A Level Maths Made Easy - Modelling with Differentiation

hey guys and welcome back to another AEV Mat made easy video this time we're going to be tackling the modeling with differentiation so a fairly tricky topic that comes up in both year 12 exams and year 13 exams and as I said it can really screw over a lot of students and I think the reason why is this is one of the brand new skills in a level so in gcsc we don't really have anything like these types of questions hence why it's pretty tricky and a level to do it so without further Ado I've gotten every past paper question that's come up so far I've left pretty much every question in the textbook for you guys to do just so you know if you join the Discord we have solutions to all of the textbook answers as well on there just for your enjoyment perusal whatever you want to call it but without further Ado let's jump straight into it so a large tank in the shape of a cuboid is to be made from a 4 54 M square of sheet metal the tank has a horizontal base and no top the height of the tank is X and two opposite faces are squares so the volume of the the tank is given by this expression cool now this might seem a bit strange here's the thing that you need to do and realistically if you're in year 12 and watching this is definitely something you need to work on year 13 you tend to be a little better at it but you need to draw diagrams always so it's a large tank in in the shape of a cuboid even if you don't know what these are just draw a good oldfashioned little cuboid and I'm going to even make it 3D although my 3D shapes are not actually you know that that's not too bad I seen worse by myself and it tells me that it has no top now sometimes when it has no top I see some students get a bit confused cuz they include a top so I like to just do a little like hashed line just to show that that side doesn't exist so to speak the two okay so the height is X so let's put X like that and the two opposite vertical faces are squares which means that means the base here is also X because this cuboid this rectangle 3D rectangle it's so are x and x because the height and the depth must be the same because they are squares right and the sides of a square all the same and it doesn't tell me how wide the shape is so I'm just going to call it y because X and Y makes 100% sense to me right and it says show that the volume is given by so here's what I would do and I'm well aware that model answers tend to not follow this exact process but I think this makes more sense is I would write down how would I work out the volume well I would do y * X x * X or x * x * y however you want to do it but it'd be x^2 y now this doesn't really look like this does it so what have I missed out well remember my golden rule that I said all throughout gcscs all throughout a levels they never give you information you don't need and if you notice there is a 54 M squar that I haven't used and this is generally where students get stuck now I want to point out you get a mark for this anyway if you get stuck at least you have one mark That's not not bad is using this other information they give you now this information can be anything sometimes it's a surface area sometimes it's the depth of water sometimes it sorry about that I had to sneeze sometimes it's even the cost per gram of metal used or something crazy like that but the idea is that you're going to somehow use this information to eliminate the variable you don't need in this case I don't need y so what does this mean and this is the tricky bit well 54 M squared that is of course some kind of area right now it's telling me that this shape is made of metal now if I think of a box right I think of a box the surface of it is the material it's made of so in short this 54 square m is actually the surface area of the cuboid take a second to to really think about that if I have a sheet of metal if I wrap it around a box that becomes the surface area area so if I work out the surface area of this shape that should be equal to 54 well let's try that the surface area is going to be well we have this front base XY or YX however you want to write it and there's two of them right because there's also one on the back okay then we have the square side so x * X that's X2 and again we have two of them boom boom then we have the bottom side which again is XY but we do not have two of them and this is where people make mistakes they put a two here because the top side is empty it's Hollow if you imagine like a fish tank and you take the lid off that's the kind of shape we're talking about right now so there is no two here and we know that that equals 54 so that means we have 3x y+ 2x^2 equals oh God 50 four so what I'm going to do is I'm going to move I want an equation for y so I can sub it in so I'm going to move everything that isn't y over so - 2x^2 from both sides 3x y = 54 - 2x^2 I'm going to divide both sides by X so let's do by by 3x right 54 divid by 3 can you do that I actually can't remember let's see it's 54 3x now I'll check out my calculator in a second and then we have 2 over 3 x that's probably where that 2 over3 comes from 18 okay yeah got you y is equal to 18 /x - 2/3 x and I'm going to Simply substitute this into this original formula for my volume so V is equal to x^2 y so equals to x^2 * 18 /x - 2/3 X so expanding this should be quite easy x^2 and then 1 /x so that's just going to leave me with 18x right because we would have 18 x^2 /x cancel the X's 18x and then we have minus 2/3 x cubed and then just do a nice check I don't know why I said nice check that is the actual answer that first part can be up to four marks pretty good now Part B and part C and this is something