CONTENTS

Introduction

Order of Topics

The Real Number Line

Data

Understanding Percents

Working with Fractions

Ratio, Proportions, and Compound Interest

Complex Fractions

The Fundamental Counting Principle

Mathematical Induction

Probability

Bernoulli's Backdoor Approach

The Binomial Expansion

Pascal's Triangle

Common Mistakes (Probability and Odds)

The Birthday Problem

Properties of Exponents and Logarithms

Popular Algebraic Techniques

Factoring

Synthetic Division

Imaginary Numbers

The XY Coordinate System

Functions

Function Composition

1-1 Functions

Wind Chill formula

Curve Fitting

The Development of Calculus

The Concept of a Limit

The Formal Definition of a Limit

Limits and Continuity

Uniform Continuity

Special Infinity Limits

All About the Derivative

Rate of Change

Getting the Derivative

Tangents and Normals

The Mean Value Theorem

Optimization Problems

Solving Max-Min Problems

The Biggest Box

The bug-Frog Problem

Curvature

Integration

Popular Integration Applications

Finding The Area Under a Curve

Finding Total Distance From a Velocity Function

Differential Equations

Solving Differential Equations

Carbon 14 Dating

Using The TI-83 for Calculus Problems

Conclusion

A JOURNEY IN MATHEMATICS TO REMEMBER: from the real numbers to calculus and beyond

Introduction

I plan to take the reader from the very basics to the world of calculus and beyond in a path that will concentrate on select topics that I chose. The topics that I included are ones that I have written for my students of statistics and calculus while teaching at Marist University. I am only assumming a minimal math background to understand them. Some may take deep thought and study while others are very straight forward. My goal is to give the average math student a better feel for math by giving them an approach that will solidify understanding. Mathematics is a state of mind and it takes a very special approach for one to feel comfortable in this wonderful domain. I kept the theory and rigor at a minimum so the average student is not overwhelmed. For those of you who are math oriented and seek a more comprehensive understanding, I refer you to the textbook I cherish greatly, "Thomas' Calculus, Eleventh Edition". In my opinion, the best calculus text ever written for the serious student.

I will start with the numbers most of us have worked with during our lives. Without the understanding of how numbers work and their relationships with each other, there cannot be much understanding of mathematics.

The Real Number Line

Integers: ...,-4,-3,-2,-1,0,1,2,3,4,... Most people deal with those numbers on the right side starting with 1. Negative numbers seem difficult for most people to understand since they rarely deal with them. An easy way to think about them as money you owe. -1 represents a debt of 1 dollar & so on. All of these numbers can be represented by a point or dot on the real number line. However, you must realize that there are gaps between these numbers where other numbers exist. If we travel to the far right on the number line the numbers get larger and conversely, they get smaller as we travel to the left. So, -25 is smaller than -3, while 25 is larger than 3. Tricky. Now, let's take a look at the numbers in the gaps between these whole numbers (integers).

Rational Numbers: Any number that can be expressed as a/b where a and b are integers (whole numbers). So, if b=1, they will be whole numbers since every whole number has a denominator of 1 (which is not written). For example, the number 14 is really 14/1. However, if the denominator is not 1, then we start to fill in the gaps. For example 7/8 is not a whole number. It is close to 1. To see how close your fraction is to a whole number just use a basic calculator and divide the top by the bottom. If you do this with 7/8, you will get 0.875. Moving the decimal two places to the right gives you the percentage 87.5%.

Try to write all mixed numbers as one fraction. That way, you can better get where it is located on the number line. For example, 3&5/8 can be written as 8x3+5 over 8 or 29/8. Doing the division, we get 3.625. This will be just a little pass the midpoint between 3 and 4. Knowing the location of numbers is important. Just remember, on the number line, a number to the right of another number is always larger. So, 3.625 is larger than 3.622 but smaller than 3.630. Special mathematical symbols are used for "greater than" and "less than". a>b means a is greater than b or saying differently, b is less than a, written b < a. So, we can state 3.6 > 3.5 and 7.88 < 7.91.

Irrational Numbers.: Not rational. Numbers like the square root of 2 (two identical numbers that multiply to give you a result of 2) is not rational. It can be approximated by 1.414, however, it goes on indefinitely (never ending & non-repeating). On the other hand, .3333...for ever can be written as 1/3, therefore is a rational number. It turns out that every repeating decimal is a rational number. This is easily proven. There are infinitely many rational and irrational numbers. One very popular irrational number you might have been exposed to is Pi. The circumference of a circle divided by its diameter regardless of the size of the circle.

Data

Discrete or Continuous VS Qualitative or Quantitative Data::

Very confusing for beginning students. Discrete Data has "gaps" between data points. There are no data points in these gaps.
Examples are: a given number of objects being counted such as a given number of bottles, number of people, or number of chemical elements in a sample.
Usually a positive whole number is associated with each object. The integers (including the negatives and zero) form a discrete set since there are no integers between any two consecutive integers.

Continuous data has no "gaps". Between any two data points there will exist another data point. Continuous implies no breaks or holes. The real number line is an example of a set that form a continuous set. Between any two real numbers there exists another real number. A curve (coming up) that is continuous can be traced with a pencil without lifting it off the paper.

Qualitative data is descriptive in nature. It has no numerical measurement associated with each data point. It is not metric in nature (measurable).
Examples are, chemical elements in cat food, your gender, your likes or dislikes, and so on.

Quantitative data is data that is measured. It is metric in nature. Any number of objects counted, your weight, your height, temperature, amount of certain chemicals in cat food and so on are examples.

Discrete data could be quantitative. For example, the number of fish you catch on a camping trip. Usually, discrete data is qualitative more often. When a measurement is involved, it becomes quantitative. Phone numbers and addresses form discrete data sets which are qualitative in nature since, without any added definitions, they do not measure anything. They are used for identification purposes and location.

Colors are tricky. If dealing with a finite number (countable) of colors, then we have a discrete set. This set is qualitative in nature (not measured). However, if dealing with the visible spectrum, then the colors are continuous (determined by their wavelengths) and this set would be quantitative in nature. So, be aware of these situations. A bit on the tricky side.

The levels of measurements are nominal, ordinal, interval, and ratio.
Nominal: Consists of names, gender, labels, or categories only. Data that cannot be arranged in any ordering scheme.(usually discrete data)
Ordinal: Data concerned with order or rank. Race results, letter grades in a course, college football rankings, and so on. The difference in data values is not measured.(usually discrete data).
Interval:(Metric data). Obtained from the measurement of quantities such as temperature and time. (usually continuous data). Uses a constant scale with differences having meaning but ratios do not (since there is no natural zero). (absence of the quantity). For example, the difference of 80 degrees & 40 degrees can be measured but you can not conclude that the ratio of 80/40=2 means it's twice as hot.
Ratio: Metric data uses a continuous, constant scale. Both differences and ratios can be measured. There is a natural zero. Examples would be the measurements of height and weight.

Scientists usually separate data (collected pieces of information) into two main types:

a) Categorical or Qualitative (data points are placed in groups)

Data in this group may contain numbers and may be ordinal in nature (i.e., house numbers), but performing mathematical operations with these numbers (adding, dividing, etc.) make no sense. Most of the time, the data sections describe a characteristic of that group (i.e., male, female, or rating scales giving different levels of satisfaction). In many cases, numbers are used for labeling in each section (i.e., 1=male, 2=female). Even though this is nominal data, it can be Ordinal (the numbers are placed in some order) (i.e., house numbers).

b) Metric or Quantitative (data points that give measure, i.e., how much )

Data of this type is mostly used in scientific fields. It may be Ordinal, as well, along with the Ratio & Interval types.

Scientists most likely use the terms Discrete & Continuous when referring to Quantitative data.

For chemistry majors, a course in Qualitative Analysis will deal with the identification of the elements in a given sample, while a separate course in Quantitative Analysis, deals with the exact amount of each element detected. I've taken both courses while in college.

Understanding Percents

First of all, here are some bascis for the non-math person

To calculate a percentage, place the number in question over the base amount representing the total or starting point.

This will give you a fraction. It could be improper (top bigger than bottom).

Then, divide the top by the bottom. This result may be a whole number or a decimal.

Convert this result to a percent by moving the decimal point two places to the right. Only now, we use the % symbol.

Example: 5 is what % of 17?

Solution: 5/17 = .2941 (four places) = 29.41 %

Example: 18 is what % of 6?

Solution: 18/6 = 3 = 300% (decimal is to the right of a whole #)

To convert a % to a decimal, move the decimal point two places to the left & DROP the % symbol.

IMPORTANT: A very common mistake beginning students make using percentages in calculations is that they fail to convert the percentage to a decimal during the calculation. So, for all operations dealing with percentages (adding, subtacting, multiplying, dividing, & powers), percentages must be converted to a real number (without the % symbol). That means, it must be in decimal form.

For example, for a problem dealing with compound interest, if the percentage is 6.7%, it must be expressed as .067 during the calculation.

Note: If no decimal is visible, it is understood to be at the far right of a number.

Example: 37 % = 37.% = .37

To convert a % to a fraction, place the percentage (WITHOUT THE % SYMBOL)over 100.

Example: 22.5 % = 22.5/100 = 225/1000 = 9/40 (divide top & bottom by 25 to get the reduced form).

Note: percentages can be probabilities but not always. If they are between 0 % and 100 %, they could represent a probability since probabilities are always numbers between 0 and 1.

However, 230 % = 2.30 can not represent a probability since this number is greater than 1.

You should be aware of how percentages are computed & how they can be very deceiving & misleading.

Always remember that percentages are computed on a base amount.

If the base amount is large, a modest increase will be a very small percentage increase. However, if the base amount is very small, a modest increase will be a large percentage increase.

For example, an amount increases from 500 to 550. That's a 50 point increase. The % increase is based on the original amount of 500, so (increase divided by the starting amount) 50/500 = 0.1 = 10 % increase.

On the other hand, if the amount is 2 and increases to 7, that's a 5 point increase but represents 5/2 =2.5= 250% increase.

In dollar amounts, the 10% increase in the first calculation ($50) is 10 times the 250% increase in the 2nd calculation ($5). So, if someone asks, "which is greater, a 10% increase or a 250% increase?", that cannot be answered unless you know the base amounts.

% increases & % decreases are used extensively in the business world (stock market & others reporting statistics) & can be very deceiving.

If you buy a stock at $100 a share & hold on to it & it declines to $50, you have just suffered a $50 dollar loss per share or 50% decrease in your investment (a decrease 50 divided by a starting amount of 100 = 0.5 or 50%).

That same stock would have to increase 100% from $50 to get back to the price you originally paid (50 divided by a starting amount of 50 = 1 or 100%).

So, a 50% loss followed by a 50% gain does not get you back to the original point.

A common mistake made by students on exams is the following: Take a stock that decreses from 100 to 50. That's a 50% decrease. To get back to 100 from 50, some students figure that since 100 is 2 times 50, the % increase must be 2=200%. They figure that 200% of 50 is 100, so give the required % increase as 200%. This is in error. The increase is not 200% of the starting point of 50. That would be an increase of 100. That would take us to 150. The proper way of looking at this problem is: an increase of 50 represents what % of the starting point of 50. That would be 50/50 = 1 = 100%. A bit confusing to say the least. One problem deals with the increase as a percentage of the starting point while the other deals with the percentage of the starting point to get you where you want to go.

The tech crash of 2000-2001 was worse than the Dow Jones crash of 1929 for many tech investers. On Oct. 21st, 1929, we had a market crash that sent the Dow Jones average from 400 to 145 (by Nov.). That's a 64% decrease. At the absolute bottom, the Dow had lost 89% of its value a few years later. It took approximately 25 years for investors to get their money back. Many think that was the worse in history. Not so. During the tech crash of 2000-2001, many popular technology stocks fell over 95%. For example, one of my favorite stocks (at that time) was AMCC (Applied Micro Curcuits). It's price went from $108/share to $3/share (97% decrease). It now trades at about $2.80/share. It would have to increase 3,757% to get back to its Oct. 2000 price (not likely). Many other tech stocks suffered similar losses. Many tech investors lost hundreds of thousands of dollars. So, for most tech investors, this was much worse than the '29 crash.

Penny stocks (stocks that usually sell for under $1 are notorious for huge percentage increases & decreases. (since base amounts are relatively small)

A stock selling for $0.25 could go to $3.00 for a $2.75 increase (2.75/.25=11= 1,100% increase), (but only $2.75 in gain), while another stock at $450 goes to $600 for a $150 gain or 150/450= 0.33 = 33% gain.

Generally speaking, traders buy large amounts of low priced stocks to capture huge % gains (or suffer huge % losses). Small gains in penny stocks can give huge returns on your investment since thousands of shares are usually purchased. However, most traders lose all or most of their money.

Stores often offer sales after secretly inflating their prices. After a quiet markup of 20%, they will offer a 15% sale price. Not a good deal. Other stores have signs that state "save 50%". Well, you don't save a penny, you never save when you spend.

Working with Fractions

In my opinion, slicing & eating pizza is an excellent way to understand the meaning of fractions.

All numbers are fractions. Whole numbers are fractions with bottoms (denominators) of one. (usually not written)

Tops (numerators) can be smaller than the bottoms (proper fractions) or larger than the bottoms (improper fractions)(not for negatives).

The number in the bottom (denominator) gives the number of identical slices of your pizza pie (before you & your friends indulge).

The number in the top (numerator) corresponds to the number of identical slices about to be eaten.

If you wish to understand 8 ths, slice your pizza pie in 8 identical slices, for 16 ths, 16 identical slices, for 5 ths, 5 identical slices, & so on. You have the freedom to slice it into as many identical slices as you wish.

For improper fractions (top bigger than bottom), you need more than one pizza pie. For example, 8/5 means you are to eat 8 identical slices but your pie has only 5. So, you need another identical pizza pie sliced up in 5 slices so that you can eat 3 more. Total eaten here would be one whole pie plus 3 slices out of 5 from the 2nd pie. The 2nd pie still has 2 slices left.

Adding pizza slices:

Adding fractions is like adding pizza slices. However, there is a rule in human nature that says you can't eat different size slices, they must all be identical. A problem lies with pies that are sliced in different numbers of slices, since some slices will be smaller or larger than other slices. So, how does one make all the slices the same shape? Necessary before adding?

Let's say you want to add 3/8 and 5/24. You might be able to do this with one pizza pie, if there is enough. You cannot slice a pie into slices more than once, so be careful the first time. For these two fractions, one person wants 8 identical slices but another person wants the same pie sliced into 24 identical slices. You need to satisfy both. Look at the bottoms & figure the number of slices that will be the LCM (least common multiple). This LCM is a number that contains all factors of 8 & 24 exactly once. (smallest number that is exactly divisible by both 8 & 24). It is 24. So, slice your pie into 24 identical slices. The 8, in the first fraction, is now 24. If you change the bottom, the top must be changed also. Since 8 was multiplied by 3 to get 24, you need to multiply the top by 3 also, to get 9. So, the first fraction is now 9/24 (it's like going in a completely opposite direction of reducing the fraction). Now the pizza pie can be sliced into 24 identical slices & you can add them. 3/8 becomes 9/24 (eat 9) & for 5/24 (eat 5 more) for a total of 14 out of the 24 slices (14/24). That leaves a lot of pizza slices left in that pie.

The LCM is know as the LCD (lowest common denominator).

Since subtraction is just adding a negative, think of subtracting slices as taking them away from the person who is about to eat them & putting them back into the pie. For example, 4/5 minus 3/7. Here again, the numbers in the bottoms indicated different size slices..a no-no. We need to find the LCM of 5 & 7. Since they do not share any factors, it will be their product, 35. So, the pie is sliced into 35 identical slices. The 4/5 becomes 28/35 and the 3/7 becomes 15/35. For our subtraction problem, the first person takes 4/5 of the pie or 28 out of the 35 slices. But you take from the person 15 of those slices & put them back into the pie. So, the person will only eat 28-15 or 13 slices out of the 35 (13/35). This is the answer to the subtraction.

Multiplying & dividing:

It's easier to multiply & divide fractions compared to adding & subtracting them. Just know the rules.

For multiplying, Tops times Tops & bottoms times bottoms. For example, (5/7)(7/2)=35/14. You can reduce (equivalent fraction with smaller numbers) this to 5/2. Notice that if you cross out any number from the bottom with the same factor in any of the tops, its OK. For (5/7)(7/2) we can cross out the 7's & get our answer quickly & not have to reduce. This can save us a lot of time. Here's another example: (11/19)(3/8)(19/5)(8/11) equals 3/5. Note that the 11's, 8's, & 19's can be crossed out.

The rule for dividing Two fractions is also simple. Just flip the fraction you are dividing by and change the operation to multiplication. Then follow the rules above. For example, 7/8 divided by 49/24 = (7/8)(24/49). Now notice that 24 contains a factor of 8 ( 3 times 8), so the 8 divides the 24 & leaves a 3. Likewise, with the 7 & 49, the 7 divides into the 49 & leaves a 7 (in the bottom). Remember, you can only do this with two numbers (one of which is in the top & the other in the bottom). So, our problem becomes (1/1)(3/7)= 3/7.

Here is a useful short-cut for dividing a fraction by a number (or quantity in higher math). Try to divide the top by the number (quantity). If it does, just write the answer over the original bottom. For example, 210/73 divided by 7. Since 7 divides the top, 210, do it. The answer is 30/73. If the number does not divide the top, just multiply the bottom & write the answer. For example, 29/63 divided by 7. Since 7 does not divide 29, just multiply the bottom & write the answer, 29/441.

When working with mixed numbers, such as 3 & 8/11, convert all of them to fractions (since all numbers are fractions) & use the rules above. For example, (3 & 9/11)(2 & 5/6) = (42/11)(17/6) = (7/11)(17/1)=119/11 Note that the 6 in the bottom divides the 42 in the top & leaves a 7 (in the top).

When working with complex fractions (tops and/or bottoms contain fractions)(see the following topic), change the top & bottom to single fractions & use the division rule. For example, [(1/3)+(3/5)] divided by [(7/8)+(1/4)] = [(5/15)+(9/15)] divided by [(7/8)+(2/8)] = (14/15) divided by (9/8) = (14/15)(8/9) = 112/135. Notice that no factors are common to the numbers of the tops & bottoms. So, when you write the final answer, it is already reduced.

Comparing the value of fractions (which is larger or smaller) is also very easy to do. Just use a calculator (or long division) & take the top & divide it by the bottom, this gives the decimal equivalent. Just look at it & you'll see which one is bigger. For example, compare 37/83 & 56/123. The decimal equivalent of the first is .446 (3 places), the decimal equivalent of the 2nd is .455, so the 2nd is slightly bigger.

Also, if two fractions have the same bottom, the one with the bigger top is larger. Similarly, if two fractions have the same top, the one with the smaller bottom is larger. (not true for negative numbers, however). They have their own set of rules.

Be extra careful with negative numbers. Any negative number is smaller than any positive number. Also, larger negative numbers are smaller than small negative numbers. For example, -3,000 is much smaller than -.001. Think of the x axis & the position of the number. A number to the right of another number is larger. So, -.001 is way to the right of -3,000 on the x axis, so it is much larger.

There are several different approaches to explaining fractions, this is just one. I find it the best for people who have fraction phobia.

Hope you enjoyed this little pizza trip. I'm hungry...for pizza.

Complex Fractions

All numbers can be considered fractions. Whole numbers have denominators of 1.

if there are smaller fractions in the numerator and also in the denominator, then the expression is called a complex fraction.

Examples: (8/9 + 1)/(7/6), (5/6 + 7/8)/(3/7 - 4/5), in arithmetic, (2/x - x/y)/(7+5/x) in algebra.

to simplify any complex fraction, we need to find the LCD (also called LCM) of all denominators of the smaller fractions involved in the top & bottom.

after finding this, we multiply the top & bottom of our complex fraction by this LCD.

if done correctly, all smaller fractions will disappear, leaving just one fraction.

