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Hospitals

On this page you can find a lot of new things about the hospitals where you can find asistance for tropical diseases and where you can research in this domain.You will find general things about the hospitals and then the link to the page of the hospital.

I.
The Hospital for Tropical Diseases in London is part of the University College Hospitals London (UCLH) NHS Trust, and associated with University College London. It is the only UK NHS hospital dedicated to the prevention, diagnosis and treatment of tropical diseases and travel-related infections.

It was founded on 8 March 1821 on board an ex-naval ship and moved onto dry land as the Dreadnought Seamen's Hospital in 1870 as part of the Royal Greenwich Hospital. The management of infectious disease moved in 1919 near to Euston Square, in central London, still under the Seamen's Hospital Society. The general in-patient wards at Greenwich continued until that hospital's closure in 1986 with special services for seamen and their families then provided by the 'Dreadnought Unit' at St Thomas's Hospital in Lambeth.

After several moves during the Second World War, The Hospital for Tropical Diseases was reestablished under the newly formed NHS in 1951 at the site of the St Pancras Hospital. Finally it moved in 1998 in to new purpose built premises within UCLH.

II.
In Romania, in Bucharest it exists a big hospital called "Spitalul Clinic"Dr. Victor Babeş" Boli infecţioase şi tropicale"(in english "Dr. Victor Babes Hospital of infectious and tropical diseases".

III.
Also in Bucharest, Romania it exists "Institutul de boli infecţioase"Dr. Matei Balş""(in english "Dr. Matei Bals Infectious diseases Institute".

IV.
CDC, ATLANTA, GEORGIA, USA

The Centers for Disease Control and Prevention (or CDC) is an agency of the United States Department of Health and Human Services based in unincorporated DeKalb County, Georgia adjacent to the campus of Emory University and east of the city of Atlanta. It works to protect public health and the safety of people, by providing information to enhance health decisions, and promotes health through partnerships with state health departments and other organizations. The CDC focuses national attention on developing and applying disease prevention and control (especially infectious diseases), environmental health, occupational safety and health, health promotion, prevention and education activities designed to improve the health of the people of the United States.

CDC Health Protection Goals
CDC is committed to achieving true improvements in people’s health. To do so, the agency is defining specific health impact goals to prioritize and focus its work and investments and measure progress.

Healthy People in Every Stage of Life
All people, and especially those at greater risk of health disparities, will achieve their optimal lifespan with the best possible quality of health in every stage of life.

Start Strong: Increase the number of infants and toddlers that have a strong start for healthy and safe lives. (Infants and Toddlers, ages 0-3 years). Grow Safe and Strong: Increase the number of children who grow up healthy, safe, and ready to learn. (Children, ages 4-11 years). Achieve Healthy Independence: Increase the number of adolescents who are prepared to be healthy, safe, independent, and productive members of society. (Adolescents, ages 12-19 years). Live a Healthy, Productive, and Satisfying Life: Increase the number of adults who are healthy and able to participate fully in life activities and enter their later years with optimum health. (Adults, ages 20-49 years). Live Better, Longer: Increase the number of older adults who live longer, high-quality, productive, and independent lives. (Older Adults, ages 50 and over).

Healthy People in Healthy Places
The places where people live, work, learn, and play will protect and promote their health and safety, especially those at greater risk of health disparities.

Healthy Communities: Increase the number of communities that protect, and promote health and safety and prevent illness and injury in all their members. Healthy Homes: Protect and promote health through safe and healthy home environments. Healthy Schools: Increase the number of schools that protect and promote the health, safety and development of all students, and protect and promote the health and safety of all staff. (e.g. – healthy food vending, physical activity programs). Healthy Workplaces: Promote and protect the health and safety of people who work by preventing workplace-related fatalities, illnesses, injuries, and personal health risks. Healthy Healthcare Settings: Increase the number of healthcare settings that provide safe, effective, and satisfying patient care. Healthy Institutions: Increase the number of institutions that provide safe, healthy, and equitable environments for their residents, clients or inmates. Healthy Travel and Recreation: Ensure that environments enhance health and prevent illness and injury during travel and recreation.

