You've chosen a good time to comment.. Basically 2 days before publishing the video.
@goodplacetostop29733 жыл бұрын
@@heliocentric1756 Michael already posted the videos for the next few weeks (he’s building a new studio so obviously he won’t be available to record videos) so I went ahead and post related homeworks under unlisted videos.
@aperson1234-t8x3 жыл бұрын
Me: this looks like something I can do! Penn: does a frontflip Me: ...
@ranpodeltapsi3 жыл бұрын
Dude that introductory backflip was dope!
@shokan71783 жыл бұрын
It was a frontflip tho :)
@jonathanho6183 жыл бұрын
length 'a' was incorrectly labelled 'b' though!
@Nah_Bohdi3 жыл бұрын
>he
@Dec381053 жыл бұрын
@@shokan7178 he meant (1/backflip)^-1
@user-cv1jb9xv2p3 жыл бұрын
0:54 I started working on the problem. 1:05 I started working on doing a front-flip. Nicely done. 👍🏼👍🏼👏🏼👏🏼
@MrRyanroberson13 жыл бұрын
unexpectedly easy for what i thought would just be outright absurd
@manucitomx3 жыл бұрын
And, when you thought nothing beats the backflip in comes 1:04. Thank you, professor.
@tahasami34093 жыл бұрын
Thank for michael penn
@jamirimaj68803 жыл бұрын
the twist is the backflip, damn P.S. if a and b are equal, then the two terms cancel out and just leaving the area of the triangle, thus the shaded area of the crescent will be EQUAL to the area of the triangle beneath it. Awesome.
@57thorns3 жыл бұрын
And that is the difference between a quarter circle and a half circle (with different radii, but still).
@michaelf.70503 жыл бұрын
@@57thorns radii?
@JohnSmith-qq7yw3 жыл бұрын
That's actually a front flip, not a backflip.
@BigDBrian3 жыл бұрын
and you can see it in the answer too, because if a and b are equal then the terms cancel out.
@goodplacetostart90993 жыл бұрын
Good Place To Start at 1:04 this time
@targetiitbcse17613 жыл бұрын
Hello, remember me from Physics galaxy channel
@goodplacetostart90993 жыл бұрын
@@targetiitbcse1761 yes I do
@mushfikaikfat3 жыл бұрын
Explanation was elegant and perfect. Made perfect sense. Nicely done Sir! Thanks for posting the video 😊😊
@bprpfast3 жыл бұрын
A front flip?!!!!!
@MichaelPennMath3 жыл бұрын
That is the twist!
@djmintyfreshful3 жыл бұрын
I’ve only seen two videos and I already love this channel
@pedropicapiedra48513 жыл бұрын
1:02 linear isometry
@GastevAleksei3 жыл бұрын
This was a quite easy problem for a Michael Penn video.
@thanosxypolytos40933 жыл бұрын
Michael you make it so easy to understand thanks for sharing the calculations are so easy to understand as well.
@Hani-ic3lh3 жыл бұрын
This was surprisingly short!
@shivansh6683 жыл бұрын
Great Personality Professor 👍
@andcivitarese3 жыл бұрын
very nice slow motion demonstration of conservation of angular momentum: the more he curls up the faster he rotates.
@rishijai3 жыл бұрын
The front flip is at 1:02
@NavyBlueMan3 жыл бұрын
Not a super difficult one but still an interesting puzzle - suppose you have a square of side length 1. Inside the square is a circle of the maximum possible radius that can fit in the square, which is shaded in. But a square has been cut out of the circle, such that it kisses the circle at 4 points, which isn't shaded in. But a circle is inside this smaller square, shaded in, with a square cut out of it, and a circle in this square, and so on to infinity. The question is, what is the total shaded area?
@davidchung16973 жыл бұрын
Love the figures on the chalkboard.
@BleachWizz3 жыл бұрын
Just a tip. You should be able to position your legs straight before you fall, honestly you're jumping too low. Be careful xP
@haibai17663 жыл бұрын
I think you should mention that a ≥ b, because otherwise the problem gets a bit more complicated
@PATRICKZWIETERING3 жыл бұрын
My answer in that case is (1/2)ab - (1/4)(b^2-a^2)arcsin(2ab/(a^2+b^2))
@demetriuspsf3 жыл бұрын
Wow that was a very low front flip. Nice!
@AndreasHontzia3 жыл бұрын
What if a is smaller than b? Some portion of the quarter circle will lay outside the semi circle. How should that be counted? Or what would be the green shaded area?