Example: simplify the first complex fraction above. Note that the smaller fractions involved are 8/9, 1/1, & 7/6. The LCD of 9,1, & 6 is 18.

Multiplying the top & bottom of our complex fraction by 18, we get:

[(18)(8/9) + (18)(1/1)]/[(18)(7/6)] =[(2)(8) + (18)]/(3)(7)= [16+18]/21 = 34/21 (reduced)

Example: simplify the 2nd complex fraction above: Note that the smaller fractions involved are 5/6, 7/8, 3/7, & 4/5.

The LCD of 6,8,7, & 5 is (2)(3)(4)(5)(7).

Multiplying the top & bottom of our complex fraction by this number, we get:

[(4)(5)(7)(5)+(3)(5)(7)(7)]/[(2)(3)(4)(5)(3)-(2)(3)(4)(7)(4)]=[700+735]/[360-672] = 1435/(-312) = -1435/312 (reduced

note: when the LCD is large (like above), it is best to leave it in factored form when multiplying the top & bottom. The smaller denominators will be part of that LCD & will cross out, leaving the rest of the factors.

note: to determine that -1435/312 is reduced, just x-ray each & you will see that they are relatively prime (i.e., no factors in common).

Example: simplify the third algebraic complex fraction above:

Notice that the smaller fractions involved are 2/x, x/y, 7/1, & 5/x.

The LCD of all denominators is (x)(y)(1) = xy. So, multiplying the top & bottom of our complex fraction by xy, we get: [(xy)(2/x) - (xy)(x/y)]/ [(xy)(7/1) + (xy)(5/x)]= [2y - x2]/[7xy + 5y] (reduced)

note: reducing an algebraic fraction is much easier, if you understand the techniques of algebra involved. (nothing is easy without understanding)

Ratio, Proportions, & Percents

A Ratio of a to b is a division, a/b. It is a way of comparing two quantities or categories. Also written a:b

~Note: A ratio does not necessarily give you the exact amounts of each category. It may or may not.

Ex: If there are 17 girls and 11 boys in a class, the ratio of girls to boys would be 17/11. Since this can not be reduced, the exact numbers appear. However, if a class consists of 18 girls and 12 boys, the ratio would be 18/12 and could be given in reduced form as 3/2, obviously, this form does not give the exact numbers of girls & boys in the class.

Use the following set-up for Ratio Word Problems: If the ratio of two categories is a to b, let the first be represented by ax and the second by bx. (since a & b are usually not the exact numbers of each). Then, in basic problems, take their sum and equate this to the total which is usually given.

Ex: The ratio of people 21 or older to those under 21 at a given event is 7 to 5. There were 228 people attending this event. Find the number of each that attended.

Solution: Let the # of people 21 or older = 7x & the # of those under 21= 5x. Since the total attendence was 228, we get 7x+5x=228 or 12x=228 or x=19. So, the # of people 21 or older would be 7x=7(19)=133 and those under 21 would be 5x=5(19)=95.

A Proportion is established when two ratios are equal. The form will look like this: a/b = c/d or a:b = c:d.

The Means in a proportion are the elements b & c (the inside elements in the second form)

The Extremes in a proportion are the elements a & d (the outside elements in the second form).

Properties of a Proportion:

1) In any proportion, the product of the Means = the product of the Extremes. This is known in common terms as Cross-Multiplication. If two ratios are not equal, then there is no proportion & these products would not be equal. This is a good way of checking whether or not you have a proportion. (or if two fractions have equal value).

2) You may interchange the elements of the means and/or the elements of the extremes at any time. This can save you time in solving some equations involving proportions.

Ex: a/b = c/d, a/c = b/d, d/b = c/a

3) You may take the reciprocal of both sides of a proportion. Note that this changes the means & extremes. Ex: a/b = c/d, b/a = d/c

4) You may take any element & multiply the other element along the means or the extremes. Ex: a/b = c/d, a = bc/d or c =ad/b or b = ad/c or d = bc/a

Ex: Solve for x: 15/7 = 37/x. Interchange 15 & x and then take the 7 & multiply the 37. We get: x/1 = (7)(37)/15 = 17.27 (two decimal places) (you could also use the conventional way of cross-multiplication)

If two quantities are related LINEARLY, proportions are useful for finding many solutions.

Ex: Under normal conditions, an individual can walk 2 miles in 22 minutes (my estimation). How far can this individual walk in an hour & a half?

Solution: Convert one hour & a half to 90 minutes (units must be the same). Set up a proportion where the tops are from the same category & the bottoms are from the other category. We get: x/90 mins = 2 miles/22 mins. The minutes will go & leave you with miles. The answer is 2(90)/22 miles = 8.18 miles (two decimal places)

~Note: Many refer to this method as the factor-label method.

Ex: A 185 lb diabetic needs 22 cc of insulin daily. Under normal conditions, how much does a 135 lb diabetic need?

Solution: x/135 lb = 22 cc/185 lb. The label of lb will go & leave you with cc's. We get x = 135(22)/185 cc's = 16.05 cc's.

Solving problems involving percentages: Translate the problem into an equation & solve.

1) Use the decimal equivalent of the percent in the equation.
2) The word of translates to multiply.
3) The word is translates to equality.
Ex: In a given town, 47.3% of the residents live in apartments. The total number of residents of the town is 3,755. (give answer to the nearest whole resident).

Solution: x = (.473)(3755)= 1776 (nearest whole resident)

Ex: 56.8% of students at a local college drink Pepsi. This number is estimated to be 2,766. How many students attend this local college?

Solution: .568 x = 2766 gives x = 2766/.568 = 4870 (nearest integer)

Percent Increase & Percent Decrease (missed often by many) (Important in business)

To find the percentage increase or decrease, find the amount of increase or decrease & place it over the STARTING POINT.

Ex: You purchased a penny stock for $0.24. In one day session it increased to $1.13. What is the percentage increase?

Solution: % increase = (1.13-0.24)/0.24 = 3.71=371% (nearest %)

Ex: You do not sell this stock (bad decision). The next day, it goes back down & closes at the original price you paid, at $0.24. What was the percentage decrease in this stock that day?

Solution: % decrease = (1.13-0.24)/1.13 = 0.79 = 79% (nearest %)

Note: The bottoms in the above problems are different since their starting points differ.

Compound Interest: P = Po (1 + r/n)^nt . Po represents the amount deposited at an APR (annual percent rate) of r %, compounded n times a year, the amount, at the end of t years, will be P.

Example: How much must be invested at the annual rate of 12%, compounded monthly, in order for the value of the investment to be $20,000 at the end of 5 years? (P=20,000, r=.12, n=12, t=5, find Po). Let Po=x. Substituting in the above formula, we get:

20000 = x(1 + .12/12)^(12)(5) = x (1.01)⁶⁰, then x = 20000/(1.01)⁶⁰ (do 20000 divided by (1.01)⁽⁶⁰⁾. Note: On scientific calculators, the symbol ^ is a power symbol. For example: 2^4 means 2 raised to the 4th power, which means (2)(2)(2)(2)=16

I recommend the TI-83 calculator for those going on to more advanced phases of math.

x = $11,009 (nearest dollar).

Mathematical Induction

This is one of the most powerful tools of mathematics. It's used quite extensively in higher mathematics for proving expressions involving n cases. (i.e., expanding formulas from 2 & 3 dimensions to the nth dimension)

It also gives us a way of working in higher dimensional spaces. Generally speaking, there are no visuals in 4-D, 5-D, 6-D, & so on, but we can still work in them with equations & formulas.

It gives a valid way of generalizing without committing the typical fallacy of "hasty generalization". It is a very powerful & remarkable method.

To understand the logic behind this method, think of an infinite row of standing dominoes, spaced perfectly, so if you tip over the 1st, it will tip over the 2nd & the 2nd will tip over the 3rd, & so on....all dominoes will fall (if the spacing between them is correct. (/->|->|->|->......)

Think of a dominoe falling as an expression being true for that case.
(i.e., 2nd dominoe falls means expression is true for n=2, 3rd dominoe falls means the expression is true for n=3, & so on)

Our goal is to have the expression true for all values of n. So, our goal is to show that all the dominoes fall.

How do we get all the dominoes to fall? First show that the 1st one falls. That would be equivalent to showing that the expression is true for n=1. We could do that just by substituting n=1 into the expression & showing that it holds (is true).

Now, for the tricky part. We must show that the spacing between any 2 dominoes is correct. What does that mean? It means, no matter where we are, let's say at the kth dominoe, if it falls, does it tip over the next one? The next one is the (k+1) dominoe. If dominoe k knocks down dominoe k+1, no matter what k is, then all will fall & the expression is true for all n. Because, if k=1, then 1st knocks down 2nd, then 2nd knocks down 3rd, & so on.

Showing k tips k+1 is the key. In words, it goes this way, "if k falls, does it
tip k+1?" This is an implicative statement. We assume the if part then we show the then part follows. (i.e., assuming the kth falls, does it knock over k+1?) Which means, if we assume the expression true for k (kth falls), can we then prove it true for k+1 (k+1 falls)? If so, then the expression will be true for all positive integers n (our goal).

Here we go. Our first induction proof.

Since 1= 1², 1+3=2², 1+3+5=3², one might guess that

1+3+5+...+(2n-1) = n² That would be a very good guess, right?

Note the expression (2n-1) gives the last term of our sum, where n stands for the number of the term. (i.e., if n=3, we have a sum of the first 3 terms & the expression (2n-1) will be 5, the last term. Then n² would be 3² or 9, the sum of the 1st 3 terms.

Now, let's prove it by math induction on n.

1) Does the 1st dominoe fall?
(i.e., is the expression true for n=1?), 1=1²? yes.

2) Assume the kth dominoe falls.
(i.e., assume the expression true for n=k)

Note: k can be anywhere along the line. Just sub k in for n.

   1+3+5+...+(2k-1)=k² (we accept this & can use it)

3) Does the kth dominoe knock down the next one, k+1?

(Must show the expression true for n=k+1).

Using the expression in 2) & any permissible math technique(s),
can we show the expression holds for k+1?

There are several different approaches to show step 3. One of which is adding the same expression to both sides of 2. It is always good to know what you are after or want to show.

Here is what we want to show. (expression true for k+1)

1+3+5+....+(2k-1)+[2(k+1)-1]=(k+1)² (can't use this, however)

Note: The term after (2k-1) would be [2(k+1)-1]
(just increase k by 1] this is (2k+1).

So, to get it 3), add the quantity (2k+1) to both sides of our assumption
in 2).

It's reasonable to do this since, by adding (2k+1) to both sides, the left side of 2) becomes the left side of what I want to show.

Now, does the right side come out to the right side of what I want?
If it does, we are done.

We get, 1+3+5+...+(2k-1)+(2k+1)= k² +(2k+1)

which is the same as: 1+3+5+...+(2k+1)= k²+2k+1

(don't need to write the term before the last one on the left side)

Since k²+2k+1 = (k+1)² , we have

1+3+5+...+(2k+1)= (k+1)² , yippee!

Note that this is the original formula when k+1 is in the formula for n.

We have just shown that the expression is true for n=k+1, if true for n=k. That means all dominoes will fall---> expression is true for all n.

This is one of the simplest induction proofs, some are lengthy & require much fooling around to get that 3rd step. However, the logical structure of all of them has been described in this essay.

Hope you enjoyed this little lesson on mathematical induction. Save it for future reference.

The Fundamental Counting Principle

If event A has n ways of occurring and a second event B m ways of occurring, then the product (n)(m) will be the number of ways both could occur in succession. A very basic multiplication rule that is very useful in solving many problems related to probability.

Example 1: Consider the digits, 1,2,3,4,5,6. How many 3-digit numbers can be made using a digit once and only once?

Solution: Use a model like, _ _ _, where the far left spot is the hundreds place, then the next spot, the tens, and the far right spot, the ones place. Since there is a restriction that the digits can only be used once, we must keep this in mind when solving this problem. Usually, no repetitions allowed is stated in the wording of the problem.

Now, we must consider the number of possibilities for choosing a digit for each spot. Since there are no restrictions on location, we can start at any spot. Starting from the left, there will be 6 choices, then only 5 for the next spot, and 4 for the remaining spot. By the fundamental principle, our answer will be the product, (6)(5)(4) = 120 numbers can be formed.

Many problems have either natural or stated restrictions, so, one must go to those positions first and fill them in before the rest. The following is an example.

Example 2: (stated restriction). Consider the digits, 1,2,3,4,5,6. Let us ask the same questions as in example 1 above but must have a 5 in the ones place.

Solution: Since the restriction that a 5 must be in the ones place, we fill that spot first. Since there is only one 5, there is only one choice for that spot, the 5. Now, go to the other spots and find the possibilities. Since no repetition is allowed, there will be 5 choices for our next spot, and 4 for the last spot. Again, using the fundamental principle, we have, (5)(4)(1) = 20 numbers that can be formed.

Example 3: (two stated restrictions. Take the digits, 1,2,3,4,5,6. How many even 4-digit numbers can be made starting with 2?

Solution: There are two restrictions (other than no repetition of digits) in this problem. The first is, starting with a 2 and the second would be that the 4-digit number be even. Since a 2 must start and is an even digit, it is eliminated for a choice for the end spot. That would only leave two choices for the end, a 4 or 6, since these numbers must end in an even digit to be an even number. So, go to these spots first and figure the possibilities, before the two spots in the middle. Remember, no repetitions allowed. So, multiplying our choices, we get, (1)(4)(3)(2) = 24 . Keep in mind, these numbers are possible ways not the digits given in the problem.

Note: Many problems allow you to repeat the digits. However, be aware, there could be restrictions as well.

Example 4: Examples with repetition allowed.

Solution: For example 1, the 6 digits may be used for each of the 3 spots. We get, (6)(6)(6) = 216 numbers.

For example 2, since our numbers must end in a 5, there is still only one choice for that spot, however, there will be 6 choices for each of the remaining spots. We get, (6)(6)(1) = 36 numbers.

For example 3, there is still only one choice for the first spot, however, there will be 3 choices for the end spot since the 2 can be repeated and is an even digit. There are no restrictions on the middle spots, so there will be 6 choices for each. We get, (1)(6)(6)(3) = 108 numbers.

PROBABILITY

The heart of Statistics (probability distributions, hypothesis testing, & so on)

Probability vs. possibility (almost anything is possible as time passes)(rolling a sum of 15 is impossible with a pair of dice)

Experiment: (procedure) the activity we are involved with (rolling dice, taking a survey, betting a horse, & so on)

Event: an outcome (getting a 7 with a pair of dice, people who fly in planes from a sample, horse A wins, & so on)

Sample space: all possible outcomes or events.

The dice triangle: 1 way of rolling a 2, 2 ways for a 3, 3 ways for a 4, & so on. Note that a 7 can be rolled 6 ways. (most likely roll)

                                7
                            6       8
                        5               9
                    4                      10
              3                                11
          2                                          12
          1 2   3 4 5 6   5 4   3   2    1 ----> 36 possible outcomes in the sample space.

Definition: The probability of an event A = P(A) = k/n, where k= the number of ways event A can occur n= the total number of ways of all possible events in the sample space. (the bottom number in the probability includes the number in the top).

Note: Venn diagrams are very popular for visualizing the events in a given sample space.

Types of probability:

1) a priori: mathematical, known in advance, does not change (dice, cards, coins, roulette, and so on)

2) Statistical or experimental: based on trials or collected data (horse racing, weather, & so on). Not fixed.

Ex. A sample of 855 randomly selected adults is taken. 710 said they have flown in an airplane. Find the Probability that a randomly selected adult has flown in an airplane.

Let F=event of flying. So, P(F)=710/855 = .83 or 83% (3 ways to express probability)

Ex. of 1): Find the probability of rolling a "hard 8" with a pair of dice. P(4 & 4)=1/36

Ex: Assuming having a male or female born is equally likely, find the probability that a couple has 2 girls & one boy when having 3 children. List the sample space for this experiment first. Sample space: GGB,GBG,BGG,GGG,BBB,BBG, BGB, GBB. P(2G&1B)= 3/8 = .375 or 37.5% chance

If P(A)=0, then event A can not occur.

If P(A)=1, then event A must occur.

If P(A)>.5, then event A is likely to occur.

If P(A)<.5, then event A is not likely to occur.

If P(A)=.5, then it is equally likely that A does or doesn’t occur

(this is know as a 50-50 chance)

Generally in most all problems, 0 < P(A) < 1 (answers >1 or <0 make no sense)

Note: The Complement of A, designated as A bar or A' would be all points (events) in the sample space outside of A. (all outcomes for which A does not occur).

Note: P(A) + P(A') = 1 so, P(A) = 1 - P(A'), this is the "backdoor" approach to probability & must be used in many problems. (see below)

Odds: Come from probabilities, so to get odds, you must get the probability first.

If P(A)=1/6 , A is not likely to occur , odds are against A occurring , 5 to 1 against or 5:1 against

If P(B)=4/7 , B is likely to occur , odds are 4 to 3 or 4:3 in favor

If P(L)=1/2 , there is a 50-50 chance of occuring, odds are even and are 1 to 1 or 1:1.

Many textbooks confuse students, i.e., they state 3 to 7 in favor?, should read 7 to 3 against)(the bigger number should always be placed first)

At many betting parlors, you will see odds like 2-7, which means that you must bet $7 to win $2. Also, more common, +120, means you need to bet 100 to win 120, or -170, bet 170 to win 100

Note: the 2 numbers for the odds always add to the bottom number of the probability.

Reduce odds, if you can ( i.e., 66 to 11 is reduced to 6 to 1, 4 to 2 is reduced to 2 to 1, & so on)

Note: know how to go from odds back to the probability. ( i.e., 3 to 2 against A, P(A)=2/5)

Note: Payoff odds (involved in games of chance) are usually not the same as the real odds. The game presenter (casino) is usually in business to make money, so odds will favor the "house". So, in the long run, the "house" will always win. But, in the short run, anything is possible. "luck" is usually used for short-term success or even a run of successes.

To compute the payoff odds use: NET PROFIT to AMOUNT OF BET

Note: A game is considered fair, if the payoff odds are the same as the real odds. (no body wins in the long run)

Ex: In roulette there are 38 numbers on the wheel. Usually, 18 red, 18 black, 0 & 00 green. Placing a bet on one position gives you a P(win) = 1/38. So, the real odds are 37 to 1 against you. However, the casino will pay 35 to 1. So, for a $1 bet on a number & it comes out, the casino will pay you $35. Your net gain is $35 (you keep the $1 you bet). There are many other bets you can make that will increase your odds of winning but will decrease the payoff. The worst odds in a casino is Keno (stay away). The best odds are from Blackjack, Craps (dice table), & Baccarat. You get the biggest "bang" for your buck at Slots (most popular casino game & produces the most revenue), although considered one of the worst games for the odds. (see my discussion on casino slot machines under the statistics link of selective topics)

Many professional gamblers play Poker, since you do not play against the house, but rather against other players. Many can make a living, if they are skilled in reading other players body language.

Probability Rules:

Addition Rules:

P(A or B) = P(A) + P(B), if A and B are mutually exclusive (disjoint)(Events A & B cannot happen at the same time)

P (getting a heart or club when one card is selected) = P(heart) + P(club) = 13/52 + 13/52 = 26/52 = 1/2

2) P(A or B) = P(A) + P(B) - P(A and B), if they are not mutually exclusive.

P (getting a heart or a King when one card is selected) = P(H) + P(K) - P(H & K) = 13/52 + 4/52 - 1/52 =16/52 = 4/13

Important Note: Use the backdoor approacch when dealing with multiple events coming from different sample spaces & repeated events from the same sample space.

