using symbolic geometry to teach secondary school ... 2 and... · the national council of teachers...

Using Symbolic Geometry to Teach Secondary School Mathematics - Geometry Expressions Activities for Algebra 2 and Precalculus

Irina Lyublinskaya,

CUNY College of Staten Island, Staten Island, NY, USA

and

Valeriy Ryzhik, Lycee-Physical Technical High School, St. Petersburg, Russia

Saltire Software Inc., Beaverton, OR USA

www.saltire.com

Copyright ” Saltire Software 2008 ISBN 1-882564-10-3

Table of Contents

Introduction ..............................................................................................................................................5

Discovering Parabolas ..........................................................................................................................7 Part 1 – Parabola by 3 Points....................................................................................................................7 Part 2 – The Existence of a Parabola Passing Through Three Arbitrary Points .....................................12 Extensions: ..............................................................................................................................................14

Solving Systems of Equations (Inequalities) with Parameters ............................................17 Part 1 – Setting Up the Problem in Geometry Expressions ....................................................................17 Part 2 – When the System Has No Solutions .........................................................................................18 Part 3 – Investigation of the Number of Solutions...................................................................................19 Part 4 – Solving Systems of Equations ...................................................................................................20 Part 5 – Solving Systems of Inequalities with Parameters......................................................................22 Extensions: ..............................................................................................................................................22

Stained Glass Design ..........................................................................................................................23 Part 1 – Setting Up Problem in Geometry Expressions ..........................................................................24 Part 2 – Creating Stained Glass Design..................................................................................................25 Part 3 – Finding Equations of the Curves in the Stained Glass Design..................................................28 Part 4 – Verification of the Equations with Geometry Expressions.........................................................29 Extensions ...............................................................................................................................................31

Translation Along Coordinate Axes ...............................................................................................35 Part 1 – Translation Along the Y-Axis .....................................................................................................35 Part 2 – Translation Along the X-Axis .....................................................................................................39 Part 3 – Commutative Property of Translation ........................................................................................43 Part 4 – Applications and Assessment Problems ...................................................................................46 Extension:................................................................................................................................................50

A Little Trig .............................................................................................................................................53 Part 1 – Investigating Area of the Triangle..............................................................................................54 Part 2 – Optimizing the Perimeter of the Rectangle................................................................................58

One Hyperbola .......................................................................................................................................61 Part 1 – Investigating Area of the Rectangle...........................................................................................62 Part 2 – Optimizing Perimeter of the Rectangle......................................................................................64 Part 3 – Optimization of the Diagonal of the Rectangle. .........................................................................66

Three Extrema (Circle) .......................................................................................................................69 Part 1 – Length of a Tangent Segment to a Circle..................................................................................70 Part 2 – Area of a Triangle Formed by a Tangent Line and the Coordinate Axes..................................75 Part 3 – Perimeter of a Rectangle whose Diagonal is a Tangent Segment............................................78

Two Parabolas .......................................................................................................................................83 Part 1 – Optimizing Perimeter of Rectangle............................................................................................84 Part 2 – Optimizing the Diagonal of the Rectangle .................................................................................88 Part 3 – Comparison of Points of Extrema for Perimeter, Diagonal, and Area. ......................................91

5

Introduction The National Council of Teachers of Mathematics (NCTM) Standards advocate a unified approach to mathematics education incorporating multiple strands in coherent focused elements. Emphasis is placed on the use of technology, visual thinking, and the connection between geometry and algebra. The notion of integrating algebra and geometry is at the forefront of mathematics education in the US today, potentially transformative of mathematics education in the 21st century when technology integration is not only reasonable because of access but also because of the nature of the impact of mathematics on knowledge and directions in this century. While geometry and algebra systems have existed separately for the last couple of decades, an application that unifies the two is available through a brand new technology. This unification facilitates a variety of well-understood goals:

• The user interface must be simple so that students can focus on the reasoning and problem solving rather than struggle with the technology.

• The application must enable exploration.

• As much as possible, the technology must blur the artificial separation between the algebra and the geometry.

• It must provide easy access but also be challenging. That is, it must be easily used by students at a wide range of performance levels while also being capable of providing challenge to the most able students.

• It must address traditional concepts in addition to facilitating the use of more modern and realistic approaches to geometry and to real applications.

• The mathematical concepts addressed must be expanded in order to make a major impact on the high school curriculum.

The technology that meets these goals is now available. Geometry Expressions developed by Saltire Software, is a computer application that, unlike other interactive geometry systems, can automatically generate algebraic expressions from geometric figures. A simple algebra system is embedded directly into Geometry Expressions. Generation and simplification of expressions along with simple algebraic manipulations are handled inside Geometry Expressions. From a pedagogical standpoint an interactive symbolic geometry system affords a remarkable opportunity to make concrete the concept of a variable in a readily identifiable real world setting. An educational tool takes advantage of this opportunity if it provides outputs in a symbolic form. These outputs are algebraic expressions using the input parameters. Our goal was to use Geometry Expressions to develop a set of problems for second year algebra and/or precalculus courses addressing the topics of geometric transformations of functions and optimization. In this volume you will find eight interactive problems with different levels of difficulty. In each problem the main focus is on the development of the students’ ability to recognize and make connections using multiple representations of the same object, such as geometric shape and function. By connecting geometric and algebraic representations the student develops a more thorough understanding of the problem.

7

Discovering Parabolas

Time required 90 minutes

Teaching Goals:

1. Students verify that a unique parabola with the equation 2y ax bx c= + + , 0a ≠ , exists through any three given points if and only if these points are not collinear.

2. Students verify that an infinite number of parabolas with the equation 2y ax bx c= + + , 0a ≠ , exist through any two given points.

3. Students should be able to construct the parabola defined by the coordinates of three points and find its equation with the help of the software.

4. Students should be able to construct this parabola by geometric transformations from the graph of the parabola y = x2 and confirm the equation given by the software.

Prior Knowledge

• Students should know the graph of quadratic functions in two forms: 2y ax bx c= + + and 2

0 0( )y a x x y= − + , 0a ≠ , and they should know this graph has the shape of a parabola.

• Students should know geometric representations and coordinate forms of the following transformations: translation along the coordinate axes, reflection about coordinate axes, and dilation along coordinate axes.

Problem:

This is a problem of the existence of a quadratic function (a parabola) as a graph. Since we consider graphs of functions, situations when two points are located on the same vertical line are not considered.

Part 1 – Parabola by 3 Points In this part of the problem students are given three specific points, the maximum, with coordinates (2, 3), and other points with coordinates (0, -1) and (3, 2). They are also given the graph of an initial parabola 2y x= and the graph of a general parabola 2

0 0( )y a x x y= − + . 1. Open a new file. If axes do not appear in the blank document, create them by clicking the

Toggle grid and axes icon on the top toolbar.

2. Select Function from the Draw toolbox. Type x^2 in the Function Type dialog.

3. Select Function from the Draw toolbox. Type y[0]+a*(X-x[0])^2 for the function. Click the graph of the function to highlight it, right click the mouse to invoke the context menu, and select Properties. From the Display Properties dialog choose Line Color – Blue.

8

4. Select Point from the Draw toolbox. Construct three points anywhere on the screen. Select point A, choose Coordinate from the Constrain toolbox and type coordinates (2, 3) for the point A, press Enter. Repeat this operation to constrain points B and C. If desired, change the font, size and color for the labels of the points.

1 2 3 4 5 6 7-1-2

1

2

3

4

-1

-2

B

C

A

Y=X2

Y=y0+a· X-x02

(3,2)

(0,-1)

(2,3)

Q1. Can you find a parabola that has an equation with integer numerical coefficients and will go through these three points? If it exists, what is the equation of this parabola? In order to do that:

Q2. Investigate which parameters in the equation of the parabola 20 0( )y y a x x= + − are

responsible for which type of transformation of the graph.

1. Click-and-drag the parabola. When you move the parabola, a small circle appears on the curve. We refer to this as the “drag handle”. A variable name labels the drag handle. Note which way you are able to move the curve with the parameterized handle (click-and-drag the handle).

2. Click another part of the parabola and another drag handle appears on the curve with another variable label. Create three drag handles for the three variables of the equation. Observe how each handle / variable is responsible for a type of transformation of the curve.

3. Record your observations in the table below:

9

Parameter in the equation 2

0 0( )y y a x x= + − Transformation (vertical translation, horizontal translation, dilation)

a

Y0

X0

A. a – dilation, y0 – vertical translation, x0 – horizontal translation

Q3. Drag and move parabola until it passes through the points A, B, and C. Construct your conjecture about the equation of the parabola (round numerical values to integers).

1 2 3 4 5 6 7-1-2

1

2

3

4

-1

-2

B

C

A

Y=X2

Y=y0+a· X-x02

(3,2)

(0,-1)

(2,3)

a

x0

y0

Students should vary parameters a, x0, and y0 to try to complete the task. Observe the values of the parameters in the Variables toolbox. When they think they have the solution they can round the numeric values of the coefficients with the help of the software and verify their equation by substitution. In the figure above, 0 00.998, 2.003, 3.009a x y≈ − ≈ ≈ .

A. The equation of the parabola is: 23 ( 2)y x= − − . Here are steps for verification:

10

1. In the Variables toolbox, enter the predicted values for the parameters, 0 01, 2, 3a x y= − = = . Change a variable value from the toolbox by clicking the variable row

and entering a new value in the data entry window below the list.

2. Click the graph of the parabola, then from the Calculate toolbox click the Real tab and the Implicit equation icon. The numeric equation for the parabola will appear on the screen

1 2 3 4 5 6 7-1-2

1

2

3

4

-1

-2

B

C

A

⇒ Y=3-(-2+X)2

(3,2)

(0,-1)

(2,3)

Y=X2

Y=y0+a· X-x02

3. Substitute values of all three points and verify that all three points lie on this curve.

Q4. Find geometric transformations that transform the graph of y = x2 into the graph of the parabola in the last problem. Is it important to preserve the order of transformations? Verify your answers with the help of the software.

A. The transformations are: translation by the vector (2, 3) and dilation by the factor -1. These are commutating. Students should complete the construction, verify coincidence of the image with the parabola and the fact that the order of transformation does not affect the result. Here are the steps of the construction:

1. Select Vector from the Draw toolbox and construct a vector from the origin to point A.

2. Click the graph of the parabola y = x2, select Translation from the Construct toolbox and then click the vector. The image of the translated parabola will appear on the graph.

11

3. Click the translated curve, select Dilation from the Construct toolbox. Select point A and leave b in the data entry box for the dilation factor (press Enter to accept the value).

4. In the Variables toolbox setup a range of values for the dilation factor b from -2 to 0. Click variable b (the row will be highlighted), enter -2 in the Minimum value box in the lower left corner and 0 in the Maximum value box in the lower right corner. Drag the slider bar to adjust the curve so that it coincides with the graph of the parabola 23 ( 2)y x= − − . The value of b should be -1.

1 2 3 4 5 6 7-1-2

1

2

3

4

-1

-2

A

C

B

E

⇒ Y=3-(-2+X)2

(0,-1)

(2,3)

(3,2)

Y=X2

Y=y0+a· X-x02

b

Q5. Do you think there exists another parabola passing through points A, B and C? Explain your answer.

A. Three non-collinear points define a parabola uniquely. This can be proved analytically.

12

Part 2 – The Existence of a Parabola Passing Through Three Arbitrary Points

In this part of the problem students investigate how the relative positions of three arbitrary points affects the existence of a parabola passing through these points. Given three points on the plane, , , and A B C , and a parabola defined by the equation 2

0 0( )y y a x x= + − .

1. Open a new file. If the axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar.

2. Select Function from the Draw toolbox. Type y[0]+a*(X-x[0])^2 for the function. Click the graph of the function to highlight it, right click the mouse to invoke the context menu, and select Properties. From the Display Properties dialog choose Line Color – Blue, Line Style – Solid 2.

3. Select Point from the Draw toolbox. Construct three points anywhere on the screen.

4. Select point A and the parabola. Select Incident from the Constrain toolbox and point A will appear on the curve. Repeat the same procedure with the points B and C.

Q1. How can you position points A, B, and C, so that the parabola no longer exists, e.g. the shape of the parabola changes into a different shape?

A. When points lie on the same line, the parabola becomes a line. Since we use the equation of parabola in the form 2

0 0( )y y a x x= + − the line will always be horizontal in the form y = constant and a = 0. Students can verify this with following the steps:

1. Move all points to lie on the line y = some constant (approximately). The parabola will look like a horizontal line.

2. Observe the value of parameter a in the Variables toolbox. The value of a is close to zero. Assume that a = 0 and enter 0 for the value of a.

13

2 4 6-2-4-6

2

4

6

-2

-4

B

C

A Y=y0+a· X-x02

3. Check the real coordinates of points A, B, and C and. Make sure to use the Real tab in the Calculate toolbox. Select each point, one at a time and choose Coordinates from the Calculate toolbox.

