How big can y be?
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3 жыл бұрынWe look at a nice geometry problem
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@MichaelPennMath 3 жыл бұрын
Hi everyone! I made a google form to suggest problems. It will be linked in every video in the description from now on.
@sohamchakote4174 3 жыл бұрын
Love your content... ❤️❤️❤️
@goodplacetostop2973 3 жыл бұрын
Sweet! Since my homeworks are less and less "hit" and more and more "miss", I’ll just go ahead and send what I’ve found there. Homeworks were great while they lasted but people moved on. It is what it is.
@goodplacetostop2973 3 жыл бұрын
Can we submit anything, even on topics like geometry, graph theory or combinatorics?
@MichaelPennMath 3 жыл бұрын
Anything is fair game!
@zeravam 3 жыл бұрын
@@goodplacetostop2973 I miss your homeworks (Miss as I want to see them again)
@sergiokorochinsky49 3 жыл бұрын
In my humble opinion, I think eliminating theta was a mistake. Having all the parameters as a function of the angle gives great insight on what is happening (yes, I'm a physicist), the equations are simpler, and the solutions are beautiful: y=1+(sin 2θ)/(1+sin θ) Optimum when (cos θ)^2=sin θ ymax=1+2/φ (φ-1)^(3/2) xmax=(φ-1)+(φ-1)^(1/2) where φ is the golden ratio.
@leif1075 3 жыл бұрын
You can solve this without calculus right?
@sergiokorochinsky49 3 жыл бұрын
@@leif1075, I doubt it. But I'm all ears... What do you propose?
@leif1075 3 жыл бұрын
@@sergiokorochinsky49 I'm thinking set up an equation with y as a function of theta as opposed to x and then plug in different values of theta until you get a maximum. I don't see why that wouldn't work?
@sergiokorochinsky49 3 жыл бұрын
@@leif1075 Oh, so you want to plot y(θ) and determine the maximum graphically. That method has a name: "brute force". It always works, it gives you a fairly good approximation with a reasonable amount of work, but you will never get the exact result. Why wouldn't you use derivatives?
@leif1075 3 жыл бұрын
@@sergiokorochinsky49 you could get the exact result sometimes. In this case i think you would.
@goodplacetostop2973 3 жыл бұрын
1:06 My body is telling me x=√2 but my mind is telling me it’s not that easy. Let’s see how wrong I am... 14:54 Good Place To Stop
@bwcbiz 3 жыл бұрын
Yeah, intuition can be misleading when it comes to rotating squares.
@skylardeslypere9909 3 жыл бұрын
Yep. Calculated the case for x=sqrt2 and got to y=3-sqrt2 but that's not correct lol
@casc3601 3 жыл бұрын
Why isn't that correct? X=√2 would give you the largest x+1 and the largest angle for the top line, no?
@freewilly1337 3 жыл бұрын
@@casc3601 The heighest angle isn't obvious at least. Let us call the heighest point of the square (a,b). If you tilt the square further such that the heighest point is closer to the left square you get a slightly smaller difference of the heights b-1, but also a slightly smaller distance (a-x) which you have to divide by. So it is not exactly clear. If the slope is maximal for x=sqrt(2) (which I would guess to without doing the math), it is the correct solution though. Noteably other comments have pointed out that the x value in the video is negative, hence doesn't solve the problem. Then x=sqrt(2) is actually the solution (that however doesn't mean the slope has to be maximal for that value of x for similar reasons!)
@casc3601 3 жыл бұрын
@@freewilly1337 thanks, I went back and what I got for the angle at the bottom of the tilted square was about 49.5 degrees, so slightly off. No idea where the negative x value came from
@paradoxicallyexcellent5138 3 жыл бұрын
Fun to see the golden ratio arising in an optimization problem (the cosecant of the tilt angle theta is the golden ratio here).
@supnava8320 3 жыл бұрын
selection of problems is getting really good. thanks for the awesome content man!
@TheProloe 3 жыл бұрын
I tried this without taking it in terms of x, and got the same result. Ended up just reducing the trigonometric equations down, which turns into just finding the max value of y = 1 + sin2θ/(cosθ + 1), not too bad at all.
@imacup5047 3 жыл бұрын
I think in the denominator it should be sinx+1
@TheProloe 3 жыл бұрын
@@imacup5047 Depends what you take θ to be, of course. I was taking it as the angle Michael marked as π/2 - θ, which of course would change things.
@imacup5047 3 жыл бұрын
@@TheProloe hmm
@mahdipourahmad3995 3 жыл бұрын
where exactly did ur solution divert from Michael's?
@TheProloe 3 жыл бұрын
@@mahdipourahmad3995 Only in the initial choice for theta and the choice to remain in terms of theta for the calculus step rather than convert to being in terms of x.
@ApresSavant 3 жыл бұрын
That was the long way for sure. I envisioned the rotated square as a cam and went right to the maximumization equation looking at the change in Y with the change of the base length X. X is bounded from 2 to 2+sqrt2. Basic engineering application of this math.
@paradoxicallyexcellent5138 3 жыл бұрын
I would call 7:57 a mistake, or at least further clarification is needed. A plus/minus sign is really needed here, and then a minus/plus sign should be used for cosine. It changes depending on whether 0
@petersievert6830 3 жыл бұрын
I agree. Others also pointed out that resulting value of x is negative, which makes no sense whatsoever given the construction and coordinates being used. So you found the bug, it seems to me.
