Free Web Hosting Provider - Web Hosting - E-commerce - High Speed Internet - Free Web Page
Search the Web

Mnemonics and acronyms for mathematics and calculus

Mathematical Mnemonics

Albert Einstein, who fancied himself as a violinist, was rehearsing a Haydn string quartet. When he failed for the fourth time to get his entry in the second movement, the cellist looked up and said, "The problem with you, Albert, is that you simply can't count."
"Do not worry about your problems with mathematics, I assure you mine are far greater." A. Einstein

Calculus overview:

Calculus was developed by both Leibnitz and Newton as a way of getting information from graphs that stalled conventional mathematics. Calculus has two forms: Integration and Differentiation. Basically, differentiation reduces data to its essentials; integration enhances information to create new products. Differentiation will reduce a 2-dimensional plot to a 1-dimensional line whose length and slope reveal performance data. Integration is used to estimate what data will arise when a 2-dimensional graph is enhanced to 3-dimensions, or what volume will result when rotated about its x-axis or y-axis.

For example, using a graph plotting distance against time, differentiation will find instantaneous velocity, point of maximum or minimum velocity. Integration will find total distance covered.

Natural Logarithms and trig functions and identities seem to give the most pain to students. I think that is because while the basic trig functions can be sketched on a graph paper, the concept of multiplying and dividing trig functions seems ridiculous. Nevertheless, it is possible and it is logical (well, from a mathematician's perspective anyway). To visualize these functions, try downloading or buying graphing software such as Equation Grapher.

Trig Identities

tan = sin/cos (tanning at sun/coast)

sec = 1/cos (my secretary has one over my cousin)

csc = 1/sin (my cousin's secretary has won over sin)

cot = cos/sin (I find cotton is cozier to wear over synthetics)

tan = sin/cos. A well-tanned-but-horny-and-cheap man, pondering whether to pay for a hooker or save money, chooses SIN OVER COST.

Trig Function in the Unit Circletrig circle

The sign of a trig function can easily be calculated by knowing that sin=y and cos=x, and that all sine is positive in the top half of the circle, all cosine is positive on the right half of the circle, and that sin + cos combine to define all trig functions. For example, the upper left quadrant has sine + and cos -, so only sine and cosecant are positive. In the lower left of the circle, both sine and cosine are negative and -- because two negatives make a positive -- cos/sin (cot) and sin/cos (tan) are positive.

Here are a few mnemonics to help recall the trig values which are positive in the quadrants of the unit circle. Starting from the upper right, we have All Sinning Czechs are on Tan Cots with Costly Secrets, and All Sing Christmas Carols wearing Tan Cottons with Cozy Secretaries. I know the x-axis is related to cosine, and the y-axis is related to sine by thinking of Cossacks in Sinai.(cos-x, sin-y)

Table of Trig Identities

This table shows trig functions in the centre (center to Americans) column are differentiated to the left, and integrated to the right. Notice the integration process always adds the Constant. The reason for the C is that when a function is differentiated, the Constant is lost, and when the process is reversed, it must come back somehow. So to compensate for any lost numbers, always include the C, even if C = zero.

Differentiated

1
-sin x
cos x
sec2 x
-csc2 x
sec x tan x
-csc x cot x

Function

x
cos x
sin x
tan x
cot x
sec x
csc x

Integrated

x + C
sin x + C
-cos x + C
Not applicable. Trig functions need upper boundaries in order to be integrable.

Trig Derivatives in a Secluded Cottage

How about a mnemonic for the trig derivatives? Try: Scot's cozy sinner of a secretary squarely and secretly co-seeks a cot. Sexy and tanned, cozy secluded cottage, casual sexual contact twice. The functions are, in order: derivative of sin = cos, cos =-sin, tan = sec squared, sec = sec tan, csc = -csc cot, cot = -csc squared. Notice that all of the co-functions have a negative value. Trying to get a mnemonic phrase to help recall all of these is difficult; I usually end up with confusing sentence fragments of secret tanning spots and cosy secretaries. Personally, I remember tanning on sun coast as the primary mnemonic, and memorize or know how to calculate the rest of them.