4. Click the line, and select Implicit equation (Real tab). The equation of the line will appear on the screen confirming the fact that the line is not a quadratic function.

14

2 4 6-2-4-6

2

4

6

-2

-4

B

C

A

⇒ ∼(1.0093458,2)

⇒ Y=2

⇒ ∼(-2.0747664,2)⇒ ∼(3.0093458,2)

Y=y0+a· X-x02

Extensions:

1. The existence of a parabola: given a parabola defined by the equation 2y ax bx c= + + and three points, A, B, C, follow steps similar to those above to determine that if three points are collinear, then a = 0, when the points lie on an inclined line with slope b, and therefore the parabola does not exist.

15

2 4 6 8 10-2-4-6-8-10

2

4

6

8

10

12

14

B

C

A

⇒ ∼(3.9874,8.9748001)

⇒ Y=1+2·X

⇒ ∼(2.4454303,5.8908605)

⇒ ∼(1.0093458,3.0186916)

Y=X2·a+X·b+c

2. The existence of a parabola through any two points: given two points, the choice of a 3rd point defines a unique parabola, so there are an infinite number of parabolas since there are an infinite number of choices for the 3rd point. In this case the teacher can provide students with conditions based on real problems, for example:

a) given the point of a basketball shot and the position of the basket, make the basket

b) make a pass with a volleyball at its highest point (vertex)

c) trace a soccer ball kick so that it hits the right upper corner of the goal.

17

Solving Systems of Equations (Inequalities) with Parameters

Time required 90 -135 minutes

Teaching Goals:

1. Students apply a graphical method for solving systems of equations (inequalities) with parameters.

2. Students apply geometric transformations of graphs of functions in order to solve systems of equations (inequalities) with parameters.

3. Students obtain solutions of systems of equations (inequalities) in symbolic form. Prior Knowledge

• Students should know the graphical method of solving equations and inequalities without parameters.

• Students should know graphs of linear functions and the equation of a circle with the center at the origin.

• Students should know that the equation y = f(x) + a corresponds to the translation of the graph of the function y = f(x) by a along the y-axis.

Problem:

Investigate the following system of equations with a parameter and find its solutions:

2 2 1y x ax y

= +

+ =

Part 1 – Setting Up the Problem in Geometry Expressions

1. Open a new file. If axes do not appear in the blank document, create them by clicking the Toggle grid and axes icon on the top toolbar.

2. Select Line Segment from the Draw toolbox and construct segment AB with point A at an arbitrary position on the y-axis. Do not put point A on the origin. Click the selection arrow to return to select mode.

3. Select the line segment and then choose Slope from the Constrain toolbox. In the data entry box type 1 for the slope followed by the Enter key.

4. Repeat steps 2-3 for the other segment, AC, with the slope -1.

As a result the graph should look like the function y = |x| on the restricted domain shown below.

18

5. Drag point A up and down the y-axis. If the construction is made correctly the shape of the graph should remain the same and the point A should remain on the y-axis.

6. Select Circle from the Draw toolbox and construct a circle with the center at the origin. Return to select mode and double-click the label of the center of the circle and re-name it to be point O.

7. Select the circle and choose Radius from the Constrain toolbox. In the edit box enter 1 to fix the radius of the circle to be constant and equal to 1.

8. Select point O, click Coordinate in the Constrain toolbox and type (0, a).

9. The drawing is shown on the picture below. The problem is now set up and ready for investigation.

Part 2 – When the System Has No Solutions

Q1: Consider a > 1. Check that the system of equations has no solutions.

A: It is expected that students will be moving the point A up and observing the values of parameter a in the Variables toolbox. Students can see that when a > 1 the function no longer intersects the circle, so there are no solutions.

Q2: Are there any other values of parameter a when the system has no solutions?

A: It is expected that students will move point A down until the graph of the function does not intersect the circle and their answer will be: yes, there are values of a when there are no solutions.

Q3. Can you find these values?

A: In order to answer this question, students should realize that they need to determine values of parameter a when the line segments are tangent to the circle. In order to do that, the distance from the center of the circle to the segment should be equal to the radius of the circle.

19

The following steps should be followed in order to do that:

1. Select the center of the circle and one of the segments (hold the CTRL key to select two objects simultaneously). Choose Distance/Length from the Calculate toolbox, Symbolic tab.

2. The distance value will appear on the screen.

1 2 3-1-2-3

1

2

-1

-2

-3

B

A

O

DC

⇒- 2 ·a

2

1

1

(0,a)

-1

3. Student should solve 2 12

a⋅= for a and decide what values of a will lead to no solutions

( 2a < − ).

4. After students solve this problem, delete the measurement from the screen.

Part 3 – Investigation of the Number of Solutions

In this part of the problem students investigate conditions when the system of equations has one, two, three, or four solutions. The main task of this part of the problem is not to solve the system of equations, but to consider the problem qualitatively.

Q4: How many solutions can this system have when parameter a is changing?

A: Expected answer: the system can have one, two, three, four or no solutions.

20

Q5: What are the values of parameter a when the system has one, two, three or four solutions.

A: Students should move point A and observe the values of parameter a (the y coordinate of point A) in the Variables toolbox or on the graph. They should have seen already that when two segments were tangent to the circle, the system had two solutions, and they already determined the value of a at this tangency to be 2a = − . The next set of values for a where the function intersects the circle in 2 points only, occurs when 1 1a− < < . When a = -1 there are three points of intersection and when a = 1 there is one point of intersection. The system has 4 solutions when 2 1a− < < − , where each segment intersects the circle twice. The teacher could expand question 5 with more specific details, such as: what are the values of parameter a when at least one of the solutions is in the 1st quadrant (2nd, 3rd, 4th); or when is the solution larger than the absolute value of 0.5 (or any other specific value), etc.

Part 4 – Solving Systems of Equations

Q6: Find the solution of the system of equations for values of parameter a in the cases where the solutions exist.

A: One of the approaches that students might take is to set up a diagram when all four solutions exist and to find intersection points and their coordinates using Geometry Expressions. Here are the steps for this procedure:

1. Select the circle and one of the line segments (hold the CTRL key to select both objects simultaneously). Choose Intersection from the Construct toolbox. Points of intersection will appear on the screen.

2. Repeat the same procedure for the second segment. For convenience you may change the label of the points of intersections.

3. Select one of the points of intersection and choose Coordinates from the Calculate (Symbolic) toolbox. The coordinates of this point (in terms of parameter a) will appear on the screen.

4. Repeat that for each point of intersection.

21

1 2 3-1-2-3

1

2

-1

-2

-3

A

N

C

KO

M

D

L

B

⇒-a2

-2-a2

2,a- 2-a2

2⇒a+ 2-a2

2,a- 2-a2

2

⇒-a+ 2-a2

2,a+ 2-a2

2⇒

a- 2-a2

2,a+ 2-a2

2

1-1 1

(0,a)

Comment: in order to find points of intersection between the curves, Geometry Expressions calculates equations for the objects on the graph. It finds the equation 2 2 1x y+ = for the circle and linear equations without restrictions for the line segments. Thus, when the function is moved up and no longer has all four points of intersection with the circle, the program will still show the points of intersection as if the line segments were infinite lines.

22

1 2 3-1-2-3

1

2

3

-1

-2

O

DB

A

C

⇒ - 2 ·X- 2·Y+ 2·a=0 ⇒ - 2 ·X+ 2·Y- 2·a=0

⇒ -1+X2+Y2=0

1

1

-1

(0,a)

Q7: Check that the number of solutions that you observed graphically in part 3 is consistent with the number of solutions you observed symbolically in part 4.

A: It is expected that students will substitute special values of a in the solutions and show that

a) when a = - 1, points L and M are the same solutions, and K and N are different;

b) when a = 1 all points K and N have the same solutions, but points L and M will evaluate to (1,0) and (-1, 0) and will look like two additional solutions; however, they are extraneous solutions due to the fact that Geometry Expressions found intersections of the circle with the extended lines and not the actual line segments. The same case applies if students substitute values in the range -1< a < 2 . Students should be able to exclude extraneous solutions supplied by the software.

Part 5 – Solving Systems of Inequalities with Parameters This problem can be expanded to solving the following inequalities or system of inequalities with a parameter:

21x a x+ ≤ ± − ; 21x a x+ ≥ ± − ; 2 21 1x x a x− − ≤ + ≤ −

Extensions: For all these extensions the approach will be similar to one considered in the problem above. 1. Given y x= on a restricted domain, 2 2( ) 1x y a+ − =

2. Given y x a= − on a restricted domain, 2 2 1x y+ =

3. Given y x= on a restricted domain, 2 2( ) 1x a y− + = More generally, both shapes translate on an arbitrary vector (a,b).

23

Stained Glass Design

Time required 45 - 90 minutes

Teaching Goals:

1. Students apply graphic methods to design various shapes on the plane. 2. Students apply geometric transformations of graphs of functions in order to design

various shapes on the plane. 3. Students find equations of all curves obtained by geometric transformations in general

form in either Cartesian or parametric form as defined by the teacher.

4. Students will determine domains of the independent variable for the equations of all curves for the design of the shape they construct in either Cartesian or parametric form as defined by the teacher.

5. Students will verify these results with the help of the software.

Prior Knowledge

• Students should know graphs of linear functions and the equation of a circle in Cartesian and parametric forms.

• Students should know the coordinate form of a translation by a vector along each coordinate axis.

• Students should know that the equation y = f(x – a) corresponds to the translation of the curve with the equation y = f(x) by vector (a, 0) along the x-axis.

• Students should know that equation y = f(x) + b corresponds to the translation of the curve with the equation y = f(x) by vector (0, b) along the y-axis.

• Students should know that a counter clockwise rotation has a positive angle and a clockwise rotation has negative angle, with angles in the range [0, π].

• Students should know that the equation y = - f(x) corresponds to the reflection of the curve with the equation y = f(x) about the x-axis.

• Students should know that equation y = f(-x) corresponds to the reflection of the curve with the equation y = f(x) about the y-axis.

Problem:

Construct a stained glass design as shown in the picture below. This design is created by using the curves given by the following equations: y x= on [-3, 3] and 2 2 2x y a+ = on [0, a ]. Try to recreate a similar design and to create other designs that could be created by these curves.

24

Determine equations of all curves on your design with their domain and verify them with the help of the software.

Comment: the idea of this problem is to be able to send this information to another person who should be able to recreate this design based on the information provided (not by mindless copying).

Part 1 – Setting Up Problem in Geometry Expressions


2. Select Function from the Draw toolbox. In the Type: box click the down arrow on the right and select Parametric from the list. In the next row, X= , enter: T ; in the next row, Y=, enter: abs(T); for Start: enter: -3; for End: enter: 3.

3. Click the equation of the function, right-click and select Hide from the context menu. Select the curve, right-click and select Properties from the context menu. Click the Line Style row and it’s down arrow to change it to Dot.

4. Select Function from the Draw toolbox. In the Type: box click the down arrow on the right and select Parametric from the list. In the next row, X= , enter: a*cos(T) ; in the next row, Y=, enter: a*sin(T); for Start: enter: 0; for End: enter: 1.57.

5. Click the equation of the function, right-click and select Hide from the context menu. Select

25

the curve, right-click and select Properties from the context menu. Click the Line Style row and it’s down arrow to change it to Dot.

1 2 3 4 5-1-2-3-4-5

1

2

3

4

-1

-2

-3

Part 2 – Creating Stained Glass Design

Q1: How can you construct a square from the given curves?

A: It is expected that students will use translation along y-axis and reflection about x-axis according to the following steps, for example:

1. Click the straight line function and select Translation from the Construct toolbox. Construct a vector of translation from the origin down along the y-axis. The translated function will appear on the screen.

2. Select the end point of the translation vector and click Coordinate in the Constrain toolbox. Enter (0,-3) for the coordinates of the endpoint of the translation vector.

3. Select the translation vector, right-click and select Hide from the context menu.

4. Select the translated function. Right-click and select Properties from the context menu. Change the Line Color to Blue and the Line Style to Solid 3.

26

5. Click the translated function and select Reflection from the Construct toolbox. Click the x- axis and the function will be reflected over the x-axis. Adjust the line color and style as in step 4 above.

1 2 3 4 5-1-2-3-4-5

1

2

3

-1

-2

-3

-4

A

B (0,-3)

Q2: How can you construct a quarter circle with its center at the bottom vertex of the square?

A: Students may suggest rotating the given quarter circle by 4π

about the origin and then

translating the center of the circle down to the point (-3, 0). Here are the steps:

1. Select the arc and click Rotation from the Construct toolbox. Click the origin and the box will appear for entering the angle. (If the Symbols toolbox is not displayed, click View / Tool Panels / Symbols) π can be found in the lower right corned of the dialog. Enter π/4.

2. Select the angle of rotation label and Hide it.

3. Select the rotated arc and click Translation from the Construct toolbox. Construct the translation vector from the origin to the point (0,-3). Hide the intermediate copy of the arc and the translation vector.

