Tutorial:Bloque de Prácticas

Resoluciones Gráficas de Sistemas de (In)Ecuacions

Un sistema de ecuaciones lineales puede quedar diseñado como un escenario para la resolución gráfica de modo tal que en cada caso se ingresen los valores correspondientes y sea posible representar ágilmente una variedad de problemas de tal tipo.

Pasos de Construcción

1 Como se anotarán dos funciones de formato m x + b, es preciso ingresar en primer lugar, desde la Barra de Entrada, cada componente necesario, empezando por...

• m_1 = 1 b_1 = 3
• m_2 = 2 b_2 = 1

... para terminar por:

• m_1 x + b_1
• m_2 x + b_2

Todo esto aparecerá registrado en la Vista Algebraica

Nota: Para que los valor de los parámetros de una y otra ecuación lineal aparezca representado como un deslizador en la Vista Gráfica, basta con tildar con un clic cada redondelito que lo encabeza en la Vista Algebraica.

2 Los valores de cada parámetro determinará la inclinación y la posición de la recta que representa la ecuación y para que todo sea más fácil de ser identificado, conviene darles los mismos colores contrastantes a los de una y otra ecuación.

Nota: En la figura se exponen...

• los deslizadores a los que se les asignó orientación Vertical en la casilla correspondiente de la pestaña Deslizador de la Caja de Diálogo de Propiedades,
• el rótulo de cada ecuación de recta a la que, además, se les cambió en nombre por f_1 y f_2 respectivamente.

3 Se puede establecer, con la herramienta , el punto de intersección entre una y otra función graficada.

Nota: Es conveniente activar la Barra de Estilo de la Vista Gráfica para agilizar cada una de las operaciones descriptas y, respecto del punto recién creado así como para cada ecuación, tener un modo sencillo de dejar expuesto del rótulo, específicamente el Valor o el Nombre y Valor.

4 Con la correspondiente herramienta , se puede establecer el texto en el lugar adecuado de la Vista Gráfica y anotar el texto estático: Solución: x = y seleccionar el nombre del punto de intersección del listado que se despliega a indicar Objetos.

Nota: Si se prefiriera identificar cada uno de los valores, se podría, en la caja de edición...

• modificar la casilla que contiene el nombre del punto de intersección, digamos A para reemplazarlo por x(A) y
• hacer nuevamente otro tanto para y(A).

5 Completar los detalles del texto dinámico y mejorar los del gráfico para que el escenario gráfico de resolución quede alistado para resolver toda una variedad de problemas.

Nota: Además de cuestiones casi cosméticas, es de importancia, darle a cada uno de los deslizadores un rando de valores y uno de incremento de modo que se puedan representar cada uno de los ejercicios que se destine a este encuadre de resolución gráfica.

6 Un desafío adicional sería el de cuestionar cuál de las inecuaciones se vincula a los sectores sombreados a uno y otro lado del punto de intersección y si hay situaciones en que se modifica esta correspondencia.

Nota:

El par de inecuaciones que se ingresan por Barra de Entrada y también se establece su texto dinámico representativo correspondiente, son...

• in_1: (y < m_2 x + b_2) ∧ (y > m_1 x + b_1)
• in_2: (y > m_2 x + b_2) ∧ (y < m_1 x + b_1)

... como puede observarse en la figura, en la que además se aprecia que los sombreados respectivos son rayados con un ángulo de 0 y 90 grados.

7 Antes de guardar el boceto del escenario creado, es conveniente dejar anotadas en un archivo de texto asociado, la serie de ejercicios y problemas a resolver empléandolo.

Challenge

Create a similar construction that allows for visualizing the graphical solution of a system of quadratic polynomials.

Aviso: Functions need to be entered using the syntax f(x) = …

Nota: Such a dynamic figure can also be used to visualize an equation in one variable by entering each side of the equation as one of the two functions.

Translating Pictures

In this activity you are going to use the following tools and commands. Make sure you know how to use each tool and command before you begin.

	Insert Image
	A = (1, 1)
	Polygon
	Vector[O, P]
	Vector Between Two Points
	Translate Object By Vector
	Move
	Text

Construction Steps

1. Open a new GeoGebra window. Show the Algebra View, Input Bar, coordinate axes, and grid. In the Options Menu set the point capturing to Fixed to Grid.

2. Insert picture A_3b_Bart.png into the first quadrant.

3. Create points A = (1, 1), B = (3, 1), and D = (1, 4).

4. Set point A as the first, B as the second, and D as the fourth corner point of the picture (Properties Dialog, tab Position).

5. Create triangle ABD.

6. Create point O = (0, 0) and point P = (3, -2).

7. Create vector u = Vector[O, P]. {{hint|You could also use [[tool Vector Between Two Points.}}

8. Translate the picture by vector u using Translate Object by Vector.

Aviso: You might want to reduce the filling of the image.

9. Translate the three corner points A, B, and D by vector u.

10. Create triangle A'B'D'.

11. Hide point O so it can’t be moved accidentally. Change the color and size of objects to enhance your construction.

Challenge

Insert dynamic text that shows

• the coordinates of points A, B, C, A', B', and D'.
• the coordinates of vector u.

Constructing a Slope Triangle

In this activity you are going to use the following tools and algebraic input. Make sure you know how to use each tool and the syntax for algebraic input before you begin.