People Prepared for Emerging Health Threats
People in all communities will be protected from infectious, occupational, environmental, and terrorist threats. Preparedness goals will address scenarios that include natural and intentional threats. The first round of these scenarios will encompass influenza, anthrax, plague, emerging infections, toxic chemical exposure, and radiation exposure.

Pre-Event

Increase the use and development of interventions known to prevent human illness from chemical, biological, radiological agents, and naturally occurring health threats.

Decrease the time needed to classify health events as terrorism or naturally occurring in partnership with other agencies.

Decrease the time needed to detect and report chemical, biological, radiological agents in tissue, food or environmental samples that cause threats to the public’s health.

Improve the timeliness and accuracy of communications regarding threats to the public’s health.

Event

Decrease the time to identify causes, risk factors, and appropriate interventions for those affected by threats to the public’s health.

Decrease the time needed to provide countermeasures and health guidance to those affected by threats to the public’s health.

Post-Event

Decrease the time needed to restore health services and environmental safety to pre-event levels.

Improve the long-term follow-up provided to those affected by threats to the public’s health.

Decrease the time needed to implement recommendations from after-action reports following threats to the public’s health.

Healthy People in a Healthy World
People around the world will live safer, healthier and longer lives through health promotion, health protection, and health diplomacy.

Health Promotion: Global health will improve by sharing knowledge, tools and other resources with people and partners around the world. Health Protection: Americans at home and abroad will be protected from health threats through a transnational prevention, detection and response network. Health Diplomacy: CDC and the United States Government will be a trusted and effective resource for health development and health protection around the globe.

CDC Structure
CDC is one of the major operating components of the Department of Health and Human Services. CDC's major organizational components respond individually in their areas of expertise and pool their resources and expertise on cross-cutting issues and specific health threats. The agency comprises these major organizational components:

Office of the Director manages and directs the activities of the Centers for Disease Control and Prevention; provides overall direction to, and coordination of, the scientific/medical programs of CDC; and provides leadership, coordination, and assessment of administrative management activities.

The CDC is under the direction of Dr. Julie Louise Gerberding, M.D., M.P.H. Dr. Gerberding has been the director of the CDC since July 2002.

Coordinating Center for Environmental Health and Injury Prevention
National Center for Environmental Health/ Agency for Toxic Substances and Disease Registry (NCEH-ATSDR) provides national leadership in preventing and controlling disease and death resulting from the interactions between people and their environment.

National Center for Injury Prevention and Control (NCIPC) prevents death and disability from non occupational injuries, including those that are unintentional and those that result from violence.

Coordinating Center for Health Information and Services
National Center for Health Statistics (NCHS) provides statistical information that guides actions and policies to improve the health of the American people.

National Center for Public Health Informatics (NCPHI) provides national leadership in the application of information technology in the pursuit of public health.

National Center for Health Marketing (NCHM) provides national leadership in health marketing science and in its application to impact public health.

Coordinating Center for Health Promotion
National Center on Birth Defects and Developmental Disabilities (NCBDDD) provides national leadership for preventing birth defects and developmental disabilities and for improving the health and wellness of people with disabilities.

National Center for Chronic Disease Prevention and Health Promotion (NCCDPHP) prevents premature death and disability from chronic diseases and promotes healthy personal behaviors.

National Office of Public Health Genomics provides national leadership in fostering understanding of human genomic discoveries and how they can be used to improve health and prevent disease.

Coordinating Center for Infectious Diseases
National Center for Infectious Diseases (NCID) prevents illness, disability, and death caused by infectious diseases in the United States and around the world.

National Immunization Program (NIP) prevents disease, disability, and death from vaccine-preventable diseases in children and adults.

National Center for HIV, STD, and TB Prevention (NCHSTP) provides national leadership in preventing and controlling human immunodeficiency virus infection, sexually transmitted diseases, and tuberculosis.

Coordinating Office for Global Health provides national leadership, coordination, and support for CDC’s global health activities in collaboration with CDC’s global health partners.

Coordinating Office for Terrorism Preparedness & Emergency Response provides strategic direction for the Agency to support terrorism preparedness and emergency response efforts.

National Institute for Occupational Safety and Health (NIOSH) ensures safety and health for all people in the workplace through research and prevention.

 * CDC performs many of the administrative functions for the Agency for Toxic Substances and Disease Registry (ATSDR), a sister agency of CDC, and one of eight federal public health agencies within the Department of Health and Human Services. The Director of CDC also serves as the Administrator of ATSDR.

