Maths Teacher Goes through YOUR SUBMITTED A LEVEL MATHS QUESTIONS

okay for okay should be good hello okay I'm just going to wait for a little longer to saveone can trickle in got quite a few questions to do today and again we'll just get as far as humanly possible as per usual hello hello okay [Music] so okay this should all be good now I'm glad you are so again I'll just give it another let's say 2 minutes and then we'll start going through some of these questions that people have given again quite a few different uh types of questions okay W stream of course every stream is a w stream Okay so we've got a few people in uh this was well again I'm just going through the the chat so if you do want to submit questions go on to the Discord and there's a stream questions little button uh little Channel you can just post an image of your question and I'll do it if I message you asking you where you got it from it's just so I can get like a clean copy like this so I can't remember who submitted this but I basically asked you know tell me where it's from and they gave me the chapter of the textbook and I copied it and it looks nicer so it's easier for everyone to see so either way we have this handy question here so a swimmer dives into a pool her position P underwater can be modeled um in relation to a horizontal distance as a quadratic equation so it's going to look like a big U so he jumps in it's like a u shape the position of the bottom of the pool is a linear equation because as you can see it's a straight line right so the actual bottom of the pool is linear determine whether this PO model predicts that the swim will touch the bottom of the pool so for this to happen these two equations we're looking for the point of intersection so if I draw this out and please forgive me is you're going to have like a u shape here and we're looking for the point of intersection if they intersect so this just means if they intersect so when we look for a point of intersection what we do is we equate the two equations don't we so that's what I'm going to do here I'm going to write half x^2 - 3x actually I'm going to make the pen a little thicker just to make it look a bit nicer is equal to 0.3x - 6 now I'm going to write this all again just in decimals just to make it a bit easier for me and I'm going to move everything over to the left hand side because remember when we do something like this we're trying to um we're trying to solve it right it's a quadratic he's already straight to the point yeah um I got told that I need to speed up a little bit by some of you guys and that's a very fair assessment so I'm going to move everything over to the left hand side we have 0.5 x^2 and then - 3x - 0.3x be - 3.3x and then + 6 = 0 now I don't actually care about solving this if you look what it says it says determine whether that's like an if right this model predicts that the swimmer will touch the bottom well if they do touch the bottom it means we could solve it so we're looking at the discriminant so I'm going to work out what B ^2 - 4 A C equals and then depending on if that's zero negative or positive I'll be able to tell if it touches okay so let's actually work out what the discriminant is well the discriminant B is going to be -3.3 so let's do that -3.3 2 - 4 a is 0.5 and C is 6 so see what that gives me and again I'm just going to shove it in my calculator because again it's all calculator exams for a levels so you don't even need to worry about doing the maths at least mentally anyway 0.5 and then multiply by six that gives me 1.11 so what does that mean well it means it's less than zero therefore there's no solutions to this quadratic so what does that mean well it's just like in my diagram it means that the swimmer and the bottom of the pool do not touch each other they do not intersect so therefore they do not um actually what is the wording the question so the swimmer will not touch uh let's write it like it says in the question the swimmer will not touch swimmer will not touch now if you notice this is a five mark question five marks and it's an exam question that's why there's an e there how there's not that many steps or one two three steps really because again that's why I like the modeling questions because if you understand the maths behind the modeling you can get five marks in just a few lines of working out it's really really good so these modeling questions I know they're harder but you get a lot of marks for maths that isn't really that crazy they just want you to be able to apply the maths to a situation okay so assuming that's all fine let's move on and I left quite a bit of space so this question is actually from mechanics so it says a particle p is moving along a straight line such that at time T seconds T is greater than equal to zero after leaving the point on the line the velocity is this find the value of T at the instant when P stops s accelerating so part A so when we have this kind of information I just want to point this out I'm going to go with this really lightning fast when you have