@TJStellmach 3 жыл бұрын
But it's easy to see that theta
@petersievert6830 3 жыл бұрын
@@TJStellmach That's the point Tim. Michael choose the sign in a way, that theta is indeed not
@petersievert6830 3 жыл бұрын
Another observation: plugging the decimal value of Michaels x (-0.168) into his y formula actually does not yield the the expected value of roughly 1.6 and so his x plugged into y(x) actually does not result in the expression he gave as final solution. It does work though, if you switch 2x² -2 to 2-x² in the y-formula.
@TJStellmach 3 жыл бұрын
@@petersievert6830 Ah, yes. You are right. I thought you meant to check both cases, and my point was just that one of them is relatively easy to reject.
@daniello4038 3 жыл бұрын
y can be expressed as 1+sin(2 theta)/(1+siin(theta)), solve: sin(theta)=(root(5)-1)/2
@PATRICKZWIETERING 3 жыл бұрын
Correct! This gives optimal value for y = 1 + sqrt(-22 + 10 sqrt(5))
@ManuelRuiz-xi7bt 3 жыл бұрын
Same here. After deriving it gets to solving s³ + 2s² - 1 = 0 with s = sin (theta) and thus s = (sqrt(5) - 1 )/2. Fun little fact: cos²(theta) = sin(theta).
@ManuelRuiz-xi7bt 3 жыл бұрын
I got to this equation y = 1 + sin(2 theta) / (1 + sin(theta)) by similarity of the two right triangles above the right square. short leg of smaller one = c + s - 1 short leg of larger one = y - 1 long leg of smaller one = s + 1 long leg of larger one = c + s + 1 with s = sin(theta) and c = cos(theta). Thus (y - 1) = (c + s - 1) (c + s + 1) / (s+1) = [(c + s)² - 1] / (s+1) = 2 c s / (s+1) = sin (2 theta) / (sin(theta) + 1)
@MarkTillotson 2 жыл бұрын
(√5-1)/2 is 1/φ, or 2/(1+√5), c is √s. Its useful to know the relationships between φ, φ^2, 1/φ, 1/φ^2, etc. For instance s^2+c^2 is 1/φ^2+1/φ, which is (2-φ) + (φ-1) = 1, quick sanity check on s and c.
@schweinmachtbree1013 3 жыл бұрын
It is interesting to note that the optimal value of x found in the video, x = 1/2 * (-1 + sqrt(5) - sqrt(2sqrt(5) - 2)), is negative! (it is about -0.168) so the optimal situation does not look like the diagram on the board, in that not everything lives in the first quadrant.
@TheMartian11 3 жыл бұрын
Who cares
@artsmith1347 3 жыл бұрын
@@TheMartian11 If x is negative, then how can the left vertex of the rotated square be on the y-axis?
@alquinn8576 3 жыл бұрын
i get x ~ +1.4
@AndreasHontzia 3 жыл бұрын
If you rotate the left square and trace the upper most point, you will get an arc. The latch is dropping onto this arc. So the touching point will be on the right half of the arc.
@artsmith1347 3 жыл бұрын
@@AndreasHontzia I don't understand the reply. The line passes through the highest vertex of the rotated square. At 8:15, the y-value at that vertex is 'x' -- which is said to be negative. If the y-value at that vertex is negative, how is the y-intercept of the line at a positive y-value? The Desmos page in the post by MathyJaphy shows 'x' at about 1.4.
@ROCCOANDROXY 3 жыл бұрын
I found it simpler to obtain y = f(theta). I obtained y = 1 + (sin(2 * theta)/(sin(theta) + 1)) and after taking the derivative and solving for sin(theta) = phi - 1 implies cos(theta) = sqrt(phi - 1), where phi is the golden ratio, implies y(max) = 1 + (2 * (phi - 1)^(3/2)/phi). Note: All your coordinates can now be expressed in terms of phi to accurately graph the problem.
@MizardXYT 3 жыл бұрын
I used Pythagoras and symmetry instead of trigonometry. Bottom corner = (a,0), Left corner = (0,b), x = a + b, a² + b² = 1. This gives y = 1 + (2ab)/(b+1). Maximum y when b² + b - 1 = 0.
@willyh.r.1216 3 жыл бұрын
Love it. Thx
@Bacony_Cakes 3 жыл бұрын
"Y" is this in my reccomended?
@demetriuspsf 3 жыл бұрын
Wouldn't the X need to be on the range [1,sqrt(2)] to be a valid solution given the way the diagram is presented?
@noelmairot 3 жыл бұрын
Thanks Michael for this beautiful problem and all contributor's posts. I focused on the fact that someones found a rational value (8/5) while others found a different answer related with the golden ratio. Both answers are beautiful and I know I'm speaking about a difference less about 1e-3, but it seems that both didn't compute the same thing. The former maximize the max of the slope of the upper line. Unfortunately, the max slope of the upper line doesn't give exactly the max value for Y. The slope can be for instance a bit larger, and Y could be a bit smaller if the right square is closer. So we first have the max of the slope for Y = 8/5 (theta about 36,87° and tan(upper line vs horizon)=1/4) and then the max of Y = 1,600566... (theta about 38,17° and sin(theta)=(sqrt(5)-1) / 2).
