The derivative can be thought of as the slope of a curve at any given point. You can calculate the derivative using the definition, which is lim (h approaches 0) [f(x + h) - f(x)]/h. Note that there are formulas to calculate the derivative which is much simpler, but will use the definition here just so you see how this works.

Suppose f(x) = x^2 + 5x

f(x + h) = (x + h)^2 + 5(x + h)

lim (h approaches 0) [(x + h)^2 + 5(x + h) - (x^2 + 5x)]/h

= (x^2 + 2xh + h^2 + 5x +5h - x^2 - 5x)/h

= (2xh + h^2 +5h)/h

= 2x + h + 5

(put in zero for h to get) 2x + 5

## Saturday, May 31, 2014

## Tuesday, May 27, 2014

Suppose we have a process where we are measuring the average ring diameter over many samples. Let's say the sample average which is the center line is 64 mm and the probability of detecting a shift from 64 mm to 64.02 mm increases as the sample size increases. If the process is large, we use smaller sample sizes than those if the shift of interest is relatively small. In general, larger samples will make it easier to detect small shifts in the process.For example, maybe a sample size of 15 will detect a shift of .01 thirty percent of the time. A sample size of 10 may only detect a shift of .01 twenty percent of the time.

As for the frequency of sampling, the most desirable situation is to take large sample sizes very frequently to best detect even the smallest shifts in the process. But that is usually not possible economically. What is general done is small sample sizes can be taken at short intervals, say sample of 10 every 15 minutes. Or take larger sample sizes at larger intervals, say sample of size 50 every hour.

Another way to handle sample size and frequency is through what is known as the "average run length" of a control chart, or ARL. It's calculation is simple, just 1 divided by p, where p is the probability that any point is out of control (outside of the upper control limit or lower control limit).

For example, if a control chart has limits of 3 standard deviations from the mean (center line), then ARL = 1/0.0027 = 370. Note that .0027 is found from a cumulative standard normal distribution chart. Many calculators will calculate this as well.

What does the ARL mean? It means that the average run length of any process in control is 370. In other words, even if the process is in control, an out of control signal will occur every 370 samples, on average.

There are other factors to be taken into account when trying to more accurately answer the questions of sampling frequency. Such factors include the cost of sampling, probabilities of various types of shifts in the process to occur, losses associated with an out of control process and more.

## Friday, May 23, 2014

## Tuesday, May 20, 2014

### Basics of a control chart

The
best way for a product to meet a customer's fitness for use, it should
be produced in a consistent and reliable manner. It must be capable of
operating with very little change or variability around the optimal
dimensions of the quality characteristics of the product. There are may
problem solving tools helpful in achieving stability and reducing
variability. One such tool is the control chart. What are the basic
principles of the control chart?

A control chart is a graphical
display of a quality characteristic that has been computed or measured
from a sample versus a sample number or time. Think of the chart as an
x-axis and y-axis, focusing only on the first quadrant values (zero and
positive values only). The x-axis would have time or sample number,
while the y-axis is the sample quality characteristic. Examples could be
"average ring diameter", "copper concentration", or "sample fraction
nonconforming". The chart contains a horizontal center line which
represents the average value of the quality characteristic measure when
the system is in the "in-control" state.

Two other horizontal lines, called the "upper control limit" (UCL) and "lower control limit" (LCL), are shown.

Sample values are plotted according to the measure of the quality characteristic. A process is said to be in control when all the points plot inside the control limits, in which case no action is required. If a point plots outside the control limits, the process is out of control and quality control experts will investigate to determine the cause of the out of control process and what course of action to take to get the process back in control.

Even if all the points plot inside the control limit, the process could be out of control if the points are not in a nonrandom manner. For example, if 17 of 20 sample points are below the center line but above the LCL and only 3 are above the center line and below the UCL, we would suspect that something is wrong. The points plotted should represent a random pattern.

