Explaining the Magical Property of Parabolic Reflectors – Part 1

3d_parabola_smThere is no coincidence that the reflective surface of car headlights, flashlights, spotlights, satellite dish antennas, solar dishes, solar cookers, microphones, reflecting telescopes, and radio telescopes have a unique curvature that can be described by the graph of the parabola y = kx2 where k = is a constant. The surface of a parabolic reflector is created by rotating the graph of a parabola about its axis of symmetry. Parabolic reflectors have a magical property in that all incoming energy (light, sound, radio waves, etc.) traveling parallel to the axis of symmetry is reflected from the surface of the reflector to the focus point of the parabolic reflector. Conversely, all energy emanating from the focus point of a parabolic reflector is reflected to the surface of the reflector, and then in a direction that is parallel to its axis of symmetry. Because parabolic reflectors have this property, they are used to collect or project energy. The graph below shows the cross section of a parabolic reflector created by rotating the graph of the parabola y = 0.125x2 about its axis of symmetry. All graphs in this post were created with my program Basic Trig Functions. Go to www.mathteachersresource.com to view multiple screen shots of the program’s modules. Click the ‘learn more’ button in the TRIGONOMETRIC FUNCTIONS section. Teachers will find useful comments at the bottom of each screen shot.

parabola_diagram1

So why do parabolic reflectors work the way they do? In this post, I will answer that question by showing how trig, pre-calculus, and calculus teachers can use modern computer graphing technology to help explain the properties of a parabolic reflector by verifying specific instances of the property. After completing the exercises themselves, students will be better prepared to follow or give a rational explanation of why this property is true for any point (x, y) on the parabola. My next post will explain why the angle of incidence equals the angle of reflection for any point (x, y) on the graph of the equation y = kx2. One of the cardinal principles of problem solving is to solve the problem for a simple or specific case of the problem before attempting to solve the general problem.

I would start my lesson by giving each student a handout that contains graphs similar to the graphs shown below. During the lesson, students would be required to use their calculators to make various calculations, verify specific facts during the lesson, and answer questions on the handout. Next, use a diagram similar to the diagram below to give a quick review of the definition of a parabola. Point (x, y) is on the parabola only if the distance from (x, y) to the focus point equals the distance from (x, y) to the directrix line. The class can check that points A, B, C, and D satisfy the definition of a parabola.

parabola_diagram2

parabola_textbx2

In the next part of the lesson, show students how they can use the slope of a line and the inverse tangent function on their calculators to find the positive angular direction of a ray. Adding + 3600 to the direction of a ray gives us an equivalent direction of a ray. Adding + 1800 to the direction of a ray gives us the direction of the opposite ray. The graph and the statements in the text box below illustrate how to determine the positive angular direction of a ray given the slope of the line that contains the ray.

parabola_diagram3

parabola_textbx3

The last part of the lesson is about demonstrating that the measure of angle ABC (angle of incidence) equals the measure of angle FBD (angle of reflection). A principle of physics states that all energy in the form of particles or waves reflects off a flat surface in such a manner that the angle of incidence equals the angle of reflection. In the diagram below, the flat surface at point B = (6, 4.5) is the tangent line to the graph of the equation y = 0.125x2 at point B. From differential calculus, we learn that the slope of the tangent line to the graph of y = kx2 at any point (x, kx2) = 2kx. Using the results presented in the previous part of the lesson, the angular direction of a side of an angle can be calculated from the slope of the side of an angle. By knowing the direction of each side of an angle, we can find the degree measure of the angle by subtracting the directions of the sides of the angle. The text box below shows the verification that the measure of angle ABC equals the measure of angle FBD. For point M = (-4, 2), the slope of the tangent at point M = 0.125*2*(-4) = -1, the direction of ray ML = 1350, the direction of ray MD = 3150, and the measure of angle LMN = the measure of angle FMD = 450.

parabola_textbx4

parabola_diagram4

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Teaching Points:

  • Provide students with a handout that contains a graph similar to the last graph in this post.
  • Have students show that the angle of incidence equals the angle of reflection for specific points on the parabola.
  • There is no reason that students can’t be given the formula for the slope of a tangent line to a point (x, y) on a parabola. Tell them that they will learn how this formula comes about when they take calculus.
  • Although it is not necessary to know the equation of the tangent line, have them find the equation of the tangent line. It’s a good review exercise and will better prepare them for calculus.
  • A curious student will want to know why this magical property is true for any point (x, y) on any parabola. Have the student find the answer to this question, and let them present his/her solution to the class. This would make a wonderful Power Point presentation. Give the student the following hint: “Use the angle difference identity for the cotangent function to show Cot(900 – β) = Cot(2700 – β) = Cot(β) = 1/Tan(β).”
  • Tell students that it’s sufficient to show that the magical property is true for parabolas of the form y = kx2 because translating, rotating, and stretching the graph of a parabola preserves the property.