@mrmathcambodia2451 3 жыл бұрын
I like this video , I like the good solution , I try to do and try to learn also.
@andrewburbidge 3 жыл бұрын
If someone might be interested in what difference is made to the maximum of y by rotating the square from where the diagonal is vertical, at the vertical, I got y = (4 + √ 2 )/(2 + √ 2 ) = 1.5858 The answer shown, 1 + √ (10√ 5 - 22 ) = 1.6006 The difference is about 0.9 % I guessed it would be more than that. As the square rotates, the lowest point slides away and the length of the whole figure declines. Eventually, I could see that from the treatment in the video.
@alexgiese6333 3 жыл бұрын
Should have kept using the angle theta instead of eliminating theta. Then you get y = 1 + 2/phi^(5/2). Much more simple.
@Shindashi 3 жыл бұрын
There IS a fun way to compute this! If you only work with the trigonometry, you get y = 1 + sin (2θ)/(1+ sin θ). The derivative of that, expressed as a polynomial in sin θ = a, has (a^2 + a - 1) as a factor, with the other factor discardable. You get y when you set a as the inverse of the golden ratio. Pretty elegant.
@redShiftish 3 жыл бұрын
I did it this way as well.
@Pengochan 2 жыл бұрын
The problem when eliminating theta is, that the quadratic equation gives two solutions. This reflects that x is the same for theta as for 90°-theta. Choosing the solution with sin(theta)>1/2, i.e. theta>45°, is IMHO the wrong one, as for that solution the "top" point of the first square is to the left compared to the (1-theta)1. It's probably better to keep theta as free variable.
@CTJ2619 2 жыл бұрын
kind of gnarly!
@kevinmartin7760 3 жыл бұрын
Interesting that x is 1/phi-sqrt(1/phi) if I calculated that right? But that's negative (about -0.168)? Perhaps the wrong root was taken for the quartic equation.
@cernejr 3 жыл бұрын
I agree. The overall result may or may not be correct. Michael suffers from pure mathematician's disease - he despises the real world. If he checked his results along the way, he would have noticed this.
@modernmirza5303 3 жыл бұрын
Nice ! But i uesd theta(no elimination) and it simplified everything and gave the equation t^3+2t^2=1 where t= sin(theta) and solving it gives t=-1,(√(5)-1)/2
@PATRICKZWIETERING 3 жыл бұрын
Correct!
@cernejr 3 жыл бұрын
So your theta is 38.2deg, that sounds plausible. Theta should be a little less than 45deg.
@CglravgHRjsksgS 2 жыл бұрын
We didn't take limitations throughout this process for x. One of them is 0≤x≤√2. An amazing problem thought!
@pratikmaity4315 3 жыл бұрын
Hi Michael. Here is a nice Diophantine Equation: (Japan 2020, Junior finals question 3) Find all tuples of positive integers (a,b,c) such that Lcm(a,b,c)=(ab+bc+ca)/4. Thanks.
@psioniC_MS 3 жыл бұрын
I get the same end result for y (≈1.601) with a slightly different approach, however, your x (≈-0.168) must be some wrong root since plugged into y the result is ≈0.399.
@tobiaskyrion6019 Жыл бұрын
The problem becomes easier, when the diagram is mirrored along the y-axis and shifted such that the non-tilted square has its lower left corner at (0,0). If we denote with x the height of the adjacent corner of the tilted square, we can find y(x) = 2*x*sqrt((1-x)/(1+x)) + 1. This has an optimum at x = (sqrt(5) - 1)/2 with gives an optimal value for y of (sqrt(5)-1)*sqrt(sqrt(5)-2) + 1.
@Etothe2iPi 3 жыл бұрын
x
@jursamaj 3 жыл бұрын
I found it simpler to set the origin at the upper right point, where the diagonal line hits the flat square (with x increasing to left for simplicity). This give 3 collinear points: (0,0), (1+sina,sina+cosa-1), (sina+cosa+1,y-1); where 0≦a≦π/4. Working out the slope, y=1+(sina+cosa-1)(sina+cosa+1)/(sina+1). I know that can be simplified, but at this point I did a numerical optimization for y, yielding y≅1.600566212, to the limit of my spreadsheet's precision. This matches the Michael's solution, again to the limit of my spreadsheet's precision.
@hgnb1001 3 жыл бұрын
I would place the origin of coordinates at an invariant (fixed point) , let say the point (X, 0) in this case would be my (0,0).
@albaihaqi4871 3 жыл бұрын
I get the same result after 2 hrs of trying, using phytgorean theorem (initially i tried to use trigonometry but i get lost) .... i even had to recall derivatives of products. Thanks for the exercise! Edit: you get new subscriber
@jarikosonen4079 3 жыл бұрын
I hoped to see also what the theta value was and numeric values of x & y... great.
@udaysrivastava1957 3 жыл бұрын
I guess the eqn for y is wrong because it should be a straight line but when i plot it on desmos i get some curve. Please explain..