that really annoys me about some students is when they don't get part A they skip Part B and C you can do Part B and C without doing part A and here's the funny thing part B and C tend to be most of the marks so let's say this question's roughly 10 marks in total you get four for the first part you get six for the next two in total which is going to be four and two marks so it's actually better for you if you just do Part B and C even if you don't know part A so it says uh given that X can vary use differentiation to find the maximum or minimum value of V here's what we're going to do we're going to differentiate this of course because it's a maximum minimum you always differentiate with respect to whatever the variable is on the right hand side okay another way I like to think of it is if you have multiple copies of a variable that's what you're differentiating with respect to so in this case with respect to X and minus 2x^2 we're going to make that equal to zero move the 2x^2 over why are we making it equal to zero we are making it equal to zero because a maximum or minimum value is a turning point or a stationary point that only happens when the gradient equals zero divide both sides by two now technically if I Square both sides I get plus or minus 3 so which ones are going to be well if I have a look at this x is a length which means it cannot be negative so I'm going to take xal 3 is that my answer no because it says work at the maximum or minimum value of V not of X you need to then resub this back in to the original equation to work out what the maximum or minimum value of V is at this value of x and and here we go I'm just going to use a calculator cuz I'm lazy but laziness is fine better than making mistakes that's for sure 2/3 3 cubed gives me 36 that's a nice number and I don't know if this is a maximum or minimum value yet uh this should be in centimet sorry Meers cubed because we have meters there I don't know if it's a maximum or Min minimum value yet because of course I haven't done the second order differential part C just justify it a maximum there you go well you differentiate one more time so differentiating this equation equals just going to be - 4x now technically to be really really rigorous you'd have to sub in this value of x into that equation which is -2 Which is less than Z therefore it's a maximum you have to be very rigorous with this the mark scheme can be very very specific but that's it for your first question these are hard questions um it comes up if you're in year 12 every other year or so on average if you're in year 13 it comes up most years so it is something you should definitely get good at but it's 10 marks pretty fat question but hopefully this makes sense again I think it's part A that most people struggle with but let's go on to a couple more with this question it is ever so slightly different and I wanted to show you this just before we move on it actually gives you the equation straight away and it doesn't say to show it but if you look at part a part two part B part C Etc they get a bit more interesting so with part A find the value of V which minimizes the cost of the journey so it's not the minimum cost of the journey that's part two so all we need to do is differentiate this first of all use your rules of indices in order to make this into a proper form so vus one uh I'm going to say 2 11 V 60 differentiate what does v stand for actually velocity I'm assuming yeah equals -500 B- 2 + 2 over 11 and again we want to make this equal to zero so if I move the 1500 V ^2 I'm actually going to write that like this but hopefully you can see it's the same thing then if I just kind of move this over so v s will go up 2 will go down 11 will go up we have B ^2 = 11 * 1500 / 2 and then square root that and that will give me my answer for part I part two is just subbing that value back into the equation for C so 11 * 1500 / 2 square root the answer and I get 90.8 really okay that seems a bit fast to be honest I heard that the most efficient speed is a bit lower than that but then again this is in kilm per hour mine was in miles per hour so actually that's actually accurate never mind so part two the minimum cost of the Journey part two just sub it straight back into the original expression I do like to actually write the whole thing out I know that might seem a bit weird but it just really shows the examiner that I know what I'm doing again the point of the exam this might sound weird is not to get the right answer is to show The Examiner you know what to do that's the important thing I know that sounds a bit weird getting the right answer should you know show the examin what to do but anyways and then we puts to the minus1 + 2 over 11 answer + 60 gives me 93 and what was the unit the unit is pound so £93 for this J if he speeds prove that it's minimized by the speed found that's actually a good point I should have said this we don't actually know it's a Minima yet although it does say in the question it's a Minima so if we want to be really rigorous we shouldn't have I shouldn't have said that so differentiate once more oky dokie dv^ 2 so that's going to be plus 3000 vus 3 yeah minus three Panic there for a second and all we need to do is prove it so we just need to sub in that value um it was this value here and you can kind of see that it's going to be positive and so obviously it's going to be a Minima but again writing the substitution out is very important it's just something that they really expect you to do 90.8 to the power of -3 it's a small number but it is a positive number greater than zero therefore Minima again can you see how the structure is the same each time right well okay