Thursday, January 30, 2014

Understanding the Basics of Polar Coordinates

The topic of polar coordinates has been confusing to many students I have worked with in my nearly 14 years of tutoring. But understanding the basics of polar coordinates can make what seems to be a very foreign concept a lot easier to understand and work with. In fact, in some situations, using polar coordinates is easier than using rectangular coordinates. What are polar coordinates? 
A scenario that can be represented in a grid-like fashion such as a layout of a city with parallel roads running north to south and parallel roads running east to west can be represented using a rectangular or Cartesian coordinate system. But how do we explore the symmetry of some of the most beautiful things in nature, things with flowing curves such as the butterfly? Trigonometric functions and a system for plotting points called the polar coordinate system is used.

The foundation of the polar coordinate system is a ray that extends to the right. The ray is called the polar axis. Just like a point on the rectangular coordinate system is represented by the ordered pair (x,y), a points P in the polar coordinate system is represented by an ordered pair (r, Θ). The directed distance from the pole, which is the endpoint of the ray, to P is r. The angle from the polar axis to the line segment from the pole to P is Θ.

For example, a clock is a good representation of a polar coordinate system. Suppose the hour hand is pointing directly to 3 on the clock. That is directly horizontal and represents the polar axis. Now, suppose the minute hand is at the 1, representing the time 3:05. Suppose the length of the minute hand is 5 inches. Therefore, r = 5. The angle from the hour hand to the minute hand is Θ. In this case since there are 12 numbers around the clock and a circle is 360 degrees, each number is 30 degrees (360 divided by 12 equals 30). There are two numbers between 3 and 1, so the angle Θ in this example is 60. The polar coordinate of this situation is (5, 60). If using radians, Θ = Pi/3.

Suppose the polar coordinates of a point P are (2, 45) or (2, Pi/4). Since the angle is positive, first draw the 45 degree angle counterclockwise from the polar axis. Then count 2 units along the terminal side of the angle to reach the point P.

The sign of r is important in plotting the polar coordinate P. If r > 0, the point lies on the terminal side of Θ. If r < 0, the point lies opposite the terminal side.

Let's take another example, P = (-3, 270) or (-3, 3Pi/2). First move counterclockwise 270 degrees, which would be a ray pointing straight down. If this was a rectangular coordinate system, the coordinate would down the y-axis. If r equaled 3, the ray would point down three units. But since r is -3, the ray points in the opposite direction 3 units in length. The ray would be pointing straight up three units along the 90 degree or Pi/2 axis. If this was a rectangular coordinate system, the point would be up the y-axis.

An interesting thing about polar coordinates is that each point can be represented infinitely many ways. You can add 2Pi or 360 degrees to Θ and it does not change it's location since it just moves the ray one complete revolution. Also you can change the sign of r and add a half a revolution, 180 degrees or Pi, and that will also not change the location. With rectangular coordinates, there is only one way each point can be represented.
There are basic relations between polar coordinates and rectangular coordinates. They are as follows:

  1. x = rcosΘ, y = rsinΘ
  2. x^2 + y^2 = r^2
  3. tanΘ = y/x
We use the relations to convert from polar coordinates to rectangular coordinates and vice versa. Suppose we have the polar coordinate (2, 3Pi/2) or (2, 270). We can convert to rectangular coordinates by substituting the given values for r and Θ into x = rcosΘ and y = rsinΘ. When doing so, we get x=0 and y = -2. Therefore the rectangular coordinates are (0, -2). 
When converting from rectangular to polar coordinates, first plot the point (x,y). Find r by using the formula x^2 + y^2 = r^2. Find Θ by using tanΘ = y/x with the terminal side passing through (x, y). For example, suppose the rectangular coordinates are (-2,0). After plotting the point we calculate r, -2^2 + 0^2 = r^2. Therefore 4 = r^2, so r = 2. Now we find the angle by taking tanΘ = 0/-2, therefore Θ = 180 or Pi. The polar coordinate P is (2, 180) or (2, Pi).

There are many other complex problems and applications with polar coordinates, but understanding the basics are crucial before moving forward on those types of problems. This guide should help you better understand the basics of polar coordinates.