Let's examine the statistical basis of the control chart. Suppose we have an x-bar control chart for piston-ring diameter. Note that x-bar is the sample average. Suppose the average ring diameter is 70 mm and the standard deviation of the process is 0.015 mm. If sample of size n = 10 were are taken, the sample standard deviation of x-bar is the standard deviation of the process divided by the square root of n. Therefore, the standard deviation of x-bar is 0.0047. If we use the 3-standard deviation control limits, the UCL would be 70 + 3(.0047) = 70.0141 and the LCL would be 70 - 3(.0047) = 69.9859. Now we can plot the sample values and see if they fall within the control limits and if they display a random pattern.

The most important use of a control chart is to improve the process. In real world applications, most processes do not operate in statistical control. Therefore, we will look for assignable causes for this out of control process. If they can be eliminated, the variability will be reduced and the process will be improved, hopefully to the point of being back in control.

This guide gives the very basics of what control charts are and how they are used. In upcoming articles, I will examine how to choose control limits, how to analyze patterns on control charts, and the other six tools of statistical process control.

Two other horizontal lines, called the "upper control limit" (UCL) and "lower control limit" (LCL), are shown.

Sample values are plotted according to the measure of the quality characteristic. A process is said to be in control when all the points plot inside the control limits, in which case no action is required. If a point plots outside the control limits, the process is out of control and quality control experts will investigate to determine the cause of the out of control process and what course of action to take to get the process back in control.

Even if all the points plot inside the control limit, the process could be out of control if the points are not in a nonrandom manner. For example, if 17 of 20 sample points are below the center line but above the LCL and only 3 are above the center line and below the UCL, we would suspect that something is wrong. The points plotted should represent a random pattern.

Let's examine the statistical basis of the control chart. Suppose we have an x-bar control chart for piston-ring diameter. Note that x-bar is the sample average. Suppose the average ring diameter is 70 mm and the standard deviation of the process is 0.015 mm. If sample of size n = 10 were are taken, the sample standard deviation of x-bar is the standard deviation of the process divided by the square root of n. Therefore, the standard deviation of x-bar is 0.0047. If we use the 3-standard deviation control limits, the UCL would be 70 + 3(.0047) = 70.0141 and the LCL would be 70 - 3(.0047) = 69.9859. Now we can plot the sample values and see if they fall within the control limits and if they display a random pattern.

The most important use of a control chart is to improve the process. In real world applications, most processes do not operate in statistical control. Therefore, we will look for assignable causes for this out of control process. If they can be eliminated, the variability will be reduced and the process will be improved, hopefully to the point of being back in control.

This guide gives the very basics of what control charts are and how they are used. In upcoming articles, I will examine how to choose control limits, how to analyze patterns on control charts, and the other six tools of statistical process control.

## Friday, May 16, 2014

### Consistent practice is the key

Remember that math is unlike many other subjects where you don't have to necessarily remember a topic when moving on to another topic. But in math, everything is a building block to the next topic. That's why it's important to build a solid foundation, learn the basics of mathematics like the back of your hand. Practice is the key to get better at simple addition, subtraction, multiplication and division. When you progress into the pre-algebra, algebra, geometry and higher mathematics you have to be able to master each or you will fall behind. Buy a good math book and work on the side. Many times I will work with students over the summer so they don't get rusty or forget what they learned the previous school year. Practice, practice and more practice is the key to keep your math skills sharp.

## Monday, May 12, 2014

### Understanding Parametric Equations

Suppose you wish to track the flight of a ball thrown and know how high
the ball is after the ball have traveled 150 feet horizontally if the
ball flight is represented by a parabolic curve. The curve can be
represented as a function of time using two equations, one related to x
(the horizontal distance) and t (representing time) and one related to y
(the vertical distance) and t.

