Problem 3 - Precalculus

Question

An angle of rotation is specified, followed by the coordinates of a point in the \(x^{\prime}-y^{\prime}\) system. Find the coordinates of each point with respect to the \(x\) -y system. $$\theta=45^{\circ} ;\left(x^{\prime}, y^{\prime}\right)=(\sqrt{2},-\sqrt{2})$$

Step-by-Step Solution

Answer

The coordinates in the \(x-y\) system are \((0, -2)\).

Step 1: Recognize the Rotation Matrix Components

To find the coordinates after rotation back to the original system, identify the components of the rotation matrix. The rotation matrix for an angle \( \theta \) is given by: \[ R = \begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix} \] Since \( \theta = 45^{\circ} \), the trigonometric components \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \) and \( \sin 45^{\circ} = \frac{\sqrt{2}}{2} \).

Step 2: Write the Inverse Rotation Matrix

To find the original coordinates \((x, y)\) from the transformed coordinates \((x', y')\), use the inverse of the rotation matrix. The inverse of an orthogonal rotation matrix is its transpose, so: \[ R^T = R^{-1} = \begin{bmatrix} \cos 45^{\circ} & \sin 45^{\circ} \ -\sin 45^{\circ} & \cos 45^{\circ} \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix} \]

Step 3: Apply the Inverse Rotation Matrix

Apply the inverse rotation matrix to the given point \((x', y') = (\sqrt{2}, -\sqrt{2})\) to find the original coordinates. Compute: \[ \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} \sqrt{2} \ -\sqrt{2} \end{bmatrix} \]

Step 4: Compute the Matrix Multiplication

Calculate the matrix multiplication: \[ \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} \frac{\sqrt{2}}{2}\sqrt{2} + \frac{\sqrt{2}}{2}(-\sqrt{2}) \ -\frac{\sqrt{2}}{2}\sqrt{2} + \frac{\sqrt{2}}{2}(-\sqrt{2}) \end{bmatrix} \] This simplifies to: \[ \begin{bmatrix} 1 + (-1) \ -1 + (-1) \end{bmatrix} = \begin{bmatrix} 0 \ -2 \end{bmatrix} \]

Step 5: Interpret the Results

The original coordinates \((x, y)\) of the point in the \(x-y\) system are \((0, -2)\). This means that post rotation of \(45^{\circ}\), the point from the \(x'-y'\) coordinate \((\sqrt{2}, -\sqrt{2})\) is situated at \((0, -2)\) in the original coordinate plane.

Key Concepts

Rotation Matrix

A rotation matrix is a mathematical tool used to rotate points in a coordinate system. When we need to rotate a point or vector in a 2D plane by an angle \( \theta \), the rotation matrix comes into play.
The formula for the rotation matrix is:

• \( R = \begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix} \)

For our problem, we are given that \( \theta = 45^{\circ} \). Using basic trigonometry, we know:

• \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \)
• \( \sin 45^{\circ} = \frac{\sqrt{2}}{2} \)

This matrix helps transform coordinates from one system to another by rotating them around the origin. It is crucial in physics, engineering, and computer graphics for rotating objects and vectors.

Inverse Rotation

To reverse a rotation that has been applied to a coordinate point, we use the inverse of the rotation matrix. Thankfully, the rotation matrices are orthogonal, which simplifies the process greatly.
An orthogonal matrix has the property that its transpose is also its inverse. Therefore, the inverse of our rotation matrix \( R \) is:

• \( R^{-1} = R^{T} = \begin{bmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{bmatrix} \)

For \( \theta = 45^{\circ} \), the inverse is:

• \( R^{-1} = \begin{bmatrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix} \)

Using this matrix, we can find the original coordinates from transformed coordinates. In our case, we went back from coordinates \((x', y')\) to \((x, y)\) using this approach.

Trigonometry

Trigonometry plays a vital role in understanding rotations in coordinate transformations. At the heart of this is the understanding of sine and cosine functions, which describe the relationship between angles and ratios in right triangles.
When dealing with rotations, the cosine of an angle gives us the horizontal component, while the sine provides the vertical component.

• For a 45-degree angle, we have: \( \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2} \)

These values are essential when constructing rotation matrices. They ensure that we properly rotate coordinates in the plane either clockwise or counterclockwise based on the given angle \( \theta \). This concept is foundational for correctly approaching any rotation problem in mathematics and physics.

Coordinate Systems

Understanding different coordinate systems and how to transform between them is key in fields like physics and computer graphics.
Coordinate systems, such as the \( x-y \) plane or its rotated counterpart \( x' - y' \), allow us to specify the position of points. By using transformation tools like rotation matrices, we can switch between these systems.

• In our exercise, the point \((\sqrt{2}, -\sqrt{2})\) is described in terms of a rotated coordinate system \( x' \)-\( y' \).
• Our goal was to find its equivalent location in the original \( x \)-\( y \) system, which we determined to be \((0, -2)\).

Thus, understanding how to navigate between coordinate systems using rotation is invaluable for analyzing and solving problems that involve movement and orientation.