CDC Workforce
CDC’s budget for 2006 is $8.5 billion. Today the staff numbers nearly 15,000 (including 6,000 contractors and 840 Commissioned Corps officers) in 170 occupations. Engineers, entomologists, epidemiologists, biologists, physicians, veterinarians, behaviorial scientists, nurses, laboratorians, economists, health communicators, toxicologists, chemists, computer scientists, and statisticians—to name only a few—each are dedicated to the pursuit of public health.

CDC is headquartered in DeKalb County, Georgia, but it has 10 other locations in the United States and Puerto Rico. Those locations include Anchorage, Alaska; Cincinnati, Ohio; Fort Collins, Colorado; Hyattsville, Maryland; Morgantown, West Virginia; Pittsburgh, Pennsylvania; Research Triangle Park, North Carolina; San Juan, Puerto Rico; Spokane, Washington; and Washington, D.C. In addition, CDC staff are located in state and local health agencies, quarantine/border health offices at ports of entry, and 45 countries around the world, from Angola to Zimbabwe.

The work force is diverse and well qualified. More than a third of CDC’s employees are members of a racial or ethnic minority group, and women account for nearly 60 percent of CDC’s workforce. Nearly 40 percent of employees have a master’s degree; 25 percent have a Ph.D.; and 10 percent have medical degrees. The average age of a CDC worker is 46.

The CDC campus in Atlanta houses facilities for the research of extremely dangerous biological agents. This setting was well represented in the Dustin Hoffman film Outbreak, although the location depicted in the film was supposed to be the United States Army Medical Research Institute of Infectious Diseases bio-research facility. The CDC labs also figure prominently in the book "The Demon in the Freezer" by Richard Preston and "Virus Hunter" by C.J. Peters, former head of the Special Pathogens Branch at the CDC.

The CDC also conducts the Behavioral Risk Factor Surveillance System,  the world’s largest, on-going telephone health survey system.

CDC Timeline
CDC Timeline

CDC: Then and Now
On July 1, 1946, the Communicable Disease Center was established. Its founder was a visionary leader in public health, Dr. Joseph Mountin. The new agency, which was established the year after World War II ended, descended from the wartime agency, Malaria Control in War Areas. Established as a small branch of the U.S. Public Health Service, the CDC was located on the sixth floor of the Volunteer Building on Peachtree Street in Atlanta, Georgia, hundreds of miles from Washington, D.C., and other federal agencies. The organization took root deep in the South, once the heart of the malaria zone.

CDC initially focused on fighting malaria by killing mosquitoes. In fact, malaria was by far CDC’s most absorbing interest; during the first year of operations, 59 percent of its personnel were engaged in this effort. Among its 369 employees, the key jobs at CDC were originally entomology and engineering. In 1946, there were only seven medical officers on duty.

Back then, CDC’s budget was about $1 million. The insecticide DDT, available since 1943, was the primary weapon in the malaria fight, and CDC’s early challenges included obtaining enough trucks, sprayers, and shovels to wage the war on mosquitoes. In CDC’s initial years, more than six and a half million homes were sprayed, and an early organization chart was even drawn—somewhat fancifully—in the shape of a mosquito.

But CDC was soon to spread its wings. CDC founder Dr. Joseph Mountin continued to advocate for public health issues and to push for CDC to extend its responsibilities to many other communicable diseases. In 1947, CDC made a token payment of $10 to Emory University for 15 acres of land on Clifton Road in Atlanta, the home of CDC headquarters today. CDC employees collected the money to make the purchase. The benefactor behind the “gift” was Robert Woodruff, Chairman of the Board of the Coca-Cola Company. Woodruff had a long-time interest in malaria control; it had been a problem in areas where he went hunting. As you can see, malaria was the catalyst for the agency’s creation. The scene was now set for CDC to expand its home, its mission, and its reach.

Today, CDC is the nation's premier health promotion, prevention, and preparedness agency and a global leader in public health. During the past 60 years, its name has changed to reflect its more complex mission. While it’s still known by the initials CDC, the agency’s name today is Centers for Disease Control and Prevention.