For example, suppose a player throws a ball from a height of six feet at an initial speed of100 feet per second at a 45 degree angle to the horizontal axis. After t seconds the location of the ball can be described as x = 100cos(45)t and y = 6 + 100sin(45) - 16t

Suppose we want to eliminate the parameter. We do this by solving for t in one of the two equations and substituting the value in for the other equation. For example, suppose we have the parametric equations x = 2t and y = t

Suppose we want to graph a curve represented by the parametric equations x = t + 1 and y = √t. First use y = √t and square both sides to solve for t. Therefore we get y

Let's try one more example. Suppose a curve is represented by the parametric equations x = 3sin t and y = 5cos t. First get the trigonometric function by itself by dividing the first equation by 3 and the second equation by 5. Doing so, gives us x/3 = sin t and y/5 = cos t. Now square and add these two quantities and you'll see this all fall into place. Squaring and adding gives x

Those familiar with the equations of conic sections will see immediately that this is an ellipse centered at (0,0) with y being the major axis of length 10 and x being the minor axis of length 6. The ellipse will be elongated vertically 10 units and horizontally 6 units.

Parametric equations are very useful in solving problems involving vertical and horizontal distance. This guide should give students the basics on understanding and using parametric equations.

For example, suppose a player throws a ball from a height of six feet at an initial speed of100 feet per second at a 45 degree angle to the horizontal axis. After t seconds the location of the ball can be described as x = 100cos(45)t and y = 6 + 100sin(45) - 16t

^{2}. The x-component is the ball's horizontal distance in feet and y is the ball's vertical distance in feet. Where is the ball located 1.5 seconds after it is thrown? We substitute 1.5 in for t in both equations to get x to be approximately 106 feet and y to be approximately 40.7 feet. We can graph the movement of the ball by substituting values for t and plotting the values of x and y as the ordered pair (x,y) and connecting the dots.Suppose we want to eliminate the parameter. We do this by solving for t in one of the two equations and substituting the value in for the other equation. For example, suppose we have the parametric equations x = 2t and y = t

^{2}+ 2. We'll solve for t in the first equation since it's easier. Therefore dividing both sides of the equation by 2, we get t = x/2. Now using t = x/2 and y = t^{2}+ 2, we get y = (x/2)^{2}+ 2. Notice we just did a simple substitution, putting x/2 in for y in for t in the second equation. Simplifying the equation, we get y = (1/4)x^{2}+ 2. One familiar with the equations of conic sections knows that this is a parabola. Written in standard form the equation is x^{2}= 4(y - 2). This is a parabola that opens up with vertex at (0, 2).Suppose we want to graph a curve represented by the parametric equations x = t + 1 and y = √t. First use y = √t and square both sides to solve for t. Therefore we get y

^{2}= t. Now substitute y^{2}for t in the first equation to get x = y^{2}+ 1. We Know this is a parabola with the vertex at (1,0) and opening to the right. You may be tempted to graph the entire parabola, but remember the restriction on t in the equation y = √t. This is only defined then t is greater than or equal to 0. Therefore, y cannot be negative and we only graph the lower half of the parabola.Let's try one more example. Suppose a curve is represented by the parametric equations x = 3sin t and y = 5cos t. First get the trigonometric function by itself by dividing the first equation by 3 and the second equation by 5. Doing so, gives us x/3 = sin t and y/5 = cos t. Now square and add these two quantities and you'll see this all fall into place. Squaring and adding gives x

^{2}/9 + y^{2}/25 = sin^{2}t + cos^{2}t. Recall from the Pythagorean identities that the right side of this equation equals 1, therefore we have x^{2}/9 + y^{2}/25 = 1.Those familiar with the equations of conic sections will see immediately that this is an ellipse centered at (0,0) with y being the major axis of length 10 and x being the minor axis of length 6. The ellipse will be elongated vertically 10 units and horizontally 6 units.

Parametric equations are very useful in solving problems involving vertical and horizontal distance. This guide should give students the basics on understanding and using parametric equations.

## Thursday, May 8, 2014

What are the fundamental trigonometric identities? The reciprocal identities are the facts that the reciprocal of sine, cosine and tangent are the three trigonometric functions cosecant, secant and cotangent, respectively. Therefore, it holds true that the reciprocal of cosecant, secant and cotangent are sine, cosine and tangent.

Since sine is opposite divided by hypotenuse and cosine is adjacent divided by hypotenuse and tangent is opposite divided by adjacent, it holds true that tangent is sine divided by cosine. Therefore, cotangent is cosine divided by sine.

