Reflection about y = 2x
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- Опубліковано 19 вер 2024
- Formula of reflection about the line y = 2x
In this video, I present a really neat application of change of coordinates: Namely, I calculate the formula of the reflection of a point about the line y = 2x. Can you guess the formula beforehand? You might be surprised! Enjoy
Change of coordinates video: • Change of Matrix
Rotation matrix video (related): • Rotation Matrix
Check out my Linear Transformations Playlist: • Linear Transformations
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Yo thumbnail is on point again, Dr. P!! Love it!
Thank you!!!!!! 😄
That one took an hour to make, I totally understand you now!
exactly!
Hello mathman
I asked my teacher about this and she said it wasn't possible. Gonna annoy her tomorrow 😂😂
I did not use linear algebra.
Me solution uses complex numbers.
Here it is:
In order to find a mapping transformation for a mirror placed on y=2x i started by making a point in the complex plane such that its angel with the "real" number line is that of the line y=2x and its magnetude is one. Then i made a second, general point that the transformation would act onto. Now, my transformation is the following: to get the output point i multiply the input point by the square of the ratio
((special point)/(input point divided by its magnitude))
In the end i can take the general coordinates of my output complex number and they would be the answer.
Woah! I used linear algebra on a school assignment once. The final for my precal class was to draw a picture using graphs. I used linear transformations to rotate functions to what I wanted. I also used calculus to come up with really accurate features. (What I did was plot the points I wanted to connect with my equation, then found the slope between the points and created a derivative graph. Then used regression and integration to smooth things out)
Integration in precalc? That's way overkill
what an absolute banger of a video
You could easily do it for any linear equation y=mx+n and a point P(a,b) by finding the perpendicular line to the linear function that also passes through the point, which will be y=((a-x)/m)+b, and then you can find the intersection, which is the midpoint of P and P', so you can find P'
Dr Peyam! Two days ago I just had an important exam which had an exercise about this. If only you would have uploaded this video earlier 😢
Awwwww!!! It was on my playlists though!
Not completely related to the video, but I found a function that describes part of the graph e^x flipped across y = 2x. It has the Lambert W function in it, here it is
[For x>0.7726, y
I've found the same matrix by simply solving a system of equations { T(1, 2) = (1, 2) , A(-2, 1) = (2, 1) }, where T is a linear operator with matrix { { a , b } , { c , d } } in Cartesian coordinate system.
Why not just solve the system of equations using the known points @4:36 ? You have a + 2b = 1 and c + 2d = 2, as well as -2a + b = 2 and -2c + d = -1. This easily becomes 5b = 4 and b = 4/5, and then a = -3/5, and similarly 5d = 3 and d = 3/5, c = 4/5
You could get the same results with 3 simple tranformations? Rotation[[cos(tan^-1(.5)), sin(tan^-1(.5))],[-sin(tan^-1(.5)),cos(tan^-1(.5))] Reflection[[-1,0],[0,1]] Rotation[[cos(-tan^-1(.5)),sin(-tan^-1(.5))],[-cos(-tan^-1(.5)),sin(-tan^-1(.5))]]
Rotate your axis of reflection so that it becomes the y axis, reflect around the y-axis, then rotate back.
I did this without linear algebra, and was honestly quite easy, but the linear algebra way is very elegant :)
Very cool!
easy to do without linear algebra. For example reflect the point (6,3) about line y = 2x
step one: calculate equation of line with (6,3) that is perpendicular to y = 2x
it is y = -1/2 x + 6
step two: use systems of equations to solve for point of intersection which is (12/5, 24/5)
third: since this is midpoint the reflected point easy to solve for (-6/5, 33/5)
Lot easier!
But systems of equations *is* linear algebra, that’s the whole point!! I’m just doing a more systematic way of what you’re doing!
Reflect about the line x sin(t) = y cos(t):
if we rotate the plane such that the line meets the x axis, then negate, then rotate back, it should work. so let's take the formula: e^-it * [1,0;0,-1] * e^it, e^it being a rotation matrix. The first two multiply to give [cos,sin;sin,-cos], and multiplying that by e^it gives [cos^2-sin^2, 2sincos; 2sincos, sin^2-cos^2] which if i recall is equivalent to the quaternion (cos^2-sin^2)i + (2sincos)k = (1-2sin^2)i + (2sincos)k for any angle input to the sines and cosines. Odd how this is supposed to map from 2d to 2d... is it really guaranteed that this quaternion works?
in this case, the input vector should be formatted as a complex number ix + ky, and multiplying this by ai+bk we get -(ax +by) -(ay+bx)j, which seems like a reasonable linear combination of x and y, though i can't prove the quaternion solution i remain confident on the matrix one. really it should work with the proper map for (x,y) -> [quaternion]
Algebraic solution: take (x,y) -> x+iy = z. c(z) = x-iy, the complex conjugate. z' = c(z * e^-it) * e^it, as the conjugate flips about the x axis, we simply need to rotate into our new basis, since the line of reflection is indeed a line which is conservative in terms of area and even the position of the origin point in this case. If, however, the line is moved H units away from the origin along a perpendicular line (the exact distance can be determined through basic algebra), simply subtract H before conjugating. The linear algebra version of this is just replacing each function with its matrix
it also seems that by your solution sincos = m/(1+m^2) and (1-m^2)/(1+m^2) = cos^2-sin^2, if 1+m^2 = sec^2, then 1-m^2 = 1 - tan^2, yes m = tangent, this makes total sense. tan/(1+tan^2) = tan/(sec^2) = sincos. our formulas totally agree under m = tangent, and your solution operates on the tangent (the point 1,M) whereas mine operates on the unit circle (a,b where aa+bb=1)
Thank you so much about your knowledge sharing
The truth I love your videos I am your fan and I would like you to share more math topics that this matery I love I sleep two hours to study it, to be a great mathematician like you and I would like you to share topics, only basic topics like theorem of pick and others I hope you share more topics: '(
The Linear Algebra in this is too hard for me :/
Is there a recommendation of where to learn this?
All my playlists!
@@drpeyam Thanks but which one should I start with?
Start with Linear Equations (Lay Chapter 1), finish all the Lay ones, then do all the Friedberg ones
Woahhh, its beautiful
T is much nicer if you use y=(tanθ)x instead of y=mx
It really do be like that sometimes
Half of that kinda flew over my head because I dont know all of the stuff that this builds off of but I think I understand
Check out my Linear Transformations Playlist: ua-cam.com/play/PLJb1qAQIrmmCGJgXVb1gt0q1weM_Bt6aE.html
I think finding the reflection about y = mx + c with algebra only is not too difficult.
Hello, I am Mexican, I dream some day to be like you, to know a lot of mathematics,
Is it bad that I'd rotate and rotate back?
Inefficient 😂
I derived the formula for reflection across y=mx using vector projections! ua-cam.com/video/JHQtA6R7fYc/v-deo.html
Sadly I never learned the basis stuff that you use in this video, but I have my own methods!
This video reflects well on you.
Hahaha, pun intended? 😉
@@drpeyam yes :-)
It's special orthogonal group 2!
Reflection about y=mx+b ?
I mentioned that in the video, you do a translation first
am whoops my mistake, I ment reflection of a function.
imgur.com/a/QuopTEn
To be fair it is easier with simple Cartesian geometry. Always love linear algebra tho
I hope and see it I'm your fan :'(💔
YAY | YAY
Oooooon!!!! 😄😄😄
@@drpeyam What's up???!!!