I've thought the following: There are infinite numbers c such that y = x² + c and x = y² find each other, but only one where it happens just one time. And when does it happen? For maximum c. So, if we find the maximum value of c, the question is solved. Then: y = x² + c y = (y²)² + c y = y^4 + c c = y - y^4 Maximum c: d/dy (y - y^4) = 0 1 - 4y³ = 0 y = 1/cbrt(4) x = y² = 1/cbrt(16) y = x² + c c = y - x² c = 1/cbrt(4) - 1/cbrt(256) c = 1/cbrt(4) - 1/(4cbrt(4)) c = 3/4(cbrt(4))
i combined them into y⁴ - y + c = 0, then reasoned as follows: i know that when c = 0 there are exactly two real intersections - (0, 0) and (1, 1) - and thus two distinct real roots to this combined equation. so if i increase c until there is exactly one distinct real root, it must be a repeated root (visually, the other two roots would merge together), so the derivative must be zero. differentiating eliminates c and leaves me with 4y³ - 1 = 0, which readily produces y = 1/³√4, and substituting back in gives c = 3/(4³√4).
I love that he's still uploading older videos. Years ago when this was recorded, I didn't know how to do the problem on my own. It's so nice now that I'm able to pause the video and give it a try and actually get it :D
I'm relearning Maths before embarking on an Engineering degree (I never made it past an A* at GCSE level) and something I underappreciated previously was how there is ALWAYS another rabbit hole to venture down. It has to be one of (if not 'THE') broadest subject out there. It's absolutely fascinating. Love the videos, they're many orders of magnitude more educational than any teacher I ever had :)
If they are tangential, they will have the same slope at the same (x,y) coordinate. Set the first derivatives of y=sqrt(x) (1/2 * x^(-1/2)) and y=x^2 + C (2x) equal to each other. This yields x = cubrt(4)/4. Plug this back into y=sqrt(x). Y=sqrt(cubrt(4)/4) = cubrt(2)/2. Plug these x and y values into y=x^2+C, and you get C=3/8 * cubrt(2).
Thank you for this nice video ! Considering the tangents is a bit tricky, however : you can SEE that it's correct, but in general two curves can have only one point in common without being tangent (e.g. two straight lines...). Thus it seems a better solution to me, to simply say that the graphs of the two functions f(x)=x^2+c and g(x)=sqrt(x) intersect only once iff the function h=f-g is zero at only one point. Considering the sign of its derivative, we get that it decreases until x=1/(third Root of 4) and increases after that. Thus h(x) takes the value zero only once iff h(1/(third Root of 4))=0, an equation which gives directly the value of c. Of course we still use the visual evidence that y is non negative, but this is less important. It can be proved as well, anyway, by the same kind of intermediate value argument. Indeed, replacing h by f+g above, we get that if c
We know that in the relevant portion, one curve is concave up while the other is concave down. If they touch at only one point, it must be tangentially.
When I was starting this, I got this step: For what values of y does y⁴ - y + c = 0 have only 1 solution? And that problem is obviously kind of hard, so it was a dead end, and then I figured out bprp's way. BUT! Suppose the original problem were in fact: For what values of y does y⁴ - y + c = 0 have only 1 solution? How would you know to turn that problem into the one given in the video?
I actually did it that way. Renaming of the variable: x^4-x+c=0 must have exactly 1 solution. Therefore the derivative on the solution must be 0 (because the polynom is of an even degree). 4x^3-1=0 and therefore x=cbrt(1/4). Back to the equation: (cbrt(1/4))^4-cbrt(1/4)+c=0, and therefore c=cbrt(1/4)-(1/4)*cbrt(1/4)=3/(4*cbrt(4))
You can also set y=x for clarity and write x^4=x-c has 1 solution. Consider then the line y =x -c only intersecting x^4 once. To solve for c we simply find when the gradient of x^4 = 1 (our value of y) and find the equation of the tangent line, which will find us our c
I saw the thumbnail and was stuck with the question since there were two unknowns (x and c) if you only considered the formulas themselves. This made me watch the video and once you mentioned the word derivative, I got all the parts I needed (forgot to consider the fact that they were tangent, so it obviously has to do with derivatives, I just wasn't considering them since I haven't done much math in the last couple of years). The thumbnail already primed me to disregard the negative sqrt(x), and the derivative wasn't that hard.
