With so much emphasis being placed on “making math meaningful”, I thought it appropriate to share my introduction to linear functions.

When introducing linear functions to my Math 10C students, I have a discussion with them describing that day's journey from my home to school.  I quickly sketch a coordinate plane on the “bat-board” with my “bat-marker”, the origin representing my point of departure.  Negotiating the trip in the “Bat-mobile”, this journey is divided into various segments, each characterized by the average velocity from one point to the next.  The students are involved throughout this process, contributing to how the graph should be labeled, first in terms of minutes followed by an “hourly” representation.   An example of these graphs appears below.

Bat-graphs:  Set 1

Minutes

A_m(0, 0)…..B_m(6, 5)….. C_m(10, 10.3333)….. D_m(30, 47)….. E_m(36, 52)

Hours

A_h(0, 0)…..B_h(0.1, 5)….. C_h(0.1667, 10.3333)….. D_h(0.5, 47)….. E_h(0.6, 52)

 

As previously mentioned, students are actively involved in determining the “time” and “displacement” components for each “leg” in the journey.  From these values, ordered pairs representing the “start point” of each segment are determined and documented.  The slopes of these individual segments is calculated and eventually linked to the velocity travelled in each leg of the journey.  This information is summarized in Table 1 below, with time being measured in “hours” and displacement in “km”.

Bat-table 1

Segment…..Domain (Interval)…..Range (Interval)…..Slope…..Velocity (^km/_h)

….A_hàB_h……..…[0, 0.1)……….….……..[0, 5)…….….….⁵/_0.1………….50

….B_hàC_h……..…[0.1, 0.1667)…….....[5, 10.3333)…....^5.3333/_0.0667…...80

….C_hàD_h……..…[0.1667, 0.5)……....[10.3333, 47)......^36.6667/_0.3333….110

….D_hàE_h……..…[0.5, 0.6]……….….…..[47, 52]……...….⁵/_0.1…………..50

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Once students have a grasp of the concepts of “domain”, “range”, “slope” and “velocity”, further discussion eventually ensues, this time focused on representing each segment algebraically, first with equations in “slope-point form” and from there, rewriting those equations in “slope y-intercept form” and finally “general form”.  These are requirements set forth in our curriculum and are very easy transitions for students to make. 

The “start points” for each segment appear below once again, this time with fractional values representing time and displacement.  Beneath those points, Table 2 summarizes the equations representing each leg in the journey.

Hours

A_h(0, 0)…..B_h(⁶/₆₀, 5)….. C_h(¹⁰/₆₀, ³¹/₃)….. D_h(³⁰/₆₀, 47)….. E_h(³⁶/₆₀, 52)

 

Bat-table 2

Segment…..   Start Point……. Slope (^km/_h)…..Bat-equation

….A_hàB_h…….. A_h (0, 0)…………..… 50…………….s = 50t

….B_hàC_h……… B_h (¹/₁₀, 5)………… 80…………...s – 5 = 50(t – ¹/₁₀)

….C_hàD_h……… C_h (¹/_6, ³¹/₃)……… 110………..... s – ³¹/₃ = 50(t – ¹/₆)

 ….D_hàE_h……... D_h (¹/_2, 47)………… 50…………... s – 47 = 50(t – ¹/₂)

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A very quick reference is also made to the average overall velocity for the entire trip.  Students quickly determine this to be represented by the slope of the line segment joining the points A_h (0, 0) and E_h (³⁶/₆₀, 52); this information is summarized in Table 3 and Graphs 1 and 2 of Set 2, both shown below.

Bat-table 3

Segment…..Start Point…..End Point……... Slope (^km/_h)…..Bat-equation

….A_hàE_h…... A_h (0, 0)……..E_h (³⁶/₆₀, 52)….…⁵²/_(36/60)….......s₁ = ²⁶⁰/₃(t)

….F_hàJ_h…... F_h (⁴²/₆₀, 52)…..J_h (⁷⁸/₆₀, 0)……^-52/_(36/60)…......s₂ = ^-260/₃(t)

 

 Bat-graphs:  Set 2

 

The point “F_h (⁴²/₆₀, 52)” represents departure from school at the end of the day.  Had the graph been drawn “to scale”, this point would have occurred much later in the day.  In order to produce a graph displaying more “definition” in the slopes of neighboring line segments, a “stop-over” of only 6 minutes was added between arrival AT and departure FROM school.

The graphs contained in Set 3 below illustrate the “round-trip” TO and FROM school on any given day. Graph 2 in Set 3 below includes line segments representing average velocity in both directions; these are shown in black.

Bat-graphs:  Set 3

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Here's a picture of the physics teacher and myself scaling the wall................. 

.........I'm Batman.

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Further Explorations

The following is an additional exploration/investigation that can be made with Math 30 students; in this section, I’m doing nothing more than throwing out some ideas.

 Refer to the graphs illustrated below in Set 4 for the following “dialogue”.

Bat-graphs:  Set 4

 

In Graph 1 of Set 4 shown above, the “6-minute stop-over” between arrival AT and departure FROM my school was removed.  The negative sloping line representing average velocity for the trip home has been slid left-ward by “6 minutes”, resulting in an absolute-value function defined as follows:

Bat-graph 1:  Set 4

 s₃(t) =  – |²⁶⁰/₃(t – 0.6)| + 52

 

Graphs 2 and 3 in Set 4 above show my attempt at modeling this “round trip” as a cyclic function; the defining equations (in function notation) and reasoning are summarized below.

Bat-graph 2:  Set4

s₃(t) =  – |²⁶⁰/₃(t – 0.6)| + 52  à  black-solid

s₄(t) =  52|sin(^2π/_2.4)t|  à  red-dotted

 

Bat-graph 3:  Set4

s₃(t) =  – |²⁶⁰/₃(t – 0.6)| + 52  à  black-solid

s₅(t) =  52 sin²(^2π/_2.4)t  à  blue-dotted

After considering the cyclic functions illustrated in Graphs 2 and 3 above, it was decided that the function defined as “s₅(t) =  52 sin²(^2π/_2.4)t”  from Graph 3 is a more accurate representation of the situation being described.  Incidentally, Graph 3 also provides an opportunity for students to investigate another trigonometric identity from a geometric perspective.

For an introductory calculus class, a rich discussion focusing on accelerated motion, stationary points and the “Mean Value Theorem” would lead very nicely to determining values of time for which instantaneous velocity of the function              “s₅(t) =  52 sin²(^2π/_2.4)t” equals the slope of the absolute value function defined as “s₃(t) =  – |²⁶⁰/₃(t – 0.6)| + 52”.  

Finding the derivative of “s₅(t) =  52 sin²(^2π/_2.4)t” will enable students to determine instantaneous velocity at any point on that function.   Calculating this function’s derivative will also show a use for a double-angle identity introduced in Math 30, as shown below.

 

s₅(t) =  52 sin²(^2π/_2.4)t

s₅^ı (t) =  104 sin(^2π/_2.4)t*cos(^2π/_2.4)t

s₅^ı (t) =  (^104π/_2.4)*2 sin(^2π/_2.4)t*cos(^2π/_2.4)t

 

Simplified version of s₅^’(t) follows:

s₅^ı (t) =  (^104π/_2.4)* sin(^2π/_1.2)t

 

s₅^ı (0.3) is evaluated below:

s₅^ı (0.3) =  (^104π/_2.4)* sin(^2π(0.3)/_1.2)

s₅^ı (0.3) =  (^104π/_2.4)* sin(^π/₂)

s₅^ı (0.3) =  (^104π/_2.4)(1)

s₅^ı (0.3) =  136.14 ^km/_h

 

Questions that will arise………

Why is this velocity so much higher than the average overall velocity of 86.67 ^km/_h? 

Is there a position on the function “s₅(t) =  52 sin²(^2π/_2.4)t” where the slope of its tangent line equals that of the absolute value function?

What changes to the graph of  “s₅(t) =  52 sin²(^2π/_2.4)t” would result in a closer approximation to the average velocity represented on the absolute value function? 

………..and many others.

 

As demonstrated above, this derivative can be evaluated for many other values of “t” and compared to the overall average velocity travelled from home to school, and back again.  Once again, students can discuss ways in which the given cyclic function is an appropriate (or inappropriate) estimation of the absolute value function, offer reasons for their beliefs and provide various improved versions along with justifications.  Integration of trigonometric functions could be introduced here as well by estimating the area, using the Riemann Sum, beneath the function defined by “s₅(t) =  52 sin²(^2π/_2.4)t”, then drawing comparisons to that of the absolute value function over the interval [0, 1.2], both bounded by the horizontal line "s = 0".

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The graphs in Set 5 below show my journey to and from school during a five-day week.  From this, extrapolating total distance driving over a 20-week semester would be very easy to calculate.  Considering the fact that the “Bat-mobile” burns 10 liters of fuel for every 100 km driven with the cost of fuel being $1.10 per liter, total cost of fuel per semester could easily be determined using related rates.

Bat-graphs:  Set 5

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As mentioned before, I introduce linear functions to my Grade 10C bat-students using this type of scenario.  Domain, range and slope are all easily determined and defined in a meaningful way using this type of example.  “Slope-point form” is the initial focus in terms of equations as this form is easily connected to the notion of ^Δy/_Δx.  Having several segments representing various velocities along the way provides the opportunity for students to see various patterns that are important in building equations, as well as in determining the domain and range for each leg of the journey.

There are also many further “explorations” available with this simple scenario that can serve to initiate all sorts of mathematical discussion and investigations with Grade 10, 11 or 12 students.

