SYMBOL OF EACH SECTION HEADER TO RETURN HERE.
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TEKS Support |
This unit contains activities that support the following knowledge and skills elements of the TEKS.
(1) (A) |
X |
(4) (A) |
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(1) (B) |
X |
(4) (B) |
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(1) (C) |
X |
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(2) (A) |
X |
(8) (A) |
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(2) (B) |
(8) (B) |
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(2) (C) |
(8) (C) |
X |
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(2) (D) |
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(3) (A) |
X |
(9) (A) |
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(3) (B) |
(9) (B) |
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(3) (C) |
X |
The mathematical prerequisites for this unit are
The mathematical topics included or taught in this unit are
The equipment list for this unit is
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Teacher Notes |
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Context Overview |
Animation serves as the context for the investigations within this unit. While many features of illusions are mathematical in nature, the focus of this unit is on analyzing their "micro" structure—the motions of single points within a stationary coordinate system. We specifically avoid investigating the relative motions of points within a moving object; that is, changes in the shape or orientation of an object as it moves.
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Mathematical Development |
The unit is designed around the concept that mathematical language models real motion in a useful way. The initial motion investigated is that of a single point moving along a line. Variables, as words or symbols, arise to identify (describe) quantities, and equations describe relationships among variables more efficiently than do tables.
The power of these mathematical models lies in their abilities to: (i) provide insight into how to predict values of some quantities when the values of others are known and (ii) extend to similar but more complex situations such as multiple objects or motion in the plane.
The early portion of the unit characterizes constant motion along a line so that the roles both of initial location and of velocity are clear. Time is measured both directly and in frames of an animation sequence. Linear functions provide the models for such situations.
Subsequent development shifts to motion in the plane and multiple representations of such motion. Parametric equations become a natural way to describe motion when the path is not on a given axis.
Graphing calculators are required in each lesson after the first, and several programs are supplied for student use. Students should modify these as their understanding of the material and of programs develops. Matrices organize the description of several points moving simultaneously. Although this provides a method by which shape changes may be rendered, we do not directly address that application in this unit.
Finally, while the unit develops the mathematical description of motion in the "discrete world" of animation, the results are equally applicable to real motion in the "continuous world."
The unit project acts as a culminating experience for students and, ideally, will reinforce their mathematical understanding of key concepts in the unit.
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Unit Project |
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The unit project is actually a final project in this case. Students will build up the knowledge as they progress through the unit. Prior to Activity 7, ask the students to decide on a figure that they will animate. Check those prior to assigning the final project to guarantee that the students have thought about what they are trying to accomplish.
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Activity 1—A Living Marquee |
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Materials needed: |
Enough construction paper in two different colors so that each pair of students can have one sheet of each color, or enough identical photos (ideally of a student or you) so that each pair of students can have one. |
One goal of this activity is to have students describe horizontal motion prior to discussing the difference between recursive and closed-form descriptions. As students work through this activity, listen for examples of both recursive and closed-form types of thinking (see the notes on Item 5 below).
Figure 1. Time = 0
Figure 2. Time = 1
Item 1
One way to approach this is to divide the class into two groups, or some even number with the number of students at least 6 in each group. One group could be the marquee, the other the observers. Take the marquee group into a separate area and have them practice. Ask them to provide the "motion" with their eyes open, then with them closed. While that group practices, discuss with the observers the idea that what they are looking for is how the marquee members decide when to "turn on" and "turn off" their points.
Item 3
Remind students that tables are usually samples of possible values and seldom display all possible values. Because the marquee will probably be from 6 to 10 frames, it is possible to list every value in the table. Notice whether students draw graphs with discrete points or continuous lines to represent location versus time (or frame).
Item 4
Encourage students to be creative in designing their own marquee motion. Listed below are several ideas that you can suggest in case students need help with designing a different marquee.
You may want to write each suggestion on a card and use them for a competitive activity. Suppose you have divided your class into 3 groups of 10 students. Group 1 draws a card. (Groups 2 and 3 can draw the cards at the same time so all three groups prepare at the same time.)
Group 1 performs its card’s marquee. The first group (either 2 or 3) to guess what is written on Group l’s card is awarded points. Group 1 may also be awarded points based on how quickly its card’s description is guessed (keep track of time).
Repeat the process with Group 2 and Group 3 presentations. You have enough cards or ideas for three rounds. You can do all three if you have the time and students enjoy the challenge.
You or your students may suggest other ideas for this activity.
Item 5
Item 5 is a key question. Pay attention to descriptions that discuss arriving at a certain number prior to turning a card (when the count got to 10, I flipped my card [closed]) or watching the person next to them (when Michelle flipped her card, I knew that I was one count later [recursive]). Both recursive and closed-form equations are important.
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Homework 1—Making a Point |
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These items could be done as a homework assignment and lead into a discussion of the two calculator programs.
These items are included to give students a sense of how to control the location and speed at which the dots can "move."
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Activity 2—Calculator Marquee |
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Materials needed: |
Graphing calculator with appropriate programs loaded CARTOON1, CARTOON2, and AADEFAUL calculator programs for your calculators Transparency 1 |
Note: If you have classroom sets of calculators, you may want to load the programs CARTOON1, CARTOON2, and AADEFAUL into all the calculators prior to this lesson. Another option is to load one calculator, then ask the students to transfer the programs to their calculators. You may also want to display these programs on a viewscreen.
The purpose of this lesson is to allow students to think about the movement of points along a horizontal axis. The second emphasis is to introduce students to calculator programming.
Items 1–5
These items should be done in class. This may be the students’ first look at programming and they will probably need some assistance.
Have students run the programs CARTOON1 and CARTOON2 and respond to Items 1 and 2. Discuss their observations as a class.
Using a transparency of the listing for CARTOON1, ask students to follow along as you step through the program. After this, have students continue through Items 3–5. Allow them to work in pairs or small groups on these items.
When students have completed these items, lead a class discussion about the results. Be sure that responses include how they decided what needed changing and the results of each change.
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Homework 2—Up in Lights |
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Handout 1 (optional)
The first part of this section relates to the previous homework on looking at individual data points and relating those to the movement of a point on the graphics screen. It is straightforward and should build students’ confidence. The second part of the section deals with the mechanics of programming a calculator. For purposes of this activity, TI commands are used. If students are using TI-83s, Handout 1 will be helpful. If some other calculator is used, it will be necessary to use different commands.
Items 1–6
These items reinforce what students did on the previous homework assignment. Students are asked to make connections between location, velocity, and time in relation to "moving" objects. Students can do this as homework with some discussion taking place the following day.
