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Orthopaedic Surgery/Cartilage

Normal Articular Cartilage Articular cartilage of diarthrodial joints are covered by hyaline cartilage, which is in turn derived from the two words - hyalose a greek word for glass and cartilago, a latin word for gristle. It is relatively acellular and has no vascular, neural or lyphatic supply. Articular cartilage by its rubber like resiliency, functions to reduce pressure and where it covers the end of a bone, its smooth surface minimises the friction effect of shearing stresses. The cartilage of our joints derives its nutrition from the movement and contact pressure of the joint for its effect on transmitting the larger solutes from the joint fluid to the cellular matrix of the cartilage; a process which is potentiated when there is slight incongruity of the opposing joint surfaces. Chrondromalacia, the softening which heralds early osteoarthritis is present in most humans after age 40 and usually starts in non-load bearing areas.

http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&dopt=Citation&list_uids=5706888

=Morphology of Articular Cartilage=
 * 1) Colour: Colour varies with age.
 * 2) *Children - transparent and white with a bluish tinge
 * 3) *Young adult - white and glossy
 * 4) *Middle aged - yellowish white
 * 5) *Advanced age - opaque and yellowish with brownish tinge
 * 6) Compressibility: Compressible under pressure but on removal of the deforming force, it springs back into its original shape (behaves as a shock absorber).
 * 7) *Children - highly compressible
 * 8) *Young adults - compressible but less than in a child
 * 9) *Middle aged - firmer and less compressible
 * 10) *Advanced age - less resilient and more resistant
 * 11) Thickness: Articular cartilage becomes thinner with disuse and in old age. In a given joint thickness varies from one area to another. In areas with concave surface, it is thicker at the periphery and thinner in the centre. Reverse is true for areas with convex surfaces. Thickness varies from joint to joint. It is thinnest in the interphalangeal joints and thickest in the patella.
 * 12) *Average thickness - 2-3 mm
 * 13) *Interphalangeal joint - 1 mm
 * 14) *Patella - 5-6 mm
 * 15) Surface: The surface appears smooth when viewed by naked eye. Microscopic examination particularly using electron microscope demonstrates surface undulations. The undulations entrap synovial fluid, facilitating the lubrication os the articular cartilage.
 * 16) *The most superficial layer of articular cartilage known as the lamina splendens is acellular measures 4 to 8 microns in thickness and consists of collagen, of a type more characteristic of synovium than articular cartilage.

http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&dopt=Citation&list_uids=15168182

=Composition of Articular Cartilage= The articular cartilage is made of cells called chondrocytes which are embedded in an extracellular matrix.

 * 1) Chondrocytes: This is one of the cells found in articular cartilage. It contributes very little to the total mass of the tissues. In a human adult it forms only 1% of the total volume. Chondrocytes differ in size and shape and metabolic activity depending on zones in the articular cartilage. Chondrocytres take active part in the synthesis of and turn over of the extra cellular matrix and replacement of degraded matrix and its components. Adult cartilage derives its nutrition from the synovial fluid. Nutrition to reach the chodrocytes must go through double diffusion barriers.
 * 2) *synovial tissue and synovial fluid
 * 3) *cartilage matrix - this system makes the chondrocytes manage with low concentration of oxygen, hence they depend more on anaerobic metabolism.
 * 4) Extracellular Matrix:It has two main components.
 * 5) *Tissue Fluid: Water contributes to 70-80% of the net weight of the articular cartilage. This tissue also contains gases, small proteins, metabolites and a high concentration of cations to balance the negatively charged proteoglycans.
 * 6) *Structural Macromolecules: These form 20-30% of the net weight of the tissue. They in turn consist of:
 * 7) *#Collagen: It contributes 60% of the dry weight. It has many types of collagen. Collagen types II, VI, IX and XI are predominant. The main collagen is type II accounting for 90-95%. Types II, IX and XI form cross-banded fibrils which form a tight mesh work. This provides the cartilage its tensile stiffness and strength and traps the large proteoglycan molecules to achieve cohesiveness. Type VI collagen is found immediately surrounding the chondrocytes assisting them to get attached to the matrix. Type X collagen is found mainly near the cells of calcified cartilage zone and suggests that it may have a role to play in the mineralization of the cartilage.
 * 8) *#Proteoglycans: These macromolecules are extremely important in affording the tissue its resiliency and in maintaining its water content. They have a linear protein core to which are attached at right angles long chains of polysaccharides known as glycosaminoglycans. Glycosaminoglycans found in articular cartilage include - hyaluronic acid, chondroitin-6 sulphate, chondroitin-4 sulphate, keratin sulphate and dermatan sulphate. Concentration of these molecules varies within the articular cartilage, and also with age, injury and disease of the cartilage. Among the glycosaminoglycans, chondroitin-6 sulphate is the principal one accounting for over 50% of the sugar component. Keratin sulphate is present only in trace amounts in immature cartilage but with advancing age, its concentration increases. Larger aggregating proteoglycan monomers are called aggregans and they fill up most of the interfibrillar space of the cartilage matrix contributing to 90% of the total cartilage matrix proteoglycan mass.
 * 9) *#Non-Collagenous Proteins and Glycoproteins: Glycoproteins have a protein core to which are attached a few chains of monosaccharides and polysaccharides.
 * 10) *#*Anchorin C-II helps to anchor chondrocyte surface protein to the collagen fibril
 * 11) *#*Cartilage oligomeric protein is found in the territorial matrix of the chondrocyte and has the capacity to bind to chondrocytes. It may have a value as a marker of cartilage turnover and may be useful in assessing the degree of degeneration in osteoarthritis.
 * 12) *#*Fibronectin and Tenascin may have a role in matrix organisation, cell-matrix interaction and the response of tissue in inflammatory arthritis and osteoarthritis.

=Structure of Articular Cartilage=
 * 1) Zones: The morphological changes in chondrocytes and matrix from articular surface to subchondral bone make it possible to identify four zones.
 * 2) *Zone I – Superficial or Gliding Zone: This is the thinnest of all the zones. It has two layers. The most superficial layer is called the ‘’’Lamina Splendens’’’ which is totally acellular. Apart from mechanical function this dense collagen skin probably reduces leakage of proteoglycans from the articular surface and protects it from the effect of harmful enzymes. Deep to Lamina Splendens, flattened ellipsoid shaped chondrocytes arrange themselves so that their major axes are parallel to the articular surface. Chondrocytes of this area produce more collagen than proteoglycan. Concentration of fibronectin and water are also highest in this zone. Collagen fibrils in this zone are numerous and lie parallel to the joint surface in the superficial zone and gives this zone great tensile strength and stiffness than the deeper layers. It also helps them to resist shearing forces generated during movement in the joint. Alteration in this zone may contribute to the development of osteoarthritis. The densely packed collagen fibrils may also limit the ingress of large molecules such as antibodies and the egress of large cartilage molecules. By behaving as a barrier this zone may isolate the remaining part of the cartilage from the immune system. Hence disruption of this zone may not only alter the structure and mechanical properties but may also release cartilage molecules that stimulate immune or inflammatory response.
 * 3) *Zone II – Transitional Zone: The cells in this zone are spheroidal in shape and have a higher concentration of synthetic organelles, such as endoplasmic reticulum and golgi apparatus than the cells in the deeper zones. The cells appear clumped into groups. They have the capacity to synthesise the matrix. The collagen fibrils in this zone are placed tangentially and latter when they traverse to the deeper aspect they change their orientation and lie perpendicular to the articular surface.
 * 4) *Zone III – Middle or Radial Zone: This is the broadest zone. The total thickness of the cartilage is determined by this zone. When well developed, this zone contributes to two-third of the thickness of the entire articular cartilage. Collagen fibrils of this zone have the largest diameter and the water concentration is the lowest. The collagen fibrils run parallel to each other and are situated perpendicular to the articular surface.
 * 5) *Zone IV – Calcified Zone: This is a thin zone that separates the Radial zone from the subchondral bone. The cells here are smaller and fewer but arranged in rows. In some areas the cells are totally buried in individual Calcified Sepulchers, suggesting that that they may have a very low level of metabolic activity. They may have a role to play in the development and progression of osteoarthritis.
 * 6) Tide Mark: The junction between the radial and calcified zones is distinctly visible as a basophilic line in stained preparations of articular cartilage This faint but distinct bluish line is known as the tide mark. It is not seen in developing individuals but in the adult cartilage The number of tide marks may increase with advancing age. This may signify that the calcified zone undergoes periodic extensions into the non-calcified zones of the cartilage with advancing age.
 * 7) Articular Cartilage – Bone Junction: At its base, articular cartilage is bordered by subchindral bone. However there is no structural continuity between the two of them. The cartilage tissue is keyed into irregular surface of the underlying bone like a jigsaw puzzle. The cartilage adjacent to the bone is calcified and has a similar rigidity like bone, hence the keying is rigid.
 * 8) Regions of the Matrix: Three distinct compartments can be delineated
 * 9) *Pericellular Region: This rim of pericellular matrix covers cell surface. It is rich in proteoglycans, anchorin C-II and non-fibrillar collagens such as Type VI collagen. Cytoplasmic extension from chondrocytes extend into this part of the matrix.
 * 10) *Territorial Region: Surrounding the pericellular region is the territorial matrix which encloses individual or clusters of  chondrocytes. In the radial zone it may even envelop each column of chondrocytes. It renders protection to the chondrocytes.
 * 11) *Interterritorial Region: It has the largest column of collagen fibres. Proteoglycans are embedded between their small little spaces.

=Interaction Between Chondrocytes and The Matrix= Chondrocytes and the matrix are interdependent on each other for maintenance. Chondrocytes secrete the macromolecules that make up the matrix and the matrix in turn protects the ch0ndrocytes from mechanical damage helping them to maintain their shape and phenotype. Synthesis and degradation of matrix molecules by is a process continuous throughout life. The mechanism that controls this balance is poorly understood but cytokines with catabolic and anabolic effects appear to have important roles. Matrix also acts as a signal transducer for the chondrocytes. It transmits signals that result from mechanical loading to chondrocytes. They in turn respond to these signals by altering the matrix. Experiments have shown that a persistent decreas4 in joint loading or immoboilisation of the joint decrease the concentration of proteoglycan. Resumption of the usage of joint leads to normalcy. Hence repetitive loading and movement helps a great deal in maintaining the homeostasis of articular cartilage.

=Aging and Osteoarthritis= Cyclic loading of the entire joint surface does not usually occur in the course of normal activities of daily living. The precursor lesion of osteoarthritis namely chondromalacia, or softening of the cartilage is observed to occur first in non-load bearing areas of the joint. Lacking the sustaining contact pressure which nutritionally replenishes the matrix and produces an anabolic mechanical stimulation,these infrequently loaded areas are depleted in sulfated mucopolysaccharides, thus they are less basophilic on haematoxylin eosin staining. As these nutritionally depleted areas from time to time do indeed sustain load they thereby break down,by a failure of normal remodeling processes, fibillation and progressive changes of osteoarthritis ensue. Secondary osseus changes then lead to increased joint congruity in the joint as a whole which lessens the nutrition enhancing effect of youthful slight incongruity in normally weight bearing areas.

