HARDEST A Level Maths Tricky Question #16

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[@hound819]

im cooked

[@lol.1296]

CHILL BROSKI

[@Bob-zo8gk]

its over for us bro

[@hound819]

Walking into hall rn, will tell how cooked I get 👦🏻👦🏽👦🏾👦🏿 or 🧑🏽‍🍳

[@lol.1296]

how was it guys

[@Thorfinn_editZ1]

@@hound819how was it

[@MintyMit]

For part b I used the discriminant = 0 for the quadratic (well quartic, hidden quadratic) to get 10/3

[@MintyMit]

Oh nvm u showed it

[@yasminh]

ty!

[@2trilltochill]

For part b you can achieve the answer by differentiating both equations and equating them so that you get -2x=18x^2-9x^3 solving gives you correct x and correct y which you can use Pythagoras for to find radius

[@MorgKev]

r^2 = x^2 + y^2 ... rewrite y^2 in terms of x^2... then get d(r^2)/dx and set to zero to find x (at max r^2). Then y^2 follows by substitution... and hence r^2.

[@Lexler34]

Great explanation, i was able to part a) but part b) i forgot about doing Rmax as a derivitive ❤

[@kausarlolz]

hi sir! i was able to do part a very easily using the trig trianlge we often do in mechanics questions (slope inclined at sinx = 4/5 etc) i set sint = x/2. made a trig triangle and subbed in cost into y = 3(2sintcost) and squared both sides :)

[@MrAstburyMaths]

Great job! I’m now a massive fan of trig triangles for parametrics!

[@AITunesUK]

at first sight, I thought this was easy enough and just found the stationary point and got distance to that which is sqrt(11) and thought it was easy enough lol, because it looks like the circle just touches the maxima on the diagram. I could not have been more wrong, this is slightly trickier than I anticipated.

[@adailyfact]

I did the exact same thing.

[@AG-ql1sy]

Any chance jack k is heading to oxford or Cambridge?😂

[@charliesmith2411]

good luck

[@14bobyroby]

You're the MATHS BOSS!

[@MrAstburyMaths]

You’re so kind! 😄

[@sulamakamyon12345]

when you do b squared- 4ac,=0 doesnt that only refer to equations where its tangent ( one intersection point), however for the circle there are multiple intersections.

[@sreenathbondilli8402]

you are findin the r value that has no solutions, which is anything greater than 10/3 if less than 10/3 you would get 4 solutions, the equation does apply as we are saying that there are no solutions when b2 -4ac

[@RingsideReels]

I did it throught modelling it as a quadratic, knowing it as 1 solution and got 10/3

[@user-bd2pk5de9c]

Hi Sir, For the 2023 maths papers, should we expect 16+ questions. Over the years Edexcel have put more and more questions in the papers., so I’m a little anxious about timing

[@stan4562]

7:50 Why is r maximised at dr/dx = 0 ? Thanks!

[@onaopemipoadeniyi]

Hi Sir, why is dr/dx = 0 when for max r at 7:39?

[@lookuptartaria7030]

I prefer the second solution

[@uvwv]

Very interesting question… dare I say even enjoyable lol

[@MrAstburyMaths]

Thanks!

[@rheajohri2706]

Hello, thank you so much for this video. I also fell into the trap of parametric differentiation in part b. I wanted to ask at 11:15 why the discriminant = 0 is used? Wouldn't it be discriminant > 0 as there are 4 solutions (not one)?

[@user-bd2pk5de9c]

I have the same question. Also why wouldn’t parametric diff be valid. Can that not be inferred from the diagram?

[@ainharamk550]

I think is because in the quadratic equation he uses ‘u’ which equals “x^2” so essentially one solution of “u” equals 2 solutions of “x”, also if the circle was smaller there’ll be 8 solutions, and ig that’s when the discriminant will be >0, So I guess 4 intersections is the same as 1 and 8 will be the same as 2, as the discriminant is normally to do with tangents to a curve, and here the circle is symmetrical in all the axis and C1 also looks symmetrical in all axis

[@billyunited1191]

As a year 13 student I got part a correct and part b as root 11 so correct to 2 sig figs lol. Very tricky

[@infintysolar1539]

Yes, this is what I got, I think we both did Pythagoras on the max point, on desmos the point is actually a little below i think, the circle is not rlly correct, I was shattered when I got it wrong tho

[@Ace__Matic]

same

[@adrianhall306]

I was also thinking of using the discriminate however why did you have to let u=x^2 to do this does the discriminate only work on highest power 2?