Items 7–14
This is the students’ introduction to actually programming a calculator. This part of the homework may be done in class or out, depending on whether the students can take the calculators home.
Regardless of the location, every student should do this part of the activity on her/his own. It is important that they become comfortable with entering and editing calculator programs so that when the final project time arrives, they will feel confident. Ask students to present their results to the class by transferring their programs to the overhead unit and displaying them for the class to see.
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Activity 3—Addressing a Letter |
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Materials needed: |
1 sheet of grid paper without coordinates 1 sheet of grid paper with coordinates |
Students begin to think about moving multiple points using the same rules.
Items 1–3
Using the grid paper without coordinates, the students should be able to create their initial(s) on the sheet, shading in the squares (pixels). Some may want to do more than a single letter. Let them know that they may want to use that for their final project. Also ask them to realize that the modeling process is designed to begin with simple examples and then continue to more complex examples. If you have enough time and resources, students could make successive pictures of their initial, using the information from Item 3. They could then create flip books that could be used to demonstrate motion.
Items 4–9
This section of the activity could be done either as a homework or an in-class assignment and is straightforward in its approach. Be sure to review the results of this second part with the entire class.
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Homework 3—Thinking About Figures |
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Materials needed: |
The completed grid sheets from Activity 3 Graphing calculator |
Items 1–2
Students can do these two whether they have access to calculators or not. With some general instruction, students are asked to organize the data they created in Activity 3 into a matrix. They are then asked to add two matrices—one that contains the data from their initial, and one that contains ones in the first row and zeros in the second. They are then asked to plot the resulting matrix. Through this they should find that the shape of the initial is retained, but the position has changed.
Item 3
If students do not have access to graphing calculators to take home, this should be done as a classroom activity, with each student working through the item individually. This item should give students an idea of how to move figures on the graphics screen. They may decide to put the program AADEFAUL into the TEST program rather than executing it from the home screen. They may also choose to set up the window values within the program (Xmin, Xmax, Ymin, Ymax).
Again, these commands are written for TI calculators. If a different type is used, the format will change depending on the difference in commands.
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Activity 4—What’s My Line? |
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The purpose of this activity is to introduce students to parametric equations. Students may work on the activity in pairs, in small groups, or individually. Have students compare results with one another as they proceed through this activity.
Items 1–7
The main emphasis in these items is the conversion between various representations of locations. The reason we can represent these in this way is that the motion is either strictly vertical (balloon) or strictly horizontal (car).
Item 7 moves students into the realm of diagonal motion.
Item 8 walks students through entering and graphing parametric equations. It also considers an appropriate viewing window. This item prepares students for the homework assignment.
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Homework 4—Escalating Motions |
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Materials needed: |
Graph paper |
This activity reinforces the idea that motion in a plane is made up of three components: vertical motion, horizontal motion, and time. Breaking the motion into its components allows one to control each separately.
Items 1–3
These items are designed to have students convert between time-series and path graphs and notations.
Students should conclude that a path graph is inadequate for determining when an object reaches a particular location. Students should also conclude that a single time-series graph does not describe the motion of an object in two dimensions. Two time-series graphs, one for each component, are necessary.
Item 4
Item 4d asks students to use the parametric mode on their calculator. This will have to be done in class if students are not able to take graphing calculators home.
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Activity 5—Going to the Movies |
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This purpose of this activity is to demonstrate how to use matrices in an animation program. Some of this was part of Homework 3, "Addressing a Letter." Many of the lines used in that homework assignment can be used as a basis for this activity.
Give students time to "play" after they have created their first program.
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Unit Project |
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In this project, students should apply all the skills and mathematical concepts learned in the unit. You may want to modify the checklist for the project to include items that you feel are necessary and/or remove others that were not emphasized.
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Supplemental Materials |
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Transparency 1 |
AADEFAUL
AxesOff:FnOff
–1® Xmin:93® Xmax
–1® Ymin:61® Ymax
ClrDraw
0®H
Pt-On(H,30)
Pause
For(T, 0, 100, 1)
Pt-Off(H–2,30)
Pt-On(H,30)
Pause
H+2® H
End
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Transparency 2 |
AADEFAUL
AxesOff:FnOff
–1® Xmin:93® Xmax
–1® Ymin:61® Ymax
ClrDraw
0®T
Pt-On(T*2,30)
Pause
For(T,1,50,1)
Pt-Off((T–1)*2,30)
Pt-On(T*2,30)
Pause
End
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Handout 1—Calculator Animation with a TI-83 |
This handout is for use with the TI-83 graphing calculator. It must be modified for use with other calculators. Assume that the ENTER key should be pressed after each keyboard procedure.
PREPARE THE CALCULATOR TO DISPLAY GRAPHICS.
TURN ON A POINT AT A SPECIFIC LOCATION.
TURN OFF A POINT AT A SPECIFIC LOCATION.
WRITE A PROGRAM TO TURN ON SEVERAL POINTS, ONE AT A TIME, TO DRAW A HORIZONTAL LINE.
0,STO, VARS, Window, Xmin
98,STO, VARS, Window, Xmax
0,STO, VARS, Window, Ymin
64,STO, VARS, Window, Ymax
Figure 1
Figure 2
Figure 3
Insert the two statements of the timing loop immediately following the command to turn on a point: PRGM, EDIT, Select LINE
Use the down arrow and move the cursor to the line :End
2nd INS
The display on the screen should have a blank line
(Figure 4): Arrow up to the blank line.
Figure 4
:End
Insert the lines:
:For(N,1,50,1)
See Figure 5.
You may save space by combining two lines in one by using a colon to separate the two statements or commands:
:For(N,1,50,1):End
Figure 5
WRITE A PROGRAM TO MAKE A SINGLE POINT FLASH ON AND OFF SEVERAL TIMES.
PRGM, EXEC, Select AADEFAUL
Include the following statements in the Blinker program (Figure 6):
0, STO, VARS, Window, Xmin
98, STO, VARS, Window, Xmax
0, STO, VARS, Window, Ymin
64, STO, VARS, Window, Ymax
PRGM, CTL, For(H,1,50,1)
2nd DRAW, POINTS, Pt-On(30,20,2)
Figure 6
PRGM, CTL, For(T,1,50,1)
PRGM, CTL, End
2nd DRAW, POINTS, Pt-Off(30,20,2)
PRGM, CTL, For(T,1,50,1)
PRGM, CTL, End
PRGM, CTL, End
See Figure 7.