In the six decades since its founding, CDC has grown dramatically: in staff, budget and mission. The world authority on communicable disease, CDC has broadened its focus to include chronic diseases, disabilities, injury control, workplace hazards, environmental health threats, and terrorism preparedness. Whereas malaria was once considered a threat to the country’s security, new threats have now emerged. CDC tackles emerging diseases and other health risks, including birth defects, West Nile virus, obesity, avian and pandemic flu, E. coli, auto wrecks, and bioterrorism, to name a few.

CDC remains committed to its vision of healthy people in a healthy world. Part of the Department of Health and Human Services, CDC applies research and findings to improve people’s daily lives and responds to health emergencies, and in 60 years, CDC has grown in size and stature, scope and science, and reputation and reach. Memories have been built and milestones achieved. World-class scientists work in world-class facilities. But while much has changed since 1946, the heart of CDC is still its people—dedicated and diligent, persevering and professional, making a difference in lives around the world.

The CDC is one of the few Bio-Safety Level 4 laboratories in the country, as well as one of only two "official" repositories of smallpox in the world. The second smallpox stores reside at the State Research Center of Virology and Biotechnology VECTOR in the Russian Federation, though it is possible that other countries may have obtained samples during the collapse of the Soviet Union.

CDC Data and Survey Systems

 * CDC Scientifc Data, Surveillance, Health Statistics, and Laboratory Information.
 * Behavioral Risk Factor Surveillance System.
 * Mortality Medical Data System.

CDC Publications

 * Comprehensive list of publications and products

 * State of CDC report

 * CDC Programs in Brief

 * Morbidity and Mortality Weekly Report

 * Emerging Infectious Disease Journal

Nosocomial Infection
Nosocomial infections, also referred to as hospital-acquired infections of health care-associated infections – are defined as infections that are not present or incubating in a patient at the time of admission to a hospital or health care facility. These infections affect 2 million Americans every year, costing 100,000 lives and adding $30.5 billion to the nation’s healthcare tab. According to the CDC, 91% of these deaths are preventable.

The most important types of nosocomial infections include urinary tract infections, pneumonia, and diarrhea, infections following surgery or invasive medical procedures, and maternal and newborn infections. • Urinary tract infections, the most prevalent type of nosocomial infection, account for approximately 35% of all nosocomial infections. Studies have demonstrated that these infections occur after urinary catheterization. • Surgical-incision-site infections account for 20% of nosocomial infections. • Pneumonia (ventilator-associated) is a prevalent type of nosocomial infection. The most susceptible individuals are those with chronic obstructive pulmonary disease and those utilizing mechanical ventilation. Pneumonia accounts for 15% of nosocomial infections. • Bloodstream infections account for another 15% of nosocomial infections.

The organisms causing most nosocomial infections usually come from the patient’s own body, known as endogenous flora. They can also come from contact with hospital staff, which occurs via cross-contamination, contaminated instruments and needles, or the environment (exogenous flora). Since patients are highly mobile and hospital stays are becoming increasingly shorter, patients are often discharged before the infection becomes apparent. As a consequence, it is often difficult to determine whether the source of the organism causing the infection is endogenous or exogenous.

Studies show that approximately 70% of bacteria that cause nosocomial infections are resistant to at least one antibiotic commonly used to treat them. The increased incidence of resistance is a common concern in health care. According to the CDC’s National Nosocomial Infection Surveillance (NNIS) System, multidrug-resistant pathogens have become increasingly problematic in recent years, especially in the critical care setting.

A 1987 study by the World Health Organization (WHO) found that the highest prevalence of nosocomial infections occur in intensive care units and acute care surgical and orthopedic wards. Not surprisingly, infection rates are higher among patients increased susceptibility because of old age and the severity of the underlying disease. Nosocomial infections add to functional disability, emotional stress, and may even lead to disabling conditions that reduce the quality of life.

Nosocomial infections can affect the quality of patient care and increase health care treatment costs. Health care institutions can prevent or reduce the incidence of many nosocomial infections by effectively incorporating an infection control program and by following the recommendations of the CDC. In an effort to improve the quality of health care for patients, it is imperative for health care professionals to understand clearly how infections can be transmitted, to focus on means of implementing preventive measures, and potentially to reduce these types of infections.