Wrapping a Piece of String Around the Earth

earth_string1My last post explains how I introduce radian angle measure to my students. That post generated three interesting questions about wrapping a piece of string around a great circle of the Earth. The first two questions (see below) are familiar to many math teachers and students of mathematics. However, I have never seen a question similar to question 3, nor was I was able to find a question similar to it discussed on the web despite abundant resources that provide solutions to math questions. Because the third question is new and interesting to me, and it relates to the content of my last post, this post focuses on solving the third question. If you would like to review detailed solutions for the first two questions, open this PDF file.

The solutions of the three questions can be explained in terms of the radius r of a circle, the length L of an arc that subtends central angle θ, and the radian measure of θ. The relationships L/r = θ and L = rθ were used to derive the equations D = 2πd and d = D/(2π). D equals the difference between the circumference of a great circle ring around a sphere and the circumference of a sphere, and d equals the uniform distance between the great circle ring around the sphere and the surface of the sphere. Because the equations D = 2πd and d = D/(2π) apply to any sphere, not just planet Earth, results derived from these equations are interesting and counter intuitive.

Question 1: Imagine that the surface of planet Earth is like the surface of a perfect sphere. Consider a piece of string, longer than the circumference of the Earth, that is used to form a great circle ring around the Earth. If the uniform distance d between the outer ring and the surface of the Earth equals 1 inch, how much longer than the circumference of the Earth must the circumference of the circular ring be?

Solution: D = 2πd = 2π(1 inch) = 6.28 inches

Question 2: Imagine that the surface of planet Earth is like the surface of a perfect sphere. Consider a piece of string, 1 inch longer than the circumference of the Earth, that is used to form a great circle ring around the Earth. What is the uniform distance d between the outer ring and the surface of the Earth?

Solution: d = D/(2π) = 1 inch /(2π) = 0.1592 inches

Question 3: Imagine that the surface of planet Earth is like the surface of a perfect sphere. Consider a loop of string, 1 inch longer than the circumference of the Earth, that is wrapped along a great circle of the Earth and pulled at a single point A so that the loop of string is taut. What is the distance between point A and the surface of the Earth? The answer to this question, 413.2237 inches, is very surprising.

The diagram below shows the setup for solving question 3. From Euclidean geometry, we learn that a radius to a point of tangency is perpendicular to the tangent line.

string_Q3_graphic

The text box below gives the key steps in solving question 3. Of course, readers will find that I skipped some steps when going from one step to the next. A suggestion for the dissatisfied or skeptical reader is to find a quiet corner of his/her dwelling, and write out the complete solution by making the appropriate substitutions and simplifying the resulting equations. The method that I used to solve the trigonometric equation is not found in typical math text books. If you find an error in my solution of question 3, or find a more elegant solution, please reply to this post, or send an email to info@mathteachersresource.com. I’m still amazed that d ≈ 413. 2237 inches!

string_Q3_solution1

A Suggestion for Teachers

After a classroom discussion of question 3, trig and pre-calculus teachers have a golden opportunity to transform their classroom into a mathematics laboratory for a day.

  • Organize the class so that each student has a lab partner.
  • Provide each lab partnership with a toy model of a sphere, a piece of string, and adhesive tape.
  • Provide the appropriate tools to measure the diameter of the sphere and string length.
  • Construct a string-sphere model similar to the model described in question 3.
  • Measure the distance from point A to the surface of the sphere.
  • Calculate the expected distance from point A to the surface of the sphere.
  • Create a written report that includes a discussion of their model, experimental measures, and a complete description of how the expected distance was calculated.
  • The written report should also include a discussion of the experimental error found in their results.

Explaining Radian Angle Measure

radian_measure_3Initially, trig students find the concept of measuring angles in radians baffling. Why start measuring angles in radians when measuring angles in degrees makes perfect sense and is so simple? Why not continue to say that a right angle measures 900 rather than π/2 radians? These are some of the thoughts I suspect that many of my beginning trig students have when they are introduced to the concept of radian angle measure. They eventually learn that radian measure simplifies a variety of calculations involving arc lengths, areas of sectors of a circle, angular velocity, and linear velocity. In calculus, they learn that it’s necessary that trigonometric function input is scaled in radians. Radian measure makes it easy to treat an input variable of a trigonometric function as a time variable or arc length variable. The purpose of this post is to show how I introduce radian angle measure to my students.