Refer to the directions given in steps 1-17 of this handout, and complete Items 18, 19, 20, 21, and 22 in Activity 3.
Figure 7
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HANDOUT 2—Listing for AADEFAUL (TI-83 version) |
Float
Radian
Func in MODE menu
Connected
Sequential
Full
RectGC
CoordOn
GridOff in FORMAT menu
AxesOff
LabelOff
Plotsoff in STATPLOT menu
FnOff in VARS, Y-VARS menu
Zstandard in ZOOM menu
ClrHome in PRGM 1/0 menu
ClrDraw in DRAW menu
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Annotated Student Materials |
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Introductory Reading |
Horror and science fiction movies, cartoons, video games, and MTV can hold your attention for hours at a time. You gasp as a person changes into an alien being, laugh as the cartoon character falls out the window, peer warily down an alley as characters search for an enemy, and sway with the images put to music.
Great advances have been made since the early days of animation. Imagination, mathematics, and advanced technology have brought to life special effects such as morphing (the gradual transformation from one shape to another) and virtual reality (the creation of the illusion that you are seeing from within the animator’s world).
This particular unit is about animation. By the end of this unit, you will be asked to create your own animation, so the central question for you to consider as you work through the unit is "How do they do that?"
To answer this question, you will need to use the modeling process. You will begin by looking at very simple animation, that of moving one dot. You will learn to move it horizontally, vertically, and, finally, diagonally. You will then add complexity by moving a figure rather than just one dot.
To think about answering the "how" question, you must first have a language for describing "what" is going on. Think about what seems simple and what seems more complicated. What features do the simple things have in common with the complex ones?
Finally, you will design and create an animation of your own. What your final product looks like will depend mostly on how well you learn the language of mathematics.
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Activity 1—A Living Marquee |
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Materials needed (per pair of students): |
Two sheets of construction paper (different colors) Staples, glue, or tape |
= 1, 2, 9, 11
Marquee lights look like a parade of dots all moving in the same direction. Each dot seems to move along in a line. In this activity, you simulate the movement of a single dot or point in a marquee.
The purpose of this activity is to determine the mathematical language needed to tell a computer or calculator how to move a single point in a horizontal direction.
Figure 1
Examples might include Current location = Previous location + 1, Start = 1 if each person flips his or her card, or Current location = Previous location + 2, Start = 1 if every other person flips his or her card. A table might look like:
Time |
Location |
0 |
1 |
1 |
2 |
2 |
3 |
for a marquee sign that starts at the first person and goes 1 card per count.
Demonstrate your creative marquee for the class.
Examples might include "I looked at the person next to me" (recursive) or "I waited for the count to reach 10" (closed form).
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Homework 1—Making a Point |
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Words seem to move across the screens of marquees. Words are made of letters, and, on marquees, letters are made of points or dots (pixels) of light. You can simplify the task of trying to describe the movement of words by analyzing the movement of one point of a letter.
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Time |
Column |
0 |
44 |
1 |
38 |
2 |
32 |
3 |
26 |
Start in column 44. Velocity is –4 columns per second.
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Time |
Column |
0 |
50 |
1 |
47 |
2 |
44 |
3 |
41 |
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Time |
Column |
0 |
5 |
1 |
8 |
2 |
11 |
3 |
14 |
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Time |
Column |
0 |
120 |
1 |
108 |
2 |
96 |
3 |
84 |
Start in column 120. Velocity is –4 col/sec or –12 col/frame (if each entry is one frame).
15 seconds
Column 36
Column values would go up by 12 instead of down.
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Time |
Row |
0 |
24 |
1 |
22 |
2 |
20 |
3 |
16 |
Motion begins in row 24 and moves down at two rows per second.
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Activity 2—Calculator Marquee |
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They seem identical. Each moves a single dot horizontally across the center of the screen at two dots per frame.
= 3, 4, 5
The calculator allows for an entire program to be copied into another one, in case you want to edit one of your existing programs to avoid writing everything from scratch. This means that a fresh copy can be edited without messing up the original. To do this for the TI calculator, go into the EDIT mode in the program that is designated to receive the copy. Then press RCL 2nd STO, then PRGM, followed by EXEC. Next, select the program to copy, then press ENTER. (Most of the time you will probably want to do this in a NEW program that does not yet have commands. If that is the case, instead of using EDIT, press PRGM, then NEW, and name the new program.) Below are two familiar programs, along with notes about what they do and how to find their commands.
CARTOON1
Command |
Description |
AADEFAUL |
Runs a program that resets calculator modes to "factory settings." |
AxesOff:FnOff |
These two commands turn off the display of coordinate axes and functions in your Y= list. A colon is used to separate commands whenever more than one command is used on a line. |
–1>Xmin:93>Xmax –1>Ymin:61>Ymax |
These four commands set the graph window so that each pixel has integer coordinates. |
ClrDraw |
This erases any previous animations and other drawn figures. |
0 > H |
H is the variable name to be used to keep track of the horizontal part of which pixel is on. |
Pt-On(H, 30) |
The pixel at horizontal location H and vertical location 30 is turned on. |
Pause |
This makes the calculator wait until ENTER is pressed before displaying the next frame. |
For(T, 0, 50, 1) |
This command goes with the END command below. T counts the "calculator seconds," going from 1 to 50, one per frame. When T reaches 50, the program and animation stop. |
Pt-Off(H-2,30) |
The pixel at horizontal location (H-2) and vertical location 30 is turned off. |
Pt-On(H,30) |
The pixel at horizontal location H and vertical location 30 is turned on. |
Pause |
Wait for next frame ENTER. |
H+2 > H |
This adds 2 to the current H value and uses the answer as the next H value; this is the recursive step. |
End |
This command goes with the For(T,1,50,1) command above. It tells the calculator how much of the program to repeat as T changes. |
Note: All pixels are required to have both horizontal and vertical location numbers. Therefore, even though we know that the vertical locations in the program above will never change, they must still be mentioned in the Pt-On and Pt-Off commands.
CARTOON2
AADEFAUL |
These commands work just as they did in CARTOON1. |
AxesOff:FnOff |
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–1>Xmin:93>Xmax –1>Ymin:61>Ymax |
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ClrDraw |
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0 > T |
Set the first time. |
Pt-On(T*2,30) |
This calculates the value of T*2 and uses that number as the horizontal location of the pixel to be turned off. The vertical location is still 30—the initial location. |
Pause |
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For(T,1,50,1) |
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Pt-Off((T–1)*2,30) |
This calculates the value of 2(T–1) and uses that number as the horizontal location of the pixel to be turned off. The vertical location is still 30. |
Pt-On(T*2,30) |
This calculates and turns on the next pixel. |
Pause |
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End |
H+2>H
Pt-On(H,30)
The corresponding portion of CARTOON2 looks like:
Pt-On(T*2,30)
The Pt-On command requires that you specify the horizontal and vertical locations of the pixel that will be turned on. The first number in the parentheses is the horizontal location, the number after the comma is the vertical location.
Points are turned on. H is the horizontal location of the "on" pixel, and it changes by 2, so the velocity is 2 pixels per frame (or per T) to the right. Likewise, T*2 is a horizontal location of an "on" pixel, so the velocity is also two pixels per frame to the right.
CARTOON1 adds 2 to the column number (H) to get the next column number (H + 2)—it’s recursive. CARTOON2 gets the column number by 2*time—it’s closed form.
These turn off the previous pixel. Since the speed is 2, H–2 is the location of the former pixel in recursive form. Likewise, T–1 is the previous time for the closed form.
Change the instructions to change the velocity of the moving point. Explain what you did, what the new velocity should be, and why your change is the way to do it.
Changing the "2" will change the velocity in either program. But remember to change both Pt-On and Pt-Off.
In CARTOON1 change the 0 in "0
> H" to a new starting column. The starting column is 0 in CARTOON2 too, so adding a different value to 2*T (and 2*(T–1) in the turn-off line) will do the job. |
Homework 2—Up in Lights |
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Time |
Column |
0 |
25 |
1 |
22 |
2 |
19 |
3 |
16 |
Complete the data chart below to represent a letter that starts at column 39 and moves 4 columns to the left each second.
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Time |
0 |
1 |
2 |
3 |
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Column |
39 |
35 |
31 |
27 |
|
Time |
0 |
2 |
4 |
6 |
8 |
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Column |
51 |
42 |
33 |
24 |
15 |
|
Time |
Column |
0 |
38 |
1 |
35 |
2 |
32 |
3 |
29 |
Starts in column 38. Velocity is 3 col/sec.
18 seconds
Column 20
The time column would remain unchanged, but the column values would differ by 6 from row to row.
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Time |
2 |
3 |
… |
5 |
6 |
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Column |
22 |
17 |
… |
7 |
2 |
Column 32. Count by 5s two steps; or velocity is –5 col/sec so displacement in first 2 sec is (2 sec)*(–5 col/sec) = –10 col so forward 10 col undoes that motion; or graph and find equation and see "y-int" = 32; ....
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Time |
2002 |
2003 |
… |
2005 |
2006 |
|
Column |
22 |
17 |
… |
7 |
2 |
Column 10032 (B I G marquee!!)
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Time |
0 |
1 |
2 |
3 |
4 |
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Column |
32 |
37 |
42 |
47 |
52 |
Column 4
Column 25
11 seconds
Find the displacement by subtracting initial location from final location. Then find velocity by dividing displacement by 5 sec.
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Programming a Calculator |
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Activity 3—Addressing a Letter |
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= 8, 10, 13, 18
In Activity 1 you simulated the movement of one point across the screen or sign. In this activity you will simulate an entire set of points moving in the same way. To do this, it is helpful to think of a letter being made up of a set of points. If you analyze the movement that takes place with one of these points, you can generalize that analysis to the entire set. Again, the idea of simplifying to understand a process arises.
Example: The left edge of my initial is in the fifteenth column from the left edge.
Numbers have been assigned to the columns of lights in the second marquee grid. Matching of columns with numbers sets up a coordinate system on the marquee that helps you communicate positions of letters. (You may already have numbered the columns when you explained where you placed your first initial.)
Left edge |
Timer |
25 |
0.0 |
26 |
0.1 |
In column 27. In column 29. In column 40.
10 columns. 10 columns per second.
L = 10t + 25 where L is the location and t is in seconds or
Location New = Location Previous + 10, Location Initial = 25
At column 35. At column 50, or the right edge. Since it moves 10 columns each second, it goes 10 columns (25 columns).
Add ten times the timer number to 25 to find the column number for the left edge.
At time 2.5 seconds. At time 1 second. Since the initial moves 10 columns each second and it has to go 25 columns, it takes 25/10 seconds (or: two and one-half blocks 10 columns wide make 25 columns, and each block takes one second).
m e rMove two columns each display change. The letter moves twice as far (twice as many columns) in the same amount of time. Faster speed is now 20 columns per second.
Add twenty times the timer number to 25 to find the column number for the left edge.
The thing that changes is that the addition changes to subtraction (or adding the opposite).
In column 5
In column 9
Two columns
7.5 seconds. Starting from column 5, it needs to go a total of 15 (or 20 – 5) more columns to reach column 20. At 2 columns per second, that’s 7.5 seconds.
Multiply speed by the elapsed time and add that to the starting position to get the location of the object. (This uses a "positive to the right, negative to the left" speed convention.)
Old Location |
New Location |
Time |
Location |
|
25 |
26 |
0.0 |
25 |
|
26 |
27 |
0.1 |
26 |
|
27 |
28 |
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Old Location |
New Location |
Time |
Location |
|
25 |
27 |
0.0 |
25 |
|
27 |
29 |
0.1 |
27 |
|
29 |
31 |
0.2 |
29 |
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31 |
33 |
0.3 |
31 |
|
33 |
35 |
0.4 |
33 |
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Old Location |
New Location |
Time |
Location |
|
25 |
24 |
0.0 |
25 |
|
24 |
23 |
0.1 |
24 |
|
23 |
22 |
0.2 |
23 |
|
22 |
21 |
0.3 |
22 |
|
21 |
20 |
0.4 |
21 |
|
Old Location |
New Location |
Time |
Location |
|
25 |
23 |
0.0 |
25 |
|
23 |
21 |
0.1 |
23 |
|
21 |
19 |
0.2 |
21 |
|
19 |
17 |
0.3 |
19 |
|
17 |
15 |
0.4 |
17 |
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Marquee Simulation #1
Marquee Simulation #2
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Homework 3—Thinking about Figures |
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A matrix can be used to help organize data. One way to use a matrix in working with data points is to have the first row of the matrix include all the horizontal components and the second row include all the vertical components. For example, the point (3,2) can be stored in a matrix as 
. Storing multiple points may be done the same way. Storing (3,2) and (4,7) in a matrix would look like: 
.
Sample: For the letter E stored on the left edge and five high, the matrix might look like:
For example: 
+ 
= 
.
A matrix of size 2 × n with all ones in the first row and zeros in the second.
Sample: 
Sample for the letter E from Item 7:
Result should be the same letter, translated to the right 10 and up 8. The shape stays the same, the position is just different.
Hint: To enter the matrices:
Hint: MATRX [A] + MATRX [B] STO MATRX [C]
The resulting matrix should match the second matrix you found in Item 2b.
Note: N is the number of points in your letter.
For(I,1,N,1)
Pt-On([A](1,I),[A](2,I))
Pt-On([C](1,I),[C](2,I))
End
The screen should display the initial twice—once in the original position and the other translated 10 to the right and 8 up from the original.
After the END statement put in the following lines:
For(I,1,N,1)
Pt-Off([A](1,I),[A](2,I))
End
Describe your results.
The only thing on the screen should be the initial in the new position.
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Activity 4—What’s My Line? |
|
= 12, 14, 15
As figures move across the screen, three different components must be addressed—Horizontal position, Vertical position, and Time. The following items are designed to help you to think about these three. Assume that you are watching a balloon drift up or down as a car moves left to right on the screen.
Car
Time |
Horizontal |
0 |
24 |
1 |
28 |
2 |
32 |
3 |
36 |
Balloon
Time |
Vertical |
0 |
50 |
1 |
45 |
2 |
40 |
3 |
35 |
Car |
Balloon |
|
Time |
Horiz. |
Vert. |
0 |
24 |
50 |
1 |
28 |
45 |
2 |
32 |
40 |
3 |
36 |
35 |
Car |
Balloon |
|
Time |
Horiz. |
Vert. |
0 |
9 |
20 |
1 |
13 |
30 |
3 |
21 |
50 |
6 |
33 |
80 |
Car
Time |
Horizontal |
0 |
9 |
1 |
13 |
3 |
21 |
6 |
33 |
Balloon
Time |
Vertical |
0 |
20 |
1 |
30 |
3 |
50 |
6 |
80 |
Car: Starts at (15,0) and moves 3 pixels to the right each frame.
Balloon: Starts at (0,28) and moves 2 pixels up each frame.
Car |
Balloon |
|
Time |
Horiz. |
Vert. |
0 |
15 |
28 |
1 |
18 |
30 |
2 |
21 |
32 |
3 |
24 |
34 |
Car |
Balloon |
|
Time |
Horiz. |
Vert. |
0 |
14 |
8 |
1 |
18 |
13 |
2 |
22 |
18 |
4 |
30 |
28 |
Frame #8 (height = 48)
Frame #13 (location = 66)
Location 42
9 pixels (difference in velocities of 1 pixel per frame, or 95 — 9*4. Note distributive law here.) *
Car horizontal = 14 + 4time, Car vertical = 0*
Balloon horizontal = 0, Balloon vertical = 8 + 5time*
Use x to represent time and y to represent the horizontal location. The slope of the graph tells the velocity of the car. The "y-int" tells the starting location. Tracing gives the time and location of other points along the motion, though in discrete frames most of this tracing information is not applicable.
T i m e H o r i z o n t a l V e r t i c a l
Car |
Balloon |
|||||
Time |
Horiz. |
Vertical |
Time |
Horiz. |
Vertical |
|
0 |
14 |
0 |
0 |
0 |
8 |
|
1 |
18 |
0 |
1 |
0 |
7 |
|
2 |
22 |
0 |
2 |
0 |
6 |
|
4 |
30 |
0 |
4 |
0 |
4 |
Car |
Balloon |
|
Time |
Horiz. |
Vert. |
0 |
14 |
8 |
|
1 |
18 |
7 |
2 |
22 |
6 |
4 |
30 |
4 |
Negative heights would mean the balloon was below the zero level, presumably the ground. Not healthy! Thus, the mathematical model—the tabular description—is valid only up through the point at which the vertical value becomes zero.
Balloon |
||
Time |
Horiz. |
Vert. |
0 |
0 |
30 |
2 |
5 |
36 |
4 |
10 |
42 |
6 |
15 |
48 |
The balloon is moving both horizontally (from 0 at 2.5 pixels per frame) and vertically (from 30 at 3 pixels per frame) at the same time—diagonal motion!
Press MODE.
"Arrow down" to the fourth line, then "arrow right" to Par and press ENTER.
"Arrow down" to the fifth line, then "arrow right" to Dot and press ENTER.
Press Y= . Now the list has places for PAIRS of equations. The Xs are for horizontal locations (in the graph, not necessarily in the motion). The Ys are for vertical locations (in the graph).
For X1T type your equation from problem 4e for the car’s horizontal motion.
(The X, T, 0 key now makes Ts instead of Xs. T stands for time, here.)
For Y1T type 0. (That is the equation for the vertical motion of the car!)
Press WINDOW. There are now Ts in the window description, too.
Set Tmin = 0, Tmax = 10, and Tstep = 1.
Set Xmin = –20, Xmax = 60, and Xscl = 5.
Set Ymin = –20, Ymax = 60, and Yscl = 5.
Press GRAPH. (You probably won’t see anything. The axis is in the way.)
Press TRACE. "Right arrow" slowly around. Notice the numbers displayed at the bottom of the screen.
Press WINDOW.
"Arrow right" to FORMAT.
"Arrow down" to the fourth line, then "arrow right" to AxesOff and press ENTER.
(Watch fast and) press GRAPH.
Now trace again.
This graph actually shows the path of the car—frame by frame! The time, horizontal location, vertical location numbers are here, just like in the table.
This graph actually shows the path of the car and the balloon—frame by frame! The time, horizontal location, vertical location numbers are here, just like in the table.
So that the trace numbers would not interfere with the graph dots.
|
Homework 4—Escalating Motions |
|
Horizontal = 2 + 3 Time*
Vertical = 3 + 6 Time*
Time |
Horizontal |
Vertical |
0 |
2 |
3 |
1 |
5 |
9 |
2 |
8 |
16 |
3 |
11 |
21 |
4 |
14 |
27 |
5 |
17 |
33 |
The object moves diagonally up and to the right, traveling up at a velocity of 6 fps and horizontally at 3 fps, so it’s relatively fast and steep.
New horizontal = Old horizontal + 3, New vertical = Old vertical + 6,
Initial horizontal = 2, Initial vertical = 3
Horizontal values would increase by 6 starting from 2.
No change in vertical. Recursive: New horizontal = old horizontal + 6. closed: Horizontal = 2 + 6 Time*
(Closed Form) Horizontal Location = 0 + 3 Frame, Vertical Location = 0 + 6 * Frame. *
(Recursive form) Equations stay the same but initial values change.
Time |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
Horizontal |
1 |
3 |
5 |
7 |
9 |
11 |
13 |
15 |
Vertical |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
13 |
Columns for horizontal and vertical locations. They tell where the object is.
New horizontal = Old horizontal + 2, New vertical = Old vertical + 1,
Initial horizontal = 1, Initial vertical = 6, horizontal velocity is 2 fps, and the vertical velocity is 1 fps.
H = 1 + 2T, V = 6 + T
Graph each of these component equations separately. (If you use your calculator, use FCN mode, not PAR.) Be sure to label your axes clearly.
(Closed form)
H = 0 + 2T
V = 0 + T
(Recursive form) Equations stay the same; initial values are zero.
Note: Since these graphs use time as their independent variable, they are called time-series graphs. They "tell about" the motion, but they do not show it directly.
The vertical locations on each of these two graphs represent the locations of the object during motion, separately. However, in these graphs, the time also has an axis, unlike in the graph of the motion.
Add the line y = 97 to the horizontal vs. time graph. Read the time from the x-coordinate of the intersection, then "look up" from that time on the vertical vs. time graph to the y-coordinate for the vertical location. (97, 183).
|
Frame 0 1 2 3 4 5 6 7 |
Horizontal 12 14 16 18 20 22 24 26 |
Vertical 4 5 6 7 8 9 10 11 |
Horizontal location = 12 + 2 * Frame
Vertical location = 4 + 1 * Frame. *
The slope of each graph shows the component velocity; the "y-intercepts" show the starting coordinates.
Horizontal values increase by 2 starting at 0. Vertical values increase by 1 starting at 0.
Final – Starting = Displacement
(Vertical displacement)/(Vertical velocity) = Duration
Total frames = Duration +1
30 – 4 = 26 units vertical displacement, thus, 26 frame changes; so 27 frames of animation are needed to see both start and finish of motion (or use graphical methods).
(Horizontal velocity)*Duration = Horizontal displacement
Displacement + Start = End location
2*26 = 52 units horizontal displacement, and 52 + 2 = 54 units horizontal end location
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Activity 5—Going to the Movies |
|
= 6, 7, 16, 17
You animated single points in previous activities. In this activity, you will make a letter move across the screen of your graphing calculator just as letters appear to move across the screen of a marquee sign. This is the next step on your way to developing more complicated animation.
The matrices used in this animation are 2 × 11 (2 rows with 11 columns) with row 1 containing all the horizontal components and row 2 containing all the vertical components. The matrix below tells the calculator to position the letter E in the lower left corner of the calculator screen.
Begin a loop with a For statement. Use the variable N; Begin with 1; Step by 1; End with 11 (because there are 11 points in matrix [C]).
The next line of the program turns on a different point with each pass through the loop. The coordinates for each point are selected from matrix [C].
Use the calculator command that selects a value from a designated position in a matrix. One calculator uses the combination [C] (2, 3) to identify the number in the second row, third column of matrix [C]. You need to use N to designate the column because you move to the coordinates in the next column every time you pass through the loop. One calculator uses the combination:
Pt-On ([C] (1,N), [C] (2,N))
Use the End statement to end the loop.
One way is to add a matrix to the current matrix, with the matrix representing how much the horizontal position changes each frame. For example, if you wanted each position to change by 2, you could define a matrix [B] that would have all twos in the first row and zeros in the second:
® [B] ; [B]+[C] ® [C] will change the matrix [C] by adding two to all the x-coordinates.
Example: Use matrices with matrix A containing the original locations, matrix B being the transforming matrix, and matrix C being the defined figure locations, P being the loop for the number of moves, N being the number of points, and M being the timing loop variable.
:prgmAADEFAUL
:0 ® Xmin:98 ® Xmax:0 ® Ymin:64 ® Ymax
:[A] ® [C]
:For(P,1,80,1)
:For(N,1,11,1)
:Pt-On([C](1,N),[C](2,N))
:End
:For(M,1,100,1):End
:For(N,1,11,1)
:Pt-Off([C](1,N),[C](2,N))
:End
:[C]+[B] ® [C]
:End
The same program as above, with the original matrix changing to reflect the location on the right of the screen and matrix B changing according to negative values.
The same program as above, with the original matrix changing to reflect the location on the right of the screen and matrix B changing according to both the first and second rows.
Answers will vary but should be similar to the program above.
|
Animation Assessment—Flash Forward |
A camera is programmed to snap a picture of boat traffic in a major harbor at 1-minute intervals. Two ships appear in a particular sequence of photos. In the first photo (t = 0), one ship is at coordinates (0, 4), and the second ship is at (8, 0). In the next photo, 1 minute later (t = 1), the locations of the ships have changed. The first is now at (3, 6), and the second is at (9, 3). If you assume that both ships continue to move at the same rates, will their paths cross? Will the ships collide?
|
Unit Project |
It’s time for you to create your own animation!
Include the following in your calculator animation.
On a separate sheet of paper, submit the following with your program:
|
Unit Summary—Mathematical Summary |
Animation is the result of small changes in location for thousands of points over several frames. The computer animator uses mathematics to keep track of the location of each point and communicate the type of change.
A reference system is necessary to locate points. When a point moves only in a horizontal direction, a single number can be used to identify the horizontal distance along a number line. Each location x is paired with a time t.
As a point moves, the location changes by an amount called displacement. The amount of displacement depends on the velocity of the point and the amount of time.
Displacement = Velocity × Time
or, expressed with symbols, D = v × t
Units of time may be measured in real time (seconds, minutes, hours) or animation time (frames). The rate at which a point appears to move depends on the displacement from one frame to the next and on the number of frames displayed per second.
The changing location of a point along a line may be represented with two types of equations.
Closed-form equations relate one quantity to another quantity. In the context of motion, closed-form equations may be used to identify the current location in terms of time or frame number. The current location is the starting location increased by the displacement that has occurred since the point started moving.
Current location = Starting location + Displacement
Current location = Starting location + Velocity × Time
xcurrent = xstart + D
xcurrent = xstart + v × t
Recursive equations relate one quantity to a previous or next value of the same quantity. In the context of motion, recursive equations identify the current location with respect to the previous location based on a corresponding change that occurs in one unit of time. The displacement is calculated from one unit of time to the next.
xinitial = Location when t = 0
xcurrent = xprevious + D1
where D1 = displacement during 1 unit of time
This can be expressed in general terms by:
x0 = a
xt = x(t – 1) + D1
A starting location must be specified when you use recursive equations.
Diagonal movement, or movement in two dimensions, requires a reference system using two variables such as (x, y) to identify the location of a point. An equation of the form y = mx + b may be used to identify the path of an object, but the equation of the path does not tell when the point passes through a particular location.
Parametric equations are used to identify the location of an object at a particular time. Location is determined by the combination of a horizontal component and a vertical component. Both closed-form and recursive equations may be used to find the horizontal and vertical coordinates of a point at a given time (remember, time may be measured in seconds or frames).
Closed form:
x = a + bt
y = c + dt
Recursive form:
xt = xt – 1 + b, x0 = a
yt = yt – 1 + d, y0 = c
The letters a, b, c, and d are called constants or control numbers. The letters a and c identify the starting location for the object (at t = 0). The letter b represents the horizontal velocity, while the letter d represents the vertical velocity.
A negative horizontal velocity means that the point is moving from right to left. A negative vertical velocity means that the point is moving downward. When b and d are both positive, the point moves upward to the right. When b and d are both negative, the point moves downward to the left.
You can alter the path of the point or object, the "where," and the location at a particular time, the "when," by changing one or more of the control numbers.
Symbolic methods have been introduced as a way to convert from one form of an equation to another and from one type of representation to another. Parametric equations with variables x, y, and t can be converted to a single equation with x and y. In the final, combined equation, horizontal location is the independent variable and vertical location is the dependent variable.
Solve the first equation of the form x = a + bt for t. Substitute the resulting expression for t in the place of t in the second equation.
Using a specific example, x = 3 + 2t and y = –1 + 4t may be converted to y = 2x – 7 by following the steps illustrated in the arrow diagrams:
x = 3 + 2t
x – 3 = 2t
t = 1/2 (x – 3)
t = 0.5x – 1.5
y = –1 + 4t
y = –1 + 4(0.5x – 1.5)
y = 2x – 7
Animation involves the movement of many points at the same time. Several points are joined together to form a letter or object. The coordinates of several related points may be organized in a matrix. In this unit you used a 2 × n matrix to list the coordinates of all the points in your animated figure.
The graphing calculator is a useful tool for understanding how important mathematics is to animation. The programming language of the calculator is similar to the programming language used by designers of animation software. The Pt-On(x, y) allows you to light up the pixel at location (x, y), and the Pt-Off(x, y) allows you to turn off the same pixel. The command Pt-On([A] (1, 3), [A] (2, 3)) would light up the point with x-coordinate found in the first row, third column of matrix [A] and y-coordinate found in the second row, third column of matrix [A].
Loops are used for steps that are repeated several times in a program or process. For most loops in programs, it is necessary to identify a "counter" variable, together with its starting and ending values and the size of the steps by which to count.
Closed-form and recursive equations have a different format when included in a calculator program.
x = a + bt becomes a + bT -> X. *
x = xt – 1 + b becomes X + b -> X.
These simple commands and equations are used to animate points and clusters of points on a calculator screen.
Mathematics is the key to computer animation. A lot is happening and a lot is changing in the field of computer animation, so get moving!
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Key Concepts |
Closed-form equation—Equations that allow you to find the value of one variable given the value of another variable. Equations of the form c = 2p + 4, y = 3x + 2, and x = 2 + 5t are examples of closed-form equations.
Counter—A variable to which a constant (usually 1) is added; it keeps track of the total times a process is performed.
Displacement—The distance in a particular direction. Displacements of –2 and +2 meters represent the same distance, but in different directions.
Frame—A picture in a series of pictures that are displayed sequentially
Iteration—Each instance that a process is repeated. A process that is repeated over and over is called an iterative process.
Loop—A programming structure that allows a process to be repeated several times. The most common loop used in this unit is the For/End loop.
Parametric equations—Two or more equations, each relating a different dependent variable to the same independent variable
Path graph—The graph of y versus x. In other contexts, the path graph is referred to as the state-space graph.
Recursive equation—Equations that indicate the relationship between the current value of a variable and the previous value of the same variable. These equations require a statement of the initial value.
Time-lapse graph—A graph of y versus x that includes sample times displayed on the graph to show when a moving object passes through a particular location on the graph
Time-series graph—Any graph or equation that uses time as the independent variable
Velocity—The ratio of the displacement of an object to the duration (time) of that displacement. It is the rate at which an object moves in a specific direction.
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Solution to Short Modeling Practice |
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Solution for A Look at Air Traffic Control |
Plot the position data for each aircraft.
The straight-line paths of the aircraft indicate that the velocity is constant for each aircraft. The difference in two consecutive positions is the change.
Aircraft 1: 
Aircraft 2: 
Aircraft 3: 
Successive positions can be calculated recursively by adding the change in position to the last position. The projected positions 20 seconds into the future are:
Aircraft 1: 
Aircraft 2: 
Aircraft 3: 
The distance between the aircraft can be measured on the graph or calculated with the distance equation as shown below.
Table 1 lists the results of the difference calculation for each aircraft, the projected positions, and the calculation of distances between the three aircraft at the end of each 20 second period. The distances that indicate a hazard are in bold type.
Sweep 2 |
Sweep 3 |
Change |
||
Acft 1 |
x |
10.5 |
11.5 |
1 |
y |
20.9 |
20.3 |
–0.6 |
|
Acft 2 |
x |
13.1 |
13.8 |
0.7 |
y |
14.3 |
14.7 |
0.4 |
|
Acft 3 |
x |
15.34 |
15.7 |
0.36 |
y |
21.1 |
19.2 |
–1.9 |
Position Projections (Hazards are Bold)
|
Last Sweep |
(= 20 sec) |
(= 40 sec) |
(= 60 sec) |
(= 80 sec) |
(= 100 sec) |
(= 120 sec) |
(= 140 sec) |
(= 160 sec) |
||
|
Acft 1 |
x |
11.50 |
12.50 |
13.50 |
14.50 |
15.50 |
16.50 |
17.50 |
18.50 |
19.50 |
|
y |
20.30 |
19.70 |
19.10 |
18.50 |
17.90 |
17.30 |
16.70 |
16.10 |
15.50 |
|
|
Acft 2 |
x |
13.80 |
14.50 |
15.20 |
15.90 |
16.60 |
17.30 |
18.00 |
18.70 |
19.40 |
|
y |
14.70 |
15.10 |
15.50 |
15.90 |
16.30 |
16.70 |
17.10 |
17.50 |
17.90 |
|
|
Acft 3 |
x |
15.70 |
16.06 |
16.42 |
16.78 |
17.14 |
17.50 |
17.86 |
18.22 |
18.58 |
|
y |
19.20 |
17.30 |
15.40 |
13.50 |
11.60 |
9.70 |
7.80 |
5.90 |
4.00 |
|
|
S1,2 |
6.05 |
5.02 |
3.98 |
2.95 |
1.94 |
1.00 |
0.64 |
1.41 |
2.40 |
|
|
S1,3 |
4.34 |
4.29 |
4.71 |
5.50 |
6.51 |
7.67 |
8.91 |
10.20 |
11.54 |
|
|
S2,3 |
4.88 |
2.70 |
1.22 |
2.56 |
4.73 |
7.00 |
9.30 |
11.61 |
13.92 |
|
Solutions to Practice and Review Problems |
Slope (m) = 
Slope (m) = 
Slope (m) = –2000
H |
S |
8 |
86.4 |
6 |
92.0 |
4 |
97.6 |
3 |
100.4 |
2 |
103.3 |
1.5 |
104.7 |
1 |
106.1 |
0.5 |
107.5 |
0.6 |
108.2 |
= 
v = or v = 8.02 
Since each packet has 5 colored pages and 27 white pages, you know that each packet’s cost is
Cost per packet = (5 × $0.065) + (27 × $0.045)
Cost per packet = $1.54
Thus, the formula for computing the cost of n packets is
C = 1.54n + 5.00
where C is the total cost in dollars and n is the number of packets produced.
C – 5.00 = 1.54n
Then divide both sides by 1.54, leaving n isolated.
n = 
So, for a total budgeted cost of $2000, the number of packets that can be produced can be computed.
n = 
n = 1295 (rounded down)
So, 1295 packets can be produced.
900x + 700y = 60,000
x + y = 75
ax + by = c ® 900x + 700y = 60,000
dx + ey = f ® x + y = 75
We see that a = 900, b = 700, c = 60,000, d = 1, e = 1, and f = 75. We can now apply these numbers to the determinant method and solve for x and y. Solving for x we get:
x = 
= 
= 
= 37.5
Similarly, solving for y we have:
x = 
= 
= 
= 37.5
Thus, Roberta should sell 37.5 acres of Type-A land and 37.5 acres of Type-B land.
900x + 700y = 60,000
900(37.5) + 700(37.5) = 60,000
60,000 = 60,000
(It checks.)
And the second equation,
x + y = 75
(37.5) + (37.5) = 75
75 = 75
(It checks.)
The answers check out.
ax + by = c ® 2x + y = 8
dx + ey = f ® 5x + 3y = 21
We see that a = 2, b = 1, c = 8, d = 5, e = 3, and f = 21. We can now apply these numbers to the determinant method and solve for x and y. Solving for x we get:
x = 
= 
= 
= 3
Similarly, solving for y we have:
y = 
= 
= 
= 2
2(3) + (2) = 8
8 = 8
(It checks.)
And then the hauling capacity . . .
5(3) + 3(2) = 21
21 = 21
(It checks, also.)
Our answers check out. You should use 3 of the five-ton trucks (x = 3) and 2 of the three-ton trucks (y = 2).
Company A: c = 5(6) = 30
Company B: c = 3(6) = 8 = 26
Company A’s charges would be $30 while Company B’s charges would be only $26.
FV = PV (1 + i)n
FV = 5000 (1 + 0.085)5
FV = 7518.28 (rounded)
So, at the end of five years, the total value of the deposit would be $7518.28.
FV = 1000 (1 + 0.00583)6
FV = 1035.49
So, at the end of the six months, the total value of the deposit will be $1035.49.
G = 2.1(19) + 3.7 or 43.6 sec
F = –1.18 A + 141
F = –1.18 (50) + 141
F = 82
For A = 65. . .
F = –1.18 (65) + 141
F = 64
Both of these seem to agree well with the graph of the data shown in the text.
Distance from start (mi)
Time (h) |
With constant speed |
With constant acceleration |
|
0 |
0 |
0 |
|
10 |
7.5 |
0.84 |
|
20 |
15.0 |
3.34 |
|
30 |
22.5 |
7.52 |
|
40 |
30.0 |
13.36 |
|
50 |
37.5 |
20.88 |
|
60 |
45.0 |
30.06 |
|
70 |
52.5 |
40.92 |
|
80 |
60.0 |
53.44 |
|
90 |
67.5 |
67.64 |
|
100 |
75.0 |
83.50 |
|
110 |
82.5 |
101.04 |
|
120 |
90.0 |
120.24 |
|
To actually find the linear equation is considerably more complicated (and not required of the students). For those who are interested, the procedure to obtain a rough estimate of the equation’s parameters is given here. An estimate of the slope can be obtained by using the endpoints of the region. For example, the ungrounded region pointed out above would have a slope estimated by using the lower point (0.188, 1.18) and the upper point (0.25, 2.18). This yields a slope m = 16.1. The intercept could be determined by extending the line and the scales on the graph, or by substituting m and one pair of points on the line into the slope-intercept form, and solving for the y-intercept, b. Using (0.25, 2.18) yields b = –1.85. Thus, the linear equation that can be used to approximate the graph in this region is y = 16.1 x – 1.85.
: x-component = 30 – 2; y-component = 23 – 4x-component = 28; y-component = 19
: x-component = 29 – 30; y-component = 24 – 23x-component = –1; y-component = 1
: x-component = –(–1); y-component = –(1)x-component = 1; y-component = –1
+ (–): x-component = 28 + 1; y-component = 19 – 1 + (–): x-component = 29; y-component = 18
AC
=
AC
=
AC
=
AC
= 6.4 grid unitsAC
= 6.4 grid units × 10 miles/grid unitAC
= 64 miles64 miles = 150 mph × time
64 ÷ 150 = Time
0.43 hr = Time
0.43 hr = 0.43 hr × 60 min/hr
0.43 hr = 25.8 minutes
Estimated time of arrival = 8:34 p.m.
BC = 
BC = 
BC = 
BC = 7.8 grid units
BC = 7.8 grid units × 10 miles/grid unit
BC = 78 miles
78 miles = 150 mph × Time
78 ÷ 150 = Time
0.52 hr = Time
0.52 hr = 0.52 hr × 60 min/hr
0.52 hr = 31.2 minutes
Estimated time of arrival = 8:37 p.m.
y-intercept = 0