very similar I should say generally the first part is making the equation in this question they gave it to us the second part is finding the cost finding the values that Maxim maximize or minimize something proving it's a maximum or minim that's the general layout sometimes they give you the first part and then part the last part will be State limitation of the model but either way so limitation of the model this is going to be different every single time just going to point that out and there's a load of different things you can say in fact in the mark scheme generally it would give you maybe five different responses and then it would say any valid answer so what would I say well one is technically speaking if I increase V more and more and more and more and more I will have infinite cost right I can make this speed like 10,000 km an hour and it make the cost bigger and bigger and bigger I can keep increasing V basically there's no limitation on what values of VR can put in that's one downside two is it doesn't take into account air resistance which is really really important three it talks about a steady speed it's not really accurate for there to be aady steady speed if you guys have been in a car before you'll know that you don't really travel at the same speed for very long there's traffic there's traffic lights there's turning Corners there's all sorts of stuff that will affect that um anything along those lines is is really important I personally like to do the mathematical way nine times out of 10 they will always predict either a negative cost or an infinite cost both of those things cannot happen so just make v a ridiculously large number does it mean the cost is ridiculous large does that even make sense that's what you need to think of so sometimes it'll be negative sometimes it would just be Infinity either way that will be a limitation that you can talk about so again hopefully seeing a very similar layout we have a cylinder very basic sometimes these can be very wordy surface area is 500 cm squ show that the volume is given by this given it varies and then the same type of thing again eight out of the 12 marks so most of the marks actually comes from just differentiating equating to zero and then double differentiating that's it so you can get most of the marks just by doing that so genuinely if you want you can even start with Part B right who cares right but I'm going to show you how to do part A cuz that's a really important bit in my opinion so we have a cylinder again I like to draw a little diagram just to help me visualize what's happening it's a beautiful cylinder look at that oh yeah beautiful has a height H radius R and it's to show that the volume of the cylinder is given by blah so the volume is of a cylinder is p Pi r^ s h again that doesn't look anything like this what letters missing well here we have R S and H here they just have R so that means I need to find an equation for H but look they've given me an expression sorry a value for the surface area so all I need to do is work out the surface area now the year 13s that my score have recently done their exams and I did hear a bunch of them asking oh what's the you know how do you work out the surface area how' you work at the surface area you don't need a formula a lot of people rely on formulas but there's no point if you unroll a cylinder you get a rectangle and you get two equal circles right that's like the net of a cylinder the circles I'm hoping are fine that's just pi r squ and there's two of them so we're going to do 2 pi r squ now I will admit that the rectangle can be a bit cheeky can be a bit annoying but if you imagine a cylinder is a circle like this and you're unraveling it so it's a straight line that wide length is just going to be 2 pi r the circumference of the circle and the height is going to be well the height so again if you get a piece of paper roll it into a cylinder and unroll it you'll see that the length of the rectangle is just the circumference of the circle so that is is going to be plus 2i r h that's all you need to do and that equals 500 cm squared so now I'm going to rearrange for H so I'm going to move the 2 pi r squ to the other side so 2 pi r h = 500 - 2K r^ 2 uh we can actually divide through by well I get 2 pi r technically so if I start with 500 ID by 2 is 250 so let's write the 250 first then it's still divided by pi r then we have 2 ID 2 of course Pi / Pi R / R so it should just be R cool subbing that into this formula that we have up here we have V is equal to Pi R 2 time H so times all of this uh juicy stuff so we have 250 over pi r minus r so the pi and the pi are going to cancel the r squ and the r is going to cancel a bit so this is going to be 200 R which sorry 250 that's why you always check the answer can I yeah there we are 250 R just like it says in the answer then we're going to have PK r s * minus r so we're going to have minus PK R cubed exactly as it says four marks pretty good right it's not too bad now Part B and C again I'm assuming this part is actually relatively okay calculate the maximum we're going to differentiate an equate to zero so DV by Dr R we have 250 minus 3 < r^ 2 equal it to zero move that over we have 250 = 3 piun r² so if I do this all in one step now one giant step we have 250 over 3 pi equal R I might sub that whole thing in yeah calate the maximum value of V so we're not actually looking at the value here so sub that back into this thing all right it's fine we're going to use the good old fashioneds button right because we're we're cool so 250 agains so whatever that answer is minus Pi a ANS cubed you are more than welcome to do that in your exam no rules against it so just do