Graphing Functions in the Kingdom

The procedure for graphing a function is long, but worthwhile. The steps should be done in sequence, and this is where a lot of students get mixed up. Briefly, the steps include: find the domain of the function, the xy intercepts, the asymptotes, the symmetry, and the first and second derivatives. To help remember the steps (which can have different terms for the same step) I came up with a story which applies memory pegs, and a single sentence based on sex (mneumonic). I couldn't make an acronym work for this one. First, the story:

In the DOMAIN of Calculus the Difficult, the King (who leads a graphic lifestyle) drives a Jensen INTERCEPTOR and greets people with "HA, VA!". The citizenry are mostly inbred and regard people having ASYMMETRICAL faces as odd. The King TOOK THE PRIME Minister and taught him the CRITICAL POINTS of running a government. It seems the economy was RISING and FALLING relative to MAXwell House coffee and MINIMUM wage. He took a SECOND PRIME Minister who had two CONCubines, one of whom was born on the CUSP between Leo and Virgo.

Of course, "taking the Prime" is another way of saying "finding the derivative", and we find the critical points by examining the derivatives of the function. HA and VA are the horizontal and vertical asymptotes. The second prime tells us if the curve is concave up or concave down, and the cusp. The first derivative indicates any maxima or minima.

The single sentence mnemonic: Does Intercourse in a Horizontal position Simply Drive Critics of Up and Down to take a Second look at Concubines? Easy to recall now??

Limits -- A Parable

Imagine an action movie actor who must run down a path dodging bullets, bombs, and debris. To make the action look real, the director has to have the projectiles come as close to the actor as possible without touching him, for as soon as the position of the bullet equals the position of actor, he dies and the whole process halts. Similarly, in calculus, the Limit x --> 0 expression means that the value for the denominator can come as close to zero as possible without actually being zero and killing the equation. Just as the movie director overcomes this problem by manipulating the images so the action appears real but the actor lives, the math student manipulates the factors to remove the formula killer and to see what happens at that point. Usually, this involves finding the zero factor hiding in the denominator and removing it so we can see what happens when x = 0. We can:

    • factor when we see a perfect square or cube in the denominator
    • multiply by conjugate when we have a root in the denominator
    • divide by highest exponent when Limit --> infinity

My prof used to say "This formula has a disease which makes it unable to function. Mathematicians are surgeons who must find this disease and remove it by cutting away misleading fat and tissue. The disease is a zero virus which is always taking new shape and hiding better." This is illustrated below, where you can see a straight substitution of x=2 will result in zero/zero. This zero virus is cured by a dose of factoring medicine to make the function healthy.

the Math Virus

Applications of Calculus

Most problems in calculus can be solved surprisingly easy -- if you follow the techniques. I solve most calculus problems by taking 4 or 5 steps from this list: find domain, find critical values, find formula, take the derivative / anti-derivative, plug in the values, plot on a number line, check your answer against the question, check your answer against the domain, and factor the numerator and denominator.

In addition to these nine, here are some sub-steps you may have to invoke to smooth over bumps in the calculation process: simplify, use the quadratic formula, factor top and bottom to remove common factors from the numerator and denominator, know when to apply the Chain Rule, U-substitution, look for horizontal or vertical asymptotes, find where f'=0, find where f''=0, know the trig identities, know how to factor the difference of squares/cubes, remember to use absolute values, and did I mention Simplify?

Oil Spill Problem pies are square

Sometimes a story is better way of remembering the procedure for solving a problem in math. Here is the famous Oil Spill Rate problem paraphrased.

I saw a BSA motorcycle parked outside of the Pies Are Square Bakery and Computer Shoppe. As usual, there was a lot of oil leaking from the motorcycle and I saw the puddle's radius was increasing at a constant rate. Inside, the clerk greets me. "I'll take the PRIME of your PIE selection, the RADIUS monitor, and I want to see your RATES." I asked the man with SQUARE FEET for the time. He replied in seconds. At that moment, the oil spill stopped.

The steps: sketch, find the two relevant formulae and merge them, simplify with respect to the variable, differentiate and solve for radius.

