Download presentation

Presentation is loading. Please wait.

Published byJace Norland Modified over 2 years ago

1
The Game of Algebra or The Other Side of Arithmetic The Game of Algebra or The Other Side of Arithmetic © 2007 Herbert I. Gross by Herbert I. Gross & Richard A. Medeiros next Lesson 14

2
Two Enrichment Examples Two Enrichment Examples © 2007 Herbert I. Gross next

3
Strange as it may seem, there is a connection between how recipes are used in cooking and how formulas are used in mathematics. It takes a special talent to create a recipe, but many people who do not have this talent are still able to follow the recipe effectively. In this context, a formula is a mathematical “recipe” that tells us how to “mix” certain variables to “create” other variables. We don't have to know the derivation of a formula in order to be able to use it. In fact, from the very first lesson in this course, we used formulas without first proving that they were correct. © 2007 Herbert I. Gross

4
next For example, it is not necessary to know why the formula A = πr 2 tells us how to “create” the area (A) of a circle once we know the radius (r) of the circle. The formula is easy to use; namely starting with the radius we multiply it by itself and then multiply the result by π. However, the actual derivation of the formula is not easy. In fact, a rigorous derivation is usually postponed until the study of calculus. © 2007 Herbert I. Gross next radius

5
While a person can follow a recipe without understanding what each ingredient does and why the ingredients are in the given proportions, there are times when this knowledge isn't enough. © 2007 Herbert I. Gross So, in the form of enrichment, this Lesson will attempt to provide the flavor of how formulas are developed by looking at two interesting examples, one of which involves a linear relationship and the other of which involves a relationship that is “almost linear”. next

6
Let's look at the linear example first. We have already defined the standard form of a linear relationship to be… © 2007 Herbert I. Gross y = mx + b next where m and b are constants.

7
© 2007 Herbert I. Gross Keep in mind that as written, we have no way of knowing whether the formula refers to an “easy” situation or a “hard” situation. For example, at the “easy” end of the spectrum, we have already seen that at a cost of $4 per pound plus a charge of $5 for shipping and handling, the cost (y) in dollars of x pounds of candy is given by the formula… next y = 4x + 5 where m = 4 and b = 5.

8
We call this an “easy” situation because most of us who were ordering the candy would know, even without seeing the formula, that we have to multiply the number of pounds we buy by $4 and then add $5 to this amount to cover the cost of shipping and handling. © 2007 Herbert I. Gross next At the “harder” end of the spectrum is the relationship between the Fahrenheit temperature scale (F) and the Celsius temperature (C).

9
Yet the method for determining the formula is the same for all linear relationships. First, however, let us explain why the relationship between the Fahrenheit scale and the Celsius scale is linear. Namely, both scales measure the height of a mercury column on a thermometer. When the mercury level is at a certain height, the thermometer doesn't know which scale we're using. © 2007 Herbert I. Gross

10
next Hence, the change in the Fahrenheit reading has to be proportional to the change in the Celsius reading. In other words, if the temperature level in the column rises by 1 inch, the change in temperature is the same even though the readings on both scales are different. © 2007 Herbert I. Gross next C F

11
As a simpler illustration, think, for example, in terms of length, if one piece of string is twice as long as another, then no matter what scale we use, the measurement of one length will be twice the measurement of the other. © 2007 Herbert I. Gross 1 yard 2 yards 3 feet 6 feet 36 inches 72 inches next = = = =

12
© 2007 Herbert I. Gross There is a basic difference between temperature scales and length scales. Namely, while the relationship between the perimeter of a square and the length of a side is a direct proportion; the relationship between Celsius and Fahrenheit is linear, but it is not a direct proportion. Linear vs. Direct Proportion

13
next For example, if a piece of string were to have no length, its length would be 0 whether we measured it in inches or whether we measured it in centimeters. On the other hand, it is well-known that when the temperature is 0° on the Celsius scale, it is 32° on the Fahrenheit scale, and that when C = 100, F = 212. At any rate, if we elect to express F in terms of C, we now know that the initial value of F (i.e., the value of F when C = 0) is 32. © 2007 Herbert I. Gross