Current research appears consistent with a hypothesis that applying load to these normally unloaded areas perhaps through a daily regimen of yoga or pilates may be a reasonable prescription for prevention of this ubiquitous degenerative process by enhancement of cartilage nutrition and anabolic mechanical stimulus through regular loading. In that cartilage has some inherent regenerative potential, a similar regimen could be helpful even once the pathologic changes have begun. Might it be that like the life forms in a tide pool, replenishment twice a day as from the rising tide may suffice to sustain our cartilage.

http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&dopt=Citation&list_uids=15952258

http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&dopt=Citation&list_uids=12721352

http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&dopt=Citation&list_uids=2383080

http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&dopt=Citation&list_uids=6493092

http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&dopt=Citation&list_uids=7226665

= Injury and Repair = http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&dopt=Citation&list_uids=15577072

http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&dopt=Citation&list_uids=14686827

title: Orthopaedic Surgery/Cartilage


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Talk:Animal Behavior

Rename proposal
I'd like to propose that the pages in this book are renamed so that they start with "Animal Behavior/". This will allow readers to see easily that all pages in the book do in fact belong in the book and will be consistent with what most wikibooks do. I trust this proposal is non-controversial, and if it happens it will be by using a bot that preserves all internal links, but am leaving a note here in case anyone has something they'd wish to add, Jguk 22:01, 25 February 2006 (UTC)

Transclusion proposal
i am proposing to ease into the section on neuroscience by building on the same section in the introduction to psychology book. the goal is to continue fleshing it out via the transclusion mechanism and to later fork it into 2 separate projects. Robert Huber 12:36, 17 January 2007 (UTC)

Orphaned pages
The following are not linked to by this book and should be linked to, merged (and marked as such), or marked for deletion. -- Adrignola talk contribs 18:35, 1 August 2009 (UTC)
 * Animal Behavior/Classical Conditioning
 * Animal Behavior/Coevolution
 * Animal Behavior/Inbreeding

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: Talk:Animal Behavior


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Greek Mythology/Beasts/Cyclops

The Cyclopes (singular Cyclops) were the three sons of Uranus and Gaia, the first king and queen of the Titans. They were imprisoned by their father into Tartarus, who feared their strength along with their three brothers, who had fifty heads and a hundred hands, also known as Hekatonchires. Their names were Brontes, Steropes and Arges.

Origin
The word Cyclops means round eye. They were born of Gaia, creator of the earth and her husband Uranus.

Appearance
Cyclopes are usually shown as having one, round eye in place of where their two eyes should be, although sometimes they are depicted as having two empty eye sockets and an eye in the middle of their forehead. Sometimes they are depicted very ogre-like, like other times depicted almost, if not all human.

Abilities
The Cyclopes were known for their great strength, and also their ability of craftsmanship. They created Zeus' lightning bolts, in return for freeing them, while he was fighting the Titans. They also forged Poseidon's trident. They were also known for going around eating humans. They worked as Hephaestus helpers under the volcano Etna making Zeus's lightning, but were killed by Apollo as revenge for Zeus's killing his son Asclepius. Cyclopes are also very well known for their powerful resistance to fire.

Other Information
One of Poseidon's sons, Polyphemus, was a cyclops. Polyphemus plays an important part in the Odyssey, as one of the monsters that beset Odysseus and his crew on their way back to Ithaca. The Cyclopes were also the helpers of one of the gods after Zeus brought them out of the underworld. In Greek Mythology Cyclops play an important role. They are found in many books including Percy Jackson, and Heroes of Olympus series. They are very massive!

title: Greek Mythology/Beasts/Cyclops


text: User:Geoking66~enwikibooks/Mac User

title: User:Geoking66~enwikibooks/Mac User


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Combinatorics

Preliminaries

 * What is Combinatorics?
 * Motivating Examples and Problems
 * Counting
 * Subsets of a set-The Binomial Coefficient
 * Binomial Theorem
 * Congruences

The Pigeonhole Principle

 * The Pigeon Hole Principle

Pairing problem

 * General principles
 * P. Hall's selection theorem
 * Applications to Latin squares and to coverings by dominoes of pruned chessboards.

The inclusion-exclusion principal

 * Applications to derangements
 * Applications to counting problems
 * Applications to rook polynomials

Partitions

 * Counting various types of partitions
 * Ferrers graphs
 * Self-conjugate partitions

Symmetric functions (and anti-symmetric functions)

 * Monomial symmetric functions
 * Elementary symmetric functions
 * Theory of equations
 * Newton's formulae and relations between symmetric functions
 * Indexing of symmetric functions by partitions.

Ramsey Theory

 * Ramsey's Theorem
 * Bounds for Ramsey numbers
 * Schur's Theorem

title: Combinatorics


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Talk:Combinatorics

Creating Material
I think that I will add some equations to some of the topics in order to at least start to populate the textbook. Please help me to create a frame for at least a few of the articles in here.Someperson (talk) 23:56, 17 December 2007 (UTC)

Welcome
Welcome to the Combinatorics wikibook. The idea is to create a basic college-level textbook on Combinatorics for Math or CS majors. It would also be useful to more advanced highschool students. Your contributions are more than welcome. Nymos 01:26, 26 February 2006 (UTC)

ToC
This book's ToC is too limited in scope. Any suggestions for improving it?--Shahab (talk) 01:47, 26 August 2008 (UTC)

title: Talk:Combinatorics


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User:Nymos

Combinatorics

title: User:Nymos


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Book of Magic/Mental Magic

Mind reading is actually not reading a person's thoughts. It is about the performer forcing the spectator to choose a particular thing or readily determining (through a variety of potential avenues) which thing or things the person has chosen. Below are two brief examples:

"Okay, you're thinking of a number between one and four. The number is three, right?"

99% of people will choose the number 3 in that scenario.

Next time when you meet a friend just show the back side of your palm and ask them to choose a finger - 99% of people would select the middle finger only.

Predicting the Future
Many methods are available for predicting the future, although it depends on what one means by that. Presenting a vague or inevitable fact as a prediction is a common method used by so-called fortune tellers, psychics and seers. Another method is to present to an audience that you will predict the future, and present a prediction but not actually explain what it is. After a notable event happens, whatever it may be, you changed the prediction after the fact and present it to everyone. Magical methods involving sleight of hand versions of this trick are an easy way to "predict the future".

Cold Readings
By mixing high quality inferences of a person's characteristics with vague or high probability statements, it is possible to make a general statement about someone's private life which is almost certainly true. With the resulting reaction, one can potentially glean more information and make their predictions more accurate and personalized, leading to subjects believing that magic, psychic abilities, or other supernatural forces are at play.

The Magician's Force Mental Effect
The Magician's Force is a term used to describe a number of things mostly dealing with cards, but can be used in almost any form of mentalism. Simply put, the Magician's Force is a way of making the spectator choose or do something that they feel was of their own doing not the magician's. The most common for this takes is in card magic. There are many different types of card forces available; you can go to the section entitled Basic Card Magic for more on card forces. This section will be dealing with forces used in mentalist effects, although one of the examples will be a card trick.

The Unknown Assistant
This is the easiest miracle effect to learn. When performing for a group of people, have a card selected and lost in the deck. Instead of revealing the card yourself, have a friend of the spectator reveal the card for you. After you leave, they will compare notes and when they realize what just happened, they will think you are the most amazing person on the planet (well maybe not the most amazing, but at least in the top two billion or so…)

As with most tricks, the patter is very important. Probably more so in a situation like this where you have to lead someone to the conclusion without making it apparent that you are doing so. Everything must look like it is on the up and up. That is to say that you have to be subtle so as not to arouse suspicions about what you are doing or saying. Let us start with a synopsis of the trick we are about to discuss. First, a card is chosen by a spectator. This can be a free choice or a force depending upon exactly what you are trying to do as well as your skill level in handling cards. The chosen card is not looked at by the spectator, and not shown to anyone else. The spectator should hold on to the card so no one else can see it, and so that it can be compared to the revealed card at the end. Another spectator, preferably a friend to the first spectator is chosen to reveal the card by receiving psychic impressions (yeah right.) After the card is revealed by the second spectator, the card is shown to match.

As the magician is going to be leading the spectators through the trick, patter and gestures must be subtle and direct. A good rapport is needed and should be established as soon as possible. This could have been done through the previous tricks done for the spectators, or through just simply talking to them to get to know them a little before beginning. Have your force ready to begin as soon as the rapport takes hold. Once begun, it is important to maintain control of the situation. If one or more of the spectators are unruly, do some other trick to gain their attention and more importantly their respect. As a performer you have a right to be respected or to choose not to perform. If your audience doesn’t give you their attention, you are better off not performing for them when the performance involves audience participation. If you get their attention and have their respect, then you can begin the presentation.

First you must have a card chosen. It is important to make the choice look like a free one, even if it is not. The only rule here is that after the choice you must know what the chosen card is. If you can give the spectator a free choice and then discern the identity of the chosen card, do so. It will make the trick seem that much more of a miracle. Otherwise you must force a card on the spectator, a card you already know the identity of. The patter used will depend on the method of card delivery. In my example, I will show doing it with a free choice of cards. This can be accomplished most easily by a stacked deck (See Basic Card Magic.) My delivery patter is as follows:

(To the first spectator) 'As I go through the deck slowly, I want you to place your finger on one of the cards, which ever you choose, and leave it there. Do you understand?'(Asking if they understand is not always necessary, but helps you know that they are following your instructions properly.) Once the spectator places their finger on a card (you can ask them if they want to change their mind or if they are sure that is the card they want, but you must maintain their finger on the card with no apparent movement of the cards.) Once you are sure that is the card they want separate the cards above their choice and as they are taking the card place the cards above it on the bottom of the deck while you are giving them the following instruction:

OK I want you to take the card and don’t look at it or let anyone else see it, simply put it up against you so no one, not even yourself can know what it is. When they have done this, you must find the identity of the card they have chosen. As I am doing this with a stacked deck, all I have to do is glimpse the bottom card (which was the card on top of the chosen card) and I will know the identity of their card. Once you have worked with a stacked deck enough you will be able to tell the next card without hesitation, which is what you must strive to do. Any hesitation will tell the audience that you are trying to remember the next card. If you need to stall long enough to figure out which card they chose, you can pick up the card box and insert the deck into the box. This gives you a little time to figure out their card while you fumble with the box, but don't take too long. Once you know the chosen card you then turn to the next spectator and begin the patter for their naming the card. I generally don’t let them know that they are going to reveal the card before the card is chosen. This doesn’t give them time to get nervous.

(To the second spectator and in the format of a setup for the reveal) Would you mind helping me with this trick? (Before they can answer, continue with) Have you ever been a magician before? Once they agree to help, then begin with: 'Well if I asked you to name their card, you would have a one in 52 chance, not very good odds. So let’s see if we can make those odds a little better shall we… In a deck of cards there are two colors, Red and Black. We can eliminate half of the cards by picking a color. So if you will please pick a color, Red or Black, Black or Red?'