I ask my students if they would be willing to bet $1,000 that the ratio of the circumference of a circle to the diameter of a circle always equals the irrational number π. Of course, I don’t allow any student to take the bet because I would be stealing their money. Why they would lose this bet involves some rather deep mathematics. In Euclidean geometry, the geometry of a flat surface, the ratio of the circumference of a circle to the diameter of a circle always equals the irrational number π, but it is not true for circles in other geometries. In spherical geometry, the geometry of the surface of a sphere, the ratio of the circumference of a circle to the diameter of a circle depends on the diameter of the circle, and this ratio can range from 2 to a limiting value of π. As the diameter of a circle approaches zero, this ratio approaches π. Because the concept of measuring angles in radians depends on the fundamental mathematical fact that the ratio of the circumference of a circle to the diameter of a circle always equals π, I tell my students that my explanation of radian angle measure applies only to angles in a flat surface geometry.

After a quick review of the definitions of basic terms relating to circles, I give my students the following informal definition of radian angle measure:

“The radian measure of central angle θ of a circle equals the number of radii, laid end-to-end on the circle that is required to subtend central angle θ.”

Therefore if L equals the length of an arc that subtends central angle θ of a circle, and r equals the radius of a circle, the radian measure of angle θ equals L/r. I then show my students a diagram similar to the angle-circle diagram shown below to demonstrate this relationship.

radian_measure_1

Angles α and β are central angles of circles with radii of 29 units and 15 units respectively. λ = 67.57 units and ω = -21.0 units are the lengths of the arcs that subtend angles α and β respectively. Of course, β and ω are negative because of the clockwise direction of angle β. By definition, the radian measure of α = 67.57/29 = 2.33 radians and the radian measure of β = -21/15 = -1.4 radians. At this point in the discussion, it is always interesting to see what degree estimates for angles α and β that students come up with.

The text box below gives a summary of the fundamental relationships between radian angle measure and degree angle measure. Using the angle-circle diagram above, you can easily explain why the fundamental relationships below are true.

radian_measure_4

The major objective of the first animation is to show the radian measure of central angle θ and the length of the arc that subtends angle θ when the radius of the circle equals 10 units. Notice that the measure of angle θ increases by 1 radian increments. This is a good example to show students because students can easily see the fundamental relationship between the radius of a circle, arc length, and radian measure of an angle. The animations in this post are representative of the dynamic presentations that you can create for your students with my program Basic Trig Functions. Of course, you have complete control over program output rate. The angle-circle diagram was also created with my program. To view multiple screen shots of the program’s modules, go to the Math Teacher’s Resource website. Click the “learn more” button in the TRIGONOMETRIC FUNCTIONS section. Teachers will find useful comments at the bottom of each screen shot.

The major objective of the second animation is to show the radian measure of central angle θ and the length of the arc that subtends angle θ when the radius of the circle equals 6 units. Notice that the measure of angle θ increases by π/12 radian increments. I have found that animations of this type help me better explain why π is involved when converting typical degree angle measures to radian angle measures.

The major objective of the third animation is to show the radian measure of central angle θ and the length of the arc that subtends angle θ when the radius of the circle equals 2 units. Notice that the measure of angle θ decreases by π/6 radian increments. I use an example of this type to remind students that angles and arc lengths can have negative values.

Readers can download the student and teacher versions of the free handout Introduction to Radian Measure. This handout has two pages of exercises and student activities that I use to introduce my students to radian angle measure. We usually work a few of each type of exercise together, and the remaining exercises are left as homework. The free handout Unit Circle is another handout that my students have found helpful.

Teaching Points and Suggestions:

    • Expressions such as π/3, -5π/4, and 11π/2 are just meaningless abstract symbols to many beginning trig students. I tell my students that they should memorize the following facts: π/2 radians = 900 ≈ 1.57 radians, π radians = 1800 ≈ 3.14 radians, 3π/2 radians = 2700 ≈ 4.71 radians, 2π radians = 3600 ≈ 6.28 radians, π/6 radians = 300 ≈ 0.52 radians, π/4 radians = 450 ≈ 0.79 radians, and π/3 radians = 600 ≈ 1.05 radians. These memorized facts give students a set of circular road signs that they can routinely reference.
    • When solving angular and linear velocity problems, it is very important to properly label numerical quantities. Example 1: 4,200 revolutions/min * 2π radians/1 revolution * 1 minute/60 seconds = 439.82 radians/second. Example 2: 2,000 revolutions/min *2π radians/1 revolution * 1 minute/60 seconds * 5.5’’ = 1,151.92 inches/second.
    • Using an angle-circle diagram similar to the angle-circle diagram above, there are an endless variety of homework and test questions that can be created. Given the radius of the circle and the length of the arc that subtends angle θ, questions that require an integration of various trigonometric concepts can easily be created. Download my free handouts Trig Exercises # 1 and Trig Exercises # 2 for some ideas.