this so we have 250 over 3 Pi gives me whatever that is so 250 brackets answer again I would always use uh brackets just cuz again paranoia it's not paranoia if it actually is like a thing so 858 yeah I'm going say 858 and I believe that was centim cubed CU I remember there being centimet squared in the area yeah cm cubed justify so that's six marks calculate the maximum value six marks crazy crazy just doing that one 2 3 four five I mean four lines really six marks pretty good so we're going to differentiate one more time just to well I mean because we have to - 6 piun r um and then subbing in the value for r that we get is of course going to give us a um oops spoilers uh let's see how I'm going to it let's do it like this and then n s it's of course going to give me a negative number so it's less than zero therefore a maximum it's all you got to do really really nice question that six marks is so juicy look at that perfect and I don't think it's too difficult so again do B and C even if you have no idea how to do a I would actually even recommend doing B and C then going back to doing a that might be a good thing for you to do so as usual in these videos the next two questions are going to kick it up a little bit so this one looks a bit more fat the one after this is very fat it's very beefy indeed so this is apparently a design for a swimming pool good um bunch of stuff semicircle rectangle cool the area of the pool is 250 M show that the perimeter so of course same as usual we're going to write an expression for the perimeter in fact I might write it on the side just to make it a little bit easier and then I'll drag it over when I send the notes at the end um well the perimeter here i' would have y + 2x + y so it's 2x + 2 y and then I would have the well the circumference of this semicircle right so the radius is going to be X so it's just going to be 2 pi r 2 pi radius so just be 2 pi x / 2 cuz it's a semicircle so it's just plus pi x right so again circumference is 2 pi r but since we're only using half a circle we're going to half that it's just pi r cool that obviously doesn't look anything like that but that's because they given me the area the area is honestly completely fine the area is equal to so with the rectangle 2x * y so it's 2x y plus < r 2 but again it's that we want half of a Circle so PK r² / 2 / 2 equal uh 250 okay and we want to rearrange for y let's move that over yeah we just just do that yeah 250 - < r^ 2 / 2 IDE 2x we get 250 ID 2 is 125 so we get 125 /x minus I just realized instead of R I can put X of course because you know it it is X there you go and then we have um so divid by 2x so that' be just p x over 4 now right because when you have 1/ two ided two again you can do the mats if you want but it'll give you 1 over4 it doubles the uh denominator subing that back into my original we get p is to 2x plus 2 * all of this in fact I'm going to I'm going to cheat 2 * 125x would be 250X and then minus 2 * all of this so that just be again it would just be pix / 2 and then plus pix now ooh this is okay some of you may be able to see that that will simplify uh to the expression that I want now with - pix / 2 +x you can treat that as minus a half + 1 which will give you plus a half right so we're just kind of ignoring the Pix for now but it will indeed give me pix over 2 now I'm going to quickly just move this underneath so we can do Part B and C Part B is a little tricky to be honest so normally when we have this kind of range what we would do is we would sub in the Big Value right and we' sub it into here and it would give me an impossible answer maybe a negative perimeter maybe zero perimeter something like that but in this case if you look if I sub in x isun 500 over Pi I get a value that's normal right it's not weird at all now there's God lost my words for a second the reason why this is two marks is because there's two sides of this I would start with the X is bigger than zero why does X have to be bigger than zero well I mean again this might be obvious because X is a length and must be positive that gives you one Mark believe it or not pretty good but with this otk 500 over Pi in my head and this is my first thought I thought well we never Square this number and that means it's probably not subbing into here we never Square it but if I look over here for my area for example or even over here whatever way you want to look at it if I sub in X into this expression I actually get something that is impossible so let's look at this 250 minus < x^2 / 2 and this is the working out that you would use to get your marks if I sub in this value it everything's actually going to cancel out so I'm going to do 250 minus I'm going to dok / 2 and then because it's x^2 I'm going to have 500 over Pi cancel 250 cancel I get 250- 250 somehow equals 2xy so in short my area becomes zero area becomes zero and what would happen if my value was greater than this it would become negative If X is bigger than Square < TK what was it 500 over Pi area becomes negative which is of course impossible because again area has to be positive so that's the actual reason that's why this question was actually very difficult because instead of subbing into the kind of main equation you actually sub it into an equation that you worked out for the area which isn't really part of the question so I thought that was a bit of an interesting one it a bit tricky part C find the minimum perimeter again change this into a form that you can actually differentiate so we have p is equal to 2x + 250 x- 1 plus < / 2 x and then DP by DX gives me 2 - 250 x - 2 plus pi / 2 yeah okay there's two values of X okay cool equals z I'll move this over can we do yeah