(It is really important that you say it that way, as this shows that you are not trying to lead the spectator to make a particular decision. They don’t know how the trick works, but if you say one choice first all the time and that one happens to be the one that the spectator says then they think you led them to that decision. Saying both choices together and both first will eliminate this thinking.)

What you have done with this statement is setup the force. You already know the color of the chosen card. By making the statement above you have said you were going to eliminate half of the cards, but you didn’t say which half. Let’s say that the chosen card was the 5 of Diamonds. We know that it is a red card, so if the second spectator says “Red” then you simply turn to the first spectator and say She says your card is a Red Card, let’s hope she is right or it isn’t going to be much of a trick. However if the second spectator says “Black” then turn to the first spectator and say She has eliminated the black cards, so your card must be “Red” (see how easy that was)

Now turn back to the second spectator and say, 'In the Red Cards there are two suits, Hearts and Diamonds, Lets see if we can cut our odds again. If you could please pick Hearts or Diamonds, Diamonds or Hearts.'

If the second spectator says “Diamonds” then turn to the first spectator and say, Now that we know it is Diamonds let’s see if we can zero in on the value. If the second spectator says Hearts, then go with 'Ok, we eliminated the Hearts, so we know your card was Diamonds. Unfortunately we can no longer eliminate half of the cards, because we have thirteen left and that isn’t evenly divisible.'

Turn back to the second spectator and say 'There are however Face Cards and Spot Cards to chose from. Face cards being the Jack, Queen, and King. Spot Cards being all of the rest. So if you could please pick Face or Spot, Spot or Face?' If the spectator says “Face” then you would say That leaves the Spot Cards. If the spectator says “Spot” then say, 'Ok there are ten Spot Cards, let’s eliminate some more. There are High Cards and Low Cards, High being the Six, Seven, Eight, Nine and Ten. Low being the Ace through Five, so please pick High or Low, Low or High?'

If the spectator says “High” then say That leaves the Low Cards. If the spectator says “Low” then say, 'There are five Low Cards, the Ace, Two, Three, Four and Five. Please pick any three of them.'

Since the Chosen card was the 5 of Diamonds, simply leave the five in play and eliminate the others. If the 5 was one of the chosen cards, then rename the three and say please chose two. If it wasn’t then you are down to two cards simply say choose one of them. By manipulating the choices in this way you end up with the chosen card rename it and have the first spectator show their card for the first time.

The main thing to remember is that you must be in control at all times. You must keep the motion going not letting the spectators have too much time between choices to think about what you are doing. If they don’t get to think about it as you are doing it, they won’t know they are being manipulated. By the time they have a chance to think about what happened, they won’t remember enough of the details to put it together.

Unpredictable Prediction
This effect is also very simple yet very powerful. Start with a piece of paper on which you will write a prediction. Fold up the prediction and display it where everyone can see it. Pick about 10 to 15 everyday items and bring them to a table. Alternate with the spectator picking items out of the group until there is only one item left. When the prediction is read, it will match the item remaining.

With this effect there are only two things to remember. The first is what the prediction is, and the second is that the magician always picks on the even numbered picks. The easiest way to see how this trick works would be to work through a sample. For the sake of time we will start at the point where there is only six items left. We will assume that the items are a candle, a book, a pencil, a mirror, a comb, a photo, and a credit card. We will also assume that the mirror is the prediction.

The magician takes the comb and the credit card and places them at the center of the table. The spectator must eliminate one of these two items, let’s say the comb. The credit card goes back with the other five items and the spectator must now choose two items and place them at the center of the table for the magician to eliminate one. Let’s say the spectator chooses the mirror and the book. These are placed at the center of the table and the magician must eliminate one of these. The magician eliminates the book and places the mirror back with the other items. The magician now chooses the candle and the pencil to place at the center. The spectator eliminates the pencil and places the candle back with the others. The spectator picks the credit card and the candle and places them at the center. The magician eliminates the credit card and places the candle back with the other items. The magician picks the photo and the candle and places them at the center. The spectator eliminates the candle and places the photo back with the mirror which is the only item left. Since there are only two items left, the magician eliminates the photo. This leaves the mirror, which matches the prediction.

It works very simply, when the magician is giving the spectator two items to choose from, the prediction is never one of the items. If the spectator puts in the prediction item as one of the choices, the magician simply chooses the other item, leaving the prediction in play. One note though don’t have too many items to choose from, limit it to about 10 to 12. More than that gives the spectators a chance to see the pattern, and can figure out what is going on.

title: Book of Magic/Mental Magic


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User:AdamGuy

I am really not sure what I am doing at the moment. I'm just learning about wikipedia... although I reference is regularly. I feel that I can too often make poor additions that get removed by wiki admins. In any event, I hope this is not in the wrong place.

My name is Adam

I am a student of all sciences. I am a man of ideas.

title: User:AdamGuy


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User:A thing

See http://en.wikipedia.org/wiki/User:AThing

title: User:A thing


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French/Lessons/Description

Mark is describing someone: 'Il est blond. Il est assez petit. Il est très intelligent. Il est sympa.'

George is describing someone: 'Elle est assez grosse. Elle est très amusante et assez patiente.'

Chantal is describing a group of her friends: 'Ils sont extrêmement amusants. Ils sont assez forts.'

Like articles, French adjectives match the nouns that they modify in gender and number.

Regular formation
Most adjective changes occur in the following manner:

 * Feminine: add an -e to the masculine form
 * un garçon intéressant → une fille intéressante
 * un ami amusant → une amie amusante
 * un camion lent → une voiture lente
 * Plural: add an -s to the singular form
 * un garçon intéressant → des garçons intéressants
 * une fille intéressante → des filles intéressantes

Pronunciation
Generally, the final consonant is pronounced only when it comes before an -e. Most adjectives, such as those above, are affected by this rule.

 * Masculine pronunciation:
 * Feminine pronunciation:
 * Feminine pronunciation:
 * Feminine pronunciation:
 * Feminine pronunciation:

With plural adjectives, the -s ending is not pronounced, so the adjective will sound exactly the same as the singular form.

Adjectives that end in e in the masculine form do not change in gender. When an adjective, such as gros, ends in -s, it does not change in the masculine plural form. Sometimes the final consonant is doubled in the feminine form.

The adjectives ce, cet, and cette mean this or that, and the adjective ces means these or those.

The adjective used must agree in gender and number with the noun it modifies:

To be more precise or to avoid ambiguity, or  can be inserted after the noun:

Natasha is describing a friend: 'Elle est très forte, elle est grande, et elle est intelligente.'

Marc is describing someone: 'Il est handicapé; il est sourd et muet.'

Jules is describing a group of his friends: 'Ils sont très polis, attachants et sincères.'

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Orthopaedic Surgery/Boutonniere Deformity

A frequently missed injury to the ligaments and central slip of the PIP joint this injury most often leads to some long term compromise in the function of the finger regardless of how it is treated. When it typically presents a few weeks after injury, presumed at first to be a "jammed" finger, the distiguishing feature is the flexed resting posture of the PIP joint which will likely still be passively correctible (extension lag rather than flexion contracture). This is seen in conjunction with a hyperextended resting posture of the distal interphalangeal joint.

Closed treatment is by an alumafoam splint applied with coban to the PIP in full extension; then active flexion exercises of the DIP joint. The PIP is splinted full time for 5 weeks then full motion is allowed. A residual flexion contracture or extension lag at the PIP is typical, but if it can be limited to 20 degrees generally does not interfere with function if active DIP flexion is near normal. Swelling of the PIP persists for the better part of a year.

If there is fixed contracture therapy is directed at achieving full extension by static or dynamic splinting or serial casting to gain full passive extension at the pip joint.

Surgical treatments include efforts to reconstruct the central slip directly, or efforts to employ release of the terminal extensor tendon to diminish hyperextension at the DIP level and hopefully inprove the active extension at the PIP level as the extensor retracts and the slack pseudotendon over the PIP receives more tension via the confluent extensor hood.

Lacerations and crush injury to the dorsum of the PIP can include central slip injuries which can also go unrecognized with resultant deformity apparent at time of follow up. Usually exploration is pursued to obtain direct repair of the central slip. A 1.3 suture anchor or a transosseus suture tied over a button on the palmar surface with a 3-0 proline are two common methods.

In the case of crush injury there may be soft tissue defects to contend with, small defects are managable with random pattern flaps if a few millimeters of coverage over the repaired tendon is needed. Alternatively a more elaborate axial pattern flap can be used where applicable. Stable coverage is needed if tendon graft is used to reconstruct the central slip.

Pinning of the PIP in extension has the advantage of lessened concern for early failure due to lack of compliance or loss of position in the splint. Access to the hand to address or rehab other injuries may be enhanced. The concerns with a transarticular pin include the prospect for excessive stiffness or pin failure requiring the problematic retrieval of the broken pin from within the bone. It is helpful in such a case to have advanced the pin all the way through the bone on the opposite side of the joint so that if it should break a protruding end is available on both sides.

If a transosseus pin can be avoided well supervised rehab can be considered early allowing short arc active flexion with passive extension or a dynamic splint with a flexion stop. In any case any pin should be removed by 4 weeks and active motion initiated. Secondary capsulotomy and tenolysis after 4 months may be needed if the joint is very stiff in extension.

At times the patient will present with a laceration over the PIP which has been closed by the ER staff and the status of the central slip is uncertain. It may seem to be intact clinically but then, within a week or two, extension lag and a tendency to hyperextension at the DIP will indicate the true diagnosis. Exploration within a four week interval from injury will usually allow for direct repair of the central slip which can then be managed as in the acutely recognized central slip injury.

http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&dopt=Citation&list_uids=16007379


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Cookbook:Basic Flapjacks

| Desserts | Cuisine of the United Kingdom

Flapjacks in the United Kingdom (not to be confused with the American pancakes also called flapjacks) are a type of bar made using rolled oats, butter, brown sugar, and golden syrup. They're great for packed lunches and the like. This is a simple version with no fruit or raisins added. Try serving with whipped cream or with melted chocolate to dip.

Ingredients

 * 110 g (½ cup or 4 oz) light brown sugar
 * 175 g (1½ cups or 6 oz) butter
 * 1 dessert spoon golden syrup
 * 175 g (1½ cups or 6 oz) rolled oats

Procedure

 * 1) Preheat the oven to 150°C (300°F or Gas Mark 2).
 * 2) Place the sugar, butter, and golden syrup together in a medium saucepan and heat until the butter has melted.
 * 3) Remove the saucepan from the heat and stir in the oats.
 * 4) Press the mixture out over the base of a lightly-greased 7½-inch (19 cm) square baking tin.
 * 5) Bake in the centre of the oven for 40 minutes.
 * 6) When baked, allow to cool in the tin for 10–15 minutes before cutting into oblong bars.
 * 7) Leave until cold before removing the flapjacks from the tin, then store in an airtight container.

title: Cookbook:Basic Flapjacks


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Educational Technology Innovation and Impact/Why use Technology in Education/Musical Intelligence

Musical Intelligence The theory of Musical Intelligence was introduced by Howard Garner as part of his theory of Multiple Intelligence. Multiple Intelligence comprises 9 areas of intelligence that people have and Gardner believes that each Intelligence can be enhanced through learning experiences. This article will look at Musical Intelligence in terms of what it is, how it is used and potential benefits within a learning environment.