 The End

In a previous blog post, I attempted to describe “Russian Peasant Multiplication” in which several “versions” of the product of “10” and “7” were illustrated from a geometric perspective.  This method involved using these two factors to represent the dimensions of a rectangle and hence, the product of the two numbers represented the rectangle’s area.  The dimensions of the rectangle were systematically manipulated, ultimately resulting in the very long and slender rectangle measuring “70 units” in length and “1-unit” in width.  Since the desired width of this final rectangle was “1-unit”, its area and length were characterized by the same value.  Hence, the product of the original two factors was considered equal to the length of the final rectangle having width of “1-unit”.

The factors of "10" and '7" will once again be investigated below, this time from the perspective of “Egyptian Multiplication”.   This form of multiplication takes on a slightly different strategy and is described below.

Let us first consider the factors of "10" and "7" as representing the dimensions of a rectangle, illustrated below.

 

Diagram 1  

When calculating “area” of such a rectangle, the diagram above can be deemed to contain “10 columns” of “squares”, with each column containing a “group of 7”.  Egyptian multiplication involves partitioning these “10 groups of 7” into a collection of “smaller” groupings of “7”, each of which is characterized by successively increasing powers of “2”.  The first example is illustrated in Table 1 below. 

 

Table 1

………..Column 1…...Column 2…...Column 3…..Column 4

Row 1…..(2⁰)(7)……….…(1)(7)………..…..7

Row 2…..(2¹)(7)……….…(2)(7)……….....14………...….14

Row 3…..(2²)(7)……….…(4)(7)……….....28

Row 4…..(2³)(7)……….…(8)(7)……….....56…….…..….56

Row 5…..(2⁴)(7)………..(16)(7)……...….112

Row 6…..(2⁵)(7)……..…(32)(7)……..…..224

Row 7.….(2⁶)(7)…….….(64)(7)………....448…....…..______

………………………………………………………….……..70

Each row illustrated above shows “groupings of 7”; these groupings are all based on increasing powers of “2”.  The premise behind this method of multiplication is to identify “powers of 2” which, when summed, equal the total “width” requirement of “10” as shown in Diagram 1 above.  Row 5 from Table 1 takes us beyond our desired “width of 10”.  Row 4, directly above that, gets us close to “10”, providing us with ”2³ or 8 groups of 7”; row 2 provides an additional “2-units of width”, thereby fulfilling the required “10-unit” total.  The individual products arising from these two rows, shown in Column 4 (Table 1) are then summed to arrive at the desired product.  A slightly varied format of this process is shown below, accompanied by its geometric representation.

(2¹)(7) + (2³)(7)

(2¹ + 2³)(7)

(2 + 8)(7)

(10)(7)

 

Diagram 2 

 

The slender rectangle to the left has an area of “14”; the area of the rectangle to its right is “56”.  When summed, these areas total “70”, which is the desired product.

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 It may be beneficial to run this scenario once again here, this time with “7 columns” of “squares”, with each “column” containing a group of “10”; this is illustrated in Diagram 3 below.

Diagram 3  

This time, these “7 groups of 10” will be partitioned into a collection of “smaller” groupings of “10”, once again characterized by successively increasing powers of “2”.  This “partitioning” is illustrated in Table 2 below.

 

 Table 2

………..Column 1…..Column 2…..Column 3…..Column 4

Row 1…..(2⁰)(10)…….…(1)(10)………....10……………10

Row 2…..(2¹)(10)…….…(2)(10)…….…...20……………20

Row 3…..(2²)(10)…….…(4)(10)….……...40……………40

Row 4…..(2³)(10)……..…(8)(10)………...80………...._____

…………………………………………..……………….…..70

Each row illustrated above shows “groupings of 10”; these groupings are all based on increasing powers of “2”.  The premise here is once again to identify “powers of 2” which, when summed, equal the total “width” requirement (that being “7-units” in this example).  Row 4 from Table 2 takes us beyond the required “width of 7”; row 3, directly above that, provides “4” of the “7” required columns, with rows 2 and 1 making up the balance.  The individual products arising from these three rows, shown in Column 4 (Table 2) are once again summed to arrive at the desired product.  A slightly varied format of this process is shown below, accompanied by its geometric representation.

 

(2⁰)(10) + (2¹)(10) + (2²)(10)

(2⁰ + 2¹ + 2²)(10)

(1 + 2 + 4)(10)

(7)(10)

Diagram 4 

 

 The left-most rectangle above has an area of “10”, followed by rectangles having areas “20” and “40” respectively.   When summed, these areas total “70”, which is the desired product.

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It may be of some interest to explore Egyptian Multiplication from a “Base 3” (or other) perspective.  I’ve included one such scenario below, once again built upon our “10” by “7” rectangle.

 

Diagram 5   

 

Table 3

………….Column 1…..Column 2…..Column 3…..Column 4

Row 1…..(3⁰)(7)……….…(1)(7)………..…..7………...….7

Row 2…..(3¹)(7)……….…(3)(7)………......21

Row 3…..(3²)(7)……….…(9)(7)……..…....63……….….63

Row 4…..(3³)(7)…………(27)(7)………....189…….…._____

……………………………………….…………..…..….…..70

 

An alternate perspective of Table 3 is shown below, along with its geometric interpretation.

 (3⁰)(10) + (3²)(7)

(3⁰ + 3² )(7)

(1 + 9)(7)

(10)(7)

Diagram 6  

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Exponential Functions Perspective

Since “powers of 2” were utilized in Egyptian Multiplication above, an opportunity presents itself to relate the “areas” explored earlier to the exponential function        “f(x)=(2)^x” and slight variations thereof.  A few such examples are illustrated below.

We will once again consider the factors of “10” and “7” as representing the dimensions of a rectangle.  The first tab below shows the original “10 by 7” rectangle; the “partitioned” version is shown in the second tab.

Illustration 1 

 

The following tabs contain exponential functions describing the “partitioned” version of our rectangle above.  Function 1 below measures the “width” of each of the partitioned rectangles; function 2 measures the “area” of each of the partitioned rectangles.

Graphs: Set 1  

Function 1:  f(x)=(2)^x

The vertical lines passing through points “B” and “D” on Function 1 represent the number of factors of “2” used earlier in Egyptian Multiplication; the horizontal lines passing through those same points represent the widths of “2” and “8”, respectively, of the partitioned rectangles.  The sum of these values represents the total width, “10”, of the original rectangle.  This width is then multiplied by “7” to obtain the rectangle’s area; this area represents the product of the original two factors, “10” and “7”.

Function 2:  g(x)=7(2)^x

While the horizontal composition of Function 2 is identical to that of Function 1, “g(x)= 7(2)^x” has been “stretched” vertically by a factor of “7” when compared to “f(x)=(2)^x”.  This factor of “7” represents the length of the rectangles referred to earlier; hence, the values of “g(x)” represent “areas” of each rectangle in question.  The horizontal lines passing through “B₇” and “D₇” therefore represent areas of “14” and “56”; these values correspond to the areas of the partitioned rectangles in Illustration 1 above and, when added, produce the product of the original factors “10” and “7”.

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Illustration 2 

 

The following tabs contain exponential functions describing the “partitioned” version of our rectangle above.  Function 1 below measures the “width” of each of the partitioned rectangles; function 3 measures the “area” of each of the partitioned rectangles.

Graphs: Set 2 

Function 1:  f(x)=(2)^x

The vertical lines passing through points “B” and “D” on Function 1 represent the number of factors of “2” used earlier in Egyptian Multiplication; the horizontal lines passing through those same points represent the widths of “2” and “8”, respectively, of the partitioned rectangles.  The sum of these values represents the total width, “10”, of the original rectangle.  This width is then multiplied by “7” to obtain the rectangle’s area; this area represents the product of the original two factors, “10” and “7”.

Function 3:  h(x)=10(2)^x

While the horizontal composition of Function 3 is identical to that of Function 1, “h(x)= 10(2)^x” has been “stretched” vertically by a factor of “10” when compared to “f(x)=(2)^x”.  This factor of “10” represents the length of the rectangles referred to earlier; hence, the values of “h(x)” represent “areas” of each rectangle in question.  The horizontal lines passing through “A_t”, “B_t” and “C_t” therefore represent areas of “10”, “20”, and “40” respectively; these values correspond to the areas of the partitioned rectangles in Illustration 2 above and, when added, produce the product of the original factors “7” and “10”.

 

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Logarithms

 Introduced by John Napier in the early 17^th century, the original purpose of  logarithms was to simplify calculations by "replacing" multiplication of numbers with "addition".  Originally referred to as "artificial numbers", Napier eventually coined the term "logarithm", a compound word with Greek origins: "Logos" and "artihmos", meaning "ratio" and "number" respectively.    

Napier developed tables of natural logarithms which, at the time, had n explicit connection to exponential functions; we now consider exponential functions as inverses to their logarithmic counterparts.  Once again, no direct connection was made between Napier's natural logarithms and the number “e" until much later on. Nevertheless, Napier did employ "anti-logarithms" in his work even though a formal definition of “e^x” had yet to be made.

 

The following links provide additional information on the origins of "Euler's constant".

 Jacob Bernoulli

 

Leonhard Euler

 

e (mathematical constant)

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“Calculating Products by Adding”

I will once again call upon my “10” by “7” rectangle below as a reference in my attempt to explain the process of using logarithms to multiply numbers. 

 

Graphs of Logarithms

Each set of graphs below each illustrate the various logarithmic functions along with their mode of usage in calculating the product of two numbers;  the second graph in each set is drawn on a larger scale than the first  in order to accommodate our desired product of “70”. 