Teaching the Pythagorean Identities

pythagorean-identity-1Only a few odd ball math types really enjoy proving trigonometric identities. Most trig students find it’s the most difficult part of the course. The three Pythagorean identities Cos2(θ) + Sin2(θ) = 1, 1 + Tan2(θ)  = Sec2(θ), and 1 + Cot2(θ)  = Csc2(θ) are the most important trigonometric identities. Most textbooks use algebraic methods to prove these identities. I will explain the Pythagorean identities from a geometric point of view by utilizing static diagrams and dynamic animations of the functions y = Tan(x), y = Sec(x), y = Csc(x). As I mentioned in a previous post, whenever possible, math concepts should be understood from both an algebraic and geometric point of view.

The diagram below shows the geometric setup for proving the identities Cos2(θ) + Sin2(θ) = 1 and 1 + Tan2(θ)  = Sec2(θ).  The boxed statements give the geometric proofs of the identities. Of course, the proofs are based on the properties of similar triangles and the theorem of Pythagoras. It is interesting to note that the line x = 1 is a tangent line, AB = Tan(θ), line OB is a secant line, and OB = Sec(θ). You can see why the functions y = Tan(x) and y = Sec(x) are so named and why these identities are called Pythagorean identities.

pythagorean_proof1

pythagorean-identity-1

The objective of the first two animations is to give dynamic geometric interpretations of the functions y = Tan(x) and y = Sec(x), and give a geometric proof of the trig identity 1 + Tan2(θ)  = Sec2(θ) where x = the radian measure of an angle that ranges from 0 to 2π radians. The cosine and sine function values are the changing x-y coordinates of a point on the unit circle as angle θ ranges from 0 to 2π radians. Students can easily see why the tangent function is undefined at x = π/2. What is shown in the animations is very similar to the dynamic presentations that you can create for your students with my program Basic Trig Functions. The program has a ‘Special Features’ checkbox which gives additional program output for the tangent, cotangent, secant, and cosecant functions. Mouse button clicks or key presses control program output rate.

The diagram below shows the setup for geometric proofs of the identities Cos2(θ) + Sin2(θ) = 1 and 1 + Cot2(θ)  = Csc2(θ). The boxed statements give the geometric proofs of the identities. Of course, the proofs are based on the properties of similar triangles and the theorem of Pythagoras. It is interesting to note that the line y = 1 is a tangent line, AB = Cot(θ), line OB is a secant line, and OB = Csc(θ). Like the identities explained above, it makes perfect sense why the functions y = Cot(x) and y = Csc(x) are so named and why these identities are called Pythagorean identities.

pythagorean_proof2

pythagorean-identity-2

The objective of the last two animations is to give dynamic geometric interpretations of the functions y = Cot(x) and y = Csc(x), and give a geometric proof of the trig identity 1 + Cot2(θ)  = Csc2(θ) where x = the radian measure of an angle that ranges from 0 to 2π radians. The cosine and sine function values are the changing x-y coordinates of a point on the unit circle as angle θ ranges from 0 to 2π radians. Students can easily see why the cotangent function is undefined at x = 0.

The graphics in this post were created with the program, Basic Trig Functions, which is offered by Math Teacher’s Resource. To view multiple screen shots of the program’s modules, go to www.mathteachersresource.com. You can use the program to easily implement the suggested activities described in the “Teaching Points and Suggestions” section below.

Teaching Points and Suggestions:

  • I have found that when presenting a lesson, it’s better to spend more time asking rather than telling.
  • You can have a student operate the program during a class presentation. This will allow you to stand near the projection screen, point out the significance of various program outputs, and ask questions as the lesson progresses.
  • Some sample questions:
    • What is the definition of a tangent line of a circle?
    • What is the definition of a secant line of a circle?
    • Why is Tan(900) undefined?
    • Why can we say, “Tan(900) equals ± ∞.”?
    • Why can we say, “The reciprocal of zero equals ± ∞.”?
  • Compare and contrast the reciprocal functions y = Cos(x) and y = Sec(x) by graphing the equations. Set the graph input parameters so that x ranges from -2π to 2π, x-tick step = π/4, and x-label step = π. Then graph the equations y = 3Cos(x), y = 3Sec(x), and Cos(x) = 0. The graphs of the equations naturally lead to a compare and contrast discussion of the three relations.
  • See what happens when the equation Sin(x)^2 + Cos(x)^2 = 1 is graphed.