	Line Through Two Points
	Perpendicular Line
	Intersect Two Objects
	Polygon
	rise = y(B) - y(A)
	run = x(B) - x(A)
	slope = rise / run
	Text
	Midpoint or Center
	Move

Construction Steps

1. Show the Algebra View, coordinate axes and the grid. Set point capturing to Fixed to Grid and the labeling to All new objects.

2. Create line a through two points A and B.

3. Construct a perpendicular line b to the y-axis through point A.

4. Construct a perpendicular line c to the x-axis through point B.

5. Intersect perpendicular lines b and c to get intersection point C.

Aviso: You might want to hide the perpendicular lines.

6. Create polygon ACB and hide the labels of the sides.

7. Calculate the rise: rise = y(B) - y(A)

Aviso: y(A) gives you the y-coordinate of point A.

8. Calculate the run: run = x(B) - x(A)

Aviso: x(B) gives you the x-coordinate of point B.

9. Enter the following equation into the input bar to calculate the slope of line a: slope = rise / run

10. Insert dynamic text: rise = and select rise from Objects, run = and select run from Objects, slope = and select slope from Objects

11. Change properties of objects in order to enhance your construction.

Challenge 1: Insert a dynamic text that contains a fraction

Using LaTeX formulas, text can be enhanced to display fractions, square roots, or other mathematical symbols.

1. Activate tool Insert text and click on the Graphics View.
2. Type slope = into the Insert text window’s input bar.
3. Check LaTeX formula and select Roots and Fractions a/b from the dropdown list.
4. Place the cursor within the first set of curly braces and replace a by number rise from the Objects drop-down list.
5. Place the cursor within the second set of curly braces and replace b by number run from the Objects drop-down list.
6. Click OK.

Challenge 2: Attach text to an object

Whenever an object changes its position, attached text adapts to the movement and follows along.

1. Create midpoint D of the vertical segment using tool Midpoint or center.
2. Create midpoint E of the horizontal segment.
3. Open the Properties Dialog and select text1 (rise = …). Click on tab Position and select point D from the drop-down list next to Starting point.
4. Select text2 (run = …) in the Properties Dialog and set point E as starting point.
5. Hide the midpoints D and E.

Exploring the Louvre Pyramid

In this activity you are going to use the following tools and some algebraic input. Make sure you know how to use every single tool and the syntax for algebraic input before you begin. Also, check if you have the picture A_3d_Louvre.jpg saved on your computer.

	Insert Picture
	Line Through Two Points
	Slope
	Angle
	New Point
	Perpendicular Line
	Intersect Two Objects
	Show/Hide Object
	Segment Between Two Points
	Move

The Louvre in Paris is one of the most visited and famous art museums in the world. The building holds some of the world's most famous works of art, such as Leonardo da Vinci's Mona Lisa. In 1989 the main entrance of the museum was renovated and a glass pyramid was built (from http://en.wikipedia.org/wiki/Louvre, February 20, 2008).

Determine the slope of the pyramid’s faces

1. Set point capturing off. Set the decimal places to 1. Change the labeling setting to All new objects (menu Options).

2. Insert the picture A_3d_Louvre.jpg into the first quadrant of the coordinate system. The left lower corner should match the origin.

3. Reduce the filling of the picture (about 50%) and set it as background image (Properties Dialog).

4. Create a line through two points with the first point at the base and the second point at the vertex of the pyramid.

Aviso: Change the properties of line to improve its visibility.

5. Use the Slope Tool to get slope triangle of line.

Aviso: Change the properties of slope triangle to improve its visibility. The slope triangle is attached to the point created first.

6. Task: Determine the slope of the pyramid’s faces in percent.

7. Create the angle between the x-axis and the line. Task: Determine the inclination angle for the pyramid’s face.

Challenge

The pyramid’s base is a square with a side length of 35 meters. Determine the height of the pyramid using similar triangles.

1. Create a new point C on the line.

2. Construct the slope triangle of the line using points C and B at the pyramid’s vertex.

Aviso: Create a line perpendicular to the y-axis through point C and a line perpendicular to the x-axis through point B at vertex of pyramid. Create the intersection point D of the two lines.

Aviso: Hide the auxiliary lines.

3. Use segments to connect point D with points B and C.

Aviso: Change the properties of the segments to increase their visibility.

Aviso: You might want to rename the vertical segment to height and the horizontal segment to halfBase

4. Move point C along the line until the horizontal segment of the triangle matches the level of the road in front of the pyramid.

5. Task: Calculate the height of the pyramid using similar triangles.

Aviso: Use the slope triangle and your new triangle. Remember that the base side length is 35 m.

Check your answer with GeoGebra

6. Show name and values of segments height and halfBase.

7. Drag point C until the horizontal segment has length 35/2 = 17.5.

Aviso: You might need to zoom out of the construction and / or move the graphics view in order to be able to do this.

8. Check if the height of the pyramid matches your answer.

Comment

By implementing the instructions above you were able to graphically determine the approximate value for the pyramid’s height. In reality, the Louvre pyramid has a base length of 35 m and a height of 21.65 m. Its faces have a slope of 118% and an inclination angle of bout 52° (from http://de.wikipedia.org/wiki/Glaspyramide_im_Innenhof_des_Louvre#Daten, February 22, 2008).