Gardner’s defines Musical Intelligence as “Musical Rhythmic Intelligence, (music smart), is the capacity to think in music, to be able to hear patterns, recognize them, and perhaps manipulate them”.

To explore Musical Intelligence further visit http://www.ibiblio.org/edweb/edref.mi.th1.html

Characteristics of musical intelligence may include someone who remembers and differentiates between tunes and sound patterns, have a good sense of tone, tap to or move rhythmically to sounds or music, and generally enjoy composing or performing music.

In a paper published by the British Journal of Music Education which looked at Musical Intelligence relating to practical aspects of musical education showed how Musical Intelligence can be incorporated into a school curriculum. The study revealed that music adds value to both the school and the pupil. Music can be included in a blend of skills and activities when performing, creating or inventing and listening or communicating. To achieve success teachers must ensure students participate in all activities including listening skills and are given opportunities to try out different activities themselves.

Schools are further enhancing Musical intelligence with the aid of Digital technology .An article by Dee Dickinson for America Tomorrow shows how students can learn elements of music that are often difficult for beginners to understand by combining digital audio with visual input. The article goes on to quote.. “One example is the Voyager Company's interactive multimedia compact disc of Beethoven's "Ninth Symphony," which enables the listener to understand the piece musically, historically, culturally, and politically.” This technology is in it’s infancy of an exciting new era of innovation for learning about and producing music.

The benefits of musical intelligence can be spread across a wide area of people’s intelligence. Gardner believes music can help people to develop in other areas such as maths, language and spatial reasoning. Dr Arthur Harvey of the University of Hawaii has shown in his paper “An Intelligence View of Music Education” how Musical intelligence has impact on all intelligences. One example of this is Hall’s study in 1952 on how musical intelligence impacts on Linguistic intelligence, the study showed that the use of background music in study halls resulted in substantially more improvement of reading comprehension than those that studied without music out of 278 eighth and ninth graders. For further examples visit http://www.menc.org/publication/articles/academic/hawaii.htm.

In conclusion Musical Intelligence can play a significant part in the development of students once teachers are given the relevant training in all aspects of the music curriculum which includes assessments and music activities.


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Knowledge Engineering for Historians

This book is on knowledge engineering for historians.

Contents

 * 1) /Preface/
 * 2) /Introduction/
 * 3) /Primer/
 * 4) /Modelling Historical Knowledge/
 * 5) /Advanced Topics/
 * 6) /Bibliography/

Links
This is a wiki textbook -- feel free to edit it, update it, correct it, and otherwise increase its teaching potential. To find out more about wikis, see the Wikipedia main page.

title: Knowledge Engineering for Historians


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Knowledge Engineering for Historians/Preface

The following is a Wikibook textbook on how historians can leverage the technology of knowledge engineering in their work.

A Wikibook can be edited by anyone which wants to expand the usability and correctness of the work, however, please do not contribute any works that you do not hold the copyright to. If you find that this textbook is incomplete, feel free to fill in the blanks.

See also: The WikiBook on Artificial Intelligence and Historiography at Wikipedia.


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Knowledge Engineering for Historians/Introduction

Introduction
This wikibook is about how historians can leverage the technical contributions that the discipline of Knowledge Engineering has made to modern research in their own work.

How We Got Here

 * historians are trying to take advantages of the technologies that computers have, but they have no money to fund any research
 * the Internet and its search engines are a great help
 * in the early 80s, in the heyday's of AI, it looked like expert systems would be the way to go, but then the AI winter hit before the historians could really learn how to build or use them
 * the "Let's use Databases!" movement fizzled in the late 90s because it was not immediately obvious how to address the issues of semantics underlying the coding of the data
 * with OWL, the knowledge engineering is getting back into the swing of things
 * ontologies such as the Council of Museum's conceptual reference model indicate that the time has come to learn this valuable tool
 * one history ontology is already available, embedded in VICODI, see also the paper by Richard Deswarte and Jan Ooesthook on their experience contributing to this effort, the ontology itself

Where we are going to go

 * in the primer, we are going to
 * find out what the fundamental benefits derivable from knowledge engineering are
 * get an idea of what the various (freely) available tools offer

 * in the basic round, we are going to
 * model some fundamental facts that historians need all the time
 * model some intermediate concepts, like sources and point of view

 * in the advanced round, we are going to
 * model more difficult stuff, like temporal projections
 * consider modelling problems that involve shifting the resolution of the knowledge we are reasoning over


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Template:Knowledge Engineering for Historians/Main

Introduction | Knowledge Engineering — A Primer | Modelling Historical Knowledge | Advanced Topics


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Knowledge Engineering for Historians/Primer

Introduction

 * give links to good KE tutorials
 * give links to good tools and their tutorials
 * give links to the AI book here on wikibooks

Discussion Plan

 * 1) The Basic Elements
 * 2) The Composite Elements
 * 3) Additional Infrastructure
 * 4) Engineering Choices


text:
Knowledge Engineering for Historians/Modelling Historical Knowledge

Modelling Historical Knowledge
Now that we are clear on the basics of knowledge modelling in general, we can turn to our task proper at hand and consider the tasks historians face in their research and how to deal with them appropriately.

We first would like to look at a main concern of historians, the proper handling of sources. Historians are nothing without their sources and they contain the knowledge we wish to represent and utilize. We then turn to the problem of how to deal appropriately with the work of their colleagues, another key element to historical research.

However, in order to discuss this topic we need to introduce some additional infrastructure, namely the notion of contexts. We will explain why we need these contexts, how they work, and what kind of problems their use poses.

 * 1) Introducing Contexts
 * 2) Modelling Sources
 * 3) Modelling Interpretations


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Wikijunior:Kings and Queens of England/The Danes

Sweyn Haraldssen (1013-1014)
Sweyn Haraldssen was born in Denmark around the year 960. His nickname is "Forkbeard" which is a nickname that probably was used during his lifetime, and refers to a long, pitchfork-like moustache, rather than a full beard. Such a moustache was fashionable at the time, particularly in England. Sweyn succeeded his father, Harold I "Blåtand" (Bluetooth), as king of Denmark, probably in late 986 or early 987. Following the death of Norway's King Olaf I Tryggvason in the Battle of Svolder in 1000, Sven established Danish control over most of Norway.

Sweyn was almost certainly involved in the raids against England in 1003 to 1005, 1006 to 1007, and 1009 to 1012, following the massacre of England's Danish inhabitants in November 1002 during the reign of Ethelred the Unready. Sven is thought to have had a personal interest in these due to his sister, Gunhilde, being amongst the victims. Sven acquired massive sums of Danegeld as a result of the raids, and in 1013 personally led the Danish fleet in a full-scale invasion.

The Laud Chronicle says that "before the month of August came king Sweyn with his fleet to Sandwich. He went very quickly about East Anglia into the Humber's mouth, and so upward along the Trent till he came to Gainsborough. Eorl Uhtred and all Northumbria quickly bowed to him, as did all the folk of Lindsey, then the folk of the Five Boroughs. (...) He was given hostages from each shire. When he understood that all the people had submitted to him, he ordered that his force should be provisioned and horsed; he went south in full force, and entrusted his ships and the hostages to his son Canute. After he came over Watling Street, they worked the most evil that a force might do. They went to Oxford, and the town-dwellers soon bowed to him, and gave hostages. From there they went to Winchester, and did the same."

However, when he came to London, the Londoners destroyed the bridges that spanned the River Thames. It is this action that is referred to in the song London Bridge is Falling Down. Sweyn suffered heavy losses and had to withdraw. King Sweyn then went to Wallingford, over the Thames to Bath, and stayed there with his troops. The leading noblemen there all bowed to Sweyn and gave hostages. London withstood the assault of the Danish army, but the city was now alone. King Ethelred the Unready fled to Normandy in late 1013. With the acceptance of the Anglo-Saxon council, the Witan, London finally surrendered to Sweyn, and he was declared "king" on Christmas Day.

Sweyn based himself in Gainsborough, Lincolnshire and began to organise his vast new kingdom, but he died there on 3 February 1014, having ruled England unopposed for only five weeks. His body was later returned to Denmark and buried at Roskilde Cathedral. He was succeeded as King of Denmark by his eldest son with his wife Gunhilde. The Danish fleet proclaimed his younger son Canute as King of England, but they and he returned to Denmark, and Ethelred the Unready became King of England again.

Canute the Great (1014, 1016-1035)
Canute (or Cnut) I, or Canute the Great was born in 994 or 995 in Denmark. He was King of England, Denmark and Norway and overlord of Schleswig and Pomerania. He was Sweyn Forkbeard's son. Canute accompanied his father on his invasion of England in August 1013, and Canute was proclaimed king by the Danish fleet on Sweyn's death the following February. However, he went back to Denmark in April 1014 once King Ethelred the Unready was restored by the Witenagemot. Canute invaded England again in August 1015, and after a series of inconclusive conflicts he won a decisive victory over the new English king, Edmund II, in October 1016. This led to a meeting with Edmund on an island in the River Severn, where they agreed to divide the kingdom, with the kingdom to be reunited again under the survivor when the first one of them died. When Edmund died in November 1016, this left Canute as sole ruler, and he was acclaimed as King of England by the Witenagemot in January 1017.

As King of England, Canute combined English and Danish institutions and personnel. His mutilation in April 1014 of the hostages taken by his father in pledge of English loyalty is remembered above all as being uncharacteristic of his rule. In 1017 Canute divided England into the four great earldoms of Wessex, Mercia, East Anglia and Northumbria, and he started the system of territorial lordships which would underlie English government for centuries. The very last Danegeld ever to be paid, a sum of £82,500, was paid to Canute in 1018. He felt secure enough to send the invasion fleet back to Denmark with a payment of £72,000 that same year.

In order to associate his line with the overthrown English dynasty and to insure himself against attack from Normandy, where Ethelred's sons, Edward and Alfred, were in exile, in July 1017 Canute married Ethelred's widow, Emma of Normandy, daughter of Richard, Duke of Normandy. Canute later made his son by Emma, Harthacanute, his heir in preference to Harold Harefoot, his illegitimate son by Aelgifu of Northampton.