@rioswim3234 3 жыл бұрын
Let the angle of rotation be θ then noting height and width of rotating square is Sinθ + Cosθ (let S = Sin(θ) etc ) Similar triangle gives us the required height is Y = 1 + 2SC/(S+1). Differentiating and equating to zero gives us the cubic S^3 +2S^2 - 1 = 0 Wolfram alpha gives us S = (√5 - 1)/2 . C is just found using pythagoras. Then carefully substitute back into expression for Y (tedious but straight forward)..
@Cheater3k 3 жыл бұрын
The top point of the tilted square is at 1/2*(x PLUS sqrt[2 - x^2]). Then you get x = [sqrt(5)-1]/2 PLUS sqrt[[sqrt(5)-1]/2].
@WiseSquash 3 жыл бұрын
this was extremely entertaining to watch. thanks Michael!
@Skandalos 2 жыл бұрын
I defined x as the x-coordinate of the bottom vertex of the tilted square and used only the pythagorean theorem to determine the coordinates of the important points, no angles involved. That way I got y = ((4x^2 - 4x^3) / (1 + x))^1/2 + 1. To determine the maximum point I knew the +1 and the ^1/2 dont change the x-value of the maximum so I simplified y to (4x^2 - 4x^3)/(1 + x) which is easy to differentiate. The maximum is at x = (-1 + sqrt(5))/4 and I was too lazy to calculate y. The fun part was that I constructed the whole situation in geogebra and also displayed the target function in the same graph which allowed simple error control.
@txikitofandango Жыл бұрын
managed this one without trig. Let A be the x-coordinate of the bottom corner of the left square, and let B be the y-coordinate of the left corner. Then A^2 + B^2 = 1. The line passes through (A, A+B) and (1+A+B, 1). The y-coordinate of the line between these two points simplifies to y = 1 + A - (B-1)^2/A. Note that dB/dA = -A/B when you use the chain rule. Setting dy/dA = 0 and simplifying a bit, you get (B-1)^2/A^2 + 2(B-1)/B + 1 = 0. It is easier to get everything in terms of B here, so you end up with the cubic B^3 - 2B + 1 = 0, which factors as (B-1)(B^2 + B - 1) = 0. Golden ratio time! B = (sqrt5 - 1)/2, and since B^2 = 1 - B, you get A^2 = B, so A = sqrt(B). From there, y = 1 + A - (B - 1)^2/A = 1 - (B^4 - B)/A = 1 - A(B^3 - 1) = sqrt((sqrt5 - 1)/2)(3 - sqrt5) + 1, which is equivalent to Michael's answer
@cernejr 3 жыл бұрын
A graph y=f(theta) would be the cherry on the cake. And also a good sanity check.
@flavioleo5100 3 жыл бұрын
Actually, results in your way gives 1.60, while computating the intersection with the y axis of the straight passing trough A(√2/2 √2) and B(1+√2,1) gives 3-√2 (≈1.59). Since the maximum y is defined for the highest y that A can assume, the process should be correct
@davidseed2939 3 жыл бұрын
Simpler to express in trig form. s,c are sin, cos . ss = sin^2 etc. y-1=2sc/(s+1) differentiating and set to zero gives.scc=(1+s)(cc-ss) Or sss+2ss-1=0 Which factorises as (ss+s-1)(s+1)=0 Or s=1-ss=cc Hence c=sqrt(s) Also s = golden ratio -1= ¥-1 0.608... y = 1 +( 2(¥-1)^1.5)÷¥ =~1.600566 the same as the video. That ¥ should be phi x =~ 1.404 ie close to 1.414 which it would be with the left square at 45°.
@keithmasumoto9698 3 жыл бұрын
A value for theta would be nice to see how far from 45 degrees it is.
@cernejr 3 жыл бұрын
I agree.
@polyhistorphilomath 3 жыл бұрын
It’s around 38.5° or ~2/3 (radians). Then y should be slightly greater than 1.60 for unit squares.
@ManuelRuiz-xi7bt 3 жыл бұрын
It's 38.17270763...°
@RonWolfHowl 2 жыл бұрын
Wow! Who would've thought a quartic polynomial would show up in a problem like this!
@vanessamichaels9512 3 жыл бұрын
shouldn't the left square hit the line at ( sqrt2/2, sqrt 2) and the right square hit the line at (1+sqrt2, 1)? If so, we have two points to make a line with, and we then just solve for when x=0.
@schrodingerbracat2927 2 жыл бұрын
y = 1 + (2sinθcosθ)/(cosθ + 1)= 1 + (sin2θ)/(cosθ + 1). considering dy/dθ = 0 leads to a trigonometric equation which boils down to t^4 + t² - 1 = 0 where t = tan θ. Optimal t = sqrt((sqrt(5)-1)/2). drawing a right-angled triangle and plugging in values for sinθ and cosθ [try not to use Mathematica], gives the result y = 1+ sqrt(10sqrt(5)-22)
@ygarasab 3 жыл бұрын
such addictive videos
@pussy2907 3 жыл бұрын
can you give me an example of Contour integration by substitution i.e. if we put z=F(Z) like integration f(x), we put x=F(t). I want to know that this type of question is valid or not
@meirkarlinsky7497 3 жыл бұрын
As other suggested - this can be solved without calculus: for max Y of left-most line, we want the top-most line to have the (acute) angle as steep as possible (while still going thru the top-most vertex of the tilted square) AND to stretch to the left as far as possible the left-most line (while still having the left-most line go thru the left-most vertex of the tilted square). Fortunately these two requirements do not contradict and they both are achieved when the tilted square has been rotated 45° - i.e. when one of its diagonals is orthogonal to the lower-most line. Then the (x,y) values of the upper-most vertex of the tilted square can be found - which enables finding the expression of the upper-most line. Inserting in this expression the x-value of the left-most line (it is also found from the achieved "optimally tilted" square) gives the desired Max Y.
