Problem 42
Question
How do you obtain the angle of rotation so that a general second-degree equation has no \(x^{\prime} y^{\prime}\) -term in a rotated \(x^{\prime} y^{\prime}\) -system?
Step-by-Step Solution
Verified Answer
The angle of rotation can be found using the equation: \( \theta = \frac{1}{2} arctan (\frac{-B}{A-C}) \) when \( A!=C \), or \( \theta = \frac{\pi}{4} \) when \( A=C \). This angle enables the change of the system to eliminate the \(x^{\prime}y^{\prime}\) term from the equation.
1Step 1: Basic understanding
A general second-degree equation can be represented in the form: \(Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0\)where \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) are constants. The objective here is to eliminate the term containing the product of \(x\) and \(y\), i.e, \(xy\) in the rotated system \(x^{\prime}y^{\prime}\). The process involves finding an angle \( \theta \) in which the equations are rotated.
2Step 2: Coordinate Transformation
Shifting to a different, rotated coordinate system involves Coordinate Transformation. This is achieved using Rotation Matrix, given by\[\(\begin{bmatrix} x^{\prime} \ y^{\prime} \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}\)\]where \(x^{\prime} = x\cos \theta - y\sin \theta\)and\(y^{\prime} = x\sin \theta + y\cos \theta\)Here, \( \theta \) is the rotation angle.
3Step 3: Find the Angle of Rotation
The angle of rotation \( \theta \) can be deduced from the equation: \(B = 2(A-C)tan 2\theta\). From this equation, we can find \( \theta \) as:\( \theta = \frac{1}{2} arctan (\frac{-B}{A-C}) \) if \(A-C!=0\),or just \( \theta = \frac{\pi}{4}\) if \(A=C\).
Key Concepts
Understanding Second-Degree EquationsCoordinate Transformation FundamentalsThe Role of the Rotation MatrixElimination of the Cross Product Term
Understanding Second-Degree Equations
A second-degree equation in two variables, such as in the form of a quadratic equation, represents a curve called a conic section. These equations can generally be expressed as \(Ax^{2} + Bxy + Cy^{2} + Dx + Ey + F = 0\), where \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) are constants. They describe various conic sections—circles, ellipses, parabolas, and hyperbolas—depending on the values of these constants.
When dealing with the equation in the context of coordinate transformation, our main goal is to simplify the equation for better understanding and ease of analysis. Simplification can often involve removing the \(xy\) term, which is where finding the angle of rotation becomes crucial. The removal of the cross product term \(xy\) simplifies the classification of the conic section represented by the equation.
When dealing with the equation in the context of coordinate transformation, our main goal is to simplify the equation for better understanding and ease of analysis. Simplification can often involve removing the \(xy\) term, which is where finding the angle of rotation becomes crucial. The removal of the cross product term \(xy\) simplifies the classification of the conic section represented by the equation.
Coordinate Transformation Fundamentals
Coordinate transformation is the process of changing the coordinates from one system to another. In our context, it refers to rotating the original \(x, y\) coordinate system to a new \(x',y'\) coordinate system. By doing this, we can often make complex equations simpler.
To perform such a transformation, we use what is known as a rotation matrix. This matrix mathematically rotates points in the coordinate plane around the origin by a certain angle, making it a powerful tool in simplifying second-degree equations by eliminating the \(xy\) term.
To perform such a transformation, we use what is known as a rotation matrix. This matrix mathematically rotates points in the coordinate plane around the origin by a certain angle, making it a powerful tool in simplifying second-degree equations by eliminating the \(xy\) term.
The Role of the Rotation Matrix
The rotation matrix is instrumental in performing rotations in the coordinate plane. It is a square matrix used to rotate vectors in a two-dimensional plane, given by the following formula: \[\begin{bmatrix} x' \ y' \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}\].
Here, \(\theta\) is the angle of rotation, and \(x\) and \(y\) represent coordinates in the original system, while \(x'\) and \(y'\) are the coordinates after rotation. This matrix is essential because it defines how each point in the plane is recalculated to its new position as a result of the rotation.
Here, \(\theta\) is the angle of rotation, and \(x\) and \(y\) represent coordinates in the original system, while \(x'\) and \(y'\) are the coordinates after rotation. This matrix is essential because it defines how each point in the plane is recalculated to its new position as a result of the rotation.
Elimination of the Cross Product Term
The elimination of the cross product term \(xy\) from the second-degree equation through rotation is essentially a method of simplifying the equation into a more recognizable form without this particular term. By doing so, the conic section can be more easily classified and analyzed.
During the rotation process, the \(B\) coefficient of the \(xy\) term can be set to zero by choosing an appropriate rotation angle \(\theta\). This angle can be calculated using the formula \(\theta = \frac{1}{2} \arctan(\frac{-B}{A-C})\) if \(A eq C\), and if \(A = C\), simply using \(\theta = \frac{\pi}{4}\), which eliminates the \(xy\) term and simplifies the equation significantly.
During the rotation process, the \(B\) coefficient of the \(xy\) term can be set to zero by choosing an appropriate rotation angle \(\theta\). This angle can be calculated using the formula \(\theta = \frac{1}{2} \arctan(\frac{-B}{A-C})\) if \(A eq C\), and if \(A = C\), simply using \(\theta = \frac{\pi}{4}\), which eliminates the \(xy\) term and simplifies the equation significantly.
Other exercises in this chapter
Problem 42
Identify the conic that each polar equation represents. Then use a graphing utility to graph the equation. $$ r=\frac{18}{6-6 \cos \theta} $$
View solution Problem 42
In Exercises \(37-50,\) graph each ellipse and give the location of its foci. $$ \frac{(x-3)^{2}}{9}+\frac{(y+1)^{2}}{16}=1 $$
View solution Problem 42
Eliminate the parameter. Write the resulting equation in standard form. An ellipse: \(x=h+a \cos t, y=k+b \sin t\)
View solution Problem 42
In Exercises 35–42, find the vertex, focus, and directrix of each parabola with the given equation. Then graph the parabola. $$ (y-1)^{2}=-8 x $$
View solution