Saw another video a few days ago with a problem like this. I tried to start solving for the "common" x-coordinate, i.e. an equation of the two functions. Well, I failed! Keyword was that they had common _tangent_ at that point whose x-coordinate could be found by solving an equation of the two derivatives of the functions. Nice👍🏻
Did you know that this is useful in doubling the cube? You can use a compass to take the reciprocal of the required constant and half it to get ∛(2), allowing you to construct a cube with a volume of *exactly* 2.
Literally just finished year 10 and this got recommended to me so I thought I’d have a look. When I tell you I was clueless I mean I stared mindlessly into the screen with not the vaguest idea of what he was talking about.
I solved it like this as well. It was very rewarding to get the correct answer, but I’m sure the quartic formula wouldn’t be provided on the actual exam.
Yeah I was like "Tangent means two things, same point and same slope." Not sure why, but I was like "this is calculus, we're definitely gonna do derivatives." So I set those equal. Plug that x into sqrt(x) and you have the y value for y=x^2 + c so c= sqrt(x) - x^2. Two assumptions and 2 formulas leading to two equations. I didn't think I'd miss calc, but the calc 3 days were fun like this. Diff eq did not go the same way.
Differentiate with respect to x or y, one might be trickier but not by much here. You get both the first derivative and the functions being equal at the tangent point:p. Just pick one to solve for, its not that hard
The way I solved was implicitly differentiate both to get 1/2y=dy/dx and 2x=dy/dx. 1/2y=2x. But because you know they are the same point you can substitute in terms x for y^2. Then 1/2y=2y^2 1/2=2y^3 1/4=y^3 cbrt(1/4)=y Now you also know what x is because what is x at that point but y^2 so x=(1/4)^2/3 Finally solve for c with those values of x and y cbrt(1/4)=((1/4)^2/3)^2 +c 1/cbrt(4)-(1/4)^4/3=c 4/4cbrt(4)-1/4^4/3=c 4/4cbrt(4)-1/4cbrt(4)=c 3/4cbrt(4)=c Not too hard a problem compared to some of those Cambridge integral ones lol
I thought you were going to do it by: y^2 = x => y = +√x plug that into y = x^2 + c put in quadratic formula solve c so that the determinant is zero, i.e. there is only one solution
@@hydrarl3869 I think it's justified here. It's one thing to not pass the application exam. It's another thing to not even know _how to look up_ how to take the application exam. One of these situations is much more "noobish" than the other.
No, when you set the function equal, you get the intersection as a function of c. But that was not the question. If you don't understand my point: Calculate c in the equation x^2+c=sqrt(x) ... you can't, because it's one function for two variables. If you think, I'm wrong: please show, how you calculate c without differentiation.
It's the easiest question in the Egyptian high school system 🥲🥲🥲 this year is going to determine whether i'm able to study engineering or i don't have enough intelligence to study engineering or other scientific degrees It's a really hard year for me.
this kind of logic puzzle constrained by arcane bizarre conventions is such a dry sterile convoluted labyrinthine mess that it is frightening to imagine devoting a lifetime to the decipherment of every kind of twisted string of lines and symbols that pops out of some paid professional clever person's frond. what possible motivation is there that drives the obsessed to ever more bizarre questions leading to ever more linguistically incomprehensible strings of lines and squiggles and digits which is then called 'the answer' ? math seems to be its own reward/punishment and to pursue these endless pointless aimless random re-iterations of the same obscure basic multi-embroidered logic puzzles seems, for some people, to be almost an opioid
I've thought the following: There are infinite numbers c such that y = x² + c and x = y² find each other, but only one where it happens just one time. And when does it happen? For maximum c. So, if we find the maximum value of c, the question is solved. Then:
y = x² + c
y = (y²)² + c
y = y^4 + c
c = y - y^4
Maximum c:
d/dy (y - y^4) = 0
1 - 4y³ = 0
y = 1/cbrt(4)
x = y² = 1/cbrt(16)
y = x² + c
c = y - x²
c = 1/cbrt(4) - 1/cbrt(256)
c = 1/cbrt(4) - 1/(4cbrt(4))
c = 3/4(cbrt(4))
Nice approach 👍
thats actually genius
Thats very neat but im not it would work all the time and you have to deal with bigger numbers
I like it
@@lih3391 why shouldn't this method work? And what do you mean with bigger numbers?