 

Natural Logarithm:  f(x) = ln(x)

…….Point…………..Addends/Sum………….Anti-Log (factors/product)

A:  f(7)=ln(7)…………..1.9459………………………….e^1.9459 = 7

B:  f(10)=ln(10)……….2.3029……….…………………e^2.3029 = 10

C:  f(70)=ln(70)……....4.2488………………………….e^4.2488 = 70

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Common Logarithm:  g(x) = log₁₀(x)

…….Point…………..Addends/Sum………….Anti-Log (factors/product)

A:  g(7)=log(7)…………..0.8451………………………….10^0.8451 = 7

B:  g(10)=log(10)……….1.0000………………………….10^1.0000 = 10

C:  g(70)=log(70)……....1.8451………………………….10^1.8451 = 70

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Base 7 Logarithm:  h(x) = log₇(x)

…….Point…………..Addends/Sum………….Anti-Log (factors/product)

A:  h(7) = log₇(7)….…...1.0000………………..……….7^1.0000 = 7

B:  h(10) = log₇(10)…...1.1833………………………….7^1.1833 = 10

C:  h(70) = log₇(70)…….2.1833……………………..….7^2.1833 = 70

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Base 2 Logarithm:  j(x) = log₂(x)

 

…….Point…………..Addends/Sum………….Anti-Log (factors/product)

A:  j(7) = log₂(7)….…...2.8074………………………….2^2.8074 = 7

B:  j(10) = log₂(10)…....3.3219………………………….2^3.3219 = 10

C:  j(70) = log₂(70)…….6.1293………………………….2^6.1293 = 70

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As mentioned earlier, John Napier used a “Table of Natural Logarithms” in his calculations by first deterniming the "addends" and then working "backwards" to find the "anti-log" of the sum of those addends.  In my examples above, I've described each "anti-log" from the perspective of "exponential functions", a luxury not available to Napier; he relied solely on the tables of logarithms. Many years after Napier's contributions, but before the advent of the electronic calculator, the slide rule was developed and used in a manner similar to that employed in reading the log tables.

 

The following link provides additional information on the use of slide rules.

Slide Rule Instructions           

 

Many would consider the examples and procedures outlined here completely useless and a colossal waste of time; I beg to differ.  I believe there IS value in exploring notions such as these and relating them to one another.  Additional explorations can be made within the context of "Egyptian Multiplication" using bases other than "2"; students can investigate why a "base 2" approach is superior to one using "base 4" factors, for example.  These additional scenarios can be represented as exponential functions, allowing comparisons to be drawn between those.  Napier's method of multiplication can be further explored with new examples and by applying a variety of different bases of logaritms to those; student will ultimately discover the "laws of logarithms" for themselves.

As an added bonus, students may learn some of the history behind much of the mathematics they are learning and gain an appreciation for the subject along with its many intricacies.

 

The End

As a Math teacher, I have always been aware of the value in connecting mathematical concepts and notions to the world in which my students live.  There are many individuals on Twitter who believe the same, some of whom provide great examples from which to build.  Unfortunately, there are many others who “talk a big game” but provide nothing to back up their “bravado”. 

I choose to continue “thrashing” around in my labyrinth of confusion; a small sample of my tangled snarl of "pedagogy" appears below.

The article that follows is a collection of a few examples I’ve used over the years at “opportune” times in the classroom.  This medium has provided me with the motivation to gather these examples and present them together as, what I hope to be, a cohesive package.

Although much of the discourse that follows is in general terms, I will relate certain aspects specifically to my home province of Alberta (because I want to).

 

A quick reference to The History of Agriculture in Alberta reveals that the farming industry in this province began to gather momentum in the mid-1870’s.  The creation of a large-scale “grid system” served to delineate the prairie land into “townships”, “sections”, and “quarter-sections”.  These divisions allowed for hardy souls of the time to homestead the land in the hopes of making a better life for themselves and their families. 

Surveyor's Basic Tools

For additional information relating specifically to the Province of Alberta, right click on the following link: 

Land Surveying History: 1905 - The Birth of Alberta

Townships measured 6-miles by 6-miles and were aligned horizontally with base-line latitudes and longitudinal “range lines”.  Base-lines occur every 24 miles and hence, four townships occur within each pair from north to south.  Since the range lines converge at the North Pole (they do not run parallel), very slight adjustments were required to account for this.  These adjustments led to each township being slightly narrower on its north boundary compared to that of its south.  A “correction line”, running parallel to and mid-way between base-line latitudes, serves as a periodic “correction” in the trapezoidal shape of the townships. 

The images below were taken from Google Earth.  The first image shows one township with each of its 36 sections outlined in black; the grey lines delineate the road allowances.  These road allowances were designed to run in the north-south direction each mile and east-west at two mile intervals; this “road system grid” allowed for relatively easy access to each quarter-section. 

The second tab below shows two sections of land; the grey lines around the perimeter denote the road allowances.  Each quarter-section measures  ¹/₂  mile north-south by  ¹/₂  mile east-west.

Google Earth

 Right click on the following links for additional information:

Dominion Land Survey

Surveying: History, Techniques and Equipment

Section: United States Land Surveying

The diagrams contained in the tabs that follow represent two sections of farmland similar to that seen in the images from Google Earth.  Important to note here is that there are 5280 feet comprising ONE mile and therefore, each quarter-section measures 2640 feet by 2640 feet.  Of equal significance here is in knowing that ONE rod measures 16.5 feet in length, each half-mile is equivalent to 160 rods and ONE ACRE is an area measure equaling 1 ROD by 160 RODS.  Therefore, each quarter-section of farmland measures 160 rods by 160 rods, producing an AREA of 160 acres.

 Quarter-Sections Measurements

 

The remainder of this article will briefly describe, from a mathematical perspective, a typical growing season in the life of a farmer as it pertains to a single quarter-section of farm-land; I’ll call him Jed. 

 Jed's Tractors

Jed pulls a 50 foot wide air seeder at an average ground speed of 5 miles per hour (Graph 1).  Considering the fact that ONE acre equals 1 rod (16.5 feet) by 160 rods (¹/₂ mile), each pass Jed completes length-wise on a quarter-section of his farm-land constitutes 3 acres (small overlap considered).  It can therefore be stated that Jed seeds at an average rate of 6 acres per mile (Graph 2).

 

Graph 1:  miles/hour seeding

m(t) = 5t

where "t" represents time measured in hours

 

Q_s(5.33, 26.67) 

This point indicates that approximately 5 hours 20 minutes

will have been spent traveling nearly 27 miles.

 

 

 

Graph 2:  acres/mile seeding

a(m) = 6m

 

Q_s(26.67, 160) 

This point indicates that nearly 27 miles

will have been travelled to seed 160 acres.

 

 

 

 Differential Calculus Perspective

The slopes of the functions shown in Graphs 1 & 2 above describe the “rate of change” occurring within each relationship described.  From a “Differential Calculus” perspective, the rates of change of each function are described as its “derivative”.  Table 1 below provides a brief summary.

 Table 1

……Function………..Slope………..Derivative

…...m(t) = 5t……......…5………….….^dm/_dt = 5

…...a(m) = 6m….…..…6………….….^da/_dm = 6

 

Graph 3, directly below, provides an illustration of these rates of change.

 Graph 3:  rates of change

^da/_dm = 6  à  blue-solid

^dm/_dt = 5  à  blue-dotted

^da/_dt = 30  à  black-solid

The rates described above can be referred to as “related rates”; they are occurring concurrently and are describing a common theme, only from different perspectives.  It is easy to reason that if Jed’s ground speed is 5 miles/hour and he is seeding 6 acres/ mile that he must then be seeding 30 acres per hour.

 

 

 

 
The “Calculus” perspective of the relationship between the three rates described above is summarized below and is referred to as the “chain rule”.

^da/_dt = (^da/_dm) ( ^dm/_dt )

where

^da/_dt represents acres seeded per hour

^da/_dm represents acres seeded per mile

^dm/_dt represents miles travelled per hour

 Therefore

^da/_dt = (6)(5)

^da/_dt = 30

The "Anti-Derivative" of  "^da/_dt"  is shown below:

a(t) = 30t + C

The value of "C" in the function above represents the number of acres seeded at "0 hours"; this value, itself, must therefore equal "0" in our scenario.

In the context of integral calculus, the value of "C" in the function above is referred to as the "constant of integration". 

 

Graph 4 below illustrates the relationship between acres seeded over hours.  The slope of this function is not surprisingly “30”, the product of the slopes of the functions shown in Graphs 1 & 2. 

 

Graph 4:  acres/hour seeding

a(t) = 30t

Q_s(5.33, 160) 

This point indicates that approximately 5 hours 20 minutes

will have been spent seeding 160 acres.

Over the course of the growing season, Jed’s barley crop will grow nicely and eventually ripen enough to harvest.

 

 

 

 

 If all goes well, Jed can pull his combines into the field and begin harvesting in August. 

 Jed has 25 foot wide straight cut headers mounted on each of his combine harvesters; with each ¹/₂ mile length travelled in his combine, Jed will have harvested around 1.5 acres.  We can therefore say that Jed’s rate of harvest is 3 acres per mile; his ground speed while harvesting is 4 miles per hour.

The four graphs contained in the tabs below illustrate the relationships between acres, miles and hours required to harvest one quarter-section of Jed’s farm.

 Graphs of harvest:   "t" again represents time in hours.