14
next That is, if we start with the standard linear form… © 2007 Herbert I. Gross next and replace F by 32 when C = 0, we see that… and since m0 = 0, the equation becomes… and if we now replace b by 32 in the formula, we get… F = mC + 32 32 = 0 + b = b 32 = m0 + b F = mC + b next

15
We also know that when C = 100, F = 212. Hence, we may replace C by 100 and F by 212 to obtain… © 2007 Herbert I. Gross next Subtracting 32 from both sides gives us… And if we now divide both sides by 100 we obtain… 180 100 = m 180 = 100m 212 = 100m + 32 F = mC + 32 9595 =

16
next If we now replace m in the formula by 9 / 5, we obtain… © 2007 Herbert I. Gross next F = 9 / 5 C + 32 F = mC + 32 Note In the linear relationship y = mx + b, notice that b always represents the value of y when the value of x is 0. That is, if we replace x by 0 in the formula y = mx + b, we see that y = m0 + b, or y = b. That's why it is helpful in starting problems such as this to know the value of y when x = 0. In particular, in this illustration it was nice to know that F = 32 when C = 0.

17
next © 2007 Herbert I. Gross Note In other words, if we had started our solution trying to use the fact that when C = 100, F = 212, the formula F = mC + b would have become 212 = m100 + b; and we would have been faced with a situation in which there was only one equation but two unknowns. Sometimes we cannot avoid having to deal with equations in which there are more than one unknown. What we do in such cases is discussed later in this course.

18
next © 2007 Herbert I. Gross Remember what a constant is. It is a number that never changes. Hence, if b = 32 when C = 0, then b = 32 no matter what the value of C is. That's why we were allowed to replace b by 32 in the formula F = mC + b even though we only demonstrated that b = 32 when C was equal to 0. Defining a Constant

19
next © 2007 Herbert I. Gross The Formula If you remembered the formula C = 5 / 9 (F – 32), you could have obtained the formula F = 9 / 5 C + 32 directly by the “undoing” method. Namely, start with C =5/9(F – 32) and multiply both sides by 9/5 to obtain 9/5C = F – 32; then add 32 to both sides to obtain 9/5C + 32 = F, which by symmetry is equivalent to F = 9/5C + 32 next

20
© 2007 Herbert I. Gross F or C? Keep in mind that F = 9 / 5 C + 32 and C = 5 / 9 (F – 32) are two different ways for expressing the same relationship. That is, we would most likely use… F = 9 / 5 C + 32 when we know the value of C and we wanted to find the value of F; and we would most likely use… C = 5 / 9 (F – 32) when we know the value of F and wanted to fine the value of C. next

21
© 2007 Herbert I. Gross Fahrenheit was a Danish physician who invented the first thermometer. Recognizing that heat caused most liquids to expand and cold caused them to contract, Fahrenheit made a tube with a narrow opening that had a well or reservoir at the bottom to hold the liquid. By making the opening very narrow the liquid was forced to move up the tube when it expanded and down the tube when it contracted. Development of the Fahrenheit and Celsius Temperature Scale

22
next © 2007 Herbert I. Gross Fahrenheit could now measure temperature simply by measuring the height of the liquid level in the tube. To do this, he had to develop a scale.

23
next © 2007 Herbert I. Gross So on a particularly cold morning in Copenhagen, he left the thermometer outside his window, and he marked the liquid level as 0°F (actually, he simply called it 0° because at the time there were no other temperature scales). He then took his body temperature and marked the liquid level as 100°F (which means that either he had a slight fever that day or else he didn't measure too accurately. Otherwise, normal body temperature today would be 100°F rather than 98.6°F)

24
next © 2007 Herbert I. Gross Celsius later raised the issue of who cared about either the temperature in Copenhagen on one particular cold morning or Fahrenheit's body temperature on a particular day? Rather he wanted to pick a scale that would be more readily recognizable to any scientist. That is, he wanted to use readings of 0°and 100° to represent temperatures that any scientist could reproduce at any time in any laboratory.