As a 17 year old boy from India studying for engineering entrance exams. I'm happy to say this video made complete sense to me.
Same here! From Austria, but still gives me confidence.
As a 15 year old boy, this video also made complete sense to me
As a 7 year old boy i cam absolutely say this video made complete sense to me
As a newly born Asian child it makes complete sense to me
As a child going to take birth in an asian family , this video makes complete sense to me
i combined them into y⁴ - y + c = 0, then reasoned as follows: i know that when c = 0 there are exactly two real intersections - (0, 0) and (1, 1) - and thus two distinct real roots to this combined equation. so if i increase c until there is exactly one distinct real root, it must be a repeated root (visually, the other two roots would merge together), so the derivative must be zero. differentiating eliminates c and leaves me with 4y³ - 1 = 0, which readily produces y = 1/³√4, and substituting back in gives c = 3/(4³√4).
I love that he's still uploading older videos.
Years ago when this was recorded, I didn't know how to do the problem on my own. It's so nice now that I'm able to pause the video and give it a try and actually get it :D
I’m quite proud of myself. I saw the thumbnail and instantly said “just set the derivatives equal!”
I'm relearning Maths before embarking on an Engineering degree (I never made it past an A* at GCSE level) and something I underappreciated previously was how there is ALWAYS another rabbit hole to venture down. It has to be one of (if not 'THE') broadest subject out there. It's absolutely fascinating. Love the videos, they're many orders of magnitude more educational than any teacher I ever had :)
Good luck for ur future engineering degree bro. I recommend other maths channels too, like 3B1B, MindYourDecisions, numberphile, and mathologer.
If they are tangential, they will have the same slope at the same (x,y) coordinate.
Set the first derivatives of y=sqrt(x) (1/2 * x^(-1/2)) and y=x^2 + C (2x) equal to each other. This yields x = cubrt(4)/4.
Plug this back into y=sqrt(x). Y=sqrt(cubrt(4)/4) = cubrt(2)/2.
Plug these x and y values into y=x^2+C, and you get C=3/8 * cubrt(2).
Thank you for this nice video ! Considering the tangents is a bit tricky, however : you can SEE that it's correct, but in general two curves can have only one point in common without being tangent (e.g. two straight lines...).
Thus it seems a better solution to me, to simply say that the graphs of the two functions f(x)=x^2+c and g(x)=sqrt(x) intersect only once iff the function h=f-g is zero at only one point. Considering the sign of its derivative, we get that it decreases until x=1/(third Root of 4) and increases after that. Thus h(x) takes the value zero only once iff h(1/(third Root of 4))=0, an equation which gives directly the value of c.
Of course we still use the visual evidence that y is non negative, but this is less important. It can be proved as well, anyway, by the same kind of intermediate value argument.
Indeed, replacing h by f+g above, we get that if c
We know that in the relevant portion, one curve is concave up while the other is concave down. If they touch at only one point, it must be tangentially.
When I was starting this, I got this step:
For what values of y does y⁴ - y + c = 0 have only 1 solution?
And that problem is obviously kind of hard, so it was a dead end, and then I figured out bprp's way.
BUT!
Suppose the original problem were in fact: For what values of y does y⁴ - y + c = 0 have only 1 solution?
How would you know to turn that problem into the one given in the video?
I actually did it that way. Renaming of the variable: x^4-x+c=0 must have exactly 1 solution. Therefore the derivative on the solution must be 0 (because the polynom is of an even degree). 4x^3-1=0 and therefore x=cbrt(1/4). Back to the equation: (cbrt(1/4))^4-cbrt(1/4)+c=0, and therefore c=cbrt(1/4)-(1/4)*cbrt(1/4)=3/(4*cbrt(4))
@@Zarunias nice solution! I hadn't thought of the derivative yet
I was about to comment something similar to this, I got this equation and then tried to solve for a repeated root
You can also set y=x for clarity and write x^4=x-c has 1 solution. Consider then the line y =x -c only intersecting x^4 once. To solve for c we simply find when the gradient of x^4 = 1 (our value of y) and find the equation of the tangent line, which will find us our c
I saw the thumbnail and was stuck with the question since there were two unknowns (x and c) if you only considered the formulas themselves. This made me watch the video and once you mentioned the word derivative, I got all the parts I needed (forgot to consider the fact that they were tangent, so it obviously has to do with derivatives, I just wasn't considering them since I haven't done much math in the last couple of years). The thumbnail already primed me to disregard the negative sqrt(x), and the derivative wasn't that hard.
I've done the most crazy solution, when I watched the vídeo to see the real solution I discovered that I was right!!!
That feeling is good!!!!!
We can also differentiate like this: dx=(2y)dx, so dy/dx=1/(2y) and we want dy/dx=2x namely 1/(2y)=2x etc.
Saw another video a few days ago with a problem like this. I tried to start solving for the "common" x-coordinate, i.e. an equation of the two functions. Well, I failed!
Keyword was that they had common _tangent_ at that point whose x-coordinate could be found by solving an equation of the two derivatives of the functions.
Nice👍🏻
Did you know that this is useful in doubling the cube?
You can use a compass to take the reciprocal of the required constant
and half it to get ∛(2), allowing you to construct a cube with a volume of *exactly* 2.
Literally just finished year 10 and this got recommended to me so I thought I’d have a look. When I tell you I was clueless I mean I stared mindlessly into the screen with not the vaguest idea of what he was talking about.
Lol I saw this problem on instagram like a week ago and managed to get it right
discriminant of y^4-y+c equals 0
256c^3-27=0
c^3=27/256
I solved it like this as well.
It was very rewarding to get the correct answer, but I’m sure the quartic formula wouldn’t be provided on the actual exam.
Yeah I was like "Tangent means two things, same point and same slope." Not sure why, but I was like "this is calculus, we're definitely gonna do derivatives." So I set those equal. Plug that x into sqrt(x) and you have the y value for y=x^2 + c so c= sqrt(x) - x^2. Two assumptions and 2 formulas leading to two equations. I didn't think I'd miss calc, but the calc 3 days were fun like this. Diff eq did not go the same way.
A similar intution to equate the equations AND the derivatives leads to Lagrange Multipliers for constrained optimization problems!
my process:
y^2=x
y=x^2+c
y=y^4+c
d/dy y=1
d/dy y^4=4y^3
tangent at:
1=4y^3
y=1/cbrt(4)
solve for c:
1/cbrt(4)=1/cbrt(4)^4+c
1/cbrt(4)=1/cbrt(4)*1/cbrt(4)^3+c
1/cbrt(4)=1/cbrt(4)*1/4+c
3/4*cbrt(4)=c
Why's this suspiciously easy?
It is a MCQ question and I think the difficulty is about right.
So simple 😊
Y=1, X=1 and C=0
Noooo😂
Lol
😂😂😂
Differentiate with respect to x or y, one might be trickier but not by much here. You get both the first derivative and the functions being equal at the tangent point:p. Just pick one to solve for, its not that hard
The way I solved was implicitly differentiate both to get 1/2y=dy/dx and 2x=dy/dx.