Graph 1:  Miles/hour  à   m(t) = 4t  à  Q_h (13.33, 53.33)

Graph 2:  Acres/mile  à  a(m) = 3m  à  Q_h (53.33, 160) 

 Graph 3:  Rates of Change....^da/_dm = 3  à  blue-solid

………..…................................ ^dm/_dt = 4  à  blue-dotted

……….................................…..^da/_dt = 12  à  black-solid

Graph 4:  Acres/hour  à  a(t) = 12t  à  Q_h (13.33, 160)

A crop’s “yield” is most often measured in “bushels per acre”.  The bushel, approximately equal to 36 liters was historically used as a “volume” measurement for dry goods.  Bushels per acre is an additional rate that could be easily analyzed in the context of "related rates" from the "Calculus perspective" above.   

 

 

 

Bushel Baskets

The first two pictures below show what grain elevators used to look like.  They have given way over the past several years to more modern versions, one of which is shown in the fourth picture below.

 Grain Elevators

 It is worthwhile noting that not all of Jed's barley will be shipped oversees; a portion may be sold domestically to cattle producers.  If his barley is deemed to be of "malting" quality, it can be sold to domestic (or U. S. based) maltsters where it is transformed into a tasty nectar.  

Quite often, there is also a local market for the straw produced.  If Jed can strike a deal with a local cattle producer, he may decide to drop the straw from his combine rather than chopping it into mulch.  In this case, the straw from Jed’s crop will be made into bales and eventually become bedding and roughage for locally grown livestock; the cost of shipping great distances has a very negative effect on profit margins.

 Straw bales

There are many other avenues related to the topic above that students can explore; I’ve only provided a catalyst to initiate the thought process.  To those of you who put the “poo-poo” on what I’ve done here……..make sure to not use anything you’ve read or learned here.  To do so, would be hypocritical.

 

 

…………and now for my reward.

The End

For the most part, trigonometric identities are proven using a “2-column” approach, even though many of these identities lend themselves very well to a more visual analysis.  This can be easily accomplished by “splitting” the identity itself into two distinct functions, graphing those functions and then proceeding from there. 

Keeping in mind that several identities do not lend themselves very well to this approach, an analysis of the geometric representations of those that DO can provide students with a more thorough understanding of what identities are and why they can be employed to simplify various scenarios. 

The identity chosen here will be one that lends itself very well to a “systems of simultaneous equations” approach.  What follows will serve as a reinforcement of the process of systems of equations as well as connecting identities to a line of reasoning already known to students.

 

The example I’ve selected to explore here is the Pythagorean Identity shown below.

sin² (x) + cos² (x) = 1

Rearranging this identity and splitting the result into two distinct functions results in the following.

sin² (x) =  - cos² (x) + 1

Therefore

f(x) = sin² (x)

g(x) = - cos² (x) + 1

 

-------------------------------------------------------------

Before showing students the graphs of these functions, it is useful to first understand WHY the graphs of “sin² (x)” and “cos² (x)” appears as they do.  A brief explanation of this is shown below.

 

Graph 1 below illustrates the function defined as “h(x) = sin (x)”.

Graph 1

Points of "significance" occuring on the function “h(x) = sin (x)” appear in Table 1 below and are again listed beneath Graph 2.  The last column in this table show the values of “sin² (x)”; these points are identified as red circles in Graph 2.

Table 1

x-value……..…..sin (x) ……….. sin² (x)

….0……….…..…….0………………...0….

….^π/₆……….……... ¹/₂…………….. ¹/₄…

….^π/₂……….……....1………………...1….

….^5π/₆……….….... ¹/₂…………….. ¹/₄…

…..π…….…………...0………….…….0….

….^7π/₆……….……. ^-1/₂………….... ¹/₄….

….^3π/₂……….……..-1……………..….1….

….^11π/₆……….…... ^-1/₂…………….. ¹/₄…

…..2π……….……....0………..……….0….

 

Graph 2

 

Points labeled on “h(x) = sin (x)”

A(0, 0), B(^π/₆, ¹/₂), C(^π/₂, 1), D(^5π/₆, ¹/₂)

E(π, 0), F(^7π/₆, ^-1/₂), G(^3π/₂, -1), H(^11π/₆, ^-1/₂), I(2π, 0)

Point labeled on “f(x) = sin² (x)”

A(0, 0), B₂(^π/₆, ¹/₄), C(^π/₂, 1), D₂(^5π/₆, ¹/₄)

E(π, 0), F₂(^7π/₆, ¹/₄), G₂(^3π/₂, 1), H₂(^11π/₆, ¹/₄), I(2π, 0)

 

The points labeled as red circles in Graph 2 represent points on the function                 “f(x) = sin² (x)”.  Graph 3 below shows both functions, “h(x) = sin (x)” and “f(x) = sin² (x)”.  Graph 4 below that illustrates the function defined as “f(x) = sin² (x)” on its own.

 

Graph 3

h(x) = sin (x)  à black

f(x) = sin² (x)  à  red

-------------------------------------------------------------

 

Graph 4

f(x) = sin² (x)

-------------------------------------------------------------

The behavior of the function defined as “j(x) = cos² (x)” can be understood using the same reasoning process as was illustrated in Table 1 above and the subsequent graphs. The functions defined by "m(x) = cos (x)" and “j(x) = cos² (x)” are shown directly below in Graph 5.

Graph 5

m(x) = cos (x)  à  black

j(x) = cos² (x)  à  grey-dotted

-------------------------------------------------------------

 

Graph 6 below provides an isolated illustration of the function “j(x) = cos² (x)”.

Graph 6

j(x) = cos² (x)

--------------------------------------------------------------------------------------------

 

It is now time to get back to the original identity and the process of proving it geometrically.

sin² (x) + cos² (x) = 1

We previously rearranged this identity and rewrote the result as two distinct functions, as shown below.

sin² (x) =  - cos² (x) + 1

Therefore

f(x) = sin² (x)

g(x) = - cos² (x) + 1

 

Graph 7 below illustrates the functions “f(x) = sin² (x)” and “j(x) = cos² (x)” on the same grid.

 

Graph 7

 

f(x) = sin² (x)  à  red

 j(x) = cos² (x)  à  black-dotted

-------------------------------------------------------------

 

Graph 8 illustrates a slightly modified version of the contents shown in Graph 7.

The function “f(x) = sin² (x)” is unchanged from Graph 7 while the function “j(x)” has been reflected in the “x-axis”, producing a new function “k(x) = - cos² (x)”; this new function, “k(x)”, is shown in blue below.

 

Graph 8

f(x) = sin² (x)  à  red

 k(x) = - cos² (x)  à  blue

 j(x) = cos² (x)  à  grey-dotted

-------------------------------------------------------------

 

The final manipulation in this geometric proof involves adding a constant of “1” to the function “k(x) = - cos² (x)”; this addition of “1” translates the function “k(x)” vertical upward by “1” unit.

This vertical translation results in the function “g(x) = - cos² (x) + 1”, shown superimposed over-top of the function defined as “f(x) = sin² (x)” in Graph 9 below.

 

Graph 9

f(x) = sin² (x)  à  red

g(x) = - cos² (x) + 1  à  blue-dotted

-------------------------------------------------------------------------------------

 

It is very obvious from the illustration in Graph 9 that the functions defined as               “f(x) = sin² (x)” and “g(x) = - cos² (x) + 1” represent the very same condition; they can therefore be considered equal, as shown below.

sin² (x) = - cos² (x) + 1 

and therefore

sin² (x) + cos² (x) = 1 

 

While more time-consuming than simply proving identities using the “2-column” method, the process described above brings in a visual approach.  Many other identities can be easily explored in this manner as well, my preferences being the "double- and half-angle" identities.

Once again, this geometric approach to proving trigonometric identities allows students to visualize what identities really are; it also provides the opportunity to re-connect with systems of equations which is, in itself, a very effective means of representing and thereby solving many “problem-based” scenarios.

 

The End

 

 

My son, Jeff, has recently become very interested in proofs by Mathematical Induction; his curiosity has led him to search out various scenarios that lend themselves to this process.  He recently challenged me to prove, by mathematical induction, some of his creations; the following serves to illustrate a few of these scenarios. 

For the record, Jeff verified my work here; consequently, I know it is accurate.

 

Example 1

The process began when Jeff was snooping around the difference of squares                “5² – 2²”; factoring this into the product of two binomial factors produces the following:

(5 – 2)(5 + 2)

This simplifies to

(3)(7)

which is a multiple of “3”

 

Curiosity set in and the following investigation was conducted to determine if other similar cases would also result in multiples of “3”.

 

Table 1

5³ – 2³  à  125 – 8 = 117…………...à (3)(39)

5⁴ – 2⁴  à  625 – 16 = 609…….…...à  (3)(203)

5⁵ – 2⁵  à  3125 – 32 = 3093……... à  (3)(1031)

5⁶ – 2⁶  à  15625 – 64 = 15561…....à  (3)(5187)

 

Each of the differences of “like” exponents above ultimately produced a multiple of “3”.  Following this observation, Jeff wanted to prove that this would always hold true for any “like” exponents chosen.  To enable this process, the following more general statement was developed.

5ⁿ – 2ⁿ  = 3m

for some integer value “m”

 

 

The goal is to prove the statement “5ⁿ – 2ⁿ  = 3m” true for all integer values “n” such that “n ≥ 1”.  The process of mathematical induction follows.

Test Case

n = 1

5¹ – 2¹

3

The result of “3” above is a multiple of “3”; therefore our text case works.

 

We now assume that n = k will also hold true; our statement is therefore re-written as follows:

5^k – 2^k  = 3m

 

------------------------------------------------------------------------

A slight manipulation of the statement above results in

5^k = 3m + 2^k  

We will use the statement directly above later on as a substitution.

---------------------------------------------------------------------------------------------------

 

If it can be shown that the original statement, “5ⁿ – 2ⁿ  = 3m“ also holds true for             “n = k + 1”, we will have proven our statement’s accuracy for all values of “n” such that “n ≥ 1”, which is the goal of the induction process.