an equation for displacement you can differentiate to get velocity which you can then differentiate again to get acceleration okay now that's a bit of mechanics so in other words I'm looking at when it stop accelerating so that's when a equals z but I don't have an equation for a so what I need to do is expand these brackets and differentiate velocity to get an equation for acceleration so let's do that velocity equals so if I expand that we get 7t + 14 - 2 T ^2 uh - 4 T Lightning Fast just collect the like terms we have -2t ^2 Bur 7 - 40 is 3T thanks nitha and that's plus 14 on the end whoops that's a four so again acceleration is just the differential of the Velocity now that might sound scary but remember differentiation is is not that bad what you do is you bring the power forward so that now becomes minus 2 * 2 which is - 4 and then you subtract one from the power well 2 - 1 is 1 so I don't need to write it do the same thing again bring the power forward that's a one so that stays as three and then that becomes T the^ of 0 which is one so that's just nothing so it's just 4 -4 t + 3 now what about the 14 well think about what differentiation is it tells you how quickly each of these terms is changing with respect to T but this doesn't change it's a constant so it's rate of change is zero so if you like you could have that as a plus Z have a GCC history marked have a nice day everyone good luck hypine so there we go so a is equal to -4 + 3 now what does the question actually say well it says when it stops accelerating so that's when acceleration is nothing so 0 = -4 + 3 so 40 = three so T is equal to 34 God hello Sky uh tenen oh God Sky King so that's four marks I know that sounds pretty scary hopefully you see it's not that bad okay go my voice is uh already giving out I had a few lessons afterwards okay so what about find the distance of P from o at the instant when P changes its direction of motion is this Su and differentiation it's not it's something it's a topic called variable acceleration it's called variable acceleration I think it's chapter something I can't remember what chapter is I'm not going to lie to you okay but it's towards the end of mechanics so B the distance of P from the origin at the instant when changes its direction of motion okay so when will that happen well if I look at V at the start here when something changes its direction of motion it means the velocity changes sign okay that's what it means the velocity changes sign the example I like to use is if I throw a ball upwards we have V when it gets to its top the V is zero and then as it starts to come down it's going to come down with negative V so can you see that the V changes sign so in other words if I want the time when it changes its direction of motion I'm solving b equal 0 so solve that so we have -2t ^2 + 3 t + 14 equal 0 now I'm going to use a calculator to work this out now some students say okay it says in this question Solutions relying on calculator technology are not acceptable now this doesn't mean you can't use a calculator it means you can't solely use a calculator so I can solve this equation using a calculator fine but if you look I still have working out that's what they're looking at they don't want you just to shop the whole thing on a calculator so menu go down to write once equation funk polom polinomial degrees is always going to be two cuz we're solving quadratics we have -2 in front of the squared term three in front of the linear term and a 14 as our constant we have 7 over2 is one time and we should get another time but you'll notice something it's a negative number now I had a Spanish map today but I can early that's a shame so if I look at the question it does say t is bigger than equal to Z but also think logically for a second can time be negative if I say I'm going to talk to you in -4 hours does that mean anything you might say it might mean 4 hours ago but doesn't really work that way so we're not going to include this value so we have our value for time brilliant okay it says find the distance of P from the origin well we need to know what s is we need to know the displacement so I said to go from s to V and V to a we differentiate to go backwards well what the opposite of differentiating it's integrating so we need to integrate velocity with respect to time so B is equal to -2t ^2 + 3 t + 14 s is going to equal the integral or sorry -2 T ^2 + 3 t + 14 with respect to time so how do you integrate well you do the opposite of differentiating when we differentiated we brought the power forward then subtracted one in integrating we add one to the power and then divide by the new power so -2 T add one to the power divide by the new power is variable acceleration on the Ed EXL spec it is yeah variable acceleration uh give me one second I'll show you in the book it's it's an ed EXL um spec question so keep going 3 T add 1 the power squar over the new power and then 14t but also you have a constant term C it is as it is as so let me show you quickly in the textbook uh I'm just