Newton's Dirty Quotient find the prime

Finding the slope of the tangent line gives us rise/run, a relationship used heavily in physics. When used with Limit h--> 0 it gives us instantaneous velocity. Here's a cute story to help remember this formula:

I was working on MY TAN when I saw this gorgeous woman jogging. Yes, I got a RISE over her RUN and felt embarrassed, so I used NEWTON's Cream to "relieve my sexual anxiety". Time was running out as it APPROACHED THE ZERO hour, but I found the VELOCITY of my hand was part of the solution.

Power Rule derive animation

The power rule is a shortcut to finding the derivative of X. Instead of applying Newton's Quotient, simply multiply the factor of X by the exponent, then subtract 1 from that exponent.

Understanding Differentiation relating area to length animation

Observe the animation just above. See how the area of a square equals the square of x? Now double the area, and the length of the side increases by only the square root of two. Increase area to three times the original size, and the length of a side increases by only the square root of three, and so on. Differentiation (and its reversed cousin, Integration) are based upon this phenomenon. Thus, differentiation allows you to calculate the changing area with respect to the changing sides of a square. This applies not only to squares, but to rectangles, triangles, cones, spheres, or just about all geometric figures.

Global Min/Max maxima / minima animation

Finding minima/maxima of a curve is a technique with many practical applications, including cost analysis, finding the most efficient way of building something, or even streamlining a process.

maximum volume animationOne application of the min/max problem is to calculate the maximum volume of a box made from a sheet of paper from which the four corners are excised and the sides folded. Let the dimensions of the paper be 5 by 3 feet, and the size of the corner we cut be x. Obviously, if x=0, the volume is zero because we still have a flat piece of paper. And if the 2 corners are just as big as the width of the paper, then we're back to zero volume. Therefore, a graph showing volume versus x will peak somewhere between 0 and 1.5 feet. The four steps to find where the graph peaks are: FORMULA, DOMAIN, DERIVATIVE, MAXIMUM.

  1. Formula: l x w x h is not 3 x 5 x X, for the length is 5-2x; width is 3-2x; and height is x. Volume is (5-2x) (3-2x) x.
  2. Find the domain by finding the critical points of the volume formula, that is, when 3-2x=0, when 5-2x=0, and when x=0. The domain is (1.5, 2.5, 0). Remove 2.5 from the list, as the domain ends when we reach 1.5.
  3. You must multiply out the volume formula before taking the derivative. (5-2x)(3-2x)x = 15x-16x2=4x3. The derivative is 15-32x+12x squared.
  4. To find the critical points, we usually factor the formula to make it easy. In this case, there is no clear factor so we apply the quadratic formula. Remember: a=12, b=32, c=15. The quadratic gives two solutions: approximately 2.06 and 0.61. Which one falls within the domain? The solution: maximum volume is achieved when x = 0.61 feet.

 

Note: in math we are exact, but in this case I rounded the solutions because exponents and roots are awkward in html, and when we are dealing with scissors on paper, we can only be accurate to a few thousandths of an inch anyway. If you'd like to graph a related rate formula easily, try the Markus Friberg Equation Grapher. I used it to make the GIF image at the top of the page. There may be some free math software available at Simtel. I'm currently investigating more downloads from RocketDownload, another good source of software evaluations.

The most basic steps are: reduce all factors to variables of x, simplify the expression as much as possible, find critical values from the derivative, plug the critical values into the formula, and decide if the solutions are maxima or minima. Often, the quadratic formula will be used to find solutions. These steps are intersperse in the following little story.

The Australian Director Steve Miller, head of GLOBAL Productions, and an expert in the DOMAIN of desert action films, REDUCES all cinema plots to VARIATIONS ON SEX. His SIMPLE FORMULA of lustful PRIME ministers DRIVEN by mathematics have made him VALUED by CRITICS from the Daily DERIVATIVE newspaper. These RECALCITRANT reviewers either love him to the MAX, or MINIMIZE his accomplishments. Sometimes, their reviews would QUADRUPLE ticket sales.

The Director established a domain, reduced factors to x, simplified the formula, derived a prime, found the critical values, recalculated the simplified formula, and used the results to define the global max or min. Sometimes the quadratic formula is used.

Integration integration animation

Integration is essentially the reverse of differentiation. Whereas differentiation is useful for reducing a curve to a tangent line, integration expands the curve to the area under the line. To integrate x, simply add 1 to the exponent, then divide by the new exponent. If you are having difficulty solving integration problems, go to The Integrator, a web page that will solve the problem for you.