@alric8 3 жыл бұрын
Using the value of Y that he got when he did the question, the square is tilted 38 degrees, not 45 degrees as you would intuitively think.
@GaryFerrao 3 жыл бұрын
13:00 "change that minus sign to an equals sign" ;-)
@zealot2147 3 жыл бұрын
I dont understand why, as others suggest, that you cannot use x=sqrt2 due to the maximum of y being related to the maximum of x, since the largest x can be is the diagonal of the rotated square of sides 1, also sin(theta) then equals cos(theta) and theta =45 with those being known values. Unless the objective is solely to solve using first term diff calculus or without the sides of the square being known
@hsjkdsgd 3 жыл бұрын
I tried the angle method. I was also stuck after differentiating for finding critical points.
@rocky171986 3 жыл бұрын
your expression for y should be (sin(2\theta))/(1+sin\theta)+1. Differentiate, then when setting the numerator to zero, convert it to an equation in sin\theta only. You should get a factorizable cubic equation in sin\theta.
@tomtheultimatepro 3 жыл бұрын
Solved it just using linear equations and pythagoras. y is the y-axis intercept of the linear function running through the two points where the square touches the slope. The resulting function is quite ugly, so I also turned to wolfram alpha to help with the derivative and setting it equal to zero. Getting the same solution felt satisfactory nonetheless.
@gibbogle 2 жыл бұрын
I did this a bit differently, using the angle of tilt (alpha) of the almost-vertical diagonal from the vertical as the parameter. Then if y = 1 + z, and c = cos(alpha), s = sin(alpha), I got: z = (2c^2 - 1)/(1 + sqrt(2)(c - s)/2) Unwilling to solve dz/dalpha = 0, I just computed z for a range of alpha and found the maximum at alpha = 0.119 to be z = 0.600566, which is sqrt(10sqrt(5) - 22). I notice that my equation for z looks a lot like yours if I put 2c^2 = x^2, s^2 = 1 - c^2 = 1 - x^2/2, s = (1 - x^2/2)^(1/2) This didn't quite give your equation, then I realised I needed to use s = - (1 - x^2/2)^(1/2). Now I get exactly your equation. :) By the way: 2c^2 - 1 = cos(2.alpha) = sin (pi/2 - 2.alpha), therefore if I put pi/2 - 2.alpha = 2.theta then alpha = pi/4 - theta. If I substitute for alpha in my equation for z I get Sergio's equation. :))
@AlbertoDalmau 3 жыл бұрын
I also found the maximum area of the trapezoid formed.
@snowfloofcathug 3 жыл бұрын
Wow that’s a cool problem
@redShiftish 3 жыл бұрын
I did this problem completely differently and ended up with 1+ [Sqrt(2)(6-2Sqrt(5))(Sqrt(Sqrt(5) -1))]/4 which calculates to the exact same number that Dr. Penn got. Anyone know how to prove that those two answers are equal??
@ZonaNordico 3 жыл бұрын
Before resolve y' = 0, shouldn't you verify the interval where denominator of y' is not zero?
@Dalroc 3 жыл бұрын
(x+2) >= 2
@robertingliskennedy 3 жыл бұрын
great pace yet great clarity - a great place to stop
@hlicj 3 жыл бұрын
My intuition was to go for y1 being the full height of the diagonal (sqrt(2)) but that was not the correct solution. It is much simpler to solve though. :-)
@geppettocollodi8945 3 жыл бұрын
So What is the angle Alpha, another interesting data point.
@dhwyll 3 жыл бұрын
Oh, pulling out the ancient calculus memories at 12:00. The derivative of the square root of a function is the derivative of the function divided by 2 times the square root of the function. That is, d sqrt(f)/dx = (df/dx) / (2 sqrt(f)) Thus, d(sqrt(2 - x^2))/dx has f = 2 - x^2 Therefore, df/dx = -2x (derivative of 2 is 0, derivative of -x^2 is -2x) Thus, (df/dx) / (2 sqrt(f)) = -2x / 2 sqrt(2 - x^2) = -x / sqrt(2 - x^2) = -x(2 - x^2)^-0.5
@lifeunitcompany9420 2 жыл бұрын
can one quickly eliminate that theta=45 is not the answer? or it is the answer
@xFuZZyGR 3 жыл бұрын
Why didn't you just say that the diagonal of the sqaure is root(2) an then solvetheproblem via similar triangles. First triangle from xE(root(2)/2,1) and the other xE(0,1)?