Note: the negative of (A OR B) is (not A AND not B)

Example: P( head on a coin OR a roll < 12 on a pair of dice). Since the sample spaces are different, you must use the backdoor approach. This gives 1 - P(no Head and a roll of 12)=1- (.5)(1/36)= .986

Example: P(when tossing a coin 3 times, getting a head on the 1st toss OR a head on the 2nd toss OR a head on the 3rd toss). You cannot use the addition rule directly. The backdoor must be used. We get, 1-P(tail & tail & tail) = 1-(1/2)(1/2)(1/2)=1-1/8 = 7/8 or .875

Multiplication Rules:

P(A and B) = P(A) P(B), if A and B are independent (one event has no effect on the other)

P (head on a coin toss and rolling a 6 on a die) = P(H) P(6) = (1/2)(1/6) = 1/12

4) P(A and B) = P(A) P(B|A), if A and B are dependent. A occurring effects the probability of B. This is conditional probability. P(B|A) means the probability of B given that A has occurred

Ex: P (getting a jack| the card is red) = 2/26 = 1/13 (since there are 2 jacks in the red suits)

Note: Experiments are done (selecting data) with replacement (so data point can be selected again) or without replacement (same data point can not be selected)(see link under topics of interest). Usually with large populations, we do not make a distinction. However, will relatively small sample spaces, a distinction must be made, since probabilities change (rule 4 takes hold).

Note: probability tree diagrams are often helpful in setting up many multistage events.

Note: multiply probabilities along one path, but add results for different paths (outcomes).

Note: Probabilities are written along each stem or limb. The probabilities along each stem horizontally must add to 1. Probabilities are multiplied along consecutive stems to get the desired result. If you have 2 different paths to the end result, the probabilities are added for each completely different path.

Example where the backdoor approach saves much time: A coin is tossed 5 times. What is the probability of getting at least 2 heads?

Translating, P(2 or more heads) = P( 2 or 3 or 4 or 5 heads). Adding probabilities does not work. Here, you must use the basic definition of probabiliy or the backdoor approach. Using the basic definition, we need to figure how many ways we can get 2 or more heads in 5 tosses. Here Pascals triangle is useful (see below). Those numbers for n=5 in Pascals triangle are: 1 5 10 10 5 1. So, for 2 or more, we add 10+10+5+1=26 ways. Then we place that number over the total number of possible outcomes 2⁵ = 32. Using the backdoor approach, P(at least 2H) = 1 - P(complement of at least 2H) = 1- P (0 or 1) = 1 - (1+5)/32 = 1-6/32 = 1 - 3/16 = 13/16 (much easier)

Bernoulli's Backdoor Approach

Here's something that will be very helpful when solving binomial distribution problems involving Bernoulli trials. I called it "Bernoulli's Backdoor", since it involves the "backdoor approach" to probability. Here's how it works.

We know, for a Bernoulli trial, we can calculate the probability of a number of successes. We need to know n (# of trials), probability of a success (p), and the number of successes we want (r). Just use 2nd, VARS, find it on the menu, insert these 3, then Enter.

However, for problems with "at least"or "at most" a given number of successes, we need to do it more than once,then add. (i.e., "at least 3" means 3 or more). ("at most 3" means 3 or less).

If the number of trials is large compared to the number of successes we want, this could force us to repeat the formula many times (i.e., at least 3 successes in 30 trials).

That could be extremely time consuming & also increase the likelihood of computational errors.

So, use Bernoulli's backdoor. That is, find the probability of the negative (complement) of what we want, then subtract it from one. (i.e., for at least 3 successes in 30 trials, it would be
1 - p(less than 3 successes in 30 trials).

That way, you only have to use the formula once (using 2nd VARS, A key on the menu...gives the cumulative probability up to 2 successes), however, you must remember to subtract the result from 1. You can compute the the final result with one calculation by doing the following:

Enter 1 - 2nd Vars, menu A, (number of trials, probability of success, upper limit of the cumulative number of trials). For this problem, enter
1 - 2nd Vars, menu A, enter (30, p, 2).

Some people might not have to use the formula at all, since the combinations are small & can be found easily & the other parts can be calculated simply by the power key.

For the "at most" problems, the backdoor approach is not used since it is already in the cumulative state (i.e., starts at 0 up to the given number of successes). So, menu A is used directly without subtracting from one. For example, to get "at most 7 heads" when flipping a fair coin 20 times, just go to menu A & insert (20, .5, 7).

The Binomial Expansion

Very useful for raising a 2-term expression (binomial) to an positive integer power. It can also be used for any type of power with an expanded interpretation of combinations.

Use for (x+h)³, (2x-5)⁴ (be careful with a negative, it belongs to the 2nd term in the formula).

The formula: (a+b)ⁿ = aⁿ +nC1 a^(n-1)b + nC2 a^(n-2)b² + nC3 a^(n-3)b³ + & so on.

An easy way to remember is to look at the pattern. The powers of a (the first term) start at n and go down one in each term proceeding until it disappears in the last term. The power on b starts in the 2nd term with a 1 and increases one in each term proceeding until it reaches the highest power (n). The combinations start in the 2nd term ( n taken one at a time, then 2 at a time, then 3 at a time, & so on until it reaches n at a time, which has a value of 1 at the end).

When n is a positive whole number, the expansion terminates. The last term will be bⁿ.

If n is relatively small, you can get these from Pascal's Triangle.

If you need to express the combinations in terms of n, use the formula nCr = n!/r!(n-r)!

If your binomial has a negative sign in the middle, this negative belongs to b, i.e., for (x-2)⁵, a=x & b= -2 in the expansion.

Basic examples:

a) Expand (x+2)⁴. We get, x⁴ + 4C1 x³ (2) + 4C2 x² (2²) + 4C3 x (2³) + 2⁴, or simplifying, x⁴ + 8x³+24x² +32x + 16

b) Expand (x+h)³. We get, x³ + 3x²(h) + 3xh² + h³

For the above examples, get the combinations quickly from Pascal's Triangle.

Pascal's Triangle

1
1    1
1    2    1
1    3    3    1
1    4    6    4    1
1    5    10 10    5    1

& SO ON....

1's on the ends. Reading a row, gives you the combinations of the 2nd number (left to right), taken 0, 1, 2, 3, ... at a time.

~For example,
in the above triangle, the 3rd number in the bottom row formed, is 10 (from the left). This will equal 5C2 . It turns out the 5C3 also = 10.

Also, we can use Pascal's triangle to calculate binomial probabilities.

~To do this, note that the sum of the numbers in any given row equals 2ⁿ, where n is always the 2nd number from the left (or right). (i.e., for the numbers 1, 5, 10, 10, 5, 1, we have, 1+5+10+10+5+1=32=2⁵) (here n = 5)

~So, if we are interested in the probability of getting at least 6 heads when tossing 7 coins, we would have to expand our triangle two more rows so the 2nd number on the right is n=7. That would give us the following row:
1 7 21 35 35 21 7 1.

Now for P(at least 6 heads in 7 tosses) = (7 + 1)/ 2⁷ = 8/128 or .0625

Notice that we took the sum of the last 2 numbers in that row since they give us the nimber of ways we can get 6 or 7 heads, respectively. Also note that 2⁷ represents the total possible number of outcomes in the sample space when tossing 7 coins.

Obviously, if the n (number of trials) is quite larger (say n > 20), using Pascal's triangle is laborious, since your triangle would be very wide & it would be a tedious job to construct. Therefore, the use of the TI-83 is recommended. However, just in case you are isolated on an island without a calculator, this could be a nice way to occupy your time.

PROBABILITIES and ODDS---COMMON MISTAKES

1) Giving answers to probabilities that don't make sense.

Probabilities are always numbers between 0 & 1, inclusively.

They can be expressed as decimals, fractions, or percentages.

2) Giving a probability when a question is calling for odds.
An answer for odds gives a whole number to a whole number where the first whole number is larger. Also, specifying that the odds are in favor or against.

3) Trying to use a complicated or an involved method of computing a probability when all that is needed is the basic definition.
The basic definition can be used to solve most problems encountered in an introductory course.

4) Failing to subtract the probability of the intersection of two events when using the addition (OR) rule with events that can occur simultaneously.
If the two events are mutually exclusive, there is no subtraction, simply add their probabilities. However, if they are not mutually exclusive, you must subtract P(A and B) from the sum of P(A) & P(B).

5) When using the backdoor approach, the complement of the event is not figured properly.
The backdoor approach is a valuable method of computing the P(A) indirectly by subtracting the P(not A) from one. Along with an error in figuring the complement, many students forget to subtract P(not A) from one.

There are many problems that cannot be solved directly and this method must be used. A very important approach for computing a probability.

6) Not considering whether or not probabilities of multiple events are computed with or without replacement.
Answers come out different in these cases. Not replacing an object after it is selected changes the size of the sample space for the next selection, consequently, its probability.

7) Misreading data from a contingency table.
Each cell of a contingency table is the intersection of the data indicated in the column heading with the data the row is indicating.

8)Failing to change the sample space in conditional probability.
For P(A|B), event B has occurred. We would like to compute P(A) under that circumstance. The new sample space used for computing P(A) is the sample space for B.

"Birthday Problem"

All students of statistics should be exposed to the very famous "Birthday Problem". It's interesting & defies common sense.

It turns out, in a group of 23 people, it is likely that 2 people share the same birthday (maybe not the same year, however).

The probability for n=23 is .505 (likely), for n=30 it increases to 70%, for n=40, it becomes 90%, at n=50 it's 97%, & at n=60, it's almost certainly the case at 99%.

Here is the math behind it.

Using the backdoor approach to probability, P(2 people have the same BD) = 1 - P(no 2 have the same BD)

So, computing the right side, we have P(no 2 have the same BD) = P(choosing a person with a BD) times P(choosing a 2nd person whose BD is different from first) times P(choosing a 3rd person
whose BD is different from first & second) times so on & so on.

P( 2 have the same BD) = 1 - (365/365)(364/365)(363/365)(362/365) & so on.

When there are 23 fractions, n=23, P=.505. Be careful computing, since trying to do 365! is too large (over flow) for your calculator, so piece by piece, multiplying & dividing will get it.

LAWS of EXPONENTS and LOGARITHMS

For those of you who will go on to calculus courses and beyond, you will be using many tools of pre-calculus while solving calculus problems. Among the many are these properties. Many calculus problems deal with functions (coming up) that are involved with exponents and logs. So, knowing how to work with them is essential. For others, you may never get the opportunity to work with them (some will be happy about this).

(a^x)(a^y)= a^(x+y), a^x/a^y=a^(x-y), (a^x)^y=a^(xy)
(ab)ⁿ=(aⁿ)(bⁿ), (a/b)ⁿ= aⁿ/bⁿ, 1/aⁿ=a^(-n)

a = b if and only if aⁿ=bⁿ raising each side of an equation
to the same power---mainly used to eliminate radicals
(fractional exponents)

LOGARITHMIC PROPERTIES

log_a(xy)=log_ax + log_ay
(the log of a product is = the sum of the logs of each factor)
(or, multiplication in the domain is equivalent to addition in the range)

  log_a(x/y)=log_ax - log_ay
  (the log of a quotient is = the difference of the logs in that order)
  (or, division in the domain is equivalent to subtraction in the range)
  (don't confuse log_ax / log_ay with this one)
  (this is the quotient of two logs) (the division of two range values)

  log_a(uⁿ)=nlog_au
  (very useful for "knocking powers down")

  A = B if and only if log_aA=log_aB
  (taking the log of both sides of an equation)

  *ln u = log_eu (natural log)
  (special symbol for the natural log) (on calculator)

   log u = log₁₀u (common log)
   (base is not written) (on calculator)

   *ln e = 1 (important feature of the number e)

   *ln A = B if and only if A = e^B (used much---very popular)
    (both ways)

    *ln(e^u)=u

    *log_aa=1

    *u= B^[log_Bu] (B raised to the power of log_Bu equals u)

    *log_AB= (ln B) / (ln A) (converting any log to natural logs)
     (could convert to any other base log by replacing ln by that base log)
     (i.e., log_AB=[log₇B] / [log₇A])

    ~Note: many students "mess up" calculus problems because of the
                misuse of properties of expos & logs...so, know them well...

~Note: Many times, these properties are used in reverse (i.e., instead of "knocking down" a power using a log property, we place the power "back up" on the quantity). So, know how to use them in both directions.

POPULAR ALGEBRAIC TECHNIQUES

Students that do poorly in higher level math courses (calculus & beyond) do so not necessarily because the concept is too difficult to apply but instead in the set-up and the steps that follow after we apply the calculus concept. For example, in a typical calculus problem, the actual calculus may call for the construction of the biggest box given a required volume with dimensional requirements (coming up in an example). Most of the work involved is not calculus, it is in the set-up of the general dimensions in terms of varible quantities and the algebra techniques involved after we apply the calculus concept. A function (coming up) is required and a derivative (coming up) needs to be calculated. Setting up the function (no calculus) could be more difficult to do than getting the derviative. Getting the derivative would not take very long but what follows to the conclusion of the problem requires a strong background in algebra, geometry, or some other technique covered in a pre-calculus course. Applying the initial calculus concept could be just 10% of the problem and the other 90% to the conclusion is centered around their pre-calculus knownledge. Their weaknesses are in their backgrounds involving techniques in these basic areas. So, for the serious student, there is no substitute for a strong pre-calculus backgound as a prerequisite for success in higher level courses.

1) Combining like terms and simplifing.

Example: Subtract -4x⁴-5x³+2x²-22 from 7x⁴-3x³+7x²-32x+17

Solution: Form the difference & simplify

7x⁴-3x³+7x²-32x+17 - (-4x⁴-5x³+2x²-22) = 7x⁴-3x³+7x²-32x+17+4x⁴+5x³-2x²+22
= 11x⁴+2x³+5x²-32x+39

2) Solving quadratic equations by factoring.

Example: Solve: 6x²-16x-70=0

Solution: divide both sides by 2 to get: 3x²-8x-35=0

factor the left side of the equaiton: (3x+7)(x-5)=0

set each factor equal to zero: 3x+7=0, x-5=0

solve each linear equation for x: x= -7/3, x=5

3) Solving simple linear inequalities.

Note: Inequalities are expressions involving the symbols for "greater than", "less than", greater than or equal to",
"less than or equal to". Solve these for x the same way you would solve equations with one exception. If you
multiply or divide by a negative quantity, the inequality reverses direction. In most basic types, you can avoid
this in the process of solving for x.

Example: Solve for x: 7-5x >0 (start by adding 5x to both sides to avoid dividing by -5 while solving for x)

Solution: 7 > 5x
7/5 > x (dividing both sides by 5) (the inequality does not have to be reversed)
reading the inequality from right to left, "x is less than 7/5"

4) Solving simple equations for a given quantity.

Example: Solve the following equation for b: 5ac - 10bc = 15c

Solution: divide both sides by 5c as long as c does not take on the value of zero.
a - 2b = 3.
add 2b to both sides: a = 2b +3
subtact 3 from both sides: a - 3 =2b
divide both sides by 2: (a - 3)/2 = b

Most other techniques used in the course will be covered in my lessons and those that involve the TI-83 calculator will be covered in detail as they come up during the course.

Factoring: Factoring an number or expression means to express it as a product of numbers or product of expressions. Prime factoring of numbers will express the number as a product of the prime numbers that make it up. For example, say you would like to factor the number 105 into primes. Start with the smallest prime (2) and see it it divides it. It does not. Then try 3 (the next prime). It does and gives 35 as a quotient. So far, we have 105=(3)(35). It is factored but not into primes. Doing the same with the quotient 35, we will get 35=(5)(7). So, combining the results, we have 105=(3)(5)(7). It is now factored into primes.

In algebra, the most common type of factoring is taking out a common factor from each term. This is just the opposite of the distributive law in mathematics. For example, take the two term expression xb-yb. We can see that b is a common factor in each term. So, take it out front and express it this way: b(x-y). Sometimes there are more expressions that are common to each term. For example, 3(x²)(y³) + 12xy will have 3xy as the greatest commom factor in both terms. So, take that out front and write it in factored form, 3xy(xy² + 4). You can always check to see if you factored it correctly by multiplying it out to see if you get the original expression back.

Difference of two squares: Expression like x² - y² can be factored into (x-y)(x+y). Some times the squares are more complicated like, (a²)(b⁴) -(a⁴)(b²).
In this case, we get take out the common factor of (a²)(b²) and write it as [(a²)(b²](b² - a²), then finally writing it as [(a²)(b²](b-a)(b+a).

Factoring the Sum and Difference of Cubes: Every once in a while, these occur in limit expressions(coming up) and is necessary to factor them in order to get the correct answer. Usually, they will occur in limits that result in 0/0 when substituting into the expression.

Both forms are factored in very similar ways. There are just sign differences.

The Sum of two Cubes:

a³ +b³ = (a+b)(a²-ab+b²)

The Difference of two Cubes

a³ -b³ = (a-b)(a²+ab+b²)

For example 8x³-27y³ = (2x-3y)(4x²+6xy+9y²)

See my section dealing with expanding binomials below. This will be very helpful in computing the derivatives(coming up), using the limit definition, for expressions containing x to powers of 3 or higher.

SYNTHETIC DIVISION

These techniques are used at all levels, including calculus & beyond.
In a basic calculus course, you will probably use synthetic division for finding certain types of limits & finding factors or zeroes of poly functions. (If they are rational numbers)

What's behind the technique?

If f(x) is a poly fcn and you divide it by a linear expression, of the form (x-b), then there will be a quotient & remainder.
The remainder, R=f(b). [i.e., f(x)/(x-b)=q(x)+ R/(x-b), where q(x) is the quotient (one degree less than f(x)].
Note that the remainder R is expressed over the divisor.

If the remainder R=0, then f(x)=q(x)(x-b), which means that (x-b) is a factor of f(x).
Also, If R=f(b)=0, then x=b would be one root of the equation f(x)=0 (also called a zero of the fcn).
The other solutions would come from the quotient, q(x)=0.
So, to find them, you would have to work with the quotient.

Synthetic division gives us an easy way to find remainders and quotients. So, when we get a R=0, we also find factors and, therefore, the roots of f(x)=0.

When you use Synthetic division, you are dividing f(x) repeatedly by expressions of the form x-b. It will let you know when a R=0 pops up.
When that does, you have found a linear factor of f(x) and a root of f(x)=0. Then working on the quotient, q(x), the same way, you can find the other factors.

If f(x) doesn't contain linear factors, forget it. Use some other method.

How does Synthetic division work? Follow these steps.

1) Write the poly fcn, f(x) in descending powers of x, making sure that any missing power is visible with a zero coefficient.
i.e., Write x⁵+4x⁴-10x²-x+6 like this: x⁵+4x⁴+0x³-10x²-x+6
note that the 3rd power of x is missing from the expression & we have to account for it with a zero coefficient.

2) Now, drop down the coefficients (just the numbers) 1   4   0   -10   -1   6

3) Now, decide what you want to divide by. Remember you will be dividing by linear expressions of the form x-b. So, a good place to start is to divide by x-1. This uses the number 1 in the process.

What's the process?

Drop down the leading coefficient of 1 (far left). Multiply the leading coefficient (which is 1 in this case) by 1 (since you are dividing by x-1) (if you were dividing by x-2, you would use 2) (if you were dividing by x+1, you would use -1) (if you were dividing by x+3, you would use -3) get it! We chose to start dividing by x-1, so we are using 1.

After you multiply the leading coefficient by 1 (in your head), then add the result to the next coefficient & write the result. Continue with this technique. i.e., multiply that result by 1 (in your head) & add it to the next coefficient & write the result. If you do it correctly, you will get the next row of numbers.

1    4    0    -10    -1    6
1    5    5    -5    -6    0   ←new row of numbers (using 1 in the process)

The last number in the row is the REMAINDER. Notice that R=0, so we know that the division by x-1 was exact, so x-1 is a factor & x=1 is a root of the poly f(x)=0 since R=f(1)=0. Also, we get the quotient which is defined by the other numbers & is one degree less than the original expression. i.e., 1x⁴+5x³+5x²-5x-6. We can now work with these coefficients to find more factors and more roots.

We can leave those numbers up there & just change our reference row to
1    5    5    -5    -6

What happens when we don't get a 0 at the end? Well, we still get the remainder, f(b) but no factors or roots, so we try again with another divisor, like x-2 (we would be using 2 in the process). We would continue to try divisors of the form x-b until a 0 pops up. When that happens, then we can change our reference row to work on the quotient for the other factors & roots.