1/2y=2x. But because you know they are the same point you can substitute in terms x for y^2. Then
1/2y=2y^2
1/2=2y^3
1/4=y^3
cbrt(1/4)=y
Now you also know what x is because what is x at that point but y^2 so
x=(1/4)^2/3
Finally solve for c with those values of x and y
cbrt(1/4)=((1/4)^2/3)^2 +c
1/cbrt(4)-(1/4)^4/3=c
4/4cbrt(4)-1/4^4/3=c
4/4cbrt(4)-1/4cbrt(4)=c
3/4cbrt(4)=c
Not too hard a problem compared to some of those Cambridge integral ones lol
Four times one is four, minus one that's three. Quick mafs
took the admissions test and got rejected today :(
noob
Never give up! You will fail your way to success!
@@FreshBeatles That's an uncool thing to say.
@@General12th But accurate
Would you have to rationalise the denominator as you are left with a root there
I thought you were going to do it by:
y^2 = x => y = +√x
plug that into y = x^2 + c
put in quadratic formula
solve c so that the determinant is zero, i.e. there is only one solution
i haven’t been taught anything like this yet but i understood the process! kinda… yeah I’m not going to Oxford
The derivate is because are tangente?
Which mic are you using?
Guess my old brain won't be going to Oxford anytime soon
How to send hard Calculus questions to you?
glad to know im basically an oxford student
Why does “being tangent” imply “same tangent line”?
Tangent means “touch”.
@@bprpcalculusbasics I’m saying why does (the parabolas being tangent at a point) imply that the tangent lines have the same slope?
A surd as a denominator??
Why c cannot be negative?
ax²+bx+c could be
x²+2x+(-8) for example..
if c < 0 then there will be 2 points of intersection. (why?)
in fact we can let m < c; there will be 2 points of intersection.
He has a habit of using ln in every question
Yes I did it!
Not hard as the exam
I solved!!
congrats!!!
It took me some 2 mins to solve mentally. Just solve 2 equations→
1/2√x = 2x and √x = x²+C. Easy deal.
Who else wouldn't get in
As a 15 year old knowing I’m gonna have to know this for my final exams hurts me so much 😭
Sir please tell me that how can I take admission in Oxford University
very difficult, you sure you can do it?
you have to apply on UCAS by October I think, you could try next year.
You have no chance of going to Oxford if you don't know how to look up the admissions process.
@@General12th
J.J. Shank
1 day ago
@light That's an uncool thing to say.
@@hydrarl3869 I think it's justified here. It's one thing to not pass the application exam. It's another thing to not even know _how to look up_ how to take the application exam.
One of these situations is much more "noobish" than the other.
Much simpler way
You forgot to take the +/- when solving y^2 = x for y ;)
He's considering x > 0 y>0 ; the parabolas intersect in 1st quadrant.
There was no need to use calculus man. Set the functions x^2 + c and sqrt(x) equal just as you eventually arrived to.
No, when you set the function equal, you get the intersection as a function of c. But that was not the question.
If you don't understand my point: Calculate c in the equation x^2+c=sqrt(x) ... you can't, because it's one function for two variables.
If you think, I'm wrong: please show, how you calculate c without differentiation.
عدت صغيرا
Oh the "beard" is gone. Now I can watch again.
But aren't they intersecting in the z-world? 🤔🤔
It's the easiest question in the Egyptian high school system 🥲🥲🥲 this year is going to determine whether i'm able to study engineering or i don't have enough intelligence to study engineering or other scientific degrees It's a really hard year for me.
I seen your every video
Why YOU have shaved?
I love men
This is high school math for us Indians. Yes its not fair 😂
this kind of logic puzzle constrained by arcane bizarre conventions is such a dry sterile convoluted labyrinthine mess that it is frightening to imagine devoting a lifetime to the decipherment of every kind of twisted string of lines and symbols that pops out of some paid professional clever person's frond.
what possible motivation is there that drives the obsessed to ever more bizarre questions leading to ever more linguistically incomprehensible strings of lines and squiggles and digits which is then called 'the answer' ?
math seems to be its own reward/punishment and to pursue these endless pointless aimless random re-iterations of the same obscure basic multi-embroidered logic puzzles seems, for some people, to be almost an opioid