Teaching the Circular Sine and Cosine Functions

Students begin the study of trigonometry by learning how to solve for the sides and angles of a right triangle. Given any two sides of a right triangle, it is possible to solve for the length of the third side and the angle measure of the two acute angles. Likewise, given the measure of an acute angle and the length of one side, it is possible to find the lengths of the other two sides and the angle measure of the other acute angle. What makes this possible are the trigonometric sine, cosine and tangent functions. Before the availability of electronic calculators, the cosine, secant and cotangent functions were used to simplify certain paper and pencil calculations because it is much easier to multiply decimal numbers than divide decimal numbers.

After mastering right triangle trigonometry, students are taught radian angle measure and the six basic trig functions in terms of a circle with center at (0, 0) and radius r > 0. It is now possible to consider trigonometric function values with input values in degrees or radians such as Sin(56°), Csc(951°), Sin(-617π/3), Cos(225°), Sec(113π), Tan(155.296°) and Cot(-7.2). In the diagram below, α is a positive quadrant II angle and β is a negative quadrant III angle. The Cos, Sin, and Tan function values of α and β are shown below. For angle α, a circle with radius = 25 units and a point at (x1, y1) on the circle is used to find the function values. For angle β, a circle with radius = 41 units and a point at (x2, y2) on the circle is used to find the function values.

Intro Sin-Cos Blog

Intro Sin-Cos Blog

In this post, I will show teachers how they can use my program, Basic Trig Functions, to dynamically present the properties of circular trig functions. Similar demonstrations in my own classes give me a more effective way to teach the properties of these functions. Static graphs fail to capture the underlying dynamic properties of the functions. Basic Trig Functions is designed to be a tool to help teach a variety of core concepts in mathematics.

The objective of the first animation is to compare and contrast the properties of the functions y = Sin(x) and y = Cos(x) where x = the radian measure of an angle that ranges from 0 to 2π radians. The functions values are the changing x-y coordinates of a point on the unit circle as angle θ ranges from 0 to 2π radians. Students easily see when the function values are positive and negative as the point (x, y) moves from one quadrant to the next on the unit circle. After a little coaching, students can find function domain values where the value of a function equals -1, 0, or 1. What is shown in the animation is exactly what can be generated in a classroom with the program Basic Trig Functions. Of course, users have a variety of choices for setting output parameters. Mouse button clicks or button presses control program output rate.

The purpose of the second animation is to compare and constrast the properties of the functions y = 2Sin(x), y = Sin(x), and y = Sin(2x) where x = the radian measure of an angle that ranges from 0 to 2π radians. The animation naturally leads to a discussion of the concepts of amplitude and period of a function. For beginning students, it is not obvious that there is a difference in how function values for y = 2Sin(x) and y = Sin(2x) are calculated.

The third animation compares and constrasts the properties of the circular functions Cos(θ), Sin(θ), and Tan(θ) where the angle mode for θ can be set to exact radian, decimal radian, or degree measure. Angle input values are controlled by clicking convenient buttons. Users have options for setting the radius of the circle, setting the angle increment/decrement value, setting angle mode, and whether or not to draw a circular arc on the circle. The circular arc feature makes it much easier to explain radian angle measure because the length of the arc is displayed as angle θ changes.

Readers can download the free handouts Trig Summary, Unit Circle, Trig Exercises 1, and Trig Exercises 2 from our instructional content page. Trig Exercises 1 and Trig Exercises 2 give suggested student exercises which are designed to reinforce and integrate a number of key concepts. On first exposure to exercises of this type, teachers and students should do some of the exercises together.

Teaching Points:

      • Instructors can have a student operate the program Basic Trig Functions during a class presentation. This allows instructors to stand near the projection screen, explain various concepts, and ask questions as the lesson progresses.

• Initially students will be confused because they are attempting to understand the generalized definitions of the trig functions in terms of right triangle trigonometry. Remind students that the generalized trig function definitions apply to both positive and negative angles of any size, not just the acute angles of a right triangle. Of course, the concept of a reference angle is useful in explaining the connection between right triangle trigonometry and the generalized definitions of the trig functions.