4. Change the new arc’s line color to Red and the line style to Solid 3.

Q3. How can you construct a quarter circle with the center in the top vertex of the square?

27

A: Students may suggest similar steps as above, but the simplest way to do this is to reflect the newly constructed arc about the x-axis. The following steps will accomplish this:

1. Click the arc with its center at the bottom vertex of the square. Select Rotation from the Construct toolbox and select the x-axis. The image will appear on the screen.

2. Select the arc and right-click Properties to change the line color to Red and the line style to Solid 3.

1 2 3 4 5-1-2-3-4-5

1

2

3

-1

-2

-3

-4

B

A

(0,-3)

Q4: How can you construct a quarter circle with the center in the left vertex of the square?

A: Students may suggest rotating the given quarter circle by -π/4 about the origin and then translating the center of the circle down to the point (-3, 0). Here are the steps:

1. Select the original arc and click Rotation from the Construct toolbox. Click the origin and the angle edit box will appear. From the Symbols toolbox, find π and type: -π/4.

2. Select the angle of rotation label and Hide it.

3. Select the rotated arc and click Translation from the Construct toolbox. Construct the translation vector from the origin to the point (-3, 0). Hide the intermediate copy of the arc and the translation vector.

4. Select the arc and right-click Properties to change the line color to Red and the line style

28

to Solid 3.

Q5. How can you construct a quarter circle with the center in the right vertex of the square?

A: Students may suggest similar steps as above, but the simplest way to do this is to reflect the arc with the center at the left vertex about the y-axis. The following steps will accomplish this:

1. Click the arc and select Reflection from the Construct toolbox. Click the y-axis and the image will appear on the screen.

2. Select the image and right-click Properties to change the line color to Red and the line style to Solid 3.

3. Click-and-drag the original quarter circle arc to adjust the transformed arcs to the desired size.

1 2 3 4 5-1-2-3-4-5

1

2

3

-1

-2

-3

-4

B

AD

(0,-3)

(-3,0)

a

Part 3 – Finding Equations of the Curves in the Stained Glass Design

In this part of the problem students have to determine the equation of each curve or function produced by transformation of the original curves. Based on the students’ prior knowledge, the teacher can determine in which form students will write the equations of the curves: Cartesian or parametric. Students should also determine the domain of the independent variable for the

29

equations of each curve. This should be completed independently and then verified with the help of the software.

A. A Square is formed by two curves: 3, [ 3,3]y x= − − produced by the equation

, [ 3,3]y x= − translated by vector (0, - 3) and 3, [ 3,3]y x= − + − produced by equation

, [ 3,3]y x= − reflected about x-axis and translated by vector (0, 3). The inner design is made of four quarter circles. When the circles are tangent, the radius a of the circle is equal to half of

the side of the square, 3 2

2a = , and the x-coordinates of the points where the arcs intersect

the squares are -1.5 and 1.5. The student can create the equations of the arcs as bounded circles centered at the origin, translated to the vertices of the square. Thus, the equations of the arcs of the circles follow.

At the bottom vertex (0,-3): the circle 2 2 92

x y+ = , translated by vector (0,-3) becomes

2 2 9( 3)2

x y+ + = , on [-1.5, 1.5] ; at the top vertex (0, 3): the bottom arc reflected about x-axis

becomes 2 2 9( 3)2

x y+ − = on [-1.5, 1.5]; at the left vertex (-3, 0): the circle 2 2 92

x y+ = ,

translated by vector (-3,0) becomes 2 2 9( 3)2

x y+ + = , on [ 1.5, 3 1.5 2]− − + ; and at the right

vertex (3, 0): the left arc reflected about y-axis becomes 2 2 9( 3)2

x y− + = , on [3 1.5 2,1.5]− .

Students may have difficulty finding the domains of the functions.

Part 4 – Verification of the Equations with Geometry Expressions

Here are the steps to find the equations using Geometry Expressions:

1. Select each curve, one at a time, and choose Implicit Equation from the Calculate toolbox. The equation of the curve will appear on the screen.

30

1 2 3 4 5-1-2-3-4-5

1

2

3

-1

-2

-3

-4

D

B

A

⇒ 9-6·X+X2+Y2-a2=0

⇒ Y=- -3+ X

⇒ 9+6·X+X2+Y2-a2=0

⇒ 9+X2+6·Y+Y2-a2=0

⇒ 9+X2-6·Y+Y2-a2=0

⇒ Y=-3+ X

(0,-3)

(-3,0)

Students should substitute the value of a and verify that the equations given by the software are the same as ones they derived. In order to verify points of intersections, students can construct a square and a circle in a new document and determine points of intersection using the following steps:


2. Select Point from the Draw toolbox and construct four points for the vertices of the square.

3. Select one of the points, choose Coordinate from the Constrain toolbox and type (0, 3). Constrain the other points with the coordinates (0, -3), (3, 0) and (-3, 0).

4. Select Line Segment from the Draw toolbox and connect the points to construct a square.

5. Select Circle from the Draw toolbox and construct a circle with the center at the point (3, 0). With the circle selected choose Radius from the Constrain toolbox and enter a for the value.

6. Holding the CTRL key, select the circle and a segment in the 1st quadrant. Choose Intersection from the Construct toolbox. The point of intersection will appear.

7. Select the intersection point, choose Coordinate from the Calculate (Symbolic) toolbox and the coordinates of the intersection point will be displayed.

31

2 4 6-2-4-6

2

4

-2

-4

H

D

A

B

C F

G

⇒ 3-2·a2

,2·a2

(0,3)

(-3,0)

(0,-3)

(3,0)a

Comment: students can substitute the value of a and find the other intersection points based on symmetry.

Extensions

1. Students can consider the following cases for the stained glass designs

a. when 3 2

2a < (arcs do not intersect)

b. when 3 2 3

2a< < (each pair of arcs intersect)

c. when 3a >

2. Investigate case b where each pair of arcs intersects. Try to make the sum of the areas of the quadrilateral formed outside the arcs and the regions formed by the overlapping sectors in a given proportion, for example, 1:4 to the total area of the square. This problem should be solved by approximation, using a square to approximate the shape formed in the center outside the arcs.

32

Students can use circles and segments and construct part of the design in a new file and then find coordinates of the intersection point for the vertex of the central “square” in terms of parameter a.

2 4 6-2-4-6

2

4

6

-2

-4

AF'

AH'

AJ' O'

S'

N' AI'

AF

V'

P

⇒32

--9+2·a2

2,32

--9+2·a2

2

a a

a

3.

33

3. Investigate the case c where 3a > . In this case concave quadrilateral shapes are formed at the vertices of the square with a convex quadrilateral at the origin. Students can again use circles and segments, construct the design and approximate the curved shapes with squares to look for specific ratio of areas for the design.

2 4 6-2-4-6

2

4

-2

-4

⇒32

--9+2·a2

2,32

--9+2·a2

2

a a

a

a

35

Translation Along Coordinate Axes


Teaching Goals:

1. Students verify that the graph of the function y = f(x) + b is derived from the graph of the function y = f(x) by translation along the y-axis by the vector (0, b).

2. Students verify that the graph of the function y = f(x – a) is derived from the graph of the function y = f(x) by translation along the x-axis by the vector (a, 0).

3. Students verify the commutative property of horizontal and vertical translations of the graph of a function.

4. Students determine that two translations by vectors (a, 0) and (0, b) is equivalent to the translation by the vector (a, b).

5. Students determine that the graph of the function y = f(x – a) + b is derived from the graph of function y = f(x) by translation by the vector (a, b)

6. Students determine the position of the translated graph of a function depending on the sign of parameters a and b.

7. Students verify these results with the help of the software.

Prior Knowledge

• Students should know the definition of a vector.

• Students should know the coordinate representation of vectors.

• Students should know how to add vectors in vector form and in coordinate form.

• Students should know the coordinate form of translation by a vector, including special cases along each coordinate axis.

Problem:

Given the function y x= , investigate translation of the graph of the function.

Part 1 – Translation Along the Y-Axis

In this part of the problem students investigate vertical translations and discover that the graph of the function y = f(x) + b is achieved by translating the graph of the function y = f(x) by vector (0, b) for any function f(x). They start working with the graph of the function y x= and then generalize for an arbitrary f(x). Here are the steps of construction and questions.


36

2. Select Function from the Draw toolbox. Type: sqrt(X) for the function.

3. Draw a second function and enter: sqrt(X)+b for the function. Select the curve, right-click and select Properties from the context menu to change the line color to Blue.

2 4 6-2-4-6

2

4

-2

-4

Y= X+b

Y= X

Q1. As you drag the graph of function y x b= + up and down along the y-axis, what do you observe about the sign of parameter b with respect to the position of this graph relative to the graph y x= ?

A: It is expected that students will make the following conjecture: b > 0 when the graph of y x b= + is above the graph of y x= ; and b < 0 when the graph of y x b= + is below the

graph of y x= .

In order to see that, students can observe the value of b in the Variables toolbox while dragging the graph of the function y x b= + .

Q2. What is the geometric transformation from the graph of y x= to y x b= + .

A. It is expected that students will bring up translation by a vertical vector.

In order to verify this with the software, follow these steps.

37

1. Select Vector from the Draw toolbox. Construct a vector AB anywhere on the plane. Click the selection arrow.

2. Click the graph of the function y x= . Select Translation from the Construct toolbox, select vector AB and the translation of the graph will appear on the screen.

3. Select the translated graph, right-click, select Properties from the context menu, and choose Line Style – Solid 2, Line Color – Red.

Q3. What are the coordinates of the vector of translation that transforms graph of y x= to

the graph of y x b= + ? Make a conjecture and verify it with the help of the software.

A. The teacher may help students come up with a conjecture by pointing to the characteristic points of the graph, in this case point (0, 0) which is the vertex of the semi-parabola. This hint should help them determine the coordinates of the vector of translation. Students will also use the software to help find the answer and verify their conjecture.

1. Select vector AB, choose Coefficients from the Constrain toolbox. The coefficients 0

0

uv

of the vector will appear in the edit box. Accept the default values by pressing Enter.

2 4 6 8-2-4

2

4

6

-2

-4

A

B

Y= X+b

Y= X

u0

v0

2. By moving the endpoint of the vector B students can overlap the translated image with the

38

graph of the function y x b= + and observe the values of u0, v0, and b in the Variables toolbox to answer the question and to verify their conjecture.

The expected answer: u0 = 0, v0 = b.

3. In order to verify this, students can double click the coefficients and substitute these values. The translated image will overlap with the graph of the function y x b= + . Drag point B to confirm that.

Q4. What vector will translate the graph of function y x b= + back to the graph of function

y x= ? Make a conjecture and verify with the software.

A. The vector of translation is (0, -b).

Q5. Generalize these results for an arbitrary function f(x) and confirm your conclusion with the help of the software.

A. Students should conclude that the graph of function y = f(x) + b is derived from graph of function y = f(x) by translation along the y-axis by the vector (0, b). They should also conclude that the graph of function y = f(x) is derived from graph of function y = f(x) + b by translation along the y-axis by the vector (0, -b). They can verify this by editing the equation of the function:

1. Double click the equation of the function and change it to: f(x). The graph of the arbitrary function will appear on the screen.

2. Click the image of the translated function and choose Implicit Equation from the Calculate toolbox. The equation y = b + f(x) will appear on the screen.

39

2 4 6 8 10-2-4-6-8-10

2

4

6

8

10

-2

-4

A

B

⇒ Y=b+f(X)

Y= X+b

Y=f(X)

0

b

Similarly, students can verify inverse translation with the help of the software.

Part 2 – Translation Along the X-Axis

In this part of the problem students investigate horizontal translations and discover that the graph of function y = f(x – a) is always derived by translation from the graph of the function y = f(x) by the vector (a, 0) for any function f(x). They start working with the graph of the function y x= and then generalize for an arbitrary f(x). Here are the steps of construction and questions.


2. Select Function from the Draw toolbox. Type: sqrt(X) for the function.

3. Select Function from the Draw toolbox. Type: sqrt(X - a) for the function. Select the function, right-click and select Properties from the context menu to change the line color to Blue.

40

1 2 3 4 5-1

1

2

3

-1

-2

Y= X

Y= X-a

Q1. As you drag the graph of the function y x a= − left and right along the x-axis, what do you observe about the sign of parameter a with respect to the position of this graph relative to the graph y x= ?

A: It is expected that students will make the following conjecture: if a > 0, the graph of y x a= − is shifted to the right of the graph of y x= ; if a < 0, the graph of y x a= − is

shifted to the left of the graph of y x= .

In order to see this, students can observe the value of a in the Variables toolbox while dragging the graph of the function y x a= − .

Q2. What is the geometric transformation from the graph of y x= to y x a= − .

A. It is expected that students will bring up translation by a horizontal vector.

In order to verify this with the software, follow these steps:

1. Select Vector from the Draw toolbox. Construct a vector AB anywhere on the plane. Click the selection arrow.

2. Click the graph of the function y x= . Select Translation from the Construct toolbox,

41

select vector AB and the image of the graph will appear on the screen.

