CSUMB College Geometry-Fall 2015

Applications of Transformations

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Whitney Chadwick, Zach Dickinson, Guadalupe Esquivel12/2/2015

Question 1

Function of Transformation Graph of Transformation Transformation Description

Unit Square Unit Square – 1 x 1 square

Vertices:A= (1, 1)B= (1, 2)C= (2, 2)D= (2, 1)

(a) f(x, y) = (x3, y3) The transformation is a bjiective function. It expands the unit square to a similar square.

Vertices:A= (1, 1)B= (1, 8)C= (8, 8)D= (8, 1)

(b) g(x, y)=(2x, 3y) The transformation is a bjiective function. It expands the square, but does not expand x and y at a constant rate, so it changes the shape to a rectangle.

Vertices:A= (2, 3)B= (2, 6)C= (4, 6)D= (4, 3)

(c) h(x, y) = (3√x, ey) The transformation is not a bijection. It expands one two sides of the square while shrinking the other two which creates a rectangle.

Vertices:A= (1, 2.72)B= (1, 7.39)C= (1.26, 7.39)D= (1.26, 2.72)

(d) i(x, y) = (cos x, sin x) The transformation is not a bijection. It changes the unit square into a line since we are only using the x values.

Vertices:A= (0.54, 0.84)B= (0.54, 0.84)C= (-0.42, 0.91)D= (-0.42, 0.91)

(e) j(x, y) = (-x, x+3) The transformation is a not a bijection. It changes the unit square into a line since we are only using the x values.

Vertices:A= (-1, 4)B= (-1, 4)C= (-2, 5)D= (-2, 5)

(f) l(x, y) = (x3-x, y) The transformation is a bijection. It expands two of the sides and keeps two the same, thus changing the shape from a square to a rectangle. Our y length value doesn’t change, but the x value is expanded.

Vertices:A= (0, 1)B= (0, 2)C= (6, 2)D= (6, 1)

(g) m(x, y) = (3y, x+2) The transformation is a bijection. The x length value is expanded because we have an exponent of 3 in our x position and our shape is shifted 2 to the right from the +2 in the y position.

Vertices:A= (3, 3)B= (6, 3)C= (6, 4)D= (3, 4)

(h) n(x, y) = (x+2, y-3) The transformation is a bijection. It shifts the original unit square to the right 2 units and down 3 units.

Vertices:A= (3, -2)B= (3, -1)C= (4, -1)D= (4, -2)

(i) p(x, y) = (0.6x-0.8y, 0.8x+0.6y) The transformation is a bijection. It rotates the unit square.

Vertices:A= (-0.2, 1.4)B= (-1, 2)C= (-0.4, 2.8)D= (0.4, 2.2)

(j) q(x, y) = (-0.6x+0.8y, 0.8x+0.6y) The transformation is a bijection. It rotates the original unit square.

Vertices:A= (0.2, 1.4)B= (1, 2)C= (0.4, 2.8)D= (-0.4, 2.2)

I would recommend using a unit square to ensure that someone develops a good intuition about these transformations because the unit square allows us to see how the shape changes in

all directions. For example, if we were to use a vertical line in transformation (g), we would only think that the shape shifts to the right instead of also noticing that it also expands horizontally.

Question 2The first student uses the vector to translate the points. He does not do it the way we have learned to do it but instead he does two different translations using the slope of the original vector.

The student is given vector AB but to do the transformation he is using vector AF when he says “over 3” and uses vector FB when he says “up 2.”

The second student does the first step exactly like the first student to translate point C. To get D’’, instead of using vector AF and FB, like student 1, this student decides to use the line segment that connects point C and D as the vector, but uses vector CG when he says over 3 and vector GD when he says “up 3.” And instead of starting with point D, he starts with C’’. The image below shows where the student chose to get “Over 3” and “Up 3”

The image below shows how C went to C’’ and how D when to D’’.

Both of these are invariants of the translation the students were supposed to do. Using Geogebra, a direct translation was done using the original vector. As shown below, all the vertices of the translated triangle lie on the same spot where the students showed they would land.

The same processes would not be useful if we are looking for an image of a figure under rotations or reflections. The image below is the reflection of the triangle about a line. The same process the students used would not be helpful in this situation because the orientation is changed which changes the slope of the line segments of the new triangle.

But if we were to reflect the triangle on the line x=c or y=c where c is a constant, then the method the students used would work with a slight adjustment. The first point would need to be found using reflecting methods but the rest of the vertices can be found using the methods of the second student by

using the line segments of the original triangles as the vector. The slopes of the corresponding sides are opposites of each other, so the direction of one of the vectors would change. In the example below the reflecting line is some x=c so the direction of the Right 3 vector changed to Left 3 in the reflection. If the reflecting line were y=c then “Up” would change to “down” in the reflection and vice versa.

The image below is the rotation of the triangle using point L. Similarly like the reflection, using the method the students used would not be helpful because the orientation of the triangle is changed; causing the slopes of the line segments to change.

Question 3

The student is incorrect the orientation is not reversed. It looks like the image is facing the opposite direction but it’s not. The student may have thought that it was reflected and if an image is reflected then the orientation is reversed. In this case, the triangle ABC can actually be rotated around the point D to get to triangle A’B’C’.

The orientation of the image is only reversed if the image is reflected. To help the student see the difference, I would show him to get to triangle A’B’C’ you need to reflect twice like in the figure below.

When you reflect an image twice, the orientation is flipped twice. Therefore the image keeps its original orientation. I would tell the student if he is confused he should use reflection to map the image onto other image to see if the orientation is reversed.