Area between two lines integration of area between curves

Calculating the area between two curves on a graph is straightforward. The theory is that if you can use integration to find the area under a curve, then the area between two curves is found by subtracting the smaller area from the larger. The steps:

  1. Find the start and finish points / upper and lower limits by finding where the two curves intersect. Simply let (curve 1) = (curve 2), or (curve 1 - curve 2) = zero. If they don't cross, then use the start / end points. Write these points on the integration symbol as shown below.
  2. If working from the x-axis [i.e., if your formulae are y=f(x)], find the upper curve. If you are working from the y-axis, find the right-hand curve. If you aren't sure which curve is what, simply replace x by a number from the domain. The formula with the larger answer is the upper or right hand curve.
  3. Write the formula and solve. Note that you solve by splitting the formula into four parts: [(anti-derivative of upper curve evaluating b) - (anti-derivative of upper curve evaluating a)] minus [(anti-derivative of lower curve evaluating b) - (anti-derivative of lower curve evaluating a)] = Area between the two curves between endpoints.

NOTE: some curves cross the axes in such as way as to lead the calculation to create a negative area. In such a case, divide the problem into two: one part for the upper side of the x-axis, and second part to calculate the area below the x-axis.

Area of the surface of a cone cone

We know it's easy to find the area of the surface of a cylinder, well how about the area of the surface of a truncated cone? Well, we find the surface area of a standard cone by multiplying the circumference by height: A = (pi)(diameter)(height). In calculus, we would see this as the area between two curves. Looking at it another way, this conic slice illustrated is what is left after a small cone is subtracted from a larger cone. This is calculus-speak for finding the difference between the upper and lower cones, and written as: Area = Integral of 2(pi)(y)[square root (1+(dy/dx)2] dx, where radius is dy, and height is not necessarily dx so we used that larger factor instead to represent the change in height. Solve as above, by plugging in values for a,b (upper and lower limits) and radius for dy.

Gravity and Acceleration Problems

2 gravity formulae

Upward motion of a projectile

This is a classic. A projectile is launched from a base in Jugoslavia 150 meters above the Adriatic sea level with an initial velocity of 50 meters per second. Find the maximum height, how long it will take to reach max height, how long it will take to hit the ocean surface, and at what speed will it be traveling at impact. (assume the Patriot missiles are malfunctioning that day and all motions are due to gravity)

Assume gravity is constant and there is no wind resistance. Measurements are in meters and seconds. The two formulae to use are shown here, where g = 9.8 m/sec; v = velocity; s = position; t = time; -4.9 is one half of the gravity constant (the negative value indicates the direction of the gravity force. For this argument, we say moving upward is positive acceleration, falling to the earth is negative acceleration). Follow these four steps in order, as the result of each step is used in the subsequent calculation.

4 steps:

  1. As the projectile rises, its initial velocity is eroded by gravity at the rate of 9.8 meters/second2. When the cumulative effect of gravity is equal to the initial velocity, we have reached the maximum height and the velocity is zero. Calculate: (initial velocity) = (gravity)(time), 50 = 9.8(t), t = 50/9.8, t = 5.1 seconds.
  2. Now that we know it takes 5.1 seconds to reach the apex, plug 5.1 into the next formula to find height. Max height is reached when s(t) = 0, that is the projectile's position is zero. In this case, s(t) = -4.9(5.1)2 + (50)(5.1) + (150) = 277.55 meters.
  3. Now that we know its height above sea level, we can calculate its drop to the water. Time to impact is determined by the gravity constant and the distance the object falls. Solving -4.9(t)2 + 0(t) + 277.55, (t)2 = 277/4.9, t = 7.5 seconds to impact. Now we can use this result to solve the other formula and find impact velocity.
  4. Impact velocity equals (gravity constant)(time) = (9.8)(7.5) = 73.5 m/sec.

The only difficulty I've had with this is getting the steps in the right order. I overcame it by writing a paragraph on how each step is calculated and why. Good luck!