@CzaroDziejCK 3 жыл бұрын
It is not correct, because, you cannot be sure that diagonals of square are paralllel or pependicular to the axes. Your way to solve this problem is only a special case. I know this pain. Long time ago I was preparing to the IMO. Almost every task was about including every case of solution. Skipping any case was painful for your score xd
@alric8 3 жыл бұрын
I thought this as well, but the diagram is misleading in that the height is actually greatest when you tilt the square slightly to the left. If you look at how he did it, and then plug in sint+cost=x (where t is theta) you will find that the value of Y is actually greatest when theta is about 38 degrees, which would mean that you can not just assume that the height is sqrt2 because the square is tilted at a different angle than expected.
@lisandro73 3 жыл бұрын
I miss why dr. Penn didn’t use the other value in the quadratic solution of sin(tetha)
@kevinmartin7760 3 жыл бұрын
Taking the other quadratic solution essentially exchanges the values for sin(theta) and cos(theta) and represents the square tilted in the other direction (with its "vertical" diagonal tilted to the left instead of to the right) but (I think) ultimately yields the same relationship between x and y. Note that both these tilted squares yield the same value for x, but the two theta values are complementary. But this probably should have been explained.
@lisandro73 3 жыл бұрын
@@kevinmartin7760 Thanks Kevin, I appreciate it. I still think he should have explained it
@gregmyles5852 3 жыл бұрын
Is the Y maximum value when the diagonal of the square is perpendicular to the slope? I'm pretty sure it is, but please enlighten me if it's not. Then you can crunch the numbers from there.
@francotello6989 3 жыл бұрын
Y max=3-sqrt(2)
@darreljones8645 3 жыл бұрын
The final maximum value of y is just a tad over 1.6. (More specifically, my calculator gives it as 1.600566212...)
@Pederzoli01 3 жыл бұрын
Great. Thanks
@umarshehzad7134 3 жыл бұрын
Sir i think you have calculated minimum value of y instead of maximum you should check it by second derivative test And also value of "y" you have found i wrong if we plug in
@walklikeamagician 3 жыл бұрын
I believe you are correct! If you plot y as a function of x, there's a minimum at the value found above, but that's the only critical point. This function actually has a max at one of the end points of its natural domain which seem to be x = 1 and x = sqrt(2). The latter looks to give you the max.
@rocky171986 3 жыл бұрын
For anyone who is confused by your answer and differs from Michael's, one simple way to check if you're on the right track is this: what is the y-coordinate of the top most corner of the tilted square? If your answer is not dependent on the tilt angle or some other variable you chose to use, and is instead just sqrt(2), you are not on the right track.
@casc3601 3 жыл бұрын
So what is the correct y value of the top coordinate?
@jezzag9739 3 жыл бұрын
@@casc3601 1.600566212
@CzaroDziejCK 3 жыл бұрын
I like watching your videos with IOM tasks. Above 8 years ago I could solve IOM tasks easily, but I regret that I can't do it right now. I fell out of a shape... :( Luckily I easily understood the essence of the problem and the solution. I know that application of coordinate system in a solution makes your life easier with this problem. :) It seems, I can feel geometry but I have problems with numbers theory' problems.
@CzaroDziejCK 3 жыл бұрын
Aaah! The most important thing: And there's a good place to stop 😁
@pietergeerkens6324 3 жыл бұрын
In the expression for x, isn't the sign wrong in front of √(2√5 - 2)? I get an evaluation for x of -0.168 with that expression; but a more reasonable value of 1.404 with the sign changed to give: x = ½ (√5 - 1 + √(2√5 - 2) ) That also is then in agreement with the results below of a maximum when cos² θ = sin θ so that sin θ = φ - 1. Also I believe the identity sin θ + cos θ = √2 sin (θ + ¼π) (well known to the electrical engineers at least) would come in quite handy in simplifying the calculations.
@dippn7047 3 жыл бұрын
Can you just make the intersection of the first square and the line (sqrt2/2, sqrt2) to find the slope of the line then use that to find out what it equals at x=0? The tallest that intersection of the first square and the line can be will be the diagonal of the square from two corners right?
@samstep4279 3 жыл бұрын
I think you mean we put the left hand square so that its diagonal is perpendicular. That's the answer, right? No... You can twist the left hand square clockwise a little bit and it will lift the slope of the line we are trying to find. It's the slope we need to find - not the height of the left hand square.
@seriously1184 3 жыл бұрын
@@samstep4279 No it won't shift he slope. If you rotate the left hand square clockwise, the bottom corner will go clockwise to, thus lowering the slope, if the left hand square is rotated that the diagonal is vertical, that point where it touches the slope is the most upper point it can get. It is a square that you rotate and not just a vertical line.
@lennyganado3975 3 жыл бұрын
@@seriously1184 and the earth is flat, right? why are you arguing with the literal established correct answer right now lol
@seriously1184 3 жыл бұрын
@@lennyganado3975 First of all I am not an flat earth believer, far from it ! Second, I am an engineer and thus I looked at this in a realistic way of problem solving (that what is often times too big of a problem for mathematicians being able to do) ! Third, the designed problem used in this video is in a closed frame ! All the outlines off thegeometrical figures are all of the same width and colour, thus making them all geometrical objects laying against each other and in one large geometrical object, making this a realistic engineering problem than anything else where the law of physics rule above anything else (a concept that is apparently unknown to mathematicians) ! So use lines of different width, colour, etc if you want to indicate what is fixed or adjustable in any design or drawing (something that apparently is also unknown to mathematicians) ! Fourth, if you start with insults, you must really be a compulsive and impulsive pric* I bid you a good day sir.