• To engage students, continually ask questions as the lesson progresses. It is more effective if a question is directed to a specific student.

• Conduct an experiment during the lesson by having students use a calculator to find Sin(20°), Sin(160°), Sin(200°), and Sin(340°). Some questions: What did you notice? Why are some sine function output values positive and other values negative? Use the inverse sine function to find six angles θ in quadrant II, 3 positive and 3 negative, such that Sin(θ) = 0.6156614753. Why does your calculator give you an error message when you enter Sin-1(1.25)? Then repeat the experiment with positive or negative angles and other trig functions.

The graphics in this blog were created with the program, Basic Trig Functions, which is offered by Math Teacher’s Resource. To view multiple screen shots of the program’s modules, go to www.mathteachersresource.com. Click the “learn more” button in the TRIGONOMETRIC FUNCTIONS section. Teachers will find useful comments at the bottom of each screen shot.

Applying the Basic Equation Transformation Rules

The ability to apply the equation transformation rules is one of the most important skills that math students can learn. When they recognize that a given graph is just a geometric transformation of the graph of some familiar basic equation, it’s relatively easy for them to find the equation of the graph. Likewise, when they recognize that a given equation is just an algebraically transformed familiar basic equation, it’s relatively easy for them to draw a sketch of the graph.

Most functions and relations beginning algebra through calculus students encounter are the result of applying an ordered series of algebraic transformations to a basic equation. Each algebraic transformation applied results in a geometric transformation of the equation’s graph. A set of basic equation transformation rules describes how graphs can be translated, reflected, and rotated.

In this post, we will see how the equation transformation rules can be used to transform the graph of the square root function. To best understand this demonstration, download the free handout Equation Transformation Rules from www.mathteachersresource.com/instructional-content.html. It contains a succinct summary of the equation transformation rules, a simple explanation of why some of the counterintuitive rules work, and examples that show how the transformation rules can be applied. This handout is a helpful resource for both students and teachers.

EquationTrans-Edit1

EquationTrans-Edit1EquationTrans-Edit1EquationTrans-Edit1

EquationTrans-Edit1 EquationTrans-Edit1

Teaching Points:

  • The biggest mistake students make is replacing x with x + k or x – k. Remind students to enclose x + k or x – k in parentheses and then simplify the equation.
  • When reflecting the graph over the y-axis, replace every x with (-x) and then simplify the equation.
  • When finding the equation of a given graph, results should be checked by picking a few key points on the given graph and then determine whether or not the x-y coordinates of these key points fit the equation.
  • When finding the equation of a graph, teach students how they should see a graph.
    Example 1: I see a line with a slope of 3 that has been slide horizontally 7 units to the right.
    Example 2: I see a cosine curve, amplitude of 4, that has been flipped over the x-axis.
    Example 3: I see a circle, radius of 4, that has been stretched vertically by a factor of 5/2.
  • Depending on the course content, students should be required to memorize the equation, the shape, and the properties of basic functions and relations. Students can make flash cards.

The above graphs, created with the program Basic Trig Functions, are offered by Math Teacher’s Resource. The program has features that facilitate learning and teaching the equation transformation rules. You can enter an equation in any of the formats shown in the examples above. Except for exponents, all equations are entered like any equation in a textbook. For example: The inequality 2x – 10Sin3(3x) + 4y2 ≤ 25 is entered as 2x -10Sin(3x)^3 + 4y^2 ≤ 25. Relationships can be implicitly or explicitly defined. The program automatically figures out how to treat an equation or inequality, and shading of all inequality relations is automatic. You can specify whether to shade the intersection or union of a system of inequalities.

The user interface, simple and intuitive for all program modules, provides numerous sample equations along with comments and suggestions for setting screen parameters in order to achieve best results. After an equation is graphed, you can plot a point on a graph near the mouse cursor and view the x-y coordinates of the plotted point. In addition to plotting points, you can find relative minimum points, relative maximum points, x-intercepts and intersection points with simple mouse control clicks. A Help menu provides a quick summary of all of the magical mouse control clicks. Of course, all graphs can be copied to the clipboard and pasted into another document. Go to www.mathteachersresource.com to view multiple screen shots of the program’s modules. Click the ‘learn more’ button in the TRIGONOMETRIC FUNCTIONS section. Teachers will find useful comments at the bottom of each screen shot.