In some problems (not here) if we try 2 numbers & the remainder, f(b) changes sign, we would know there is a root between them, since the graph of the poly fcn would have to cross the x axis & it is a continuous function (the intermediate value theorem for continuous fcns). So, we can then try a rational fraction in the process. (between those numbers). Yes, this process works for rational fractions also. If you try a rational fraction & get a fractional result (when multiplying, then adding) you can stop. It won't work. Also, if the leading coefficient is A and the last coefficient is B, all rational fractions that will work must have tops which are factors of B & bottoms which are factors of A. Just a little time saving device.

So, we have found one factor of the expression, namely, x-1.
Using Synthetic division on the new row, try 1 again (it is possible for an expression to have repeated factors), So, here we go.

1    5    5    -5    -6
1    6    11    6    0 ←yes, another 0, so x-1 is a factor again.

Change to the new row & try it again (it's possible for an expression to have
the same factor multiple number of times)

1    6    11    6
1    7    18    24 ←no such luck, x-1 is not a factor for the 3rd time

Now, forget about that row (when there is no zero) & try x-2 (using 2)

1    6   11   6
1    8   27   60 ←we can see, no matter what positive number I use, I
                          will never get a zero here, so lets try some negatives.
                          lets test x+1 (using -1 in the process)
1    6   11   6
1    5    6   0 ←yes, so x+1 is a factor also. Now, the new quotient with
                         the coefficients 1,5, & 6 is quadratic, since we started
                         with the 5th power & found 3 factors (remember, every
                         time we divide, the quotient drops a degree)

Since the 3rd quotient is quadratic, take it out of the problem & try to factor
this expression by conventional means.

x²+5x+6 → (x+2)(x+3)

So, in summary, we have factored x⁵+4x⁴-10x²-x+6 into 5 linear factors
(x-1)(x-1)(x+1)(x+2)(x+3). And, if we set this poly=0 we find the 5 roots,
namely, 1,1,-1,-2,-3. These are the zeroes of the poly (x intercepts).

In some cases, the poly might not have all linear factors. It could have a factor like x² +1 that can't be factored any further. Hopefully, these factors can be found after finding the linear ones by Synthetic division (When the last quotient is quadratic).

The concept of a Limit is coming up. You will be able to understand this example much better after it is covered or you could go to it now then come back to this example.

Example in calculus:   Let's say you want the answer to the following limit:
                               lim     (x³+64)/(x+4) , it's that 0/0 again
                               x→ -4

So, you'll need to fool with it. How do you factor x³+64? Well, if you forgot, use Synthetic division to find them.

Write x³+64    like 1    0    0     64   & try 4
                                 1    4    16    128 nope!, no 0, useless to try positives,
try -4                         1    -4    16    0 ←yes, a 0, so x+4 is a factor.
                                now, back to the limit

            lim      [(x+4)(x²-4x+16)]/(x+4)   = lim (x²-4x+16) = 48, nice!
           x→-4                                            x→-4

Notice that the 2nd factor was formulated by the other numbers, 1,-4,16 above.
That gives, x²-4x+16 for the other factor (which happens to be the quotient)

Hope you enjoyed this lesson & it helps out on future problems.
Save it for reference.

IMAGINARY NUMBERS

Note: We will be dealing with real numbers in our discussions. However, be aware that mathematical analysis can be applied to non-real numbers (imaginary numbers) as well. Before I continue with the real numbers, let me give my readers a brief description of Imaginary numbers and how to work with them. They are used extensively in higher math & in many engineering related problems.

To understand them, you must understand the basic unit of the system,
namely the square root of -1. (which is given its own symbol)

The square root of 1 is 1, since (1) (1) = 1, the square root of 4 is 2, since (2)(2)=4, the square root of 9 is 3, since (3)(3)=9, & so on. These are some of the "nice ones". (come out very nice).

Notice that to be a square root of a number, the number times itself must equal the number. The square root of 2 doesn't come out nice, but is real. You can find it on your calculator. It is irrational (unending, non-repeating decimal). It can be approximated by 1.414. Likewise for the square roots of 3,5,6,7, & so on. All of these numbers are real, since they can be found on the real number line.

However, the square root of -1 is not. You will not be able to find a real number such that when multiplied by itself gives -1. So, this number does not exist as a point on the real number line. So, we call it imaginary & give it the symbol i.

This is the basic imaginary unit. All other imaginary numbers can be expressed in terms of i.

The Complex Plane

A 2-D system (similar to the x & y axis system) is invented to represent these types of numbers. The horizontal axis consists of the real number line & the vertical axis is the imaginary axis, I.

Since all numbers are complex numbers, i.e., form a + bi, where a is real part & bi is the imaginary part, they can be represented using this axis system.

If b=0, then the complex number is real (no i's involved) & each number would be on the horizontal axis (real axis).

If b is not 0, then the complex number is not real & is off the real axis.

For example, to locate 3+2i on the axis system, you would start at the origin & move 3 units to the right on the real axis then move 2 units up in the imaginary direction.

Any number off the horizontal real axis is imaginary. To locate -7-4i, you would move 7 units to the left of the origin on the real axis then move straight down 4 units in the imaginary direction. Therefore, all numbers of the form a+bi can be represented by a point in this system (called the complex plane).

If a=0, then the complex number is of the form bi, or pure imaginary.

To locate 3i, just move straight up from the origin 3 units. It will be on the imaginary axis. Likewise, -8i would be down 8 units from the origin on the imaginary axis.

So, all pure imaginary numbers are on the imaginary axis, while the numbers 2+3i, -7-5i, -4+7i, 8-3i, & so on, are still imaginary, but not on the imaginary (vertical) axis.

Working with complex numbers

The rules for working with complex numbers follow the same principles as we used for the real numbers, with few exceptions. Some new rules must be established for multiplication:

(i)(i) = i²= -1, i³ = (i)(i)(i) = (i)²(i) = (-1)(i) = -i, i⁴ = (i)²(i)² = (-1)(-1) = 1

Since i²= -1 & i⁴= 1, imaginary numbers can give a real number result when multiplied.

Note that i³=(i)(i²) = (i)(-1) = - i. Odd powers of i give an imaginary result.

We see that there are only 4 powers of i, namely, i,-1,-i,1. Every power of i higher than 4 can be reduced to one of these, just divide the power by 4 & take the remainder for the power.

If the remainder is 1, then the answer is i,
If the remainder is 2, then the answer is -1,
If the remainder is 3, then the answer is -i
If the remainder is 0, then the answer is 1

These are know as the 4 powers of i (i,-1,-i,1)

For example, take i²³⁷. Divide 237 by 4 to get a quotient of 59 with a remainder of 1, So, i²³⁷ = i¹ = i

Here are some examples of adding , multiplying, subtracting, & dividing complex numbers.

Ex: (7-3i) + (8+23i) = 15 +20i (just add the real parts & the immy parts separately)

Ex: (2-3i)(5+2i) = 10 +4i -15i -6i²= 10 -11i -6(-1) = 10 -11i +6 = 16 -11i

Ex: (4-9i)-(-5-6i) = 4-9i+5+6i=9-3i

Ex: (2-5i)/(1-3i) = (2-5i)(1+3i)/(1-3i)(1+3i) [multiplying top & bottom by the conjugate of the bottom, i.e., (1+3i)].
That gives (2+6i-5i-15i²)/(1+3i-3i-9i²) = (2+i-15(-1))/(1-9(-1)) = (17+i)/10

The uses for these numbers is beyond the scope of this course, however, for those of you who take more advance math & science courses, you will certainly encounter them.

Now, back to the real numbers.

The XY Coordinate System

I first started our discussion with the introduction of the number line where all real numbers are located. (Rational & Irrational numbers)(we will call this line the X-axis). Imagine yourself at the zero point. This I will call the origin. You have the freedom to travel to the right (greater than 0 and beyond) or to the left (less than 0 in the negative direction). For each position you located, there will be one number on the X-axis associated with it.

Now, imagine another number line (exactly the same as the first at a 90 degree intersection of our previous line (X-axis) where the zero of that line is exactly the same as the zero of the X-axis. We will call this line the Y-axis. So, if you are positioned at the point where the lines cross, there will be a 0 for the X-axis and a 0 for the Y-axis. We will designate this location using the following symbol (0,0) where the first zero MUST come from the X-axis always. So now your freedom of travel has increased from left or right (on the X-axis) along with up or down (in the Y-axis direction). This way, you can travel anywhere in the plane determend by X & Y axes. But always remember, the first number in the ordered pair is always from the X-axis direction.

Let's take the ordered pair (3,2). To get there from your starting point (0,0)(the origin), you must travel 3 units to the right, then go up 2 units in the Y-axis direction.
Let's take the ordered pair (-5,7). To get to that point, we must move 5 units in the negative direction on the X-axis, then up 7 in the Y-axis direction.
Let's take the ordered pair (-3,-5). To get there, we must travel 3 units in the negative direction on the X-axis, then move down 5 units in the Y-axis direction.
Let's take the ordered pair (0,-10). To get to this point we do not move at all on the X-axis and move directly down 10 units in the Y-axis direction. You will be a point on the Y-axis.
Every point in the plane determined by these two axes can be located. This is called The XY coordinate plane. This plane will serve as visual for many of the important concepts of the calculus. Later, for the serious math student, it will be expanded to 3-D by introducing another axis (Z-axis) with the development of Multivariable Calculus.

Functions

Calculus deals with the study of Functions and their behavior. So, it's very important that the student have a good understanding of these important class of curves.

I mentioned that functions are special curves, so what is a curve? A curve is nothing more than a visual for an infinite grouping of points (for most cases) in a given pattern. Remember, each point is an ordered pair of numbers, first from the X-axis, 2nd from the Y-axis.

There is usually a rule (equation) that dictates how the x and y coordinates are related. For example, if the y coordinate is always twice the x coordinate, the set of points will line up in a definite pattern. Here are some of them: (1,2), (2,4), (3,6), ....,(100,200),....,(-20,-40), & so on. There will be an infinite number of these ordered pairs that will determine the pattern of the curve. In this case, all of these points will lie on a straight line determined by the rule (equation), y=2x. This will be the equation of this line. So, in this case, y=2x describes a curve which is a straight line.

We can get a visual of this line by plotting (placing a dot at many points determined by the equation) on the XY plane. This is called a graph of the curve determined by its equation.

A different equation would, most likely, produce a different kind of line or curve.

For example, y=3x-5 would produce different ordered pairs but would still line up straight. So, it will be an equation of a different line. However, y=x^2 (x squared) would not produce a straight line. It would be curved up with its lowest point at the orgin. This curve is called a Parabola. Fortunately, we have calculators that will give us a graph of any equation we enter into the input. A great time-saving device.

For the serious math student who looks into this more deeply, they will notice that the XY coordinate system is not convenient nor possible for locating points on some curves. So, other types of movements are invented that make it much easier or possible for analyzing different types of curves. For example, instead of traveling left or right then up or down to locate a point, it may be more convenient to rotate a certain number of degrees (think of the orgin as a little circle) about the origin then going outward a given distant. This is the basis of a new coordinate system based on "polar coordinates". So, each order pair would not have an x and y, but instead a distance out and an angle rotation. These will be the polar coodinates for the curve. Many very interesting looking curves are developed in the polar system, many of which are very difficult to describe or impossible to get in the XY system. So, when someone mentions a different coordinate system, just think of it as a different path of locating points on the curve. In multivarible calculus (Calculus III) there are several that are developed and used.

Back to our XY system, certain curves are Functions and some are not. So, in short, what makes a curve a Function. Keeping it in simple terms for the beginner, just think of a curve that will satisfy the vertical line test. That is, if all vertical lines that cut the curve do so in exactly one point only, it's a Function. This will eliminate circles, ellipses, and many other curves from the class of functions. The serious math student will be exposed to other methods and ways of determining the class of Functions & techniques that can be used to study non-functional curves.
Another simple way of looking at this property will be with ordered pairs. If verial lines cut our curve in exactly one point only, then there will never be two or more ordered pairs (points on the curve) on the same vertical. Say it another way, the graph of a function will never have two points on its graph with the same first element (x coordinate with different y-coordinates). For example, (7,3) and (7,5) cannot be on the graph of a function (these points are on the same vertical line where all x's are 7). Which, in turn, illustrates a vertical line cutting the graph in more than one point.

Functional Notation: Special notation is used when working with functions. Let's go back to our equation y=2x which determined a straight line when graphed. In this equation, the y coordinate will always be twice the x coordinate. Since this curve determines a function, special notation is is used. We say, y is a function of x. In symbols, y=f(x). This does not mean f times (x). It's just a way of stating that this curve is a function. It does not give the equation unless it is stated what the relationship is specifically. If I say f(x)=2x, then I know the functional equation. The important feature is that f(x) is taking the place of y. We could state all the ordered pairs on this function as (x,f(x)). So, y and f(x) have the same meaning. However, by using the functional notation, it becomes very convenient when the equation is in the form f(x)=2x. That way, we can input any x and get instant results. For example, for this function, I can say f(0)=2(0)=0, which means that when x=0, y=0. I can describe the ordered pairs in just one line equation instead of two. I can say quickly, f(1)=2, f(5)=10, f(-4)=-8, & so on. Much more concise then saying when x=-4, y=-8. Of course, you always have the option of not using functional notation and going back to the, so called, y notation. But always remember, when using functional notation, f(x) is replacing y.
Let's take a few more examples:

f(x)=3x²-5x+4, f(0)=4 (when x=0 is substitued, the y-value or functional value is 4). f(1)=3(1)²-5(1)+4=2, f(-1)=3(-1)²-5(-1)+4=12 (tricky).

f(x)= 4/(x²-4), f(0)=4/-4=-1, f(1)=4/(1²-4)= -4/3. Note: x=2 or x=-2 cannot be used since it creats a zero in the denominator (bottom). We say, that 2 and -2 are not in the domain of the function. The Domain of a function consists of all x-values that produce a y-value. These do not. The y values produced by the acceptable x-values are called the range of the function. The domain & range are a way of stating the scope of the function (values for which the function has meaning).

Since a functional value (y value in the range) depends upon an x value in the domain, then x is the independent variable while y is the dependent variable. In most applied problems in calculus, there is meaning to the functions used. For example, if we wanted to analyize the movement of a particle, we would consider the distance traveled as a function of time. In this case, we would use s for distance and t for time and the function would be described as s=f(t), where t is the independent variable. In business, if we were interested in profit as a function of items produced (x), we would consider a profit function P(x). So, depending on what we are analyizing the varibles may change.

Composition of Functions

This is a way of combining 2 or more functions in a special way. Think of a function as a mapping that takes a point in one set (pre-image) to a point in another set (image). (i.e., set A to image points in set B)(x to f(x))

The domain (accepted values) of the function f are the points in set A. A point in the domain of f, x, is taken from set A to its image under f to f(x) in set B.

The set of all images under f are points in set B. The set of these image points is the range of f.

There might be points in set B that are not image points under f, thus not in the range of f.

Let g be a function that takes image points in B to points in set C, f(x) to g[f(x)], where f(x) is a point in set B & g[f(x)] is it's image point in set C.

The function g takes points in the range of f (image points in set B) to their images in set C.

It is necessary that the image of points under f (these are the f(x)’s) are accepted by g (in the domain of g). If not, they will not have images in set C & the composition is not defined.

Let f(x) = -x²-1 and g(x) = √x. We see here that the composition is not defined since all points in the range of f are not accepted by g. That is, the range of f consists of negative numbers and the domain of g are just positive numbers. We will only consider compositions that are defined.

Let f(x) = x+3 and g(x) = x² . Find the function composition (fog)(x). By definition, (fog)(x) = f [g(x)] . This means, replace x in function f by g(x).
Therefore, (fog)(x)= f[g(x)] = f[x²]= x²+3.

If we reverse the composition and calculate (gof)(x) = g[f(x)] (this means, replace x in function g by f(x)), we get, (gof)(x) = g[f(x)] = g(x+3) = (x+3)². We notice that (fog)(x) does not equal (gof)(x).

Generally, function composition IS NOT a commutative operation.

The beginning of a function compositon does not always start with an operation on a single variable. For example, to do the composition (gof)(3x-1) using the above functions, we need to calculate g[f(3x-1)]=g[(3x-1)+3]=g(3x+2)=(3x+2)².

1-1 Functions

A function is 1-1, if for every image there is EXACTLY one pre-image. That is, for every value of f(x) there is exactly one value of x that corresponds to that value. (for each y-value there is exactly one x-value).

f(x) = x² is not 1-1. Since f(2)=4 and (-2)=4. So, we have a y-value of 4 with two x-values.

f(x) = x³ is 1-1. You cannot have 2 different numbers giving the same result when cubed.

Graphically, 1-1 functions satisfy the HORIZONTAL LINE TEST. That is, all horizontal lines that you can draw that intersect the graph, must do so in EXACTLY ONE POINT.

f(x)=x⁴ (not 1-1), g(x)=x³ (yes),   y=sqr(x)(yes),   y=2x+3 (yes)

1-1 functions are very special, they have inverses.

~How to find the inverse of a 1-1 function. Use the procedure below.

  1.   Replace function notation with y-notation. (i.e., replace f(x) by y).
  2.   Interchange x and y.
  3.   Solve the new equation for y (this is the inverse function).
  4.   Special notation is used for the inverse. f^(-1)(x) is the symbol for the inverse of f.

If f(x) = 3x - 2, find f^(-1)(x), the inverse of f. Note: f is 1-1, so f^(-1)(x) will exist

   1.   Replace f(x) by y:   y = 3x-2
   2.   Interchange x and y:   x=3y-2
   3.   Solve for y:    x+2=3y gives (x + 2)/3 = y   , so, f^(-1)(x) = (x + 2)/3


If f(x) = x³ , find f^(-1)(x). Note: f is 1-1, so f^-1 will exist.

          1.   Replace f(x) by y:   y = x³   .
          2.   Interchange x and y: x = y³
          3.   Solve for y:   y = x^(1/3) (cube root of x)

Inverses are very useful for finding pre-images if their images are known. (i.e., finding x, given f(x) or finding x, given y.)

Visualizing the inverse of a function

If we graph a 1-1 function and its inverse on the same set of axes, we will notice that they are reflections of each other about the line y = x. This reflection can easily be performed by folding a piece of paper.

The line y = x is a line with slope of one passing through the origin. To get the sketch of the inverse with a piece of paper, sketch the curve for which the inverse function is desired. Make sure it is a 1-1 function or the inverse will not exist. Also, do not label the axes. Then, you must make a crease along this line. Then fold the top part of the crease down and bottom part of crease back and up. Do not rotate the paper. You will then be looking at the back of the paper at the graph of the inverse function. Hold it up to a light source for a better view.

"WIND CHILL"

This is an example of a piecewise defined functions in 2 variables. The function (Wind Chill)(W) depends on 2 variables, T and V. So, depending on their values they are subsituted into the equation to determine the value of W. Many situations in every day life are modeled by these types of functions. The number of variables Involved may be more than 1. Calculus of 2 or more variables is studied in Calculus III. One popular everyday example is a model for determining the WIND CHILL (how cold it feels considering both the temperature & the velocity of the wind. This is a complicated model. Other popular 1-variable models are for postal & telephone rates. The latter are in a class called "step functions." The following is a model used for Wind Chill.

W(T,V) gives the effective temperature taking into consideration the present temperature, T and the velocity of the wind, V.

(Adapted from UMAP)

                   T                                                            for V between 0 & 4 (inclusive)
W(T,V)= {  0.0817(5.81+3.71(√V) -0.25V)(T-91.4)+91.4   for 4 < V < 45
                   1.60T-55                                                   for V = 45 or greater

Example: W(40,20) would represent the perceived temperature (degrees F) with a present temperature of 40 degrees F and wind velocity of 20 mph. V=20 will use the middle line in the above defined function, since 20 falls between 4 & 45.