Denmark and Norway
In 1018 or 1019 Canute succeeded his elder brother, Harold II of Denmark, as King of Denmark, and appointed his brother-in-law Ulf Jarl as the earl of Denmark. When the Swedish king, Anund Jakob, and the Norwegian king, Olaf, took advantage of Canute's absence and attacked Denmark, Ulf convinced the freemen to elect Harthacanute king, since they were miscontent with Canute's absenteeism. This was a ruse from Ulf since his role as the caretaker of Harthacanute would make him the ruler of Denmark. In 1026, Canute returned to Denmark and with Ulf Jarl's help, he defeated the Swedes and the Norwegians at the Battle of Helgeå. However, Canute had not forgiven Ulf Jarl for his earlier actions, and at a banquet on 24 December 1026, the two started arguing with each other whilst playing chess. The next day, Canute had one of his household troops kill Ulf Jarl in the church of Trinity. In 1028, Canute conquered Norway with a fleet of fifty ships from England, though his attempt to govern Norway through Aelgifu and his other son by her, Sweyn, ended in rebellion and the restoration of the former Norwegian dynasty under Magnus I.

Commanding the waves to go back
He is perhaps best remembered for the story of how he commanded the waves to go back. According to the legend, he grew tired of flattery from his courtiers. When one such flatterer gushed that the king could even command the obedience of the sea, Canute proved him wrong by actually going into the sea at Thorney Island and proving that he couldn't; a king's powers have limits. This legend is now usually misunderstood to mean that he believed himself so powerful that the natural elements would obey him, and that his failure to command the tides only made him look foolish. Whether this event really happened or not is unknown.

Death and legacy
Canute is generally regarded as a wise and successful king of England, although this view may in part be attributable to his good treatment of the church, which controlled the history writers of the day. The image that has come down from them is that he was a religious man, despite the fact that he lived openly in what was effectively a bigamous relationship, and despite his responsibility for many political murders.

Canute died in 1035, at Shaftesbury in Dorset, and was buried at Winchester. On his death, Canute was succeeded as King of Denmark by Harthacanute, who reigned there as Canute III. Harold Harefoot became King of England, then after his death in 1040, Harthacanute became King of England too.

Harold Harefoot (1035-1040)

Harold I Harefoot was born in Denmark around the year 1012. earned the name "Harefoot" for his speed and skill at hunting. He was the illegitimate son of King Canute by his concubine Aelgifu. Harold's younger half-brother Harthacanute, the son of Canute and his queen, Emma of Normandy, was the legitimate heir to the thrones of both Denmark and England at Canute's death in 1035. However, because Denmark was threatened with invasion from Norway, Harthacanute was unable to travel to England and instead sent as regents Emma and Harold Harefoot. Harold took effective power in England and in 1036 secured recognition by Harthacanute as regent during the latter's absence in Denmark. Harold and Emma argued over who should govern the kingdom. The powerful Earl Godwin sided with Harold, and in 1037, after Emma had fled, Harold seized the treasury at Winchester and thus the throne, and was crowned at Oxford.

In general little is known about his reign and he appears to have been a colourless and weak character. His period of rule is associated with the blinding and death of Alfred the Aethling, Emma's son by King Ethelred the Unready, following Alfred's return to the kingdom (possibly in an attempt to take the throne) with his brother Edward the Confessor. Harold never married, but he had an illegitimate son, Elfwine, who became a monk on the continent. Harold died in Oxford in 1040, just as Harthacanute was preparing an invasion. He was buried at St Clement Danes Church, Westminster, but Harthacanute later had his body dug up, beheaded, and thrown into a fen bordering the River Thames.

Harthacanute (1035-1037, 1040-1042)
Harthacanute (sometimes Hardicanute or Hardecanute) was born in 1018 or 1019. He was the only son of Canute the Great and his queen, Emma of Normandy. His name means Canute the Hardy. He succeeded his father as King of Denmark in 1035, reigning as Canute III, but conflict with Magnus I of Norway prevented him from sailing to England to secure his position there so it was agreed that his elder illegitimate half-brother Harold Harefoot would be regent in charge of England.

Harold, after Harthacanute's continued absence, took the English crown for himself in 1037. In 1038 or 1039 Harthacanute settled the situation in Scandinavia through an agreement with Magnus in which they agreed that if either of them should die without an heir, the other would be his successor. He then prepared an invasion of England to depose Harold, and in 1039 arrived in Bruges in Flanders (modern-day Belgium), where his exiled mother was. An invasion was not necessary though as Harold died in March 1040 before it could occur. According to the Anglo-Saxon Chronicle, Harthacanute then landed at Sandwich in June with a fleet of 62 warships. Being unable to exact revenge on Harold while he was still alive, he had the dead Harold dragged up and thrown into a fen.

Harthacanute was a harsh and very unpopular ruler. He severely increased the rate of taxation to pay for his fleet, and perhaps the most notable event of his reign in England was a revolt at Worcester in 1041 against these high taxes. This revolt was crushed, with the near destruction of Worcester. The story of Lady Godiva riding naked through the streets of Coventry to persuade the local earl to lower taxes may come from the reign of Harthacanute.

Harthacanute never married and had no children. In 1041, Harthacanute invited his half-brother, Edward the Confessor, who was Emma's son by King Ethelred the Unready, back from exile in Normandy to become his co-ruler and heir. In June 1042, Harthacanute died at Lambeth and was buried at Winchester. Edward became king on Harthacanute's death, thereby restoring the Anglo-Saxon royal line.


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Talk:Calculus/Higher Order Derivatives

2/26/06

Hi,

Just trying to reorganize this page and provide more examples. I think it would be better if we created an entire section within the applications of derivatives page for kinematics.

technochef techno_chef [att] swissinfo [dot] orgg [remove last g]

Just added a section about finding the nth derivative of x^m - font size is too small and I'm very tired. So, someone please increase the font size in that section, and check the math to make sure I didn't screw something up. -64.30.204.249 08:13, 15 March 2006 (UTC)

Is concave up really convex? Cyrus Jones (talk) 00:47, 16 May 2008 (UTC)

Is it really Newton's notation?
From the page:
 * With the Newtonian notation, the derivative of the function $$f(x)$$ is denoted by $$f^\prime(x)$$

Isn't that Lagrange's notation (apparently from "Théorie des fonctions analytiques")? (Wikipedia) --Mrwojo (talk) 02:46, 29 December 2008 (UTC)

Removed text
I removed the following text from the page:

One can derive a general expression for the nth derivative of $$x^{m}$$ where n and m are integers and $$n<m$$.

...
 * $$f (x)=x^{m} \ $$
 * $$f' (x)=(m)x^{m-1} \ $$
 * $$f'' (x)=(m)(m-1)x^{m-2} \ $$
 * $$f^{\prime\prime\prime} (x)=(m)(m-1)(m-2)x^{m-3} \ $$
 * $$f^{(n)} (x)=\frac{m!}{(m-n)!}x^{m-n}$$

Remember that $$0!=1$$, by definition.

Power, MacLaurin and Taylor series make use of this expression.

The current examples don't use this approach, so it seemed a bit out of place. Perhaps it could be expanded on or used in a later module. --Mrwojo (talk) 22:52, 12 September 2009 (UTC)

title: Talk:Calculus/Higher Order Derivatives


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Calculus/Kinematics

Introduction
Kinematics or the study of motion is a very relevant topic in calculus.

If $$x$$ is the position of some moving object, and $$t$$ is time, this section uses the following conventions:

 * $$x(t)$$ is its position function
 * $$v(t)=x'(t)$$ is its velocity function
 * $$a(t)=x''(t)$$ is its acceleration function

Average Velocity and Acceleration
Average velocity and acceleration problems use the algebraic definitions of velocity and acceleration.

 * $$v_{\rm avg}=\frac{\Delta x}{\Delta t}$$
 * $$a_{\rm avg}=\frac{\Delta v}{\Delta t}$$

Examples
Example 1:

A particle's position is defined by the equation $$x(t)=t^3-2t^2+t$$. Find the average velocity over the interval $$[2,7]$$.

 * Find the average velocity over the interval $$[2,7]$$ :

 * $$v_{\rm avg}$$
 * $$=\frac{x(7)-x(2)}{7-2}$$
 * $$=\frac{252-2}{5}$$
 * $$=50$$
 * }
 * $$=50$$
 * }
 * $$=50$$
 * }

Answer: $$v_{\rm avg}=50$$.

Instantaneous Velocity and Acceleration
Instantaneous velocity and acceleration problems use the derivative definitions of velocity and acceleration.

 * $$v(t)=\frac{dx}{dt}$$
 * $$a(t)=\frac{dv}{dt}$$

Examples
Example 2:

A particle moves along a path with a position that can be determined by the function $$x(t)=4t^3+e^t$$. Determine the acceleration when $$t=3$$.

 * Find $$v(t)=\frac{ds}{dt}$$.

 * $$\frac{ds}{dt}=12t^2+e^t$$

 * Find $$a(t)=\frac{dv}{dt}=\frac{d^2s}{dt^2}$$.

 * $$\frac{d^2s}{dt^2}=24t+e^t$$

 * Find $$a(3)=\frac{d^2s}{dt^2}\bigg|_{t=3}$$

 * $$\frac{d^2s}{dt^2}\bigg|_{t=3}$$
 * $$=24(3)+e^3$$
 * $$=72+e^3$$
 * $$=92.08553692\dots$$
 * }
 * $$=92.08553692\dots$$
 * }
 * $$=92.08553692\dots$$
 * }

Answer: $$a(3)=92.08553692\dots$$

Integration

 * $$x_2-x_1=\int\limits_{t_1}^{t_2} v(t)dt$$
 * $$v_2-v_1=\int\limits_{t_1}^{t_2} a(t)dt$$

authors: יהודה שמחה ולדמן

title: Calculus/Kinematics


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Talk:Electronics/Adders

Reason for Change

 * $$A \wedge B \wedge \neg D \vee A \wedge \neg B \wedge D \vee \neg A \wedge B \wedge D $$

Or
 * $$A.B. \bar {D} + A. \bar {B} .D + \bar {A} .B.D $$

The above boolean is wrong if we look at the last entry in the table. The above expression gives 1.1.0 +1.0.1+ 0.1.1= 0 which is wrong. This means we should add + A.B.D to it. But it then reduces to A.B + A.D + B.D.

title: Talk:Electronics/Adders


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Calculus/Optimization

Introduction
Optimization is one of the uses of calculus in the real world. Perhaps we are a pizza parlor and wish to maximize profit. Perhaps we have a flat piece of cardboard and we need to make a box with the greatest volume. How does one go about this process?

This requires the use of maximums and minimums. We know that we find maximums and minimums via derivatives. Therefore, one can conclude that calculus will be a useful tool for maximizing or minimizing (collectively known as "optimizing") a situation.

In general, an optimization problem has a constraint that changes how we view the problem. The wording of the problem (whether subtle or not) can also drastically change how we view the problem. The constraint is the thing we will absolutely not change (such as the general shape, or the amount that we are willing to pay for the entire operation, etc.). If we can identify what we are going to change and what we are going to require stay the same, we will be on our way to solve the problem.

