Rotation of Axes - bye bye xy

[@skylerpretto1221]

This really is the only video one needs to watch about rotation of axes.

[@ForPubg-g6g]

Unbelievable 😍, I understand 3 hrs lecture in 15 minutes. Thank you sir.

[@hairyfiddler2274]

U helped group of engineering students, thanks sir .

[@WatermelonMoon260]

Thanks for this informative and quick but dense video. This supplemented my original understanding of rotations and helped me understand them better.

[@Huiando]

Lovely way to covert hairy equation to a nice ellipse, parabola or hyperbola forms. Haven't thought rotations could be so useful. Thank you!

[@itsmmdoha]

This really is the only video one needs to watch about rotation of axes. (2)

[@savannahconlin6865]

This video is the only thing that has helped me understand this subject. Thank you!!

[@yolo1867]

wow what an amazing lecture on rotation/conics. covered everything i wanted. thanks sir.

[@a__f45yay72]

This concept became crystal clear to me,thank you very much for explaining and make it so much easier to me

[@lucascruz3977]

Thank you so much! Finally I understand how the conversion between the systems of coordinates is done.

[@michaelpallmer819]

I believe the rotation angle should be about 26.57 degrees and the sine of theta should be 1/(5)^.5

[@gkdgnomer]

GREATEST INTRO EVER SIR

[@jr01theweeb]

Idk why its so hard to find the formula on 3:51 on the internet. Thank you for helping me convert from and to rotated axis!

[@lepakshikumar1154]

Thanks sir The method u provided for rotation of axis is quite easy to understand as compared to the traditional method

[@zombockerman]

Very informative, loved the intro. Many thanks

[@kartikeyayadav9088]

Plz solve 5x^2 +5y^2 -8xy-9=0 On solving we get circle rotated by 45° Which does not change anything But on wolframalpha it shows an ellipse rotated by 45°

[@math4every1]

I am not sure how you got a circle. You do have A = C = 5 initially but if you have an xy middle term the graph could still be anything. If you look at the test at the end of the video B^2 - 4AC determines the shape. in this case B = -8 so B^2-4AC < 0 which makes it an ellipse. However if say B = 20, B^2 - 4AC > 0 and the graph would be a hyperbola and if B=10 then B^2 - 4AC = 0 and the graph is a parablola. When I made the substitution for 45 degrees my new equation had: A' = 5 - 8 + 5 = 2 B' = -10 + 10 = 0 C' = 5 + 8 + 5 = 18 after dividing by 2 x^2 + 9y^2 = 9 , which is an ellipse.

[@thomasolson7447]

I'm guessing all the algebra associated with x^2±x*y+y^2=1 is just the circle. It somehow exists because of the cosine law. So, I'm guessing there is an algebra for the ellipsoid to. Which is a rotated/skewed sphere from conics. Or maybe the cosine law only works in 2d. I would need to think about this.

[@alexzubkov9551]

got a precalc test tomorrow, wish me luck

[@That-Green-guy73]

It's been 3 years, how'd it go?

[@Lylia-rk7ft]

Rotate the coordinate axes to remove the xy-term and sketch the graph of 3x+6√3xy+9y+(2√3 -4)x-(4√3 +2)y+20 . How can I solve it?

[@math4every1]

cot (2*theta) = (A - C)/B = (3 - 9)/6*sqrt(3) = -1/sqrt(3). This implies 2*theta = -60, hence theta = -30 degrees is the angle you need to rotate through to get rid of the xy term. Hence plug in x = x'cos(-30) - y'sin(-30) and y = x'sin(-30) + y'cos(-30). This should give you a conic section with no xy term. Then just graph that on an axis system rotated -30 degrees.

[@alejandroalvarez4814]

Thanks! Have a test that includes this tomorrow!

[@boss-fh6oz]

Hello, thanks for making this video. What if my answer for the last question was (x'^2/4) + (y'^2/16) = 1. Would it be wrong? (Whether the answer is Yes or No, please state why)

[@issammaine1349]

That was beautiful, thank you.

[@吴陈炜]

Thank you for your video, sir. However, I've got a problem. When we do the transformation to 3x^2 + 2xy+y^2 = 1, if I choose cot(2theta)=1-3/2=-1, I would got a ellipse, while I choose cot(2theta)= 3-1/2=1, I would not get the the xy item eliminated. It is really a headache. Would you please explain it. I will appreciate it.

[@math4every1]

I am assuming an algebraic mistake. cot(2theta) = 1 implies theta = 22.5 = 45/2. So cos(theta) = sqrt(2 +sqrt(2))/2 and sin(theta) = sqrt(2 - sqrt(2))/2. If you call these C and S respectively you can see that 2*C*S = sqrt(2)/2 and C^2 - S^2 = sqrt(2)/2. If you use these for your substitution your x'y' term should have a coefficient of -3(2CS) + 2(C^2 - S^2) + (2CS) = (-3 + 2 +1)sqrt(2)/2 = 0.

[@吴陈炜]

@@math4every1 Thank you for you reply. I recognized that I've made the the mistake. I got the proof at www.maa.org/external_archive/joma/Volume8/Kalman/General.html

[@codewithmeenakshi]

Thank you so much Sir. I needed it... now my doubt is clear.. Thanks Sir!!

[@wahabfiles6260]

Is this similar to eigen vector transformation to remove correlation? Does xy represent correlation? What is the significance of 'xy' why remove it?