Recall the Pythagorean Theorem which states that the sum of the square of the legs of a right triangle equals the hypotenuse squared (a

^{2}+ b

^{2}= c

^{2}). From this relationship we know that sin

^{2}x + cos

^{2}x = 1, 1 + tan

^{2}x = sec

^{2}x and 1 + cot

^{2}x = csc

^{2}x. The pythagorean identities can be proven, but it's more important to know the identities, as they are used frequently in verifying other identities.

Let's verify some identities by changing to sines and cosines. That is often a method used to verify identities. Always work from the side of the equation that is most complicated, proving it equals the side of the equation that is less complex.

**For example: Verify that (cscx)(tanx) = secx**.

Change every function in terms of sine and cosine. Therefore we have

(1/sinx)(sinx/cosx) = secx

Notice the sinx in the denominator and the sinx in the numerator cancel each other, so we are left with 1/cosx and we know from the reciprocal identities that it equals secx, so we have verified the identity.

Let's try a more complex example.

**Verify that (cosx)(cotx) + sinx = cscx**

Again, work with the left side of the equation since it's the more complex side and change everything in terms of sine and cosine. Therefore, we get

cosx(cosx/sinx) + six = cscx

cos

^{2}x/sinx + sinx = cscx

Getting a common denominator of sin

^{2}x, the equation becomes

(cos

^{2}x + sin

^{2}x)/sinx = cscx

Notice that cos

^{2}x + sin

^{2}x = 1, therefore the equation becomes 1/sinx which we know from the reciprocal identities is cscx. The identity is verified!

Let's try one more example. This is the most difficult of the three.

**Verify that sinx/(1 + cosx) = (1- cosx)/sinx**

The method used in the previous example does not apply here, since everything is already in terms of sine and cosine. So now we have to get creative and find out how we can use a pythagorean identity to solve this. The pythagorean identities have squared terms in them, in particular in this case sin

^{2}x + cos

^{2}x = 1. Notice we can maneuver the equation to get 1 - sin

^{2}x = cos

^{2}x and 1 - cos

^{2}x = sin

^{2}x. If we work on the left side of the equation and multiply the numerator and denominator by 1 + cosx, we will get a desired result.

sinx/(1 + cosx) * (1 - cosx)/(1 - cosx) = sinx(1 - cosx)/(1 - cos

^{2}x)

Notice the 1 - cos

^{2}x in the denominator, which is perfect as it matches up with the manipulated version of the pythagorean identity above. The value of 1 - cos

^{2}x = sin

^{2}x. The equation now becomes sinx(1 - cosx)/sin

^{2}x. The sinx in the numerator cancels with a sinx in the denominator, leaving us with (1 - cosx)/sinx. The identity is now verified!

There are many examples of verifying identities in any trigonometry or pre-calculus book. Learn the identities and strategies for manipulating equations. Practice and you'll become a master and solving such problems

## Tuesday, May 6, 2014

http://www.consultingfact.com/blog/how-to-improve-your-mental-math-for-the-management-consulting-application/

## Saturday, May 3, 2014

During
my nearly thirteen years of tutoring math, there has been confusion
among students when it comes to prime numbers. Part of the reason is the
definition of prime numbers seen in most textbooks, which is that a
prime number is any natural number whose only factors are 1 and itself.
Using that definition, almost everyone will say that 1 is a prime
number, when in fact it is not.

The true definition is that a prime
number is any natural number that has only two factors, which generally
is 1 and itself. That definition excludes 1, which only has one factor,
because the only numbers that multiply to give you 1 is 1 and 1. All
other natural numbers that are not prime are called composite numbers.

I have a way of teaching students the prime numbers from 1 to 100 using a few rules and tricks which make the explanation a lot clearer. You can start eliminating numbers as possible prime numbers based on the following:

*Any number that ends in 0 is divisible by 10 and 5.

*Any number ending in 5 is divisible by 5.

*Any even number is divisible by 2.

*Any number whose digits added is divisible by 3.