References: Nibley, M. (2007). Hospital infection control saves lives, cuts costs. Medical News Today. Retrieved September 30, 2007, from [www.medicalnewstoday.com]. Preventing nosocomial infections. Infection Prevention Guidelines. Retrieved September 24, 2007, from [www.reproline.jhu.edu/English/4morerh/4ip/IP_manual/20_Nosocomial.pdf]. Terrie, Y. C. (2006). Nosocomial infections: Impact on patient care. Pharmacy Times. Retrieved September 30, 2007, from

Infection Prevention in Hospitals
About 5 percent of hospital patients end up having a longer hospital stay because of nosocomial infections. According to the Centers for Disease Control and Prevention, each year about 2 million patients get at least one unwanted visitor -- an infection. What's more, each year nearly 90,000 people die from hospital-acquired infections.

In recent years, hospitals have worked hard to reduce these statistics, resulting in a 30 percent drop in the number of infections over the last decade. However, it still makes sense to do what you can to protect yourself. If you're worried about not feeling well enough to stay on top of things, ask a friend or relative to stay with you in the hospital and serve as your advocate. Paying attention to cleanliness and asking questions can help you have a shorter and healthier hospital stay. Here are some other tips: • Make sure doctors and nurses wash their hands before examining you. The CDC cites hand washing as the single most effective way to control the spread of disease. • Wash your own hands carefully after using the bathroom or handling soiled materials: Scrub for at least 15 seconds with warm, soapy water. • If you're receiving fluids through an intravenous catheter, let your nurse know if the dressing around it becomes wet or soiled. • Infections from urinary catheters are common, so these dressings should be clean and dry, and the catheter should not remain in longer than necessary. • Keep an eye on wound dressings and drainage tubes, and let your nurse know if they become loose or wet. • Ask friends and family not to visit if they're feeling ill.

• If you need surgery and are overweight, losing a few pounds before you go in the hospital can help reduce your risk of post-surgery infection. • Be extra vigilant if you have diabetes because high blood sugar increases your risk of infection. Work with your doctor to control your blood sugar before, during, and after your hospital stay. • Don't be afraid to ask questions! Understanding your treatment plan will make it easier for you to be involved in your own recovery.

Paying attention to cleanliness and asking questions can help you have a shorter and healthier hospital stay.

Hand hygiene Cleansing hands is the easiest way to reduce the risk of spreading germs that cause infections. •	Wash your hands after using the bathroom, blowing your nose, coughing, sneezing, and before eating. •	If you cannot get to a sink, please ask your health care provider for a waterless alcohol handrub. •	Please ask your health care team members if they have cleansed their hands. They remove germs from their hands with a waterless alcohol handrub or with soap and water to protect you.

How to wash your hands Use soap and warm, running water. 1.	Wet your hands. 2.	Put some soap on your hands. 3.	Rub your hands together for at least 15 to 20 seconds. Cover all surfaces, including between your fingers and under fingernails. 4.	Rinse with running water. 5.	Dry your hands with a paper towel. 6.	Use a paper towel to turn off the faucet.

Surgery site infections To help prevent infection at the site of your surgery, you may be given an antibiotic within 60 minutes before surgery which should be stopped within 24 hours after surgery (48 hours after heart surgery).

If you need hair removed from the surgery site, a member of the health care team should use clippers instead of a razor. This will help avoid cuts.

RERFERENCES: Montgomery, N. (2006). How to Prevent Hospital Infections. “A Healthy Me.” Retrieved October 1, 2007 from http://www.ahealthyme.com/topic/hospitalinfections. A Healthy Me. 2006.

How to Prevent Infections during Your Hospital Stay. (2007). Allina Health Systems Press. Retrieved October 1, 2007 from http://www.allinahealth.org/ac/transcript.nsf/alltopics/InfectionPreventionDuringHospitalStay

Further Information

 * CDC Emerging Infectious Diseases Laboratory -- Atlanta, Georgia

title: Hospitals


text: Template:German courses

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

title: Template:German courses


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BCP/Lilium orientale/Logs

2008 Logs

 * (no archived logs yet)

Recent Logs

 * --SB_Johnny | talk 11:39, 5 July 2007 (UTC)
 * --SB_Johnny | talk 10:15, 10 July 2007 (UTC)
 * --SB_Johnny | talk 10:04, 20 July 2008 (UTC)

title: BCP/Lilium orientale/Logs


text:
BCP/Salvia verticillata/Logs

2008 Logs

 * (no archived logs yet)