How to Solve
These general steps should be taken in order to complete an optimization problem. The reason why this algorithm works comes from a few mathematical theorems which you will probably not need to know when completing these problems. Usually the problems given will be mathematically simple (in other words, there are not a lot of cases to test). However, if you wish to know, they work like this:
 * 1) Write out necessary formulas and other pieces of information given by the problem.
 * 2) * The problems should have a variable you control and a variable that you want to maximize/minimize.
 * 3) * The formulas you find may contain extra variables. Depending on how the question works out, they may be substituted out or can be ignored (which will be explained later).
 * 4) Combine the formulas together so that the variable you want to maximize/minimize is on one side of the equation and everything else on the other.
 * 5) Differentiate the formula. If your equation has multiple variables, pick any variable to differentiate as long as it is not the one you control (i.e. pick the variable that you could not get rid of from the formula).
 * 6) * Note that during differentiation, if you come across a variable that you have not picked, imagine it as a number and apply the necessary differentiation rule. Do not treat it as a variable in this case.
 * 7) Set the differentiated formula to equal 0 and solve for the variable you control.
 * 8) The value you get here is your answer. If you instead have another formula, that means that your answer depends on those other variables, which would usually be what the question asked for if you have such a situation that you have another variable to juggle to begin with.
 * A derivative of 0 is either a global or local maximum or minimum. Usually the question will tend towards answering that question without much difficulty (like always positive numbers, for example)

Volume Example
A box manufacturer desires to create a closed box with a surface area of 100 inches squared and with a square base yet rectangular sides. What is the maximum volume that can be formed by bending this material into a box?
 * Write out known formulas and information
 * $$A_{base}=x^2$$
 * $$A_{side}=x\cdot h$$
 * $$A_{total}=2x^2+4x\cdot h=100$$
 * $$V=l\cdot w\cdot h=x^2\cdot h$$

 * Write the variable $$h$$ in terms of $$x$$ in the volume equation.
 * $$2x^2+4xh=100$$
 * $$x^2+2xh=50$$
 * $$2xh=50-x^2$$
 * $$h=\frac{50-x^{2}}{2x}$$
 * $$\begin{align}

V &=\left(x^{2}\right)\left(\frac{50-x^2}{2x}\right)\\ &=\frac{50x-x^3}{2} \end{align}$$
 * Find the derivative of the volume equation in order to maximize the volume
 * $$\frac{dV}{dx}=\frac{50-3x^2}{2}$$

 * Set $$\frac{dV}{dx}=0$$ and solve for $$x$$
 * $$\frac{50-3x^2}{2}=0$$
 * $$50-3x^2=0$$
 * $$3x^2=50$$
 * $$x=\pm\frac{\sqrt{50}}{\sqrt3}$$

 * Plug-in the $$x$$ value into the volume equation and simplify
 * $$\begin{align}

V &=\frac{1}{2}\left[50\sqrt{\frac{50}{3}}-\left(\sqrt{\frac{50}{3}}\right)^3\right]\\ &=68.04138174... \end{align}$$ Answer: $$V_{max}=68.04138174...$$

Volume Example II
It is desired to make an open-top box of greatest possible volume from a square piece of tin whose side is $$\alpha$$ by cutting equal squares out of the corners and then folding up the tin to form the sides. What should be the length of the side of a cut-out square? If we call the side length of the cut-out squares $$x$$, then each side of the base of the folded box is $$\alpha-2x$$, and the height is $$x$$. Therefore, the volume function is $$V(x)=x(\alpha-2x)^2=x(\alpha^2-4\alpha x+4x^2)=\alpha^2x-4\alpha x^2+4x^3$$. We must optimize the volume by taking the derivative of the volume function and setting it equal to 0. Since it does not change, $$\alpha$$ is treated as a constant, not a variable.
 * $$V(x)=\alpha^2x-4\alpha x^2+4x^3$$
 * $$\begin{align}

0&=V'(x)=\alpha^2-8\alpha x+12x^2\\ &= 12x^{2}-8\alpha x+\alpha^{2} \end{align}$$ We can now use the quadratic formula to solve for $$x$$:
 * $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
 * $$\begin{align}

x &=\frac{-(-8\alpha)\pm\sqrt{(-8\alpha)^2-4(12)(\alpha^2)}}{2(12)}\\ &=\frac{8\alpha\pm\sqrt{64\alpha^2-48\alpha^2}}{24}\\ &=\frac{8\alpha\pm\sqrt{16\alpha^2}}{24}\\ &=\frac{8\alpha\pm 4\alpha}{24} \end{align}$$
 * $$x=\frac{\alpha}{6},\frac{\alpha}{2}$$

We reject $$x=\frac{\alpha}{2}$$, since it is a minimum (it results in the base length $$\alpha-2x$$ being 0, making the volume 0). Therefore, the answer is $$x=\frac{\alpha}{6}$$.

Volume Example III
A cylindrical can is required to hold $$1.024\text{ L}$$ of liquid. Determine the measurements that will minimize the amount of material required to construct the can of specified volume. This is a classic problem for optimization. The problem is we do not have any specified measurements. Luckily enough, we are given SI units, so we can write what we are given accordingly, Recall what we are trying to minimize: "the amount of material required to construct the can of specified volume." This means the surface area needs to be minimized! The unfortunate part is we have two variables (it is not constant). Luckily, we can use the volume to write $$h$$ in terms of $$r$$:
 * Conversion factor: $$1\text{ L}=1000\text{ cm}^{3}$$. This means that through dimensional analysis (or basic algebra): $$1.024\text{ L}=1024\text{ cm}^{3}$$. Recall that this is the volume!
 * Volume of a cylinder: $$V=\pi r^{2}h$$
 * Surface area of a cylinder: $$S=2\pi rh+2\pi r^{2}$$
 * $$1024\text{ cm}^{3}=\pi r^{2}h$$
 * $$\Rightarrow \frac{1024\text{ cm}^{3}}{\pi {r\text{ cm}}^{2}}=\frac{1024}{\pi r^{2}}\text{ cm}=h$$

We can the substitute this information into the surface area formula and minimize (set the first derivative to zero):
 * $$S(r)=2\pi r\left(\frac{1024}{\pi r^{2}}\right)+2\pi r^{2}$$. Eliminate the $$\pi$$ and $$r$$.
 * $$\Rightarrow S(r)=2\cdot\frac{1024}{r}+2\pi r^{2}$$ Set the first derivative equal to zero.
 * $$\begin{align}

\Rightarrow0=S'(r)&=\frac{d}{dr}\left(\frac{2048}{r}+2\pi r^{2}\right)\\ &=\frac{d}{dr}\left(\frac{2048}{r}\right)+\frac{d}{dr}\left(2\pi r^{2}\right)\\ &=\frac{d}{dr}\left(2048\cdot{r}^{-1}\right)+2\pi\frac{d}{dr}\left(r^{2}\right)\\ &=2048\cdot\frac{d}{dr}\left({r}^{-1}\right)+2\pi\left(2r\right)\\ &=2048\cdot(-1){r}^{-2}+4\pi r\\ &=-\frac{2048}{{r}^{2}}+4\pi r\\ &=\frac{4\pi r^{3}-2048}{{r}^{2}} \end{align}$$ All that is left to do is find the critical values of the operation. This ultimately means to let the numerator equal zero and find what values of $$r$$ make the denominator equal to zero. Here is the operation for the numerator:
 * $$4\pi r^{3}-2048=0$$
 * $$\Leftrightarrow 4\pi r^{3}=2048$$
 * $$\Leftrightarrow r^{3}=\frac{2048}{4\pi}=\frac{512}{\pi}$$
 * $$\Leftrightarrow r=\sqrt[3]{\frac{512}{\pi}}=\frac{8}{\sqrt[3]{\pi}}\approx5.4623$$

Here is the operation for the denominator:
 * $${r}^{2}=0$$
 * $$r=\pm0$$

Let us note that we do not care about one of the critical values being equal to zero. This problem is asking us to find the value that makes the condition true but also minimized. While zero does minimize the amount of material used to construct the can, we cannot make a can if we do not show one. This solution would be interesting, however, if we were asked about the graph of $$S(r)$$. Nevertheless, our final answer may not necessarily be $$r\approx5.4623\text{ cm}$$ and $$h\approx10.9245\text{ cm}$$. We need to verify that the critical value we found is the correct answer. That is where we substitute values for $$r\in\left(0,\,\frac{8}{\sqrt[3]{\pi}}\right)$$ and $$r\in\left(\frac{8}{\sqrt[3]{\pi}},\,\infty\right)$$ into the first derivative:
 * For $$r\in\left(0,\,\frac{8}{\sqrt[3]{\pi}}\right)$$, let $$r=5$$: $$S'(5)=\frac{4\pi(5)^{3}-2048}{(5)^{2}}<0$$. For $$r\in\left(\frac{8}{\sqrt[3]{\pi}},\,\infty\right)$$, let $$r=6$$: $$S'(6)=\frac{4\pi(6)^{3}-2048}{(6)^{2}}>0$$.

Based on these results above, it is clear that $$r=\frac{8}{\sqrt[3]{\pi}}$$ is a value that minimizes the function such that it allows us to make a can. Keep in mind, if you do not show that the value we found is a minimum, then this value could also be a maximum. Imagine if that were the case, and we suggested the company to produce a can that would maximize the amount of material used. Potentially millions of dollars of material costs would offset the profit from the optimum. Either way, we have found our answers: Answer: the radius is $$5.4623\text{ cm}$$ and the height is $$10.9245\text{ cm}$$.

Sales Example
A small retailer can sell $$n$$ units of a product for a revenue of $$r(n)=8.1n$$ and at a cost of $$c(n)=n^3-7n^2+18n$$, with all amounts in thousands. How many units does it need to sell to maximize its profit? The retailer's profit is defined by the equation $$p(n)=r(n)-c(n)$$, which is the net revenue generated. The question asks for the maximum amount of profit, or the maximum of the above equation. As previously discussed, the maxima and minima of a graph are found when the slope of the said graph is equal to zero. To find the slope, one finds the derivative of the function of interest – here, $$p(n)$$. By using the subtraction rule $$p'(n)=r'(n)-c'(n)$$:
 * $$p(n)=r(n)-c(n)$$
 * $$p'(n)=\frac{d}{dn}[8.1n]-\frac{d}{dn}\left[n^3-7n^2+18n\right]=-3n^2+14n-9.9$$

Therefore, when $$-3n^{2}+14n-9.9=0$$ the profit will be maximized or minimized. Use the quadratic formula to find the roots, giving $$n=0.869,3.798$$. To find which of these is the maximum and minimum the function can be tested:
 * $$p(0.869)=-3.97321,p(3.798)=8.58802$$

Because we only consider the functions for all $$n\ge 0$$ (i.e., you can't have $$n=-5$$ units), the only points that can be minima or maxima are those two listed above. To show that $$3.798$$ is in fact a maximum (and that the function doesn't remain constant past this point) check if the sign of $$p'(n)$$ changes at this point.