3. Select the translated image, right-click and select Properties from the context menu to make Line Style – Solid 2, and Line Color – Red.

Q3. What are the coordinates of the vector of translation that transform the graph of y x= to

the graph of y x a= − ? Make a conjecture and verify it with the help of the software.

A. The teacher may help students come up with a conjecture by pointing to the characteristic points of the graph, in this case point (0, 0) which is the vertex of the semi-parabola. This hint should help them determine the coordinates of the vector of translation. Students will also use the software to help find the answer and verify their conjecture.

1. Select vector AB, choose Coefficients from the Constrain toolbox. The coefficients 0

0

uv

of the vector will appear in the edit box. Accept the default values by pressing Enter.

1 2 3 4 5 6-1

1

2

3

-1

-2

A B

u0

v0

Y= X

Y= X-a

2. By moving the endpoint of the vector B students can overlap the translated image with the graph of the function y x a= − and observe the values of u0, v0, and a in the Variables toolbox to answer the question and to verify their conjecture.

42

The expected answer: u0 = a, v0 = 0.

3. In order to verify this, students can double click the coefficients and substitute these values. The translated image will overlap with the graph of the function y x a= − . Drag point B to confirm this.

Q4. What vector will translate the graph of the function y x a= − back to the graph of the

function y x= ? Make a conjecture and verify with the software.

A. The vector of translation is (-a, 0).

Q5. Generalize these results for an arbitrary function f(x) and confirm your conclusion with the help of the software.

A. Students should conclude that the graph of function y = f(x – a) is derived from the graph of function y = f(x) by translation along the y-axis by a vector (a, 0). They should also conclude that the graph of function y = f(x) is derived from the graph of function y = f(x – a) by translation along the x-axis by a vector (-a, 0). They can verify this by editing the equation of the function:

1. Double click the equation of the function and change it to: f(x). The graph of the arbitrary function will appear on the screen

2. Click the image of the translated function and choose Implicit Equation from the Calculate toolbox. The equation y = f(x - a) will appear on the screen.

43

2 4 6 8 10 12-2-4-68

2

4

6

8

-2

-4

-6

A B

⇒ Y=f(X-a)

a

0

Y=f(X)

Y= X-a

Similarly, students can verify the inverse translation with the help of the software

Part 3 – Commutative Property of Translation In this part of the problem students will investigate the commutative property of translations given by vectors (0, b) and (a, 0) along the coordinate axes of the graph of an arbitrary function y = f(x).

Here are the steps of construction and questions.


2. Select Function from the Draw toolbox. Type f(X) for the function.

3. Select Vector from the Draw toolbox. Construct two vectors, AB and CD. Click the selection arrow. Select vector AB, choose Coefficients from the Constrain toolbox, type: (0, b) for the coefficients and press Enter. Select vector CD, choose Coefficients from the Constrain toolbox, type: (a, 0) for the coefficients and press Enter.

Q1. Make a conjecture about relative position of the curves transformed by two methods 1) translate the graph of the function y = f(x) by vector AB and then translate the resulting graph by the vector CD, or 2) translate the graph of the function y = f(x) by vector CD and then translate the resulting graph by the vector AB. Verify your conjecture with the help of the

44

software.

A. Students will formulate their conjecture prior to working on the computer, then complete constructions on the computer and determine coincidence of the graphs, confirming or rejecting their conjectures. Here are the steps of the constructions.

At this point, it’s useful to adjust the default Curve setting to the thinnest black line so that you can see when two curves are the same. From the drop-down Edit menu at the top of the screen select Settings and the Geometry tab. In the Curve box check that the Line Color is Black and the Line Style is Solid 1.

Method 1:

1. Click the graph of the function y = f(x), select Translation from the Construct toolbox, then click vector AB. Click the resulting image, right-click and select Properties from the context menu to change the line style to Dot.

2. Click the translated image, select Translation from the Construct toolbox. Then click the vector CD. Click the translated image, right-click and select Properties from the context menu and choose Line Style – Solid 4, Line Color – Light Blue (or any pastel shade).

Method 2:

3. Click the graph of the function y = f(x) and select Translation from the Construct toolbox, then click vector CD. Click the resulting image, right-click and select Properties from the context menu to change the line style to Dot.

4. With the graph still selected, choose Translation from the Construct toolbox. Then click vector AB.

Observe the image constructed by method 2 coincides with the image constructed by method 1.

Students should drag point B up and down and point D left and right to verify that this conjecture works for vectors of various lengths and directions.

45

Q2. What is the equation of this curve?

A. It is expected that students will apply prior knowledge from parts 1 and 2, and will be able to write in either order:

(0, ) ( ,0)( ) ( ) ( )b ay f x y f x b y f x a b= → = + → = − + , or

( ,0) (0, )( ) ( ) ( )a by f x y f x a y f x a b= → = − → = − + . Students then can check their findings using the software – click the curve and select Implicit Equation from the Calculate toolbox.

Q3. Is it possible to derive a graph of the function y = f(x – a) + b from the graph of the function y = f(x) in one step?

A. It is expected that students will conjecture that a translation by a vector (a, b) will transform one graph into another in one step. Students can check that using the software. In order to do that students will need to add two vectors of translation and then translate the original graph of function by the resultant vector to verify that it will coincide with the function y = f(x – a) + b

46

Part 4 – Applications and Assessment Problems

Q1. Given two functions, ( )f x x a= − and ( )g x x b= + .

a. Can graph of one of the functions be derived from the graph of the other function by translations?

b. If yes, by how many ways?

c. If yes, write the vector of translation for the graph of f to be transformed to the graph of g and visa versa.

d. Confirm your results with the help of the software

A. a. The graph of one of the functions can be derived from the other by translation along the

x-axis

b. There are two translations, one from graph of f to g, and another one from g to f, if a + b ≠ 0. If a + b = 0 and thus f(x) = g(x) then there is only one translation.

c. The vectors of translation are (±|a + b|, 0)

d. Here are the steps of construction:

1. Click Function from the Draw toolbox and type sqrt(X-a) for the function f(x). Click OK. Select the graph of the function, right-click and select Properties from the context menu, select Line Style – Solid 2, Line Color – Blue.

2. Click Function from the Draw toolbox and type sqrt(X+b) for the function g(x). Click OK. Select graph of the function, right-click and select Properties from the context

47

menu and select Line Style – Solid 2, Line Color – Red.

3. Click Vector from the Draw toolbox and construct a vector AB anywhere on the positive y coordinate plane. (The translation vector needs to reside in the valid domain of the function; otherwise Geometry Expressions will map the vector’s endpoint to an indeterminate location.) Click the select arrow. Select the vector, choose Coefficients from the Constrain toolbox and enter coefficients |a + b|, 0 for the vector.

4. Select point A and the graph of ( )g x x b= + . Select Incident from the Constrain toolbox. Point A will lie on the graph of the function. Point B will automatically be placed on the graph of ( )f x x a= − . Students can drag vector AB along the curve to see that the endpoints of the vector lie on both curves for all points on both graphs.

Likewise, construct the opposite vector, (-|a+b|, 0) and attach it to function f(x).

Q2. Given two functions, ( ) ln( )f x ax= and ( ) ln( )g x bx= , 0, 0, 0, 0a x b x> > > >

a. Can the graph of one of the functions be derived from the graph of the other function by translations?

b. If yes, by how many ways? c. If yes, write the vector of translation for the graph of f to be transformed to the graph of

g and visa versa. d. Confirm your results with the help of the software

A. a. The graph of one function can be derived from the other one by translation along the y-

axis

b. There are two translations, one from graph of f to g, and another one from g to f, if a ≠ b If a = b and thus f(x) = g(x) then there is only one translation.

c. The vectors of translation are 0, ln ab

±

48


1. Click Function from the Draw toolbox and type ln(a*X) for the function f(x). Click OK. Select the graph of f(x), right-click and select Properties from the context menu, select Line Style – Solid 2, Line Color – Blue.

2. Click Function from the Draw toolbox and type ln(b*X) for the function g(x). Click OK. Note: If the default values of a and b are set to the same value, (check the Variables toolbox), the two functions will be superimposed. Select one of the variables in the Variables list and change its value in the edit window below to separate the curves. Select the graph of g(x), right-click and select Properties from the context menu, select Line Style – Solid 2, Line Color – Red.

3. Click Vector from the Draw toolbox and construct a vector AB anywhere on the positive x coordinate plane. (The translation vector needs to reside in the valid domain of the function; otherwise Geometry Expressions will map the vector’s endpoint to an indeterminate location.) Click the selection arrow. Select the vector, choose

Coefficients from the Constrain toolbox and enter for the vector.

4. Select point A and the graph of g(x). Select Incident from the Constrain toolbox. Point A will lie on the graph of g(x). Point B will automatically be placed on the graph of f(x). Students can drag vector AB along the curve to see that endpoints of the vector lie on both curves for all points on both graphs.

Likewise, construct the opposite vector, and attach it to function f(x).

Q3. Given two functions, ( )f x ax b= + and ( )g x cx d= + , 0, 0a c≠ ≠

a. Can the graph of one of the functions be derived from the graph of the other function by

49

translations?

b. If yes, by how many ways?

c. Can you specify a translation vector for the graph of f to be transformed to the graph of g and visa versa?

d. Confirm your results with the help of the software

A. a. Yes if a = c.

b. There are an infinite number of vectors that connect any two points on the two lines.

c. Students can specify any vector with endpoints on these two lines or give the general form of the vector between these two lines


1. Click Function from the Draw toolbox and type a*X+b for the function f(x). Click OK. Select the graph of the function, right-click and select Properties from the context menu, select Line Style – Solid 2, Line Color – Blue.

Comment: if the student already determined that lines should be parallel, he/she should type the line with the same slope. If not, the student will start with the general equations of the lines and will arrive at the conclusion that their slopes are the same and edit the equation of the function to fit this conclusion.

2. Click Function from the Draw toolbox and type a*X+d for the function g(x). Click OK. Select the graph of the function, right-click and select Properties from the context menu, select Line Style – Solid 2, Line Color – Red.

3. Click Vector in the Draw toolbox and construct a vector AB anywhere on the plane. Click the selection arrow.

4. Select point A and the graph of f(x). Select Incident from the Constrain toolbox. The point A will appear on the graph of f(x).

5. Select graph of f(x), choose Translation from the Construct toolbox and click vector AB. The translated image of graph of f(x) will appear on the screen.

50

2 4 6 8 10 12-2-4-6-8

2

4

6

8

10

-2

-4

A

B

Y=X·a+bY=X·a+d

6. Move point B around to overlap the translation with the graph of function g to see various ways that the graph of f can be translated to graph of g.

Extension:

Q1. Given two functions, 1( )f xx

= and ( )1

x ag xx+

=+

. What is value of parameter a and the

translation vector needed to equate these two functions. Investigate this problem with the help of the software

A. Here are the steps of construction:

1. Click Function from the Draw toolbox and type 1/X for the function f(x). Click OK. Select the graph of f(x), right-click and select Properties from the context menu, select Line Style – Solid 2, Line Color – Blue.

2. Click Function from the Draw toolbox and type (X+a)/(X+1) for the function g(x). Click OK. Select the graph of g(x), right-click and select Properties from the context menu, select Line Style – Solid 2, Line Color – Red.

51

2 4 6 8 10 12 14-2-4-6-8-10-12-14

2

4

6

8

10

12

14

-2

-4

-6

Y=1X

Y=X+a1+X

Note: The students’ graphs may look different to the image above. Click a in the Variables list, then manipulate the slider to change the shape of g(x).

3. Click Vector in the Draw toolbox and construct a vector AB anywhere on the plane. Click the selection arrow.

4. Select the graph of function g(x) (red), click Translation in the Construct toolbox, then click the vector AB. The image of the graph of g(x) will appear on the screen.

5. Vary parameter a and vector AB until the translated image coincides with the graph of the function f(x) (blue). In order to vary parameter a and change the shape of the function g, click-and-drag the graph. A drag handle, labeled a, appears on the curve. (Alternatively, you can vary the value of a in the Variables toolbox as described above.) Vary the direction of the translation by movinging points A and B.

52

2 4 6 8 10-2-4-6-8-10

2

4

6

8

10

-2

-4

A B

⇒ Y=2.0406168+X

1+X

⇒1.0526318-1.0219269

Y=1X Y=

X+a1+X

6. When the desired position is found, students can observe the value of parameter a in the Variables list. They can also calculate the Real equation for the function g(x) by selecting the graph of the function, choosing the Real tab in the Calculate toolbox and selecting Implicit equation.

7. In order to calculate the translation vector, students select the vector, choose the Real tab in the Calculate toolbox and select Coefficients. The numerical coefficients of the vector will appear on the screen. All answers shown on the screen are numerical approximations and allow students to make a conjecture about exact solutions.