 

Induction Process

n = k + 1

5^{k + 1}  – 2^{k + 1} = 5(5^k) – 2(2^k)

The substitution mentioned earlier can now be made; “3m + 2^k“ in for “5^k”.

5^{k + 1}  – 2^{k + 1} = 5(3m + 2^k) – 2(2^k)

Expanding, factoring and simplifying the right-hand side results in the following:

5^{k + 1}  – 2^{k + 1} = 5(3m) + 5(2^k) – 2(2^k)

5^{k + 1}  – 2^{k + 1} = 5(3m) + 3(2^k)

 

5^{k + 1}  – 2^{k + 1} = 3(5m + 2^k)

 

Since “3(5m + 2^k)” is multiple of “3” for integer values of “k” such that “k ≥ 1”, our inductive proof is complete.   It can therefore be stated that “5ⁿ – 2ⁿ” will equal some multiple of “3” for all values of “n” such that “n ≥ 1”.

 

 

Example 2

This second scenario is identical to the first, only having different bases, which ultimately result in multiples of “5”.  The induction process is identical to that shown in Example 1; I will nevertheless include it here as illustrating the same process a second time brings forth valuable patterns for students to see.

 

Prove the following to be true for all values of “n”, such that “n ≥ 1”.

7ⁿ – 2ⁿ  = 5m

for some integer value “m”

In other words, we must prove that “7ⁿ – 2ⁿ” will result in a multiple of “5” for all values of “n”, such that “n ≥ 1”.  The induction process follows.

 

Test Case

n = 1

7¹ – 2¹

5

The result of “5” above is a multiple of “5”; therefore our text case works.

 

We now assume that n = k will also hold true; our statement is therefore re-written as follows:

7^k – 2^k  = 5m

 

------------------------------------------------------------------------

A slight manipulation of the statement above results in

7^k = 5m + 2^k  

We will use the statement directly above later on as a substitution.

---------------------------------------------------------------------------------------------------

 

If it can be shown that the original statement, “7ⁿ – 2ⁿ  = 5m“ also holds true for               “n = k + 1”, we will have proven our statement’s accuracy for all values of “n” such that “n ≥ 1”, which is the goal of the induction process.

 

Induction Process

n = k + 1

7^{k + 1}  – 2^{k + 1} = 7(7^k) – 2(2^k)

The substitution mentioned earlier can now be made; “5m + 2^k“ in for “7^k”.

7^{k + 1}  – 2^{k + 1} = 7(5m + 2^k) – 2(2^k)

Expanding, factoring and simplifying the right-hand side results in the following:

7^{k + 1}  – 2^{k + 1} = 7(5m) + 7(2^k) – 2(2^k)

7^{k + 1}  – 2^{k + 1} = 7(5m) + 5(2^k)

 

7^{k + 1}  – 2^{k + 1} = 5(7m + 2^k)

 

Since “5(7m + 2^k)” is multiple of “5” for integer values of “k” such that “k ≥ 1”, our inductive proof is complete.  It can therefore be stated that “7ⁿ – 2ⁿ” will equal some multiple of “5” for all values of “n” such that “n ≥ 1”.

 

 

Example 3

The distinct pattern that emerged from the first two examples shown above prompted Jeff to develop a more “general” case of what he was observing; this general case is shown below.

Prove the following to be true for all values of “n”, such that “n ≥ 1”.

aⁿ – bⁿ  = (a – b)m

for some integer value “m”

In other words, we must prove that “aⁿ – bⁿ” will result in a multiple of “(a – b)” for all values of “n”, such that “n ≥ 1”.  The induction process follows.

 

Test Case

n = 1

a¹ – b¹

(a - b)

The result of “(a – b)” above is a multiple of “a – b”; therefore our text case works.

 

We now assume that n = k will also hold true; our statement is therefore re-written as follows:

a^k – b^k  = (a – b)m

 

------------------------------------------------------------------------

A slight manipulation of the statement above results in

a^k = (a – b)m + b^k  

We will use the statement directly above later on as a substitution.

---------------------------------------------------------------------------------------------------

 

If it can be shown that the original statement, “aⁿ – bⁿ  = (a – b)m“ also holds true for      “n = k + 1”, we will have proven our statement’s accuracy for all values of “n” such that “n ≥ 1”, which is the goal of the induction process.

 

Induction Process

n = k + 1

a^{k + 1}  – b^{k + 1} = a(a^k) – b(b^k)

The substitution mentioned earlier can now be made; “(a - b)m + b^k“ in for “a^k”.

a^{k + 1}  – b^{k + 1} = a((a - b)m + b^k) – b(b^k)

Expanding, factoring and simplifying the right-hand side results in the following:

a^{k + 1}  – b^{k + 1} = a(a - b)m + a(b^k) – b(b^k)

a^{k + 1}  – b^{k + 1} = a(a - b)m + (a – b)(b^k)

 

a^{k + 1}  – b^{k + 1} = (a – b)(am + b^k)

 

Since “(a – b)(am + b^k)” is multiple of “a - b” for integer values of “k” such that “k ≥ 1”, our inductive proof is complete.  It can therefore be stated that “aⁿ – bⁿ” will equal some multiple of “(a – b)” for all values of “n” such that “n ≥ 1”.

 

 

The repetitive process above sparked additional curiosity in Jeff’s mind.  The general case illustrated in Example 3 above allowed for some quick number substitutions for the variables “a” and “b”.  One such “trial” explored is shown below.

 

 

 

Connections

Having just proven the statement “aⁿ – bⁿ  = (a – b)m” for all values of “n ≥ 1” and for some integer “m”, Jeff decided to investigate this expression further by making a series of alterations to it.  This progression is shown below.

aⁿ – bⁿ  = (a – b)m

b = 1

Therefore

bⁿ = 1

 This substitution led to the following statement

 

aⁿ – 1  = (a – 1)m

 

This statement could be proven true for all values of “n” using induction; this process is not required for proof here, however, as the general case shown in Example 3 above takes care of all such cases involving the statement “aⁿ – bⁿ  = (a – b)m”.  

 

-------------------------------------------------------------------------------------

Several "trials" below serve to document various substitutions of integer values for the variable "a" in the statement “aⁿ – 1  = (a – 1)m”. 

Trial 1

In this first trial, it was decided to substitute the integer "3" for “a” in the statement          “aⁿ – 1  = (a – 1)m”; the result of this substitution is shown below.

aⁿ – 1  = (a – 1)m

3ⁿ – 1  = (3 – 1)m

3ⁿ – 1  = 2m

 

Jeff quickly verified, using technology, that the function “f(n) = 3ⁿ – 1” produced multiples of “2” for values of “n”.  An illustration of one such use of technology is shown in Graph 1 below.

Graph 1

 

The four points highlighted on the graph of “f(n) = 3ⁿ – 1” are listed directly below.

 f(n) = 3ⁿ – 1

A_E(1, 2)…….B_E(2, 8)…….C_E(3, 26)…….D_E(4, 80)

 

  

These same function values occur on the linear function defined as "f(m) = 2m" in Graph 2 below.  By comparing the manipulated variables of the corresponding points shown in Graphs 1 and 2, one can more easily appreciate the impact of the accelerating nature of exponential versus linear functions.

Graph 2

f(m) = 2m

A_L(1, 2)…….B_L(4, 8)…….C_L(13, 26)…….D_L(40, 80)

 

 

It is worthwhile to note here that the comparison of these two sets of points is entirely similar to the relationship existing between the graphs representing frequencies of notes in the "Octave Series" (exponential) and "Harmonic Series" (linear) from The Mathematics of Music.

 

-------------------------------------------------------------------------------------

 

Trial 2

In this trial, it was decided to substitute “22” for “a” in the statement “aⁿ – 1  = (a – 1)m”.

 

aⁿ – 1  = (a – 1)m

22ⁿ – 1  = (22 – 1)m

22ⁿ – 1  = 21m

 

Some additional explorations from this point eventually led to the following query; What would happen if “14” was subtracted from both sides of the statement                            “22ⁿ – 1  = (21)m”?  What Jeff discovered is shown below.

22ⁿ – 1  –  14 = 21m – 14

 

It was immediately recognized that “7” was a factor of both terms on the right-hand side of the statement directly above.  The version below shows the statement in factored form.

 

22ⁿ – 15 = 7(3m – 2)

 

The statement directly above indicates that the function “f(n) = 22ⁿ – 15” produces values which are multiples of “7”; this was once again quickly verified using technology.

 

-------------------------------------------------------------------------------------

Trial 3

The final trial here shows “43” substituted for “a” in the statement “aⁿ – 1 = (a – 1)m”; the progression of this alteration is shown below.

aⁿ – 1 = (a – 1)m

43ⁿ – 1 = (43 – 1)m

43ⁿ – 1 = 42m

According to the statement directly above, the function defined as f(n) = 43ⁿ – 1 will produce multiples of “42” for all values of “n” such that “n ≥ 1”.  A quick check with technology will verify this.

 -------------------------------------------------------------------------------------

 

 

Additional modifications can be made to the statement “43ⁿ – 1 = 42m” to initiate further explorations; two such cases are shown below.

Case 1

Adding “35” to both sides of the statement above will result in the following:

43ⁿ – 1 + 35 = 42m + 35

43ⁿ + 34 = 7(6m + 5)

According to the result directly above, the function defined by f(n) = 43ⁿ + 34 should produce values that are multiples of “7”.  Again, a quick check with technology verifies this.