The Genius of René Descartes – Part 2 (The Parabola)

Utrecht_Weenix_Descartes-300In my previous blog, I discussed how René Descartes (1596 – 1650) discovered a way to synthesize geometry and algebra, which resulted in a revolution in mathematics. This synthesis is the reason Descartes is credited as the father of analytic of geometry. Because of Descartes’s discovery, we can derive an x-y variable equation that describes the relationship between x and y for every point (x, y) on a conic curve.

In this blog, I will discuss how teachers can use modern computer graphing technology to help students gain a better understanding of the definition of a parabola and the equation of a parabola. In a future blog, I will discuss some of the magical properties and applications of the parabola. This discussion is intended to provide teachers with a general approach to teach the parabola. The specific approach is left to the discretion of the individual teacher. Teachers may want to provide a handout that students complete as the lesson progresses. Students should be required to use their calculators to make various calculations and verify specific facts during the lesson. When graphs are projected on a screen, presentations can be dynamic, especially with a moving trace mark and mouse drawn line segments.

Now for a quick review of the parabola. All parabolas are defined in terms of a fixed point, the focus of the parabola, and a fixed line, the directrix of the parabola. Point (x, y) is on the parabola if and only if the distance from (x, y) to the focus point equals the distance from (x, y) to the directrix line. All parabolas have an axis of symmetry and the directrix of the parabola is perpendicular to the axis of symmetry. The vertex and focus are on the axis of symmetry, and the vertex point is equidistant from the focus and directrix. Study the basic parabolic graph below.

The Genius of RenŽ Descartes Blog - 2

Teaching Points: (Depending on the class, teachers need to give appropriate coaching.)

  • Similar to the graph above, show the graph of a parabola and its directrix, in a handout and or on a projection screen.
  • Students should be told something about a focus point and the directrix line but not the definition of a parabola. The definition of a parabola and the derivation of the equation will come at a later time.
  • For at least four points (x, y) on the parabola, have students use the theorem of Pythagoras to calculate the distance from (x, y) to the focus and from (x, y) to the directrix line.
  • Hopefully, most students will realize an amazing property of the parabola. For every point (x, y) on the parabola, the distance from (x, y) to the focus always equals the distance from (x, y) to the directrix. The class can experiment with other points on the parabola.
  • Repeat the experiment with the equation y = 0.75x2. Does this new parabola have the same amazing property?
  • Consider the parabola with equation y = kx2. How does changing k change the focus point? In general, how does changing k affect the shape of the graph? If we know the focus point of the parabola, can we find the equation of the parabola? Teachers and students can answer these questions by experimenting and just fiddling around with a computer graphing program.
  • In another lesson, teachers can show students how the equation y = kx2 comes about. Since all parabolas are geometrically similar figures, the equation of most parabolas can be found by applying the standard equation transformations rules.
  • Special equation transformation rotation rules:
    Rotate graph 90 degrees counterclockwise about (0, 0): Replace x with –y and y with x.
    Rotate graph 90 degrees clockwise about (0, 0): Replace x with y and y with –x.
    Rotate graph 180 degrees about (0, 0): Replace x with -x and y with –y.
  • Homework practice problems and exam questions.
    Given the vertex and focus of a parabola, have students find the equation of the parabola and sketch the graph of the parabola and the directrix.
    Given the graph of a parabola with key points, find the equation of the parabola, find the x-y coordinates of the focus and find the equation of the directrix line.

Teachers might consider allowing students to have some fun by having them do a project in which they are told to experiment and fiddle around with a computer graphing program to see what interesting relation graphs they can come up with. I can guarantee that an inquisitive student will come up with a graph that no human in the history of mankind as ever seen. Other than myself and some of my students, no human has ever seen the graph of the relation 2xSin(3x) + 2y = 3yCos(x + 2y) + 1.

The above graphic, created with the program Basic Trig Functions, is offered by Math Teacher’s Resource. Except for exponents, all equations are entered like any equation in a text book. Example: The inequality 2x – 10Sin3(3x) + 4y2 ≤ 25 is entered as 2x -10Sin(3x)^3 + 4y^2 ≤ 25. Relationships can be implicitly or explicitly defined. The program automatically figures out how to treat an equation or inequality, and shading of all inequality relations is automatic. Users can specify whether to shade the intersection or union of a system of inequalities. The user interface provides numerous sample equations along with comments and suggestions for setting screen parameters in order to achieve best results. The user interface for all program modules is simple and intuitive. After an equation is graphed, users can plot a point on a graph near the mouse cursor and view the x-y coordinates of the plotted point. In addition to plotting points, relative minimum points, relative maximum points, x-intercepts and intersection points can be found with simple mouse control clicks. A Help menu gives users a quick summary of all of the magical mouse control clicks. Of course, all graphs can be copied to the clipboard and pasted into another document. Go to www.mathteachersresource.com to view multiple screen shots of the program’s modules. Click the ‘learn more’ button in the TRIGONOMETRIC FUNCTIONS section (or click here). Teachers will find useful comments at the bottom of each screen shot.