25
next © 2007 Herbert I. Gross Accordingly he used 0°C to represent the temperature at which water froze and he used 100°C to represent the temperature at which water boiled. Using these two scales, it turned out that… -- when the Celsius reading was 0°, the Fahrenheit reading was 32°; -- and when the Celsius reading was 100°, the Fahrenheit reading was 212°. next

26
© 2007 Herbert I. Gross In other words, there are 180º F per 100º C. 100º C212º F Written as a fraction 180º F 9ºF 100º C 5ºC = This explains why m = 9/5 0º C 32º F 100º C180º F next

27
© 2007 Herbert I. Gross Let's now turn our attention to an interesting relationship that is “almost linear”. More specifically there are times when the rate of change is not constant, but the rate of change of the rate of change is constant. “the rate of change of the rate of change” might sound a bit like a tongue twister, but it is something that we deal with quite often. Rate of Change

28
next © 2007 Herbert I. Gross For example, the speed of an object is already a rate of change. Namely, it is the rate of change of the distance the object travels with respect to the time it took to travel this distance. The acceleration of the object is the rate of change of its speed with respect to time. Thus, acceleration may be viewed as a rate of change of a rate of change. next For example, if the speed is measured in feet per second, the acceleration could be measured in feet per second per second or feet per second per minute, etc.

29
© 2007 Herbert I. Gross So, for example, suppose the acceleration of an object is 2 feet per second per second. In other words, suppose that the speed of the object increases by 2 feet per second every second. If the object starts from a stopped position, at the end of 1 second its speed is 2 feet per second; at the end of 2 seconds its speed is 4 feet per second; at the end of 3 seconds its speed is 6 feet per second etc. Thus, the speed of the object is not constant, but its rate of change (acceleration) is constant.

30
next © 2007 Herbert I. Gross Another example is the rate of change of the area (A) of a square with respect to the length (L) of one of its sides. The formula is A = L 2. In the chart below the third column represents the rate of change of A with respect to L when L increases by 1; and the fourth column represents the rate of change of the third column with respect to L when L increases by 1.

31
next © 2007 Herbert I. Gross In terms of a chart… LA 11 24 39 416 525 636 749 Rate of Change 3 5 7 9 11 13 Rate of Change of the Rate of Change 2 2 2 2 2 next A = L 2

32
next © 2007 Herbert I. Gross There is an interesting pattern that seems to be indicated in the previous chart. For example: 1 + 3 = 4 1 + 3 + 5 = 9 1 + 3 + 5 + 7 = 16 1 + 3 + 5 + 7 + 9 = 25 1 + 3 + 5 + 7 + 9 + 11 = 36 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 next = 2 2 = 3 2 = 4 2 = 5 2 = 6 2 = 7 2 next

33
© 2007 Herbert I. Gross To see the last equality more visually, we may think of 49 (that is, 7 2 ) as the area of a square, each of whose side is 7 units. In the diagram below notice that the shaded L-shaped regions represent the odd numbers, 1, 3, 5, 7, 9,11, and 13. 1 3 5 7 9 11 13

34
next © 2007 Herbert I. Gross Another example is to suppose that we want to form the sum of consecutive integers starting with 1. For example, if we let S n denote the sum of the first n integers starting with 1, we see that… S 1 = 1 S 3 = 1 + 2 + 3 = 6 S 2 = 1 + 2 = 3 S 4 = 1 + 2 + 3 + 4 = 10 next

35
© 2007 Herbert I. Gross -- In forming S 2 we added 2 to the previous sum. S 1 = 1 S 3 = 1 + 2 + 3 = 6 S 2 = 1 + 2 = 3 S 4 = 1 + 2 + 3 + 4 = 10 next -- In forming S 3 we added 3 to the previous sum. -- In forming S 4 we added 4 to the previous sum. -- In general in forming S n we add n to the previous sum. -- Thus, each term in the sum is 1 more than the previous term.