Substituting T=40 & V=20 in the above (use middle line), we would get 18.3 degrees F. Needless to say, most people using the wind chill temp would not use the formula (many do not know it or once they see it, they would like to forget it quickly). Instead, charts are given showing the results for a wide range of combinations of T & V.

In the Fall of 2001, the National Weather Service revised the formula based on new research on how cold air affects people. Consequently, new charts have been used for wind chill and the numbers will not be consistent with the above formula. They will turn out to be significantly different.

The revised formula for all combinations of T & V is W(T,V)= 91.4 - (0.474677 - 0.020425V + 0.303107(√V)(91.4 - T)
Not a piecewise defined function.

Curve Fitting

In many practical applications, models (equation types) are chosen to fit data points (either collected or given in the problem). Once the model equation is determined, it can be used for describing all possible data points from that problem and predicting or estimating these points. It will be up to you to determine which model seems to be appropriate for your data points.

Plotting these points (scatter plot) will give you a good idea of the type of model to choose. The independent variable (x) is on the horizontal axis while the dependent variable (y) is on the vertical axis.

Now, you will undertake the problem of finding the best model (equation type) that best fits your data.

Once a model is chosen and a equation is determined, predictions can be made.

The model used may or may not be appropriate for the data. Your knowledge of model types and the scatter plot is important to determine this result.

Popular model types (equation forms) are as follows:

1) Linear: y=ax+b or y=a+bx, depending on which menu you choose on the TI-83.

2) Quadratic: y=ax²+bx+c

3) Cubic: y=ax³+bx²+cx+d

4) 4th Power (Quartic): y=ax⁴+bx³+cx²+dx+e

5) Natural log (ln): y=a+lnx

6) Exponential: y=a+b^x

7) Logistic curve: y= c / (1+ae^-bx)

8) Trig curve (sine curve): y=a sin(bx+c)

~Values of the constants a,b,c,d, & e are determined by the TI-83

TI-83 Procedures

1) Collect the data or data may be given.

2) Determine the dependent & independent variables.

3) Enter your data in lists using, STAT, edit, enter, move to L₁, enter independent variable values (x), move to L₂, enter dependent variable values (y).

4) Choose a friendly viewing window compatible with your data. Clear all equations from y= menu.

5) Get a scatter plot by pressing 2nd STAT PLOT, enter, ON, enter, move curser to top left plot, enter, move curser down, L₁ for X List, L₂ for Y List, pick type of mark for your plot, enter, press graph.

6) Make a decision on the type of equation model you will use.

7) Press STAT, move to CALC, move curser down to one of the menus 4 to end, enter, 2nd L₁, 2nd L₂, enter.

8) The TI-83 will give the values for the constants associated with the equation model chosen.

9) Substitute these values into the model equation.

10) This equation may now be used for describing your data set and for prediction purposes.

11) Many different models can be found for the same data set, but usually, one will be the best.

The development of Calculus

Slope of a Curve: The graph of a linear function is always a straight line. That means between any two points on the line (if we connect them), the slant or better known as Slope between any two points is always the same. Slope is defined as the change in y-values divided by the change in x-values between any two points. For a staight line this is always the same. Take for example our line y=2x. If we take any two points, say, (1,2) and (3,6) and calculate the y change it would be 6-2 or 4. The x change between the same two points is 3-1 or 2. Dividing the y change by the x change we ge 4/2 or 2. If we did this many times changing the points we would always get the same result 2. We can say that the slope of this line is 2. It is visible in the equation in front of the x. This is an important feature of the equation of a staight line in the form of y=mx+b that most of you were exposed to in lower grades. The popular symbol for slope is m.

However, if our curve is not a straight line, then the curve will not have the same slope between any two points. We must rethink what we mean for the slope of the curve.
In those cases, we use a tangent line (touches the curve in exactly one point) and concentrate how we can get the slope of that tangent line. This will lead us into the most important concept of Calculus. That is the concept of a "LIMIT".

This concept is the foundation of Calculus. For the serious math student, much time is spent here and much analysis is given in a formal course. This is one concept students often find very difficult to comprehend since it involves much theory and many formal proofs. Very difficult for beginning calculus students. No worries, I will skip all of that and try to give you a very informal view.

Slope of the Secant line: A secant line is a line connecting two points on a curve. If the curve is not a straight line, it will only intersect the curve at those points in the neighborhood of those points. The rest of the secant line is off the curve. If we "nail" down one of those two points (keep it stationary) and let the other point move closer to that stationary point, the x and y change between those points become smaller. The closer we allow the movable point on the curve to get to the stationary point, the smaller & smaller the the x and y changes become. As that point moves closer to the stationary point secant lines lines are drawn, we will notice that these secant lines get closer and closer to the tangent line drawn at the stationary point. Using math language, we say that as the movable point approaches the stationary point, the secant lines approach the tangent line. So, that would mean that the slopes of these secant lines will approach the slope of the tangent line (this is what we are after). By formulating an expression that represents the slopes of the secant lines all we have to do is to let the change in x approach 0 and see if this expression gets close to a value. If it does, we have it. Easier said then done.

From our previous discussion, we know that the slope of any line between two points is defined by the change in y over the change in x. Writing this in mathematical symbols, this is represented by Δy/Δx. The Δ symbol reads "change in".

As the movable point on the curve approaches (symbol for approaches is →) the stationary point Δx →0 and the slopes of the secant lines → the slope of the tangent line. Since this slope is so important to us, we give it a special symbol and a special name. The symbol is dy/dx and we call it the Derivative of the function at this point. So, your first interpretation (and most basic) of the derivative of a function is an expression that will give the slope of the tangent line to the graph of the function at any point you desire, just substitute the x value of the point into the derivative expression to get it. There will be other important interpretations that will be be given as we continue our discussion. Getting the actual expression for the derivative may take effort and a good knowlege of algebra, but is presented carefully in a formal course.

The Concept of a Limit

The limit concept is the basis of calculus. Without it, there is no derivative, hence no calculus. Mathmaticians have refined this concept to a precise definition so as too confuse many non-math people and make it totally non understandable for many others. This formal definition is still taught today in a basic calculus I course. In my opinion, many get so confused by it, they lose motivation for the subject. As an RA during my undergraduate time, this was the #1 question students would come to my door with. That forced me to keep studying the concept and learn it well. Any teacher would tell you that you can not explain something to someone if you do not understand it yourself.

I will state the formal definition of a limit in the next article and show you a simple proof.

Basically, we want the f(x) values to get close to L when the x values get close to a on the x-axis. I think what confuses most students is the use of the absolute values along with the inequalities. Think of L as a level & (0,L) is on that level on the y-axis. Think of Î as a small positive number which measures the distance away from L both above & below L. This is call a neighborhood of L. For a given Î, we should be able to find a neighborhood of a on the x-axis (a positive distance d on both sides of a that will produce y values (f(x) values) that lie in the e neighborhood of L, no matter how small e becomes. If you can find a d (neighborhoods of a) for each Î (neighborhood of L), then the statement is justified (the limit is indeed L).

Don't worry about this, 85% of math majors don't really understand it either.

The only reason I'm spending time on it is because it's the basis of getting the derivative of a function. The limit process is used. Recall, we looked at the slopes of many secant lines on the graph as the moveable point approached the stationary point and used the limit in determining the limit of the slopes to be the slope of the tangent line to the curve at the stationary point. This is the derivative of the function at that point.

In a formal course in calculus much time is spent on limits and the analysis is very intensive. Many theorems are proven using that formal definition and rarely understood by many. However, the very best students have no problem with this and usually are very successful as math majors.

Cauchy's Definition of a Limit

This is the most feared and dreaded by most beginning calculus students

It is, however, the most important definitions in Calculus.

It defines a limit of a function in precise mathematical terms.
In this form it is used to prove many limit statements (including most limit theorems).

Also used to define other concepts which have limits for definitions

Unfortunately, this is a big "stubbling block" for most beginning calculus students. Even experienced students have much difficulty using it.

In informal words, to justify a limit is L as x approaches a fixed number a, one must demonstrate that no matter what neighborhood size Î (epsilon) of L you choose, you will always be able to find a neighborhood size d (delta) of a, such that, all x's in that d-neighborhood of a, will give functional values within the Î (epsilon) distance of L (are mapped into that Î-neighborhood of L)

Note: Informally, think of yourself as working at delta corp ($1,000/day). Your job is to respond to callers that send in different size epsilons (neighborhoods of L, the limit of the day). You need to respond to each caller by giving them a delta (how close to a, the value x is approaching) that will satisfy each epsilon. If the phone does not ring, you have nothing to do. Doing you homework & knowing the limit of the day, you can figure out what response will work in advance. This way, your job is made quite easy. Figuring this correct response requires you to prove the limit of the day (which I will show in this discussion)

We do not want x to equal a, so the definition eliminates x=a by the statement 0<|x-a|. So, we are dealing with deleted neighborhoods of a

a limit, alone, does not depend upon what happens at the point it approaches only close to it on both sides.(for basic limits)

however, some concepts that are defined in terms of limits will. (i.e., continuity) (we need the limit to equal the functional value there)

Here is the full statement of Cauchy's definition:

To show (or prove) that lim f(x)= L as x → a, one must show the following:

"(For all)Î>0, there exists ($) d>0, such that ('), if 0< |x - a| < d, then
then |f(x) - L| <Î

Special symbols, included, are used for "for all" (universal quantifier), "there exists"(existential quantifier), and "such that". Also, an if - then statement can be symbolized with a right arrow.

Example of a simple proof:

Prove: lim (3x-5) = 7 as x→4 using Cauchy's definition.

Proof: To Show: "Î>0, $d>0, ', if 0< |x - 4| <d, then |(3x-5) - 7| <Î.

Analysis:

Let Î>0 be given.(phone rings). Consider |(3x-5) - 7| = |3x-12| = |3(x-4)| = 3 |x-4|

By forcing 3 |x-4| <Î, we see that |x-4| <Î/3

Claim: So, choose d ≤ Î/3.

Synthesis: if |x-4| < Î/3, then 3|x-4| <Î, or |(3x-5) -7| <Î.

Which satisfies the conditions of Cauchy's definition

~Note: So, at Delta corp, just divide the incoming call number by 3 & all callers will be satisfied & you should get a raise in pay. Also note that your response is the largest delta that will work. If this neighborhood size gets you where you want, then any smaller one would also.

Most teachers do not require the synthesis when testing students

This is a simple proof. Many proofs are not and require other conditions

LIMITS & CONTINUITY

GENERAL OVERVIEW

1) LIMITS must be understood first. They are separate concepts, however, CONTINUITY depends upon the understanding of LIMITS.

2) A limit looks like this: lim f(x)
x→a

It reads: the limit of f(x) as x approaches a

f(x) is a functional expression (could be simple or complex) involving x, a is a fixed point on the x-axis or +∞ or -∞. For the infinity types, this means the x-values move to the far right on the x-axis forever (+∞) or to the far left on the x-axis forever (-∞).

3) For a limit to exist, the expression for f(x) must get closer & closer to ONE & ONLY ONE VALUE
If not, then there is no limit.

4) You must examine the x-values close to a on BOTH SIDES of a, since f(x) could get close to different values when approaching a from either side of a. They are called the right & left side limits. These must be the same for a limit to exist. A little plus sign in the exponent spot on a is used for the symbol for the limit from the right & a little negative sign for the limit from the left.

5) The TI-83 can be used to get a limit or to verify that there is no limit (will show). However, this procedure is lengthy & may not be practical for time limit testing. So, other much shorter methods will be covered. To use your calculator, code in f(x) & go to 2nd Graph (table) & start inserting x-values close to the a-value. Look at the y-values [values for f(x)] & note if they are getting closer to one value. Make sure you use values for x on both sides of a.

6) Remember, when dealing with a limit, we are not concerned about x reaching a, just the values of f(x) as x gets closer & closer to a from both sides of a. We may or may not reach our limit, we simply don’t care about that. We can have a limit without reaching it.
(the distance to a wall cut in half forever, the limit is the wall, but you will never reach it).

7) Once limits are understood, one can analyze CONTINUITY. Basically, this means CONNECTED. So, a connected curve would be a continuous curve. That means NO BREAKS in the curve. Which means, NO HOLES or JUMPS or SHOOTING UP or DOWN forever when x-values approach the point a on the x-axis or increase to the right or left forever (infinity moves).

8) To be connected at a point a, we need a functional value f(a) to be defined on the function. Also, to eliminate any JUMPS, we need the f(x) values getting close to this functional value as we approach a from both sides. That is, we need the limit of f(x) as x approaches a to equal f(a).

9) So, we see that the concept of CONTINUITY at a point depends upon the limit existing at a & being equal to the y value or functional value at a, namely f(a). If a function is not continuous a point a, we say it is DISCONTINUOUS there or the function has a DISCONTINUITY at a.

10) To understand certain concepts clearly, studying cases where they fail to hold only strengthens our understanding. So, these will be covered.

11) As in most math concepts, doing many different types of problems, related to the concepts, would be the main path for total understanding.

UNIFORM CONTINUITY

Being connected is a nice way of thinking of continuity.

Question: "Is there only one type of continuity?"
  Answer: "no"

A function can be connected (continuous) in a "stronger way". To understand this, first review basic continuity at my review link.

You will see that being continuous at a point depends on a special limit holding:

If lim f(x) = f(a), then f is continuous at x = a.
    x→a

Using Cauchy's definition for this limit (see link in review section), for any positive e, there exist a positive d, such that if x is in the d-neighborhood of a, then the f(x) is within e of f(a). In this case, the d depends on the choice of e and the specific point a.

In essence, uniform continuity states that the choice of d will depend on e alone and not any specific point in the interval.

Informally, this means that for any e>0 and any two x values in the interval, we can find a d >0, such that, whenever these x values are within d, their functional values are within e.

Saying it another way, this means that x values close to each other, give functional values close to each other. This is a stronger type of continuity and a more rigid requirement than regular continuity.

Of course, if a function meets this requirement, it certainly will meet the lesser one (being uniformly continuous implies regular continuity).

It is important to note that the same d must work throughout the interval for x values in the interval. Hence, d depends on e alone not where we are in the interval.

To illustrate this with an example, let's take the classic example often given in advanced calculus & real analysis courses. That is , f(x) = 1/x over the open interval (0;1).

First, let me prove that it is continuous over (0;1). The easy way to do this is to establish the existence of a derivative at each point in (0;1).
Since f '(x)=- 1/x², we can see that f is differentiable there, so, f is continuous there (see link, differentiability & continuity in the review section).

Now, to show f(x) = 1/x is not uniformly continuous in (0;1). This will take a bit more work.

I will be using an indirect proof (very important type & used widely in advanced courses). It's a proof of contradiction. That is, we will assume f is uniformly continuous over (0;1) and show that this is impossible (this is the contradiction). So, our assumption is false & f will not be uniformly continuous there.

Note: In Latin it's called "Reductio Ad Absurdum" or RAA for short. Basically meaning, "reduction to the absurd". The absurd is the resulting contradiction achieved in this "indirect proof".

Here we go. Assume f(x) = 1/x is uniformly continuous in (0;1). That means for any e>0 we should be able to find a d>0, such that , for x₁, x₂ in (0;1), whenever these values are within d of each other, then f(x₁) and f(x₂) are within e of each other.

Using more mathematical formulation, let e>0 be given, and since we are assuming f uniformly continuous, there is a d>0, such that, |x₁ - x₂| < d
gives |f(x₁) - f(x₂)| <e, by definition.

But |f(x₁) - f(x₂)| = | (1/x₁) - (1/x₂)| = | (x₂ - x₁)/x₁x₂ | = |x₁ - x₂ |/|x₁x₂| < d/x₁x₂ < e. (remember e has been chosen already and the d must work for that choice).

This is impossible, since x₁ and x₂ taken sufficiently close to zero will make that fraction as large as we want (denominator can be made as small as we like by taking very small x values)(the product of two numbers near zero can be made arbitrarily small). So, it can be made to be greater than e, a contradiction of the uniform continuity property. So, we reject our assumption that f is uniformly continuous on (0;1) and conclude that is isn't.

In essence, the functional values can be made to separate as far apart as we like by taking x values very close to zero.

Generally speaking, any function which is unbounded on an open interval, cannot be uniformly continuous on that interval.

Also, if a function is continuous on a closed interval, it will automatically be uniformly continuous on that interval.

A very important place where this is used is in the proof of the fundamental tie between the limit of a Riemann Sum and the area under a curve.

This is an advanced concept and is useful in proving theorems (as noted) and other important properties in real analysis.

Special Infinity Limits

Short-cuts for limits of quotients of polynomials functions where x →∞

Use for these types:

lim     Axⁿ + terms containing lower powers of x
x→∞ Bx^m + terms containing lower powers of x

3 possible results:

1) If n=m:     Answer: A/B (highest powers in top & bottom are equal)

2) If n>m:     Answer: no limit (highest power in top greater than that of bottom)

3) If n<m:     Amswer: 0 (highest power in bottom is greater than that of the top)

Note: Short-cuts are also valid if x→ -∞

ALL ABOUT THE DERIVATIVE

The slope of any secant line connecting two points on the function represents the average rate of change of the function with repect to x between those points, while the slope of the tangent line at an isolated point is an instantaneous rate of change at that point. There lies the power of the calculus. Our ability to describe a function at any isolated point by means of its tangent line slope. No where else in mathematics are we able to do this.

The derivative of a function is developed in the early stages of Calculus. It serves as the foundation of one of the two major operations of calculus, namely differentiation. The 2nd major operation of calculus is integration which is developed later.

It is called the derivative since it is derived from a parent function. It will give very important information about the parent function that leads to the solution to many problems in science and business applications.

Some of those properties include the following:
The rate of change of the dependent variable of the parent function with respect to the independent variable. For example, if the equation of the parent function takes the form R=f(x), where R is the revenue collected when selling x units of an item, the derivative will tell us how revenue changes at any level, x, of sales. Another popular way to phrase the problem would be, "how fast is the revenue changing when the sales level, x, is at a certain value"?

In the above example, the units for the derivative could be in dollars per item for a specific number of items. When rate of change is mention, it assumes an instantaneous rate. Average rate of change requires no knowledge of calculus and is simply the slope of the line connecting two points on a function (secant). The word Average will be use, if this is what we want, otherwise, it will be assumed that the rate of change is instantaneous(tangent line). This will require getting the derivative.

As we noted above, the derivative gives us the slope of any tangent line drawn to the parent function at a specified point. Unlike the average rate of change between two points (the slope of a line connecting these points), the derivative gives the slope at one point. This will serve as a description of behavior of the parent function at this isolated point (there lies the power of the derivative, as noted).

Different points on the parent function have different sloped tangent lines (assuming a nonlinear parent function) so the derivative will give us a way of analyzing the parent function at any isolated point desired.

A linear parent function will have a derivative which is the same at all points on that curve since the slope between any two points, no matter where you take them, is always the same. So, in this case, the average rate of change between any two points = the rate of change at any point. The slope of a linear function is constant.
For nonlinear functions, we need the derivative to find the slope at different points since the slopes of the tangent lines drawn at these points change as we travel along the curve.

In business applications, the derivative gives a marginal value at an indicated value of x (items produced or sold). So, to find the marginal revenue at a sales level x =20, you would substitute x=20 into the derivative of the revenue function, which would be given. Marginal values usually deal with revenue, cost, and profit tied to the parent function used in the specific problem. The in-dept study of this is called Marginal Analysis (covered in a basic business calculus course).

So, in summary, the derivative is usually interpreted in 4 ways:
1) rate of change of one quantity (dependent variable) with respect to another (independent variable).
2) the slope of tangent lines to the graph of the parent function at various places.
3) for business applications, it will give marginal values of the quantity examined.
4) for distance (s) - time (t) functions, the derivative gives velocity (v) since it measures how distance changes wrt time (i.e.,ft/sec or miles/hr). Please note that velocity is directed speed. Speed is a scalar quantity (magnitude only) while, with a direction, gives velocity, a vector quantity.