[@adrianhall306]

Never mind I answered my question makes sense😂

[@BBK583]

which paper is this from?? great video btw

[@user-bl2vl2bq7h]

Mock set 2 paper 1 pure I believe

[@UNKNOWN-jk6wv]

which year pls? @@user-bl2vl2bq7h

[@user-bd2pk5de9c]

Hi Sir, wouldn’t you make the discriminant greater than 0 since C1 and C2 are intersecting multiple times ?

[@sentheaS]

Old comment I know but no you wouldn't . The quadratic you are setting equal to zero is not the quadratic that draws the graph in the diagram (there isn't a quadratic that draws a graph in the diagram, it has to be defined parametrically). If you go ahead and look at the graph of the quadratic with r substituted into it, you will see that there is one (repeated) real solution. Another way you can understand it is the following (I'm not sure if this is necesarily why or if I'm just guilty of confirmation bias), if you graph the cartesian y = 3x(1-(x/2)^2)^(1/2), you will get one half of the curve C1. You get the parts of C1 occupying the bottom left and top right quadrants. When graphing the cartesian form of a circle (C2), you will only ever get one half of the circle, otherwise you wouldn't be graphing a function (functions can not have one-to-many mappings). As such, you cannot graph the cartesian form of C1 & C2 such that they have more than one intersection. cartesian C1 will always occupy two diagonally opposed quadrants, whether you graph C1 or -C1. C2 will always occupy either the top two quadrants or the bottom two. Just a note as to how I got that function for y: I did the first part differently, I got to y = 3xcost, then to get cost in terms of x, I did x/2 = sint, (x/2)^2 = (sint)^2, (x/2)^2 = 1 - (cost)^2, (cost)^2 = 1 - (x/2)^2, cost = sqrt(1-(x/2)^2), then plugged it into y = 3xcost. As I square rooted (cost)^2, it's actually y= +/- 3x(1-(x/2)^2)^(1/2). If you just square root the funtion y^2 from part a, then you will end up with a graph for modulus y, destroying this information.

[@gaurighosalkar4294]

Sir, which year paper is this?

[@2cutzaky055]

It’s a Mock Set

[@ramens]

GOAT

[@highimdamon7101]

p Goat

[@yourmum2839]

Dear Mr Astbury, I am writing to apologize for my behavior during our recent conversation about the maths question. I realize that I was wrong to backbite against you and that my ego got in the way of the truth. I should have listened to your explanation more carefully and acknowledged that you were right about the answer. I want you to know that I deeply regret my actions and I understand that they were hurtful and disrespectful. I value our friendship and I hope that we can move past this incident and continue to work together in a positive and supportive manner. Please accept my sincere apologies for any hurt or harm that I may have caused you. I promise to be more mindful of my words and actions in the future, and to work towards building a stronger and more respectful relationship with you. Thank you for taking the time to read this comment. I look forward to hearing from you soon. Sincerely, A year 13 A-level student

[@JAM-wn5uc]

bro what

[@RobloxGuardian]

💀

[@ClintonNjoku]

Chat GPT

[@ahmedmustafa7753]

Why can you not stake dy/dx for part 2 and find out the value of t for when the grad is 0

[@nabil1x]

do you mean to diff. parametrically then set dy/dx=0? the points of intersection btwn C1 and C2 do not occur at a flat gradient

[@user-bd2pk5de9c]

I took that approach as well. Can’t it be inferred from the diagram that the points if intersection are the maximum points of C1?

[@user-bd2pk5de9c]

I understand that its not stated explicitly, but allot of maths is inferring from diagrams/wording

[@UNKNOWN-jk6wv]

@@user-bd2pk5de9c it's never just inferring there's always something that tells you explicitly