 

Case 2

Subtracting “12” from both sides of the statement above will result in the following:

43ⁿ – 1 – 12 = 42m – 12

43ⁿ – 13 = 6(7m – 2)

According to the result directly above, the function defined by f(n) = 43ⁿ – 13 should produce values that are multiples of “6”.  Again, a quick check with technology will verify this.

 

-------------------------------------------------------------------------------------

 

There are many additional explorations that can be made into these types of statements, many of which Jeff has already connected to.  While much of the process outlined here has no “practical” applications to many, knowledge of this process will serve Jeff very well in his future studies.

The End

 

 

Sometimes referred to as the “Russian Peasant Algorithm”, this method of multiplication involves a systematic “halving” of one factor (in column 1) with a corresponding “doubling” of a second factor (in column 2).  This process is repeated until the column being “halved” reaches a value of “1”; the corresponding value occurring in “column 2” at this point represents the product of the original two factors.

A simple example of this process is shown below and will show the pattern occurring for the factors “8” and “6”. 

 

Table 1

…………Column 1………..Column 2

Row 1: …….8………….……..….6

Row 2: …….4……………..……12

Row 3: …….2…..…………..….24

Row 4: …….1……..……..….…48.........(1 by 48)

 

Sketching a rectangular grid and “rearranging” its contents easily turns this into a visual representation of multiplication.  The dimensions of each rectangle are determined by the values shown in Columns 1 & 2; the product of those dimensions represent the “area" of each corresponding rectangle.

Row 1:  “8” groups of “6” results in “48”.

Row 2:  “4” groups of “12” results in “48”.

Row 3:  “2” groups of “24” results in “48”.

Row 1:  “1” groups of “48” results in “48”.

 

 

Issues quickly arise, however, it the values being halved do not divide evenly by “2”.  The example directly below shows such a scenario.

 

Table 2

…………Column 1………..Column 2

Row 1: …….10………….……..….7

Row 2: ….….5……………………14

 

The “5” in Column 1 above is not divisible by “2” so “1” group of “14” is removed and set aside.  Geometrically speaking, this can be considered a rectangle having dimensions “1 by 14” and is represented in Row 3 below.  Row 4 directly beneath that represents a “4 by 14” rectangle, that which remains of the “5 by 14” rectangle above after having removed and set aside its “1 by 14” portion. 

Row 3: …….1……………………14.........(1 by 14)

 

Row 4: …….4……………………14

The “halving” process can now continue on until Column 1 reaches a “dimension of 1”.  Any additional “odd” numbered factors occurring in Column 1 along the way are dealt with in the same manner as described above.

 

Row 5: …….2…..…………..….28

Row 6: …….1…..…………..….56.........(1 by 56)

 

 

Geometrically speaking, we now have two rectangles having a “width” of “1” unit, one measuring “14” units in length, the other “56”.  When these rectangles are joined end-to-end, the result is one longer rectangle having dimensions “1 by 70”.  The length of this final rectangle represents the product of the original two factors, “10” and “7” is this example.

 

 

 

Two additional examples will be shown below (without related diagrams).  The example in Table 3 directly below re-visits the first example illustrated earlier with the factors reversed.

 

Table 3

…………Column 1………..Column 2

Row 1: …….6………….……..….8

Row 2: …….3……………………16

 

The “3” in Column 1 above is not divisible by “2” so “1” group of “16” is removed and set aside.  Geometrically speaking, this can be considered a rectangle having dimensions “1 by 16” and is represented in Row 3 below.  Row 4 directly beneath that represents a “2 by 16” rectangle, that which remains of the “3 by 16” rectangle above after having removed and set aside its “1 by 16” portion. 

Row 3: …….1……………………16……(1 by 16)

Row 4: …….2……………………16

 

The “halving” process can now continue on until Column 1 reaches a “dimension of 1”.

 

Row 5: …….1…..…………..….32……(1 by 32)

 

Geometrically speaking, we now have two rectangles having a “width” of “1” unit, one measuring “16” units in length, the other “32”.  When these rectangles are joined end-to-end, the result is one longer rectangle having dimensions “1 by 48”.  The length of this final rectangle, and its area, each represent the product of the original two factors, “6” and “8” in this example.

 

 

Table 4

…………Column 1………..Column 2

Row 1: …….26………….……..….7

Row 2: …….13……………………14

 

The “13” in Column 1 above is not divisible by “2” so “1” group of “14” is removed and set aside.  Geometrically speaking, this can be considered a rectangle having dimensions “1 by 14” and is represented in Row 3 below.  Row 4 directly beneath that represents a “12 by 14” rectangle, that which remains of the “13 by 14” rectangle above after having removed and set aside its “1 by 14” portion. 

Row 3: …….1……………..........…14.........(1 by 14)

Row 4: …….12……………………14

 

The “halving” process can now continue on until Column 1 reaches a “dimension of 1”.

 

Row 5: …….6……………………28

Row 6: …….3……………………56

 

Row 7: …….1…...…........………56.........(1 by 56)

Row 8: …….2……………………56

 

Row 9: …….1………………...…112.........(1 by 112)

 

 

Geometrically speaking, we now have three rectangles having a “width” of “1” unit, measuring “14” units in length (Row 3), “56” units in length (Row 7), and finally “112” units in length (Row 9).  When these rectangles are joined end-to-end, the result is one longer rectangle having dimensions “1 by 182”.  The length of this final rectangle, and hence its area, each represent the product of the original two factors, “26” and “7” in this example.

 

This will no doubt be considered a complete waste of time by many individuals.  Those same individuals, in all likelihood, favor handing small children a calculator to find products such as these (because it’s much faster). 

There is no question that using technology IS much quicker to find products such as these.  I’m completely in favor of using technology for “crunching” large numbers.

The process outlined above, however, has much value in showing how factors of a given product relate to one another.  Relating this process to “area” of a series of geometric shapes adds another valuable dimension to the exploration (in my opinion). 

There is NO need to use insanely large number in this process; more can be gained by repeating the process several times using numbers that are “manageable” and “recognizable” to students.

 

The End

 

 

When students explore many of their observations more deeply, they often discover patterns emerging.  In many cases, these patterns ultimatey lead them to the very formulae they've been using in various contexts and disciplines.  

My son does this on a very regular basis; one such example is outlined below.  I've chosen to begin the process here as an induction proof of the sum formula for geometric series.

 

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I was recently asked by a former student to explain Mathematical Induction and its use in proving Bernoulli’s Inequality.  Here is my attempt at this.

 

When using induction to prove a mathematical statement to be true, the statement itself must first be given. 

A “base case” is shown initially to prove that the statement holds true for our starting point; this is either “0” or “1” (right click on the link to Mathematical Induction  to open a new window for additional information).

The inductive proof itself follows and is similar to a recursive pattern.  If the statement is proven true for a value “k” and then again for a value “k+1”, then it can be assumed that the statement will hold true for “k+2”, “k+3”, “k+4”, “k+5”, “k+6”, and so on. 

 

Example 1

 Prove, by induction, that the expression “7ⁿ – 4” is divisible by “3” for all values of       “n ≥ 0”.

 

First, we will set our expression equal to “3m”, such that “m є ℤ”  à ℤ denotes integers. 

 

7ⁿ – 4 = 3m

 

Since "3m" is an integral multiple of "3", and our expression "7ⁿ – 4" has been set equal TO that multiple of "3", the expression itself must be divisible by "3".

 

 

Prior to the actual induction process this statement will first be tested for accuracy, using values “n = 0” and “n = 1”.  If these tests show that the expression is a multiple of “3” (and therefore divisible by “3”), the induction process will proceed.

 

Basis

n = 0

7⁰ – 4

- 3

Since “- 3” is an integral multiple of “3”, our requirement of divisibility is met.

 

2^ndTest

n = 1

7¹ – 4

3

Once again, since “3” is an integral multiple of “3”, our requirement of divisibility is met in our second test; this second test isn’t required but can be helpful to observe.  Additional test values could also be executed but to prove divisibility by “3” for ALL values of “n”, the induction process will be employed.

 

 Essentially, we are asked to prove that the exponential function “f(n) = 7ⁿ – 4” will be divisible by “3” for all values of “n”.  That is to say, if we can show that the function “f(n) = 7ⁿ – 4” is a multiple of “3” for all values of “n”, we will have proven its divisibility by “3”.

In other words, we need to prove the following for all values of “n”:

 

7ⁿ – 4 = 3m

Where “m” is some integer

 

We first assume the statement to hold true for “n = k“; then we work towards proving its accuracy for  “n = k+1”.

 

Induction Process

n = k

7^k – 4 = 3m

A minor rearrangement of the equation above results in a version that makes our process very obvious.

 

7^k  = 3m + 4

Multiplying both sides of the equation directly above by “7” will ultimately lead to         “7^{k + 1}”, which is what we are pursuing.

 

7(7^k)  = 7(3m + 4)

7^{k + 1}  = 7(3m) + 7(4)

 

Our goal is to make this new version above resemble the original statement used in our test cases.  Once again, that original statement was “7ⁿ – 4 = 3m”.  Subtracting “4” from both sides of this revised equation takes us one step closer to meeting our objective.

 

7^{k + 1}  - 4 = 7(3m) + 7(4) – 4

 

Grouping the last two terms on the right-hand side of the equation above, followed by some simple factoring leads to yet another version of our equation.

 

7^{k + 1}  - 4 = 7(3m) + (7(4) – 4)

7^{k + 1}  - 4 = 7(3m) + 4(7 – 1)

7^{k + 1}  - 4 = 7(3m) + 4(6)

 

One final procedure puts a “lock” on this proof.  A multiple of “3” can now be factored out of each of the two terms on the right-hand side of the equation directly above.  What results from having completed this factoring is “3(7m + 8)”, an “integral multiple of 3”; this was our goal from the outset and the proof will be complete.