AVERAGE AND INSTANTANEOUS RATE OF CHANGE

Generally, average rate of change, between 2 values of the independent variable, is the slope of the secant line connecting these points on the graph of the curve & instantaneous rate of change, at one value, is the slope of the tangent to the curve at that point.

For s = f(t), this was average velocity & instantaneous velocity (or just velocity) at a pt.

Well, these concepts can also be applied to any function, as long as we know its equation.

Having said all that, here is a review of the procedures involved.

Finding the average rate of change of a function between 2 values of the independent variable.

Given a function (an equation)---this could be a formula for some geometric figure

1) use functional notation---i.e., if Volume is a function of x, use f(x) or V(x) in place of V

2) for average rate of change, we have 2 fixed values of the independent variable, say a & b, where b>a

3) compute the slope of the secant line to the function connecting the pts (a,f(a)) and (b,f(b))

   i.e., f(b)-f(a)
             b-a

Example: A circle is expanding. Find the average rate of change in the area with respect to the radius as the radius changes from 2 inches to 5 inches.

Solution: The formula used is A=pr². Let f( r)= pr², r=the radius

Compute [f(5)-f(2)]/(5-2) = [p(25)-p(4)] / 3 = 21p/3 = 7p sqr "/ inch


Finding the instantaneous rate of change at a value of the independent variable, say at r=a [the fixed pt is a,f(a)]

1) use functional notation (as in above)

2) choose a variable pt, say x,f(x)

3) formulate the slope of the secant lines as in 3) above, i.e., [f(x)-f(a)]/(x-a)

4) then take the limit as x approaches a,

i.e., lim [f(x)-f(a)]/(x-a)
x→a

Example: In the above problem, compute the instantaneous rate of change of area with respect to the radius when the radius is 3 inches.

Solution: compute: lim   f(r)-f(3) = lim p[(r²-9)]/(r-3)
                                r→3    (r-3)        r→3

                            lim p(r-3)(r+3) = 6p square inches/inch
                           r→3       r-3
~Use the same exact procedure on any formula (any figure), but
make sure you have the formula in functional notation.

~Note: This is not the only approach. There are others that work as well, however, for basic problems, this one is easiest to understand, in my opinion.

How to get the derivative of a function

By definition: Initially, in most calculus I courses the student is required to compute the derivative by use of it's definition. In words, it would be the limit of the secant line slopes as the change in x between the two points approach 0. This does not help us get it. We need a workable mathematical way of getting it. So, the following is what this looks like:

dy/dx = lim Δy/Δx
Δx→0

Note: In many cases it's more convenient to use the functional notation equivalent of the above to calculate the derivative. In that case, we use a new symbol, f '(x) for dy/dx and one letter h for Δx and f(x+h)-f(x) for Δy. It will have the following look:

f '(x) = lim [f(x+h)-f(x)]/h

h→0

There is a third form for the derivative that was used in the previous section that concentrates on getting the derivative at a fixed point [a,f(a)] by letting the variable point [x,f(x)] approach the fixed point on the curve. It has the following form:

i.e., f '(a) = lim[f(x)-f(a)]/(x-a)

x→a

Example of using the limit definition of the derivative to find its formula

Let's take the function f(x) = -2x²-3x-5. We wiil formulate the derivative in steps using the above definition.

1) f(x+h)= -2(x+h)²-3(x+h)-5 (just replace x by x+h where you see an x)

= -2(x²+2hx+h²) -3(x+h)-5 (square the x+h)

= -2x²-4hx-2h²-3x-3h-5 (eliminate the parentheses)

2) f(x+h)-f(x)= -2x²-4hx-2h²-3x-3h-5-(-2x²-3x-5) (subtract the f(x) expression)
= -2x²-4hx-2h²-3x-3h-5+2x²+3x+5 (eliminate the parentheses)

= -4hx-2h2-3h (simplify) all terms should contain a least one factor of h)

3) [f(x+h)-f(x)]/h =-4x-2h-3 (divide each term by h)

4) lim [f(x+h)-f(x)]/h h->0 = -4x-3 (sub in 0 for h) all terms with h will disappear)

this is f '(x)=-4x-3 (this is your answer)

Not all derivatives are this easy. Some require quite a bit of algebra before the division in step 3

Note: Since we need to calculate many derivatives in calculus, this process can be time-consuming every time we need a derivative. So, short-cuts are developed that will enable us to quickly compute them. Usually, calculating the derivative by definition is only used at the very beginning during the development of the derivative for theorical purposes. Short-cuts are coming up.

DERIVATIVE FORMULAS

Note: In a typical calculus I course, these formulas are proven using the basic definition of the derivative. Once proven, getting the derivative by definition is rarely used and often forgotten. I've included a comprehensive list of formulas that involve different classes of functions. Many of you many not be familar with these and will need to study much of pre-calculus in order to feel comfortable working with them. During our application discussion (how to use the derivative to solve basic problems), we will confine our discussion to the very basic functions. However, realize that persons in the scientific world deal with most all of these functions on a continuous basis.

Constant: If y = k, any constant, then dy/dx = 0

Example: If f(x)=7, then f '(x)=0

Power Rule: If y=uⁿ, then dy/dx = n u^(n-1)du/dx

Example: If g(t)= (t² + t - 5)⁷, then g '(t)= 7(t²+t-5)⁶ (2t+1)

General Expo: If y = a^u, then dy/dx = a^u (ln a) du/dx

Example: If y=2^(x²), then y'=[2^(x²)](ln2)(2x)

The special Expo: If y = e^u, then dy/dx = e^u du/dx

Example: If C=e^(1-2t), then C ' = [e^(1-2t)](-2)

The natural log: If y = ln(u), then dy/dx = (1/u) du/dx

Example: If P(x)= ln(x²-4x+33), then P '(x)= [1/(x²-4x+33)] (2x-4)

The trig fcns: If y = sin(u), then dy/dx = cos(u) du/dx

If y = cos(u), then dy/dx = - sin(u) du/dx

If y = tan(u), then dy/dx = [sec²(u)] du/dx

If y = sec(u), then dy/dx = sec(u)tan(u) du/dx

If y = csc(u), then dy/dx = - csc(u)cot(u) du/dx

If y = cot(u), then dy/dx = [- csc²(u)] du/dx

Sums: If y = u + v, then dy/dx = du/dx + dv/dx

Product Rule: If y = uv, then dy/dx = u dv/dx + v du/dx

Quotient Rule: If y = u/v, then dy/dx = [v du/dx - u dv/dx] / v²

Note: I've eliminated many other types of functions along with advanced techniques used for the more complicated types. A formal calculus course would cover them in dept. The best way to remember these is to do many problems dealing with them. The more you do, the easier it will be to recall them.

TANGENTS & NORMALS

The linear function: there are many equation forms for a straight line. The one which is most popular & convenient (in my opinion), in higher mathematics, is point-slope form: y - y₁ = m (x - x₁), where (x₁,y₁) is any point on the line and m is the slope of the line.

~Other forms, such as y = mx + b (slope-intercept), Ax + By + C = 0 (general form), two-pt form, determinant form, & normal form are also used (on occasions, when required by a specific problem), but not as much.

~Notice that all linear equations do not have exponents on x & y greater than one. However, be careful, an equation could contain higher powers of x & y and could define more than one line.
For example: (x-y+2)(3x+5y-7)=0 defines two intersecting lines. If the left side were multiplied out, terms with exponents of 2 would appear and the 2nd degree term xy would also be present.

TANGENT LINES

~To write an equation of a tangent line, one needs to know its slope & a fixed point on the line (using point-slope form).

~The slope is found by calculating the derivative of y with respect to x and substituting the coordinates (usually just x, sometimes just y, & sometimes both x & y) into the derivative.

~A major mistake made by beginning calculus students is to place the derivative (as a variable expression) into the equation form for the line. This is very wrong & destroys the linear form. So, make sure you realize that the slope is a constant (derivative must be evaluated at that point).

Ex: Find an equation of the tangent line to the curve f(x) = x² at the point (-2,4). First, find f'(x)=2x. Then, substitute x=-2 into the derivative to find the slope of the line: f'(-2)= -4. Now, use point-slope form: y - 4 = -4 [x - (-2)] or y-4=-4(x+2). Beginning students seem to have a difficult time leaving the answer this way. Many want to multiply across the parenthesis & solve for y. (completely unnecessary, in most cases)

NORMAL LINES

~Once you have mastered writing the equation of a tangent line, you are ready for the normal line.

~Simply stated, it is a line perpendicular to the tangent line at the point of tangency. That means it contains the same point as the tangent line. Therefore, the only difference is its slope.

~From basic math, the slopes of perpendicular lines are related by negative reciprocals (some teachers prefer to say , "the product of their slopes is -1"). Either way is fine. So, if the tangent slope is 2, then the normal slope is -1/2. Be careful when the tangent slope is 1 or -1, the normal would seem to be its negative (but actually is the negative reciprocal).

~So, once the derivative is calculated at that point, you can use point-slope form to state both equations quickly (they will look the same except for their slopes).

~Normal lines have wide applications in math & science and usually are used in conjunction with vectors (calculus II & specially calculus III). Normal components of forces play an important part in the motion of particles and normals to surfaces are also studied in calculus III.

~Two curves are said to be Orthogonal if they intersect at right angles. Which means, their tangent lines at their intersection points are perpendicular. To show this, one needs to calculate the derivatives of each curve and substitute the coordinates of each point of intersection into these derivatives. If the resultant numbers are negative reciprocals, then the curves are said to be orthogonal. "Orthogonal Trajectories" are related to force fields & energy and are studied in differential equations.

The Mean Value Theorem

Also called the Law of the Mean.

First, let's discuss Rolle's Theorem (it is used to prove the Mean Value Theorem.

In simple terms, Rolle's Theorem states that if a function is continuous over a closed interval, [a;b] and has a derivative over the open interval, (a;b) and the values of the function at the endpoints, f(a)=f(b)=0, then we can conclude that there must be at least one place between a and b for which the derivative is zero. Think about it...makes sense, yes?

The Mean Value Theorem is even simpler. It states that, if a function is continuous over a closed interval [a;b] and differentiable over the open interval (a;b), then we can conclude that there will exist at least one value of c between a and b, for which the derivative at c (slope of the tangent line at x=c) has the same value as the slope of the secant line connecting the points a,f(a) and b,f(b). Putting it another way, we can say that there will be at least one value between a and b where the instantaneous rate of change is equal to the average rate of change.

This knowledge could save you money and points on your license

Let's see how.

Ex: Say you get on the NYS thruway at 9 am in New Paltz and travel (nonstop) to Buffalo. You arrive at Buffalo's toll plaza at exactly 2 pm. A state trooper looks at your card and decides to give you a ticket for speeding. You, of course, are in denial. The trooper takes out a pad and pencil & explains the Mean Value Theorem to you. He says that if you left New paltz at 9 am and arrived at Buffalo at 2 pm, that's 5 hours covering the exact mileage of 400 miles. That means you averaged 80 mph. The speed limit is 70 mph. So, somewhere you had to be going 80 if you averaged 80. In essence, that's what the Mean Value Theorem states. The figures I'm using are for the sake of argument and the actual mileage and times could vary. However, since you listened carefully to the Trooper & was quite polite, he did give you a break.

I will leave the proof of Rolle's theorem & the Mean Value theorem to a formal calculus class.

Why is this Theorem considered very important?

Well, it is used to prove many other important properties of functions that we have been assuming from day one.

~Here are just a few:

1) If a function has a zero derivative over an interval, then it is constant over that interval.

2) If a function has a positive derivative over an interval, then it is increasing over that interval.

3) If a function has a negative derivative over an interval, then it is decreasing over that interval.

4) If two functions have equal derivatives over an interval, then they differ by a constant.

So, know the Mean Value Theorem well, it can save you some grief.

Many other intertesting properties of functions can be established by the use of this theorem.

It is sometimes referred to, at mathematical tea parties, as a "tool theorem".

Example: Find the value(s) of c prescribed by the Mean Value Theorem for y= (1/3)x³ -2x² +5x -7 over [3;6].

Solution: Since the function y is continuous over the closed interval [3;6] and dy/dx exists over the open interval (3;6), y satisfies the hypotheses of the Mean Value Theorem.
(i.e., connected over the closed interval & smooth over the open interval). Note: y = f(x)

Therefore, we may conclude the following (conclusion of the MVThm):

There exists a value c in (3;6), such that f '(c) = [f(6) - f(3)]/(6-3) (equation 1)

Let us find the value(s) of c in this problem.

Code in y or f(x) into your calculator using the menu y=

Go to 2nd graph (table) to find the values of f(6) = 23 and f(3) = -1.

We need f '(x) (which is also dy/dx): f '(x) = x² -4x +5, so f '(c) = c² -4c+5
Substitute into (equation 1) above to get the equation:

c² -4c+5 = (23 - (-1))/3 or, equivalently, c² -4c +5 =8 which gives c²-4c -3 = 0

By the quadratic formla, we get, c = [-(-4)+√(28)]/2 and c = [-(-4)-√(28)]/2

Since the 2nd value of c = [-(-4)-√(28)]/2 is not in (3;6), we reject it.

So, c = [-(-4)+√(28)]/2 = 2+√(7) (exact answer)(approximately 4.646 to 3 decimal places).

Popular applications of the derivative

The Nature of Optimization Problems: Optimization problems (also known as max-min problems) have a wide variety of applications in many different fields.

Among the most popular are those in science & engineering.

Here is a brief list of certain qualities we seek in these fields. This list is, by no means, complete

1) Constructing the largest box with given dimensions (maximizing volume)(example will be given).

2) Using the least amount of material in constructing a figure that covers a fixed surface area. (minimizing amount of usable material)

3) Finding the largest 2-D geometric figure having certain properties. (maximizing area)

4) Finding the shortest path (minimizing distance) or path of least time.(minimizing time)

5) Finding smallest length or perimeter that encloses a given area. (minimizing length)

6) Constructing an object that weighs the least. (minimizing weight)

7) Constructing the strongest object. (maximizing strength)

8) Finding the fastest velocity or speed. (maximizing speed or velocity)

9) Finding the quickest reaction time in chemistry. (maximizing reaction rate)

10) In business & economics, we are concerned mainly with Maximizing Profit & Minimizing Costs.

Note: In all of the above, we need mathematical models (mathematical formulations)(functions) that we can use to apply the calculus of optimization analysis.
Some may be very good (laws of physics) & others just good approximations (usually in nonscientific areas). The following are some popular examples.

How to Solve Max/Min Problems

These types of problems are also called, "mini/max", "optimization", "extreme value" problems. Basically, there are two approaches. If numbers are "nasty" or zeroes of certain types of equations can not be found by easy methods, the TI-83 or other computer software must be used. Let's take the first category:

1) The derivative method: Extreme values of functions (high/low points on the curve) will occur at one of 3 places:

a) where the first derivative = 0. (tangent line is horizontal) (most common)

b) At the end points of the interval when a domain is specified or implied (must check these)

c) Where the derivative fails to exist. (rare, but does happen sometimes) (i.e., y=x^2/3 at x=0) (cusp with vertical tangent, but function is continuous there) Take a look with your TI-83. Use x in [-1;1], y in [-1;1].

To find these points in a), take the derivative, set it = 0, then solve for x. (assuming the numbers are nice & you can solve that equation).

If not, the TI-83 is used. Don't forget, we are looking for points, (a,b), so pick up the y-value from the table after you code in the original function.

The y-values of these points must be compared to the others (if any) from parts b) & c). To get the end points, just sub in the x-values there & use the table.

Once you have the derivative, check & see if there are any x-values that make the derivative undefined (usually where the bottom is zero or a negative under an even root). If there is a defined y-value for the original function there, find it & compare it to the other points found. Most of the time, there will not be one. (example given above is an exception)

Usually, we are after absolute max/min over the entire domain, so, in most instances, you can disregard the local or relative max/min. Make sure you read the wording of the problem & see if a closed interval is stated. If not, assume you are after the absolute max/min for the entire curve. If you are asked to find ALL extreme values for the function, include relative along with absolute values of the function. If the word points is included, then your answers will include both x and y values for the function.

A sketch would consolidate your understanding, so look at the y-values & set a friendly window for viewing the curve. If you did things correctly, you'll see their locations.

2) The TI-83 method: In the event that the numbers are "nasty" or the derivative equation can not be solved easily, use this technique.

a) code in the equation for which you wish to find max/min points

b) usually, a specified domain for x is indicated, so, go to the table & insert values in small increments from left to right that are in the interval of the domain. Take a look at their y-values in the 2nd column. This will give you a good idea of how to set up your window. For the x values, just use values that include the domain.

c) Now you are ready to view the curve. Make any necessary adjustments to your window so that you can clearly view all the high & low points of your curve over the domain indicated.

d) press 2nd trace (Calc). Go to appropriate menu for Maximum (4) or Minimum (3). Press enter. The curve should reappear & you'll be prompted to give a left guess (x-value) around a high point (if you are after a max pt). To get an idea of values on the left side of this point, use trace & move the marker with the cursor along the curve & read the x-value at the bottom of your viewing window. After entering the left guess, you will be prompted for a right guess. Do similar steps. After you enter that, you will be prompted for a guess (any number between the left & right guesses). Enter that, & walla walla, you answer comes out at the bottom of the screen. It may not be exact, since "nasty" numbers include repeating decimals & irrationals, so round off may be necessary. Always check the directions of the problem. It should be stated there.

Read your problem carefully & give the answer(s) that the problem calls for. Sometimes, only the value of the independent variable is needed (i.e., what production level (x) produces the largest profit (y) ) Most of the time both are asked for in problem (max/min points) (answers are in the form (x,y)). Also, in most cases, a graph is called for in a problem, so read the directions carefully.

1) Finding the highest point on the graph of a function:

Given y= -2x²-4x+7. If you graphed this function (use a TI-83 calculator), you will notice that it's a parabola openning downward. At this highest point, the tangent line to this curve is horizontal. This means that its slope = 0. Since the derivative is an expression that gives the slope of tangent lines, we need to know where the derivative = 0. So, computing the derivative using the short-cut formula, we get dy/dx =-4x-4. Setting dy/dx=0, we get -4x-4=0 or x=-1. We need to pick up the y coordinate of the point so substitue x=-1 into the equation of the function to get y=-2+4+7=9. So, the highest point on this curve is at (-1,9).

2) Maximizing Profit:

A company is in business of manufacturing and selling luxury yachts. Since these are big ticket items, they estimate the maximum number they can manufacture per year is 100. Based on historical data points and trends, they have estimated that their profit in each year would be modeled by the following equation: P(x)= -2x³+210x²+7200x, where x represents the number of yachts sold. How many must be sold each year for maximum profit? (nearest integer)

Get P'(x)= -6x²+420x+7200. For our discussion, the maximum or minimum values of the functions used will occur where the derivative=0. In typical calculus courses, there are other places we need to look. I will not consider these in our discussion. So, setting P'(x)=0 gives the following -6x²+420x+7200=0. Dividing both sides by -6 simplifies our equation and does not alter the solution, we get x²-70x-1200=0. Checking the discriminant b²-4ac to see if it factors, we get b²-4ac = √9700, not a perfect square. So, the quadratic formula is needed.(you may need to go back to your basic algebra to look it up) x=[-(-70)+√9700]/2 and x=[-(-70)-√9700]/2 We can reject the 2nd value since it is a negative. We get, x=84.24 to two decimal places. Insert x=84.24 to get a profit value of 901170. So, x=84.24 produces the greatest profit for x values between 0 and 100. Since we cannot produce a fraction of a yacht, the answer is 84.

3)Finding the rate of change of one quantity with respect to another quantity

A cube is expanding. How fast is the volume changing with respect to its surface area when the edge of the cube is 2 units in length?

