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Magazine Chapter 4: Problem 17
In parts (a) through (e), find an equation of the image of the line \(y=2 x\) under (a) a shear of factor 3 in the \(x\) -direction. (b) a compression of factor \(\frac{1}{2}\) in the \(y\) -direction. (c) a reflection about \(y=x\) (d) a reflection about the \(y\) -axis. (e) a rotation of \(60^{\circ}\) about the origin.
Short Answer
Expert verified
(a) \(y = \frac{2}{7}x'\) (b) \(y' = x\) (c) \(y = \frac{1}{2}x\) (d) \(y = -2x\) (e) Complex, involves solving \(y' = mx'\).
Step by step solution
01
Shear in the x-direction
To find the new equation of the line after a shear of factor 3 in the x-direction, we apply the transformation \( x' = x + 3y \). Substitute \( y = 2x \) into the shear equation:\[ x' = x + 3(2x) = x + 6x = 7x \]Thus, the new equation remains \( y = 2x \), but in terms of \( x' \) and \( y \), we have \( x' = 7x \). Therefore, the transformed line in terms of \( x' \) translates to:\[ y = \frac{2}{7}x' \].
02
Compression in the y-direction
Apply a compression of factor \(\frac{1}{2}\) in the y-direction on the line \(y = 2x\). This means \( y' = \frac{1}{2}y \).Substitute \( y = 2x \) into this expression:\[ y' = \frac{1}{2}(2x) = x \]Thus, the equation of the line after the compression is:\[ y' = x \].
03
Reflection about y=x
To reflect the line \(y = 2x\) about the line \(y = x\), we swap the coordinates \((x, y)\) to \((y, x)\).Thus, the equation of the transformed line will be:\[ x = 2y \]This can be rewritten as:\[ y = \frac{1}{2}x \].
04
Reflection about the y-axis
To perform a reflection about the y-axis, change \(x\) to \(-x\). Starting with \(y = 2x\), replace \(x\) with \(-x\). The new equation will be:\[ y = 2(-x) = -2x \]Thus, the equation of the line after the reflection is:\[ y = -2x \].
05
Rotation by 60 degrees
To rotate the line \(y = 2x\) by 60 degrees around the origin, use the rotation matrix.The rotation matrix for 60 degrees is:\[\begin{bmatrix}\cos 60^{\circ} & -\sin 60^{\circ} \\sin 60^{\circ} & \cos 60^{\circ}\end{bmatrix}= \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}\]Apply this to the line equation:If \(y = 2x\), let \((x', y')\) be the rotated coordinates:\[x' = \frac{1}{2}x - \frac{\sqrt{3}}{2}y \quad \text{and} \quad y' = \frac{\sqrt{3}}{2}x + \frac{1}{2}y\]Using \(y = 2x\), we solve:\[x' = \frac{1}{2}x - \frac{\sqrt{3}}{2}(2x) = \frac{1}{2}x - \sqrt{3}x = \left( \frac{1}{2} - \sqrt{3} \right)x \]\[y' = \frac{\sqrt{3}}{2}x + \frac{1}{2}(2x) = \frac{\sqrt{3}}{2}x + x = \left( \frac{\sqrt{3}}{2} + 1 \right)x \]Solving \( y' = m x' \) gives us the slope:\[ m = \frac{\left( \frac{\sqrt{3}}{2} + 1 \right)}{\left( \frac{1}{2} - \sqrt{3} \right)} \]Thus, the equation is quite complex and typically solved numerically in practical scenarios.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Shear Transformation
A shear transformation alters the shape of an object by sliding its layers sideways. Picture it like pushing the top of a rectangle sideways while keeping the bottom fixed. This action distorts the object without changing its area or parallel lines. In the context of our exercise, we're applying a shear transformation to the line equation \( y = 2x \).
When we shear in the \( x \)-direction with a factor of 3, the transformation equation becomes \( x' = x + 3y \). Plugging \( y = 2x \) into this transformation gives us \( x' = 7x \). Therefore, the line after shear appears as \( y = \frac{2}{7}x' \). This new line shows how the vertical movement of each point on the line causes a horizontal displacement by a factor of 3 per unit of \( y \).