Solving a Problem of Varying Salt Concentration using the integration factor

Saltwater dilution problem

Problem: A pipe is supplying 4 pounds of salt per gallon of water to a barrel at a rate of 2 gallons per minute. The barrel originally held 25 pounds of salt dissolved in 50 gallons of water. The barrel is losing 2 gallons of brine per minute through a drain. Calculate how much salt will be in the barrel after 25 minutes.

This type of problem is suitable for integration by calculus because the salt concentration changes with time and volume. On a graph, we see the original salt concentration (25/50, or 0.5) changes quickly at first, then levels off as it nears the concentration of the inflow (4 pounds per gallon). The curve is similar to a Michelson-Menton curve (resembling a quarter ellipse) where the initial rate (slope) is very steep but the curve flattens more and more slowly as the slope approaches zero. Thus, the change in salt concentration is not linear and so requires integration techniques to solve the area under the curve. The general formula we use is derived from (rate = inflow - outflow) and written as [dy/dt + p(x)y = q(x)], where:

  • dy/dt = change in salt concentration over time
  • p(x) = outflow of salt solution, or p(x) = (pounds of salt/gallons of water in the barrel)•(gallons per minute drained from barrel)
  • q(x) = inflow, or q(x) = (pounds of salt/gallons of water)•(gallons flowing in / minute)
  • y = pounds of salt
  • y(t) = pounds of salt at some point in time
  • t = time

The key an easy solution is use something called integration factor (IF). The integration factor we use is designated IF in the general formula and is used to remove a lot of ugly terms. The IF is the number e with an exponent equal to the integral of p(x):the Integration Factor

General Steps

  1. Assign letters to appropriate quantities given in the problem: dy/dt + p(x)y = q(x).
  2. Calculate the integration factor by solving the integral of p(x) and make the result an exponent of the number e. It should be e to the power of (t/some number)
  3. Let the integration factor = IF, write formula as: d/dt (IF•y) = IF•q(x).
  4. Integrate this formula to get: IF•y = IF•qx + C
  5. Divide by IF to get: y = qx + C•IF-1.
  6. Calculate C by letting t = zero, and y = initial mass of salt in barrel. (e0 = 1).
  7. Use this value of C to write the formula specific for this problem and solve for y by letting t = whatever time has passed ... or solve for t by letting y = whatever salt concentration is wanted.

Specific Steps

  1. Name the variables and constants to satisfy time (t), rate (dy/dt), and initial salt mass (y0).
  2. Make a general formula to describe rate = inflow - outflow, where y = mass of salt, t = time, dy/dt = change of salt mass over time, y(o) = initial salt mass, and (y)(rate)/gallons = outflow of salt.
  3. Plug in values: dy/dt = ( 42) - 2 y/50 = 8 - y/25.
  4. To make it easier to find y-value, move the y-factors to the left: dy/dt + y/25 = 8.
  5. Create an integrating factor: integrate 1/25 and make it an exponent of e. et/25 is added to the formula to enable us to solve it by integration. This factor emerges unchanged by integration process.
  6. Integrate both sides and solve. The result is et/25 y = (8 • 25)e t/25 + C = 200et/25 + C.
  7. Divide both sides by e t/25. This removes the integration factor but adds e -t/25 to the Constant. Note: when you divide C by e t/25, the result is Ce with a negative exponent.
  8. Solve for t = 0, y = 25. These are the initial conditions of the water in the barrel and will solve for C. In other words, 25 = 200 - C because when t = 0, et/25 becomes e 0/25 which equals one. C = 25 - 200, C = -175.
  9. Rewrite the original formula using the value for C. y = 200 - 175e -t/25. Note the negative exponent.
  10. To find the amount of salt (y) after 25 minutes (t), plug in the values and use a calculator to solve. You should have 25 = 200 - 175e -25/25, or y = 200 - 175e-1 , which is approximately 136 pounds. Would you like to see The Graph?

 

Radioactive Decayradioactive decay GIF

Radioactive decay occurs when an atom of an unstable isotope emits particles. The time it takes for half of the atoms to decay is called the "half-life" of that element. The half life of carbon-14 is 5750 years, which makes it an excellent clue to the age of organic materials between 1,000 and 10,000 years old. The radioactive carbon is formed in the upper atmosphere from nitrogen, and combined with oxygen to make carbon dioxide. This radioactive carbon dioxide is absorbed by plants, which are eaten by herbivores, which in turn are devoured by carnivores. That is how the carbon-14 enters our bodies.