remembering Which chapter it is variable accelerations page 181 in your textbook what was that bit further down my God h okay so it's page 181 in your textbook so if I show you this chapter here is called variable acceleration and yeah it's covering everything we're talking about here so we have these equations it won't show you the differentiation just yet using differentiation Etc so yeah it is 100% a um as topic and it is at Excel okay so God brilliant now I can say that c is going to be uh zero I'm good at Stats should I do S2 or M1 um up to you if you like stats then do S2 so now I need to know what that constant term is well I know that tal 0 the displacement should be zero right it should be at its original spot so if I sub in t0 s0 can you see that c is going to be equal to zero cuz all of these terms will cancel out so my equation for S is -2 over3 T cubed + 3 /2 T ^2 + 14 T and all it's asking me is the distance from the origin so I just Sub in my time 3/4 sorry not 3/4 7/2 whoopsies spoilers T = 7 /2 so we get S = -2 over 3 7 /2 cubed + 3 over 2 7 / 2 squared + 14 7 over2 good Lord shove it all in the calculator job done easy peasy lemon squeezy uh 7 over2 is just 3.5 that' be easier for me oops are normals as and if so what chapter normals what do you mean by normals is in like normal lines then yes yeah so uh equations of normals come up in the circles chapter so coordinate geometry and we get 38.8 M you're very welcome to 3sf okay so far is everyone okay we'll keep going we've got some really spicy questions coming up in a bit so we're getting through a lot okay here's an equation of the normal have a read through that just while I drink some water cuz I'm dying okay so find the equation of the normal to this long expression and this is where a lot of students turn off and they just go I don't get it I don't get it relax what is a normal a normal is a line that's perpendicular to a curve so what do we need I need the gradient of this thing and then the normal will be - 1 over the gr and then I have my coordinate remember to find an equation of a line you need two things you need the gradient and you need a coordinate we you can get the gradient from the curve -1 over it and the coordinate they've given us so how do we work out the gradient well how do you work out the gradient of anything we differentiate so Y is equal to and now I want to write this in index notation because we can't really differentiate fractions yet so I'm going to make that xus one - 3 x - 2 - 4x to -3 + 74 now we differentiate and this gives me an equation for the gradient because the gradient of a curve will change over time that that's the definition of a curve remember what I said bring the power forward first so negative X subtract one from the power bring the power forward plus 6 because 9 is 3 * - 2 X the - 3 + 12 - 4 * - 3 right x - four and constants go away so that's gone now that is our equation for the gradient now that is hideous so how do I work out the gradient well you sub in the coordinate they give you so at x = -2 Dy by DX and you're just subbing it into that expression and then you that is your gradient okay so get our calculator out we get to do something fun yay whoops negative -2 to the -2 you could make it a fraction again if you really wanted to but it's not necessary thank you Sky King but again that's kind of my job right that's what a teacher does well hopefully breaks things down into make it easy make it easier maths is all about logical steps okay so it looks scary break it down okay that curve makes me want to vomit too slow down negative a quar nice and easy further math isn't easy no so now we can actually work on this the gradient of the normal I'm going to call that MN is the negative reciprocal which by the way if you're ever unsure do it on your calculator do -1 over - 1 4 and you should get four so if you ever unsure just do it on a calculator okay now we have y equals I like to use MX plus C but we have 4x + C if we Sub in these coordinates now we can work out all of the values okay when is a hard AAR question video coming out I'm editing it now the first one I've started with is sua and then I want to do logarithm okay so we have 1 = 4 * - 2 + C so we have 1 = - 8 + C so C is equal to 9 so we have y = 4x + 9 now it says to give it in this form ax + b y = c okay fine I will do that as move this over we have -4x + y = 9 that f FS the criteria so yeah I am I have recorded a bunch of really hard AAR questions I've started with suvat and I might do each question gets its own video because they're quite long they're about 10 markers at 12 markers and then I'm going to move on to logarithms because logarithms is the next hardest one so people hate suat people hate logarithms I do it in that order but again if you want to let me know which one you find difficult I'll let you know from WME what's w Wait do you mean mme math made easy or okay next question and this is the last one quite small but this is really common first principles a lot of students suck at okay and that's fine I got