Students now can make a conjecture about the values of parameter a and the translation vector. They should confirm this through analytical work:

1 1 1( ) 11 1 1

x a x a ag xx x x+ + + − −

= = = ++ + +

. Translation exists only when a = 2, so that

1 11 1

ax x−

=+ +

. Then, 1( ) 1

1g x

x= +

+. Thus the graph of g(x) can be derived from the graph

of 1( )f xx

= by translation by the vector (-1, 1). Thus, the inverse translation that

transforms graph of g(x) into the graph of f(x) is accomplished by the vector (1, -1). In order to verify that, students can substitute these values for the parameter a and vector AB into software.

By analogy with given examples the teacher can point out the associative property of translation. Students can verify that the order of translation does not affect the result. This property can be combined with the commutative property of translation.

53

A Little Trig


Teaching Goals:

1. Students interpret a given word problem and create geometric constructions with which to solve the problem.

2. Students choose an independent variable and define it as a constraint in the geometric construction.

3. Students analyze expressions for the optimized quantities, such as length, sum of lengths (perimeter), and area by doing the following:

o Determine the domain of functions

o Explore minima or maxima visually using the software

o Graph an expression as a function of a chosen independent variable to determine the existence of a minimum or maximum value

o Determine expressions for the extrema of functions using the software

o Confirm solutions for extrema analytically

Prior Knowledge

• Students should know the definition of sine and cosine.

• Students should know the law of cosines.

• Students should know half-angle identities:

=+

2cos2)cos(1 2 θθ ;

( )

=−

2sin2cos1 2 θθ .

• Students should know that any angle inscribed in a semi-circle has a measure of 90 degrees.

• Students should know the formula for the area of a triangle.

Problem:

A triangle is inscribed in a semi-circle with one of the legs the diameter.

a. Find the extremal area of this triangle. Determine if this is a minimum or a maximum.

b. Find the extremal value for the perimeter. Determine if this is a minimum or a maximum.

c. Will these two extremal values occur at the same point?

54

Part 1 – Investigating Area of the Triangle

In this part of the problem students analyze the word problem and complete the geometric construction of the semi-circle and the inscribed triangle described in the problem. Students should define an independent variable for the angle measure of the vertex opposite to the diameter. Students can use the dynamic features of the software to analyze the problem qualitatively. Then they can determine a symbolic expression for the area of the triangle using Geometry Expressions’ symbolic calculation features and analyze the problem graphically and analytically.

Note: You should check the Edit / Settings / Math dialog to make sure that in the Math box, the Angle Mode is set to Radians (this is the default). If it is set to Degrees, click the down arrow on the right and change it.

Here are the steps of construction and questions.


2. Select Function from the Draw toolbox. In the Function Type dialog choose Parametric for the Type, enter a*cos(t) for X, enter a*sin(t) for Y, 0 for the Start value and 3.14 for the End value of the parameter. The semi-circle will appear on the graph.

3. Adjust the radius of the circle by dragging the circle to fill most of the graph.

4. Select Point from the Draw toolbox and draw two points on the x-axis and one on the circle.

5. Select one of the points on the x-axis and choose Coordinate from the Constrain toolbox. Type (a, 0) for the coordinates. Select the second point on the x-axis, and constrain it to (-a, 0).

6. Select the point on the semi-circle and the semi-circle. With both of them selected, choose Point proportional along curve from the Constrain toolbox. Type t in the data entry box.

7. Choose Polygon from the Draw toolbox and draw a triangle connecting the constructed points.

55

Q1. Drag the point on the circle and observe the changes in the area of the triangle. Do you think the area has a minimum? A maximum?

A. From the observations and ideas of symmetry students should conclude that since the area of the triangle approaches zero when the point on the circle approaches the endpoints of the diameter, the minimum does not exist and a maximum occurs right in the middle, when this point is located at an equal distance from the endpoints of the diameter (i.e. the triangle is isosceles).

Q2. Find an expression for the area of the triangle. Explain your work

A. The area of the triangle can be found as:

A= Ya *221

=a*a sin(t)

=a2sin(t) where t is the angle measure of the point on the circle relative to the origin (in radians)

Students can confirm this calculation using Area from the Calculate toolbox.

0.5 1.0 1.5 2.0 2.5-0.5-1.0-1.5-2.0

0.5

1.0

1.5

2.0

2.5

3.0

-0.5

B

C

A

t

(-a,0)(a,0)

56

0.5 1.0 1.5 2.0 2.5-0.5-1.0-1.5-2.0

0.5

1.0

1.5

2.0

2.5

3.0

-0.5

B

C

A

⇒ a2·sin(t)t

(-a,0)(a,0)

Q3. Plot this expression as a function of x and investigate if this function has a maximum value.

A. The function has maximum when 2π

=x . Here are the steps of construction:

1. Right click the expression for the area and choose Copy As / String from the context menu.

2. Choose Function from the Draw toolbox. Paste the expression into the Y= line of the Function Type dialog. Replace (t) with (x) in the expression (the function, Y, must be in terms of x) and click OK. The graph of the function will be displayed.

3. Select the Point tool from the Draw toolbox and draw a point on the new graph.

4. Select both the point and the graph. Choose Point proportional along curve from the Constrain toolbox and enter t for the value. Now, the x-coordinate of the point on the plot is equal to the radian measure of the angle for the point on the circle.

Q5. Drag the point along the circle and observe how the point on the graph is moving. Can you confirm your conjecture about the maximum of the area of the triangle?

A. When the vertex of the triangle on the semi-circle divides the arc into two equal parts, the

57

area of the triangle is at the maximum.

0.5 1.0 1.5 2.0 2.5-0.5-1.0-1.5-2.0

0.5

1.0

1.5

2.0

2.5

3.0

-0.5

B

C

A

E

⇒ a2·sin(t)

t

t

Y= a2·sin(X)

(-a,0)(a,0)

Q6. Can you analytically justify the position of the maximum using the expression for the area of the triangle?

A. Students can state that sin(t) has a maximum at 2

t π= , thus the area will be maximum at

this point.

Q7. Can you achieve the same result by geometric considerations?

A. The area of the triangle is defined as:

A= bh21

= ah221

=ah

Since radius (a) of the circle remains constant, the maximum is achieved when height is maximal. This occurs when vertex of the triangle is at the top of the semi-circle, h = a and

2t π=

.

58

Part 2 – Optimizing the Perimeter of the Rectangle

In this part of the problem students explore how the perimeter of the triangle changes using the dynamic features of the software. They can then determine the symbolic expression for the perimeter of the triangle and analyze its function analytically and graphically. Since the base of the triangle remains constant, it is sufficient to consider summing the other two sides in order to answer the optimization question. The teacher may ask students to save their existing file that includes constructions for the triangle, and delete all unnecessary calculations. If the teacher decides to start with the blank file, repeat steps 1 – 7 from part 1. Here are the steps of construction and questions from this point forward.

Q1. Drag the point on the semicircle and observe the changes in the perimeter of the triangle. Do you think the perimeter has a minimum? A maximum?

A. From the observations and ideas of symmetry students may conclude that when the point on the semicircle is approaching either of the endpoints of the curve, the perimeter is approaching the same constant value of 4a, so there should be an extremum of the perimeter, most probably right in the middle. However, it is impossible to determine if the extremum is a maximum or a minimum solely from the observations.

Q2. Find an expression for the perimeter using Geometry Expressions. Plot this expression as a function of x and investigate if this function has a minimum or maximum value.

A. As noted above, it is sufficient to consider the sum of two sides of the triangle in order to find the extremum. Students can follow the steps described below to find expressions for each side using the software and then add them. The analytical solution comes from the theorem of cosines: ));cos(1(2)cos(2 22222 tataaab +=−−+= π

))cos(1(2)cos(2 22222 tataaac −=−+= . The perimeter has maximum at 2π

=t , the same

maximum as the area’s maximum. Here are the steps of finding expression for the sum and constructing a graph.

1. Change the default settings by selecting: Edit / Settings / Math, the Output box, click the Show Name data line, to select True for this value.

2. Click one side of the triangle and choose Distance/Length from the Calculate toolbox, Symbolic tab. The software will produce an expression for the length in terms of a and x. Repeat that for the 2nd side.

3. Select Expression from the Draw toolbox and type the expression representing the sum of two lengths: z0 +z1. The software will produce the expression for the sum.

59

0.5 1.0 1.5 2.0 2.5-0.5-1.0-1.5-2.0

0.5

1.0

1.5

2.0

2.5

3.0

-0.5

B

C

A

z0⇒ 2 · 1+cos(t)· a

z0+z1⇒ 2· 1+cos(t)+ 2· 1-cos(t) · a

z1⇒ 2 · 1-cos(t)· a

t

(-a,0)(a,0)

4. Right click the expression, choose Copy As / String.

5. Select Function from the Draw toolbox. In the Function Type dialog copy the expression into the Y= line. Replace (t) with (x) in this expression. Click OK. The graph of the function will appear on the screen. If you can’t see the function, click the Scale Up or the Scale to Fit icon until you can see the plot of the function.

6. Select Point from the Draw toolbox and draw a point on the plot.

7. Select both, the point and the plot of the graph. Choose Point proportional along curve from the Constrain toolbox and enter t for the value. Now, the x-coordinate of the point on the plot is equal to the radian measure of the angle for the vertex on the semi-circle.

Students can drag the point on the semi-circle and observe that the maximum is achieved when the point on the semicircle divides the arc equally.

60

1 2 3 4 5-1-2-3-4

1

2

3

4

5

6

-1

B

C

A

E

z0⇒ 2 · 1+cos(t)· a

z0+z1⇒ 2· 1+cos(t)+ 2· 1-cos(t) · a

z1⇒ 2 · 1-cos(t)· a

t

t

Y= 2 · 1+cos(X)+ 2 · 1-cos(X) · a

(-a,0)(a,0)

Q3. Simplify expression for the sum of two sides and confirm your findings algebraically.

A. Students can use the Half-Angle identity to simplify the expression for each length:

( ) ( )2 20 1 2 2cos ( / 2) 2sin ( / 2) 2 cos( / 2) sin( / 2)z z a t t a t t+ = + = + when 0 t π< < . The

maximum is achieved when 2cos( / 2) sin( / 2)

2t t= = and

2t π= . The perimeter at this point

is: 2 2 2 2 (1 2)P a a a= + = + .

61

One Hyperbola


Teaching Goals:

1. Students interpret the given word problem and complete geometric constructions according to the condition of the problem.



o Determine the domain of a function based on conditions of the problem

o With the help of the software, visually determine if a quantity reaches a minimum or maximum.

o Graph expression as a function of a chosen independent variable to determine the existence of a minimum or maximum

o Determine expressions for the extrema of functions with the help of the software


Prior Knowledge

• Students should know the properties of a rectangle.

• Students should know basic properties of an hyperbola, including its equation, graph, and asymptotes.

• Students should know that if the product of two positive quantities is constant, then the minimum of the sum of these quantities is reached when these two quantities are equal:

If a*b=constant, then a+b=min when a = b; or geometrically, if the area of a rectangle is constant, the perimeter is minimal when it is a square.

Problem:

Given a branch of an hyperbola in the 1st quadrant defined by the equation ayx

= . Consider the

open region bounded by the coordinate axes and the hyperbola, containing a rectangle with two sides on the coordinate axes and one vertex on the hyperbola.

a. Investigate if the area of this rectangle can have a minimum or maximum value and justify your answer.

b. Find the extremal perimeter of this rectangle. Determine if this is a minimum or a maximum.

62

c. Find the extremal length of the diagonal of this rectangle. Determine if this is a minimum or a maximum

Part 1 – Investigating Area of the Rectangle

In this part of the problem students analyze the word problem and complete a geometric construction of the hyperbola and an inscribed rectangle according to the specifications of the problem. Students then choose an independent variable and define it in the construction. Students can use the dynamic features of the software to analyze the problem qualitatively first; then they can determine the symbolic expression for the perimeter of the rectangle using the symbolic calculation features of Geometry Expressions and analyze the problem graphically and analytically. Here are the steps of construction and questions.


2. Select Function from the Draw toolbox. In the Function Type dialog enter a/x for the Y value. The hyperbola will appear on the graph. If needed, adjust the window by using one of the scaling icons or drag the graph. Students only need to see the 1st quadrant of the coordinate plane

3. Select Point from the Draw toolbox and draw a point on the hyperbola in the 1st quadrant.

4. Select the point and the hyperbola. With both of them selected, choose Point proportional along curve from the Constrain toolbox. Enter x in the data entry box. Here x is the x-coordinate of the point on the parabola. This is an independent variable.

5. Select Polygon from the Draw toolbox and draw a rectangle with one vertex on the point on the hyperbola, one on the origin, and two on the coordinate axes.

6. Click the segment connecting the point on the hyperbola with the point on the Y axis and the Y axis outside the segment boundary. (If you try to constrain the top segment of the box perpendicular to the segment on the Y axis you will receive an over constrained message, so make the constraint with the axis itself, not the segment lying on the axis.) With both of them selected choose Perpendicular from the Constrain toolbox. Repeat for the segment connecting the hyperbola to the X axis.