Why is Division by Zero Forbidden?

Zero1What is 5/0? When I ask my beginning algebra students that question, the most popular incorrect answer they give me is 0. The next most popular incorrect answer is 5. After repeated reminders by their math teachers, students eventually learn that 5/0 is undefined, has no value, or is meaningless. (I once told a class of 9th grade algebra students that if they use their calculator to divide a number by zero, the calculator will explode in their face. One student looked at me and said, “Really?” I forgot how literal 9th graders can be. At least I got the student’s attention.) When I ask college algebra, trigonometry, statistics, technical math or calculus students why a number divided by zero is undefined, I either get an answer that begs the question or students say it’s simply a mathematical fact that they learned in a previous course.

So how do you explain division by zero? There are two ways. The first depends on a basic understanding of division of two numbers. It goes something like this: Students learn that a / b = c if and only if a = b*c. Therefore 986 / 58 = 17 because 58*17 = 986. Is 5 / 0 = 0? No, because 0 * 0 ≠ 5.   Is 5 / 0 = 5? No, because 0*5 ≠ 5. Since 0 times any number never equals 5, 5 / 0 is NOTHING or undefined. So what about 0 / 0? The problem here is that 0 times any number equals 0, and therefore 0 / 0 would have infinitely many answers, which in turn would be rather confusing. So we say that any number divided by zero is undefined.

The second explanation involves a deep mathematical insight from the 12th century Indian mathematician and astronomer, Bhāskara II, who developed the basic concepts of differential calculus. The 17th century European mathematicians, Newton and Leibniz, independently rediscovered differential calculus. This second explanation due to Bhāskara II goes something like this. Consider a single piece of fruit. If we divide 1 piece of fruit by ¼, we get 4 pieces of fruit. If we divide 1 piece of fruit by 1/10,000, we get 10,000 pieces of fruit. As 1 is divided by smaller and smaller numbers that approach zero, the number of pieces of fruit increases without bound. Therefore 1/0 = ∞ and, in general, n/0 = ±∞ if n does not equal 0.

Bhāskara II, Newton and Leibniz discovered the revolutionary concept of a limit of a function at a point, which enabled them to get around the problem of division by zero. Once that problem was solved, it was a relatively easy task to find methods to calculate a rate of change over a time interval of length zero, rate of change over a fleeting instant of time, or rate of change over a flux of time, as Newton would say. In The Ascent of Man, Dr. Bronowski tells the viewer, “In it, mathematics becomes a dynamic mode of thought, and that is a major mental step in the ascent of man.” Differential calculus is all about the mathematics of variable rates of change. I should mention that differential calculus students learn a slick technique for finding the limiting value of an x-variable expression as x approaches a constant k and the value of the expression when x = k is 0/0 or ∞/∞.

The graphic below shows the graphs of the functions y = 2Sin(x) and y = 2Csc(x) along with its vertical asymptotes. The graphs are color coded green, blue and red respectively. Because Csc(x) = 1 / Sin(x), the Csc(x) function is undefined at precisely those values of x where Sin(x) = 0. It’s interesting and fun to advance a trace mark cursor on the graphs of these functions. On both graphs, the horizontal velocity of the trace mark is constant, but the vertical velocity of the trace mark changes as the value of the x changes. As x approaches a vertical asymptote, the trace mark races towards ± ∞. Differential calculus gives us a complete understanding of the phenomena of the moving trace cursor.

Blog12graphic

The above graphic, created with the program Basic Trig Functions, is offered by Math Teacher’s Resource. The equations entered into the program were: y = 2Sin(x), y = 2Csc(x), and Sin(x) = 0. Go to www.mathteachersresource.com to view multiple screen shots of the program’s modules. Click the ‘learn more’ button in the TRIGONOMETRIC FUNCTIONS section. Teachers will find useful comments at the bottom of each screen shot.

Differential calculus is not only interesting and fun, but it can also be a stress reliever. At least it was for Omar Bradley, the famous American WWII general. He took a calculus book with him on battle campaigns, and when opportunity allowed, he worked differential calculus problems to relieve the stress of a battle campaign.