Friday, January 24, 2014

Did you know that there is a complex number whose cube is 8? Most people forget about complex numbers as a cube root. If someone asks, "What is the cube root of 8 ?",  the majority of people will say 2. But -1 + sqrt(-3) is also the cube root of 8.

Let's verify by cubing -1+sqrt(-3)

(-1 + sqrt(-3))^3 = 1(-1)^3 + 3(-1)^2(sqrt(-3)) + 3(-1)(sqrt(-3)^2) + (sqrt(-3))^3

                          = -1 + 3(3i) - 3(3i)^2 + 3i^2(3i)

                          = -1 + 3(3i) + 9 -3(3i)

                          = 8

So if someone wants to wear #8 on a uniform, they could also be clever and say, "I want to wear number (-1 + sqrt(-3))^3."

Saturday, January 18, 2014

Review of my second book, unfortunately the person spelled my name Kauggman, oh well... Kerry Kauggman’s Algebra Simplified Intermediate & Advanced is a mathematics textbook. It cannot be read like a typical novel or a non-fiction narrative. Algebra Simplified Intermediate & Advanced follows Kauggman’s previous book Algebra Simplified Basic & Intermediate. Not having read Kauggman’s earlier book, I am unsure as to what has and has not been covered in the first volume, although given the thoroughness with which this textbook is written, I am fairly confident that the previous volume prepares the reader for the content of the sequel.

The target audience of Algebra Simplified Intermediate & Advanced consists mainly of the middle school and high school students who will be using it for their mathematics coursework, although some older readers might need it for a GRE exam, or some similar standardized test. I would also recommend that parents look at Algebra Simplified Intermediate & Advanced in order to brush up on the subject so they can help their children with their homework.

Given my extensive background with algebra in school, I feel that I cannot evaluate this book on one very important point– how does it teach a challenging subject to someone with no prior knowledge of that topic? For me, reading this book was essentially a refresher course, though to be honest, I needed to review some topics quite a bit. However, a student who has never wrestled with quadratic equations or inverse functions before might have a slightly harder time grasping the concepts. Since this textbook will often be used in conjunction with a class, the teacher ought to be able to answer any lingering questions students might have behind these algebraic concepts.

Each chapter ends with numerous sample questions for the student to answer. Unlike many textbooks, Algebra Simplified Intermediate & Advanced provides step-by-step explanations of some of the more difficult equations. These further examples help the learning process, although one hopes that unscrupulous students will not simply copy the extended answers instead of actually trying to reason through the equations themselves.

Crafting a quality textbook is a great challenge, not simply because it is difficult to convey information, but one also has to keep the attention of the readers. Unfortunately, many young people are studying algebra not because they want to, but because they have to, and though Algebra Simplified Intermediate & Advanced does a very good job of explaining and presenting relevant information, there are only minimal attempts to show how learning algebra has applications outside of the classroom, and there are no attempts whatsoever at humor (though in all fairness, it is challenging to see where would be the best place to make a joke). Some details about the history of algebra and the development of the discipline might have been interesting, but their absence really ought not to be considered a major fault. Perhaps instilling adolescents with a love of algebra is too daunting a task to be realistic, but it would have been interesting if Algebra Simplified Intermediate & Advanced had made some effort to be entertaining as well as factual. Textbooks need to do something to set them apart from other tomes in their fields, and snappy writing and readability can be great assets if they do not come at the cost of lucid information. Three stars out of four.

If anyone is interested in this book, you can get it at

Wednesday, January 15, 2014

review of my first book

This book can be purchased at Follow the link to my page.

Official Review: Algebra Simplified Basic & Intermediate

[Following is the official review of "Algebra Simplified Basic and Intermediate" by Kerry Kauffman.]

This is a textbook format Algebra book intended to assist the student in basic to intermediate algebra, and beginning college algebra. The author is Kerry Kauffman who has a Bachelor of Science in statistics and has tutored students such as you in many different subjects. He also assists students online at Kaplan University.

The preface explains in easy to read terms the intention of the book for example who and how it intends to help, how to use this book effectively and shows the author’s understanding of the needs of the student. This book also explains how to use this new found knowledge in everyday life which is always the million dollar question when it comes to algebra.