Problem

A bone is tested and discovered to have a ratio of carbon-14 to carbon-12 that is 60% that of a modern bone. Using calculus, estimate the age of the bone, assuming carbon-14 levels have remained steady for the past 10,000 years.

Solution

  1. Use two formula to solve this problem: the first formula generates the decay constant, k; the second formula gives us the age and mass of carbon-14.
  2. k = (-1/t) ln2, where t = half-life of element and the natural log of 2 is about 0.69. In the case of carbon-14, k = (-1/5750)(0.69) = -0.0002.
  3. Age of the sample is related to the initial mass and ekt, or (remaining mass) = (initial mass)(ekt). Mass is 0.6; k = -0.0002; so 0.6 = (1.0)e-0.0002(t). ln 0.6 = (0.0002)t, t = 3000 years.

Pi Numbers

Poetic Pi

Word-length pi mnemonics have been around a long time. An old chestnut, but still a favorite, is:

How I need a drink, alcoholic in nature, after the heavy lectures involving quantum mechanics.

A much less well-known example is this nice poem by Joseph Shipley (1960):

But a time I spent wandering in bloomy night;
Yon tower, tinkling chimewise, loftily opportune.
Out, up, and together came sudden to Sunday rite,
The one solemnly off to correct plenilune.

Math Links

Lessons

Hofstra University has on-line math tutorials. University of Texas offer Calculus for Idiots. University of Illinois, Urbana, has calculus lessons using a new teaching method. Students wanting to look at potential theses projects may want to look at Famous Unsolved Math problems. (funny, my math text has hundreds of them :-)

Historical

Animations

  • The Virtual Calculus Page has dozens of good links.
  • The Calculus Page offers many charts and Java animations to illustrate principles of analysis.
  • The Calculus Carnival is heavy on the Java images, but interesting.
  • Calculus animations from Douglas N. Arnold. "These are excerpts from a collection of graphical demonstrations I developed for first year calculus. I use many such demonstrations to illustrate and enrich my classes in the McAllister Technology Classroom. Those interested in higher math may also want to visit my page of graphics for complex analysis."

Entertainment

Math TRIVIA

Each king in a deck of playing cards represents a great king from history. Spades = King David, Clubs = Alexander the Great, Hearts = Charlemagne, and Diamonds = Julius Caesar.

111,111,111 x 111,111,111 = 12,345,678,987,654,321

One day while playing with my new calculator, I wanted to see what would happen when I multiplied 666 (the Number of the Beast, Scriptures?) by itself. Surprise! 6.66 to the 4th power yields 1967.4192. To me, that suggested a particular day in 1967, so I multiplied 0.4192 x 365 = 153.008. The 153rd day of that year (153-31-28-31-30-31=) is June 02. Checking the history book, I discover that was the day Israeli air force attacked Egypt. Even more coincidental is the time. If you multiply 0.4192 by 365.25 instead of 365 (as the real year is almost a quarter of a day longer than the calendar year) you get 153.1128 days into 1967. To find the exact hour, multiply 0.1128 by 24 hours = 2.7072, or 02:43 a.m., which is about the local time of the air attack on Egypt, 02 June 1967.
So ... did the Bible predict this? Did the Israeli military commander read Scriptures with a calculator nearby? Or is it a coincidence?

Two trains 200 miles apart are moving toward each other; each one is going at a speed of 50 miles per hour. A fly starting on the front of one of them flies back and forth between them at a rate of 75 miles per hour. It does this until the trains collide and crush the fly to death. What is the total distance the fly has flown?

The fly actually hits each train an infinite number of times before it gets crushed, and one could solve the problem the hard way with pencil and paper by summing an infinite series of distances. The easy way is as follows: Since the trains are 200 miles apart and each train is going 50 miles an hour, it takes 2 hours for the trains to collide. Therefore the fly was flying for two hours. Since the fly was flying at a rate of 75 miles per hour, the fly must have flown 150 miles. That's all there is to it.

blood | bone | brain | cardio | cellbiology | chemistry | chem links | embryology | genetics | learn | math | physiology | psychology | quotes