quite a few questions from from that um exam board or like website I can't remember what the hell it's called anyways so you learned this in the lesson yeah you need to be able to do this for any derivative they give you okay let me show you how it works let me take it nice and slowly so we're told that f ofx = x cub and to work out the derivative we would do the limb as H tends to zero of f ofx + hus F ofx over H yeah no you cannot differentiate it normally because it's Pro from first principles so this is the definition of a a derivative okay you have to do it the examiner will laugh at you yeah so first things first I want to write out what f of x plus h is that would be well think about it it's just like a graph transformation right I'm replacing x with X Plus H I'm going to show you the common pitfalls that I see in this question okay so the first one is well let's write out what it's asking us to do so limb as H TS to zero that is going to be x + H cubed minus X cubed over H the thing that I see some students do which it's crazy is say Well when H tends to zero and they get rid of all of the hes what what do you mean don't the limit is the very last thing you do because if I T this to zero technically this could be an infinite thing right because I'm dividing by zero but the problem is is functions reach zero at different speeds so if I make h10 towards zero it will not reach zero as fast as H cubed H cubed reach there reaches there faster so you can't do it that way so what you're going to do some algebra in fact it's mostly algebra here's what you do so I hate having to write this out each time but I don't want to confuse you expand the brackets simplify see what happens so let's do X+ H X+ H and X Plus H I can't remember what this is off the top of my head so we get to do this fully what I'm going to do now I'm going to do something a bit cheeky I'm not going to write the limit just for a bit because otherwise I'm going to get carpal tunnel so that is going to be x^2 + 2 xh or 2 HX either way around plus h^ 2 - x cubed H expand that again we get x * x x Cub plus 2 x^2 h plus h^2 X you know what I should write this the other way around really so technically variables should go at the end so technically I should write it like this I'm being a bit picky here not going to lie but I'm just feeling in that mood you know technically constants come first let's do that uh where was I okay and then plus h x^ 2 plus 2 h^ 2 X+ H cubed - x cub and then you're simplifying so X cubed and X cubed go away which is nice let's do the HX s first so that's 2 hx^ 2+ hx2 so that is 3 H x^2 so I can tick those two off then I go to the X's I have h^2 x + 2 H2 X so that's 3 h^ 2 x and then plus h cubed all over H divide through by H so again the H goes here here here here and you get 3x^2 + 3 HX + h s okay now now let's bring back the limit let's look at the limit one more time we have limb H TS to zero this now that is not a fraction which means if I make h z nothing naughty happens right nothing bad happens here we might get infinite that's scary but here if I make H 10 towards zero well this ter will go away and this term will go away so what am I left with I'm left with 3x^2 which is what the question asked me to show you need to be able to do this for any they can throw at you so they could have changed this to be x ^ 4 2 x^2 3x cubed you need to be able to do it so follow along the steps the limit is the very last thing if you do the limit and get the answer then you're good if you do the limit and you get something weird you're not good God my voice how's that so far difficulty to eh you have to be able to do this with s and cosine soon as well you don't need a final statement if you write this so if you write T to zero TS to zero you can also write like QED at the end which is Latin for as stated in the question but yeah you just need to to show show that the limit's going to zero so we've got a couple more stream questions um they want to help with part C specifically but I might do the whole thing CU it's logarithms so let's go so I can zoom out little bit cool logarithms uh sucks normally so oh I see okay anyways anyway sorry I got distracted by something here is another application question for exponentials now once you've done the exponentials chapter you'll find that exponentials and logarithms are the same thing in the sense that if you have a question about exponentials you're going to have logarithms if you have a question about logarithms you're going to have exponentials okay they're Inseparable so the first few parts and I just want to point this out this is a very typical question structure get that form in Part F Part F anyways okay oh part c yeah no worries we're going to have a look at it so x equal to D e^ 0.2 T where t x is amount of blah blah blah yada yada yada first of 15 Mig is given so that means D is 15 use the model to find how much is in there cool this is easy X is the amount in the bloodstream 15 is our initial dosage to e^ minus 0.2 and it's 4 hours God e to the.2 4 that should