y right here ok ha I thought you enjoyed integrals you sounds like I thought this would be exactly what you would love to try and calculate ok so do the easy poppers as I usually would like to start with put up 2 pi in the front and then integral 1 to infinity this is 1 over X and we multiply by the square root all right this is 1 plus 1 over X to the fourth power so right yes let's just add so this is going to be multiplied with X to a fourth and also x to a fours so all we get X to the fourth plus 1 over X to the fourth and we have the DX right here and I think I don't know if I can actually integrate this but I will argue that it might not converge because if I take the square root of x to the fourth on the bottom that will give us x squared on the bottom and then on the top we have square root of x to the fourth plus 1 DX yes yes and then this right here I will just say 2 pi integral 1 to infinity this is just square root of x to the 4th plus 1 over X to the 3rd power DX and we will have to use some comparison test to argue that this right here unfortunately diverges so very good yeah yeah I will just say by computers and by comparison okay computer weight comparison there we go a comparison with the integral 1 to infinity this right here just X square if you ignore the 1 so it's x squared and this is extra power reduced that is 1 over X this right here I purchase so on oh this right here also diverges to infinity because this is actually bigger not yes yes very good you spotted that very early I'm very impressed so what we've actually shown if we take stock of the whole problem is that we've calculated for this infinite horned called Gabriel's horn we've calculated that the volume is equal to PI just and the surface area is infinity yes does that make sense no it kind of like a contradiction because how can you have something that's finite volume by like meaning that you can fill in with that finite amount of stuff like paint or like water but you can never finish painting the corn so yeah this is yeah this is this is why it's favor this is why it's called Gabriel's horn because exactly as you've said you could fill this with PI units of paint and it would completely fill to the top and then if you tried to paint the outside you would be there forever which i think is quite a neat anita example or a neat idea involving convergence and divergent integrals and series so i think we can all agree there that steve has performed excellently in our interview question there we tackled Gabriel's horn and we've sort of resolve the paradox maybe maybe we have maybe we haven't but we certainly got to the right answers and hopefully had a bit of fun while doing it so we're going to do a second question but this is going to come in it another video so do look out for that one but the first half of the interview is now over Steve is sitting in a very good position I think we could all agree he did well and it's all resting on the second question no in the meantime do go and check out Steve's awesome channel black pen red pen if you haven't already like I mentioned it's really great loads of really nice calculus marathons have been stevedore six hours of doing integrals it's crazy and if you have enjoyed this video as well please subscribe to my channel that would be super awesome and just to say thank you to Steve and we'll be back again soon with another question thank you so much Tom you
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Channel: Tom Rocks Maths
Views: 230,910
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Keywords: oxford maths interview question, oxford maths, oxford admissions interview, oxford maths interview, oxford interview question, oxford mock interview, oxford practice interview, oxford mathematician, maths interview, Oxbridge interview question, Oxbridge maths interview, tom rocks maths, tomrocksmaths, tom crawford, tom crawford numberphile, steve chow, blackpenredpen, black pen red pen, gabriels horn, surface of revolution, volume of revolution, toricellis trumpet, mathematics
Id: htB_NGmPKVI
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Length: 18min 44sec (1124 seconds)
Published: Thu Jul 02 2020
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