Here, it does, and for $$n$$ greater than $$3.798$$, $$P'(n)$$ will continue to decrease. This demonstrates that the firm will be maximizing its profits at $$n=3.798$$. The retailer selling $$3.798$$ units would return a profit of $$$8,588.02$$. However, the answer is $$n=3.798$$.

Revenue Example
A widget firm has a linear demand function $$D(q)=-\frac{1}{5}q+10$$ where $$q$$ kilograms of widgets (in hundreds) are shipped and sold to customers. At what price should the widgets be sold such that the revenue is maximized? The revenue of a firm is calculated by multiplying the quantity sold to the price. Since $$D(q)$$ is the price, and $$q$$ is the quantity sold, $$R(q)=q\cdot D(q)$$. The question is asking us to find the price that will maximize the quantity. This means finding the quantity is the first goal. The final goal is the price found.

Our first goal is finding the quantity that will maximize the revenue. The revenue function is maximized if the slope of the tangent line is zero at a point and the sign of the slope around those points change sign from positive to negative. This is achieved through taking the derivative and applying the first derivative test. This is made easier when the revenue function distributes the $$q$$:
 * $$\begin{align}

R(q)&=q\cdot D(q)\\ &=q\cdot\left(-\frac{1}{5}q+10\right)\\ &=-\frac{1}{5}q^{2}+10q \end{align}$$ Let us now take the first derivative:
 * $$\frac{d}{dq}\left[R(q)\right]=\frac{d}{dq}\left[-\frac{1}{5}q^{2}+10q\right]$$
 * $$\begin{align}

R'(q)&=\frac{d}{dq}\left[-\frac{1}{5}q^{2}\right]+\frac{d}{dq}\left[10q\right]\\ &=-\frac{1}{5}\frac{d}{dq}\left[q^{2}\right]+10\frac{d}{dq}\left[q\right]\\ &=-\frac{1}{5}\left[2q\right]+10\left[1\right]\\ &=-\frac{2}{5}q+10 \end{align}$$ Set the first derivative equal to zero to find the critical values:
 * $$R'(q)=0=-\frac{2}{5}q+10$$
 * $$\Leftrightarrow-10=-\frac{2}{5}q$$
 * $$\Leftrightarrow25=q$$

Take note of the original revenue function. Because the revenue function is a quadratic that has its $$q^{2}$$ term multiplied by a factor less than 0, it is guaranteed there is only one critical value, and it must be a maximum. The calculus-based approach simply confirms this truth (using the first derivate test):
 * For any $$q<25$$, $$R'(q)>0$$. For any $$q>25$$, $$R'(q)<0$$. Therefore, the critical value $$q=25$$ must be a maximum.

All that is left to do is input this domain into the demand function to find the price the widgets should be sold at. This is a simple substitution exercise:
 * $$\begin{align}

D(25)&=-\frac{1}{5}(25)+10\\ &=-5+10\\ &=5 \end{align}$$ The widgets should be sold at $$$5$$ to maximize the revenue.

Distance of Function to Origin
The function $$f(x)=\sqrt{4-x}$$ is graphed in the $$xy$$ plane. At what point of the function should a particle living on it stop so that the straight line distance from that point to the origin is minimized? Let us write all the relevant equations we might want to keep in mind: The problem with optimization problems is that most of the time, you work with two variables as opposed to one. However, the main idea and focus is to write the two-variable equation into a single-variable equation. It is important to relate the variables here. The best way to do this is by using the set-up of the problem to your advantage.
 * $$D=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$$
 * $$y=f(x)$$

We want to leave the function at $$(x,y)$$ and minimize the distance required to reach the origin. This means we let $$\left(x_{2},y_{2}\right)=(x,y)$$. Recall that the origin is at $$(0,0)$$, so $$\left(x_{1},y_{1}\right)=(0,0)$$. With all the information we have here, let us go ahead and write the distance function in terms of $$x$$ the singular variable. Because $$y=f(x)$$:
 * $$\begin{align}

D(x)&=\sqrt{\left(x-0\right)^{2}+\left(f(x)-0\right)^{2}}\\ &=\sqrt{x^{2}+\left(\sqrt{4-x}\right)^{2}}\\ &=\sqrt{x^{2}+4-x}\\ &=\sqrt{x^{2}-x+4} \end{align}$$ We will show two ways to take the derivative: the standard way and the non-standard way. The standard way will require us to use the chain rule, so let us demonstrate this procedure:

Standard way

 * $$D(x)=\sqrt{x^{2}-x+4}$$
 * $$\Rightarrow\frac{d}{dx}\left[D(x)\right]=\frac{d}{dx}\left(\sqrt{x^{2}-x+4}\right)$$
 * $$\begin{align}

\Rightarrow D'(x)&=\frac{d}{dx}\left(\sqrt{x^{2}-x+4}\right)&\left(\text{Recall the chain rule: }\frac{d}{dx}\left[f\left(g(x)\right)\right]=\frac{df}{dg}\cdot\frac{dg}{dx}\right)\\ &=\frac{d}{dx}\left(x^{2}-x+4\right)\cdot\frac{df}{dg}\left(\sqrt{x^{2}-x+4}\right)\\ &=\left(2x-1\right)\cdot\frac{1}{2\sqrt{x^{2}-x+4}}\\ &=\frac{2x-1}{2\sqrt{x^{2}-x+4}} \end{align}$$ We now have the first derivative of the distance formula. To minimize the distance between $$(x,y)$$ and the origin, set the first derivate equal to zero. By now, you should know why we have to do this:
 * Let $$D'(x)=0$$.
 * $$\frac{2x-1}{2\sqrt{x^{2}-x+4}}=0$$ implies the numerator is zero. As such:
 * Let $$2x-1=0$$.
 * $$2x-1=0$$
 * $$\Leftrightarrow2x=1$$
 * $$\Leftrightarrow x=\frac{1}{2}$$
 * Recall that if the denominator can be set equal to zero, then the value for $$x$$ which makes that true is a critical value. As such, let $$2\sqrt{x^{2}-x+4}=0$$:
 * $$2\sqrt{x^{2}-x+4}=0$$
 * $$\Rightarrow\sqrt{x^{2}-x+4}=0$$
 * $$\Rightarrow x^{2}-x+4=0$$
 * There is no real value for $$x$$ which makes the above equation true. This can be shown using the determinant is less than zero: $$b^{2}-4ac<0$$. As such,

The only critical value is $$x=\frac{1}{2}$$. By now, you should be saying that we need to use the first derivative test:
 * For any $$x<\frac{1}{2}$$, $$D'(x)<0$$. For any $$x>\frac{1}{2}$$, $$D'(x)>0$$. This implies the function before the critical value is decreasing, and the function after the critical value is increasing. This implies the function at the critical value is a local minimum.

As such, the $$x$$-value that will minimize the distance between $$(x,y)$$ and the origin is $$x=\frac{1}{2}$$.

However, you have not finished the question yet. You still have to find the point of the function that the particle needs to stop at. This is simply a matter of substitution into the function $$f(x)$$:
 * $$\begin{align}

f\left(\frac{1}{2}\right)&=\sqrt{4-\left(\frac{1}{2}\right)}\\ &=\sqrt{\frac{8}{2}-\frac{1}{2}}\\ &=\sqrt{\frac{7}{2}} \end{align}$$ Most teachers in calculus classes don't mind if you leave the answer as $$\left(\frac{1}{2},\sqrt{\frac{7}{2}}\right)$$. However, for the minuscule number of teachers or testing organizations that care about rationalizing the denominator, the work for $$\sqrt{\frac{7}{2}}$$ is done below:
 * $$\begin{align}

\sqrt{\frac{7}{2}}&=\frac{\sqrt{7}}{\sqrt{2}}\\ &=\frac{\sqrt{7}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\\ &=\frac{\sqrt{7}\cdot\sqrt{2}}{2}\\ &=\frac{\sqrt{14}}{2}\\ \end{align}$$ This is the reason most teachers don't mind if the denominator is left as a square root; it distracts from the calculus (although slightly). This leaves us with the final answer: The particle must leave the function at $$\left(\frac{1}{2},\frac{\sqrt{14}}{2}\right)$$ to minimize the distance required to get to the origin.$$\blacksquare$$

Non-standard way: implicit differentiation
Only the implicit differentiation will be shown, since any calculations beyond that would result in the same procedures we outlined before. Recall $$D(x)=\sqrt{x^{2}-x+4}$$. Square both sides to return the function seen below:
 * $$\left[D(x)\right]^{2}=x^{2}-x+4$$

Taking the derivative of the above function with respect to $$x$$ will require us to implicitly differentiate (although this differentiation is neatly disguised). The calculations are demonstrated below:
 * $$\frac{d}{dx}\left[\left(D(x)\right)^{2}\right]=\frac{d}{dx}\left(x^{2}-x+4\right)$$
 * $$\begin{align}

\Rightarrow2D(x)\frac{dD}{dx}&=\frac{d}{dx}\left(x^{2}\right)-\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(4\right)\\ &=2x-1+0\\ \end{align}$$
 * $$2D(x)\frac{dD}{dx}=2x-1$$
 * $$\begin{align}

\Leftrightarrow\frac{dD}{dx}&=\frac{2x-1}{2D(x)}\\ &=\frac{2x-1}{2\sqrt{x^{2}-x+4}} \end{align}$$ Notice how the calculation above resulted in the same first derivative when done the standard way, meaning the calculations from herein will be the same. With this let us move on to the next problem.

Maximizing the Rectangular Area Formed by a Curved Shelf
Let there be a function $$f(x)=15\log(x)$$ that describes the curve of one part of a stylized shelf. The curve ends $$100$$ centimeters horizontal of the vertical wall. The manufacturer wants the front face of this portion to have one large maximum rectangular indentation whereby the top-left corner of the indentation forms a horizontal part that is parallel to the shelf's bottom. How far from the left-hand side of the shelving unit must the vertical wall formed in between be such that the condition set by the manufacturer is true? A calculator may be used here. This problem seems very weird. However, in actuality, this is a disguised variation of another common optimization problem, the maximum rectangular area formed by a function problem. The problem becomes much more manageable when thinking about it in those terms. What the manufacturer wants, after all, is a maximized indentation size. It is not asking about the volume of that size, only the part where you can find this is actually true.

The image shown on the right is actually the maximum area formed. However, if the problem did not provide an image (and no numbers on the axes), then one would not be approximating.