[@embo1502]

What do you do if there are neither A or C terms?

[@math4every1]

That was my very first example: xy = 18. You do the same thing.

[@arongil]

Thanks for the comprehensive video.

[@arongil]

I made a graph on Desmos (a great online graphing site) demonstrating the math you explained. www.desmos.com/calculator/6hys5gkvtq

[@Happen2Bme]

Thank you so very much for giving us knowledge.

[@jaysontran6307]

Hello, Glen Gray. Can you help me find the points of intersection between an ellipse and a line? The line is defined through two points, while the ellipse knows the center coordinates, the radius of the short axis, the radius of the long axis, and an angle of inclination compared to the Ox axis. Thank you very much.

[@math4every1]

Jayson, You should be able to figure the equation in X and Y for the ellipse using the info from this video. The next step is usually to find the parametric representation of the line through the two points: X = x1 + (x2 - x1)*t , Y = y1 + (y2 - y1)*t . If you then substitute these into your equation for an ellipse you will get a 2nd degree equation in the single variable t. Solve this for t and plug back into your parametric equations to find (x, y) at your intersection.

[@jaysontran6307]

@@math4every1 Thank you very much

[@jaysontran6307]

I have 1 more question. How to calculate the swept angle of the center point between 2 direction points 0 o'clock and 3 o'clock. Does it have anything to do with the two-axis lengths. Thanks. I have 1 more question. How to calculate the swept angle of the center point between 2 direction points 0 o'clock and 3 o'clock. Does it have anything to do with the two-axis lengths. Thanks. i.imgur.com/iEKxi6Q.png

[@math4every1]

@@jaysontran6307 This is just a vector calculation. If P1 and P2 are two points on your ellipse and C is the center point then let V1 = P1 - C and V2 = P2 - C . Then just use the dot product V1*V2 = |V1|*|V2|*cos(theta) where theta is the angle between the two vectors.

[@jaysontran6307]

I'm really sorry, my ultimate goal is to find the coordinates P3. Point P3 is the central point on the arc P1, P2. Up to now I have only calculated the coordinates P1, P2 and C. imgur.com/WZBKiJ8

[@nz7467]

THANKS!!! This helped so much

[@chileshechitabanta9995]

Its all clear now, i have just found out the whole idea about conic sections

[@ozzyfromspace]

This was awesome! Thanks 😊🙏🏽🙌🏽

[@عطاالدهامشة-ع8ح]

Dear sir How can rotate y=cos(x) With best wishes

[@math4every1]

Just use the substitution described at the 6:07 time stamp. Is this really what you want to do?

[@عطاالدهامشة-ع8ح]

@@math4every1I mean, after rotation, a difficult equation appears, so how do we find the value of y through Lambert's equation?

[@satishsaini2551]

excellent work sir...

[@FernandoBelloDeveloper]

GREAT VIDEO! Thanks! :D

[@hamzamajid8539]

I actually can't get how we derive the angle of rotation's equation.

[@math4every1]

We plug our rotation transformation into the original equation which has the xy term we wish to remove. At about the 8:51 mark you can see the result of this substitution and the coefficients of the new equation. We use the coefficient of the new XY term and set that equal to 0 because we want it to disappear. This is around the 9:19 mark. This gives us the desired angle in terms of the old A, B, C coefficients. In particular cot(theta) = (A - C)/B.

[@procerpat9223]

outstanding !

[@akshatjangra4167]

Nice song taste !!!

[@0enzogamer0]

Thanks a lot

[@ikarus1955]

Thankfully

[@bishalsarmah7920]

Why coefficient of x'y'=0

[@math4every1]

We are trying to convert this back into our old form Ax**2 + Cy**2 + Dx + Ey + F = 0 where it is easy classify and graph the equation as one of our standard conics (parablola, ellipse, hyperbola).

[@nashvaughan3175]

Excellent video. Thank you!

[@renecabrera7224]

Thanks

[@jdcurrie1974]

is there a way to rotate solid 3d graphs by rotation of axes.

[@math4every1]

Yes of course, but that would take another video. Essentially you do a double rotation, like once in the xz plane and then around the z axis (xy plane). You typically go to a 3x3 matrix which is a product of smaller transforms.

[@jeatig]

What would (x', y') be if the point (-2, 6) was rotated counterclockwise 30°??

[@math4every1]

This would be the same as rotating the axes clockwise 30 degrees. x' = (-2)cos(-30) + 6sin(-30) = -sqrt(3) - 3 y' = -(-2)sin(-30) + 6cos(-30) = -1 + 3sqrt(3)

[@pedronpetrus826]

I owe you an A-level

[@feruruguayo]

muchas gracias.

[@golden_fpv9023]

Well made video, needs much better explained steps.

[@cklecture9203]

very good

[@muhammadseyab9032]

helping video

[@Snoo29293]

I'm studying web development and apparently math and physics seem to be important so I got to study those stuff now.

[@krizz-404]

omg thank you soooo much

[@strugglingcollegestudent]

Thank you (:

[@toddbiesel4288]

It appears that you are saying cosine when you mean cotangent.

[@math4every1]

Yes, sorry about that. The slide is correct. Sometimes my mouth moves along quicker than my brain. It is more used to saying cosine than cotangent and I do hurry a bit to keep the videos reasonably short.

[@ahmedsherif2182]

You r awesome

[@vaishnavipal2298]

Your voice is not clear there are disturbances and noise

[@summykumar8891]

are u a form of god ?..