@RexxSchneider 2 жыл бұрын
@@seriously1184 It's a pity that an engineer such as yourself missed the point that as you rotate the first square a little clockwise from its highest point, not only does the bottom corner move left, but so does the rightmost corner and so _the second square moves left in its entirety._ That movement left of the "hinge" of the slope increases the slope initially, and that is why the value of y will in fact increase a little as the first square rotates clockwise from its highest point.
@carly09et 3 жыл бұрын
Two points on a line (root 2 /2 , root 2) and (1 + root 2 , 1), so solve for y intercept. Calculus is only needed to demonstrate this is the max via perturbation.
@vanessamichaels9512 3 жыл бұрын
That's what I thought. When I do that though, I get 1.58ish, though
@mandaparajosue 3 жыл бұрын
Using theta you'll find y -1 = 2(cos θ)(sin θ)/(sin θ + 1). Max for sin θ = φ (the golden ratio! Wow!) so cos θ = φ^(1/2). Wolfram Alpha is not necessary.
@adambelyea3340 3 жыл бұрын
Would appreciate a more detailed explanation of how the quadratic equation was done. Which value was a, b and c.
@General12th 3 жыл бұрын
a = 2, b = -2x, and c = x^2 - 1. The quadratic formula is letting us solve for sin(theta).
@dimy931 3 жыл бұрын
For those of you that did the calculus in theta- what's the optimal value for theta? My intuition says 45 degrees but I would expect the answer to be nicer if it were the case
@giotadetsi45 3 жыл бұрын
θ~38.17° max [sinθ*cot(θ/2)]
@martinschulz6832 3 жыл бұрын
Why is the value of x negative?
@joaopalrinhas5242 3 жыл бұрын
Why is sin(theta) = 1/2 (x+sqrt(2-x^2))? Why not ±?
@saidtanji 3 жыл бұрын
Brcause theta is
@artsmith1347 3 жыл бұрын
The negative result would make the y-value for every point on the rotated square less than 1. The straight line would then cross the y-axis at a value less than 1 -- which is lower than when theta = 0, so there is no maximum y-value to be found when theta is between 0 and -pi.
@ademaupsilon 3 жыл бұрын
Put x=1 to the equation
@nellvincervantes6233 3 жыл бұрын
Another challenge sir. Solve some difficult physics problems.
@mrminer071166 3 жыл бұрын
Why no animation to show all the possibilities? With parametrization according to the distance between the bottoms of the squares?
@mrminer071166 3 жыл бұрын
Why not set up Penn's (x,0) AS the origin?
@paolamassetti4426 3 жыл бұрын
Solution is exactly y=8/5; the method used is approximate and abstruse, you can solve it in easier way considering other similar triangles
@MarkTillotson 2 жыл бұрын
Alas you fell into the trap. y depends on theta in a more intricate manner.
@claymusic2205 3 жыл бұрын
almost exactly 1.6
@CauchyIntegralFormula 3 жыл бұрын
If we stick to parametrizing by theta, we get the top point as (cos(θ), cos(θ) + sin(θ)) and the upper right point as (cos(θ)+sin(θ)+1,1). Then, the slope of the line segment is (cos(θ)+sin(θ)-1)/(-sin(θ)-1), so y = (cos(θ)+sin(θ)-1)/(sin(θ)+1) * (cos(θ) + sin(θ) + 1) + 1 from the upper right point and the point (0,y). This simplifies as y = 2sin(θ)cos(θ)/(sin(θ)+1) + 1, and its derivative is 2[ ( (cos^2(θ) - sin^2(θ))(sin(θ) + 1) - sin(θ)cos(θ)cos(θ) )/(sin(θ)+1)^2 ]. y is maximized only when this derivative is 0, so we need that the numerator of this fraction is 0. Using cos^2(θ) = 1 - sin^2(θ) and letting s = sin(θ) for brevity, we get (1 - 2s^2)(s+1) - s(1 - s^2) = 0, or s - 2s^3 + 1 - 2s^2 - s + s^3 = 0, or -s^3 + 1 - 2s^2 = 0, or s^3 + 2s^2 - 1 = 0. This factors as (s+1)(s^2 + s - 1) = 0; since s = sin(θ) is between 0 and 1 as θ is between 0 and π/2, the only solution is s = (sqrt(5) - 1)/2. Curiously, for this angle, cos^2(θ) = (sqrt(5) - 1)/2 and 1/(sin(θ) + 1) = (sqrt(5) - 1)/2, so we get that y = (3 - sqrt(5)) sqrt( (sqrt(5) - 1)/2 ) + 1. sqrt( (sqrt(5) - 1)/2 ) doesn't seem to have a nice form, so that's our answer. This is equivalent to the answer Dr. Penn gets, so that's good. tl;dr: θ = sin^(-1)(1/ϕ)
@Ablatius 3 жыл бұрын
I find this answer in term of theta more apropriate, as we would naturally measured the rotation of the first square instead of the measure of x (length which doesn't have a real physical signification for the square)
@holyshit922 Жыл бұрын
This equation is not so difficult to solve by hand After squaring both sides you can check for rational roots or get rig of repeated root with differentiation Remaing quartic can be easily solved by rewriting it as difference of two squares and then as a product of two quadratics After solving polynomial equation check the solution because squaring both sides may produce extraneous solutions
@Walczyk 2 жыл бұрын
i just assumed the square was rotated by 45 degrees and then i made two points and wrote the straight line. This could be solved more simply with euler's calculus of variations imo c
@myrrito Жыл бұрын
I love your videos in general but this solution is a bad one, simply because I haven't heard of a math competition where in the solution you can plug in wolfram alpha. I would recommend a little simpler solution which is based on labeling the same angle theta. Label the big trapezoid (whose base y we need to calculate) ABCD by starting from the bottom left corner. Label the topmost point of the tilted square as P. Then in terms of theta you can easily calculate the distances from P to the lines AB and BC. Let F be a point on the line segment AD such that FC is perpendicular to AD. Again you can easily calculate FC and angle MCD in terms of theta. In the end you will find that y = AD = 1 + sin (2 * theta) / (sin (theta) + 1). Calculating the first derivative and its roots will bring you to a cubic equation sin^3 (theta) + 2 * sin^2 (theta) - 1 = 0 which is very easy to solve because it transforms to t^3 + 2*t^2 - 1 = 0 which has one integer root and only one positive root. From here you can find sin (theta), cos (theta), sin (2 * theta) and therefore AD.