63

0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5-0.5

0.5

1.0

1.5

2.0

2.5

3.0

-0.5

B

DC

A

x

Y=aX

Q1. Drag the point on parabola and observe the changes in the area of the rectangle. Do you think the area has a minimum? A maximum?

A. From the observations students may not conclude an exact answer, since as one side of the rectangle increases, another side decreases. The correct answer is that the area of this rectangle remains constant and does not depend on the position of the point on the hyperbola.

Q2. Find an expression for the area of the rectangle. Explain your work

A. The area of the rectangle can be found as aA X Y X a constX

= ⋅ = ⋅ = = . Students can

confirm this calculation using the Area tool from the Calculate toolbox.

64

0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5-0.5

0.5

1.0

1.5

2.0

2.5

3.0

-0.5

B

DC

A

⇒ a x

Y=aX

Note: it is important for students to be exposed to optimization problems with functions that do not necessarily have extremal values, such as this problem.

Part 2 – Optimizing Perimeter of the Rectangle

In this part of the problem students explore how the perimeter changes using dynamic features of the software first, then they can determine the symbolic expression for the perimeter of the rectangle and analyze its function analytically and graphically. Use the constructed rectangle in Part 1 to answer these questions.

Q1. Drag the point on the hyperbola and observe the changes in the perimeter of the rectangle. Do you think the perimeter has a minimum? A maximum?

A. From their observations students may conclude that when the point on the hyperbola is approaching either of the coordinate axes, the perimeter is getting infinitely large, so there is no maximum value. However, that means there should be a minimum value. From ideas of symmetry of hyperbola they can then conjecture that the minimum of the perimeter will occur when the rectangle becomes a square, and the vertex of the rectangle that is on the hyperbola is located on the vertex of the hyperbola.

Q2. Find an expression for the perimeter algebraically. Then check your work using the

65

software. Plot this expression as a function of x and investigate if this function has a minimum value.

A. The perimeter can be found as 2( ) 2 aP x y xx

= + = +

. Students can find the expression

for the perimeter using the software to confirm their answer. Using the fact that the area of the rectangle is constant, students can conclude that the minimal perimeter will be achieved when the rectangle becomes a square. Thus, the perimeter reaches its minimum when

,ax x ax

= = . So the minimum value of the perimeter is 4P a= . Students then graph this

function and determine that the point they found is indeed the minimum of the perimeter. Here are the steps of finding the expression for the perimeter and constructing the graph.

1. Click the rectangle and choose Perimeter from the Calculate toolbox, Symbolic tab. The software will produce an expression for the perimeter in terms of a and x.

2. Right click the expression for the perimeter, choose Copy As/ String.

3. Select Function from the Draw toolbox. In the Function Type dialog, Cartesian Type, paste the expression into the Y= line. Press OK. The graph of the function will appear on the screen. If you can’t see the function, use one of the scaling options to see the plot of the function.

4. Select Point from the Draw toolbox and draw a point anywhere in the blank space. Select the point and choose Coordinate tool from the Constrain toolbox. Type ( )sqrt a for x0 and 4* ( )sqrt a for y0. Observe the point jumping to the function minimum.

5. Select both, point A and the x-axis. Choose Perpendicular from the Construct toolbox. A vertical line will appear on the graph.

6. Move point A to the minimum and observe that the rectangle becomes a square.

66

1 2 3 4 5 6

1

2

3

4

5

B

G

DC

A

⇒ 2· x +ax

Y=2· X +aX

x

a,4· a

Y=aX

Part 3 – Optimization of the Diagonal of the Rectangle.

In this part of the problem students explore how the diagonal of the rectangle (with endpoints B and D above) changes using the dynamic features of the software. Then they can determine a symbolic expression for the diagonal of the rectangle and analyze the function graphically and analytically. Note: since the diagonals of the rectangle are equal, it is the same problem to consider either diagonal; however, by choosing diagonal BD it may be more obvious for students how the length of this diagonal changes with the movement of point X along the hyperbola. The teacher may ask students to save their existing file that includes constructions and computations for the rectangle, and delete all unnecessary constructions and calculations. If the teacher decides to start with a blank file, repeat steps 1 – 6 from Part 1. Here are the steps on of construction and questions from this point forward.

Select Line Segment tool from the Draw toolbox. Draw a segment BD (connecting vertices on the coordinate axes).

Q1. Drag point X along the hyperbola and observe the changes in the length of the diagonal of the rectangle. Do you think the diagonal has minimum length? A maximum length?

A. From the observations students may conclude that when the point on the hyperbola approaches either of the coordinate axes, the diagonal becomes infinitely long, so there is no

67

maximum value. However, that means there should be minimum value. From ideas of symmetry of hyperbolas they can then conjecture that the minimum of the diagonal length will occur when the rectangle becomes a square, and the vertex of the rectangle that is on the hyperbola is located in the vertex of the hyperbola.

Q2. Find an expression for the length of the diagonal algebraically. Then confirm your expression using software. Plot this expression as a function of x and investigate if this function has minimum value.

A. The length of the diagonal can be found as 2

2 2 2 aL x y xx

= + = +

. Students can find

the expression for the diagonal using the software to confirm their equation. Using the fact that area of the rectangle is constant, it also means that the square of the area is constant. Therefore the sum of squares of the sides is minimal when the rectangle becomes a square.

Thus, the length of the diagonal is minimum when ,ax x ax

= = . So the minimum value of the

diagonal is 2L a a a= + = . Students then graph this function and determine that the point they found is indeed the minimum of the perimeter. Here are the steps of finding the expression for the diagonal and constructing the graph.

1. Click the diagonal and choose Distance/Length from the Calculate toolbox, Symbolic tab. The software will produce an expression for the diagonal in terms of a and x.

2. Right click the expression for the perimeter, choose Copy As / String.

3. Select Function from the Draw toolbox. . In the Function Type dialog, Cartesian Type, paste the expression into the Y= line and click OK. The graph of the function will appear on the screen. If you can’t see the function, choose one of the scaling options to adjust the plot.

4. Select Point from the Draw toolbox and draw a point anywhere in the blank space. Select the point and choose Coordinate from the Constrain toolbox. Type ( )sqrt a for x0 and

(2* )sqrt a for y0. Observe the point jumping to the function minimum.

5. Select both, point A and the x-axis. Choose Perpendicular from the Construct toolbox. The vertical line will appear on the graph.

6. Move point A to its minimum and observe that the rectangle becomes a square.

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0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5-0.5

0.5

1.0

1.5

2.0

2.5

3.0

-0.5

C D

G

AB

⇒a2

x2+x2

Y=aX

xa, 2 · a

Y= X2+a2

X2

69

Three Extrema (Circle)


Teaching Goals:

1. Students interpret given word problem and complete geometric constructions according to the condition of the problem.


3. Students analyze expressions for the optimized quantities, such as length, sum of lengths (perimeter), ratio of lengths, and area by doing the following:

o Determine the domain of functions based on conditions of the problem

o Use the software to visually explore if a quantity reaches a minimum or a maximum

o Graph an expression as a function of a chosen independent variable to confirm the existence of a minimum or a maximum

o Determine expressions for the extrema of received functions based on geometric considerations


Prior Knowledge

• Students should know that the tangent to a circle is perpendicular to the radius of a circle.

• Students should know that if the sum of two positive quantities is constant, the maximum of the product of these quantities is reached when these two quantities are equal, i.e.:

if a b const+ = , then maxa b⋅ = when a = b;

or geometrically, the area of a rectangle of constant perimeter is maximal when it is a square.

• Students should know that a positive quantity and its square reaches a maximum at the same value of the independent variable.

Problem:

Given a circle with its center at the origin and radius R, a tangent is drawn to an arbitrary point on the circle, so that the point is located in the 1st quadrant of the coordinate plane.

a. Find the extremal length of a tangent segment with the end points lying on the coordinate axes. Determine if this is a minimum or a maximum.

b. The point on the circle divides the tangent segment into two segments. Find the the ratio of the lengths of these segments with the tangent segment at its extremal length.

70

c. Find the extremal area of a triangle formed by the tangent line and coordinate axes. Determine if this is a minimum or a maximum.

d. Consider a rectangle in the 1st quadrant with two adjacent sides on the coordinate axes and with the tangent segment of the circle described above being a diagonal. Find the extremal perimeter of this rectangle. Determine if this is a minimum or a maximum.

Part 1 – Length of a Tangent Segment to a Circle

In this part of the problem students analyze the word problem and complete the geometric construction of a circle and a tangent segment as described in the problem. Students then choose the independent variable and define it in the construction. Students can use the dynamic features of the software to analyze the problem qualitatively first, then they can determine the symbolic expression for the length of the segment using the symbolic calculation feature and analyze the problem graphically and analytically. Here are the steps of construction and the questions.


2. Select Circle from the Draw toolbox. Draw a circle with the center at the origin of the system of coordinates.

3. Click the circle, select Radius from the Constrain toolbox and type R for the radius.

4. Click the circle, select Tangent from the Construct toolbox. Construct a tangent line on the circle in the 1st quadrant.

5. Click the tangent line, right-click and select Properties from the context menu and change Line Style to Dot.

6. Select both, the tangent line and x – axis, and choose Intersection from Construct toolbox. Repeat the same steps and construct the intersection of the tangent line with the y – axis.

7. Select Line Segment from the Draw toolbox and draw a segment with endpoints being constructed at the intersection points.

71

2 4 6-2-4-6

2

4

-2

-4

E

C

B

D

A

F

R

Q1. Drag one of the endpoints of the segment and observe the changes in the length of the tangent segment. Construct your conjecture.

A. From the observations and ideas of symmetry students may conclude that the length of the tangent segment reaches its minimum when the point of tangency divides the arc of the circle equally in two parts.

Q2. What is your choice for the independent variable in this problem?

A. Students may choose different variables, for example, the x-coordinate of the point of tangency, or the y – coordinate of the point of tangency, or one of coordinates of the endpoints of the tangent segment. If they do not choose any variables, the program will select the angle between the radius to the point of tangency and x – axis for calculations with the tangent segment.

Note: Depending on what constraint is chosen as the independent variable, the software will limit the choice of “draggable” points. For example, if you constrain the x-coordinate of the point of tangency, only this point can be dragged to change the position of the tangent segment. The endpoints of the segment are no longer “draggable”.

Q3. What is the range of values for the chosen independent variable if the point of tangency only moves along the arc in the 1st quadrant?

72

A. For the x-coordinate of the point of tangency, the range of values is [0, R].

For this case the steps of constructions are shown below.

1. Select the point of tangency and y – axis. Choose Distance/Length from the Constrain toolbox. Mark the distance x.

2. Click the tangent segment and choose Distance/Length from the symbolic part of “Calculate” toolbox. The software will produce a function for the length of the segment in terms of radius of the circle R and x.

Q3. Plot the expression for the length as a function of x and investigate if this function has a minimum value on the domain.

A. Students can graph the function and confirm that the function plot has a minimum for 0 x R≤ ≤ . Here are the steps for constructing the graph.

1. Right click the expression for the length of the tangent segment, choose Copy As/ String.

2. Select Function from the Draw toolbox. Copy the expression into the Y= line of the Cartesian Function Type dialog, and click OK. The graph of the function will appear on the screen. If you can’t see the function, choose one of the scaling options to adjust the plot.

3. Select the point of tangency on the circle and the x – axis. Choose Perpendicular from the Construct toolbox. Drag the point of tangency and observe where the function has a minimum.

73

2 4 6 8 10 12 14 16-2-4-6

2

4

6

8

10

12

-2

-4

A B

C

E

D

⇒R3

x· R2-x2

x

Y=R3

X· R2-X2

R

Q4. What is the value of x where the function has a minimum?

A. At this stage of the problem students already conjectured that the point of tangency divides the arc of the circle into two equal parts. Thus, the radius to the point of tangency at this case is an angule bisector of the angle formed by the coordinate axes, and thus divides this angle into two 45° angles. Now students can consider the right isosceles triangle with side x and

hypotenuse R, in order to find x. The answer is 2

2x R= .

Q5. Find the minimal length of the tangent segment.

A. To solve this algebraically students can substitute the found value of x into the function: 3

22

2

22

RY RR RR

= =

−

. The answer can be determined right away from geometric

considerations: the hypotenuse of the right triangle is equal to twice the length of the median, so the length of the tangent segment is 2R.

Q6. To confirm that this value is the minimal value of the function, plot the point with

74

coordinates 2 ,2

2R R

. Here are the steps of construction:

Select Point from the Draw toolbox and draw a point anywhere in the blank space. Select the point and choose Coordinate from the Constrain toolbox.

1. Using the Symbols palette (or the sqrt function), type 2

R for x0 and 2 R∗ for y0. Observe

the point jumping to the function minimum.

2 4 6 8 10 12 14 16-2-4-6

2

4

6

8

10

12

-2

-4

A B

C

E

D

G

⇒R3

x· R2-x2

x

R

Y=R3

X· R2-X2

2·R2

,2·R

Q7. The point on a circle divides the tangent segment into two segments. Find the ratio of the lengths of these segments when the tangent segment has its minimal length.