The entire number is divisible by 3. An example of this is 39. 3+9 = 12, which is divisible by 3, so 39 is also divisible by 3.

*Any 2 digit number where both digits are the same is divisible by 11.

*Also any number that has a square that is a whole number is not prime.

By applying these rules you can eliminate the following as possible prime numbers:

4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30,32,33,34,35,36,38,39,40,42,44,45,46,48,49,

50,51,52,54,55,56,57,58,60,62,63,64,65,66,68,69,70,72,74,75,76,77,78,80,81,82,84,85,86,88,90,

92,93,94,95,96,98,99, and 100.

The numbers remaining (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,87,89,91,97) are prime.

What if one wants to know the prime numbers greater than 100? There are a few good methods to read and study. The Sieve of Eratosthenes is an ancient method to find any prime number up to a specified number. For a faster, although more complex method to find prime numbers up to a specified number, one can use the Sieve of Atkin. The Sieve of Atkin is an optimized version of the Sieve of Eratosthenes and involves dividing numbers by 60, getting the remainder, and using the remainder to solve more complex quadratic equations. Most of the time a computer or calculator would be needed to solve such equations.

The rules I give are enough to determine prime numbers between 1 and 100, which is generally enough for all practical purposes. I hope this information will be useful for any student to find the prime numbers between 1 and 100.

I have a way of teaching students the prime numbers from 1 to 100 using a few rules and tricks which make the explanation a lot clearer. You can start eliminating numbers as possible prime numbers based on the following:

*Any number that ends in 0 is divisible by 10 and 5.

*Any number ending in 5 is divisible by 5.

*Any even number is divisible by 2.

*Any number whose digits added is divisible by 3.

The entire number is divisible by 3. An example of this is 39. 3+9 = 12, which is divisible by 3, so 39 is also divisible by 3.

*Any 2 digit number where both digits are the same is divisible by 11.

*Also any number that has a square that is a whole number is not prime.

By applying these rules you can eliminate the following as possible prime numbers:

4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30,32,33,34,35,36,38,39,40,42,44,45,46,48,49,

50,51,52,54,55,56,57,58,60,62,63,64,65,66,68,69,70,72,74,75,76,77,78,80,81,82,84,85,86,88,90,

92,93,94,95,96,98,99, and 100.

The numbers remaining (2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,87,89,91,97) are prime.

What if one wants to know the prime numbers greater than 100? There are a few good methods to read and study. The Sieve of Eratosthenes is an ancient method to find any prime number up to a specified number. For a faster, although more complex method to find prime numbers up to a specified number, one can use the Sieve of Atkin. The Sieve of Atkin is an optimized version of the Sieve of Eratosthenes and involves dividing numbers by 60, getting the remainder, and using the remainder to solve more complex quadratic equations. Most of the time a computer or calculator would be needed to solve such equations.

The rules I give are enough to determine prime numbers between 1 and 100, which is generally enough for all practical purposes. I hope this information will be useful for any student to find the prime numbers between 1 and 100.

## Friday, May 2, 2014

Trigonometry, measurement of triangles, is used in various
occupations including engineering, architecture and navigation. We start
with the study of trigonometry by examining six functions. These
trigonometric functions are determined by inputs that are he measures of
acute angles of a right triangle. The outputs are the ratios of the
length of the sides of right triangles.

If you draw a right triangle, that is a triangle with one right (90
degree) angle and two acute angles, note a base angle as Ө. We note the
sides of the triangle based off this angle and the right angle. The side
opposite the right angle, which is the longest side of a right triangle
is the

The trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. If you have a scientific calculator, you may have seen buttons with sin, cos and tan on them. Those are the trigonometric functions sine, cosine and tangent. Sine is the length of the side opposite Ө divided by the length of the hypotenuse. Cosine is the length of the side adjacent to Ө divided by the length of the hypotenuse. Tangent is the length of the side opposite of Ө divided by the length of the side adjacent to Ө. To figure out cotangent, secant and cosecant, they are simply the reciprocals of tangent, cosine and sine, respectively.