 

7^{k + 1}  - 4 = 3(7m + 4(2))

 

 

Final Statement

7^{k + 1}  - 4 = 3(7m + 8)

 

Original Statement

7ⁿ – 4 = 3m

 

We have now proven, by induction, that the expression “7ⁿ – 4” is divisible by “3” for all values of “n”.  Students can “verify” by testing values of “n” with a calculator.  Some additional discoveries can be made by graphing the function “f(n) = 7ⁿ – 4” and exploring its relationship with the linear function , “g(m) = 3m”,representing multiples of “3”.  This linear function will eventually “get there” but not nearly as quickly as its exponential counterpart.

 

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Optimization problems are typically solved at the high school level by representing the scenario as a quadratic function, then completing the square to find that function’s vertex.  The vertex contains the “maximum” or “minimum” value and a value of the functions manipulated variable for which this “optimal” value occurs.

 

Several examples of this are shown below.

 

      Example 1

Design a rectangular enclosure of maximum area given 16 meters of fencing; all four sides require fencing.

 

 

 

Given the conditions stated here, the perimeter of our enclosure cannot exceed 16 m; since we want to maximize area, we will assign all 16 meters of fencing to the perimeter.  The following relationship therefore emerges from this requisite.

 

2(length + width) = 16

length + width = 8

 

From this, two inter-related functions can be defined, both in terms of “x”; one representing the rectangle’s length, the other, its width.  These functions appear below and are illustrate in Graph 1.

 

w(x) = x à  blue

l(x) = -x + 8  à  red

 

Graph 1

 

 

 

 

The point of intersection of these two linear functions is shown as point K (4, 4) along with several other points on each function.

We know that “area” is a function of the product of a rectangle’s length and width.  It is therefore very easy to determine several different “area” possibilities by simply multiplying the values of each function for various “x” positions.  This is summarized in Table 1 below.

 

Table 1

Width:  w(x) = x……….Length:  l(x) = -x + 8…….Area àPoint

A:  w(0) = 0………………....C:  l(0) = 8…………………(0)(8)=0  àA

M:  w(1) = 1.……………......N:  l(1) = 7…………………(1)(7)=7  àN

E:  w(2) = 2………………....G:  l(2) = 6…………………(2)(6)=12àI

Q:  w(3) = 3………………....R:  l(3) = 5…………………(3)(5)=15àU

K:  w(4) = 4………………....K:  l(4) = 4…………………(4)(4)=16àL

S:  w(5) = 5………………....T:  l(5) = 3…………………(5)(3)=15àV

H:  w(6) = 6………………....F:  l(6) = 2…………………(6)(2)=12àJ

O:  w(7) = 7………………....P:  l(7) = 1…………………(7)(1)=7  àO

D:  w(8) = 8………………....B:  l(8) = 0…………………(8)(0)=0  àB

 

The points identified in bold saltire represent the area values for each integral value of “x” in the interval [0, 8]; these points clearly form the shape of a parabolic curve.

The equation of this parabola can be found very easily, either through a quadratic regression command using technology or by simply using the vertex and one other point on its locus to uniquely determine the defining function in vertex form.

Either method results in the defining function shown below.

 

A(x) = - (x - 4)² + 16

 

A slightly altered version of the first illustration appears below in Graph 2.  From the “dotted” version of the quadratic function shown here, it is easy to see that its axis of symmetry is drawn through the midpoint of the “x-intercepts” of the original linear functions representing length and width.

Graph 2

 

 

 

 

The axis of symmetry passes through the point of intersection of the linear functions in this first example.  This occurs here since the functions representing “length” and “width” have slopes of equal magnitude; they are guaranteed to intersect midway through the interval [0, 8].  This interval is defined by the respective zeros of the linear functions representing “width” and “length”.

The maximum area of our original rectangle, and therefore the maximum value of the parabola representing that area,  is simply the product of the values  of these two linear function at the “axis of symmetry”; in this example, that maximum area is “(4)(4) = 16”.

 

The defining function of this parabolic curve representing the area of our rectangle is determined in a different manner below.  This is shown below as a product of the functions “w(x)” and “l(x)”.

 

A(x) = x (-x + 8)

A(x) = -x² + 8x

 

Completing the square naturally results in an identical version (in vertex form) of this function as was seen earlier, as shown below.

A(x) = - (x - 4)² + 16

 

 

 

 

Example 2

Design a rectangular enclosure of maximum area given 16 meters of fencing.  Only three sides require fencing this time as the fourth side is bordered by a building.

 

 

 

Once again, the perimeter of our enclosure cannot exceed 16 m; since we want to maximize area, we will assign all 16 meters of fencing to the perimeter.  Since only three sides require fencing this time around, the following relationship will emerge.

Assigning variables to represent length (y) and width (x) leads to the following statement:

2x + y = 16

 

I will take a slightly different approach to the problem this time.

Since our goal is to maximize “area”, we first need an expression the represents the area of our rectangle.  This expression is shown below.

 

A(?) = (x)(y)

 

In order to determine a “maximum” value for area here, we know the graph of “A(?)” must be a downward opening parabola.  We also know that “A(?)” will be measured on the vertical axis; this leaves only one remaining axis to represent “(x)(y)” from the function statement above.

 

We have two choices from here; we must either express “y” in terms of “x” or vice versa.

 

From the equation “2x + y = 16” above, it is obvious that isolating “y” would be an easier option as fractions are avoided.  The manipulated version of our equation here is shown below.

 

y = - 2x + 16

 

Substituting this expression of “y” in place of that variable in our function statement is shown below.

 

A(?) = (x)(y)

A(x) = (x)(-2x + 16)

A(x) = -2x² + 16x

 

Completing the square and re-writing the function above in vertex form results in the following:

 

A(x) = -2(x - 4)² + 32

 

According to the function above, the vertex has coordinates (4, 32), implying a maximum are of “32” occurring when “x=4”.  This is verified in the illustration below.

 

Graph 3

 

 

 

 

Also shown in the Graph 3 illustration are the two linear functions representing length and width of the rectangle, both as functions of “x”; their algebraic definitions are shown below.

 

w(x) = x à  blue

l(x) = -2x + 16  à  red

 

The axis of symmetry of the parabola can once again be determined easily as being the midpoint of the “x-intercepts” of the two linear functions representing length and width.  After all, these appear as the factors of the parabolic function defining area below, and therefore determine the “x-intercepts” of that parabola itself.

 

A(x) = (x)(-2x + 16)

 

Once again, this axis of symmetry will pass through the midpoint of the interval [0, 8], defined by those zeros.  The maximum area can be easily determined by evaluating the product of the functions defining length and width at “x=4”.  This is very quickly verified below.

 

w(4) =4

l(4) = 8

 

Maximum Area

(4)(8) = 32 

 

 

Worthwhile noting here is that the point of intersection is “offset” from the axis of symmetry in this scenario.  This occurred because of the differing magnitudes in the slopes of the functions representing length and width.

It is easy to see that if the function representing width had been “stretched” by a factor of “2”, point A in the illustration above would have merged with point B precisely on the axis of symmetry.

Had this happened, however, the maximum area would also have increased by a factor of “2”.  Given the restrictions in the scenario put forth here, that would be impossible since only 16 meters of fencing was made available; doubling the width while holding everything else constant would require an increase in the amount of fencing required.  The resulting rectangle is shown below.

 

 

 

The rectangle above has an area of 64 square meters.  The perimeter of this larger, more spacious rectangle is 32 meters, 24 of which require fencing.  The shape of this new enclosure is once again a perfect square.

 

 

 

Partitions

 

Example 3

Design a rectangular enclosure of maximum area given 16 meters of fencing.  Only three sides require fencing this time as the fourth side is bordered by a building, but two partitions are added on the interior running perpendicular to the building.  

There will now be less fencing available for the length of our structure; as a result, our maximum are will be reduced from that in the previous example.

 

 

 

The functions below represent width and length as functions of “x” given the conditions established above.  Graph 4 below gives the geometric interpretation for this new scenario.

w(x) = x à  blue

l(x) = -4x + 16  à  red

 

Graph 4

 

 

 

 

The midpoint of the interval [0, 4] determined by the x-intercepts of the linear functions representing our rectangle’s width and length, defines the axis of symmetry for the related parabola.  The locus of points on this parabola represents all possible areas for the rectangle in Example 3.

 

 

 

Example 4

Once again, design a rectangular enclosure of maximum area given 16 meters of fencing; again, only three sides require fencing as the fourth side is bordered by a building.  This time, however, six partitions are added on the interior running perpendicular to the building.  

There will now be much less fencing available for the length of our structure since there are many “multiples” of our widths required; as a result, our maximum are will be further reduced from that in the previous examples.

 

 

 

The functions below represent width and length as functions of “x” given the conditions established above.  Graph 5 below gives the geometric interpretation for this new scenario.

w(x) = x à  blue

l(x) = -8x + 16  à  red

Graph 5

 

 

 

Again, the midpoint of the interval determined by the x-intercepts of the linear functions representing our rectangle’s width and length defines the axis of symmetry for the related parabola. 

In this last example, the midpoint of the interval [0, 2] is “x = 1”; this defines the axis of symmetry for the related parabola.  The locus of points on this parabola represents all possible areas for the rectangle in Example 4.

 

 

To further illustrate the effect of adding extra partitions on the maximum possible area of a rectangular enclosure, I’ve included two additional illustrations below.  Each of these shows the functions from examples 2, 3 & 4, drawn on the same scale for a more direct comparison. 

