Quadratic equations in rotated form

Some long, time-consuming problems

The last three weeks have been very busy for me. I have been tutoring all math subjects, from fractions to Statistics and multivariate calculus.
Looking back over these past weeks it all seems kind of blurred but one topic stands out from the rest because, by coincidence, I had two sessions on the same topic with two different students, both during last week.
The topic in question is the rotation of quadratic equations in the two-dimensional coordinate (x,y)-plane. It had been a long time since I last taught this subject. It does not come up very often in my tutoring sessions, so I noticed the coincidence when I had two different students independently reviewing with me these geometrical transformations in the same week.
Also, each student separately made the same comment after we worked out problems of this type about quadratic equations: “Wow! This is a lot of work!”
They are right, it is a lot of work. The general problem starts with a quadratic equation like, for example, 5x²+2xy+10y²-12x-22y+17=0,
with a non-zero coefficient in the “xy” term.
The goal of the exercise is to find a specific angle, let’s call it θ, so that the transformed (rotated) equation in the alternate variables x’ and y’ lacks the x’y’ term.
The variables x and y are connected to x’ and y’ by means of these two equations:
x = x’ cos θ – y’ sin θ
y = x’ sin θ + y’ cos θ
Solving these problems requires several steps. I list them here, hopefully without going into too much detail:
First, finding the value of tan(2θ), the tangent of the angle double of θ.
Second, finding the measure of the angle θ itself.
Third, finding the values for cos θ, sin θ, and their squares.
Fourth, plugging those trigonometric values into the formulas below to find the new coefficients for the transformed quadratic equation:
A’ = A cos² θ + B sin θ cos θ + C sin² θ
B’ = 0
C’ = A sin² θ – B sin θ cos θ + C cos² θ
D’ = D cos θ + E sin θ
E’ = E cos θ – D sin θ
F’ = F
where A, B, C, D, E, and F are the coefficients of the original equation.
So you can see each one of these problems involves a lot of algebraic and trigonometric calculations. These problems are long, time-consuming, and you have to pay very close attention to all details to ensure an accurate result.
Anyway, in the video below you can see a room-size metallic structure (some kind of architectural sculpture) where Richard Serra, the artist, incorporated two congruent ellipses, one at the base of the room, and the other formed by the upper edge of the wall. The two ellipses are identical in shape but they are rotated with respect to each other. This is a real, tangible example of the rotation of a conic section. It is relevant to this post because quadratic equations represent conic sections, like the ellipses we see in the video. It is a very interesting structure. Take a look:

Solving the Rubik's Cube Puzzle

Step-by-step solution in a couple of YouTube videos by Dan Brown.

I just signed-up to YouTube yesterday, and this post is mostly meant as practice for myself posting videos into my blog.
While exploring YouTube’s archives I found a few videos about solving the Rubik’s Cube puzzle. Rubik’s cube is one of my favorites puzzles because it is closely related to both Group Theory and Graph Theory, branches of modern math. Playing with Rubik’s cube also helps somehow develop one’s intuition about the Cartesian (x, y, z) coordinate system in 3-D space.
In the two videos below, Dan Brown incorporates a little algebraic notation to precisely describe a few sequences he uses in his general solution of the Rubik’s cube.
So far I have not used Rubik’s cube as a teaching aid in any of my tutoring sessions, so this post really does not necessarily have a lot to do with tutoring but I decided to include it anyway because the puzzle does have to do with math, and it is fun.
I hope you will enjoy the videos!

P.S. After loading these first videos I decided to search for other videos with content related to that of my previous posts, so I will be including some more videos in those older posts too.

Solving Equations For a Particular Variable

A very basic principle

An equation has one equal sign.

The equal sign divides the equation into left hand side and right hand side.

The two sides may look totally different from each other as expressions but the equal sign says their numerical value has to be the same.

The fundamental principle of equations says that, when two expressions have the same numerical value, if we apply one operation to both expressions, the resulting expressions after the operation is performed will also be equal in value. They will be equal not to the original expressions, but to each other.

So, if A, B and C are three algebraic expressions, and we have the equation A = B, then all of the following will also be valid equations:

A + C = B + C

A - C = B - C

(A)(C) = (B)(C)

A/C = B/C [provided C is not zero]

A^2 = B^2

Square root of A = Square root of B

This fundamental principle is used over and over to solve equations for specific variables, one step at a time.

For example, in solving for x the equation (3x + 1)/2 = 5y - 4, we can do it like this:

1) Multiply both sides by 2 and we get

3x + 1 = 2(5y - 4)

2) Subtract 1 from both sides and we get

3x = 2(5y - 4) - 1

3) Divide both sides by 3 and we get

x = ( 2(5y - 4) - 1)/3

Now the equation has been solved for x in a series of steps, where each step consists of applying one and the same operation to BOTH sides of the equation.

The fact that the resulting expression for x can be simplified to

x = (10y -9)/3

is not relevant here. I am only illustrating the process we use to isolate x one step at a time by applying the same operation to both sides of the equation.

The following YouTube video from InterAlgebra12 shows several more examples: