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PHS 542 (Summer, 2018, 5 weeks) Daily Details

This file contains daily handouts, including theory & activities, as well as homework assignments.

A separate file is basically an instructors’ solution manual, containing answers to selected activities and homework keys, the quizzes used during the course, and quiz keys.

Week 1 --- Day 1

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Syllabus: PHS 542: Integrated Math and Physics

Summer at ASU (2018: Bob Rowley, Robert.J.Rowley@asu.edu)

Catalog description: Mathematical models and modeling as an integrating theme for secondary mathematics and physics. Enrollment by teams of mathematics and physics teachers encouraged. COURSE DESCRIPTION: A. Overview:

Week 1: Teaching theory with Dr. Hestenes, introduction to the four math models, introductory activities with Geogebra and GlowScript.

Week 2: Math Model 1 (MM1: constant rate of change), similarity & ratio problems, linear graphs, slopes. Introduction to Geometric Algebra (GA) in 2D.

Introduction to MM2 (constant change in rate), parabolas and their graphs. Week 3: Continue MM2, parabolas in projectile motion, GA Primer treatment. MM3 (rate proportional to amount), exponentials and their graphs, Professor Bartlett’s treatment of exponential growth and non-renewable resource lifetime. Week 4: MM4 (change in rate proportional to amount), trig functions, Euler’s formula

as used in GA, oscillatory motion in physics applications. GA Primer topics in classical physics, 3D rotations.

Week 5: Continue GA Primer topics. Beginning relativity using GA treatment. Some standard relativity examples from the GA viewpoint. B. Course plan and rationale: Daily handouts. Some theory, many cooperative practice activities, homework assignments, and either a final exam or final project. One goal is to deepen teacher understanding of the importance of the mathematical models which span a large portion of physics topics. The other goal is to familiarize teachers with the benefits of Geometric Algebra, so that they may be encouraged to introduce GA basics to high school students, since it can aid in understanding of geometry and vector operations. STUDENT LEARNING OUTCOMES: At successful course completion, students will have

- deepened their understanding of four basic mathematical models in physics, - become acquainted with Geometric Algebra for math and physics, - practiced teamwork and communication to improve their instruction, - improved their instructional pedagogy by incorporating inquiry methods, critical and

creative thinking, and cooperative learning, in the context of mathematical modeling. LISTING OF ASSIGNMENTS: This course meets for 47.5 hours (19 days), and ABOR requires that you do at least 90 hours of work outside of class. Assignments will be listed by the instructor, daily. GRADING POLICIES: A-B-C grades: B means average; a 3.0 GPA is minimum requirement for MNS and other graduate degrees. Grades are based on attendance, participation, homework, and final exam.

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Policies of Arizona Board of Regents (ABOR), ASU, and Department of Physics: * ABOR: Each student is expected to work a minimum of 45 hours per semester hour of credit. * Pass-fail is not an option for graduate courses. https://students.asu.edu/grades-grading-policies. * 3.0 grade point average (GPA) is minimum requirement for MNS & other graduate degrees. * Incomplete: only for special circumstances. Must finish course within 1 year, or it becomes “E”. * An instructor may drop a student for non-attendance during the first two class days (in summer). * An instructor may withdraw a student with a mark of "W" or a grade of "E" only in cases of "disruptive classroom behavior". * The ASU Department of Physics is critical of giving all A's, because it indicates a lack of discrimination. A grade of "B" (3.0) is an average graduate course grade, and obviously not all students do above-average work compared to their peers. Some of you can expect to earn a "B", and those who are below average but do acceptable work will earn a "C". Academic dishonesty policy: Academic honesty is expected of all students in all examinations, papers, laboratory work, academic transactions, and records. The possible sanctions include, but are not limited to, appropriate grade penalties, course failure (indicated on the transcript as a grade of E), course failure due to academic dishonesty (indicated on the transcript as a grade of XE), loss of registration privileges, disqualification and dismissal. For more information, see http://provost.asu.edu/academicintegrity. Disability policy: Qualified students with disabilities who require disability accommodations in this course are encouraged to make their requests to the instructor on the first class day or before. Note: Prior to receiving disability accommodations, verification of eligibility from the Disability Resource Center (DRC) is required. Disability information is confidential. REQUIRED INSTRUCTIONAL MATERIALS: No textbook. Optional is a 3-ring binder (preferably 1 inch thick); 6 tab inserts. Either access to the GA Primer (http://geocalc.clas.asu.edu/GA_Primer/GA_Primer/) or a personal copy of the GA Primer as PDF. Access to the internet for free math software tools Geogebra and GlowScript. REQUIRED MEDIA: None. REQUIRED READING: Daily handouts, selected topics from the GA Primer. RECOMMENDED READING: None.

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Glowscript (2018) Intro

Python is an interpreted, general-purpose programming language, around since 1991 and making a resurgence now. VPython was created in 2000, adding visual libraries to Python to help make 3D modeling much easier, allowing creation of 3D shapes and motions with a minimum of programming knowledge. As an interpreted language, no compiling is necessary, although each user needed to have the software installed on their computer. That’s the downside of an interpreted language, but the upside is that you can progam an idea, test it, and jump quickly back and forth between editing and testing.

Web-based programming has the popular benefits of not needing software versions for each operating system, and of maintaining revisions and corrections of the software in one place available to all users immediately. The trade off is requiring access to the web.

Glowscript is the web-based descendent of VPython, created in 2011 by the creators of VPython. It should run in any browser. They say Chrome offers the most helpful error statements. The simplest way to explain:

Go to http://www.glowscript.org/.

Click on one of the Example programs links, which opens in a new tab. Notice the tabs for examples in Glowscript, JavaScript, RapydScript, and examples for the books Matter and Interactions (I & II), by Chabay & Sherwood. Pick any example from the Glowscript tab, either Run or View to begin to get acquainted.

Click on Help on the right side of the blue banner, which opens in a new tab. Explore.

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Some Rowley examples can be found at: http://www.glowscript.org/#/user/Capaii/folder/Examples/

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GlowScript Activity (objects & motions):

1. Go to http://www.glowscript.org/. Click on Sign in on the right side of the blue banner. If you already have a Google account (gmail), you can use that. If not, or if you want a different account just for GlowScript, use the Create account option.

2. You will see one or two folder tabs, plus one Add Folder tab, and you’ll automatically be in one folder. You should see PUBLIC Create New Program Download options. We can toggle between PUBLIC and PRIVATE for the folder type. Pick a folder and choose Create New Program. Name your program something like “FlyingBee”, the “.py” extension will be automatically generated.

3. The programming window opens with the first line already filled in. Since the button function is a little tricky to jump right into, start with these instructions (change comment lines as you like, as long as they start with a #):

Notice again the Help link on the right side of the blue banner. It opens in a separate tab and will by your main source of information on objects and syntax. Notice the drop

GlowScript 2.7 VPython ################################################################################ # # Name: ________________________________________________________________ # # Part 1: set up green disk, red radius vector, blue force vector, # black torque vector, and yellow/orange bee. # Part 2: make bee fly in circle, parallel to disk, around torque vector. ################################################################################ scene.background = color.white scene.title = "Circling bee.\n" #------------------------------------------------------------------------------- flying = True def toggle(b): global flying flying = not flying if flying: b.text = "Hover" else: b.text = "Fly" button(text="Hover", pos=scene.title_anchor, bind=toggle) #-------------------------------------------------------------------------------

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down menus on the left side in the Help tab. There are example snippets and sometimes links to whole example programs. Cut & Paste, in general, is a wonderful tool.

4. Test your program with Run this program on the left side of the blue banner. When ready, click on Edit this program in the same location. As you go, feel free to Run and Edit as often as you like. It’s much easier to catch little mistakes as they happen.

5. With liberal use of Help, Run, and Edit, work your way through the rest of the program, filling in the details:

Notice that vec(2,0,0) is legitimate shorthand for vector (2,0,0).

# Green disk, radius=2, centered at origin, 0.02 units tall, lying in x-y plane: cylinder(……….)

r = vector(2,0,0) # Define a radius vector on the disk. F = vec(…..) # Make the force vector point in the y-direction, length=1. torque = ….. # Make the vector cross product, r x F.

# Show the r vector (red, placed at origin, width=0.1): arrow(……….)

# Show the F vector (blue, placed at tip of r, width=0.1): arrow(……….)

# Show torque vector (black, placed at origin, width=0.1): arrow(……….)

# Yellow bee, placed at (-1,1,1) to start, radius=0.2): body = ellipsoid(……….) wings = ellipsoid(……….)

bee = compound([body, wings]) #------------------------------------------------------------------------------- while True: # Repeat forever.

rate(50) # Try different rates to speed up or slow down the bee.

if flying: # Rotate 1 deg. (π/180), about origin, axis = torque vector: bee.rotate(angle=pi/180, origin=vec(0,0,0), axis=torque) # Try different origins and axes to see how rotate works.

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Your final run should look something like this.

OK, it looks more like a blimp with a surfboard on top, but you get the idea. The compound([…,…,…]) statement allows combining several objects into one. The bee.rotate(…..) statement then rotates the whole compound object.

6. In the rotate command, change the origin so that the bee rotates in place.

7. In the rotate command, change the origin to vec(-0.5,0.5,0). Describe the motion. Now, change the origin to vec(-0.5,0.5,8). How does the motion change?

8. Put the origin back to vec(0,0,0). Now, try axis=vec(1,1,0) in the rotate command. How can we describe the motion? Can you see the plane of rotation?

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PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 1 --- Day 2

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From “Modeling Software” paper (Hestenes)

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(Ramp, String,

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From “Modeling for STEM Education Reform” 2017 paper (Hestenes)

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Geogebra Intro

Either download, install, and open Geogebra Classic https://www.geogebra.org/download

or choose to run it online https://www.geogebra.org/classic.

The online help manual can be found at https://wiki.geogebra.org/en/Manual.

When Geogebra first opens, you should see an algebra/input window on the left and a 2D graphics window on the right. By right clicking in the graphics area, you can show/hide axes and the graphing grid.

Depending on the version, drop down menus may be in different places, but experimentation is easy, and there is an undo button just in case. A couple of useful menu options to find are for choosing 2D versus 3D graphing, and managing tools shown on the toolbar:

The toolbar has drop down options for placing points, lines, vectors, circles, etc. You’ll also find options for finding intersections, drawing perpendiculars, making measurements, adding sliders, etc. No need to memorize, just explore when searching for something you’d like to try. Almost all options can be performed either with a command in the input window (with helpful command completion hints, again to avoid memorization) or by using the toolbar.

A common mistake I can’t seem to avoid: after performing some operation from the toolbar, remember that operation is still in effect until you click on the pointer arrow on the left of the toolbar. For example, maybe I just placed a circle, then want to scroll the view, so I click the left mouse button to drag the scene, but instead, another circle is begun. So, then I have to hit the undo button and click on the pointer arrow before proceeding with something new.

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Geobebra Activity 1 (reflection, translation, and dilation):

1. Place 3 points: (-2,1), (-4,5), and (-8,3), using the Point drop down toolbar button. You could also use a command, such as P1=Point({2,3}). The syntax is a little harder to remember. With the toolbars, objects are automatically labeled alphabetically, although you can right click on any object and change its many parameters (name, extra label, color, size, etc.).

2. Use the Polygon toolbar option to connect the 3 points with a triangle object. A triangle and 3 line segments are made and assigned labels.

3. Select the Reflect about Line toolbar option. Click in the triangle to select it, then click on the y-axis to draw the reflected objects (triangle, points, segments).

4. Draw the point (1,-4).

5. Draw a Vector from the reflected original triangle point, that is (2,1), to this new point.

6. Select the Translate by Vector command from the toolbar. To complete the command, select the point (2,1) again, then select the vector just created. This will make a copy of the reflected triangle, translated by this vector. Once again, besides the second triangle, we get copies of the triangle vertices and its side segments.

7. Select the Dilate from Point command from the toolbar. To complete the command, select the third triangle vertex (1,-4) as “the center point”, and type in 0.5 as the dilation factor. That should shrink the third triangle, leaving this vertex unchanged.

8. Notice the big, filled in dots to the left of each object’s name in the algebra window on the left. If you click on any of these dots, see how you can hide/see that object. Hide the following objects: the second triangle (reflection of the original), its 3 vertex points, the point made in step 4, the vector made in step 5, the third triangle made in step 6, and its 3 vertex points. We should now be left with only the original triangle and the final triangle (after reflection, translation, and dilation).

9. Let’s make some measurements. Select the Angle command from the toolbar. To identify which angle we want, select the vertices of the first triangle, in order, points (-4,5), then (-2,1), then (-8,3).

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10. Use the Angle command again to view and measure the top angle of the first triangle and the two corresponding angles in the final triangle. We now have proof that the two triangles are similar. Now, it may seem that all the numbers and letters are getting in the way of each other. But, we can grab each label with the left mouse button and move them to more desirable locations. They cannot be moved too far, but they can be nudged enough for clarity.

11. Select the Distance or Length command from the toolbar. Select the long side of the first triangle. Do the same for the long side of the final triangle. We could have already seen these lengths listed in the algebra window, but now these two are shown in the diagram, also. Of course, the smaller length had better be half the larger.

12. Finally, select the Area command from the toolbar, then select the first triangle to see its area labeled. Do the same for the final triangle. Right click in an empty space on the graph, and hide the grid, to see which view looks better to you. Does the ratio of areas for the two triangles seem correct?

13. Notice the first triangle happened to be a right triangle. But, notice that you can drag any point defining the first triangle at will, and everything correctly follows in the image (final) triangle. Can you drag the points of the final triangle?

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Your final graphs should look something like this.

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PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 1 --- Day 3

July 4th holiday --- no class.

Week 1 --- Day 4

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Ponderings on Brain Power

What’s the “best” way to teach a beginner how the knight moves?

Geometric: Operational: See the 8-pointed star of influence? Moves like an L (mimics shape of knight)? See it as the corners of an octagon? 1 lateral + 1 diagonal move (commutative)?

What does “best” mean? Easiest to visualize? Quickest to absorb? Surest for memory recall? Easiest to apply in practice? Most likely to avoid oversights (knights are notorious for surprise attacks --- odd, hopping motion --- witness the sneaky fork in 2nd diagram)?

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MIT historical site: https://libraries.mit.edu/archives/exhibits/exam/. This site shows entrance exams (and solutions) around 1870 when, by counting heads and schools, there were probably less than 1000 engineers in the U.S., whereas there were about 1.6 million U.S. engineers in 2015, according to the Bureau of Labor Statistics.

Let’s look at some of their entrance exam questions from arithmetic/algebra/geometry.

What if we gave them to 2 groups of students, telling one group they are are simple, old 8th grade math questions to see how much they’ve learned, and telling the other group they are some old college entrance questions. Will students’ results be affected by anticipation?

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Will the first group do better because these should be easy (confidence) or worse because they feel they should know this and panic if they struggle (fear). Will the second group do better because it’s OK if they don’t know it all (relaxation) or worse because they aren’t supposed to know all this (submission). It’s well known (or believed) that in two-person sports there is an extra advantage when the players mutually think one of them should win.

1. Find the sum, then the difference, then the product of 35/9 and 17/24. 2. Find the value in decimals of 1/5 + 3/4. 3. How many degrees is .01 of a circumference? 4. By selling a house and lot for $5,790, the owner lost 31/2 per cent. What was their cost?

--------------------------------------------------------------------------------------------------------- 5. Simplify the following expression by removing the brackets and collecting like terms:

3a – [b + (2a – b) – (a – b)]. 6. Multiply 3a2 + ab – b2 by a2 - 2ab – 3b2 , and divide the product by a + b. 7. Reduce the following fraction to its lowest terms:

6 2 3

6 4 2

x a x yx a y+−

.

8. Solve 7x – 5y = 24, 4x – 3y = 11. ---------------------------------------------------------------------------------------------------------

9. Prove that sum of the three angles of a plane triangle equals two right triangles. 10. Prove that the side of a regular hexagon inscribed in a circle is equal to its radius. 11. The radius of a circle equals 10. Find its area. 12. Define similar polygons. To what are their areas proportional? 13. The perpendicular dropped from the vertex of the right angle upon the hypotenuse

divides it into two segments of 9 and 16 feet respectively. Find the lengths of the perpendicular, and the two legs of the triangle.

The last problem is popular on the web. Nerds love to show their solutions. It’s a word problem, so the first helpful tool is a good (and correct) diagram. You may then see a successful attack using the Pythagorean theorem.

But, similarity arguments avoid squaring numbers, and perhaps offer more insight into the various relationships. When three similar triangles have been identified, notice how they could be nested inside each other.

And, when this last diagram is drawn, notice how students might go wild with ratios, some of which look good, but are faulty. Beware (of this and knight forks)!

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From “Modeling for STEM Education Reform”, 2017 paper (Hestenes)

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Geogebra Activity 2 (Can we make a Descartes multiplying sliderule?):

1. Put a point at the origin, label it O. Put a point A somewhere in the first quadrant.

2. Make a segment from O to A, name it SegA. In the input box in the algebra window, enter the formula a=Length(SegA).

3. This time make a point by a formula in the input box: I=Point({x(A),y(A)})/a. This point corresponds to the label “1” in the Descartes diagram on the previous page. As we drag point A around, this point I is always 1 unit away from the origin.

4. To emphasize the last point, use the Circle with Center and Radius command in the toolbar to put a circle centered at the origin with radius = 1. Right click on the circle to change its settings. Under the Style tab, make it a dotted line.

5. Place a Slider using the toolbar, named b, for a number from 0 to 1. Make a point using the input box: B=Point({b,0}). Notice how B moves as we move the slider.

6. Draw a segment from B to I.

7. Use the Parallel Line command in the toolbar. To complete the command, we need to select A for the point the line goes through, and the segment drawn in step 6 for the line our new line will be parallel to.

8. Now, use the Intersect command in the toolbar. To complete the command, we need to select the line made in step 7 and also the x-axis. Rename the intersection point C.

9. Make a segment from O to C, labeled c. Make a segment from C to A. The construction is basically finished. We can move the point A around and use the slider to watch the value of c match a times b. But, we may as well show some measurements in the diagram and practice making a text box.

10. Use the Distance or Length command in the toolbar to label the length of SegA. Repeat the same command for segment c.

11. Use the Text command from the toolbar. Type C = AB in the edit box and hit OK. This feature in Geogebra Version 6 seems flakier than in Version 5. Move the text to your desired position in the graphics window. Right click on the text and click on Settings, then go to the Text tab. We see C = AB in the edit box, and wish to alter it as follows. Below the edit box are some tabs, click the one that looks like the Geogebra icon. Try making it look like:

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To get the orange boxes for a, b, and c, click on the appropriate variables shown in the lower section. This is the way to get Geogebra to fill in values for variables in the text box. The flakey part is that clicking on the variables in the lower section was hit and miss for me. It didn’t work for several attempts, then started to work properly.

Of course, feel free to try changing fonts, using boldface or italics, or changing font size or adding in some LaTeX symbols.

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Your final graphs should look something like this.

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PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 2 --- Day 1

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Math Model 1: Constant Rates (Ratios, Similarities)

The constant rate dy kdx

=

math model shows up in many applications (constant

{ }, , , , ,...Ev a f p , approximations from Taylor series, etc.). Understanding of straight line graphs

are aided by understanding of ratios and similarities (especially for triangles).

What are similar triangles? We’ve had some visual examples of them in the Geogebra introductory worksheets. Did we adequately work through the 1870 MIT entrance exam triangle question (including enticing, but incorrect ratios)?

A popular modeling activity is to have students calculate the height of a building using similar triangles of shadows from the building and from a shorter, easier to measure object. We can ask students why this works. They might ask themselves “What if geometry works differently for a tall building and a shorter object?” or “What if light bends differently at the lower altitude of the object?” Such questions aren’t as silly as some students might think (general relativity?). We’ve had thousands of years to convince most of us that light rays (a model) seem to travel in straight lines (a model) and make shadows that can be compared to similar triangles. We stick with as simple a model as we can to get our solutions. And every time we actually go to the trouble of directly measuring the height of the building, our solutions can be verified. Of course, we can have students measure a couple of smaller objects which they can also directly measure to convince themselves that similarity works.

In general, when forming similarity ratios, the compared quantities need have no intrinsic relation to each other:

Quantity 1 Quantity 2 Connected by Length on ground Height of building Shadow Lengths or radii Areas or volumes Sidewalk construction, density, etc. Distance Time Motion of objects Number of chickens Number of corn ears Amount chickens eat, or amount of both we eat

The equatorial belt problem (see worksheet) can be part of (or preceded by) a modeling activity on the circle circumference to radius ratio. The formula is easy, the constant ratio seems obvious, but intuition often settles on absolutes rather than ratios.

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From Dr. Richard Clawson:

Interesting: we can represent mixtures by vectors (horizontal component represents total volume, vertical component represents acid volume). Adding mixtures is represented by adding vectors. Concentration is slope (m = tan θ = vertical-component / horizontal-component).

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Equatorial Belt Investigations

We’ve increased radii of the small and large belts by the same amount (0.2 units).

Estimate how much larger the red circumference is than the blue circumference: _________

Estimate how much larger the orange circumference is than the green circumference: _________

Calculate ΔC for the red/blue circles: ____________

Calculate ΔC for the orange/green circles: ____________

Discussions: Were our guesses close? Were the answers surprising?

Can we see what’s happening based on the circumference formula?

Explain results based on the graph which follows.

Earth: R ~ 6,370,000 m, C ~ 40,000,000 m.

We want to increase the circumference belt by only 12.6 m.

How much space is there (ΔR) between the belt and Earth’s equator?

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Plot circumference vs radius (same straight line for any circle).

Show ΔR on the graph from R=1.0 to 1.5, then draw and read off the value for ΔC.

Repeat for a larger radius (R=6.0 to 6.5).

Is ΔC the same for both, or larger for larger radii? Why?

0

10

20

30

40

50Ci

rcum

fere

nce

Radius

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Continuing the Earth example, let’s examine the harder (and even less intuitive) question:

When we pull the belt taut at one point, what is the distance of that point from the equator (h in the diagram)?

Remember, R ~ 6,370,000 m, C ~ 40,000,000 m, and we chose ΔC = 12.6 m. We want to find h in the diagram (of course, very much NOT to scale).

Obviously h should be larger than the ΔR we found for the first belt problem.

Should h be larger than the 2ΔR?

Should h be smaller than the 6.4 m? Think about the difference between pulling the belt as described and shown above versus wrapping the belt tight all the way around and pulling the extra slack to a height above the ground.

Can you find formulas for calculating h in general? Try on your own for awhile. Then, whether successful or not, try searching for a decent solution on the internet. If you find any solution not requiring solving a transcendental equation, you deserve a major award!

Resulting value for h in this example: _______________ m.

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High school students can find the equations (perhaps with the proper nudging) after learning sine, cosine, and tangent ratio definitions, as well as the relation between arclength and angle for circles.

Once the correct equations are found, there are various possibilities for seeking a solution:

• For values as shown in the previous diagram, graphical means can be used.

• Give students hints on series approximations, and how to look up Taylor series for tangent and secant on the internet, so that the equations can be reduced to algebra.

• Use an online equation solver, such as Wolfram’s.

• Many calculators have equation solvers.

It would be easy to turn this into a simple classroom project where students model the setup, develop ideas & formulas, and experimentally check their predictions.

For example, find a large circle (or cylinder or hoop), say about 50 cm in diameter. Use interlinking pull chains (or different colored string). Make a belt of equal radius to the circle. Then attach several cm of a different color string to make the belt longer. Now, both questions above can be measured.

The reason for the string of length ΔC to be of different color is that when we now pull from the center of that added string to get the belt taut (and maybe fasten it down with a pin or tack or tape), it will be obvious where that ΔC ends up in our second diagram and why it can be so much smaller than h.

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More Ratio & Similarity Exercises

I recently missed this trivia question: What is the volume of a pyramid? I ruled out the correct multiple choice answer Ah/3 because I remembered that as the volume of a cone. I hate missing math problems, so I spent the rest of the evening trying to calculate it with various methods, including calculus. I was unsuccessful that night (perhaps beer, loud conversations, and more trivia questions played a role).

It doesn’t matter what shape the base has, as long as the cross sections are similar to each other as we climb the height. On the right is a curve 4 2cosr θ= + in the z=0 plane, with all points on the curve connected by straight line segments to point A.

Below is a nice diagram from the web:

The idea is easy enough. We want to add the volumes of the sections, where the areas of the geometrically similar sections are decreasing regularly with height. And, that height is in the

direction perpendicular to each area section.

A non-calculus method:

Label the sections, bottom up: n = 1, 2, …, N. Each section has a height: Δh = h/N. Any linear dimension d (perpendicular to h) of a section decreases in a similarity ratio as n increases, and therefore, the section areas decrease as the square of that ratio:

sec sec

2

2

1

12

bott of top oftion tion

n

N n N nd d d dN N

N nA d A A

N

+ − −= =

+ − ∝ ⇒

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[ ]

[ ] ( ) ( )( )

[ ] [ ]

( )

22

21 1

2

3

3 2

2

2 2 2

1 2 1 22

1 1 1 21 2 1 22 2 6

1 2 1 1 1 2 11 2 1 24 2 6 4 3

2 1 4 1 12 1 112 12 3 4

N N

nn n

N

h A NVol h A N n nN N

N N N N NAh N N NN

Ah N N N Ah N NN N NN NAh N N AhN AhN N N

= =

→

+ ∆ = − + + + + ++ = − + +

+ + + + + = + − + = + − − − = + = = −

∑ ∑

3Ah∞→

It’s all algebra, although not particularly easy, and we have to know (or look up) some standard summation formulas. For example, to derive the 2n∑ formula:

( ) ( ) ( )

( ) ( ) ( )

( )( )

3 3 33 3 2

1 1 1 1 1

32 2

1

22

1

1 1 1 3 3 1 1

1 33 1 1 3 1 22 2

1 1 21 23 2 6

N N N N N

n n n n n

N

n

N

n

n n N n n n N N

N Nn N N N N N N

N N NN N Nn

= = = = =

=

=

= + − + + = + + + − + +

+ = + − − − = + + −

+ + + += =

∑ ∑ ∑ ∑ ∑

∑

∑

And, even for that, we needed to know the 1∑ and n∑ formulas, as well as the

binomial power expansion (that is, Pascal’s triangle) for ( )31N + .

A calculus method:

The ideas are the same. We need to add the sections again. Let our vertical parameter vary from z=0 to z=h. The height of each section is dz. And, again, the area of sections decrease in the square of the similarity ratio as z increases:

( )32

200

3 3

hh h zh z A AhV Adz

h h−− = = − =

∫

That’s quite a bit easier. We still need to know the integration power rule and the chain rule (which some have said is the most important rule in calculus). By going from the top

down, the integrand square becomes ( )2/z h , and then we don’t even need the chain rule.

Well, the point of all that is that the results depend on ratios of similar objects. Similarity is a powerful tool (and, calculus can be easier than algebra?).

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Why the factor of 1/3? For a triangle, the factor in the area formula is 1/2. In these cases, we might notice the factors come from averaging along the similarity dimension:

14. 1 12 3triangle avg avg cone avgA bh hb bh V hA Ah= = = = =

For the triangle, the similarity dimension (width decreasing along h) decreases linearly. For the cone, the similarity dimension (any cross sectional width decreasing along h) leads to the base area decreasing quadratically. These averages can be understood and approximated graphically, or can be pretty easily found with calculus. To get the average height with calculus, we divide the area under the curve by the interval width.

( ) ( )2

0

5 , 2 52

1 5 5 12 2 2 2avg

xf x h f

xh dx h

= = =

= = =∫

( ) ( )2

3 2

0

, 3 91 133 3avg

f x x h f

h x dx h

= = =

= = =∫

( ) ( ) ( )

( )0

sin , / 2 11 2 2sinavg

f x x h f

h x dx hπ

π

π π π

= = =

= = =∫

( ) ( ) ( )

( )

2

2

0

sin , / 2 11 1 1sin

2 2avg

f x x h f

h x dx hπ

π

π

= = =

= = =∫

39

Parabola Average Height with Geogebra

1. Open Geogebra in 2D graphing mode (default) and enter ( ) 2f x x= .

2. Make points A=(3,9) and B=(0,4). 3. Make a horizontal line function ( ) ( )g x y B= .

4. Make a line through A and perpendicular to the x-axis. 5. Find the intersections of f(x) and g(x) {intersect tool from the toolbar menu}. Hide the

left intersection. Let’s assume here the right intersection is called point D. 6. Begin to type the command Inegr… and you should see a pop-up list offered. Choose

the option IntegralBetween(<Function>, <Function>, <Start x-value>, <End x-value>). Fill in appropriate values to find the areas between the f & g functions for the section between the y-axis and point D, then for the section between points D and A. Feel free to choose different colors for these areas.

Notice that the IntegralBetween( ) function shades in the desired area and also shows the integral value (area value) in its label. The two areas are probably different, but we can grab point B and move it along the y-axis until the areas match. That means the area above the line matches the area below the line, so the line height shows the average height. Think of it as a parabolically shaped ice cube melting. Then the final water level shows the average height.

The Golden Ratio

Euclid (~ 300 B.C.) defined the Golden Ratio. Point C divides segment AB into two smaller segments. If we choose C such that the ratio of AB to the larger segment (AC) equals the ratio of the larger segment to the smaller segment, those matching ratios are called the Golden Ratio:

AB AC f g g horAC CB g h g

f f +≡ = ≡ = =

Since the ratios are what we care about, there’s no reason not to let 1h ≡ . Then gf = and the

last equation leads to the quadratic equation: 2 1 0f f− − = . That’s easy enough to solve, and we’ll want the postitive root to fit with our segment ratio idea above.

40

Make the last diagram interactive in Geogebra:

1. Make points A=(0,0) and B=(10,0). In their settings, make them fixed points. 2. Make point C anywhere between, such as (6,0). 3. Make segments AB, AC, and CB. Label them to match our diagram above. 4. Make the ratios 1 /r f g= and 2 /r g h= . 5. Make a text box so we can watch

the ratios change as we move point C back & forth. Then adjust point C to find that Golden Ratio approximation. ϕ = _____________________

Solve the quadratic 2 1 0f f− − = for the positive, exact solution: ϕ = _____________________

What is the calculator approximation to 10 significant digits? ϕ = _____________________

What does ( )1 1f f − = tell you to expect for the decimal value: 1/ ϕ = ____________________

What does 2 1f f= + tell you to expect for the decimal value: ϕ2 = _____________________

Start with 1 1 1 1 ...x = + + + + . Now, if you square both sides and relabel what you are looking at, you should be able to quickly solve for x: x = _____________________

The Fibonacci sequence: F(n) = 1, 1, 2, 3, 5, 8, … {n ≥ 1}, F(n) = F(n-1) + F(n-2) {n ≥ 3}. Using 2 1f f= + , we can reduce all powers of ϕ and (1- ϕ) to linear functions of ϕ. ϕ2 = ____________ ϕ3 = ____________ ϕ4 = ____________

(1- ϕ)2 = ____________ (1- ϕ)3 = ____________ (1- ϕ)4 = ____________

Perhaps you recognize the coefficients, which leads to a formula for the Fibonacci terms:

15. ( ) ( )15

nn

F nf f− −

= . Using this and your answers above, find: F(4) = __________

41

Geometers over the millennia have gone crazy over all the spooky reoccurrences of the Golden Ratio. The diagonals of a regular pentagon make a regular pentagram (5-star). Nested inside a regular pentagram is another pentagon, etc.

ABD is called a Golden Triangle, and ADE is called a Golden Gnomon.

Given the dimensions in the diagram, convince yourself that the listed angles are correct. Fill in some other angles to convince yourself that triangles ABD and ABF are similar. Then the extra labels should be verifiable. Finally, use

11 1x

x=

− to solve for the diagonal of the pentagon with unit sides: x = ____________________

The Golden Triangle is isosceles with ratio of long sides to short side equal to ϕ.

The Golden Gnomon is isosceles with ratio of long sides to short side equal to 1 1ff= − .

In terms of ϕ:

cos(36°) = sin(54°) = __________ ≈ __________

cos(72°) = sin(18°) = __________ ≈ __________

42

PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 2 --- Day 2

43

Math Model 1: Constant Rates (Straight Line Graphs)

Clearly, the straight line is the most common math model with which to model relationships between physical quantities. Why?

Straight lines are easy to work with, easy on the intuition.

They are a good fit for many applications, such as: Distance covered versus time (with constant velocity): 0x vt x= +

Hooke’s law (restoring force for springs): ( )0f k x x= − −

Kinetic friction force versus applied normal force: k kf Nµ=

Ohm’s law (voltage dropped across resistors versus current): RV iR= Forensic anthropology (human height versus femur length): h aF b= +

Conversion formulas (Fahrenheit to Celsius): ( )5 329

C F= −

They are useful approximations over some parameter range in almost all applications. They’re the first terms in any Taylor series: 0 1

22( ) ...a a xf x a x+ ++=

Well before calculus, students can understand (by explanation, videos, experiments) that many things behave linearly as long as motions aren’t too big or stresses too high.

Students can be shown a Taylor series (no need yet to know where it came from), and then can compare the original function graph with graphs of early terms of the series.

Activity 1: height versus femur length.

From the book Forensic Anthropology: What We Learn from Bones, in centimeters:

Caucasian Males: CM = 2.32 F + 66 (±4) Caucasian Females: CF = 2.47 F + 54 (±4) African (Am.) Males: AM = 2.10 F + 72 (±4) African (Am.) Females: AF = 2.28 F + 60 (±3)

Keeping in mind that 254 cm (= 100 in = 8.33 ft) might be just about our maximum

height, pick a reasonable value for maximum femur length: Fmax = __________ cm.

Plot these four lines in Geogebra. To start, enter CM = func and, from the drop down list, choose Function(<Function>, <Start x-value>, <End x-value>), and replace the place holders with Function(2.32x+66, 0, Fmax).

44

Naturally, the curves are all very similar, but by panning & zooming, and perhaps even by adding intersection points, we can ask:

At what femur length are Caucasian and African women about the same height? ____________ cm

At what femur length are Caucasian and African men about the same height? ____________ cm

At what femur length are Caucasian men and women about the same height? ____________ cm

It appears from the graph that at some femur length women surpass men in height. Can you think of any possible reasons for that?

Activity 2: the Celsius versus Fahrenheit formula.

In Geogebra, enter ( )5( ) 329

C F F= − . Then plot y x= .

From the graph:

F = 212 C = ______ F = 32 C = ______

Plot the intersection of the two straight lines.

Intersection coordinates: (_____, _____).

What does that mean?

From the formula ( )5 329

C F= − , find the temperature where both scales read the same

value:

Matching temperature numerical value = __________

45

Activity 3: two cars appoaching.

Two cars, 120 m apart when the timer is set to 0 s, are approaching each other at constant speeds. The car heading to the right is going at 4 m/s and the one heading to the left is going at 8 m/s (with respect to the ground, of course). Find where they meet, and when they meet.

Related question: Do we think beginning students get a better understanding of this problem from the graphical method or by solving equations? (Geogebra hint: hold Shift-key and drag mouse on horizontal axis to stretch it.)

As is often the case, there are many ways to attack this problem. With Geogebra, students can set up the straight line representations of the motion even without knowing the equations.

First, put points A & B at the t=0 positions of the cars, that is at A = (t,d) = (0,0) and B = (0s,120m).

As long as we can get students convinced that 4 m/s means in 1 s the first car goes 4 m, in two seconds it has gone 8 m, etc., we can place two more points: C = (5s,20m), and for the second car, D = (2s,120-16m) = (2s,104m). We could use any other times, but picking two points is enough to draw the position versus time straight lines.

Let Geogebra draw those two lines, green for the first car, red for the second.

Tell Geogebra to find the intersection between these two lines, and label that intersection point Meet. At that point, we can read off the coordinates:

Where the cars meet: __________ m When the cars meet: __________ s

We can calculate from the graph (or let Geogebra do it) the line slopes. Of course the slopes are the 1D velocities, which is why the second car shows a negative slope (it’s going backwards at 8 m/s).

46

Some students will still be a little baffled. “I thought you said the cars are going right and left, why are they shown on the vertical axis?” “If they’re going in two directions on the same line, why are these graphs showing diagonal lines?”

Let’s hope we (and other students in a modeling setting) can clear up these questions (or else the confusions will last and propagate). We could say we’re looking from above, we could twist our heads (or redraw the graphs) so that the cars move in the right-left directions. We’re used to placing time in a horizontal direction and we can think of it as taking a lot of film shots at regular time intervals, so that the graph represents a film strip of the action. The cars aren’t moving sideways, but their captured positions are caught on the film strip laid out on the horizontal axis.

And, if they start to get that idea, but would like a little more reassurance, the diagram above has added three more points. The point time is just placed on the time axis (x-axis to Geogebra), and acts as a slider. We can drag it left or right, and of course its x-coordinate is the time in our problem.

Then two more points are created to represent the cars, moving towards each other on the vertical axis. Green point: car1 = Point({0, 4 x(time)}) [Curly braces needed here.] Red point: car2 = Point({0, 120 – 8 x(time)})

Now, when the time slider point is dragged from 0 to 10 on the horizontal axis, we can see the green and red cars approaching in a straight line at the appropriate speeds. And, at any time, we can see for each car that its position coordinate and the time coordinate match the corresponding point on that car’s straight line (its film strip image). If we’re successful, all students should now get the idea and believe in the utility of position versus time graphs.

One algebraic solution: When they meet, they both have the same time value, so

distance 120speed 4 / 8 /

d dt dm s m s

−= = = ⇒ = __________ m

t = __________ s

Another algebraic solution:

Relative speed of approach = 12 m/s 12012 /

tm s

= = __________ s

4 md ts

= =

__________ m

47

Activity 4: faster car chasing slower car.

This time both cars start off at the zero position, but the slower car (4 m/s) gets a 6 s head start over the faster car (8 m/s). Eventually, the faster car catches up. Find where they meet, and when they meet.

As in the previous example, place two points for each car and connect their position versus time straight lines. The first car takes off at (0,0). Using its speed, pick another point, such as where it is at 4 s. Then, draw its line.

The second car is still sitting at the starting position until 6 s, so it takes off at (6,0). Using its speed, pick another point, such as where it is at 8 s. Then draw its line.

Now find the intersection point. When the fast car catches up,

t = __________ s d = __________ m

One algebraic solution:

Distance (d) versus time (t) equation for slow car: ________________________

Distance (d) versus time (t) equation for fast car: ________________________

When they meet, they both have traveled the same distance, so set their distance formulas equal and solve for the clock time, then find that distance:

t = __________ s d = __________ s

Another algebraic solution: Head start distance of first car after 6 s, running at 4 m/s: Δd = __________ m

Relative speed of approach = 4 m/s. Starting at 6 s, the second car has to use this relative speed to close the Δd gap. You could now calculate that Δt, then add 6 s to get the clock time. Then calculate the total distance traveled.

t = __________ s d = __________ s

48

Taylor series: “everything’s” linear!

As we’ve seen, it’s really easy to plot functions in Geogebra (and in many other programs, calculators, etc.).

2

1 ...2!

x xe x= + + +

We can see the exponential function is approximated by

1xe x+

for 0.2 0.2x− ≤ ≤ .

3 5

sin( ) ...3! 5!x xx x= − + −

We can see the sine function is approximated by

sin( )x x

for 0.5 0.5x− ≤ ≤ .

49

PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 2 --- Day 3

50

Intro to Geometric Algebra (GA)

Dr. David Hestenes wrote his Primer on Geometric Algebra years ago as a “quick” introduction. A few years after that, he asked me (Bob Rowley) to make a web version. All the original text was left intact, and solutions to suggested exercises were added, as well as many interactive diagrams created with Geogebra: http://geocalc.clas.asu.edu/GA_Primer/GA_Primer/index.html. If some solutions seem too long, I am to blame, as I was having fun and may have gotten carried away here and there.

Dr. Hestenes has spent around 50 years, after his original theoretical invention of GA, explaining and expanding its applications to many areas of physics in his books and hundreds of published articles. There are now many followers worldwide, publishing their own articles and books, one set of followers and expert theoreticians being located at Cambridge University in England.

As an aside, I’ll mention some favorite books. I agree with most physics enthusiasts that The Feynman Lectures on Physics (3 volumes by Nobel Prize winner Dr. Richard Feynman) are so enjoyable for their beauty and clarity. Also very enjoyable, with a fresh and unusual delivery, are Matter & Interactions I & II, by wife and husband Drs. Ruth Chabay and Bruce Sherwood, co-inventors of VPython and Glowscript. Both of those are intended as college freshman courses. I’m still astonished by the breadth and depth of my personal favorite, New Foundations for Classical Mechanics, 2nd Edition, by Dr. Hestenes.

So, we’ll get into GA, and we can follow the website for much of that, trying various ideas along the way. Vectors have been around for 150 years, and all of vector analysis is still part of GA, although things can often be done in a simpler, cleaner way. By the way, notations and viewpoints in GA have changed a little over the years, and, as always, different authors have different preferences and quirks. That used to bother me (as a student), yet no theory or math formalism is born completely finished and perfect, we all learn as we go. Any of us might in the future find a more elegant way to do this or that.

One of the great attractions of GA is that it is a single mathematical framework to handle everything in physics. As a preview of this, here are examples of a variety of math tools used in physics and their equivalents in GA:

In standard literature, if a and b are vectors (sometimes called polar vectors), the cross product, ×a b , is called an axial vector, because it behaves differently under space inversion than polar vectors.

But, in GA, we use the wedge product, i∧ = ×a b a b (we’ll learn what it is later). Then, not surprisingly, it acts differently than vectors, because it is different. It’s essence is represented by a bivector, not a vector.

51

2D rotations:

i is imaginary. 2 1i = − . i is a real bivector. 2 1= −i . Multiplying complex numbers The idea is very similar in GA, acts like rotation/dilation. we’ll cover that in class.

Using a rotation matrix: In GA:

( ) 1

2

cos sin,

sin cosaa

θ θθ

θ θ−

=

a ( ), e θθ = ia a

3D rotations:

Hamilton (1843) invented He had invented an important aspect quaternions as a 3D extension of GA, but his “complex” vector of complex numbers in order is just a GA bivector, and there’s no to do rotations in 3D. need for imaginary numbers. Using a rotation matrix: In GA: ( ) ˆ ˆˆ, , i ie eθ θθ −′= = n na n a a

( )( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )

21 1 2 3 1 3 2

122 1 3 2 2 3 1

223 1 2 3 2 1 3

ˆ, ,

cos 1 cos 1 cos sin 1 cos sin1 cos sin cos 1 cos 1 cos sin1 cos sin 1 cos sin cos 1 cos

R

n n n n n n na

n n n n n n na

n n n n n n n

θ

θ θ θ θ θ θθ θ θ θ θ θθ θ θ θ θ θ

′= = ≡

+ − − − − +

− + + − − − − − − + + −

a n a a

Pauli matrices from non-relativistic quantum mechanics,

16. 1 2 3

0 1 0 1 01 0 0 0 1

ii

σ σ σ−

≡ ≡ ≡ −

behave exactly like GA’s 3D orthonormal (unit vectors perpendicular to each

other) basis. So, instead of the standard notations { }ˆ ˆ ˆ, ,i j k or { }1 2 3ˆ ˆ ˆ, ,e e e , which

are just fine, we often use { }1 2 3ˆ ˆ ˆ, ,σ σ σ . I tend to avoid the first set because how

many I’s do we need ( i = unit vector, i = unit bivector, i = unit trivector)? And, I often use the middle set simply because they seem easier for me to freehand rather than sigmas (even though they are almost the same muscle movements).

One last tidbit. Maxwell’s equations (2 scalar and 2 vector equations) look like this in GA: 0F cJµ=

52

GA Practice in 2D

Given: ( ) ( )1 2 1 2 2 1ˆ ˆ ˆ ˆ ˆ ˆ6 2≡ ≡ + ≡ − ≡ + ≡ −i e e a e e b e e c a b d a b

2 ______=a 2 ______a ≡ = =a a ˆ ____________=a 1 ____________− =a

2 ______=b 2 ______b ≡ = =b b ˆ ____________=b 1 ____________− =b

__________=ab 2 __________a= + =ac ab 2 __________a= − =ad ab

___1ˆ____ e= ia e ____

1ˆ____ e= ib e ________ e= iab

How can we tell whether &a b are perpendicular to each other?

________ e= iac What is the angle from a to c ? __________ °

Sketch , , , &a b c d all starting from the origin:

53

Sketch , , , , &−a b b c d . Start with a . Attach & −b b to the end of a . Then attach c from the start of a to the end of b , and attach d from the start of a to the end of −b . Label all angles and lengths of sides.

54

Activity: faster car car chasing slower car, with different directions in 2D.

Starting at a time of t = 0, car 1 has a position given by ( ) 1 2ˆ ˆ1 4 24t t= +x e e . Car 2 doesn’t

take off from the origin until t = 3 s and has a speed of 8 m/s, but we don’t know what direction to take in order to catch up with car 1. So, we can write its position is

( ) ( ) 1ˆ2 8 3t t e θ= − ix e . Find t, θ, and 1 2=x x for the case where car 2 catches car 2.

Let’s see how car 1 is moving:

( ) 1 2ˆ ˆ1 0 ______ ______= +x e e ( ) 1 2ˆ ˆ1 1 ______ ______= +x e e

( ) 1 2ˆ ˆ1 2 ______ ______= +x e e ( ) 1 2ˆ ˆ1 3 ______ ______= +x e e

What is the velocity of car 1? 1 2ˆ ˆ1 ______ ______= +v e e

How about car 2 (remember, θ is unknown, but constant)?

( ) 1 2ˆ ˆ2 3 ______ ______= +x e e 1ˆ2 ______ e θ= iv e

What is the square of 2 ' sx direction? ( )2

1 1 1 1 1ˆ ˆ ˆ ˆ ˆ ______e e e e eθ θ θ θ θ−= = =i i i i ie e e e e

When car 2 catches car 1, their position vectors should be identical, so set them equal to each other, then divide by 4 to simplify: ( )1 2 1ˆ ˆ ˆ6 2 3t t e θ+ = − ie e e . Now, square both sides

and notice that the angle disappears, so now you can solve for the clock reading:

t = ______ s.

Now you can plug that time into the last equation and solve for the angle:

cos ______θ = sin ______θ = ______θ = °

What is the position vector when the cars intersect? 1 2ˆ ˆ______ ______= +x e e

55

Geogebra simulation:

Make a slider for time θ varying between 0 and 15 s.

Make a slider for angle θ varying between 30° and 45°.

Make a point x1=Point({4t, 24}).

Make a point x2=Point(8(t-3){cos(θ), sin(θ)}).

Vary the sliders and watch the motion. Turn on the trace of the points and try some more simulations. Turn on animation for the t slider. You can also vary the speed of animation.

Why pick a { }1 2ˆ ˆ,e e coordinate frame?

Another advantage of GA is that we typically don’t need to pick a frame. Usually, certain vectors or directions are important in a problem, and why not use them as our guides? In this case, everything of interest centers around the motion of car 1, and its velocity is constant. We don’t even need to make it a unit vector. Let’s see if we can solve the whole thing in terms of { }1, , ,t θi v .

( ) ( )( ) ( ) ( )( )( ) ( )

( ) ( ) ( ) ( )

1 1 1 1 1

2 1 1 2

22

36.91 1 1 1 2 2 1

24 4 6 6

3 2 6 2 3

6 6 36 4 3 3 84 38 6 10 cos sin 36.95 5

ˆ ˆ ˆ ˆ8 8 6 4 8 6 32 24 8 40

d v t t t

t t e t t e

t t t t and t t

e and

e

θ θ

θ θ θ θ

°

= = = = + = +

= − = ⇒ + = −

+ − = + = − ≤ ⇒ =

+ = ⇒ = = ⇒ °

= + = + = + = =

i i

i

i

d v i x v d v i

x v x x i

i i

i

x v i e i e e x e

Success! You see we can put things in terms of { } { } { }1 2ˆ ˆ, , ,E N right up= =e e if

and when we like, but we had no need to start there. In fact, the solution is a little quicker without adding them to the mix.

What did we do, in summary?

We saw almost all quantities could be easily expressed in terms of 1 &v i . We set the intersection position vectors equal to each other. By multiplying that equation by its reverse, θ was temporarily eliminated, so we could easily calculate the time of flight. With t=8 known, we could go back and find θ and ( )1 8x .

56

More Viewpoints

First, people had to keep track of quantity for bartering. They could count fingers, make scratches, argue about is an apple worth 2 eggs, or vice versa?

OK, we need to count. Scratches (abacus) Roman numerals Arabic symbols 1, 2, 3, … The invention of these new symbols blew some ancient minds, until it became commonplace.

The concept of half a banana seems easy enough. New notations always took some mind stretching: HALF = ½ = 0.5 (each notation was a struggle to get used to). Initial skeptics might ask how we trade half a raw egg, for example. Bananas fine, but you surely can’t imagine half an egg?!

Fine, we’ll accept counting numbers, and the nerds can even imagine fractions. But what’s this “zero” thing? If I’m out of apples, I have nothing, not a “thing called zero.” Yet, a symbol was invented: “0”. Some may have asked, why do I need a symbol? Where would I ever use it?

Now for a real leap: negative numbers. I can’t have negative two apples?! If I have 7, I can give you 2, so I only have 5 left. But some math nerds wanted to say “I’ll add -2 apples to your 7, leaving you 5.” Who knows, that may have taken a generation or two to swallow.

So, now we have “rational numbers” (whole numbers, 0, negative numbers, and fractions). The Pythagoreans (wizard of Oz cult society of math worshippers) believed rational numbers were pure, and anything of value could be measured in appropriate ratios. They even knew there was a major problem with the hypotenuse of a right triangle whose other sides had length one.

But eventually --- “irrational” numbers and weird new symbols: 1 52, 2.718..., ,2

e π f += = .

The next advance: direction matters in some situations, so define a vector as an “object” with both magnitude and direction. Vector notations in use here & there: a= =a a .

Vector magnitude notations: 2a a= = = = =a a aa a a

.

Vector direction notations: ˆPoint finger, draw arrow of length 1, / a=a a . How can we to specify vectors with numbers? Draw the vector, label its length. Imagine axes: {East, North, Up (out of paper)}, {x, y, z}, { 1 2 3ˆ ˆ ˆ, ,e e e },

then add magnitudes in those directions: ( ) 1 2 3 2 3 1ˆ ˆ ˆ ˆ ˆ ˆ2,3,1 2 3 3 2= + + = + +e e e e e e .

Notice order matters with the (2,3,1) notation, but not with unit vector notation.

If these are vectors, what do we call regular numbers? Scalars.

57

Next step: what kind of math can we do with vectors? Multiplying by a scalar ( 5 , 2.3 , ...−a b ) or combining by addition ( , 2 1.5 , ...+ −a b a b ) are pretty easy to understand. It’s easy enough (after some practice) to draw additions of 2D vectors, but in 3D, algebra is more efficient. Given: 1 2 3 1 2 3ˆ ˆ ˆ ˆ ˆ ˆ2 2 & 4 6 2= − + = + +a e e e b e e e

( ) ( )

1 2 3

1 2 3 1 2 3 1 2 3

ˆ ˆ ˆ6 5 4ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 1.5 4 2 4 6 9 3 2 11

+ = + +

− = − + − + + = − − +

a b e e ea b e e e e e e e e e

(very easy with algebra, pretty hard to draw)

But, we can also multiply vectors:

Dot product: ( ) ( ) ( )cos shadow of on shadow of on ab a bθ≡ = =aba b b a a b

( ) ( ) ( )

Wedge product: bivector (or plane element) with orientation given by sweeping towards (CCW or CW)

and magnitude = sin sin sinab a b b aθ θ θ

∧ ≡

= =ab

a ba b

Geometric product: ab

GA product in terms of dot & wedge: = + ∧ab a b a b

Dot product in terms of GA product: ( )12

= + =a b ab ba b a

Wedge product in terms of GA product: ( )12

∧ = − = − ∧a b ab ba b a

Special cases: ( )parallel vectors commute⇒ = =a b ab a b ba

( )perpendicular vectors anti-commute⊥ ⇒ = ∧ = −a b ab a b ba

2 2 2

1 1 1 1 1 2 3

1 2 2 1 2 3 3 2 3 1 1 3

ˆ ˆ ˆ ˆ ˆ ˆ ˆ1ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ:unit bivectors

= = = = == − = − = −

e e e e e e ee e e e e e e e e e e e

If there’s only one plane involved (as in almost all HS physics problems), we usually use a shorthand notation for the only needed bivector:

( ) ( )21 2 1 2 1 2 1 2 1 2 1 1 2 2 1 1 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1≡ = = = − = − = −i e e i e e e e e e e e e e e e e e e e

Unit trivector: 21 2 3 1 2 3 1 2 3ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1i i≡ = = −e e e e e e e e e

58

PHS 542 Name: ________________________________________

Homework: Sliding block on inclined plane with kinetic friction. _____________________________________________________________________________

(Situation map) (System Schema)

(Interaction Map = FBD = Free Body Diagram)

The typical question is to find (the formula for) the acceleration of the block down the plane.

17. ( )Newton's 2 Law: In this case:

nd m All vector forces acting on the objectm m

=

= + +∑a

a g N f

We’ll try three different ways to solve it. The problem is easy (at least after we’ve seen it a hundred times). We’re more interested in practicing and comparing methods.

m

System

gR

Ramp

Earth

Environment

m Ramp

This is a standard problem in beginning physics classes. Gravity (action at a distance) pulls the object towards the center of the earth with the force mg . The inclined plane pushes back with the normal force N (a contact force, the word normal meaning perpendicular to the surface). The magnitude of the kinetic (or sliding) friction force is found experimentally to be approximated (roughly) by f Nµ= . The friction force increases (another straight line example) as the surfaces are pushed harder together. The coefficient of kinetic friction (usually denoted by kµ , but we can drop the subscript here) depends on surface materials and conditions, but is constant in any particular case.

59

Method 1: the standard treatment.

We break forces into components down the incline and components perpendicular to the incline. Then we have two scalar equations, which can be used to solve for N and a.

The first confusion students typically have is which of the angles shown correspond to θ?

1. Which angle represents θ in the diagram? a) α b) β

2. What is the component of mg parallel to N? a) sinmg θ− b) cosmg θ−

3. Use Newton’s law in the direction parallel to N to find its value. N = a) sinmg θ b) cosmg θ

4. Now write the value of the friction force. f = a) sinmgµ θ b) cosmgµ θ

5. Use Newton’s law in the direction down the incline to find the formula for acceleration in terms of { }, ,g µ θ . a = ______________________________

6. Why doesn’t the acceleration depend on the mass also? Wouldn’t more mass mean the object is pushed harder into the surface, therefore increasing the friction?

Explanation:

60

Method 2: using 1e down the incline and 2e perpendicular to the incline.

That makes three of the vectors pretty easy to write down:

1 2 1ˆ ˆ ˆa N f N Nµ µ µ= = = = − ≠a e N e f e N

The last two are a reminder that even though the friction force magnitude is proportional to the normal force magnitude, the two forces point in different directions.

7. Write the acceleration due to gravity force in terms of { }1 2ˆ ˆ, , ,g θ e e :

g = ____________________ 1e + ____________________ 2e

8. Now fill the representations above into m m= + +a g N f :

ma = ____________________________________________________

9. Use the 2e coefficients to solve for N. N = __________________

10. Plug that in and use the 1e coefficients to solve for: a = __________________

As you see, the two methods are basically the same.

61

Method 3: using some GA notations and the natural directions g and a .

Two natural directions are the direction of the gravity force (the basic cause of the motion) and the direction of the acceleration (the resulting motion).

We don’t really have to bother with finding two perpendicular directions.

We know that in GA, the product of two unit vectors gives a rotor which can be used to rotate any vector in the plane of the two

vectors by the angle between them. In this case:

( ) ( )/2 /2ˆ ˆ ˆ ˆ ˆ ˆsin cose e e e eπ θ πf θ θ θ θ− − −= = = = ⇒ = ∧ =i ii i iga i g a g a i

Remember also that right-multiplying any vector by i rotates that vector 90° CCW. So, for example: ˆN=N ai and ( )ˆ ˆN Nµ µ µ= = = −f Ni ai i a , as we would expect.

Notice in the system schema, on page 1 above, that there is one bubble for the Ramp interaction, yet we keep writing apparently two interactions, normal force and friction force. It isn’t necessary, but it is interesting that with GA we can write those forces as one geometric product: ( ) ( ) ( )ˆ ˆ1 1N Nµ µ µ µ+ = + = + = + = −N f N Ni N i ai i a i .

Our force equation now looks like: ( ) ( )ˆ ˆm m N m Nµ µ= + − = − +a g a i g i a . For the last

part, remember that vectors in a plane anti-commute with the bivector of the plane, so ˆ ˆ= −ai ia . Multiplying both sides of the force equation by a gives:

( ) ( )ˆ ˆ ˆma m N m m Nµ µ= − + = + ∧ − +ga i g a g a i

11. The left side is a scalar, so the right side is, also. The bivector terms must add up to zero. Use that fact to solve for N: N = __________________

12. Now use the scalar terms to find a: a = __________________

62

PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 2 --- Day 4

63

Math Model 2: Constant Change in Rate (Parabolic Graphs)

The basic math models we’re emphasizing are really about helping pre-calculus students get acquainted with solutions to some basic differential equations before calling them by that name. We’re familiar with the logical and basically historical progression of mathematical insight:

• Pre-HS: arithmetic (concept of numbers, zero, negatives, rational/irrational, beginnings of algebra).

• Algebra: improving skills, factoring, the abstraction of using variables. • Geometry: visualization of our 3D world, regularities, similarities, etc. • Trigonometry: finding & using ways to handle geometrical ideas with algebra. • Pre-calculus: functional relationships between variable quantities, rates, graphs. • Calculus: slopes (derivatives), areas (anti-derivatives = integrals), the chain rule.

MM1 was the constant rate dy kdx

=

model. That is, the slope of the function’s graph was

constant, so that when plotted on its own graph, the slope is a horizontal line.

MM2 is where the rate changes linearly, 0dy k kxdx

= +

. That is, the slope of the function’s

graph changes in such a way that when it is plotted on its own graph, the slope is a non-horizontal straight line. That means that when the function itself is graphed, it can’t be a straight line because its slope is changing, but that slope changes in a nice, simple way. Of course, we know the function we’re talking about is a parabola, and we have many practical examples:

• Distance covered versus time (with constant acceleration): 20 0

12

x x v t at= + + .

• Projectile vertical position versus horizontal position: ( )0

20 2

0

tan2 x

gy y x xv

θ= + −v .

• Area versus linear dimension of closed plane figures.

• Kinetic energy versus speed: 212

K mv= .

• Spring potential energy versus stretch: 212

V kx= .

• Crop yield versus fertilizer (Clawson): 200 0 2 peak

Iy y I f ff

+ −

• The first three terms in many Taylor series: 20 1 2

33( ) ...a a x af x a xx+ += +

64

Activity 1: use Geogebra to convince students that the slope of a parabola changes linearly.

1. Plot ( )25f ( ) 12 416

x x x= + − .

2. Place point A on the parabola. 3. Draw a tangent line through A on f. 4. Define m to be the slope of that line. 5. Make red point { }( )B Point (A),x m= .

Turn on the trace for B. 6. Watch what happens as A is dragged along the parabola, and look and the slope and B values. 7. Pick two particular B values and use them to write the equation of its line, then draw that line. Drag A some more.

In practice, maybe students should write down several slope points from the parabola and plot these rate points on a separate graph. In some way like this, we need to remind them that the parabola and its rate graph are probably different units, so that different axes should be used. For example, if the parabola is distance versus time, then the rate graph is really velocity versus time.

Parabola magic: the slope along the curve varies linearly. Students may wonder, what’s the big deal, is that so unusual? If so, let them try different functions:

The slope of the semicircle is not Of course ( )f xx e= has its own

tracing a straight line. What is it? magic. At every point on its curve, Compare to the graph of cot(θ). its slope value = its function value!

How much of the circle closely simulates a parabola?

65

Activity 2: conversely, if the rate of a function changes linearly, the function is a parabola.

As a concrete example for students to ponder, say we have a container of water filled to an initial volume (at time t=0) V0. Then, we turn on a special filler hose. It’s special in a number of ways:

• When it is turned on, there is an immediate, initial fill rate, F0 (gal/sec). • There is an electronic flow valve set to linearly increase the fill rate from there. • So, the fill rate is increasing, following a straight line formula: 0( )F t F tα= + .

Our goal is to find the function for the total volume as a function of time.

1. If you had a fixed fill rate, the volume added would simply be the fill rate times the time elapsed. Notice in that case, the fill rate function would be a horizontal straight line, 0( )F t F= , and the volume added would be the area under that curve.

2. But, the rate itself is increasing with time. However, now we can do a typical trick for handling curvy functions. Break our time interval into many, very small time increments. For each tiny time increment, the fill rate wouldn’t have time to change much, and we could estimate that its value as roughly constant over that increment. (This kind of argument goes back at least as far as Newton).

3. For each tiny time increment, the volume added is therefore just the area of that approximately rectangular sliver of time increment under the rate function. We still have to add up all those slivers to get the total volume added, but that amounts to adding the area under our rate function. That’s integration in calculus! Our added volume is simply the area under our rate function, no matter what that function looks like.

4. Now for the easy part. We are only interested here in a straight line rate function. That means calculating the area under the straight line is extremely easy. Now that we know it’s this area we’re after, we don’t need to actually break our function into time slivers, because our particular function is a straight line. It’s the area of a trapezoid, or the area of the rectangle, 0F t , plus the area of the triangle above it. Or, even easier, again because it’s a straight line rate, just take the average height times the time!

( ) ( )0 0(0) ( )

2F F tV t V avg fill rate t V t+

= + = + = ________________________

And, that’s a parabola! Can we try duplicating these ideas in Geogebra? Pick values for ( )0 & 0V F and let α be a slider: 3 3α− ≤ ≤ . Geogebra can show the area under our

straight line rate function. Then, plot that area versus x (representing t here).

66

Activity 3: parabola equations, standard/polar form, vertex/focus/directrix, reflection properties.

As shown by Dr. Hestenes in NFCM2 (New Foundations for Classical Mechanics, 2nd ed.), we can write an equation for the parabola in terms of the parameters shown here.

Curves of conic sections can be defined in terms of a focus and a directrix line (with or without vector notation). We call the focus our origin. Then the radius vector points from there

to a point on the curve. And, the conic can be defined by cosr

d rε

θ=

−, where ε is

called the eccentricity. This works as a defining polar equation for circles ( )0 dε = ⇔ = ∞ , ellipses ( )0 1ε< < , parabolas ( )1ε = , and hyperbolas ( )1ε > . We

usually label the semi-latus rectum by the script . We can see from the definitions and diagram, in general dε= , and so we can notice in our parabola in the diagram,

1 dε = ⇒ = . We use vector d as our polar axis. In general, we can define the eccentricity vector in a similar way:

ˆˆ ˆcos

:1 cos

:1 cos

r

Conic r r

Parabola r

ε θ

ε θ

θ

≡ = =

= − ⇔ =+

=+

ε d ε ε r ε

r ε

1. In Geogebra, make a slider named slr varying between -5 and 5.

2. Geogebra does a nice job with polar equations. Type in the formula ( )1 cos

slrrθ

=+

.

Notice the standard (default) polar axis is the x-axis, so in which

direction is the open side of your parabola? Right. Left. Up. Down.

3. Modify the argument in ( )cos θ so that it opens down, as in our diagram above.

4. Enter the function ( )d x slr= . Now, vary the slider and verify predictions.

5. What is happening when the slider goes negative?

67

Keep in mind, with the form ˆˆ1

r =+ ε r

, our polar axis, ε , can be in any direction we

like. To compare with our rectangular parabola formulas (and our above diagram and Activity 1), let’s choose our y-axis direction to correspond to the ε direction. So, our parabola will open down, but we’ll still allow our vertex to be at any general point (h,k).

Rearranging our polar equation and remembering d= ,

( )2

2

cos

2

2

r d r

dx h y k

dd y k

θ= − ⇒

− + − +

= − − +

Squaring and simplifying gives us a standard form of the rectangular equation:

( ) ( ) ( )2 2122

x h d y k or y k x hd

− = − − = − −

6. Our parabola from Activity 1 was ( )25 12 416

y x x= + − . Complete the square and put

it in the second of these standard forms.

y = ___________________

7. Use this result to read off d: d = __________

8. Read off the vertex coordinates: (h,k) = (________, ________)

68

One of the intriguing properties of the parabola is that all incoming (light) rays bounce off the walls of the (silvered) parabola in such a way as to pass through the focus.

Standard proof:

There are some convoluted proofs out there, but one of the clearest was found at https://amsi.org.au/ESA_Senior_Years/SeniorTopic2/2a/2a_2content_13.html from the Australian Mathematical Science Institute. They choose axes so that the formulas are easiest. We can fill in some gaps of their proof.

They use an upward opening parabola and use symbol a for our d/2, so their parabola formula fits our first standard form on the last page.

Their parameter p can be thought of as a Geogebra slider. The point P(2ap,ap2) satisfies their parabola formula and varying p allows P to slide along the parabola.

9. Use 2

4xya

= to find the slope of

the tangent line at x=2ap:

m = __________

10. Now use y=mx+b for the tangent line, use the m just found and point P to find b, so that we can get the equation of the tangent line:

y = __________________

11. Use the tangent line formula to verify the value of intersection point T.

12. Because Q is an incoming parallel ray, QPB STP∠ = ∠ . The goal is to prove

this also equals SPT∠ . Do this by proving SP ST= .

69

GA proof:

We can transform our scalar polar equation into a vector equation for the curve:

( ) ( )

( )

ˆˆ:1 cos

ˆˆ ˆ1 cos

:1 cos

dConic r and e

d er

eParabola

θ

θ

θ

εε θ

θε θ

θθ

= =+

⇒ = =+

=+

i

i

i

εr

εr ε εr

dr

Then we can get an expression for the unit tangent vector along the curve. That is perhaps too hard for high school students to do, although interested students might be able to follow along while you do it. We could just present the answer to HS students, but you can try now to see if you get the same answer:

( ) ( ) ( )ˆ 1ˆ2 2cos

edd

θ

θ θθ θ

+≡ =

+

id it r

Since it is a unit vector, tangent to the curve, pointing towards increasing θ, we can predict ( )ˆ 0t and check a couple of other places against our earlier diagrams:

( ) ( ) ( )ˆ ˆ1 1ˆˆ ˆ ˆ02 22 2π π− + = − = =

d i i d i it d i t t

The general argument and proof of a parabola’s reflection property is really easy and beautiful to express in GA. The incoming parallel ray can be represented by d . The tangent vector acts like a mirror and reflects like this:

( )ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆreflected off ⊥ ⊥= = + = −d t t d t t d d t d d

So, to prove the parabola’s reflective property is simply to show

ˆˆ ˆ ˆParallel rays reflect through focus ⇔ = −t d t r

The hard part was getting expressions for the various unit vectors. To finish the proof:

( ) ( ) ( )( )ˆ ˆ ˆ1 1 1 1ˆˆ ˆ ˆˆ ˆˆ

2 2cos 2 2cose e e e

e eθ θ θ θ

θ θ

θ θ

−+ + + += = = −

+= −

+

i i i ii i

d i ddt d t

ir d d r

70

PHS 542 Name: ________________________________________

Homework: Parabolic functions. _____________________________________________________________________________

In selling a certain product, a company has a testing period where they try various selling prices and find how many units are sold. We can think of using either s or q as our independent variable, but this table determines how they are related to each other. The cost to make each item is fixed at c = 10. We’d like to see what the profit functions, P(s) or P(q), look like and can tell us. In general, the profit function should will be ( )P q s c= − .

1. What does the table tell us is the relationship between quantity sold in terms of selling price? q(s) = __________________

2. What is the profit function in terms of q only? P(q) = __________________________

3. Use P(q) to find what quantity corresponds to maximum profit. You can figure this out by graphing the function, or by using calculus, or by putting the parabolic function in a standard form and reading off the vertex.

qmax = ________________

4. What would that maximum profit be? P(qmax) = ________________

5. What is the (linear) rate of change of profit

with respect change in quantity sold? ( ) dPr qdq

= = ________________

6. What is the profit function in terms of s only? P(s) = __________________________

7. Use P(s) to find what quantity corresponds to maximum profit. smax = ________________

s = selling price

q = quantity of units sold

15 700 30 400 45 100

71

8. Now, what would the maximum profit be? P(smax) = ________________

9. What is the (linear) rate of change of profit

with respect to change in selling price? ( ) dPR sds

= = ________________

10. Can you get this R(s) another way, by plugging q(s) into r(q)? That is, is it true that ( )( ) ( )r q s R s= ? a) Yes. b) No.

One of the most important calculus concepts is the chain rule for composite functions. Let’s practice it here.

11. Use the formula found earlier for q(s) and for dPdq

to find dPds

using the chain rule:

Write the final answer in terms of s and show your work.

( ) ( )dP dP dq dqR s r qds dq ds ds

= = = = ___________________

72

PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 3 --- Day 1

73

Math Model 2: Constant Change in Rate (Projectile Motion)

The GA Primer has a section, Constant Acceleration Model, which deals with an object under the influence of a constant force (main application: gravity). What makes up the model?

• The object (or projectile) is considered to be a point particle of mass m. • The net force & acceleration (due to gravity = g ) are approximated as constant ( gravF

really varies with altitude & curvature of Earth, and local mass density of the ground). • We assume there is no other non-constant force acting on the particle (ignoring air

friction, air buoyancy, gravity from other objects, etc.).

Of course, being a model, our conclusions should apply equally well to any constant force application (gravity is not the only approximately constant force in nature). Here, we can mention a few other ideas not explicitly spelled out in the GA Primer.

Newton’s 2nd Law: dm mdt

= = ⇒ = =∑ vF g a g a .

What are we saying here? Our rate is constant. That’s our Math Model 1! That means our velocity function should be a linear equation (linear in t). It doesn’t matter that our coefficients

in the equation are vectors, just that they are constants. Thus: ( ) 0dt tdt

= + =xv v g .

We could have easily integrated the acceleration equation to get the velocity equation. But, we didn’t have to, since we knew what to do from our Constant Rate Math Model. But notice, we now have a rate which is changing linearly. That’s our Math Model 2! That means our position function should be a parabola (in variable t). Again, we could use calculus to integrate this equation to get the position equation. But, we don’t have to, since we’ve already found another way. Remember that in Activity 2 from our introduction to Math Model 2, we worked out this method for finding the total volume when the fill rate was increasing linearly:

( ) ( )0 0 0(0) ( )( )

2F F tF t F t V t V avg fill rate t V tα +

= + ⇒ = + = +

Again, we emphasize the point of working with models. When you’ve worked out details in one model, you only have to change symbols to see the equivalent details for an equivalent model. Therefore:

( ) ( ) ( ) ( )

( )

00 0 0

20 0

212

tt t t avg velocity t t

t t t

+= + ⇒ = + = +

= + +

v vv v g x x x

x x v g

74

Since 0x represents the initial position of our projectile, we could just call that initial position our origin, and drop it from the equation. Or, equivalently, the GA Primer defines the range vector: 0 ˆ ˆx y≡ − ≡ −r x x gi g . So, the GA Primer starts with these equations:

2 00 0

12 2 avgt t t t t+

= + = + = =v vv v g r v g v

Some other ideas useful in physics:

preferred because alsondvalid in relativityNewton's 2 Law: dm

dt= → =∑ ∑ pF a F

2

1

2

2

Work done bya force

1Power input2

1Kinetic energy2

W d W

dW d d dKP m mvdt dt dt dt

K mv W K

= ≡ ⇔ ∆ = ∆

= ≡ = = = ≡

= = ∆ = ∆ = ∆

∫x

xF x F x

vF v v

F x

Work is a scalar quantity, but can still be plus or minus. Doing positive work on an object increases its kinetic energy. Each of the forces acting on a particle can do positive, negative, or zero work. It’s the total (net) force which counts. If the net force acting on the particle is zero, no work is done, the kinetic energy doesn’t change. In that case, we say the kinetic energy is conserved (remains constant).

Some forces in nature have a nice property. If the work a force does along one path of motion is the same as that done along any other path of motion, the force is called conservative (notice that’s not the same word or meaning as conserved). In other words, the work done by such a force depends only on the initial and final positions, not on the path (position, velocity, acceleration) taken during the motion. It also means that if the particle returns to the same initial position, the work is reversed (if K is gained along the path, it is lost on the way back). Gravity is one such force. Friction is not (along any path, back or forth, work done by friction is always negative).

( ) ( )2

1

if axis2 1points up

yby gravity gravW m d W mgy mgy mgy−= → = − − = −∆∫

x

xg x

If the particle moves downward, y∆ is negative, the work done by gravity is positive, and, as we know, the particle speeds up, increasing K. If, as in our projectile examples, gravity is the only force acting,

( )( ) ( )defining gravitational

potential energy

0

0 0all forces other grav

other

W W W mgy K

W K mgy K GPE

∆ = ∆ + ∆ = −∆ = ∆

∆ = = ∆ + ∆ → = ∆ +

75

We have our basic ideas and equations from above (notice we can express our relative position in terms of initial velocity and time, and also in terms of current velocity and time).

2 2 00 0

1 12 2 2 avgt t t t t t t+

= + = + = − = =v vv v g r v g v g v

Notice the second expression is a parabolic equation in time. It is also the parametric orbit equation of the projectile (by simulating it in software, we see not only the path of motion, but the correct timing, how its speed changes as the particle moves along the path).

If we use the first expression to eliminate t from the second, we get a parabolic equation in speed:

( ) ( ) ( )

( )( ) ( )( )

21 1 1 10

2 20 0 0 0 0

2 20 0

2 20

12

2 2 2

2 21 10 constant2 2

t

v v

and gy v v

mv mv mgy K GPE

− − − −= ∆ = ∆ ⇒ = ∆ + ∆

= ∆ + ∆ = − + = − + ∧

∧ = ∧ = − = −

+ = + = + =

g v v g r g v v g v

gr v v v v v v v v v

g r v v g r

We’ve now seen three ways to get the energy equation for ideal projectile motion: from the definitions on the previous page, from the hodograph/trajectory diagram in the GA Primer, and now by eliminating t in the motion equations. The key idea is that knowing the altitude (y=0 at the launch point) of the particle is enough to know its speed. It doesn’t matter if the particle happens to be moving up, down, or sideways, its speed varies only with altitude.

It is also a common idea to eliminate t from the horizontal and vertical parts of the position equation to get a parabolic equation in coordinates:

2 20 0

0 0 02

020 2

0 0 0 0

ˆ1 ˆ ˆ2 2 (horiz.) :ˆ2

ˆ ˆ (vert.) : 2 2 22

x x

y

x x x x

x xt t t gt x tv v

vx x gy y g y x xv v v v

∧= + ⇒ = + ⇒ = = =

∧

= − = + ⇒ = − +

g r ir v g gr gvg v i

g r g v

This shows the parabolic path, but we’ve lost all timing information. We’ve looked at parabolic equations before. There doesn’t seem to be much point in transforming these into the polar form for projectile examples (although feel free to think about that). However, it might be useful to put the first (time parabola) and third (the path parabola) equations into standard form, in order to read off vertex values. The energy equation (speed parabola) is already simple enough to read off what we need.

76

The time parabola:

{ }2 20 0 0 0 0

220 0 02

ˆ ˆ ˆ2 2 2 2 sin

2Completing the square:

2 2 2

y y

y y y

y t gt v t gt v v

v v vg gy t t tg g g

θ= − = + = − + − = =

= − − = − −

g r g v g v

Vertex of (y,t) parabola: 2 2

0 0 0 0,2 2

y y y yvertex vertex

v v v vt y

g g g g

⇒ = =

That’s cute, but remember, we might have fired downward, in which case the time to reach the vertex is negative (before we fire) and the vertex of the parabolic path is not on the trajectory at all. Also, we could have gotten both of these in simpler ways than by completing the square. Velocity vectors are always tangent to the path of the particle, which means at the peak, ( ) ˆ0y vertexv at vertex = = −g v .

20

20

20 0

1ˆ210 02

2

y vertex vertex vertex

y yvertex vertex

t t t

v gt y gt

v vt y

g g

= + = −

= − + − = −

= =

g v v g r v g

And, of course, the last equation can be reached even more quickly from the energy equation.

The coordinate parabola:

It’s easy enough to complete the square:

220 0 0 02

2 20 0 02 2 2

y y x y

x x x

v v v vg gy x x xv v g v g

= − + = − −

But, notice: 0 00 0

0

x yx vertex x vertex

x

v v xx v t x v t tg v

= ⇒ = = =

and we can plug these (which we got when developing the coordinate parabola equation) directly into the time parabola equation:

22 22 2 20 0 0 0 0 0 0 0

20 0 02 2 2 2 2 2

y y y x y y x y

x x x

v v v v v v v vg g x gy t xg g g v v g g v g

= − − = − − = − −

77

Well, those are more equations than we typically need, but we may occasionally find the vertex information useful:

( ) ( )

( )

0 100 0 0 0 0 0 0

22 2 20 0 00 20

0

sinˆ , cos , sin

sin 2 sin 1cos2 2 2

yx y vertex

x y yvertex vertex vertex vertex

v vv e v v v v tg g

v v vv vx t v y gtg g g g

θ θθ θ

θ θθ

−= ↔ = = = = −

= = = = = =

iv gi g v

Again, these might be useful, but keep in mind if we’re firing down, or if we hit the target before reaching the vertex of the parabola, then these formulas are not helping (except in our imaginations about what might have happened before the launch or what might have happened if the target were placed elsewhere).

Can we say anything about the final time of flight? Remember the GA Primer gets a formula for the total time of flight in terms of launch speed ( 0v ), firing angle (θ), and target sighting angle (ϕ):

( ) ( )

1 2

0 0

ˆ ˆ ˆ ˆ ˆcos sin

sin2 2 sincos

final

final

R e R R X Y

v Rvt Tg gX

f f f

θ fθ f

f

≡ = = − ≡ +

−≡ = = −

ir R gi gi g e e

Finally, remember that too many equations can have detrimental effects. It can be overwhelming. It’s too much to remember, which one should be used? And, you have to keep asking if each equation applies to your particular problem. If you were going to make a career out of projectile analysis, you might eventually memorize these out of habit. But, in general, it’s best to keep a handle on the fundamentals and to realize all these other equations are consequences of the fundamental (linear) velocity and (quadratic) position expressions.

Activity 1: time of flight using GA & fundamental position equation.

The GA Primer used the hodograph/trajectory parallelogram to find the formula above:

( ) ( )0 0sin2 2 sincos

v RvTg gX

θ fθ f

f−

= = −

If we know the launch velocity and the target range, we can find the time of flight. But, we can get this and other forms directly from the position equation:

78

20

1 ˆˆ sin cos2 2

T T gR gR gRπ f f = + ∧ = − ∧ = ∧ = + =

R v g g R R g g R i i

1. As in the expression on the right, write the following wedge product in terms of i and angle(s):

0∧ =g v ____________________

2. In a similar way, find:

0∧ =R v ____________________

A beautiful use of the wedge product is to eliminate terms from vector equations, because the wedge product of parallel vectors is zero.

3. Perform the following wedge product on both sides and solve for T:

20

12

T T ∧ = + R R v g

T = _______________________

4. Perform the following wedge product on both sides and solve for T 2:

20 0

12

T T ∧ = + v R v g

T 2 = _______________________

79

5. Perform the following wedge product on both sides and solve for T:

20

12

T T ∧ = + g R v g

T = _______________________

6. Multiply the answer in question 3 by the answer in question 5 to verify the result obtained in question 4.

7. Perhaps we don’t know the launch angle or the time of flight, but do know the launch speed and the target range. Set the answer in question 3 equal to the answer in question 5 (eliminating T). Then get an expression for ( )sin 2θ f− , showing how

we can find the launch angle in terms of speed and target range:

( ) ( ) ( ) ( )2sin cos siHint s n: n iα β α β α β= + + −

( )sin 2θ f− = ____________________________________________________

80

Activity 2: a cannon-target problem.

The cannon muzzle speed is fixed, all we can do is change its firing angle. Given a target within range, we might find two workable angles, a line drive or a lob. First, let’s develop an “experimental” tool in Geogebra.

1. Put a point, labeled target, somewhere in the first quadrant, such as (9,4). We can always drag it to any other place.

2. Make a slider, labeled v0, for the initial speed, varying between 5 and 25.

3. Make a slider, labeled θ, for the firing angle, varying between 0 and π/2.

4. Define the horizontal position: X(t) = v0 * cos(θ) * t.

5. Define the vertical position: Y(t) = v0 * sin(θ) * t – 4.9 * t^2.

6. Define: path = Sequence(Point({X(t),Y(t)}), t, 0, 2.5, 0.1). You can play around with these parameters. The time (t) variable goes from 0 to 2 in increments of 0.1.

OK, we now have a tool to test our predictions. With a large enough v0, we can vary the angle to make direct shot or a lob. Now, we can try to solve a cannon problem on paper, then test our numbers. For example:

Given: ( ) ( )0 14 9.8 10 30 , 8.66,5v g X Y≡ = = ∠ ° = =R

7. Define: myV0=14. We don’t want to confuse this constant with our v0 slider value.

8. Use your answer from Activity 1, question 7 to find launch angle θ in radians. Or, better yet, let Geogebra do the formula work. Either way, make a definition myθ = your answer for launch angle. We don’t want to confuse this prediction with the launch angle slider value we created above.

9. Make definitions for your predicted launch:

myX(t) = myV0 * cos(myθ) * t , myY(t) = myV0 * sin(myθ) * t – 4.9 * t^2 .

10. Define: myPath = Sequence(Point({myX(t),myY(t)}), t, 0, 2.5, 0.1).

11. If your prediction worked correctly, was it for the direct shot or the lob? a) Direct. b) Lob.

12. We should be able to get the other solution, and show it with myθ2, myX2(t), myY2(t), and myPath2. Then we’ll see both predictions (direct and lob). Finally, we can vary the sliders, and check that parameters which hit the target match ours. Two solutions are from taking solutions to asin() in both 1st & 2nd quadrants.

81

PHS 542 Name: ________________________________________

Homework: Constant Acceleration Model, projectile motion. _____________________________________________________________________________

We’ve looked before at this block sliding with friction down an incline. We saw there were three forces acting on the block as it slides, ˆ, cos , Km mg θ µ= =g N ai f N i , where the

direction down the incline is designated by a .

We (should have) found an expression for the constant acceleration down the plane:

( )ˆ sin cosKg θ µ θ= −a a

Let’s put some numbers in.

210Kg 30 9.8m/sm gθ≡ ≡ ° =

1. If the coefficient of kinetic friction is too high, this angle won’t be large enough to cause the block to slide. What is the maximum Kµ above which sliding cannot occur?

(max) ____________Kµ =

Now, let’s say 0.45Kµ = . Also, assume the block starts from rest and will slide 0.40 m down the incline before hitting the bottom. Since the block starts from rest, with no initial velocity, the equations of motion are particularly easy, only one direction of interest, down the incline.

21Speed down incline: Distance down incline: 2

v at s at= =

2. What is the acceleration (m/s2) down the plane? a) 0.11 b) 1.93 c) 1.08

3. How long does the trip down take? a) 0.86 s b) 1.93 s c) 2.69 s

82

4. What is the speed (m/s) of the block when it reaches the bottom? a) 0.93 b) 0.30 c) 1.08

According to the work-energy principle applied to this case:

( )

( ) ( ) ( )20

1 0 02

friction normalforce

bott

W W K GPE

mv mgy

+ = ∆ +

∆ + ∆ = + − +

f s N s

5. What is the magnitude of the friction force (N)? a) 15.3 b) 84.8 c) 38.2

6. What is the magnitude of the normal force (N)? a) 19.6 b) 84.8 c) 38.2

7. What is the work (J) done by the friction force during the trip down the incline? a) -15.3 b) -84.8 c) 38.2

8. What is the work (J) done by the normal force during the trip down the incline? a) -15.3 b) 0 c) 15.3

9. What is the change in kinetic energy (J) during the trip down the incline? a) -15.3 b) 19.6 c) 4.32

10. What is the change in potential energy (J) during the trip down the incline? a) -15.3 b) -19.6 c) -4.32

83

The TV show More Than Human (Discovery Channel, 11/14/2003) talked about and showed the video of the accidental "miss" of Ermes Zamperla, third generation "human cannonball" performer, at the Florida State Fair in Tampa, in February 2002. He landed on the ground beyond the cushion. These cannonball performers have always sensibly flipped in the air to hit a cushion or water on their back. Less sensibly, Ermes wore no helmet. He broke both feet, damaged several vertebrae in his back, had fluid on his brain, was unconscious for a week, but fortunately recovered pretty well (except for getting arrested a week later after crawling to the couch to slug his girlfriend during a fight --- bad month for everybody).

The cannon was aimed at 45º for maximum range. Cannonball performers say they take off at speeds more than 40 MPH (their cannon mechanisms are proprietary), stay in the air around 3 seconds, and might reach speeds approaching 60 MPH.

The facts known from observers and the video record in this case are that the horizontal range was 140 feet, the take-off angle was 45º, and the maximum height was 57 feet. We'll use feet instead of meters this time. You might also want to know that 60 MPH = 88 ft/s exactly.

There's not much information there. We don't know the starting height (looks like the end of the cannon is 20 or 30 feet above the ground in the video), so, if we call that take-off height h above the ground, the final vector is ˆ ˆ140 h= +R gi g . We know the cannon angle, but not the initial speed. Still, we have just enough information to find out everything about the flight:

( )

( )

2

00

20

2 20

ˆ ˆ57 45 32.2 / 140

: 140 cos 451 22 1 1: sin 45 140

2 2

vertexy h g ft s h

vHorizontal v T TT T

Vertical h v T gT gT

θ= − = ° = = +

= ° == + ⇒ − = ° − = −

R gi g

R v g

11. Solve the Vertical equation for T2:

T2 = _______________________ s2

12. Solve the Horizontal equation for 20v , then use the last result to eliminate T2 from this

equation, so that now you have an equation for 20v in terms of h:

20v = _______________________ (ft/s)2

84

13. You now have an equation for 20v in terms of h. That should sound familiar. Since, during

flight, there are no forces (according to our ideal model) acting on the cannonball other than gravity, we can write our energy conservation equation as:

2 2

take-off

1 12 2 vertex

mv mgy mv mgy + = +

Use this to get another equation for 20v in terms of h:

20v = _______________________ (ft/s)2

14. Now, set the two expressions for 20v equal to each other, so that we have a quadratic

equation for h. Solve that equation, getting the take-off height above ground:

h = _______________________ ft

15. The rest is easy. Plug this h into our equation for T2 and find the time of flight:

T = _____________________ s

16. Plug this into our original Horizontal equation to find the take-off speed:

0v = _____________________ ft/s

17. Finally, convert this to MPH:

0v = _____________________ MPH

85

PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 3 --- Day 2

86

Math Model 2: Constant Change in Rate (The Momentum Principle)

We’ve had a good look now at our ideal projectile model. Depending on the initial data given, we might have more or less trouble solving for missing data and getting a complete picture of how the projectile moves.

In the cannon-target problem, we knew the initial speed and target position, but we had to

calculate the firing angle. Then we could use 20

12

t t= +r v g to plot the trajectory. We used

Geogebra to do this in one activity, but we could use many other kinds of software, including graphing calculators or GlowScript.

In the human cannonball problem, we were given the initial firing angle (and some other data, including the target position) and had to work hard to find the initial speed.

Now, we’d like to play some more with GlowScript. As a reminder you can go to http://www.glowscript.org/ and login with your gmail account (or make a new one). You might also want to open (in another browser tab) http://www.glowscript.org/#/user/Capaii/folder/Examples/ in order to see some of my GlowScript example files. Then, you can cut & paste from my files to yours as needed. The least painful way to learn a new language is to start with some working program and try modifications to try new ideas.

Activity 1: projectile fired from a mountain top.

1. Look at the GlowScript program ProjectileMotion01.py. You can run it, and click Edit to see the program itself.

2. In your own GlowScript account, select Create New Program. Then use cut & paste to copy the program lines from ProjectileMotion01.py to your own program.

3. We don’t have a target in this program, we are just seeing what the motion looks like when we are given the initial velocity vector. Try using the velocity vectors found in the cannon-target activity and human cannonball homework mentioned above.

87

The Momentum Principle

The co-inventors of GlowScript and authors of Matter and Interactions (I & II), Chabay & Sherwood, introduce their Momentum Principle (MP) algorithm early on to their students. With this algorithm and the quick access to visual 3D programming with GlowScript, they hope to excite students by showing how easy it is to test various physical models.

It comes from Newton’s 2nd Law:

2 2

: :1 /

netd mNewton m Einsteindt v c

= = =

−

p vF p v p .

This is a vector differential equation, where we would need to integrate the net force with respect to time. But, there are many kinds of forces, with different formulas, some of which are too hard to integrate. The MP algorithm avoids the calculus problem by approximating solutions in small steps. The basic algorithm is very simple:

new oldt t∆ ∆ ⇒ = + ∆p F p p F

That tells us if we know a starting momentum and applied net force, then we can get a good approximation for the next momentum. How good? The smaller the time increment, the better, yet then it takes longer to get anywhere because you have to do many more time increments in order to view a decent amount of the trajectory. We have to find a balance.

We’ve been looking at constant forces, but most forces in nature vary with position (Newton’s Universal Gravity law, Hooke’s law for spring motion, electric forces and forces from magnets, etc.) and/or velocity of the particle (magnetism, viscous friction, etc.). That means on each time increment, the force may need to be updated, also. And, finally, remember the main goal is to see the trajectory of the particle, so obviously its position needs to be updated on each time increment.

So, that means the full algorithm might need to update three things on each time increment. The full (Newtonian) MP algorithm looks like this:

{ }

( ) { }max

See discDefine init

ussion beloial values for position, momentum, and force.

Define the size of the time increment, .For 0 t

w.

Define/update neto

force.

The Momentum

,

Pnew new

new old old

tt t

t

∆

=

=

= + ∆

F F r p

p p F { }

{ }

rinciple.

Update position using .

Update particle image to see next step in trajectory.

newnew old tt

m∆ = ∆= + ∆ r vpr r

88

How do we know we’ve picked a good time increment? Pick any amount and see what trajectory takes shape. Then pick a smaller increment, perhaps one tenth as much. If the new trajectory looks different, we’re still making poor approximations. When you get to the point where the previous trajectory looks like the new trajectory, you can be confident you have a good approximation.

Activity 2: momentum principle.

1. Look at the GlowScript program MomentumPrincipleTemplate.py. Test it out. It’s just the constant gravity projectile motion we’ve been studying.

2. In your own GlowScript account, select Create New Program. Then use cut & paste to copy the program lines from MomentumPrincipleTemplate.py to your own program.

3. Try different initial momentum vectors.

4. Try different time increments. Make the increment large enough that the trajectory is obviously incorrect. Make the increment small enough to notice the program dragging due to speed of the microprocessor trying to keep up with calculations.

Activity 3: viscous frictions.

1. Create a new program and copy your last program lines into it.

2. Add another ball, blue, and with postion: pos=r+vec(0,0,4*ball.radius). Make an orbit for this ball, also.

3. For the new ball, copy & replicate an MP algorithm for the viscous force (uv) in the listing. Its force line should look like: F2 = m * g - uv * b2.p / m. This adds an air friction force proportional to speed. The faster the ball goes, the greater the friction.

4. Test it out until any bugs are cleared out. You should see trajectories for the ideal yellow projectile and for the blue projectile responding to air friction.

5. Repeat steps 2-4 for a friction force proportional to speed squared (non-laminar flow): F3 = m * g - uv2 * (mag(b3.p/m)**2) * norm(b3.p). This kind of friction simulates higher speed friction, typically used for calculating skydiver terminal speed.

6. We should see the trajectories of all three projectiles.

89

Of course, aerodynamics is a very complex subject. For example, wind resistance varies with shape of the object, smoothness of the skin, mass density of the fluid (air), and quite drastically with the speed of the object, amongst other things. There are so many theoretical regimes, that about the best we can do here is to summarize that at “low” speeds, air friction is proportional to speed, while at “high” speeds, air friction is proportional to speed squared. Here, the uv parameter in the GlowScript program will be replaced by α, and the uv2 parameter by β, for simplicity of notation.

“Low” speed (free fall, start from rest, s = distance fallen):

` ( )

0 0

/ /1 1

Terminal speed

v t

net

t m t m

dv dvF ma m mg v dtmgdt mv

mg mg mv e s t e

mg

α α

αα

α

α α α

α

− −

= = = − ⇒ = −−

= − ⇒ = + −

=

∫ ∫

We can perform or look up the integrals to derive the speed and distance formulas shown. But, to get the terminal speed, we could just look at the original boxed equation of motion, and ask when the rate of change of speed is zero, what is the now constant, and therefore terminal, speed?

A 200 lb skydiver has a terminal speed of 210 MPH. Assuming this formula holds for skydivers:

200 lb = ________ N 210 MPH = ________ m/s α = ________ Kg/s

“High” speed (free fall, start from rest, s = distance fallen):

`

2

0 02

tanh ln cosh

Terminal speed

v t

netdv dv gF ma m mg v dtmgdt mgv

mg mv gt s gtmg mg

mg

ββ

β

β ββ β

β

= = = − ⇒ = −−

= ⇒ =

=

∫ ∫

Again, just look at the original boxed equation of motion, and read off the terminal speed.

A 200 lb skydiver has a terminal speed of 210 MPH. Assuming this formula holds for skydivers:

200 lb = ________ N 210 MPH = ________ m/s β = ________ Kg/s

90

Notes for Large Angle Pendulum using Momentum Principle

Newton’s Law: ˆnetd dm m m m Tdt dt

= = = = + = −v pa F g T g r

Circular motion (but speed varies). Call length of cord = L. Use the pivot point as the origin. Let i represent the CCW bivector of the plane of motion. Let θ = 0 be the downward direction.

2

ˆ

ctr tang

L e θ θ ω

ω ω ω ω α ω α

= ⇒ = = ≡

⇒ = = = + ≡ + = − + = +

ir g v r ri ria v r ri ri ri i ri r ri a a

Since T is parallel to r we can separate, if desired, the centripetal and tangential parts:

( ) { }

2

2

22

ˆˆ ˆ ˆ ˆ ˆ: cos

co

as J.J.P. tho

s cos

ˆ ˆ ˆ ˆ ˆ: sin sin

si

ught

n

ctr

tang

small

m m m m T mCentripetal m L m T T mg

vT m L mg m mgL

Tangential m L m

MM

mg mgg gL L

θ

ω α

ω θ

ω θ θ

α θ θ

α ω θ θ θ

= − + = − = +

= − = − = − −

= + = +

= = ∧ = = −

= = = − →≈ −

a r ri g r g TF r g rr r r

F ri g rr ir ri

{ }4

We still have a conserved energy equation because T is perpendicular to ∆r , and so can do no work, so we just have kinetic energy (KE) and gravitational potential energy (GPE). Here, we’ll define zero potential energy to be when θ = 90°:

( ) ( )

2max

2

max max max

1 cos 0 cos2

2 cos cos 2 1 cos

E KE GPE constant mv mgL mgL

vm mg v gLL

θ θ

θ θ θ

= + = = − = −

= − = −

We can use this to eliminate the 2v term in our Tension formula in the Centripetal section above, to write T in terms of just the angle.

( )

( ) ( ) ( )

2

max

min max max max max min

cos 3cos 2cos

cos 0 3 2cos 3 2

vT m mg mgL

T mg T mg mg T

θ θ θ

θ θ θ θ θ

= + = −

= = = = − = −

91

Well, the goal was just to see what to use for net m= +F g T in our Momentum Principle in a GlowScript program. We just want an expression for this in terms of either r or θ.

( )

( ) { }

( )

max

max

2 2 2 2

ˆassuming corre

ˆ ˆ ˆ3cos 2cos cosˆ ˆ ˆ3 2

2 3 2 3 * ,

sponds to

2

3

init

init init init

init

T mg and remember g

m m

m m m mdotL L L L

θ θ θ

θ

= − = − − =

= − +

= − = − = −

T r r g r

T g r g r r

gg r g r r r r r r g r

r

r

The last form is the way we might program that formula into GlowScript.

2

** /

* ,2 3

new old old

new old new

new new init new

dtdt m

mm dotL

= += +

= + −

p p Fr r p

F g r g r r

92

PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 3 --- Day 3

Show 1st half of Al Bartlett talk.

93

Math Model 3: Rate Proportional to Amount

MM1 was about constant rates of change, MM2 was about rates of change linear in the independent variable (typically x or t in our examples), and now MM3 is the model in which the rate of change is linear in the dependent variable:

( )ax xa

dy y y y edx

βα αα ββ β

− = + ⇔ = + −

There are many examples of this functional behavior, and we’ve just seen the first one listed here.

• Low speed, free fall from rest ( 0 0v = ) through a liquid (viscous friction):

( )/ /1t m t mdv mg mg mgg v v e edt m

α ααα α α

− − = − ⇔ = − + = −

• Simple model of exponential (as opposed to logistic) population growth:

0rtdP rP P P e

dt= ⇔ =

• This last model also fits a good approximation to the compound bank interest

formula, where t is the number of years of compounding, and n is the number of compounding events per year:

( )1 lim 1tnt n

tn r rt

n

r rA P A P P e Pen n

→∞

→∞

= + → = + = =

• N = number of remaining radioactive particles when random particles emit:

0tdN N N N e

dtλλ −= − ⇔ =

• Voltage across a discharged capacitor after a switch is closed in an RC circuit:

( ) ( ) ( )/ /1 1t RC t RCbattbatt batt batt

Vdv v v V e V V edt RC RC

− − = − ⇔ = − + = −

• Current through an unenergized inductor after a switch is closed in an RL circuit:

/ /1Rt L Rt Lbatt batt batt battV V V Vdi R i i e edt L L R R R

− − = − ⇔ = − + = −

94

• The rocket equation. The rocket works by expelling propellant backwards. We’ll define the rocket heading as the positive direction, and we measure velocities with respect to the frame in which the rocket started from rest (that is, for example, the earth frame). Then, if v is the rocket speed, the speed of propellant exhaust with respect to the rocket is usually called Ev , but we need to know its

velocity in our earth frame, so that is Ev v− . The total mass of the rocket (frame + fuel) is called m . And dm is the change in mass of the rocket. (Some call this the bit of propellant exhausted, but here it refers to the change in rocket mass, so we expect it to come out negative, but we’ll let the equations take care of that.) If we consider our rocket plus exhausted fuel as our system, and if there are no other forces acting on that system, then the momentum of our system should be the same before dm is exhausted as after:

( )( ) ( )( )

0

0

E

E

E

mv m dm v dv dm v vm dv v dm dm dv v dm v dm

dm dv v dm m dv

= + + + − −

= + + − +

≈ ⇒ = −

Thus, our rocket equation fits the MM3 model:

( )/1 a Ev v va

E

dm m m m edv v

− −= − ⇔ =

That tells us how the mass of the rocket varies with rocket speed (of course, it decreases with increasing speed). If we prefer to see how speed varies with remaining total mass, just take the logarithm of both sides and rearrange:

{ }l . .n aa E Non relativistic v c speed of lighmv v

mtv − = = +

Finally, we might be more interested in how the speed of the rocket varies with time. If we let µ represent the constant rate of propellant ejection, then the total rocket mass can be written as 0 /m m t dm dtµ µ= − ⇔ = − .

E E EE

dv dm dt mm dv v dm m v vdt dt dv v

µµ

= − ⇒ = − = ⇒ =

Now it’s once again in the form of the MM3 model:

( ) ( )0 0/ /0 0 0 0

00

1 1

ln 1

E Ev v v v v v

E E

E

m m m mdt t t e edv v v

v v v tm

µ µ µ µ

µ

− − − − = − ⇔ = − + = −

⇒ = − −

95

It should be clear from all these examples that when the rate of change of a variable is a linear function of that same variable (or, rate of change is proportional to amount), then the variable will be a function of the exponential function. So, let’s look at exponential functions.

We’ve played with Geogebra enough to know how to plot functions.

In the left diagram, we’ve plotted ( )& lnxe x . We can see graphically they are inverses of each

other (in fact, the natural logarithm is defined to be the inverse of the exponential). In the right diagram, we’ve plotted ( )22 & logx x , which are also inverses of each other. We’ve also had

Geogebra calculate the slopes and the areas under the curves from -∞ to the x-value of our moveable point on each graph.

We can see here the beauty of the xe function. The slope is its derivative, which is always equal to the value of the function. The green area is its integral, which is also equal to the value of the function (not so surprising after the previous statement). In symbols:

xx x x x x xd e e e dx e e dx e C

dx −∞= = ⇒ = +∫ ∫

And, as mutual inverses: ( ) ( )ln lnx xe x e= = .

96

Perhaps I should have learned better ways, but to this day, when I see ( )ln x , I think “What

power do I raise e to in order to get x ?” But, between using graphs for intuition and calculators for manual labor, we can understand this function pretty well. Here, as a reminder, are about all the things we need to know about this function:

( )

( )

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

1

11ln 1ln 1

lnln ln ln ln ln ln log

ln

n

n

bb

x C for nd x x dx ndx x x C for n

aab a b c a b a ac b

++ ≠ −= = +

+ =

= + − = =

∫

We can use these properties to find derivatives and integrals for exponentials and logarithms to any base:

( ) ( ) ( )ln ln

ln

xx ax x xd d aa e a a a dx C

dx dx a= = = +∫

In particular, we can now understand why the right-hand and left-hand diagrams are so similar, and where the numbers come from in the right-hand diagram.

( ) ( ) ( )( )

( )

1.7 1.71.7

2 2 ln 2 2 ln 2 0.693 3.25 2.25

2 22 0 4.69ln 2 0.693

x x x

xx

dAmount y m ydx

Green area dx−∞

−∞

= ≡ = = = = =

= = = − =

∫

Activity: examining some properties of exponentials and logarithms.

1. On the same graph, plot 22y x− = .

2. On the same graph, plot ( )23y x= − .

3. On the same graph,

plot ( )2/ 2y x=

97

4. We never quite answered why the functions 2 &x xe are so similar. Notice:

( )( ) ( )ln 2 0.6932 exp ln 2 xx x xe e= = = . Comparing to step 3 above, we are replacing

x by 0.693x in the exponential function. No wonder they are similar! They are the same basic graph, except that one of them is squeezed skinnier in the horizontal dimension with respect to the other.

Which one is skinnier, horizontally? a) 2x b) xe

We started with ( )ax xa

dy y y y edx

βα αα ββ β

− = + ⇔ = + −

. For the record, we’re

solving a first order, linear differential equation (sounds scary!). But we can also use the method of separation of variables, which means get all the x-variables on one side, all the y-variables on the other, and then integrate (still scary?). But when you integrate, there’s an arbitrary integration constant, unless you are also given a little more information, such as the value of our function at a particular point. That kind of information is usually called an initial condition. In the general solution above for that kind of equation, I’ve assumed we were given an initial condition, ( )a ay x y= , or, you can think of it as being given a point ( ),a ax y .

5. Let’s check. Given: ( )ax xay y eβα α

β β−

= + −

dydx

=

From the 1st equation, rewrite

this expression in terms of y: ( )ax xay eβα

β−

+ =

Now plug that into the

derivative expression: dydx

=

You should be back to the original differential equation.

From our solution, find: ( )ay x =

98

6. Given our solution for the voltage across a charging capacitor, ( ) ( )/1 t RC

battv t V e− = − ,

what is the initial voltage? ( )0v = _______________

7. For the charging capacitor, find: ( )v ∞ = _______________

8. What do we get at t RC= ? ( )v RC = _______________

9. Write the capacitor solution again, but now assuming the capacitor has a voltage initv already stored on it when the switch closes at t=0. Use our solution template to find the new solution:

( )0 initv v≡

( )1battVdv v v tdt RC RC

= − ⇔ =

99

PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 3 --- Day 4

Show 2nd half of Al Bartlett talk.

100

Math Model 3: More on Exponentials

We’ll summarize some ideas from Professor Al Bartlett’s talk “Arithmetic, Population and Energy: Sustainability 101” (http://www.albartlett.org/presentations/bartlett_presentations.html).

We’ve mentioned the compound interest formula: 01 1nt ntr rA P A

n n = + = +

. Banks play an

advertising game: they’ll offer you a 5.0% interest (in the good old days) rate AND (the gimmick) compound it daily for you, so that your annual yield (APR or APY) is really 5.13%.

( )( )365 10.051 1.05127 5.13%

365 + = →

And, we’ve noted: lim 1n

r

n

r en→∞

+ =

. For example, 0.05 1.05127e = , which shows with these

formulas that 365 is a pretty good approximation for infinity!

Professor Bartlett uses the symbol k instead of the bank notation r:

{ }0 0

growth rate {such as, 0.05} 100 % growth rate {5%}

Continuous growth rate model: :kt t

k R k

N N e Radioactivity N N e λ−

≡ ≡ =

= =

The growth rate can be positive, zero, or negative. In the case of radioactivity, it is customary to make the decay clear with a negative sign in the exponent (so that λ is positive). We might use P for Population, or N for Number of items, or A for Amount. We can use the formula to represent any kind of growth (population, inflation, amount of oil used annually, national food consumption, etc.).

What does the growth rate factor do to the natural logarithm function? See the diagram.

101

In real life, growth rates can change over time. If the growth rate is constant, we call that steady growth. If 0 constantk< = , the amount still grows exponentially. Many people hear the word steady and think staying the same, but staying the same would be zero growth.

Professor Bartlett describes how we can do a lot with these growth exponentials using simple arithmetic by learning just a few tricks and useful approximations:

( ) ( )

( )

( ) ( ) ( )

0 00

2

1/2

ln / ln /

ln 2 0.693 69.3 70: doubling time100

ln 1/ 2 ln 2 ln 2 0.693. : half life

kt N N N NN N e t or k

k t

Growth Tk k k R

Rad decay Tλ λ λ λ

= ⇔ = =

≡ = = ≈

−≡ = = =

− −

The middle term shows his rule of 70, and as special cases, the investors’ 7-10 rule:

7% interest doubles your investment in 10 years, while 10% interest doubles your investment in only 7 years.

The rule of 70 let’s that work for any other growth rate:

A 2% growth rate doubles the amount in 35 years, a 5% growth rate doubles the amount in 14 years, etc.

One of the most important (and fascinating) things about MM3:

We say rate is proportional to amount, and now we know amount grows (or decreases) exponentially, so that means rate grows (or decreases) exponentially, too! The simplest way to really get a feel for this is probably with radioactivity.

1/2/

0 012

t TtdN N N N e N

dtλλ − = − ⇔ = =

This represents the Number of undecayed particles (amount) as a function of time, which is decreasing with time as particles randomly decay. But, in practice this is hard to use, because who knows the number of particles at any time?

We don’t count (or weigh & calculate) the number of particles. What we do is hold up a Geiger counter and see how many clicks we get per second (and machines can detect and display that rate, called Activity). That means we’re keeping track of rate, not amount.

102

And yet, because MM3 is the case where rate is proportional to amount, we get the same formula for rate as we did for amount! That’s a beautiful and useful consequence.

( )

( ) { }0

0 0 0 0

Geiger counter ctivity (decays/sec)

0

t

t t

dNA A t A N N edt

A A N A A e just as N N e

λ

λ λ

λ λ

λ

−

− −

= ≡ = − = =

= = ⇒ = =

N is the amount left undecayed (in particles), while A is the Geiger counter Activity (rate in decays/s), and both of them are decreasing with the same decay rate along an exponential curve.

Professor Bartlett uses the same argument to find the time left (time of expiration, TE) until we run out of a non-renewable resource, given the current consumption rate (r0).

( ) ( )

( ) { }0

0 0 0 0

slope of the exponential curve

0

kt

kt kt

dNr t N t kN kN edt

r r kN r r e just as N N e

= = = =

= = ⇒ = =

This can be a little more confusing than the Geiger counter example, because we’re using the word “rate” in different ways. Maybe we should use a different symbol for consumption rate. The symbol k represents the constant growth rate, describing the shape of the amount (of population or average income or tons of coal dug up) function, which is a rising exponential curve. But the slope of that exponential curve keeps increasing, also, and ( )r t tells us that the

slope also increases like an exponential (remember, that’s the magical property of the exponential). Now, just as for any rate function, if we find the area under the rate curve between two points, we’ll have the total amount between those two points. That is, we’ll have the amount consumed. So, his derivation (in an Appendix) is:

( ) ( )000 0

1t t kt ktrAmount consumed r t dt r e dt e

k= = = −∫ ∫

Professor Bartlett uses R for the amount of available Resources left, and TE for the time until all those resources are used up (Expired):

( )0

0

11 ln 1EkTE

r kRR e Tk k r

= − ⇒ = +

Why all that trouble? Why not use: 0 00

1 ln 1EkTE

RR N N e Tk N

+ = ⇒ = +

.

103

Well, good question. The integration isn’t necessary at all. At first, we might think it’s because we don’t know 0N , while we do know (or can make a good guess in the case of oil

reserves) how much is left, R, and our current rate of consumption, 0r . True, 0N (how much we’ve already consumed) isn’t relevant at all to the question of how long will it take to consume what’s left. Yet, we did find a relation, 0 0r kN= , and when we use this to eliminate 0N from our equation, we get Professor Bartlett’s equation.

From the NY Times in December, 2016:

“The United States population grew by 0.7 percent in the last year, its smallest annual expansion in 80 years, the Census Bureau said this week. The nation added about 2.2 million people from July 2015 to July 2016, bringing the total population to just over 323 million.”

0 323 __________ __________N million k R= = =

At this rate of growth, how long will it take to double our population to 646 million? T2 = _______________ yrs

At this rate of growth, what population can you predict for July of 2026?

( )10P = ______________ million

104

PHS 542 Name: ________________________________________

Homework: Exponential functions. _____________________________________________________________________________

1. Use a calculator to get these answers to 5 significant digits:

100 1000

100001

1 11 ____________ 1 ____________100 1000

11 ____________ ____________10000

e e

+ = + =

+ = = =

2. Remember, we found ( )ln 2 0.6932 xx xe e= = . Then, we should be able to evaluate these without a calculator:

( )____2.079 0.693 6.93__________ __________e e e= = =

3. Use ( )ax xa

dy y y y edx

βα αα ββ β

− = + ⇔ = + −

to write the solution to this

rate equation: 12 3dy ydx

= − , with the requirement that the point (x,y)=(0,5) satisfies our

solution (that is, lies on the solution curve).

y = _____________________________

4. If the annual bank interest of 1.4 % is compounded daily, how many years will it take to double your money?

T2 = ______________ yrs

105

5. While measuring a certain radioactive source, a Geiger counter’s Activity drops to 90.3 % of its starting value after 3.00 minutes. What is the source’s half-life?

1/2T = ______________

6. The current world population is about 7.6 billion, with a growth rate of about 1.09 % / yr. At that rate, what will it be in 10 years?

P(10) = _________ billion

7. (Modified from Al Bartlett): with the data given in question 6, let’s assume that population growth rate held for all time in the past. If that population grew from just 2 people, Adam Eve, how long ago did they start things off?

ΔT = _____________ yrs

106

PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 4 --- Day 1

107

Math Model 4: Change in Rate Proportional to Amount (Trig Functions)

More thoughts on the four math models: why these four? Wouldn’t the following six be the logical group of the simplest rates and changes in rate?

0

222 2

22 2

1 2 3

64 5

dy dy dyk k kx ydx dx dx

d y yd y d yk k Cx dxdx dx

α β

αβ β

= = + = +

= ±= = +

← +← −

MM1 MM2 MM3

3MMMM24

The first three equations are equivalent to the first three Math Models (that is, the rates are either constant, proportional to the independent variable, or proportional to the dependent variable). The next three equations show how the change in rate (2nd derivative) might also vary in the same ways (be constant, proportional to the independent variable, or proportional to the dependent variable).

Of course there are all kinds of other possible differential equations, so why focus on only these (which, by the way are all called linear differential equations with constant coefficients, which have well known standard methods of solution)? Well, we wanted to think about math used in physics in high school (and beyond) and to convince ourselves (and students) that most problems, even seemingly different problems, can be handled with just a few basic ideas:

• Our math models. We don’t expect to teach differential equations in HS, but we can seek solid understanding of rates (slopes), and see how the rate behaviors listed above can produce solutions with just the use of algebra and trig.

• Geometric Algebra. It unifies vector operations, rotations, allows division in the algebra, adds planes & volumes to the algebra, and can be introduced to HS students.

So, how did we get from the six kinds of equation above to four math models?

Equations 2 and 4 are easy: they’re the same equation! The derivative of equation 2 is equation 4. We can phrase MM2 as rate changing linearly, or as change in rate being constant.

Why not make a math model for equation 5? For one thing, it’s a very easy equation to solve. For another, its solution has nothing new to offer that hasn’t already been studied in MM1 & MM2, its simpler relatives. Let’s remind ourselves how this set of equations can be solved with just the power rule in calculus:

108

2

2 2 3

:1:2

1 1 1:2 2 6

y k y A kx

y k y B kx y A Bx kx

y k Cx y B kx Cx y A Bx kx Cx

′ = ⇒ = +

′′ ′= ⇒ = + ⇒ = + +

′′ ′= + ⇒ = + + ⇒ = + + +

MM1

MM2

Eq. 5

Let’s also remind ourselves why these kinds of equations keep showing up in physics studies. It’s because Newton’s 2nd Law has been such a successful presenter of cause & effect.

2

2Newtonian net

net Mechanics

d d ddt m dt dt

= → = = =Fp v rF a

Newton’s law tells us that a cause ( )netF produces an effect ( )a . But it doesn’t tell us what that

cause looks like or where it comes from. For that, we need the heavy lifting of physics experiments, models, and theories. For example, we need to figure out that friction is propor-tional to normal force, or that gravity near Earth’s surface is a constant vector, or that gravity from the sun weakens as the inverse square of sun-Earth distance, or that the force exerted by a spring is proportional to how much we stretch it. Once we figure out a formula for the net force in our particular problem, then we need to solve for our usual desired solution (how position varies with time). And our solving method usually fits into one of the four math models.

2

02

22 2

2

0 0net

net

net

net

d ddt dt

d d dm tdt dt dt

dmdt

d dmdt dt

β β

β β

= ⇒ = = ⇒ = = ←

= ⇒ = = = ⇒ = + = ←

=− ⇒ = = − ←

=− ⇒ = = = − ←

v rF a v k MM1

v r rF k a k v v k MM2

vF v a v MM3

v rF r a r MM4

One more reason to ignore equation 5 is that in terms of Newton’s law, it would be ( ) ˆnet k C t= +F n , which describes a force increasing linearly in time in a certain direction. Such

a force doesn’t seem to exist in nature (can you think of a case?). We could maybe intentionally create such a force (a rocket sled with a computer controlled thrust?), but then Newton’s equation of motion would be easy to solve, anyway, so it isn’t worth extra time spent.

Finally, notice that equation 6 is broken into two possibilities. However, the plus sign corresponds to MM3, because:

( ) 2: 0y y y y y yα β β β α β αβ β′ ′′ ′= + ⇒ = + = + = +MM3

109

And, notice that it wouldn’t matter what the sign of β itself was, our last term would still be 2 yβ+ . Remember, for MM3, our solutions were exponential functions, rising or falling

depending on the sign of β. For the plus sign in equation 6, because it’s a second order differential equation, our general solution could have both rising and falling exponentials (with the same magnitude of growth rate):

{ }

( )

( )

21 2

1 2

2 2 21 2

:

x x

x x x x

x x

x x

y y y C e C e

d dCheck e e e e the chain ruledx dx

y C e C e

y C e C e y y

β β

β β

β β

β β

ααβ ββ

β

β

αβ β αβ ββ

−

−

−

′′ = + ⇒ = + −

≡ ⇒ =

′⇒ = −

′′⇒ = + = + = +

Now, we might find situations where this kind of behavior happens (the amount exponentially decays for awhile, then exponentially grows with the same magnitude of growth rate), but, as we see, it’s not really a distinct model.

After all those preliminary thoughts, let’s finally move on to MM4:

( ) ( )

( ) ( )

( ) ( )

22

1 22

1 2

2 2 21 2

cos sin

: sin cos 0

cos sin

d y y y C x C xdx

Check y C x C x

y C x C x y y

ααβ β β ββ

β β β

αβ β β β αβ ββ

= − ⇔ = + +

′ = − + − ′′ = − + = − − = −

Now, we can see the difference in the plus or minus signs in equation 6. Taking two derivatives of an exponential gives a plus exponential again, but taking two derivatives of a sine (or cosine) function gives a minus sine (or cosine).

From the point of view of Newton’s 2nd Law, MM4 is the model where our applied net force is a negative force proportional to the amount of position displacement; that is, a restoring force. Such forces are EXTREMELY common in nature. Everything vibrates! We see the vibration in the sine & cosine terms, which are oscillatory functions. Every time you push or pull on some solid object enough to distort it a little, then let go, the natural restoring force in the solid tries to undo the distortion, overshoots, gets distorted the other way, resulting in the natural vibration. All we need is a restoring force linearly proportional to displacement. And, any function of position can be written as a Taylor series:

( ) 20 1 2 0 1... when x

restoringis smallf x a a x a x a a x F kx= + + + →≈ + → ≈ −

110

Some typical examples of MM4 in physics:

• Hooke’s Law for springs: 2

spring 2

d x kF kx ma xdt m

= − = → = − .

• Small angle pendulum: 2

2sin sind g gF mg ma mLdt L Lθθ θ θ θ≡ − = = ⇒ = − −

• Constant inward radial force, such as from tension in the string attached to an object on a frictionless table and given an initial speed

perpendicular to the string (circular motion): 2

2ˆnet

d TT m mdt mR

= − = = ⇒ = −rF r a r r

As a result of MM4 solutions involving sines & cosines, it’s time to review trig functions.

Activity 1: trig functions in terms of sides of a right triangle.

Given the right triangle and labels shown here, fill in the definitions of the six basic trig functions (sinθ, cosθ, …):

opphyp

= adjhyp

=

oppadj

= hypopp

= hypadj

= adjopp

=

Pythagorean theorem: 2 2 2

2 __________________________opp adj hyphyp

+ =⇒

Activity 2: with Geogebra, check MM4 ( )2y yβ′′ = − for the function ( )sin 2y x= .

1. Plot ( ) ( )sin 2f x x= and put a point A on the curve.

2. Draw a tangent line on the curve through A. It should be named g (or rename it).

3. Make a point: B=Point({x(A),Slope(g)}). Turn on its trace. This shows the derivative (that is, plots the slopes) of our original sine curve.

4. Let Geogebra do the calculus. Enter the command: Derivative(f). Also plot the second derivative using Derivative(f,2).

hyp opp

θ adj

111

5. Drag point A around. Make sure point B (our hand-made derivative) follows the Geogebra created derivative curve.

6. Hide other objects as desired, but focus on the original sine curve and its first

derivative functions. Are you convinced that ( ) ( )sin 2 2cos 2d x xdx

= ?

7. Now focus on the original sine curve and its second derivative functions.

Does the sine function satisfy MM4 ( ) ( )24sin 2 siny x xβ β′′ = − = − ?

8. Change the original sine function to ( ) ( )cos 2f x x= . You might have to retype it,

but all the other statements should still work as they did. Use the graphs to convince yourself that:

( ) ( ) ( ) ( )2cos 2 2sin 2 4cos 2 cosy x y x y x xβ β′ ′′= ⇒ = − ⇒ = − = −

Activity 3: finding the inverse function for cos(x).

1. Plot ( ) ( )cosf x x= . Now plot its inverse function, ( ) ( )acosg x x= . It only shows

the curve corresponding to the first quadrant (0 < θ < 90°), so that the inverse function is never allowed to have an infinite slope.

2. For (x,y) graphs, the inverse function can also be drawn by reflecting the original function about the line y = x. That’s because:

( ) ( ) ( )( ) ( )( )1 1 1... 1 ...y f x f y x f f f f− − −= ⇔ = ⇒ = =

But, then, in order to plot the inverse function, ( )1f − , we trade the coordinates,

so that we can plot it like any other function, ( )1f x− :

( )

( ) ( )

1

cos 2.0 0.42 acos 0.42 2.0

y x Plot y f x−⇒ =

= − ⇔ − =

o

Check out the last line on your Geogebra graph. Plot the line y = x. Now, from the menu options, choose Reflect about Line, then select the original cosine function and the line y = x. You’ll get the extended acos() function.

3. In order to clearly see the various curves, make the original cosine function red, make the reflected cosine green, and make the acos() function blue, with thick width.

112

Activity 4: all six trig functions on the unit circle.

You’re probably familiar with showing the sine & cosine functions as coordinates of a point on the unit circle, ( ) ( ), cos , sinx y θ θ≡ . Some treatments take that as the official

definition of those two trig functions. Do you also know that all six trig functions can be shown as the value of line segments in the diagram below?

1. Finish labeling all the heavy segments above (red, green, 2 more blacks, and 2 more purples).

2. There are many right triangles and similar triangles in the diagram. See if you can spot the three pythagorean identities:

2 2 2 2 2 21 cos sin sec tan csc cotθ θ θ θ θ θ= + = − = −

113

PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 4 --- Day 2

114

Math Model 4: Some Physics Applications of ( )2y yβ′′ = −

We’ve reviewed sines, cosines, exponentials, and mentioned Taylor series. To actually work out a Taylor series is a calculus problem. But for pre-calculus high school students, it is reasonable to show them one and let Geogebra graphs (first term, then the first two, then the first three, etc.) convince them how we can approximate weird functions as sums of powers of x (straight lines, parabolas, etc.). We showed how to do this in some week 2 activities. We could memorize the big three (but in this era, we can just search on a smart phone to find them):

{ }

( ){ }

( ){ }

2 3 4 5 6 7

2 4 6

3 5 7

1 ...2! 3! 4! 5! 6! 7!

cos 1 ... : cos cos2! 4! 6!

sin ... : sin sin3! 5! 7!

x x x x x x xe x Neither even nor odd

x x xx Even function x x

x x xx x Odd function x x

= + + + + + + + +

= − + − + − =

= − + − + − = −

The patterns are beautifully easy to remember. If we can think of any kind of object which squares to minus one (originally people thought of the unit imaginary, but now we have a unit bivector i and the 3D psuedoscalar i, which are real objects), then these beautiful patterns can be related to each other, as Leonhard Euler did 300 years ago:

{ }{ }

2 3 4

2 3 4

cos sin : 1, , 1, ...

cos sin : 1, , 1, ...

x

ix

e x x because

e x i x because i i i i

= + = − = − =

= + = − = − =

i i i i i i

Because of these useful notations, we can write the geometric product of two unit vectors in a number of ways, and choose which way might work better in different situations:

ˆ ˆ ˆˆ ˆ ˆcos sine θ

θ θ= + ∧= +

= i

ab a b a bi

ˆab is the rotor which, when multiplied on the right, rotates any vector in the i plane by angle θ. And, we can once more see the Pythagorean identity:

( )( ) 2 2ˆ ˆˆ ˆ1 cos sin cos sin cos sine eθ θ θ θ θ θ θ θ−= = = + − = +i iabba i i

If we haven’t yet seen how to easily get the law of cosines and law of sines, see the GA Primer.

115

Activity 1: Hooke’s Law, 1D spring motion on a horizontal frictionless table.

1. The spring constant, k (N/m), can be determined by hanging the spring vertically, hanging a mass from it, and measuring the stretch. Let’s say we find the spring stretches 5 cm when a 20 gm mass is hung from it. What is the spring constant?

k = _________ N/m

We want to look at a block (mass m) attached to a spring on a frictionless table. The position where the block is comfortably at rest is called x = 0. We’ll pull it out (0 < x) a certain distance (x = x0) and let go, watching the oscillation.

2. If we take our mass as our system, being acted on by external forces of Earth, table, and spring, draw the System Schema and Interaction Map (Free Body Diagram):

System Schema FBD

3. Since there’s no friction, the normal force doesn’t tell us much, so we can concentrate on the 1D motion:

( ) ( )

2 2' ' 2

2 2 2

1 2

:

: cos sin

Hooke s Newton snet Spring Law nd Law

dx dx kF kx m As x xdt dt m

solution x C t C t

ω

ω ω

→− → ⇒ = − ≡ −

= +

MM4

MM4

Use this to find the formula for velocity as a function of time:

v = ___________________________ m/s

116

4. Using the standard notation for initial values: ( ) ( )0 00 0x t x and v t v= ≡ = ≡ ,

use our last two equations (for position and velocity) to find 1 2&C C in terms of

0 0, , &x vω , and then write the two equations in these terms:

( ) ( ) ( ) ( )______ cos ______ sin , ______ cos ______ sinx t t v t tω ω ω ω= + = −

5. For the k found in question 1, and a 20 gm mass, work out the following quantities:

1________ / ________ ________2

rad s f Hz T sf

ωωπ

= = = = =

6. Now, we’ll pull the mass 4 cm from equilibrium and let it go from rest. Find:

0 ________x m=

0 ________ /v m s=

( ) ( )______ cos ______ sinx t tω ω= +

( ) ( )______ cos ______ sinv t tω ω= −

7. What are the position and velocity at t = 2 s?

x(2) = ________ m

v(2) = ________ m/s

117

Remember the work-energy principle. It turns out springs obeying Hooke’s Law are also conservative forces, and we can define a potential energy for springs (SPE) the same way we defined potential energy for gravity (GPE) near Earth:

( )2 2

1 1

2 2 22 1

1 12 2by spring springW d k d k x x kx = = − = − − = −∆

∫ ∫x x

x xF x x x

( )

( ) ( )

2

2

102

102

all forces other grav spting

other

W W W W mgy kx K

W K mgy kx K GPE SPE

∆ = ∆ + ∆ + ∆ = −∆ −∆ = ∆

∆ = = ∆ + ∆ + ∆ = ∆ + +

So, for our frictionless spring, where there are no other net forces acting on the mass than the spring force, we should have:

2 21 1constant2 2

E K SPE mv kx= = + = +

The energy of this system is constant, as it converts kinetic to spring potential energy, then back, and repeats.

8. When x = 0.04 m, find:

K = __________ J

SPE = __________ J

E = __________ J

9. When t = 2 s, find:

K = __________ J

SPE = __________ J

E = __________ J

118

Activity 2: circular motion with a constant force (string Tension) towards the center, ctrT .

2

2ˆ ctr

net ctrTdT m m

dt mR= − = = ⇒ = −

rF r a r r {a vector form of MM4}

Uniform (constant speed) circular (constant radius) motion is a major topic in physics, with some interesting features. With constant speed, the kinetic energy is constant and the only force ( ctrT ) is perpendicular to the motion. That in itself is an interesting feature. For example, when during projectile (parabolic) motion might this be true?

Uniform circular motion first of all, of course, is circular motion, and we know how to rotate a vector in a plane, so all we have to do is rotate it at a constant rate:

( ) ( )

0 0

22 2

0 2

& uniform constant t

t

de edt

d d dedt dt dt

θ ω

ω

θ ω

ω ω ω ω

= = = ≡ ⇒ =

= ≡ = = = = = = −

i i

i

r r r r

r v rv r r i r i a r i r

So, we have an easy formula for the motion itself (in the red box), and we can use these formulas and compare with the force equation above to get:

( )

222 2 2 2 2

2 22 2

ctrctr

T mvv R T m R mamR R

v mvv R a R Centripetal force m RR R

ω ω ω ω

ω ω ω

= = = ⇒ = = =

= = = = =

r i

Of course, the tension is called centripetal force because it pulls towards the center. We’ve been trying hard not to confuse tension ctrT with period of revolution 1/T f= .

1. Let’s see if we can calculate the centripetal force of the sun pulling Earth into (almost) uniform circular motion. We know the period, 365.25T days= , so we’ll just need to convert into seconds, and then find the angular velocity.

T = ____________________ s

2 ________________ /rad sTπω = =

119

2. Now we can look up the mass of Earth (in Kg) and the distance of the sun from Earth (in m) to find the centripetal force. Let’s also find our rotational speed:

____________________Em m Kg= =

____________________SER R m= =

2 ____________________ctrT m R Nω= =

_______________ / _______________v R m s MPHω= = =

3. Well, now we supposedly know the force of attraction between Earth and the sun, so we could check Newton’s universal gravity law:

( )( )11 30

2 2

6.674 10 1.989 10

____________________

ES ESE

mGM mFR R

N

−× ×= =

=

The homework assignment (on another handout) involves a combination of Activities 1 & 2. We have a frictionless table, and roughly circular motion, but the string is replaced by a stretched spring, which really messes up the circular motion (the forces aren’t constant or simple). And, we give the mass a kick (initial tangential velocity). So, it obviously isn’t a 1D spring vibration, nor is it uniform circular motion.

An unfortunate reality is that we can make models, balancing that they’re close enough to reality while still easy enough to solve and predict results, but as soon as we make the model a little more complicated, we might make it too hard to solve. That’s the case here, but Momentum Principle to the rescue! As we’ve seen, it is a simple algorithm (based on Newton’s 2nd Law) for allowing a computer to step through the motion and show us an orbit. But, the great thing about it is that we can use it just as simply for ANY force formula at all.

In this case, we can write our vector, net, spring force as ( eqr is the equilibrium length of the

spring):

( ) ( )ˆ ˆ 1 eqspring eq

rk stretch k r r k

r

= − = − − = − −

F r r r

That shows the formula used in the GlowScript MP algorithm.

120

Activity 3: the simple pendulum (bob of mass L) with small angular deviation (down: θ = 0).

1. If we take the mass as our system, being acted on by external forces of Earth and string of length L, draw the System Schema and Interaction Map (FBD):

System Schema FBD

Since the string is assumed inextensible, there can only be motion perpendicular to the radius of motion. We can write Newton’s 2nd Law in vector (GA) form and work with the bivector part of the equation, but, even with GA, sometimes it’s easier just to write a component equation. Thus, we can deal just with the force equation parallel to the arc:

2

2sin sind gF mg ma mLdt Lθθ θ θ≡ − = = ⇒ = −

Unfortunately, that doesn’t fit any of our math models. To simplify the model, we now assume the angle is small, so that:

{ }

( ) ( )

2 32

2

1 2

... &3!

cos sin

d g g in tdt L L

C t C t

θ θθ θ ω θ θ

θ ω ω

= − − + − ≡ −

= +

MM4

2. How small is small? Finish the table: θ (degrees) θ (radians) sin θ % error 5° 10° 15°

3. For a pendulum of length 1.5 m, find: __________ /rad sω =

__________f Hz=

__________T s=

121

PHS 542 Name: ___________________________________ Homework: Momentum Principle (distorted circular motion). ________________________________________________________________________ Consider a 400 gram particle on a frictionless table, attached to an ideal spring. Assume the spring constant is 50 N/m, and its equilibrium length is 20 cm. To start things off, we pull the particle out to 25 cm (a stretch of 5 cm), and give it a sideways push (perpendicular to the radius) of 2.5 m/s. Our goal is to find out how the particle moves from there on. Finish & use this GlowScript program: ________________________________________________________________________ scene.background = color.white scene.title="Motion due to a Linear Restoring Central Force " #-------------------------------------------------------------------------------------------------- running = True def toggle(b): global running running = not running if running: b.text = "Pause" else: b.text = "Run" button(text="Pause", pos=scene.title_anchor, bind=toggle) #-------------------------------------------------------------------------------------------------- k = 50 # Spring constant (N/m). r_eq = 0.2 # Equilibrium, unstretched length (m). dt = 0.0001 # Time increments, in seconds. #-------------------------------------------------------------------------------------------------- ball = sphere(radius=0.02, color=color.yellow, pos=vec(0.25,0,0)) ball.m = 0.4 # Mass of ball (Kg). ball.v = vec(0,2.5,0) # Initial velocity = 2.5 m/s in the y-direction. ball.p = ball.m * ball.v # Initial momentum (Kg-m/s). #-------------------------------------------------------------------------------------------------- # Put it on a green round table: cylinder(pos=vec(0,0,-0.02), axis=vec(0,0,-0.04), radius=0.6, color=color.green) #-------------------------------------------------------------------------------------------------- # Put a little red peg in the middle: cylinder(pos=vec(0,0,-0.02), axis=vec(0,0,+0.04), radius=0.01, color=color.red) #-------------------------------------------------------------------------------------------------- orbit = curve(color=ball.color, radius = ball.radius/4) while True: # Run until key hit, then stop for screen capture. rate(8000) if running: F = - k * ball.pos * (1 - r_eq / mag(ball.pos)) # Hooke’s spring law. ball.p = ball.p + F * dt # New momentum. ball.pos = ball.pos + (ball.p / ball.m) * dt # New position. orbit.append(pos=ball.pos) # Tracing the orbit.

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1. Try changing your initial velocity to vec(0,0,0). Now describe the motion:

a) Single closed loop ellipse centered on the peg. b) Single closed loop ellipse, peg is at one focus.

c) A short straight line back & forth motion.

2. Now, try an initial velocity of vec(0,0.5,0). One loop now (that is, about 360°) looks most like which shape? a) Triangle. b) Square. c) Hexagon. d) Octagon. 3. The natural angular frequency of a spring is given by /spring k mω = . For our spring, this is: a) 14.14 rad/s b) 12.50 rad/s c) 11.18 rad/s d) 10.00 rad/s 4. If the particle were just attached to a cord of length 0.20 m, and given tangential speed of 0.559 m/s, its angular velocity is given by /circle v Rω = . What is the value of circleω ? a) 0.358 rad/s b) 0.716 rad/s c) 1.398 rad/s d) 2.795 5. For your answers in questions 3 & 4, what is the ratio /spring circleω ω ? a) 2 b) 4 c) 8 d) 16 6. Now, try an initial velocity of vec(0,1.118,0). One loop now (that is, about 360°) looks most like which shape? a) Triangle. b) Square. c) Hexagon. d) Circle. 7. Work out the new value for circleω . What is the new value for /spring circleω ω ? a) 2 b) 4 c) 8 d) 16 8. Finally, let's try the prediction of Robert Hooke (Hooke's Law for spring force). Make r_eq = 0 and choose which orbit results: a) A rotating ellipse centered b) A stationary ellipse centered at the origin. at the origin.

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PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 4 --- Day 3

124

The GA Primer: Geometric Product, Inverse, 2D Rotation

Review the first sections on vector definitions and rules, look at the Geogebra downloaded files, make sure all is understood through the parametric equation examples.

Review the geometric product (and dot product & wedge products). Show the relationship between wedge product and cross product in 3D (4D differences). Talk about bivectors, parallel & perpendicular projections, commutivity of bivectors with parallel/perpendicular vectors. Laws of cosines & sines. We can now divide by a vector. Review commutivity for vectors which are parallel or perpendicular to each other. Squares of simple bivectors (in 3D, they are all simple) are negative scalars.

Review 2D rotations (mention 3D rotations). Discuss the Primer’s left multiplying of rotors using a CW unit bivector versus right multiplying using a CCW unit bivector. You could get comfortable with either convention (pros & cons of each?). Show how complex numbers and their operations relate to GA in 2D. Mention how Hamilton invented quaternions as a 3D extension of 2D complex numbers in order to create a method for calculating rotations in 3D, yet this method is already part of 3D GA naturally, and GA has a more meaningful notation.

Think about scalars, vectors, bivectors, and pseudoscalars. Then introduce the grade operator. Next the Primer shows some plane trigonometry theorems, including sum of angles formulas. The Primer does two such trig identities in one derivation, or we could prove one at a time using the grade operator.

The Barycentric Coordinates discussion in the Primer is probably unnecessarily wordy. Simplify and just point out the major ideas.

Activity 1: rotations in 2D

The diagram at the right shows our typical inclined plane for motion down (or up if we pull our object left) the incline.

It makes sense to think about the minimum number of symbols we can use to work with such problems. That’s because the more symbols we use, the more extra equations we need to relate those symbols to each other. In this diagram, we have 5 symbols listed, but we can express all in terms of just these three: { }ˆ , ,θg i . We see that

1ˆ ˆ=e g i because 1e is what we get when we rotate g CCW by 90°. We can also see from the GA Primer’s discussion of rotors that one of the following is true.

θ i

125

1. Circle the true statement:

1 1 1 1ˆˆ ˆˆ ˆˆ ˆˆ) ) ) )a e b e c e d eθ θ θ θ− − −= = = =i i i ise i se i se se

2. Using 1ˆ ˆ=e g i , write s in terms of { }ˆ , ,θg i .

ˆ ____________=s

3. If θ = 30°, write s in terms of { }ˆ ,g i .

( )

ˆ ˆ ˆ______ ______

ˆ ______ ______

= +

= +

s g g i

g i

Activity 2: beam held up by wall pivot and rope.

This is a standard beam problem (statics) in physics textbooks. The beam is attached to a pivot on a wall, at an angle of 45° with respect to the vertical above the pivot point. The rope is tied so that it makes an angle of 60° at the wall. The mass of the beam is M = 10 Kg, acting through its center. Another mass, m = 4 Kg, is attached to the end of the beam. We’d like to find T

, the tension in the rope (we know the direction, but not the magnitude) and the vector reaction force, R

, of the wall on the pivot point of the beam.

126

Let our system be the beam. Draw the forces { , , ,M mg g T R } in the Free Body Diagram at the right. Show angles (either values or labels) for &T R .

1. Write down Newton’s 2nd Law for this situation. The net forces should add up to zero because there is no acceleration of the bar.

0 Forces= =∑ ______________________________________________________

This gives us enough to solve for two unknowns, but we have three. So, we need another equation. Those teaching physics will know we can get another equation by writing a torque equation. And, if we calculate the torque about the pivot point, which R

passes through, then those two unknowns will be eliminated from the equation, and we can solve the resulting equation for T

. And once we find T

, we can plug it back into the equation above to find R

.

Definitions of torque: Vector Torque Bivector Torque

≡ ×τ r F { }Tq i i= = ∧ = ×τ r F r F

Vec ~ plane⊥ r F . Bivec ~ planer F . Dir by RH rule. Orientation by ~ sweepr F .

Applicable Physics laws: Forces Δ(momentum) Torques Δ(ang. Momen.)

i

d dmdt dt

= =∑ p vF i idI idt

∧ =∑ ωr F

Statics (fixed beam): 0i=∑F 0i i∧ =∑r F

127

2. Using the pivot point as our torque origin, write the torque equation in this case. Each ir will be the radius vector pointing from that pivot point to the point of

application of each force. For example: 0 0∧ = ∧ =Rr R R .

0 i i= ∧ =∑r F ______________________________________________________

3. Solve that equation for the tension:

T = ____________________ N

/3 /3ˆ ˆ__________T e e Nπ π= − = −i iT g g

4. Plug this vector tension back into your force equation from question 1. Solve for R

.

ˆ ˆ____________ ____________ N= +R g i g

______ˆ____________ e N= iR g i

5. Does R

point along the beam? a) Yes. b) No.

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PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 4 --- Day 4

129

Summary Review of GA

Vectors have magnitude and direction: ˆa=a a

We can multiply vectors. Geometric product: , , .etc+ab abc de

Any nonzero vector has an inverse: 2

1 1 12 2

ˆ1

a a a− − −= = ⇔ = = =

a a aa aa a a

Standard perpendicular, unit vectors: { } { }1 2 3ˆ ˆ ˆ, , , ,x y z direcs → e e e

Geometric product of two vectors produces a scalar + a bivector: = + ∧ab a b a b

Dot product:

{ }( ) { }( )

( )0

cos cos cos

12

shadow of on shadow of onab a b b a

scalar part of symmetrical part of

θ θ θ= = =

= = = +

=

=

=a b

ab

b

ab

a a b

ab ab ba

Wedge product:

{ }( ) { }( ){ }( )( )

( )2

sin sin sin

12

,

the part of the part of

unit bivector in plane orient

ab a b b a

area of parallelogram

bivector pa

ed as

rt of antisymmetrical part of

to

θ θ θ=∧ = = =

=

= = =

= ⊥

=

=

⊥

−

a b i i i

i ab

ab ab ab

b a a b

ab b

ab

a

ba

Scalars commute with any geometric product: ( )( ) ( ) ( )5 5 5 5= = =ab a b a b ab

Parallel vectors commute with each other: ( ) ( ) 23 3 3g− = − = −g g g g

Perpendicular vectors anticommute with each other: 1 2 2 1ˆ ˆ ˆ ˆ= −e e e e

Vectors in a plane anticommute with the plane’s bivector:

( ) ( )

1 2 2 1 2 1 1 1 1 2 2 2 1 1 1

2 2 1 1 2 2 1 1 2 2

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ:Similarly a a a a

≡ = − ⇒ = − ⇒ = = = = −

= − ⇒ = + = − − = −

i e e e e e e i e i e e e e e e e iee i ie ai e e i i e e ia

ˆab is called a rotor because it rotates a into b : ( ) ( )ˆ ˆ ˆˆ ˆ ˆ ˆ ˆmultiplied by rotor = = =a a ab aa b b .

130

Because i squares to minus one ( )21 2 1 2 1 1 2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1= = − = −i e e e e e e e e , it makes sense to use Euler’s

formula, which is a useful mathematical shorthand (derived from Taylors series):

ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ ˆ ˆcos sin e e and eθ θ θθ θ −= + ∧ = + ≡ → = =i i iab a b a b i a b a b

We can see this because if the rotor with angle θ rotates a into b , then the rotor with angle minus θ should rotate b back into a . Another way to see this (algebraically):

( ) ( )ˆ ˆ ˆ ˆ ˆˆ ˆ cos sin cos sine e eθ θ θθ θ θ θ −= ⇒ = = + = − =i i iab a b i b b i b

So, we see e θi rotates vector a CCW in the i -plane by angle θ. But, we haven’t specified a in any way. It could be any vector. Therefore, e θi will rotate any vector in the i -plane CCW by θ.

The exponential notation for the rotor is cleaner. In e θi , we right away see the plane and angle. Look at: ( ) ( )/3 cos / 3 sin / 3 0.5 0.866e π π π= + = +i i i . It’s often easier to multiply a single

favtor rather than a sum of two factors. And, as mentioned, if you saw just the right side, you’d have to work to figure out the angle being used.

We can see a rotor in the product of any two vectors: ˆ ˆˆ ˆa b ab abe θ= = = iab a b ab .

Keep in mind that { }1 2ˆ ˆ,e e are our choice of directions. So, in any physics problem stuck in a

single plane, we are free to make this choice for ourselves. Sometimes we choose them to stand for horizontal and up. Other times we might choose down an incline and perpendicular to the incline.

Why do some of us prefer { }ˆ ˆ,g gi ? They are two perpendicular directions and i is the same

CCW unit bivector as before. But, you don’t have to ask somebody which direction they want to choose for 1e (for example, horizontal or down the incline). Wherever you are, g has one and

only one direction, towards the center of the earth. And, by using gi , we don’t need another symbol (like 2e ), and we don’t need subscripts. As a result, this notation just seems more well-defined and more efficient. But, it’s always your choice.

131

PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 5 --- Day 1

132

Circular Motion

For the plane of the circle: 1 2ˆ ˆCCW unit bivector (90 rotor) = ≡ °i e e .

We can write the position vector and velocity vector as a functions of time:

( ) ( ) ( ) ( ) ( )0 0t d dt e t e t t

dt dtθ θ θ ω= = = ≡i irr r v r i r i

Notice the velocity vector is always rotated 90° from the position vector. This should make sense, because the velocity gives the direction of motion, and if it’s circular motion, that direction should always be tangent to the circle.

One more time derivative will give the general circular motion acceleration:

( ) { } ( ) { }

( ) [ ] ( ) ( ) ( )

22

2

2

2

2rad tang

Ang. speed / Ang. accel. /d d dt rad s t rad sdt dt dt

d d dt t t tdt dt dt

θ ω θω α

ω ω α

ω ω α ω α

≡ = ≡ = =

= = = = +

= + = − + = +

r va r i v i r i

r i i r i r ri a a

Notice that, in general, the acceleration may have a component towards the center (centripetal acceleration) and a component along the circle (tangential acceleration).

There are a number of commonly used relationships hiding in those equations:

2

2rad tangconstant 2 2 /vr v r a r v a r f T

rω ω ω α ω π π= = = = = = = =

Activity 1: old LP record player, with a riding dirt particle.

Assume we have an old 78 RPM record spinning, with a dirt particle of mass m riding at a radius of 8 cm.

1. What is the frequency in Hertz (rev/sec)?

f =______________ Hz

2. What is the angular speed?

ω = ____________ rad/s

133

3. What is the linear speed of the dirt particle?

v = ______________ m/s

4. What is the magnitude of the angular acceleration?

α = ____________ rad/s2

5. What is the magnitude of the centripetal acceleration of the dirt particle?

rada = ___________ m/s2

6. What force is causing this centripetal acceleration?

________________________________________________________________

Activity 2: the vertical circle (in gravity).

The results shown on page 1 (expressions for 2, , andω ω α= = − +r v ri a r ri ) are valid for any circular motion. By definition, circular means r constant= .

For Uniform Circular Motion, 0constant andω α= = , as in the case of the old record player in Activity 1.

Now we want to consider a vertical circle in the presence of gravity. For example, we might have a carnival motorcycle driver in a vertical, circular hoop. As long as the speed is great enough, the motorcycle will be able to keep its wheels on the hoop.

We can see this isn’t Uniform Circular Motion, because speed will be smaller at the top and larger at the bottom, thanks to the help of gravity.

A ball of mass m = 0.080 Kg is attached to a cord of length r = 0.6 m, and we swing it in a vertical circle. At the top of the circle, the speed is min max 3.5 /v v m s= = .

1. What is the angular speed at the top of the circle?

_______________ /top rad sω =

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2. What is the centripetal force ( )2ctr top topF m rω= at the top of the circle?

_______________ctr topF N=

3. Draw the Free Body Diagram when the ball is at the top of the circle:

4. What is the weight of the ball? mg = ______________ N

5. What is the tension in the cord at the top of the circle?

Ttop = ________________ N

Because the tension is always perpendicular to the direction of motion, the tension can do no work, only gravity can. Therefore, we still have conservation of energy. If we define y = 0 at the center of the circle, then:

2 2 21 1 12 2 2top bottE mv mgy mv mgr mv mgr= + = + = −

6. What is the constant total energy?

E = _______________ J

7. Use that to find the (maximum) speed at the bottom of the circle.

_____________ /bottv m s=

135

8. What is the angular speed at the bottom of the circle?

_______________ /bott rad sω =

9. What is the centripetal force ( )2

ctr bott bottF m rω= at the bottom of the circle?

_______________ctr bottF N=

10. Draw the Free Body Diagram when the ball is at the bottom of the circle:

11. What is the tension in the cord at the bottom of the circle?

Tbott = ________________ N

12. Find ctr bott ctr topF F− . _____________ctrF N∆ =

13. Fill in the factor in this formula: ____ctr bott ctr topF F mg= + .

14. Find bott topT T− . _____________T N∆ =

15. Fill in the factor in this formula: ____bott topT T mg= + .

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PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 5 --- Day 2

137

Short Note on Maxwell’s Equations

T-shirt comedy: Future Geometric Algebra version. Integral form of Maxwell’s equations. Differential form of Maxwell’s equations.

c ≡ 299,729,458 m/s = speed of light in vacuum. ρ = charge density ( )3/C m .

0µ ≡ 724 10 Kg m

Cπ −× = permeability of free space. J = current density ( )2/A m .

0ε ≡ 2

2 20

1 Cc N mµ

= permitivity of free space. 1 2 3ˆ ˆ ˆi = e e e = GA unit pseudoscalar.

F ic= + =E B the Faraday bivector (in 4D). N VC m = =

E electric field, [ ]T =B magnetic field.

( )

( )

( )0 0

1F

1

1

t icc

i icc t t

c i i ic cJ c cc t t

µ µ ρ

≡ ∂ +∇ + ∂ ∂

= +∇ +∇∧ + + ∇ +∇∧∂ ∂

∂ ∂= ∇ + − ∇× + ∇× + + ∇ = ≡ −

∂ ∂

E B

E BE E B B

E BE B E B J

00/ρ ε→ ∇ =E Gauss’s Law for E (flux proportional to enclosed charge).

0 21

1c t

µ ∂→ ∇× = +

∂EB J Ampere’s Law (induced B fields from J or changing E ).

2 t∂

→ ∇× = −∂BE Faraday’s Law (induced EMF from changing B , Lenz’s Law).

30→ ∇ =B Gauss’s Law for B (no net flux, no magnetic monopoles).

138

GA Special Relativity Intro

From Wikipedia: “An inertial frame of reference in classical physics and special relativity is a frame of reference in which a body with zero net force acting upon it is not accelerating; that is, such a body is at rest or it is moving at a constant speed in a straight line.” There is plenty of room for philosophical discussion here, and we might sometimes be fooled, but we say an inertial frame of reference is not experiencing acceleration.

We often talk of Earth as one (approximately) inertial system (or we could imagine instead a free floating space station). And we often think of a rocket ship moving at constant velocity with respect to Earth as another. We could perform experiments in the rocket lab, or a lab on Earth, or they could both observe another thing, such as a comet hitting another planet.

An Event (explosion, collision, passing a marker, …) can be characterized by a time and a place of happening. In GA, we use this notation to record events:

Event recorded by Earth: X ct= + x {Scalar + vector, not a rotor,} Same event recorded by rocket ship: X ct′ ′ ′= + x { called a Paravector. }

So far, there’s nothing magical or mysterious about this. We’re just writing down where and when something happened as seen by various “inertial” observers. On Earth, we imagine our set of coordinate axes and have our synchronized clocks to measure events. The rocket ship can imagine its coordinate axes attached to it (riding along with it) and its own set of synchronized clocks. Earth and the rocket may, quite logically, record different measurements (after all they are moving with respect to each other). And, it would be nice if, when the rocket lab transmits its data for an event (the primed numbers), we could translate that data into Earth coordinates (the unprimed numbers), or vice versa.

In fact, if these are isolated inertial systems, we can’t say which one is stopped (if either) or which one is moving. That is a statement of relativity. But what kind of relativity (Galileo’s or Einstein’s or …) we’re talking about determines how we translate from primed to unprimed coordinates or vice versa.

Until Einstein’s paper in 1905, it was assumed that time was universal. Assuming all our measuring clocks are synchronized (a bad assumption), Earth clocks and rocket ship clocks should be the same. Seeing the fault in this assumption was Einstein’s great revelation.

For example, how do you synchronized Earth clocks with Moon clocks? Carry them back & forth? How do we keep checking? Even if we did, might clocks run faster on the Moon? How would we know? It (now) seems obvious that we must synchronize with light signals and some kind of communication.

139

That concept probably made sense even before 1905. But, in the years before, physicists were running into a problem (which turned out to really be a solution) when measuring the speed of light. You can look up the Michelson-Morley experiment and related experiments. The speed of light in empty space seems to be a universal constant. If you think that’s as it should be, think about light coming at us from celestial bright objects which are approaching us compared with light from receding objects. Yes, you may know about the Doppler effect, but that relates to the frequency of light, not its speed, which seems to be the same to us for approaching or receding objects. It’s so universal, it has now become a standard to lean on:

Speed of light in empty space ≡ c ≡ 299,729,458 m/s. ct represents time meters. 1 time meter corresponds to about 3.34 ns.

Given that, imagine the problem that Einstein had to solve. Let’s say when the rocket ship passes by Earth in a near encounter, we set both our clock systems to zero and make sure our axes are lined up and our axes markers all coincide. Now, we set off a flash of light (or a series of flashes), which expands in a spherical wave. Here’s the conundrum: both the Earth scientists and the rocket ship scientists will measure that they are at the center of the spherical wave, yet they are moving with respect to each other, so how can that be?

Both systems must agree that the light wave expanding along the x-axis must obey:

( ) ( ) ( )2 2 22x ct and x ct ct x ct x′ ′ ′ ′= = ⇒ − = −

In GA notation: &X ct X ct X X X X′ ′= + ≡ − ⇒ =x x

Yet, the only way that can be true, since we know the x-coordinate of the wave seen from the rocket is different than the x-coordinate of the wave seen from the Earth, is if the time measured by a rocket clock as the wave passes by is different than the time measured by the corresponding Earth clock --- that is, there is no universal time.

Einstein’s 1905 paper (readable by HS students, since it is practically all fairly simple algebra) used these ideas to develop the formulas we can use to translate event numbers from the primed frame to the unprimed numbers, or vice versa. Equations of this nature were known before (when it was thought time was universal), but those equations were altered to allow for different time measurements as well as different space measurements. In fact, even Einstein’s equations had be found before (that’s why they are called Lorentz transformations), but he gave the new logic needed to use the universal speed of light and to throw out the idea of universal time.

For the Lorentz transformations, we assume Earth and rocket ship coordinates coincide when their coordinate origins pass by each other at 0t t′= = . Also we assume that the velocity of the rocket with respect to Earth is a constant, v , pointing along the common, positive x-axes.

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Lorentz transformations:

( )

( ) ( )( ) ( )

2

2 2

111 /

/ /

ubiquitous relativity factorc

t t c or t t c

t t

γ

γ γ

γ γ

⊥ ⊥ ⊥ ⊥

≤ = = < ∞−

′ ′ ′= + = −

′ ′ ′= + = −

′ ′= =

v

x v x v

x x v x x v

x x x x

The GA form of these:

( )

( )21 / 1 1 /2 1

cB Boost B B and B cγ γ γγ

+ += = ⇒ = = +

+

v v

X BX B or X BX B′ ′= =

If we stick to easier examples where both Earth and the rocket just see events on their corresponding x-axes (that is, ′x x v ), these become:

( ) ( )( ) ( )

( )( ) ( )( )( ) ( ) ( ) ( )

2 2

22

/ /

1 / 1 /

/ /

t t x v c or t t xv c

x x vt x x vt

X BX B B X X BX B B Xct c ct ct c ct

ct x v c t ct xv c t

γ γ

γ γ

γ γ

γ γ γ γ

′ ′ ′= + = −

′ ′ ′= + = −

′ ′ ′= = = =′ ′ ′ ′+ = + + + = − +

′ ′ ′ ′= + + + = − + −

x v x x v x

x v x v

Because all these equations (even the general vector ones) are linear, they also apply to differences between two events. In other words, if we have events A and B, then all of our above formulas work for AX or BX or even for BA B AX X X≡ − .

In any typical relativity problem, our first job is to clearly identify two events (sometimes the

first event is when the Earth and rocket coordinates coincide, i.e., at 0A AX X ′= = ). Then we can plug into formulas above to solve for various unknowns.

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Activity 1: time dilation.

Assume the rocket speed is 0.8c with respect to Earth. When the Earth scientist records that 100 s have elapsed since the rocket passed Earth, our goal is to find how much time has elapsed for the scientist on board the rocket.

Event A: rocket passes Earth, which means 0A AX X ′= = .

Event B: Event B is the rocket passing an Earth clock which reads 100 s, somewhere along Earth’s x-axis.

1. What is the relativity factor for this case?

__________γ =

2. What is Earth’s x-coordinate for Event B? That is, how far does Earth say the rocket has traveled in 100 s? Leave the answer in terms of c (don’t waste time multiplying out the value for c).

x = __________ c

3. Use ( )2/t t xv cγ′ = − to find the elapsed time on the astronaut’s clock.

__________t′ = s

Time Dilation: Moving clocks seem to slow down. The astronaut sees less time elapsed on the rocket clock than the time elapsed on Earth clocks.

Let’s try this again using GA notation. All our vectors are in the same direction, so

1 1 1ˆ ˆ ˆ, , 0.8x x and c′ ′= = =x e x e v e .

4. Write X for Event B. BX = ___________________________

5. Write X ′ for Event B. BX ′ = __________________________

6. Write 2B for this problem. 2B = ___________________________

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7. Write out 2B BX B X ′= .

____________________________________________________

8. From the scalar part of this last equation, find the rocket’s clock reading.

__________t′ = s

Activity 2: length contraction.

Use the same example as in the last section, but let’s see what the astronaut thinks about the distance markings attached to Earth. To do this, think about the origin (Event A). The astronaut sees the Earth origin moving backwards. When Event B happens, we know the reading on the astronaut’s clock (answer to question 8). At that moment, Earth’s origin is passing one of the astronaut’s clocks with the same reading because the astronaut knows all clocks in the rocket system are synchronized. Let’s call this happening Event C.

1. In the time t′ (from question 8), how far does the astronaut think Earth’s origin has moved?

____________vt m′ =

2. This last answer is what the rocket scientist thinks is the distance between Earth’s distance markers at the origin and at Event B. What does an Earth scientist record as this distance?

Bx = ____________ m

That is our “rest” distance between markers. As Earth flies past the astronaut, that length seems shorter (contracted). Check: (moving length) = (rest length) / γ .

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PHS 542 (Summer, 2018, 5 weeks) Daily Details

Week 5 --- Day 3

Project Presentations

Week 5 --- Day 4

Project Presentations