we can't really add those two together sadly 5 two equals and again we're going to write it in terms of X2 and then moving this around we get X is equal to the square of 250 over 2 +un / 2 so going to give you a second just to think about that we've swapped the x s and this whole thing and then we've square rooted it then all we do is we sub that back into the original to find the perimeter and again we're going to use the good oldfashioned ANS button plus 250 a n s-1 plus P / 2 a NS and that'll give me my final answer so let's quickly sub this in so we have 250 over 2+ and then pi over 2 yeah that's my answer God I don't know why I brain froze there for a second plus 250 whoops that's not 50 ANS to the minus1 plus pi / 2 gives me let's call it 59.8 and I think this is centimet no it's met that actually makes a lot more sense that's quite a massive swimming pool to be honest um yeah 3sf job done not too bad once we got started now for probably the hardest one that I've seen okay so a company makes drinks containers out of metal probably called a can uh they model as closed cylinders with a base R height H and the capacity is 355 cubic cm yes so it's a can cool I mean can they be any more obscure maybe so height radius so we've already done this one a similar we've done the same shape before so we should hopefully be experts the capacity is 355 cubic cm however the metal used for the circular base and the curved side is 04 P per square cm and the circular top is 9 P per square meter centimeter sorry both metals are of negligible thickness so they don't affect the capacity is the idea there so that the total cost of the metal is given by that cool let's work out an expression now just like in the very first question this Pence per square cm is all to do with surface area right so you have a surface area that you're going to create out of metal in order to make this can shape now the circular base that's going to be p piun r s so what I'm going to do is I'm going to put it in Brackets p piun r s and the curved side remember that's going to be the circumference of the circle so 2i R * H that's all times by 0.04 P right cuz 0 04 * < r^ 2 and 0.04 * 2K r h adding on the circular top is 0.09 PK r s uh we could expand and simplify so that's what I'm going to do so 0.04 pi r s plus that' be 0.08 now if we times it by two plus 0.09 PK r s we can put the p r squ together that gives me 01 1 3 pi r 2 plus 0.08 pi r h but again we have that 0.13 PK r s we've got that already but we need to get rid of the H and it says the capacity is that well capacity is the same as volume so it's piun r^ 2 H equals 355 so rearranging this for H we get 355 PK R 2 so subbing that into our equation we can now get our answer from here I'm willing to bet that you guys would be able to do Parts uh b c and d in fact I would be willing to bet it's definitely the first part that is the really tough challenging part 355 p squ so we can cancel the pies one of the RS 355 over 0.08 sorry times 0.08 that's the difference it's probably going to be 28.4 but why risk it you know why not just do it so 0.08 * 355 gives me 28.4 as expected uh 28.4 over ah job done so again the cost thing definitely does throw some people um but again basically you're just doing what the question says and I hate saying that cuz it sounds really patronizing but that is effectively what you're doing um if you really simplify every A- Lev question down if you follow the steps the process that's generally what they want H even if it seems like something's very challenging or very weird like we've never learned how to work out total cost yeah I know but they kind of tell you how to do it right it's a cost per square me met so you need to find out surface area try and use that now on to Part B and C and D so differentiate equal to zero cool um yeah Part B C by Dr R we get 0.26 piun R that' be - 28.4 / R 2 I took a bit of a shortcut there see if you can work out how I did that move the r squ over oh good point okay times both sides by R 2 so we get 0.26 < R cubed = 28.4 then all we're going to do is move that all over 28.4 over 0.26 Pi Cube rooted and did I do anything wrong I don't think so but it says to actually work out the value for C so we're going to sub that back into our original equation back up here so again using the good old fashion answer button we can work out our value so cube root which is that button over is 28.4 over 0.26 Pi gives me that subing into there we get 0.13 Pi answer squared plus 28.4 over answer gives me 30 does it say how many significant figures 3 SF so 13.1 um is that just going to be in Pence I assume right yeah pens 13.1 P to make a can interesting verify that it's a Minima yeah figured verify it to Minima so differentiate this thing again we get uh 0.26 Pius plus even uh that would be 56.8 over R cubed and if we sub in the value subing that into here you will get that it is greater than zero therefore it's a Minima and what's the last thing oh my bad yeah that was just the value of R for Part B So my answer here is actually the answer to Part D that gives us the final thing um oops but that's fine um since I noticed it's all good and I didn't lose any marks that brings us to the end of modeling with differentiation fairly tricky it's not the easiest thing in the world but we're going to do some stats and mechanics next but I guess that depends on when you're watching this video we're going to do some stats mechanics next and get through all of that this is probably one of the hard hardest topics in year 12 maths but in a level maths with the whole modeling and everything but again integration vectors then some stats and mechanics and we're done with the year 12 content