We need the formulas for the volume of a cube and its surface area. You may have to look these up. V=x³ (length x width x height) and S=6x² (6 sides all of which have an area of x²). Since we want to compare the volume V with the surface Area S, we need to translate our question to mathematical terms. Since the derivative measures rate of change, we need to calculate dV/dS when x=2. Looking at the formulas for V and S, we see that both are in terms of x. So, dV/dx and dS/dx can easily be calulated by short cuts. We get, dV/dx = 3x² and dS/dx =12x. So, how does one get dV/dS from these two quantities? Just divide them: [dV/dx} divided by [dS/dx] will give you dV/dS since the dx quantities in both divide away. This property is know as a form of the chain rule for derivatives. So, dV/dS =[3x²]/12x=x/4 (reduced). Now, substitute x=2 into this derivative to get our answer of 2/4 or 1/2. To make sense of this answer, the units will be 1/2 cubic units increase in volume per square unit increase of surface area at the instant when x=2 units. For example, if the units are in inches, then, at the instant when the edge of the cube is 2", the volume is increasing at a rate of 1/2 cubic inch for every square inch of surface area increase.

4) The biggest box problem

This is a classic problem in most Calculus I courses.

I will give you a variation with my own numbers.

At the end of this article I will show you how to solve it using your TI calculator, but the understanding of the basic concepts involved are lost. So, I will be solving it the long way. That way, important concepts are stressed and it would help you apply them to other problems.

Here is your problem: You would like to make a box from a piece of rectangular cardboard measuring 50" in length and 30" in width by cutting identical size squares from the corners then folding up the sides.

The problem you have is the size of the squares removed from the corners. Obviously, different size squares will produce different size boxes. Some with very low heights having relatively large bases and others with higher heights having smaller bases. The base consists of the resultant length & width of the box after the cut is made.

What you want to find is the size of the square cut from the corners that will give you the largest possible box.

In theory, there are an infinite number of boxes you can make. Some with very little space inside (small volumes) and others with more space (bigger volumes).

Is there a square cut that produces the largest box (maximum volume)?

Let's find out. To apply the optimization (max/min) techniques we need to formulate an equation for the quantity to be optimized in terms of the independent variable.

So, let x represent the side of the square that is cut from each corner. The values of x are obviously restricted between 0 and 15", since it would make no sense out side of this range. Our cardboard is only 30" wide. Of course, a positive value for x is necessary as well.

After the x" by x" squares are removed from the corners, the length of the resultant box can will be expressed as 50-2x and the width would be represented by 30-2x. The height, when folded upwards would be represented by x.

So, the volume, V would be represented by the equation, V=(50-2x)(30-2x)x, the product of the three dimensions which would give us all possible volumes for any x in the acceptable range for x.

At this point in the problem we can take two paths. The derivative method or the calculator method.

I will show you both ways, but first the derivative method.

In mathematical terms, we would like to find the value(s) for x between 0 and 15 that maximizes V.

Critical values for x are found by solving dV/dx = 0. We do not need to look elsewhere since there are no end points for x to check and the equation for V is a polynomial. Polynomials are smooth curves which means that derivatives exist everywhere. So, let's find where dV/dx = 0.

Differentiating V using the generalized product rule for 3 factors, we get

dV/dx = (-2)(30-2x)x+(-2)(50-2x)x+(1)(50-2x)(30-2x).

For those of you not familiar with this product rule extension, it goes like this: derivative of first factor times the other factors + the derivative of the second factor times the others + the derivative of the third factor times the others, and so on..

We need to solve dV/dx=(-2)(30-2x)x+(-2)(50-2x)x+(1)(50-2x)(30-2x) = 0

~To make the numbers smaller, divide both sides by 4, we get

(-1)(15-x)x+(-1)(25-x)x+(1)(25-x)(15-x)=0. Of course, 0 divided by 4 = 0

Eliminating all symbols of inclusion, we get,

-15x+x²-25x+x²+375-40x+x²=0 or 3x²-80x+375=0, a quadratic equation. Let's see if it factors by computing the discriminant, b²-4ac. We have a=3, b=-80, and c=375, so b²-4ac=1900 (not a perfect square)

So, we need to find the solutions using the quadratic formula

So, x=[-(-80)+√1900]/2(3) and x=[-(-80)-√1900]/2(3)

Which gives, x=20.60 and x=6.07 to two decimal places. We can reject x=20.60 since it is outside of the possible values for x. So, x=6.07" seems to be the answer. How can we be sure?

We can use our common sense and conclude that it must give a maximum volume since there is no minimum. (i.e., taking x very close to zero in the formula for V will make V as small as we want.

The other way is to graph V and locate the maximum by using our TI calculator. Let me show you this procedure.

First, code in the equation for V in the y= menu (upper left). Make sure you code in V not dV/dx.

Go to the table of values for V (2nd Graph). Make sure your table is set up properly so you can enter various x values.
If not, go to 2nd window (table set) and see if the Indpnt is on ASK and the Depend is on AUTO.

Now enter a few values between 0 and 15 & observe the V values. Notice that no V value is greater than 5000. Now, set up your window with x_min=0. x_max=15, y_min=0, y_max=5000

Now, press graph. Hopefully, you should be viewing the Volume curve and also observing that the curve reaches a max point somewhere

We hope that the x value at that max point is consistent with what we got earlier, namely x=6.07

Let's find out. Press 2nd trace (calc) and go to menu 4 (Maximum), enter. Now, you will be prompted to give a guess for x to the left of the high point. If you are not sure where you are, use your courser to move the marker along the curve & see those x values. I entered x=4 as a left guess. Any x to the left of the high point will work. Then press enter. Now, you will be prompted for a right guess. I entered x=8. Again, any guess to the right of the high point will work. Press enter. Now, enter any guess between your left guess & your right guess. I entered x=7. Press enter and the solution for x should appear at the bottom of the window, Yippie!, x=6.07.

Therefore, you would cut 6.07" by 6.07" squares from the corners of your cardboard to get the box (with an open top) that holds the most. (i.e., Maximum Volume)

Note: These example are intended to give a very brief taste of just a few places where the derivative is useful. Just the very "tip of the iceberg". The scientific and business communities often deal with many more different types of problems where derivative are used. A formal course in calculus will expose the student to many more very interesting applicaions. One of my favorites is where to stand on viewing a painting or picture on the floor for the best view. Unfortunately, this invoves more complicated functions (inverse trig functions) along with maximizing the angle of view. A bit too advanced for our discussion, however, the general procedures are very similar.

The Bug-Frog Problem

The Problem: A bug is traveling north along the curve x = 0.25y². This is a basic parabola opening right with a vertex at (0,0). Since the y is squared, there will be symmetry with respect to the x-axis. A hungry frog is situated at the point (3,0) on the x-axis. This frog is very lazy and will not move from this point. The frog needs to be patient and wait for the bug to come within reach of his tongue, which is 2.9 inches. At this moment, the bug is located at the point (9,-6) on the curve. Using the distance formula between two points, it can be calculated to be 8.49 inches away from the frog. Therefore, well out of reach of that hungry frog. However, as it travels upward along this curve, will there be a point on the curve that the frog's tongue captures the bug? Let's find out.

The problem reduces to the problem of finding the point(s) on the curve closest to (3,0), the position of the frog.
So, we need to set-up a general distance function that will give the distance from any point (x,y) on the curve to the point (3,0), the position of the frog. Let's call this distance s.

Again, use the formula for the distance between the two points (x,y) and (3,0).

We get, s = √[(x-3)²+ (y-0)²] which simply becomes s = √[(x-3)²+ y²].

We want to find the x and y value(s) the make s a minimum.

Since s is a positive distance, if we minimize s², then that will automatically make s a minimum. That way, we can make our work simpler by eliminating the square root.

So, Let Q = s². So, Q = (x-3)² + y²
We need the right side in terms of one variable. No problem, since x = 0.25y² gives y² = 4x. So substitute 4x for y² in the equation for Q.

We get, Q = (x-3)² + 4x. Now, we are ready to take the derivative of Q and find the critical value(s) where it is 0

dQ/dx = 2(x-3) + 4. Setting dQ/dx = 0, we get, 2x-6 +4 = 0 or x =1. Since the 2nd derivative is always positive (2), we know that x = 1 produces a minimum Q, consequently, a minimum s, hence, the x value on the curve closest to the point (3,0).

when x=1, y² = 4, which gives y = 2 and -2. So, the points (1, -2) and (1, 2) are the points on the curve (the bug's path) closest to (3,0), the frog.

Since the bug is at (9,-6) presently and is safe, it needs to escape being eaten by making it past (1,-2). If it survives this point, it will not be eaten since no other point on the curve is closer.

So, let's use the distance formula between two points once again to find that distance from (1,-2) to (3,0)

we get, √[(1-3)² + (-2-0)²] = √[4 + 4} = √(8) = 2.83 inches. Bad news for the bug, since the frog's tongue can reach 2.9 inches. So, the frog gets a meal.

CURVATURE

Question: How curved is a curve?

Answer: It varies, depends on where we are on the curve.

Question: Is there a number that measures the degree of "curvedness"?

Answer: Yes, it's called curvature.

Question: Is it a positive or negative number?

Answer: Could be positive or negative, depends on the shape of the curve at that point.

Question: What do you mean by shape?

Answer: If the curve is concave down at that point, curvature will be negative. If the curve is concave up there, the curvature will be positive.

Question: Why?

Answer: The definition of curvature is given by the derivative (instantaneous rate of change) of the angle the tangent line makes with the positive directed horizontal axis (inclination angle) with respect to arc length along the curve (s), so, this will be negative when drawn to a point where the curve is concave down (inclination angle decreases) and positive where the curve is concave up (inclination angle increases).

Note: There is an equivalent way of defining curvature using vectors. In calculus II and III, parametric representations of curves are mainly used to analyze the motion of a point on plane and space curves.

Question: How does one calculate the curvature at a point?

Answer: By use of a formula which I will derive.

Let us formalize the definition in mathematical terminology.
Curvature = k = dq/ds (the rate of change of the inclination angle (q) wrt arc length)

The curve can be in parametric form (x and y considered functions of time t) or by a direct relationship (i.e., y=f(x) or x=f(y)).

In calculus II, you will derive the integral that gives arc length for all 3 cases.
In parametric form s = ∫ [(dx/dt)² + (dy/dt)²]^(1/2) dt.

Then ds/dt = [(dx/dt)² + (dy/dt)²]^1/2 by the fundamental theorem.

Since dy/dx = tan(q) by definition, q = tan^-1(dy/dx)

But in parametric form dy/dx = [(dy/dt) / (dx/dt)] by the chain rule.

So, q = tan^-1[(dy/dt) / (dx/dt)]. hang in there.

So, dq/dt = {[(dx/dt)(d²y/dt² - (dy/dt)(d²x/dt²)]/(dx/dt)²]} divided by
{1+[(dy/dt)/(dx/dt)]²}....ugh!

~Finally, k = dq/ds = (dq/dt) / (ds/dt) = (the above mess) divided by
[(dx/dt)² + (dy/dt)²]^(1/2)....even worse!

Let us make this easier to read. Let dx/dt = x*, dy/dt = y*, d²x/dt² = x**, d²y/dt² = y**.

So, our formula becomes: k = (x*y**-y*x**) / (x*² + y*²)^(3/2)...a little better, but still quite involved. (a dot is usually placed over the x and y in conventional courses)

So, there we have it. All we have to do is apply it & get our curvature number. Remember, we need parametric form for the curve to use this formula. There are other formulas for curvature that do not need parametric form (coming up).

Let us do some examples:

Example: Find the curvature of a straight line. Guess first.

Let x=t, y=mt+b represent the line in parametric form. By the way,
there are many parametric forms of a curve that can be formulated,
this is just one.

x*=1, y*=m, x**=0, y**=0, so, k = [(1)(0)-(m)(0)] / [(1²)+m²]^(3/2)=0
Did you guess right? makes sense, yes?

Question: Is there a formula for curvature if we have a direct relationship, y = f(x) or x = f(y)?

Answer: yes. Let us derive it.

We use arc length for these types of functions. For y=f(x) , ds/dx = [1+(dy/dx)²]^(1/2) and for x=f(y), ds/dy = [1+ (dx/dy)²]^(1/2).

So, since k = dq/ds and q = tan^-1(dy/dx) or cot^-1(dx/dy), we get , k=(dq/dx)/(ds/dx) or (dq/dy)/(ds/dy) giving us the following;

k= d²y/dx² / [1+(dy/dx)²]^(3/2) or d²x/dy² / [1+(dx/dy)²]^(3/2)

~So, applying this to y=mx+b, using the first, we get, k = 0/ (1+m²)^(3/2) = 0, much easier.

Example: Find the point on y=lnx where there is maximum curvature.

Use the formula for y=f(x). Since y' = 1/x, y''= -1/x².
So, k= (-1/x²)/[1+(1/x)²]^(3/2).
Note that for all x in the domain of this function, the curvature is negative. Makes sense since this curve is always concave down.

We need to maximize this expression. So, find where dk/dx = 0. Use the quotient rule to find dk/dx and simplify the expression. Then find where the numerator of dk/dx = 0 and you will get x=1/√2. Then pick up the y value of (-1/2)ln2. (I will leave this as an exercise for you)...nice of me.

Using a parametric representation for a circle of radius a, one can show that the curvature will be constant and equal to 1/a. makes sense.

If we draw a circle with "closest contact" to the curve at the point in question (on the concave side), it will be called the "circle of curvature" for that point. The radius of this circle is called the "radius of curvature" for that point & the center of this circle is called the "center of curvature for that point". Note that there are many circles that can be drawn tangent to the curve at that point, but only one has the "closest contact". This means the first & second derivatives of the curve agree with the first & second derivatives of the circle at that point. (the more derivatives having the same value for two curves, the closer the curves approximate each other)

So, a sharp turn on a curve, has a small circle of curvature, thus a small radius of curvature. On the other hand, a straighter curve has a very large circle of curvature & radius of curvature. The circle of curvature & radius of curvature become undefined for a straight line.

Much of this is all tied together in calculus II & III when analyzing the motion of a particle on a plane or space curve. Tangent & normal Components of the vectors involved at a point on the curve include the velocity, speed, acceleration, and the curvature of the path. By analyzing these forces, one can see what it takes to keep a point on a given path (important when driving your car along a curved road).

Integration

Integrals:

The process of finding derivatives and using them to solve problems is called differentiation. Think about derivatives as analyizing a problem at various instants, since it will describe the parent function at particular points of interest. The process of Integration does exactly the opposite. It will take instants and bring them together as a whole, so to speak. When dealing with integration, we generally know the derivatives and seek the parent function. So implied in the name of the process. By going from instants to the whole gives us a path to an unknown parent function that we seek in our problem. Generally speaking, most students find this reverse process much more difficult since it involves a indept knowlege of derivative properties and more precise decision making for success. I will give you a very brief and informal description on how this works and give some basic examples of a few types of problems that uses integation to solve them. The integration process is also referred to as anti-differention.

The integral(or anti-derivative) of a function, f(x), is symbolized by ∫f(x)dx. The differential dx indicates the variable of integral is always written to the far right. We are seeking a parent function with a derivative of f(x).

The expression between the integral sign and the differential is the expression to be integrated. This representation is for a function of a single variable, usually studied in Calc I & II.

In multivariable calculus (calc III), the function to be integrated is usually dependent on several variables, i.e., f(x,y) or f(x,y,z) and multiple integrals appear. For example, ∫∫f(x,y)dxdy and ∫∫∫f(x,y,z)dxdydz represent a double and triple integral in calculus III and can be done by two and three separate integrations, respectively. They are done by separate integrations starting from the inside out. For example, in the case of the triple integral, the first integration is done with respect to x, then with respect to y, and finally with respect to z. The other variables in the expression f(x,y,z) are treated as constants when performing the integration with respect to one of them. For multiple integrals, the order of integrations is always done from the inner most variable out to the variable on the extreme right. (These variables are indicated by their differentials). Don't worry about these. Only those math majors who survive Calculus I & II will get the opportunity to be exposed to them.

These are examples of Indefinite Integrals, since no limits (numbers or symbols that appear at the bottom and top of the integal symbol) appear on the integration symbol. So, the results will be variable expressions with unknown constants (assuming no conditions on the problem are given). Definite Integrals have limits on the integration symbol (lower & higher).

So, going from a derivative to its parent function will involve integrating the derivative. However, for indefinite integrals, the result will contain many answers since an unknown constant will be present. In symbols, ∫f(x)dx=F(x) + C. I use F(x) in the result since that function will differ from f(x). The constant C is arbitrary and could take on an infinite number of values, if no other information is given in the problem.

It will always be the case that F'(x) = f(x). This is a fundamental relationship that ties both processes together. This is one of the fundamental theorems of calculus. For example, ∫2xdx=x² + C. Note that the derivative of x^2 is 2x, which is the expression in the integral. However, only the simplest of integrals can be done by guess work. Rules are established for integration so more complicated integrals can be evaluated. There are expressions that are very complicated & special techniques are studied (calc II) and other integrals that can’t be solved. However, most applications deal with definite integrals. Their values can be approximated very closely. The calculator will do this for us very quickly (the TI-83 calculator).

Some Integration formulas: There are numerous integration forms that are listed in many books. It's virtually impossible to know them all. I will list just a few popular ones that seem to be used in the most basic of problems. What makes this difficult is that the student needs to pick the correct form for their integral. Not a very easy task. If the wrong form is chosen, the problem will not be able to be solved. The following are a few of the basic rules.

Basic integration formulas with examples

1) ∫du = u + c
EXAMPLE: ∫dx = x + c

2) ∫k f(u)du = k∫f(u)du (k any constant)
EXAMPLE: ∫2pxdx = 2p∫xdx

3) ∫[f(u) + g(u)]du = ∫f(u)du + ∫g(u)du (integrate a sum over each term)
EXAMPLE: ∫(x² + 2x + 3)dx = ∫x2dx + ∫2xdx + ∫3dx= ∫x2dx + 2∫xdx + 3∫dx

4) (POWER RULE): ∫u⁽ⁿ⁾du = u⁽ⁿ⁺¹⁾ / (n+1) + C (Must make sure du is present)
EXAMPLE: ∫x²dx = x³ / 3 +C, since u=x, n=2, du=dx & is present.
EXAMPLE: ∫(5x²+ 7)4xdx, since u=5x²+7, du=10xdx (need to multiply inside by 10 & compensate by dividing on the outside by 10)
(1/10)∫(5x²+7)⁽⁴⁾10xdx = (1/10)[(5x²+7)⁵]/5 + C = (5x²+7)⁵/50 + C

5) ∫e^udu = e^u+ C
EXAMPLE: ∫e^5xdx = (1/5)∫e>^5x5dx = (1/5)e^5x+ C

6) ∫du/u = ln |u| + C
EXAMPLE: ∫xdx/(x² +7) = 1/2 ∫2xdx(x²+7) = (1/2)ln(x²+7)+C
EXAMPLE: ∫(sin3x)dx/cos3x = (-1/3)∫(-sin3x)3dx/cos3x =(-1/3)ln |cos3x|+C

7) ∫cos(u)du = sin(u) +c
EXAMPLE: ∫ {[cos(lnx)] /x} dx = ∫ [cos(lnx)] [1/x] dx = sin(lnx) + C

8) ∫sin(u)du = -cos(u) +C

9) ∫sec²udu = tan(u) +C

10) ∫ sec(u)tan(u)du = sec(u) +C
EXAMPLE: ∫xsec(x²)tan(x²)dx = (1/2)∫ sec(x²)tan(x²) 2xdx= (1/2)sec(x²) + C

11) ∫ csc(u)cot(u)du = -csc(u) + C

12) ∫ csc²udu = -cot(u)+C

~NOTE: An integral without an differential is meaningless.

In multivariable calculus, derivatives & integrals are taken with respect several different variables. Partial derivatives & multiple integrals are used. So, when differentiating, one must state reference to the variable. (on occasions, in single varaible calculus, it is understood).

However, not always. For example, the derivative of x wrt to x is 1, but the derivative of x wrt to t is dx/dt. Likewise for integration. The differential to the far right of an integration gives the variable of integration. In multiple integrations (double & triple integrals), several different differentials are present. So, one would integrate wrt one, then integrate again, wrt another, & so on. These techniques are performed by treating all other varaibles as constants (temporally) until the operations are completed. Don't be concerned with this unless you find yourself in a Calculus III course.