An important thing to note is that the value of the trigonometric functions depend only of the size of the angle Ө and not on the length of the sides. For example, a triangle with opposite side 2 and adjacent side 1 has the same trigonometric functions as a triangle with opposite side 30 and adjacent side 15, since the ratio between the two is 2:1 in both cases.

Now that we know the basic trigonometric functions, let's evaluate them in an example. Suppose we have a right triangle with the length of the opposite side of angle Ө to be 4 and the length of the adjacent side to angle Ө to be 3. We can find the values of the six trigonometric functions once we know the length of the hypotenuse. We can get that by using the Pythagorean Theorem, which is "a squared" plus "b squared" equals "c squared", if a, b and c are the sides of the triangle, c being the hypotenuse. Using the equation, we get c = 5. Therefore sine is opposite divided by hypotenuse, which is 4/5. Cosine is adjacent divided by hypotenuse, which is 3/5 and tangent is opposite over adjacent, which is 4/3. Cotangent is the reciprocal of tangent, so cotangent is 3/4. Secant is reciprocal of cosine, so secant is 5/3. Cosecant is reciprocal of sine, so cosecant is 5/4.

We can also find the angle measures of a right triangle using trigonometric functions. If you've ever noticed sin

These are just some of the basics with right triangle trigonometry. There is much more to learn, but understanding these basics are key to progressing to more difficult topics and applications of trigonometry.

**hypotenuse**. The other sides are**opposite**of Ө and**adjacent**to Ө. It's important to know the distinction between the sides because the six trigonometric functions are based on the ratio between these sides.The trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. If you have a scientific calculator, you may have seen buttons with sin, cos and tan on them. Those are the trigonometric functions sine, cosine and tangent. Sine is the length of the side opposite Ө divided by the length of the hypotenuse. Cosine is the length of the side adjacent to Ө divided by the length of the hypotenuse. Tangent is the length of the side opposite of Ө divided by the length of the side adjacent to Ө. To figure out cotangent, secant and cosecant, they are simply the reciprocals of tangent, cosine and sine, respectively.

An important thing to note is that the value of the trigonometric functions depend only of the size of the angle Ө and not on the length of the sides. For example, a triangle with opposite side 2 and adjacent side 1 has the same trigonometric functions as a triangle with opposite side 30 and adjacent side 15, since the ratio between the two is 2:1 in both cases.

Now that we know the basic trigonometric functions, let's evaluate them in an example. Suppose we have a right triangle with the length of the opposite side of angle Ө to be 4 and the length of the adjacent side to angle Ө to be 3. We can find the values of the six trigonometric functions once we know the length of the hypotenuse. We can get that by using the Pythagorean Theorem, which is "a squared" plus "b squared" equals "c squared", if a, b and c are the sides of the triangle, c being the hypotenuse. Using the equation, we get c = 5. Therefore sine is opposite divided by hypotenuse, which is 4/5. Cosine is adjacent divided by hypotenuse, which is 3/5 and tangent is opposite over adjacent, which is 4/3. Cotangent is the reciprocal of tangent, so cotangent is 3/4. Secant is reciprocal of cosine, so secant is 5/3. Cosecant is reciprocal of sine, so cosecant is 5/4.

We can also find the angle measures of a right triangle using trigonometric functions. If you've ever noticed sin

^{-1}, cos^{-1}, and tan^{-1}on your calculator those are the inverse sine, inverse cosine and inverse tangent buttons and are used to find the angle given a the trigonometric ratio. For example, in our above problem we know sine = 4/5, cosine = 3/5 and tangent = 4/3. Knowing these ratios we can get the value of Ө and therefore the other angles of the triangle. Take inverse sine of 4/5 and we get 53.1 degrees, which will be the same as taking inverse cosine of 3/5 and inverse tangent of 4/3. All will give the value of Ө at 53.1 degrees. Since we have a right triangle, we know one angle is 90 degrees and since the angles of a triangle add to 180 degrees, the third angle is 36.9 degrees.These are just some of the basics with right triangle trigonometry. There is much more to learn, but understanding these basics are key to progressing to more difficult topics and applications of trigonometry.

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