36
© 2007 Herbert I. Gross As the number of terms in the sum increases the process of adding term by term becomes more and more tedious. For example, suppose we want to compute the value of S 200 ; that is, 1 + 2 + 3 + 4 + …+ 198 + 199 + 200. Certainly, as we've already seen it wasn't tedious to compute, say S 4, but imagine trying to keep on going this way until we got to 200. More specifically, the main problem with this approach is that in order to find the sum of the first 200 whole numbers, we first have to find the sum of the first 199, etc. next

37
© 2007 Herbert I. Gross However, there is a pattern to the terms that gives us a way to form the sum without having to add the terms successively. The pattern is based on the fact that the difference between successive terms is always the same. For example, the logic we are going to use would have been the same if we wanted to find the sum of the first 200 even positive integers; that is… 2 + 4 + 6 + 8 + 10 +....... + 400

38
next © 2007 Herbert I. Gross To see the “trick” let's look at a simpler case such as S 5 1 + 2 + 3 + 4 + 5 -- Notice that if we start at 1 and look at the terms from left to right, each term is 1 more than the previous one; but if we start at 5 and look at the terms from right to left each term is 1 less than the previous one. next

39
© 2007 Herbert I. Gross Since the increase of 1 balances a decrease of 1, it means that if we start at both ends and add the terms in pairs, each of the pairs will have the same sum. This might be more visual if we represented the above in the form… S 5 = 1 + 2 + 3 + 4 + 5 S 5 = 5 + 4 + 3 + 2 + 1 2S 5 = 6 + 6 + 6 + 6 + 6 = 5 × 6 2 2 S 5 = 5 × 3 And if we now divide both sides of the equation by 2, we see that… + next

40
© 2007 Herbert I. Gross The idea is the same whether there are 5 terms or 200 terms or 2 million terms! Namely, when we add any number of consecutive whole numbers starting with 1, the sum of the first term and the last term will be 1 more than the last term (because no matter what the last term is, the first term is always 1); and if we pair the terms starting from both ends, each of the pairs will have the same sum. In the case of adding the first 200 whole numbers, the sum of the first term and the last term is 201; and since there are 200 terms there are 100 pairs; so the sum will be 201 × 100, or 20,100.

41
next © 2007 Herbert I. Gross More generally, if we let S n stand for the sum of the first “n” positive whole numbers… S n = 1 + 2 + 3 +........... + n, next S n = (n + 1) n 2 Then… = (n + 1)n 2 = (n + 1) n 1 2

42
next © 2007 Herbert I. Gross A Geometric Interpretation To visualize why, for example, S 5 = 5 × 6, 2 look at the figures below… next Think of square tiles where each tile represents 1 unit of area. The shaded tiles in the above diagram represent 15 units of area, which may also be viewed as 1 + 2 + 3 + 4 + 5 = S 5

43
© 2007 Herbert I. Gross next In a similar way the shaded tiles below also indicate 15 units of area as well as the sum 1 + 2 + 3 + 4 + 5 = S 5

44
© 2007 Herbert I. Gross next We may put these two diagrams together to form the rectangle below. On the one hand the area of the rectangle is 6 × 5 or 30 units, and on the other hand it is 2 × S 5. Hence S 5 = 6 × 5. 2

45
© 2007 Herbert I. Gross Another Application Suppose there are 6 people in a room (whom we shall name simply by 1, 2, 3, 4, 5, and 6), and they all exchange handshakes. We can compute the number of handshakes in 2 ways.

46
next © 2007 Herbert I. Gross Each of the 6 people shakes hands with each of the other 5 people. Thus it would seem that there were 6 × 5 or 30 hand shakes. However, this method counted each handshake twice; for example, when 1 shakes hands with 2 it's the same handshake as when 2 shakes hands with 1. Therefore the total number of handshakes is 6 × 5. 2 One Way

47
next © 2007 Herbert I. Gross Another way to think of it is to visualize that the room is initially empty and the people come in one at a time; say in the order of 1, 2, 3, 4, 5 and 6. The Second Way 1 comes in first and there is no one to shake hands with. next 2 shakes hands with 1; which we will denote by (2,1) 3 shakes hands with 1 and 2; that is (3,1) (3,2) 4 shakes hands with 1, 2 and 3; that is (4,1) (4,2) (4,3) 5 shakes hands with 1, 2, 3 and 4; that is (5,1) (5,2) (5,3) (5,4) 6 shakes hands with 1, 2, 3, 4, 5 and 6; that is (6,1) (6,2) (6,3) (6,4) (6,5)

48
© 2007 Herbert I. Gross The number of handshakes on the one hand is given by… 6 × 5, 2 Handshake Summary and on the other hand by… 1 + 2 + 3 + 4 + 5. Hence, 1 + 2 + 3 + 4 + 5 = 6 × 5 2 5 × 6 2 = next