be good 6 point and it says to three decimal places so 6.74 okay 6.74 perfect two marks bad a second dose given 5 hours after the first dose using the same model for the second dose show that the total amount is 13.75 4 well what we're going to do is for the first dose X = to 15 eus 0.2 and that's 7 hours after right cuz look 2 hours after the first dose well sorry after the second dose would be 5 hours after the first dose plus another two hours right because it's 7 hours after the initial one so put in seven equals once again let's just put this in seven boom um it says to three decimal places let's just do this to three decimal places 3699 and then the second dose well that's only two hours after the second dose so we're going to sub in 15 e to- 0.2 I realized I probably shouldn't have rounded just yet but oops mistakes happen probably should have kept a bit more exact 10.55 so in total you would add them together which can you see would give you approximately 13.75 4 so if any of you want to become doctors that's the type of stuff you get to do although to be honest the computer probably does it for you so don't worry too much about [Music] it okay so this one's going to be interesting no more doses are given at time T hours after the second dose the total amount of the anti iotic in the bloodstream is 7.5 mg so let's write this out X actually no hang on okay does it make sense that I could have written this all on one line I could have written this as 15 e to .27 plus 15 e to 0.22 does that make sense you could have written that all on one line same thing right so let's do the same thing our total in the bloodstream is 7.5 and that equals 15 e- 0.2 now the time is where things are going to be a bit tricky because it says T hours after the second dose so that means the first dose was t + 5 hours ago plus 15 eus 0.2 T I hope that makes sense hopefully what's a good way to revise year 12 I can't even see it there's emojis in the way year 12 maths uh I mean do what we're doing right now right go through as many hard questions as you can the harder the better yeah you need to go through more questions more actual difficult questions don't stick with the easy stuff your brain should hurt after you revise okay so there is a bunch of different ways we could do this and I'm just thinking about what the most efficient way would be so I could take the log CU it says then and it has over e that's one thing I could do but I'm going to start factorizing just a little bit because I believe if I factorize my life is going to be a little bit easier so if I factorize out first of all I'm going to divide both sides by 15 that and the next thing I'm going to do is I'm going to expand this bracket as well um 7.5 over 15 is just a half right that's easy e and there's a reason why I'm doing this I promise I'm not just trying to kill time oops there shouldn't be a 15 here my apologies e to .2 T like that now if you remember your rules of indices you add powers when you when you multiply or if you want to say vice versa when you're subtracting you're dividing so in other words I could write this as e to .2t divided by E factorize out the e to the negative blah blah blah blah okay move that over to the other side which is going to look hideous but is what it is it will clean itself up soon to take the natural log of both sides should we clean this up first ah okay I see what they've done okay never mind we won't need to clean it up so Ln one over two they've done something very clever which is um good equals .2t can you see we're getting closer and closer to this little thing here hey deviation so I just want to show you guys something I going to write it over here just CU it it is related but it's not directly related 1 e + 1 I could put that all um I could write that as uh how do I want to write this to negative one that's what I want to do and that is going to give me the form right so and was there a number in front of here there is it's the same number two oh we've done it we've done it okay so we have two one again I'm ignoring the left hand right hand side for now because if I expand this bracket now watch what's going to happen in fact I'm going to write it slightly differently as well because I've seen what's going to happen we put two first plus if you look at this inside the bracket B plus b over e b is two using the power L we're going to bring the minus the front divide both sides by 0.2 so the negatives will cancel that brings me to 5 Ln 2 + 2 e = t so uh a was 5 and B is 2 they're both integers and B is the same in both cases cow Bunger hopefully that was okay there's someone called Nosh send that in um I'm not someone did say so no I'm assuming that was you hope that makes sense um oh sorry Dylan yes I see your question how did I miss that okay you wanted me to do was it question 16 did you say is it question six yeah okay question 16 let's bring that over it's kind of hard to see I'm not going to lie but I'll do my best God I might have to draw this out again okay good now okay let's uh do this