@TJStellmach3 жыл бұрын
"How should that be counted" seems like a definitional problem. What area are you even looking for at that point?
@AndreasHontzia3 жыл бұрын
@@TJStellmach You could try to match the solution from the video to the problem. But the easiest would be to say that a ≥ b.
@PATRICKZWIETERING3 жыл бұрын
I found the case where a
@trueriver19503 жыл бұрын
Depending on what you want the result for, you can then put the last two terms together and factor them using difference if two squares. That might, or might not, be a better place to stop
@peglor3 жыл бұрын
A better place to stop? What is this heresy? :-P
@aviralsood81413 жыл бұрын
A frontflip! This is amazing!
@DrWeselcouch3 жыл бұрын
Nice video Dr. Penn! Very impressive flip!!!
@HecticHector3 жыл бұрын
This is one of the few videos of yours that I could easily do on my own
@goodplacetostop29733 жыл бұрын
HOMEWORK : The square ABCD has side length 2√2. A circle with centre A and radius 1 is drawn. A second circle with centre C is drawn so that it just touches the first circle at point P on AC. Determine the total area of the regions inside the square but outside the two circles. SOURCE : Mathematical Mayhem from Crux Mathematicorum Vol. 37, No. 6
@athelstanrex3 жыл бұрын
Michael.
@goodplacetostop29733 жыл бұрын
@@athelstanrex Penn.
@claudio99913 жыл бұрын
8-5pi/2
@goodplacetostart90993 жыл бұрын
Ans 8-5π/2
@goodplacetostop29733 жыл бұрын
SOLUTION *8 − 2√2 − 5π/2 +9arccos((2√2)/3)* From the Pythagorean Theorem we get AC = √((2√2)² + (2√2)²) = √(8 + 8) = √16 = 4. Since CP is a radius of the larger circle, and AP = 1 we get CP = CA − PA= 4−1 = 3, therefore the radius of the large circle is 3. Now let's introduce more points : - Point E is the intersection of the circle with centre C and segment AB - Point F is the intersection of the circle with centre C and segment AD - Point G is on the circle with centre C such that the segment GDC is a radius of that circle - Point H is on the circle with centre C such that the segment HBC is a radius of that circle - Point I is the intersection of the circle with centre A and segment AB - Point J is the intersection of the circle with centre A and segment AD We're looking for the areas of the figures JPF and IPE. Using [A] to represent the area of figure A we get, [GDF] = [FCG] − [FCD] = α/2π · π(3)² - (√(3² - (2√2)²) · 2√2)/2 = 9α/2 - √2 where α = ∠FCG = arccos((2√2)/3). Next, we have [PCDF] = [PCG] − [GDF] = 9π/8 − 9α/2 + √2. Thus [JPF] = [ACD] − [APJ] − [PCDF] = 4 − π/8 − (9π/8 − 9α/2 + √2) = 4 − √2 − 5π/4 +9α/2. By symmetry [IPE] = [JPF] thus the area of the regions inside the square but not inside a circle is 2[JPE], or 8 − 2√2 − 5π/2 +9α = 8 − 2√2 − 5π/2 +9arccos((2√2)/3) ≈ 0.38 units.
@alainbarnier19953 жыл бұрын
Michael Penn I like this small problem... and the big jump ;-)
@schmetterlingsjaeger3 жыл бұрын
Where was the big twist? At 1:03 ?
@TJStellmach3 жыл бұрын
Couldn't the entire shaded figure be described as a kind of big twist?
@rajdeeplahiri88423 жыл бұрын
Please bring in some more problems based on combinatorics, also if you could take up some calculus problems from the jee advanced exam.
@biohoo223 жыл бұрын
The front-flip flex is why I will never be a stellar mathematician.
@wschmrdr3 жыл бұрын
3:35 Last I checked, pie are round.
@呂永志3 жыл бұрын
One thing may be proved first: arc AC and arc PC don't crossover. If a≤b, what will happen?
@PATRICKZWIETERING3 жыл бұрын
My answer in that case is (1/2)ab - (1/4)(b^2-a^2)arcsin(2ab/(a^2+b^2))
@hmmmm61743 жыл бұрын
Easier than what it looks ngl, great stuff!
@StephanvanIngen3 жыл бұрын
Your vids are just enjoyable to watch, thanks:)
@kujmous3 жыл бұрын
Can every individual area be determined?
@kujmous3 жыл бұрын
I can't find a way myself.
@MYSBRZ3 жыл бұрын
Thank you Michael, geometry is good, Tnx😍❤
@JSSTyger3 жыл бұрын
(1/8)(4ab+πa²-πb²)
@rijubhatt83663 жыл бұрын
Dude, did you really Somersault standing? Wow!