Key points:
When we shear in the \( x \)-direction with a factor of 3, the transformation equation becomes \( x' = x + 3y \). Plugging \( y = 2x \) into this transformation gives us \( x' = 7x \). Therefore, the line after shear appears as \( y = \frac{2}{7}x' \). This new line shows how the vertical movement of each point on the line causes a horizontal displacement by a factor of 3 per unit of \( y \).
Key points:
- Shear does not change line slopes directly in the original coordinate system, but affects their expression when using the new coordinates.
- The angle between lines may change.
Compression Transformation
Compression transformation is like squeezing an object along one axis, reducing its size in that dimension while leaving the other dimensions unaffected. Imagine pressing down a spring; that鈥檚 the visual effect a compression can create. For the line \( y = 2x \), applying a compression in the \( y \)-direction at factor \( \frac{1}{2} \) means \( y' = \frac{1}{2}y \).
By substituting \( y = 2x \) into \( y' = \frac{1}{2}y \), we get \( y' = x \). The transformation yields a line that swings closer to the \( x \)-axis, effectively halving the vertical stretch of the original line.
Core insights include:
By substituting \( y = 2x \) into \( y' = \frac{1}{2}y \), we get \( y' = x \). The transformation yields a line that swings closer to the \( x \)-axis, effectively halving the vertical stretch of the original line.
Core insights include:
- Compression impacts the line slope by altering its steepness.
- The direction of the line remains unchanged; only its angular position to the axes is adjusted.
Reflection Transformation
Reflection is a familiar transformation that flips an object over a certain line, much like creating a mirror image. In mathematics, a reflection can occur over different axes or lines.
In this exercise, the first reflection occurs about the line \( y = x \). This operation swaps the coordinates \( (x, y) \) to \( (y, x) \), transforming \( y = 2x \) into \( x = 2y \). Thus, re-expressing this as \( y = \frac{1}{2}x \). The second reflection we consider is about the \( y \)-axis, where we substitute \( x \) with \( -x \) in \( y = 2x \), leading to \( y = -2x \).
Reflections hinge on:
In this exercise, the first reflection occurs about the line \( y = x \). This operation swaps the coordinates \( (x, y) \) to \( (y, x) \), transforming \( y = 2x \) into \( x = 2y \). Thus, re-expressing this as \( y = \frac{1}{2}x \). The second reflection we consider is about the \( y \)-axis, where we substitute \( x \) with \( -x \) in \( y = 2x \), leading to \( y = -2x \).
Reflections hinge on:
- Reversing the equality side for line reflection also inverting or switching coordinates.
- Changing sign transforms the line slope across the axis being reflected.
Rotation Transformation
A rotation transformation consists of spinning a shape around a point, commonly the origin, changing its orientation but preserving its size and shape. It's akin to turning a wheel on an axle. Mathematically, this involves a rotation matrix.
For the line \( y = 2x \), rotating it 60 degrees about the origin uses a matrix: \[\begin{bmatrix}\cos 60^{\circ} & -\sin 60^{\circ} \\sin 60^{\circ} & \cos 60^{\circ}\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}\] Applying this matrix to each point on the line changes the coordinate position, producing \((x', y')\) based on \( x = \cos(60^{\circ})x - \sin(60^{\circ})2x \) and \( y = \sin(60^{\circ})x + \cos(60^{\circ})2x \). The new line equation emerges from balancing these coordinates in a new space.
Consider these points:
For the line \( y = 2x \), rotating it 60 degrees about the origin uses a matrix: \[\begin{bmatrix}\cos 60^{\circ} & -\sin 60^{\circ} \\sin 60^{\circ} & \cos 60^{\circ}\end{bmatrix} = \begin{bmatrix}\frac{1}{2} & -\frac{\sqrt{3}}{2} \\frac{\sqrt{3}}{2} & \frac{1}{2}\end{bmatrix}\] Applying this matrix to each point on the line changes the coordinate position, producing \((x', y')\) based on \( x = \cos(60^{\circ})x - \sin(60^{\circ})2x \) and \( y = \sin(60^{\circ})x + \cos(60^{\circ})2x \). The new line equation emerges from balancing these coordinates in a new space.
Consider these points:
- Rotation preserves the magnitude of vectors鈥攐nly their direction changes.
- With trigonometric functions, the angle of rotation is crucial in determining the post-rotation position.