THE DEFINITE INTEGRAL

Question: "what is a definite integral?"
Answer: "A number" (that's it!)

Question: "what does it represent?"
Answer: "Possibly an area, volume, length, or some other numerical value of some quantity."

Question: "can it be a negative number?"
Answer:"sure can"

To understand how this can happen, we go to the basic area problem.

Some simple areas can be found by inscribing or circumscribing rectangles between the curve and the x axis and then letting the number increase to infinity. If the sum is formulated properly, the limit of the sum will equal the area in question. Be careful, since sums below the x axis are negative.

These sums are called Riemann Sums.

The tie between the limit of a Reimann Sum and the area in question can be shown in a very involved proof. It is lengthy and requires much background in calculus. It is omitted here.

It also can be proven, in a less involved way, that the area under a curve and between x=a and x=b can be found by integration. (uses one of the fundamental theorems, namely, the tie between differentialion and integration) . All you have to do is to set up the definite integral from a to b and evaluate it.

The definite integral from x=a to x=b is denoted as ∫_a^{^{^b}}f(x)dx and is evaluated by integrating f(x), then substituing the upper limit in the expression, then subtracting the lower limit substituted in the expression. If we denote F(x) as the integral (without the constant), it would look this way: F(b) - F(a).

~Well, since the limit of a Riemann Sum equals the area and a definite integral equals the area, we define the definite integral as the limit of a Riemann Sum. Therefore, many problems are solved by setting up Riemann Sums.

~Consequently, defining the definite integral as this limit, we have a direct tie to the area problem and other problems that can be expressed and formulated as limits of these types of Reimann Sums. This will give us a way to solve all kinds of applied problems (more complicated area problems, volume problems, surface area problems, work problems, arc length problems, fluid pressure problems, just to name a few).

~Therefore, if a problem is subject to analysis by using a limit of a Reimann Sum, it can (theoretically) be solved by a definite integral.

~Basically, Here is how the problem must be formulated

lim ∑f(kx)Δx= ∫_a^{^{^b}}f(x)dx
Δx→ 0

So, informally, if a problem involves a solution from x=a to x=b, we partition the interval into k subintervals, each of length Δx, form the Riemann sum, then set up it's limit. In actuality, only one representative term of the sum is needed since they are all formed the same way (this is known as a "specimen" piece).(i.e., to find the volume of a solid, we can think of the solid as the sum of an infinite number of smooth disks of different sizes. However, the volume of each one is represented the same way, so all we need is a specimen disk in the formulation that represents each one. This would be called our specimen volume piece) However, we don't need to evaluate the limit, just set it up so that we can recognize the funcion f. Then all we have to do is integrate f(x) from a to b for our answer. Solid disks may not be appropiate for all volumes. Depending on their cross-section and how they were generated, a different specimen piece must be chosen (i.e., washers, shells, or a special one with a given cross-sectional area to fit our problem).

~Informally, one can think of the symbol ∫_a^{^{^b}}f(x)dx as the lim∑over [a;b] and the Δx in the sum becoming dx in the integral.

~It turns out that many applied problems can be "set up" by using Riemann Sums of these types, therefore, can be solved by a definite integral.

~For those of you who go on to calculus II, you will see many.

~The above mention problems are at the heart of calculus II.

Thank goodness for our TI-83 calculator since the number associated with any definite integral can be found easily. You do not need to know any integration rules. It will be done automatically. An added feature along with the answer will be a sketch of the curve inticating an area that the value of the definite integral represents. Very, very, Nice!

Popular applications of integration

1) Area under a curve: Let's take the function f(x)=x² (a parabola openning up from the origin). Let's say we wanted to find the area under this curve and above the x-axis from x=1 to x=3. To get this we set up the definite integral of x² from x=1 to x=3. It will look like this:

∫₁^{^³}f(x)dx

To evaluate this definite integral we first integrate x². Get the result (without the constant of integration), then substitute the upper limit of 3 into it and the lower limit of 1 into it and subtract them. It looks like this:

∫₁^{^³}(x²)dx

=x³/3. Now, substitute the limits into this expression and subtract them:(3³/3)-(1³/3)=9-1/3=27/3-1/3=26/3 square units of area. The TI-83 calculator will complete this in a matter of seconds and give you a graph of the area in question.

2)Finding total distance traveled by a particle given its velocity: Let's say a particle travels according the the velocity V=t²-4t-1. The total distance traveled will be the amount of area between the horizontal axis (t-axis in this case) and the velocity curve. Let's find the total distance traveled between t=2 and t=4.

We need to set-up the definite integral from 2 to 4 of the velocity function. It looks like this:

∫₂^{^⁴}(t²-4t-1)dt

=(t³/3-4t²/2-t). Now, substitute the limits into this expression and subtract, we get
[(4³/3-4(4²)/2-4]-[1³/3-4(1)^{2

Differential Equations

Courses in differential equations usually follow the calculus
sequence of I, II, and III.

In calculus I, you are required to solve
only one type where the variables involved can be separated. This is called the
"variables separable" method.

Solving all types of differential eqs is essential for most
fields of engineering, specially mechanical and electrical. In all disciplines (including business related fields), whenever rates of change (derivatives) are involved, you will, most likely, encounter differential equations. (see example on Radiocarbon dating).

Differential
eqs are equations that involve derivatives and/or differentials.

They
occur in many areas. Growth/decay, motion of masses, electrical circuit
theory, acoustics, among many.

First order diff. eqs. involve
derivatives no higher than the first. These are categorized by types and require
different techniques for each type.

Second order diff. eqs. involve
derivatives no higher than the second.
Along with the second derivatives,
they could also involve the first derivatives. Techniques here are a little more
sophisticated.

Certain higher order diff. eqs. (with derivatives higher
than two) are also covered. Also, systems of differential equations are studied.
This requires much knowledge of matrices & how to work with
them.

More advanced types involve partial derivatives (covered in
calculus III) and are necessary for the analysis of sound waves, water waves,
and related acoustics (i.e., music).

The techniques for solving partial
differential eqs. are quite involved and much calculus background is
necessary.

One of the most interesting techniques for solving many of
the basic types is by the use of Laplace Transformations. In simple terms, it
transforms a calculus differential eq into an algebraic equation with no
calculus. It is then solved in the algebra world then transformed back (using
inverse Laplace Transformations) to the calculus world where the solution is
given. (Many a calculus student would like that to happen to all of their
calculus problems)

Another interesting technique for the second order
type is by the use of series (covered in calculus II). These are referred to as
series solutions. The technique can be very long and involved but is a great
application of series.

In my opinion, one of the most interesting
situations encountered in differential equations is the analogy between the
simple LRC electrical circuits and the mechanical system counterparts. These
occur in the study of the second order linear types with constant coefficients.
Let me give you a brief description.

The differential equation for the
electrical system is
Ld²q/dt² +Rdq/dt +(1/c)q = E(t),

where q=charge (coulombs), c=capacitance (farads), dq/dt=current (amps),
R=resistance (ohms), d²q/dt²=rate of change in current,
L=inductance (henries) and E(t) is the total electromotive force of the circuit
(volts).

In the mechanical system, we have an object moving through a
medium and has a very similar type of differential equation looking like this:

Md²y/dt² + Rdy/dt +ky=F(t),
where
d²y/dt²=acceleration, M=mass of the object (weight/g),
dy/dt=velocily, R=resistance (proportional to the velocity), y=displacement,
K=constant (proportional to the distance y traveled) (for a spring, it's Hooke's
constant) and F(t) is total force acting on the system.

~One can
therefore model a mechanical system using the similar electrical system and
predict behavior quite nicely. Since the differential equations of both systems
are quite alike, their solutions will be also. A very interesting
analogy.

A course in differential equations is usually not required for
math majors (depending on the school). It certainly would be at MIT or
Cal.Tech., since it is vital to most engineering programs.

3) Solving Basic differential equations by the variables separable method: In my opinion, the most useful use of the integration process is in solving basic differential equations where the variables of the equation can be separated. Having taught differential equations for a good number of years, I can tell you, first hand,
that only a select few can be solved this way. Most others require specialize techniques that must be used. Courses in differential equations are extremely important to engineering in various fields. For our discussion, we will concentrate on solving acceleration, velocity, and distance problems using the method of variables separable.
Let's take a look at an example.

EXAMPLE: A particle moves in a straight line with acceleration a=12t²+6t. It starts from rest with an initial velocity of -3 ft/sec (traveling in our designated negative direction). Find the equation for distance s at any time t.

Since acceleration measures the rate of change in velocity with time, it is a derviative, a=dv/dt. Also, velocity measures the rate of change in distance with time, it also is a derivative, v=ds/dt. These quantities need to be substituted into the equation to give the desired differential equation to solve. An equation which involves derivatives is called a differential equation.

Since a=dv/dt, we have the differential equation dv/dt = 12t²+6t. Separating the variables we get, dv = (12t²+6t)dt.(notice the separation of the variable on opposite sides of the equation)

Integrating both sides, ∫ dv = ∫ (12t²+6t)dt. Which gives us,

v = 12t³/3 + 6t²/2 + C. Since v = -3 when t=0, substituting these values in the equation for v determines C = -3.

Substitute -3 for C and we have, v = 4t³+3t²-3.

Since v = ds/dt, we have another diff. Eq., ds/dt=4t³+3t²-3.

Separating the variables & integrating, we get, s=t⁴+t³-3t+c.

Since the particle starts from rest, s=0 when t=0, so c=0.

Therefore, the equation for distance is, s = t⁴+t³-3t.

~NOTE: When separating the variables, make sure the differentials are "up stairs" and to the right of the expressions to be integrated.

Carbon 14 Dating

Regular carbon or Carbon 12, as researchers call it, is an
element which is found in all living things. But elements can come
in unstable forms, called isotopes. Isotopes have extra neutons in their nuclei, and are often
radioactive.

Carbon14 is an isotope of the regular stable element Carbon
12. This
is due to the interaction of cosmic rays in the upper
atmosphere. The earth consists of approximately 99% of regular Carbon
12 and trace amounts of carbon 14. Carbon 12 has 6 neutrons & 6
protons while carbon 14 has 8 neutrons & 6 protons. This makes
carbon 14 unstable and radioactive. Organic life absorbs Carbon 14
and Carbon 12 (stable form of carbon) through the atmosphere as
long as it is living. Once it dies, both forms of carbon are no
longer absorbed. The level of Carbon 12 in the dead matter remains
constant while the level of Carbon 14 decreases due to its unstable
nature (it decays). Therefore, the percentage of Carbon 14 found
decreases and the constant ratio of C14 to C12 decreases from
this natural constant amount. By measuring this decrease, the age of
the organic sample can be estimated.

This is the method of Carbon 14 dating (technically
called Radiocarbon Dating). How this is
done mathematically will be shown in a simple example at the end of
this discussion.

The "half-life" of a radioactive substance is the amount of time it
takes for half of the amount present (at any given time) to
dissapear. For Carbon 14, it
takes quite a long time, some 5,600 years. Many other radioactive
materials have very long half-lives as well. This is a major problem
facing our
present civilization today (what to do & how to store this
material). It turns out, only a vast amount time will eventually
eliminate this radioactive material...much too long to wait. For carbon
14, it will take roughly 60,000 years to eliminate all
significant amounts. That means, life-forms existing at that time or
before are not subject to Carbon 14 dating. There are similar dating methods using isotopes of other elements
that have longer half-lives to date artifacts older than the limits of the Carbon 14 method.

Knowing the constant ratio of Carbon 14 to regular Carbon 12 during the
life of the item and by measuring the lower ratio of Carbon14 to
regular Carbon 12 when an artifact is found, lab techniques can be used to
determine how long the Carbon 14 has been decaying, consequently, how
much time has passed since it's death.

For example, bones and other organic material from an ancient
civilization could be dated to
determine their ages. Researchers, particularly archaeologists, often
use
this technique for dating, usually within 50,000 years. Beyond that
time, other radioactive elements are used that have a longer half-life
since the amount of C14 left is insignificant.

Some historians believe The Shroud of Turin
(claimed to be the burial cloth of Christ), is a fake. Carbon14 testing
has dated it to the middle 13th century-14th century. That's the same
time it was claimed to be found, suggesting that it was a forgery. This
controversy still remains today.

UPDATE: The latest scientific experiments using carbon dating and additional blood spatter patterns have confirmed that the Shroud is a fake.
Not all, including religious sectors, disagree with these results and continue to believe in its authenticity.

The mathematics behind simple growth/decay is not
that involved. From the rate of change (derivative of the amount of
Carbon 14 present with respect to time) we can find an exponential
equation relating the quantities. Knowing the half-life of the
substance, we can estimate the unknown constant in the equation.
Consequently, relating the variables directly. Hence, time can be
estimatied. All of this is programed in the lab techniques used for
this method.

Simple example:: A bone
is found during an excavation to contain 54% of the carbon 14 normally existing naturally (i.e., 54%
of the original amount of Carbon 14 is remaining). Using
the half-life of 5600 years for Carbon 14, estimate the age of this
bone.

Since the time rate of change of the amount of Carbon 14 is
proportional to the amount present at any given time t, we have this
simple differential equation:

dN/dt = kN, where N=amount of carbon 14 present at any time t.

Separating the variables and integrating, we get: ln(N)=kt +C

At t=0, let No represent the amount of carbon 14 present, so we

can represent the constant C as ln(No). Replacing this constant and

solving for N, we get: (I'm calling this equation A)

(A) N = No e^(kt) (this is the basic form for simple growth/decay)

or N/No = e^(kt). Since the ratio N/No = .54, we
substitute this value into the above equation and solve for t in terms
of k. We get:

(I'm calling this equation B)

(B) t= (ln.54)/k or approximately, t= -0.616186/k. Now, for the half-life.

Substituting (1/2)No for N and 5600 for t in equation (A),

we get k = -0.0001238 (this is called the decay constant for C14)

Note: Decay constants are determined by the half-life of the radioactive substance used in the procedure.

Therefore, Using this value in equation (B), we get, t = 4977 years approximately.

~SOME CALCULATOR PROCEDURES FOR THE TI-83~

FINDING FUNCTIONAL VALUES

Table Method:
Mode: Rad & function

Enter equation: Y₁=

Press 2nd window to set up your table (Table Set)

(move to auto for integer values of x)

(move to ask for any value of x you want) (keep this setting)

(keep y at auto)

press 2nd Graph

enter desired x-value

y-value appears in 2nd column

Note: For graphing purposes, Zoom Fit (menu 0), will graph the

equation. However, if a more complete graph is desired,

change the x & y ranges in the window. It's best to experiment

with various ranges to see the effect on the graph. Looking at

the range of values from your table will give you a good indication

how to set up values in your window for a real good graph.

Using the trace button

When the graph is visible, clicking the trace button will place a
symbol on the graph. Moving the cursor in different directions along
the curve will give the coordinates of its location at the bottom of
the screen. Error messages will occur is the y values are too large
(overflow error) or if the x value entered is not in the range of the x
values of the window. You make have to adjust the window ranges to get
them. This is a very convenient way of finding points on the graph.

Alternative Method

Mode: Rad & function

Enter equation: Y₁=

Press 2nd QUIT

Press VARS

Move to Y-VARS

Cursor at 1: Function, Enter

Cursor at Y₁, Enter

Insert value in the domain of Y₁, i.e., Y₁(a), Enter

The functional value at a will appear

Repeat for other values.

NOTE: When graphing functions, make sure your plots are off in the y= menu. If not, you will get an error message.

STORING & RECALLING A GIVEN VALUE

1) Take any number or a result of a computation you wish to store & proceed to the next step.

2) Press STO-> , ALPHA (blue) & any key with a small blue letter.

3) To recall this value, press 2nd, RCL, ALPHA, & the key where you stored the value, enter.

Note: Storing values is very handy in complex operations where a particular complex operation is repeated several times.

DRAWING TANGENT LINES TO FUNCTIONS & GETTING THEIR EQUATIONS

Mode: Rad & function

Enter equation: y₁=

Zoom 7 (this will graph equation)

2nd draw

Select 5: will get Tangent(

Enter in x-value of point

Tangent line will be drawn & equation will be given.

EQUATION SOLVER   [for solving f(x)=g(x) in the form 0=g(x)-f(x)]

Click mode to get in Func mode

Click Math

Click cursor up to Math, then click enter

Move cursor up, then click enter

Should see: Eqn: 0=

if there is another equation there, click clear

Should see Eqn: 0 =

Enter the right side of the equation (after the = sign)

Click Alpha then Enter

Enter a guess (near a solution)

Click Alpha then Enter

Solution for x is shown

repeat with another guess for other intersections

Here's how to find the MAXIMUM & MINIMUM pts using your TI-83.

Understand that these are approximations & not exact values, if

they do not come out exactly. For exact values, use the procedures

outlined in class. (most likely, you'll have to find exact values on

tests). However, the following procedure is a nice way to check your

answers.

1) Code in your equation at Y₁.

2) Set your viewing window for x & y (this might have
to be adjusted, if the entire curve is not visible). To set a friendly
window, look at functional values (see Finding functional values, above)

3) click graph (graph of f should appear)

4) click 2nd trace

5) enter (3) Minimum or (4) Maximum

6) Enter an x value to the left of the min/max pt

7) Enter a value to the right of the min/max pt

8) enter any value between as a guess

9) the coordinates of your min/max pt will appear at the
bottom of the
window

(you could also just use trace & zoom to get there)

GUESSING LIMITS USING YOUR CALCULATOR

Choose values close to the limit.(on either side, when appropriate)

Calculate the value of the expression

See if the expression is getting closer to one number.

Note: There will be better ways to get the exact limit without the use of a calculator.

GRAPHING THE DERIVATIVE OF A FUNCTION

go to the
y= menu to Y1=, click the MATH menu to 8, enter, then code in the
equation of the function, x,x), then press graph. Make sure you have a
friendly viewing window.

NUMERICAL INTEGRATION (for evaluating Definite Integrals).

First, enter your function (to be integrated) by pressing Y= (top left).

Make sure you set the proper viewing window for X. The range (Xmin to
Xmax) must include the limits of integration or you will get an error
message. The Graph of the function, with the shaded area in question,
is not necessary to get the final answer to the definite integral.

Next, press 2nd, Trace for the Calc menu, go to menu 7, enter.

The display will prompt you to enter the lower limit of integration.

Press enter, then enter the upper limit, press enter.

The result is the numerical value corresponding to the definite integral.

It may or may not be exact, depending on the function & the limits of integration.

It represents the area between the x-axis & the function in
question between those limits. It may be negative, if the area is below
the axis.

~Note: Common mistakes leading to wrong answers are mostly
likely associated with the improper use of parentheses, especially with
calculations involving e^x. Finding e^x with the
power key(^) will give an incorrect result unless the entire term is in
parenthesis along with its power (need two sets of parentheses). So,
when confronted with calculations involving e^x, it is best to use the 2nd e^x key directly.

~NOTE: if you're having much difficulty with your calculator & seem
to be getting "junk" most of the time with all kinds of nasty messages,
do the following:

click 2nd MEM ENTER

7 Reset ENTER

2 Defaults ENTER

2 Reset ENTER

Clear

This will reset you calculator

you might lose saved programs, so take note

Conclusion

I hope all readers enjoyed this Journey in mathematics. Much has been omitted. I just wanted to give the readers a taste of some topics I feel are essential. Most calculus I,II,and III courses will cover many more topics which include lots of theory (a big stumbling block for most students). Some of you will continue this journey to more advanced grounds and others, most likely not. For those of you who do, your minds will experience an expansion only a select few are fortunate to achieve. If you got anything out of this journey, I hope you realize that mathematics requires a special mind set and a unique approach. Understand and follow the rules in this discipline and you will be on your way. Look for simple solutions to problems first before you assume more complex approaches. By learning and understanding more mathematics, I guarantee you will become a much better thinker. This will lead to better decision making, consequently, a better life. My thanks to all that
took the time to travel with me on this journey.}