49
© 2007 Herbert I. Gross A Note on Arithmetic Series The sum 1 + 2 + 3 +... + n is a special case of what mathematicians refer to as an arithmetic series. An arithmetic series is any sum in which the difference between two consecutive terms is constant. next Definition

50
next © 2007 Herbert I. Gross The first term can be any number we choose and we then pick the remaining terms so that the difference between consecutive terms is always the same. (The sum 1 + 2 + 3 + 4 + 5 + 6 +... + n is the special case of an arithmetic sum in which the first term is 1 and the difference between consecutive terms in the sum is also 1). next

51
© 2007 Herbert I. Gross For example, suppose we choose the first term to be 6 and we then choose each new term to be 4 more than the previous one. Thus, the arithmetic series would look like… T 1 = 6 next T 2 = 6 +10 T 3 = 6 + 10 + 14 T 4 = 6 + 10 + 14 + 18 T 5 = 6 + 10 + 14 + 18 + 22 T 6 = 6 + 10 + 14 + 18 + 22 + 26 T 7 = 6 + 10 + 14 + 18 + 22 + 26 + 30

52
next © 2007 Herbert I. Gross The important point is that as we move from left to right each term is 4 more than the previous term; and this is equivalent to saying that as we move from right to left each term is 4 less than the previous term. Point

53
next © 2007 Herbert I. Gross So, for example, one way to compute, say, T 6 is to mimic what we did previously. Namely, we write the sum twice, once from left to right and once from right to left. In this way we obtain… T 6 = 6 + 10 + 14 + 18 + 22 + 26 T 6 = 26 + 22 + 18 + 14 + 10 + 6 2T 6 = 32 + 32 + 32 + 32 + 32 + 32= 6 × 32 2 2 T 6 = 3 × 32 And if we now divide both sides of the equation by 2, we see that… + next = 96

54
next © 2007 Herbert I. Gross Notice that to find the sum T 6 … -- First, we added the first term (6) and the last term (26). -- Then, we multiplied this sum by the number of terms in the sum (6). -- Finally, we divided the above answer by 2.

55
next © 2007 Herbert I. Gross The above procedure would work for finding the sum of any arithmetic series. Namely… Step 1: Add the first term (F) and the last term (L). Step 2: Multiply the result in step one by the number of terms in the sum (n). next Step 3: Divide the answer in Step 2 by 2.

56
© 2007 Herbert I. Gross In terms of a formula… T n = n(F + L) 2 In the special case, T n = 1 + 2 + 3 + 4 + 5 +... + n, F = 1 and L = n next

57
© 2007 Herbert I. Gross An interesting closing example might be to revisit our earlier example concerning the fact that the sum of consecutive odd numbers, starting with 1, is always a perfect square. By way of review… 1 = 1 2 1 + 3 = 4 = 2 2 1 + 3 + 5 = 9 = 3 2 In general, the sum of the first n consecutive odd numbers is n 2. Thus for example, the sum of the first 30 odd numbers is 30 2 or 900. next

58
© 2007 Herbert I. Gross However, suppose we hadn't noticed this fact and wanted to find the sum of the first 30 odd numbers; that is: 1 + 3 + 5 + 7 + 9 +... + 59. The above is an arithmetic series because each term always exceeds the previous term by 2. next

59
© 2007 Herbert I. Gross -- The sum of the first term (1) and the 30th term (59) is 60. -- 1,800 ÷ 2 = 900 next -- The number of terms is 30 and 30 × 60 = 1,800. Applying the Formula to the Sum of the First 30 Odd Numbers. next = 30 2

Similar presentations

OK

Key Stone Problem… Key Stone Problem… Set 17 Part 2 © 2007 Herbert I. Gross next.

Key Stone Problem… Key Stone Problem… Set 17 Part 2 © 2007 Herbert I. Gross next.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on articles of association sample Ppt on omission of articles in grammar Dentist appt on saturday Ppt on old age homes Ppt on flash animation tutorial Ppt on construction company profile Ppt on solar system for class 6 download Ppt on lines and angles powerpoint Ppt on supply chain management of nokia windows Ppt on abs system in bikes