A. Since the point of tangency is at the midpoint of the segment, the ratio is 1:1.

The analytical solution to the problem consists of two parts: a) finding the length of the segment; b) finding the minimum value. The teacher may decide to ask more advanced students to find both or one of the parts of the analytical solution. Here is the analytical solution for the teacher’s reference:

a) 2 2 2 2L DE AD AE= = + . The equation of the tangent line to the circle 2 2 2x y R+ = can

75

be written as 2ax by R+ = , where (a, b) are coordinates of the point of tangency. If 0x = ,

then 2Ry

b= , so

2RADb

= . If y = 0, then 2Rx

a= , so

2RAEa

= . By the Pythagorean

theorem: 4 4 4 2 2

22 2 2 2

( )R R R a bLa b a b

+= + = . Since 2 2 2a b R+ = , then

3RLab

= . Substituting for b

in terms of a, we get: 3

2 2

RLa R a

=−

b) In order to achieve the minimum lenght L, and since R = const, we should maximize the denominator. We know that ab = max, when a2b2 = max, so we are looking for maximum of product of two quantities such that the sum of these two quantities is constant: 2 2 2a b R+ = .

Thus, the maximum is reached when 2

2 2

2Ra b= = . Then, L = 2R, agreeing with our

answer in Q5 above.

Part 2 – Area of a Triangle Formed by a Tangent Line and the Coordinate Axes

In this part of the problem students construct a triangle over the already constructed points and explore how the area of the triangle changes using the dynamic features of the software. Then they can determine symbolic expression for the area of the triangle and analyze this function graphically and analytically. The teacher may ask students to save the existing file that includes constructions and computations for the tangent segment, deleting all unnecessary constructions and calculations. If the teacher decides to start with the blank file, repeat steps 1 – 7 from Part 1. Here are the steps of construction and questions from this point forward.

Select Polygon from the Draw toolbox. Draw a triangle by connecting the origin and the endpoints of the tanget segment already constructed.

Q1. Drag the point of tangency along the circle in the 1st quadrant and observe changes in the area of the triangle. Construct your conjecture about the minimum and maximum area.

A. From their observations and ideas of symmetry students may conclude that the area of the triangle reaches its minimum when the point of tangency divides the arc of the circle equally in two parts, as in the case of the tangent segment discussed in Part 1.

Q2. Find an expression for the area of the triangle algebraically. You can use the software to determine expressions for the lengths of segments.

A. Most students will probably decide to find expressions for the lengths of legs in a right triangle, and then use the area formula. A few students may realize that they can use the length of the hypotenuse as the base and the radius as the height of the triangle. In either case, the formula for the area can be calculated using the software or by hand:

4

2 22RA

x R x=

−. Here are the steps for the calculations with the help of the software:

1. Check the default settings for Show Name by selecting: Edit / Settings / Math. If the Show Name value in the Output box is set to False, click the data line, and select the True entry.

76

2. Click one of the legs of the triangle and choose Distance/Length from the Calculate (Symbolic) toolbox. The software will produce an expression for the length of the segment in terms of the radius of the circle, R, and x.

3. Repeat these steps for the other two sides of the triangle.

4. Select Expression from the Draw toolbox. Click a blank space on the screen and the expression edit box will appear. Using the Symbols palette (or put subscripts in square brackets [ ] ), type an expression for the area of the triangle using the given variables for

the legs: 3 4

2z z⋅ . (Subscripts may differ.) The output expression will be calculated.

5. Repeat step 4 and use a different expression to find the area: 5

2R z⋅

. The same output

expression should appear for the area.

2 4 6 8 10-2-4-6-8-10

2

4

6

8

10

-2

-4

C

E

D

A B

z5⇒R3

x· R2-x2

z4⇒R2

x

z3·z4

2⇒

R4

2·x· R2-x2

R·z5

2⇒

R4

2·x· R2-x2

z3⇒R2

R2-x2

x

R

To confirm their calculations, students can measure the area of the triangle using Area from the Calculate (Symbolic) toolbox.

77

2 4 6 8 10 12 14-2-4-6

2

4

6

8

10

12

-2

-4

E

D

A B

C ⇒R3

x· R2-x2

⇒R2

x

⇒R4

2·x· R2-x2

⇒R2

R2-x2

x

R

Q3. What is the value of x where the function for the area of the triangle has a minimum?

A. At this stage of the problem students already conjectured that the point of tangency divides

the arc of the circle into two equal parts. Thus, 2

2x R= .

Q4. Find the minimal area of the triangle.

A. Students may substitute the value found for x into the function: 2A R= . From geometric considerations, the hypotenuse of the right triangle is equal to twice the length of its height, and the height is equal to the radius in this case, so students can easily calculate the answer:

21 22

A R R R= ⋅ = .

Q5. To confirm that this value is the minimal value of the function, plot a function of the area,

plot the point with coordinates 22 ,2

R R

, and see if this point is at the minimum of the

function. Here are the steps of construction:

1. Right click the expression for the area of the triangle and choose Copy As/ String.

78

2. Select Function from the Draw toolbox. Copy the expression into the Y= line of the Cartesian Function Type dialog, and click OK. The graph of the function will appear on the screen. If you can’t see the function, choose one of the scaling options to adjust the plot.

3. Select Point from the Draw toolbox and draw a point anywhere in the blank space. Select the point and choose Coordinat from the Constrain toolbox.

4. Using the Symbols palette, type 2

R for x0 and R2 for y0. Observe the point jumping to the

function minimum.

5 10 15 20 25-5-10-15

5

10

15

20

-5

-10

H

E

D

A B

C⇒

R3

x· R2-x2

⇒R2

x

⇒R4

2·x· R2-x2

⇒R2

R2-x2

Y=R4

2·X· R2-X2

2 ·R2

,R2

x

R

Part 3 – Perimeter of a Rectangle whose Diagonal is a Tangent Segment

In this part of the problem students consider a rectangle in the 1st quadrant with two adjacent sides lying on the coordinate axes and with the tangent segment used in Parts 1 & 2 above, being a diagonal. Students will explore changes in the perimeter of the rectangle when the point of tangency moves along the arc of the circle. They then determine that the perimeter is minimal when the point of tangency is in the middle of the tangent segment. Students can again use the previous sketch or start with the blank sketch and repeat steps 1 – 7 from Part 1. In this part of the problem students should be able to work independently with minimum

79

teacher’s help. Here are the steps of construction and questions from this point forward.

1. Select Polygon from the Draw toolbox. Draw a quadrilateral by clicking the vertices of the rectangle in the following order: the origin, the point of intersection of the tangent line and the y–axis, an arbitrary point in the 1st quadrant, the point of intersection of the tangent line and x–axis, and the origin.

2. Select the y-axis (outside of the rectangle) and the side of the rectangle opposite to the side lying on the y-axis, and choose Parallel from the Constrain toolbox.

3. Select the x–axis (outside of the rectangle) and the side of the rectangle opposite to the side lying on the x-axis and choose Parallel from the Constrain toolbox.

Q1. Drag the point of tangency along the circle in the 1st quadrant and observe the changes in the perimeter of the rectangle. Construct your conjecture.

A. From the observations and ideas of symmetry students may conclude that the perimeter of the rectangle reaches its minimum when the rectangle becomes square and point of tangency divides the arc of the circle equally in two parts, as in the two previous problems.

Q2. Find an expression for the perimeter of the rectangle algebraically. You can use the software to determine expressions for the lengths of the rectangle’s sides.

A. It is sufficient to find expressions for two adjacent sides of the rectangle using the software. If they have already found these in the previous problem, they only need to use the software to calculate the expression for the sum of the lengths. In either case, the formula for the perimeter

can be calculated using the software or by hand as follows: 2

2 2

1 12P Rx R x

= +

− . Here are

the calculation steps using the software:

1. Check the default settings for Show Name by selecting: Edit / Settings / Math. If the Show Name value in the Output box is set to False, click the data line, and select the True entry.

2. Click one of the sides of the rectangle and choose Distance/Length from the Calculate (Symbolic) toolbox. The software will produce an expression for the length of the side in terms of radius of the circle R and x.

3. Repeat these steps for the adjacent side to the one above.

4. Select Expression from the Draw toolbox. Click a blank space on the screen and the expression edit box will appear. Using the Symbols palette, type an expression for the perimeter of the rectangle using the variables for the sides: 1 22 ( )z z∗ + (subscripts may differ). The output expression will be calculated.

To confirm their calculations, students can measure the area of the rectangle using Perimeter in the Calculate (Symbolic) toolbox.

80

2 4 6 8 10-2-4-6-8-10

2

4

6

8

10

-2

-4

-6

H

C

E

D

A B

⇒R3

x· R2-x2

z2⇒R2

x

z1⇒R2

R2-x2

⇒ 2·R2·1x

+1

R2-x2

2· z1+z2 ⇒ 2·R2·1x

+1

R2-x2

R

x

Q3. What is the value of x where the function for the perimeter of the rectangle has a minimum? A. At this stage of the problem students already conjectured that the point of tangency divides

the arc of the circle into two equal parts. Thus,2

2x R= .

Q4. Find the minimal perimeter of the rectangle.

A. Students can substitute the value found for x into the function: 4 2P R= . From geometric considerations, when the rectangle is a square, its diagonal is 2R, so each side is 2R and the perimeter is four times that value.

Q5. To confirm that this value is a minimal value of the function, plot the function of the

perimeter, plot the point with coordinates 2 ,4 2

2R R

and see if this point is at the

minimum of the function. Here are the steps of construction:

Right click the expression for the perimeter of the rectangle and choose Copy As / String.

1. Select Function from the Draw toolbox. In the Function Type dialog, Cartesian Type,

81

paste the expression into the Y= line and press OK. The graph of the function will appear on the screen. If you can’t see the function, choose one of the scaling options to adjust the plot.

Select Point from the Draw toolbox and draw a point anywhere in the blank space. Select the point and choose Coordinate from the Constrain toolbox.

2. Using the Symbols palette, type 2

R for x0 and 4 2R for y0. Observe that the point

moves to the function minimum.

5 10 15 20 25 30 35-5-10-15-20-25

5

10

15

20

25

30

35

-5

-10

H

C

E

J

D

A B

⇒R3

x· R2-x2

z2⇒R2

x

z1⇒R2

R2-x2

⇒ 2·R2·1x

+1

R2-x2

2· z1+z2 ⇒ 2·R2·1x

+1

R2-x2

Y=2·R2·1X

+1

R2-X2

2 ·R2

,4· 2·R

R

x

83

Two Parabolas


Teaching Goals:

1. Students interpret the given word problem and complete geometric constructions according to the condition of the problem.



o Determine the domain of calculated functions based on conditions of the problem

o Use the software to determine visually if a quantity reaches a minimum or maximum

o Graph an expression as a function of the chosen independent variable to determine the existence of a minimum or maximum

o Determine expressions for the extrema of calculated functions with the help of the software


Prior Knowledge

• Students should know the properties of a rectangle.

• Students should know the equation and graph of a parabola.

• Students should know the formula for finding the vertex of a parabola.

• Students should be able to solve quadratic equations.

Problem:

Given two parabolas defined by the equations 2y x b= − and 2y b x= − . Consider the region enclosed by these two curves. A rectangle with sides parallel to the parabolas’ axis of symmetry is inscribed in this region.

a. Find the extremal perimeter of this rectangle. Determine if this is a minimum or a maximum.

b. Find the extremal length of the diagonal of this rectangle. Determine if this is a minimum or a maximum

c. Will the area of this rectangle be extremal when perimeter and / or diagonal are extremal?

84

Part 1 – Optimizing Perimeter of Rectangle

In this part of the problem students analyze the word problem and complete the geometric construction of two parabolas and an inscribed rectangle according to the condition of the problem. Students then choose an independent variable and define it in the construction. Students can use the dynamic features of the software to analyze the problem qualitatively first, then they can determine the symbolic expression for the perimeter of the rectangle using the symbolic calculation feature and analyze the problem graphically and analytically. Here are the steps of construction and questions.


2. Select Function from the Draw toolbox. In the Function Type dialog enter X^2 – b. The parabola will appear on the graph. Repeat this step and enter function b – X^2. If needed adjust the window using one of the scaling options.

3. Select Point from the Draw toolbox and draw two points on each parabola, one in each quadrant.

4. Click a point in the 1st quadrant on the upper parabola and on the plot of the parabola. With both of them selected, choose Point proportional along curve from the Constrain toolbox and enter x for the value. Here x is the x-coordinate of the point on the parabola.

5. Click a point in the 2nd quadrant on the upper parabola and on the plot of the parabola. With both of them selected, choose Point proportional along curve from the Constrain toolbox and enter -x for the value.

6. Repeat the same steps for the points on the lower parabola, choosing constraint x for the point in the 4th quadrant, and –x for the point in the 3rd quadrant.