The index beautifully outlines each chapter making searching for your trouble areas a breeze. All key terms are in a bold text and are again defined in the glossary in the back of the book. Usually you don’t think you’ll do much reading in an algebra textbook, or even if you do you probably won’t understand anything anyway, but I found the chapter to be organized and easy to understand. All key terms are explained very well. I wish I had this book in high school.

Throughout the book the reader gets a sense of understanding from the author that is rare. The author understands the common troubles that are faced by algebra students. For example; word problems. I've met quite a few people who have told me that they never understood them and I even had trouble with them myself in school. The author has dedicated a whole section to them in the back of the book again using real life situations. Unlike many of our textbooks we had to use years ago the answer key in the back not only has the solutions, but the steps on how to get there. No more pulling a number out of thin air.

The book covers everything from the basic descriptions of numbers and what they are to basic college algebra. I would give this book a recommendation to be used in any school curriculum. I rate it a 4 out or 4 stars.

Friday, January 10, 2014

I've posted this in the past but have had a lot of interest in this, so will repost for those that missed it the first time.

Over the course of mathematics, we learn how to raise a quantity to an exponent. We know, for example, that xn means we multiply x by itself n times. But suppose n is zero. How do we multiply a number by itself zero times? We are taught that any number raised to the zero power equals one, but rarely does anyone explain why this is the case.

One way to explain the zero exponent phenomenon is to use rules for exponents. When multiplying like bases, exponents are added. For example, 35 x 30 = 3(5 + 0). Therefore 35 x 30 = 35, so 30 must equal 1.
When dividing like bases, exponents are subtracted. For example, (46)/(40) = 4(6 - 0). Therefore (46)/(40) = 46, so 40 must equal 1.
Another way to show that any base raised to the zero power is one is to examine some patterns. Notice the pattern in the following:
  • 24 = 16
  • 23 = 8
  • 22 = 4
  • 21 = 2
  • 20 = ?
Notice that as exponents decrease by 1, the result is divided by the base number 2. Continuing this pattern, 20 = 1. This will hold true no matter what the base number is.
Finally, I can use the concept of limits to show that any number raised to the zero power is one. Suppose we take 5n, and we start with n = 1. We know that 51 = 5. Take the square root of 5, which is equivalent of 5(1/2), which is approximately 2.24. Now take a smaller value for n, such as 1/3, which gives us 5(1/3) = 1.71. Continue to take values for n smaller and smaller but not less than or equal to zero. You'll start to notice what is happening, 5(1/10) = 1.17 , 5(1/1000) = 1.002, 5(1/100000) = 1.00002. Notice how the result is getting closer and closer to 1. We say that the limit as n approaches 0 of 5n = 1.
I just explained three methods that clarify why any number raised to the zero power equals one. Next time someone is puzzled by this fact, you can explain the reason behind this somewhat vague and often unexplained topic in mathematics.

Tuesday, January 7, 2014

Suppose X and Y are independent. X and Y are both normally distributed with mean 100 and 96, respectively. The standard deviation of X is 6 and the standard deviation of Y is 8. What is the probability that X is greater than Y?

The standard deviation of X+Y = sqrt(8^2 + 6^2) = sqrt(100) = 10

Now we need the Z score for difference of X and Y..  

Z = (X - Y)/(st dev) = (100 - 96)/10

Z = .4

Using a Z table you get probability  = .6554

Saturday, January 4, 2014

Suppose you want the equation of the line tangent to the curve y = x^3 - 3x^2 + 2x + 5 at x - 4.  For the equation of a line you need the slope and y-intercept. You can get the slope at x=4  by differentiating and substituting 4 for x.

y' = 3x^2 -6x + 2

y' at x = 2 is 3(2)^2 -6(2) + 2 = 2

We know the slope at x= 4 is 2.

plug 4 for x into the original equation to get the y-coordinate

y = 4^3 - 3(4)^2 + 2(4) + 5 = 29

no put 2 for m, 4 for x and 29 for y into y = mx + b and solve for b

29 = 2(4) + b

b = 21

The equation is y = 2x + 21