one so the diagram shows a sector um of a circle so that's o a b where's a oh a is over here okay oh jeez okay I need to thin that up a there blah blah blah blah blah blah okay line AC is a tangent which means that is going to be 90° and OBC is a straight line OBC oh yeah OBC is a straight line that's fine the region R is Bound By The Arc Bound By The Arc ab and the lines AC and yeah okay that's God that's so much information all right find the area of R okay so easiest way to do that is find the area of the triangle here and then um subtract the area of the sector that's what I would do um don't think we have do they give me all the information so R is 12 is C also on the point of sector no okay so and DB it doesn't say anything okay all right so this is a right angle triangle and we want to find the area um so we want to do base time height so base is this height is this radians is just a unit for angles it's really not that bad so if you have a look that's the opposite and the adjacent side so we can whoa okay that was careful we're trying to work out X right because we need the base and the height because opposite the right angle is the hypotenuse which is OC so X would be so well let's just do it tan 1.2 equals opposite which is X over adjacent so X is equal to 12 tan 1.2 again even if you don't know what radians are it's a good question to look at the only difference is that if you notice there is a d on my calculator can you see that that means I'm in degrees mode I need to go whoops into radians mode which is angle unit radian and I do the same thing it's no different boom 30.9 that seems quite big all right that seems very big doesn't it all right I mean is what it is okay so the area of the triangle O A so area o a c is just a half base which is 12 * 30.9 let's do that so answer * 12 * 0.5 gives me 18519 and then wait this is eight marks oh what on Earth just happened all right hold on oh okay this is eight marks that's really generous okay so as you can see I've got the error of the big triangle yeah and I'm just going to subtract just a random YouTuber great stuff thank you very much I'm just going to subtract the sector here okay I'm just going to subtract it away have a second just to make sure that makes sense uh I'm just responding to something yeah make sure that makes sense to you and all we need to do now is sub it into the formula now I'm not sure if you know this but area of a sector normally is angle over 360k R 2 okay usually but when we use radians this bit just becomes Theta so it becomes quite short which is nice so so instead we just have a half R 2 Theta sorry I should clarify all of this becomes um a half Theta basically so the area of OB is just a half R 2 * Theta which was 1.2 so 0.5 * 12 s * 1.2 gives me 86.4 so all you need to do is subtract them 18519 minus 86.4 gives me 98.79% units are cm squared eight marks bad God anyways um wait I just want to tip this one off um where is it where am I looking that is the sh area of the Shaded region so the last B is find the perimeter of Da okay ah okay so Part B quickly go through this so da is how do I want to write this how do I want to write this CU things now I can't see the image properly um I'm going to rub some of this stuff out is going to be this Arc Length here okay and arc length is just R Theta so I'll call it ba = 12 multiplied by 1.2 which I'm going to use a calculator for just to be completely honest is 14.4 good we know the first one and to Da a so the other things we we need is we need this point here and this point here now I'm looking that is all the way up to B and it's kind of annoying I can't see properly okay so we have a smaller triangle and a bigger one now because they're parallel I'm not sure if there's more information but if they're parallel then we can use scale factors to work them out but I'm not sure a BC okay I me unless they want you to yeah so that is all the information they give you so yeah from there all you need to do is so we've worked out ba if you're working out da which as you can see would be there to there to there we've worked out the curved area all you need to do now is work out these two sides here using scale factors or trigonometry um just going to check one thing quickly okay okay that was a hell of a lot I hope that makes a bit more sense um yeah we got through quite a bit and I've checked them all off so again don't forget you can in the Discord on the stream question chat you can just post any questions that you want me to do and I'll do them in the kind of following weeks um next week there won't be a stream because I have a dentist appointment and sadly the only time they had was 6 p.m. which sucks but yeah I most likely won't be here for this which is a shame but that is just the way it is um yeah so anyways I hope you guys enjoyed uh tree if you submit it on Discord that's the best way to do it because email gets too messy I missed some of the emails yeah hope you guys enjoyed and see you guys not next week the following week when I'm finished with my medical stuff