7. Select Polygon from the Draw toolbox and draw a rectangle with vertices being these four points.

85

0.5 1.0 1.5 2.0 2.5-0.5-1.0-1.5-2.0-2.5

0.5

1.0

1.5

2.0

-0.5

-1.0

-1.5

C

AB

D

Y=-X2+b

x-x

x

Y=X2-b

-x

Q1. Drag one of the vertices of the rectangle and observe the changes in the perimeter of the rectangle. Do you think the perimeter has a minimum? A maximum?

A. From the observations students may conclude that when vertices of rectangle get closer to the coordinate axes, the perimeter is getting smaller, so students may make a conjecture about existence of maximum. The existence of minimum cannot be established from the visual observations due to the fact that it is not clear if the function for the perimeter is monotonic and since the rectangle does not exist when vertices collapse on the axes and it becomes a segment.

Q2. What is the range of values for the independent variable, x, if the rectangle is inscribed in the region encompassed by two parabolas?

A. Since rectangle does not exist when vertices are on the coordinate axes, the domain is ( ,0) (0, )b b− U .

Q3. Find an expression for the perimeter of the rectangle using software. Plot this expression as a function of x and investigate if this function has an extremal value on the domain.

A. Students can graph the function and determine that the function’s plot has a maximum on the domain. Here are the steps for finding an expression for the perimeter and constructing its graph.

1. Click the rectangle and choose Perimeter in the Calculate (Symbolic) toolbox. The software

86

will produce an expression for the perimeter in terms of b and x.

2. Right click the expression for the perimeter and choose Copy As/ String.

3. Select Function from the Draw toolbox. In the Function Type dialog, Cartesian Type, paste the expression into the Y= line and press OK. The graph of the function will appear on the screen. If you can’t see the function, choose one of the scaling options to adjust the plot.

4. Select point A and the x-axis. With both of them selected choose Perpendicular from the Construct toolbox. The vertical line through point A will be constructed. Students can now move point A and observe on the graph where the perimeter has its maximal value.

1 2 3 4 5-1-2-3-4-5

1

2

3

4

5

6

-1C

AB

D

⇒ 4· x + -2·b+2·x2 + 2·b-2·x2

Y=4· X + -2·X2+2·b + 2·X2-2·b

-x x

Y=X2-b

x-xY=-X2+b

Q4. What are the values of x at the function’s maxima?

A. Students can observe values of x in the Variables toolbox while moving point A and conjecture that x = 0.5 and -0.5. They can also enter the value of 0.5 for x in the Variables toolbox and observe that the vertical line passes through the maximum of the perimeter function.

Q5. Confirm your prediction algebraically.

A. Students should use the expression for the perimeter given by the software. Due to symmetry about y-axis, it is sufficient to consider the case when x > 0. Since x b< , then 22 2 0b x− > , so the expression for the perimeter becomes: 2 24 2(2 2 ) 4( )P x b x x x b= + − = − − − . The

87

expression is quadratic, so the maximum is reached at the vertex of the parabola. Using the

formula for the coordinates of the vertex, we find: 4 18 2

x −= =−

. The value of the 2nd maximum is

12

x = − .

Q5. Find the maximum perimeter of the rectangle.

A. To solve this algebraically students can substitute the found value of x into the function: 21 14 4 1

2 2P b b

= − − − = + . Due to symmetry, at

12

x = − we obtain the same result.

Q6. To confirm that this value is the maximal value of the function, plot the point with coordinates 1 ,4 12

b +

. Here are the steps of the construction:

1. Select Point from the Draw toolbox and draw a point anywhere in the blank space. Select the point and choose Coordinat from the Constrain toolbox.

2. Enter 1/ 2 for x0 and 4* 1b + for y0. Observe the point jumping to the function maximum.

1 2 3 4 5 6-1-2-3-4

1

2

3

4

5

-1

-2

C

AB

F

D

⇒ 4· x + -2·b+2·x2 + 2·b-2·x2

Y=4· X + -2·X2+2·b + 2·X2-2·b

-x x

Y=X2-b

12

,1+4·b

x-xY=-X2+b

88

Part 2 – Optimizing the Diagonal of the Rectangle

In this part of the problem students explore how the diagonal of the rectangle changes using the dynamic features of the software first, then they can determine the symbolic expression for the diagonal of the rectangle and analyze the function graphically and analytically. The teacher may ask students to save their existing file, including constructions and computations for the rectangle, and delete all unnecessary constructions and calculations. If the teacher decides to start with the blank file, repeat steps 1 – 7 from Part 1. Here are the steps of construction and questions from this point forward.

Select Line Segment from the Draw toolbox. Draw a segment AC (connecting the vertex in the 1st quadrant with the vertex in the 3rd quadrant).

Q1. Drag one of the vertices of the rectangle and observe the changes in the length of the diagonal of the rectangle. Do you think this length has a minimum? A maximum?

A. From the observations students may conclude that when vertices of the rectangle get closer to the coordinate axes, the diagonal length is getting larger, so students may make a conjecture about existence of minimum. The maximum does not exist since diagonal is largest when the vertices of the rectangle lie on the y-axis, depending on the value of b, but the rectangle does not exist at these points.

Q2. Find an expression for the length of the diagonal using the software. Plot this expression as a function of x and investigate if this function has an extremal value on the domain.

A. Students can graph the function and determine that the function plot has a minimum on the domain. Here are the steps for finding an expression for the length of the diagonal and constructing a graph.

1. Click the diagonal and choose Distance/Length from the Calculate (Symbolic) toolbox. The software will produce an expression for the length of the diagonal in terms of b and x.

2. Right click the expression for the perimeter, choose Copy As/ String.

3. Select Function from the Draw toolbox. In the Function Type dialog, Cartesian Type, paste the expression into the Y= line and press OK. The graph of the function will appear on the screen. If you can’t see the function, choose one of the scaling options to adjust the plot.

89

0.5 1.0 1.5 2.0 2.5-0.5-1.0-1.5-2.0-2.5

0.5

1.0

1.5

2.0

2.5

-0.5

-1.0

D

B

C

A⇒ 4·x2+ -2·b+2·x2 2

-x

Y= 4·X2+ 2·X2-2·b2

Y=-X2+b

x-x

x

Y=X2-b

Q3. What is the value of x where the function has minimum?

A. Students should use the expression for the length of the diagonal given by the software. Consider the expression under the square root. 2 2 2 2 4 2 24 (2 2 ) 4 4 8 4x x b x x x b b+ − = + − + . This is a biquadratic expression, so we can use 2z x= and write the expression in terms of z:

2 24( (1 2 ) )z b z b+ − + . The minimal value is achieved at the vertex of the parabola, at

22 12

bz x−= = , and thus the function has a minimum at

12

x b= ± − .

Q5. Find the minimum length of the diagonal.

A. To solve this algebraically students can substitute the value of x found above into the function: 2

21 14 2 2 4 2 ( 1) 4 12 2

L b b b b b = − + − − = − + − = − .

Q6. To confirm that this value is the maximal value of the function, plot the point with coordinates 1 , 4 12

b b

− −

. Adjust the position of point A and confirm your results numerically using

90

specific values for b. Here are the steps of construction:

1. Select Point from the Draw toolbox and draw a point anywhere in the blank space.

2. Select the point and choose Coordinat from the Constrain toolbox.

3. Enter ( 0.5)sqrt b − for x0 and (4* 1)sqrt b − for y0. Observe the point jumping to the function minimum.

4. Select both point A and the x-axis. Choose Perpendicular from the Construct toolbox. The vertical line will appear on the graph.

5. Move point A so that the perpendicular intersects the minimum point on the function. Observe the numerical values of b and x in the Variables toolbox. Use the formulas that you found to verify that point A is at the minimum.

0.5 1.0 1.5 2.0 2.5-0.5-1.0-1.5-2.0-2.5

0.5

1.0

1.5

2.0

2.5

-0.5

-1.0

F

D

B

C

A

⇒ 4·x2+ -2·b+2·x2 2

-x

-0.5+b, -1+4·b

Y= 4·X2+ 2·X2-2·b2

Y=-X2+b

x-x

x

Y=X2-b

A. For b = 1, using the formula for x we get: 0.5 1 0.5 0.5 0.707x b= − = − = = , which corresponds the value of x = 0.703 (or a value very close to this) shown in the Variables toolbox with high precision.

91

Part 3 – Comparison of Points of Extrema for Perimeter, Diagonal, and Area.

In this part of the problem students investigate if points of extrema for the perimeter, diagonal, and area of the previously inscribed rectangle can be the same. Using dynamic and symbolic features of the software, students will plot expressions for all three quantities and compare graphs of these functions on the domain. The teacher may ask students to save the existing file that already includes constructions for the rectangle, and delete all unnecessary constructions and calculations. If the teacher decides to start with the blank file, repeat steps 1 – 7 from Part 1. Here are the steps of construction and questions from this point forward.

1. Select Line Segment from the Draw toolbox. Draw a segment AC (connecting the vertex in the 1st quadrant with the vertex in the 3rd quadrant).

2. Check the Edit / Settings / Math dialog to make sure that the Use Assumptions value in the Output box is set to False. (Otherwise your area and perimeter graphs will be quite different.)

3. Plot the expression for the perimeter, diagonal, and area of the rectangle with the following steps:

a. Click the object (rectangle or diagonal of the rectangle) and choose the appropriate tool from the Calculate (Symbolic) toolbox. The software will produce an expression for the corresponding function in terms of b and x.

b. Right click the expression for the perimeter, choose Copy As / String.

c. Select Function from the Draw toolbox. In the Function Type dialog, Cartesian Type, paste the expression into the Y= line and press OK. The graph of the function will appear on the screen. If you can’t see the function, choose one of the scaling options to adjust the plot.

4. Choose different colors for each function plot by right clicking on the plot, and using Properties from the context menu.

5. Hide expressions for the functions by right clicking the expression and choosing Hide from the context menu.

6. Select both, point A and the x-axis. Choose Perpendicular from the Construct toolbox. A vertical line will appear on the graph.

92

1 2 3 4-1-2-3-4

1

2

3

4

5

-1D

B

C

A

⇒ 4·x2+ -2·b+2·x2 2

⇒ 4· x + -2·b+2·x2 + 2·b-2·x2

⇒ 4·b·x-4·x3

-x

x-x

x

Q3. Does area of the rectangle have a maximum or minimum on the domain?

A. As it is seen from the plot of the function, the area has a maximum value at two points that are symmetrical.

Q4. Are points of extremum for the perimeter and the area the same? For the diagonal and the area?

A. No, these points do not necessarily coincide. Students can drag point A and observe the vertical line through all three functions.

Q5. Is it possible for the points of extremum to coincide for all three functions? Explore using dynamic features of the software.

A1. Students can vary the numerical value of b in the Variables toolbox and try to achieve this condition. The steps of construction are the following:

1. In the Variables toolbox select parameter b.

2. Enter a minimum and maximum value for b (in the data entry boxes at the bottom left and right corners of the dialog).

3. Use the slider to adjust the value of b and observe the graphs of the functions.

93

4. Move point A so that the perpendicular intersects the extremum of one of the functions and compare it with the extrema on the other functions for a particular value of b.

Q6. Using expressions for points of extrema found in parts 1 and 2, find a value of b where the rectangle of maximum perimeter is also the rectangle with the minimum diagonal. Will this rectangle’s area be at a maximum?

A. Algebraically, students can set equal the point of maximum for the perimeter function and the point of minimum for the diagonal length function they found in Parts 1 and 2 of the

problem:1 12 2

b − = , so 3 0.754

b = = . Now, they can enter this value for b in the Variables

toolbox to see if the area function for this value of b has a maximum at the same point. It appears on the graph that the area does have a maximum at this point.

Q7. To confirm that this value is the maximal value of the function, plot a tangent line at this point and determine the slope of the tangent line. What is the value of the slope?

A. The real calculation of the slope should have a value of 0, which confirms that when 34

b = , 21

is the x value for the maximum area of the rectangle. Here are the steps of construction:

1. Select Point from the Draw toolbox and draw a point on the function of the area.

2. Select the point and the area function and choose Point proportional along curve from the Constrain toolbox. Enter 0.5 for the constraint value and the point should move to the intersection between the graph of the function and the vertical line.

3. Select this point and the plot of the function and choose Tangent from the Construct toolbox. The tangent line will be constructed.

4. Click the tangent line, choose Slope from the Calculate (Real tab) toolbox. The value of the slope will appear on the screen.

5. To avoid accidental changes in the values of b and x, enter these values in the Variable toolbox and lock them (select the variable and click the lock icon just below the variables list box on the right).

94

1 2 3 4-1-2-3-4

1

2

3

4

5

-1

D

B

F

C

A

⇒ 4·x2+ -2·b+2·x2 2

⇒ 4· x + -2·b+2·x2 + 2·b-2·x2

⇒ 4·b·x-4·x3

⇒∼0

x-x

x

0.5

-x

using symbolic geometry to teach secondary school ... 2 and... · the national council of teachers...

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