@qobilruzmatov48 3 жыл бұрын
Nice good
@ronleblanc1094 3 жыл бұрын
I stand corrected.Slightly nudging the left hand square to the right increases the height of y by 0.0147 units.Good job Michael.
@rafael7696 3 жыл бұрын
Very nice problem. Thanks.
@arshsverma 3 жыл бұрын
4 weeks ago?
@goodplacetostop2973 3 жыл бұрын
@@arshsverma If you dig down enough, you will notice some Michael videos are unlisted. Which means you can watch them, post comments but these videos will be public in few weeks.
@goodplacetostop2973 3 жыл бұрын
@@arshsverma Michael usally posts a ton of videos at the same time and schedules them to go public one or two per day.
@tusharchetal 3 жыл бұрын
Why cant we just say the diagonal is √2 and since the two diagonals bisect each other, x is √2/2 or even the diagonal bisects the angle so theta is 90/2 of 45?
@sebastianw.6637 3 жыл бұрын
Because the tilted square isn't tilted by 45°
@tusharchetal 3 жыл бұрын
That doesn't change the length of the internal diagonal?
@a_llama 3 жыл бұрын
Symmetry seems to suggest the maximum is achieved at theta = pi/2
@xCorvus7x 3 жыл бұрын
Consider that after some angle < π/4 any further increase in the angle (and thus in x) decreases the slope more than it is lifted meanwhile, which leads to a net decrease in y.
@dhanishsrinivas 3 жыл бұрын
Do you mean pi/4?
@Pederzoli01 3 жыл бұрын
Really close to that 👌
@Etothe2iPi 3 жыл бұрын
I just realized, x is wrong. It should be g+sqrt(g)=1.40418... g is the golden ratio. Btw it's much easier using the angle as the variable.
@Daniel-ef6gg 3 жыл бұрын
Why do you assume sin(theta) > cos(theta)? It looks like you should assume the opposite.
@wesleypinchot5294 Жыл бұрын
Hi, Michael! I love your videos because you demonstrate clear solutions to challenging problems, skip the tedious too-obvious steps, and admit your mistakes. BUT I don't understand how at 7:30 you got sin theta = 0.5(x+sqrt(2-x^2)). According to the quadratic formula, shouldn't it be 0.5(2x +- sqrt(4x^2 -4•2(x^2 -1))) = x +- sqrt(2-x^2) ? I see that my result CAN'T be right because the sqrt must be tiny and sin theta must be close to 0.5x, but I don't understand what I'm doing wrong. Help!
@mathunt1130 3 жыл бұрын
It seems to me that it is intuitively obvious that y will be largest if the second square is rotated 45 degrees giving the width to be 1+2^1/2.
@lennyganado3975 3 жыл бұрын
this will get you within 1% of the correct answer, but still incorrect. It is not 45 degrees. it is 45 plus or minus 4 degrees to either side.
@sharpmind2869 3 жыл бұрын
I got the answer 3-√2, as the y will be maximum if the tilted square's diagonal is perpendicular to the base line
@craftylord3336 3 жыл бұрын
Same. Haven't watched the video but it was fun to attempt.
@davethesid8960 3 жыл бұрын
It would've been a LOT easier to solve this by similarity of triangles since y clearly has maximum when θ=π/4. Or am I missing something? Someone please explain!
@alexgiese6333 3 жыл бұрын
y would be maximized at pi/4 if the right square didn't move as the left square rotates, but the right square moves as the left square rotates.
@davethesid8960 3 жыл бұрын
@@alexgiese6333 Thank you very much, I didn't think of that
@dannybodros5180 Жыл бұрын
7:23 I think I found a mistake. That quadratic equation should be - 2sin^2θ + 2xsinθ - x^2 + 1 = 0
@cxeazn 3 жыл бұрын
The math in this problem might be first year level calculus but if this were on my final exam I would have gotten a 0.
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