@ramtindorostkar7653 жыл бұрын
math with parcore!
@bibeksubedii00013 жыл бұрын
Easiest one 1⃣
@liab-qc5sk3 жыл бұрын
Homework for timetravlers again :prove or disprove that pi+e is transdental
@DoDO-zr5sk3 жыл бұрын
why the secound circle calculation becomes plz give reason 🙏
My dream is to have a blackboard as big as Michael's
@yahav8973 жыл бұрын
Would you like to make a video on the brachistochrone problem? I think it'd be neat :)
@rishijai3 жыл бұрын
I thought this problem required some heavy Integral calculus, I was fooled.
@nasim090219753 жыл бұрын
Hmm, how about finding the area of that small patch on the upper left side of the red right-angled triangle? lol
@astrolad2933 жыл бұрын
It's not conceptually hard, just tedious, and probably very messy. Call the intersection of the quarter circle and line AC Z. Calculate the coordinates of Z. From those you can compute the area of the triangle ZBC and the circular segment PBZ (get the angle PBZ from the dot product of vectors BP and BZ). The area of the patch is then the area of triangle ABC minus the the areas calculated. There's another geometric solution using the area of the circular cap to the right of AC, the area of triangle ABC and the quarter circle. Using calculus you can calculate the area under the circular arc PZ and the associated smaller triangle. Take your pick. I think the first one is the easiest.
@noeticresearch3 жыл бұрын
A frontal somersault in slow motion with jazzy music playing in the background?? Is there anything you can't do Michael?? :-)
@targetiitbcse17613 жыл бұрын
Maybe he can't do organic chemistry videos
@aryakaranjkar9393 жыл бұрын
I love the chalk and the flip 😂😁
@jackhandma10113 жыл бұрын
Michael Penn is gonna be the next Ronald Graham with that acrobatics.
@hyperboloidofonesheet10363 жыл бұрын
Mathrobatics.
@rorydaulton68583 жыл бұрын
Why did you go against the naming convention for the sides of a right triangle? The usual convention is that vertex C is at the right angle, side c is the hypotenuse, side a is opposite vertex A and side b is opposite vertex B. Following that convention, your ending formula be correct if you replaced your labels A, B, and C with the labels B, C, and A respectively. Or am I missing something?
@trueriver19503 жыл бұрын
I have noticed before that Michael doesn't use that convention. But that's all it is, a convention: nobody says you have to follow it
@wikipediaboyful3 жыл бұрын
Motion to change QED for "And that's a good place to stop".
@billbusen3 жыл бұрын
Seconded.
@That_One_Guy...3 жыл бұрын
Oh yeah nice Monke Flip lol
@pwmiles563 жыл бұрын
Hi Michael. Did you know, a 3D body rotating under inertia describes a rotation about the axis of angular momentum which is a function of the time difference only?
@ramakrishnasen43863 жыл бұрын
Well well well
@EthnHDmlle3 жыл бұрын
I did it the unnecessary way, with trigonometric functions. Over complicating it as usual.
@MathElite3 жыл бұрын
Awesome video
@zhusan2dui3 жыл бұрын
I will use integration 😀
@CM63_France3 жыл бұрын
Hi, For fun: 1 "and so on and so forth".
@r75shell3 жыл бұрын
where is a big twist?
@vibhanarayan96683 жыл бұрын
The way u did is bit lengthy , it would have been simpler if u assign diff parts diff variables then add and subtract
@denimator56513 жыл бұрын
normal people: 8 him: o o
@KarlHsia3 жыл бұрын
这个问题就有点水了,但是这个前空翻。。。我靠
@jaikiraj20753 жыл бұрын
Class 10 th mathematics in cbse
@ishaansharma64283 жыл бұрын
Ikr
@Ankit-xc1hd3 жыл бұрын
Yep
@raider87093 жыл бұрын
Penn Flip
@morancium3 жыл бұрын
EZ
@dimitritome51183 жыл бұрын
Man you're such a hottie. You should open an onlyfans
@JohnSmith-qq7yw3 жыл бұрын
By far, one of the most annoying things about youtube videos is the audio mixing. There needs to be a standard for youtube audio levels. The vocals audio levels are very low in this video, then there is blasting music that comes in at some point (flip performance). It's so annoying. KZbinrs, please put a little more effort into making sure your audio levels are consistent across the entire video. Otherwise it makes it so that the watcher has to constantly be adjusting the volume up and down throughout the video. This is so irritating!