 

Graph 6

 

 

 

Consider the graphs above as they may relate to Linear Programming.  Such a scenario would include inequalities such as those shown below.

w(x) ≤ x à  blue

l(x) ≤ -2x + 16  à  red

 

These inequalities relate to the largest parabola shown in Graph 6.  Because of the “less-than-or-equal-to” requirement, the point of intersection of the two linear functions, “E₁”, represents the optimum point for such a condition.  The “maximum” area that would be acceptable given the conditions stated would occur at point D₁.

 

 

Graph 7, below, shows an additional linear function which passes through the vertices of each of the quadratic functions.  This function is defined as follows:

g(x) = 8x

 

 

Graph 7

 

 

 

 

Additional relationships of the functions and their connections to the number of partitions in the corresponding rectangles can be made here.  Have students to explore cubic functions by finding the products of three linear functions, etc.

 

One last mathematical connection to make here would be to derivatives.  The area functions for each of the graphs shown above could be explored further by finding their derivatives; students could identify “stationary” points using the first derivative, determine the nature of those points with the second derivative (rather trivial here) and then verify their findings by connecting back to the illustrations provided.

 

Supply and Demand curves could even be related to the linear functions represented in the examples above.

 

 

Once again, the problem stated at the outset is “contrived” and largely meaningless (on its own).  I believe the purpose of questions such as these is to provide an avenue to explore how functions interact with one another, discover patterns, make predictions and then verify those predictions.  Much of what is learned through these explorations CAN be applied to scenarios that ARE “real-life”, whatever they may be.

Expecting students to discover these patterns and connections, through their own explorations only, is extremely short-sighted.  I believe students learn more effectively by making connections on their own but they DO need a starting point from which to work.

 

The End

 

 

Included below are links to related "Iso-Perimetric" problems, for further understanding; thanks to Pat @OnThisDayInMath for these.  Right click to open each in a new window.

Iso-Perimetric Problems (1)

Iso-Perimetric Problems (2)

Iso-Perimetric Problems (3)

 

 

 

 

 

The clock face above has several divisions that are very clearly displayed; each of these represents a portion (fraction) of a 24 hour period, representing one full rotation of the Earth on its axis.

The largest of these divisions can be considered the clock itself as it represents exactly “¹/₂” of this 24-hour period (¹²/₂₄). 

 

Since the 12-hour clock is displayed here, the remainder of discussion below will be based on the assumption that one trip around the clock face will constitute one entire “unit”.  High-noon will be considered that starting point; twelve o’clock midnight will represent the completion of the day’s “Post Meridiem” journey.

 

Ante Meridiem (A.M.):  “before midday”

Post Meridiem (P.M.):  “after midday”

 

 

Modular Arithmetic

When tracking time on a 12-hour clock, people are immersed in “Modular Arithmetic”; "mod 12" to be exact.  On the 24-hour clock, 14 hours after midnight appears as “14:00”.  On our 12-hour clock, this is referred to as “2:00 pm”, which is to say “2 hours after midday” (or 2 hours after 12-noon).

The generally accepted manner of interpreting modular arithmetic is summarized very briefly below.

 

14 – 2 = 12

Since the difference of 14 and 2 is a multiple of “12”, the statement below can be made.

 14 Ξ 2  (mod 12)

 

15 – 3 = 12

Since the difference of 15 and 3 is also a multiple of “12”, the statement below can once again be made.

15 Ξ 3  (mod 12)

 

 

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Another very common example of modular arithmetic occurs when tracking days of the week.  For example, October 1, 2011 falls on a Saturday; it is easily determined that October 29 also falls on a Saturday by looking at a Calendar.  However, modular arithmetic, “mod 7” this time, can also be employed in coming to this conclusion, as shown below.

We know that October 1 falls on a Saturday; we would like to determine the day on which October 29 falls.

29 – 1 = 28

The difference of 29 and 1 is “28”, which is a multiple of “7”.

We can therefore make the following statement:

29 Ξ 1  (mod 7)

 

Since 29 Ξ 1  (mod 7), we can be sure that October 29 and October 1 fall on the same day of the week; more precisely they are exactly 4 weeks apart, since there are 4 multiples of 7 in 28 (28 being the difference between the two numbers being compared).  

The day on which October 31 falls can be determined very easily as well.  Since "31" is 2 days past "29", Halloween Day falls on Monday (2 days after Saturday).  

 

 

 

 

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......back to our clock

One other consideration here will eliminate the need for a duplication of reasoning.  Since seconds relate to minutes in the same way that minutes relate to hours, there is no need to describe both cases; what applies to one relationship can be directly transferred to the other by simply changing the parameters. 

The relationship between minutes, hours and their representation on the clock face will be detailed below.

 

 

 

Fractions

The clock face shows two obvious sets of divisions, each of which represents a fraction (portion) of the entire face. 

The smallest divisions each represent the passage of “one minute”; there are 60 of these in total on the clock face.  Each interval of one minute can therefore be considered to be ¹/₆₀^th  of a full hour.

 

Because these divisions are so numerous, it is helpful to divide these 60 increments into sub-groups to more easily track time.

This is entirely similar to keeping a “tally” when counting large numbers of items in which four vertical lines followed by a fifth line, drawn diagonally through the previous four,  produces a series of “five-bar gates”, each of which symbolizes sub-groups of five items. 

 

On the clock face, there are 12 groupings of five-minute intervals, accounting for a total of sixty minutes on one full trip around the clock face.  Each sub-group of five-minute intervals is delineated by a symbol of some sort; the symbol chosen represents the set of natural (counting) numbers. 

This lends itself very nicely to tracking the number of hours past 12-noon as well as how many multiples of five-minute intervals the minute hand has travelled from its starting position at the top of the hour.

 

 

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We can credit the Ancient Babylonians for much of our present day time-keeping strategies.  Because “60” can be broken down into many factors (it is highly composite), “fractional” components are very numerous and easy to come by.  

This civilization divided the night and day into 12 hours each; hours were divided further into 60 one-minute intervals and each minute into 60 seconds.  

Similarily, each solar year was approximately 360 days in length; this was, in turn, divided into 12 months to approximate the number of lunar cycles in one year. 

It is for this ease of fractional representation that the Ancient Babylonians favored the “Base-60” number system.

 

In addition, the 360 degree connection to circles is directly related to the 360 day solar year approximation mentioned above.

 

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Design a Clock

Suppose for a moment that we are in charge of designing a clock that will help us track the passage of time. 

The first clock we design has only a minute hand along with 60 divisions on its face, each representing a one-minute interval.  One complete rotation of this minute-hand (360 degrees) represents the passing of 60 minutes, also known as one hour. 

It is soon decided that this clock would function much more effectively if it could also keep track of the number of hours passed.  Knowing that there are 12 hours in each of a day and night, it is decided to add a second set of partitions to the clock face, this time to represent the number of hours in each day/night cycle.  

There are 12 increments in this second set of partitions, each occurring at five-minute intervals.  An “hour-hand”, slightly shorter than the minute-hand to set it apart, is created and attached to a gear designed to make one complete rotation for every 12 rotations of the minute-hand.  This will ensure that 720 minutes tracked by the minute-hand in one "day" (12*360 degrees) will result in exactly 360 degrees of rotation (12 hours) by the hour-hand.  The gear ratio for this requirement must be 1:12.

 

Modern clocks employ a harmonic oscillator that maintains a constant frequency, resulting in higher precision time keeping; this serves the same purpose as does the flywheel in your vehicle’s engine. 

We would no doubt have discovered the need for this (and invented it) had someone else not beat us to it.

 

 

One complete rotation of the minute-hand (360 degrees) results in an advancement of one hour, or “¹/₁₂^th” of a full rotation on the clock face by the hour-hand.  Therefore, each “one-hour increment” on the clock face represents ¹/₁₂^thof 360 degrees or 30 degrees (¹/₁₂ * 360).  Table 1 below summarizes this relationship; the table begins at 12-noon.

 

Table 1

.….360-Degree Rotations………….............………Time Interval……..

Minute-hand…... Hour-hand………………. Minutes........…..Hours (as time)

…..1……….……....1*¹/₁₂=¹/₁₂……………….60 min………..…60*¹/₆₀= 1 o’clock

…..2……….……....2*¹/₁₂=²/₁₂…………..….120 min……….…120*¹/₆₀= 2 o’clock

…..3………....…....3*¹/₁₂=³/₁₂…………...….180 min……….…180*¹/₆₀= 3 o’clock

…..4……….……....4*¹/₁₂=⁴/₁₂………….…...240 min………..…240*¹/₆₀= 4 o’clock

…..5……….……....5*¹/₁₂=⁵/₁₂………......….300 min………..…300*¹/₆₀= 5 o’clock

…..6……….……....6*¹/₁₂=⁶/₁₂…………...….360 min…….....…360*¹/₆₀= 6 o’clock

…..7……….……....7*¹/₁₂=⁷/₁₂…………...….420 min……..……420*¹/₆₀= 7 o’clock

…..8……….……....8*¹/₁₂=⁸/₁₂……………....480 min……......…480*¹/₆₀= 8 o’clock

…..9……….…..…..9*¹/₁₂=⁹/₁₂……………....540 min………..….540*¹/₆₀= 9 o’clock

….10………..…....10*¹/₁₂=¹⁰/₁₂………..…....600 min………..…600*¹/₆₀= 10 o’clock

....11………...…....11*¹/₁₂=¹¹/₁₂………….….660 min………...…660*¹/₆₀= 11 o’clock

 ...12………..……..12*¹/₁₂=¹²/₁₂………….....720 min………...…720*¹/₆₀= 12 o’clock

 

The graph below shows a geometric interpretation of the contents of Table 1.

 

Graph 1