Let $$x$$ be the distance from the left-hand side of the shelving unit, and let the height of the rectangle formed be $$y=f(x)$$. Such a construction means that $$(100-x)$$ is the length of the rectangle while the height is $$15\log(x)$$. Therefore, the area function is
 * $$A(x)=15(100-x)\log(x)$$

We want to find the maximum of the function given herein. Let us find the derivative of the area and set it equal to zero:
 * $$A(x)=15(100-x)\log(x)$$
 * $$\Rightarrow\frac{d}{dx}\left[A(x)\right]=\frac{d}{dx}\left[15(100-x)\log(x)\right]$$
 * $$\begin{align}

\Rightarrow A'(x)&=15\frac{d}{dx}\left((100-x)\log(x)\right)&\left(\text{Recall the product rule: }\frac{d}{dx}\left[f(x)\cdot g(x)\right]=f\cdot\frac{dg}{dx}+g\cdot\frac{df}{dx}\right)\\ &=15\left[(100-x)\cdot\frac{d}{dx}\left(\log(x)\right)+\log(x)\cdot\frac{d}{dx}\left(100-x\right)\right]\\ &=15\left[\frac{100-x}{x\ln(10)}-\log(x)\right]\\ &=15\left[\frac{100-x}{x\ln(10)}-\frac{\ln(x)}{\ln(10)}\right]&\left(\text{Multiply by }\frac{x}{x}\text{ to the right fraction.}\right)\\ &=15\left[\frac{100-x}{x\ln(10)}-\frac{x\ln(x)}{x\ln(10)}\right]\\ &=15\left[\frac{100-x-x\ln(x)}{x\ln(10)}\right]=0\\ \end{align}$$ However, after finding the first derivative, it is evident that finding the value for $$x$$ in this specific situation is unknown to us at this time because we do not know about the Lambert-W Function. Therefore, a graphing calculator is acceptable to use now (typically, many teachers want to see as much work as possible before allowing you to use the calculator for solutions). If one does not have a graphing calculator, use Desmos or Geogebra or the many other online graphing calculators.

The answer obtained by those websites is $$x\approx 23.94721$$ or an exact answer of $$x=e^{W(100e)-1}$$. We know this is also the maximum that we are looking for because the derivative function is positive for $$x<23.94721$$ but negative after it. For those curious, the maximum area is $$1573.44277$$ centimeters-squared. Answer: $$x\approx 23.94721$$

Velocity in Different Regions
A light ray, according to Fermat's Principle of Least Time, travels in the fastest time over the least amount of distance (i.e. a light ray travels the path that results in the least amount of time wasted). The speed of a light wave is defined as the speed of light to the refractive index of the material traveled, or $$v=\frac{c}{n}$$. Let a light wave travel through two materials, starting at $$n_{1}$$ and ending at $$n_{2}$$ where $$n_{1}<n_{2}$$. The angle of incidence and refraction is given on the diagram to the right. What is the relationship between the light ray traveling through the two different materials? This does not seem like an optimization problem. In fact, it seems that trying to make this into an optimization problem would be unintuitive. However, this beautiful result in physics is but a clever discovery found by making this into an optimization problem.

Let us write what we know: All that is left to do is put it all together. We want to find the distance that light travelled. Let us review what we know. We are trying to manipulate $$\Lambda$$ so that the time travelled is as small as possible to reflect and model how light ravels through two different materials. Everything else that is not a function of $$\Lambda$$ is a constant. Since we are trying to model the amount of time it takes light to travel through two different materials, we should manipulate the distances so that it models time.
 * The distance between the point of incidence and the change in the material is $$\Upsilon$$ (The diagram has an italicized capital upsilon).
 * The distance that minimizes the travel time the most is $$\Lambda$$ (The diagram has an italicized capital lambda). This means the remaining horizontal distance, according to the diagram, is $$d-\Lambda$$.
 * The distance between the change in materials and the end of the refractive material is $$\Gamma$$ (The diagram has an italicized capital gamma).
 * The distance travelled by the incident ray is $$\sqrt{(\Lambda)^{2}+(\Upsilon)^{2}}$$.
 * The distance travelled by the refractive ray is $$\sqrt{(d-\Lambda)^{2}+(\Gamma)^{2}}$$.

Recall that speed is a ratio between distance and time $$\left(v=\frac{D}{\Delta t}\right)$$, so to model the amount of time light took to travel with the given distances, all it means is some rearrangements. Because light travelled through two different materials, the speeds are different. Let $$v_{1}>v_{2}$$ where $$v_{1}$$ is the speed of light traveling through the first material of index of refraction $$n_{1}$$ (and similar for $$v_{2}$$). The time light took to travel through the two materials is given as a function of $$\Lambda$$, so We now finally have an optimization problem. We want to minimize the amount of time travelled. Set the first derivative of $$T(\Lambda)$$ equal to zero. Remember, everything else besides $$\Lambda$$ is a constant.
 * $$t_{1}=\frac{\sqrt{\Lambda^{2}+\Upsilon^{2}}}{v_{1}}$$
 * $$t_{2}=\frac{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}{v_{2}}$$
 * $$T(\Lambda)=t_{1}+t_{2}=\frac{\sqrt{\Lambda^{2}+\Upsilon^{2}}}{v_{1}}+\frac{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}{v_{2}}$$
 * $$T(\Lambda)=\frac{\sqrt{\Lambda^{2}+\Upsilon^{2}}}{v_{1}}+\frac{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}{v_{2}}$$
 * $$\frac{d}{d\Lambda}\left[T(\Lambda)\right]=\frac{d}{d\Lambda}\left(\frac{\sqrt{\Lambda^{2}+\Upsilon^{2}}}{v_{1}}+\frac{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}{v_{2}}\right)$$
 * $$\begin{align}

0=T'(\Lambda)&=\frac{d}{d\Lambda}\left(\frac{\sqrt{\Lambda^{2}+\Upsilon^{2}}}{v_{1}}+\frac{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}{v_{2}}\right)\\ &=\frac{d}{d\Lambda}\left(\frac{\sqrt{\Lambda^{2}+\Upsilon^{2}}}{v_{1}}\right)+\frac{d}{d\Lambda}\left(\frac{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}{v_{2}}\right)\\ &=\frac{1}{v_{1}}\cdot\frac{d}{d\Lambda}\left(\sqrt{\Lambda^{2}+\Upsilon^{2}}\right)+\frac{1}{v_{2}}\cdot\frac{d}{d\Lambda}\left(\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}\right)\\ &=\frac{1}{v_{1}}\cdot\left(\frac{2\Lambda}{2\sqrt{\Lambda^{2}+\Upsilon^{2}}}\right)+\frac{1}{v_{2}}\cdot\left(\frac{2(d-\Lambda)\cdot(-1)}{2\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}\right)\\ &=\frac{1}{v_{1}}\cdot\left(\frac{\Lambda}{\sqrt{\Lambda^{2}+\Upsilon^{2}}}\right)-\frac{1}{v_{2}}\cdot\left(\frac{d-\Lambda}{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}\right) \end{align}$$ Recall what we are trying to solve for. We are not trying to solve for $$\Lambda$$; we are trying to model the situation. If we did have to solve for $$\Lambda$$, it would be a nightmare (we may even leave it as a non-trivial exercise for the reader), but nevertheless, we can begin modeling the situation. First, move the negative term to the other side:
 * $$\frac{1}{v_{1}}\cdot\left(\frac{\Lambda}{\sqrt{\Lambda^{2}+\Upsilon^{2}}}\right)-\frac{1}{v_{2}}\cdot\left(\frac{d-\Lambda}{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}\right)=0$$
 * $$\frac{1}{v_{1}}\cdot\left(\frac{\Lambda}{\sqrt{\Lambda^{2}+\Upsilon^{2}}}\right)=\frac{1}{v_{2}}\cdot\left(\frac{d-\Lambda}{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}\right)$$

The next thing we should attempt to do is rewrite the expression so that we clean-up those fractions (they are intimidating after all, and it would not be practical to model).

We can use the trigonometric functions to help us here in this department. Because the trigonometric functions are simply relations of the ratio of a right triangle, this would be a very good solution.

The normal is defined as the line perpendicular a given "surface". Because the distance $$\Upsilon$$ is parallel to the normal, the angle generated by that is also $$90^{\circ}$$. Because the horizontal distance $$d$$ light travelled is parallel to the "surface," the incident angle of the light, $$\theta_{1}$$, is an angle between the distance $$\Upsilon$$ and the incident light vector $$\mathbf{I}$$, and the light vector $$\mathbf{I}$$ is a transversal, by the corresponding angles theorem, the angle between the distance $$\Upsilon$$ and the incident light vector $$\mathbf{I}$$ is also $$\theta_{1}$$ (see the upper-left corner of the diagram for reference).

The result of this geometric derivation is simply this: we have found an angle of part of a right triangle. As a result, we can write the distance as a trigonometric equivalent We can use similar logic to show that As a result, we can rewrite the equations involved into something much more manageable and easy to calculate without measuring ratios:
 * $$\sin(\theta_{1})=\frac{\Lambda}{\sqrt{\Lambda^{2}+\Upsilon^{2}}}$$
 * $$\sin(\theta_{2})=\frac{d-\Lambda}{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}$$
 * $$\frac{1}{v_{1}}\cdot\left(\frac{\Lambda}{\sqrt{\Lambda^{2}+\Upsilon^{2}}}\right)=\frac{1}{v_{2}}\cdot\left(\frac{d-\Lambda}{\sqrt{(d-\Lambda)^{2}+\Gamma^{2}}}\right)$$
 * $$\Leftrightarrow\frac{1}{v_{1}}\cdot\sin(\theta_{1})=\frac{1}{v_{2}}\sin(\theta_{2})$$

We are technically not done with the problem. Although we may have related the two different light vectors, we did not model the situation in terms of index of refractions. This leaves us with the two final steps. Recall the speed of light going through a certain material of index of refraction $$n$$ is $$v=\frac{c}{n}$$. This is equivalent to saying $$\frac{1}{v}=\frac{n}{c}$$. Since the speeds relate to the index of refraction, we may rewrite the above equation as
 * $$\frac{n_{1}\cdot\sin(\theta_{1})}{c}=\frac{n_{2}\cdot\sin(\theta_{2})}{c}$$

Multiply both sides by $$c$$ and we finally found Snell's Law: Answer: $$n_{1}\sin(\theta_{1})=n_{2}\sin(\theta_{2})$$ Keep in mind, this result is still a derivative. However, we are allowed to model real life scenarios in terms of differential equations. Sometimes, it may be impossible to model situations without differential equations. We are allowed to write our equation like this (physics certainly does not care here). Finally, although this is not strictly an optimization problem, thinking about in that sense allowed us to obtain a nice rule of the world (a law in science). We proved this behavior, so it seems light will be consistent according to this equation. This in effect has allowed us to confirm in 2003 that negative index of refraction exists (even through experimental observations) and allowed us to discover properties of spin waves.

title: Calculus/Optimization


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User:Westfall~enwikibooks

User:Westfall

title: User:Westfall~enwikibooks


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User talk:Westfall~enwikibooks

User talk:Westfall

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Westfall. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Westfall~enwikibooks that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:51, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 05:38, 19 April 2015 (UTC)

title: User talk:Westfall~enwikibooks


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Talk:Linux Guide/Getting Help

You can also get help at the website of the Linux distribution. It usually contains F.A.Q and docs. Can also contain forums, mailing lists, etc. There are also IRC channels on many IRC networks such as freenode ( irc.freenode.net #linux )

title: Talk:Linux Guide/Getting Help