Recent Logs

 * SB Johnny-LM 11:14, 7 July 2007 (UTC)
 * SB Johnny-LM 00:11, 6 August 2007 (UTC) (from a datestamped photograph)
 * SB Johnny-LM 09:43, 8 September 2007 (UTC)
 * SB Johnny-LM 08:48, 27 September 2007 (UTC)
 * SB Johnny-LM 18:28, 17 November 2007 (UTC)
 * SB Johnny-LM 11:31, 20 June 2008 (UTC)
 * SB Johnny-LM 10:29, 14 July 2008 (UTC)

title: BCP/Salvia verticillata/Logs


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BCP/Salvia guaranitica/Logs

This page contains recent bloom logs for the plant Salvia guaranitica (Anise Sage).

Recent Logs

 * SB Johnny-LM 11:14, 7 July 2007 (UTC)
 * SB Johnny-LM 00:11, 6 August 2007 (UTC) (from a datestamped photograph)
 * SB Johnny-LM 10:40, 25 August 2007 (UTC)
 * SB Johnny-LM 09:41, 8 September 2007 (UTC)
 * SB Johnny-LM 07:36, 19 June 2008 (UTC)
 * SB Johnny-LM 10:27, 14 July 2008 (UTC)

=====Archived Logs===== (If there are more than 5 signatures under the recent logs heading, move the oldest logs to under this heading (generally leave the 3 most recent).
 * (no archived logs yet)

title: BCP/Salvia guaranitica/Logs


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BCP/Salvia vanhouttei

Salvia vanhouttei

title: BCP/Salvia vanhouttei


text:
BCP/Salvia vanhouttei/Logs

This page contains recent bloom logs for the plant Salvia vanhouttei (Salvia).

Recent Logs

 * --SB_Johnny | PA! 19:28, 28 July 2007 (UTC)
 * SB Johnny-LM 00:11, 6 August 2007 (UTC) (from a datestamped photograph)
 * SB Johnny-LM 10:41, 25 August 2007 (UTC)
 * --SB_Johnny | talk 09:41, 1 September 2007 (UTC)
 * --SB_Johnny | talk 21:04, 17 September 2007 (UTC)

=====Archived Logs=====

(If there are more than 5 signatures under the recent logs heading, move the oldest logs to under this heading (generally leave the 3 most recent).
 * (no archived logs yet)

title: BCP/Salvia vanhouttei/Logs


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BCP/Melilotus alba

Melilotus alba

title: BCP/Melilotus alba


text:
BCP/Melilotus alba/Logs

2007 Logs

 * Hazel 18:01, 5 August 2007 (UTC)
 * --Jomegat 00:43, 10 August 2007 (UTC)
 * --SB_Johnny | PA! 10:57, 11 August 2007 (UTC)
 * --SB_Johnny | talk 21:41, 17 September 2007 (UTC)

2008 Logs

 * Jomegat 03:29, 20 June 2008 (UTC)
 * Jomegat 23:07, 1 July 2008 (UTC)
 * Jomegat 04:45, 15 July 2008 (UTC)
 * Jomegat 19:45, 26 July 2008 (UTC)
 * --SB_Johnny | talk 10:55, 29 July 2008 (UTC)
 * Jomegat 19:08, 6 August 2008 (UTC)
 * Jomegat 02:35, 13 August 2008 (UTC)
 * Jomegat 20 August, 2008
 * Jomegat 23:16, 3 October 2008 (UTC)

2009 Logs

 * Jomegat 00:14, 26 June 2009 (UTC)
 * Jomegat 00:45, 15 July 2009 (UTC)
 * Jomegat 23:44, 18 July 2009 (UTC)
 * Jomegat 03:51, 2 August 2009 (UTC)

Recent Logs

 * Jomegat 19:44, 26 June 2010 (UTC)

title: BCP/Melilotus alba/Logs


text:
BCP/Monarda didyma/Logs

2007 Logs

 * --SB_Johnny | talk 12:29, 25 June 2007 (UTC)
 * --SB_Johnny | talk 10:51, 1 July 2007 (UTC)
 * SB Johnny-LM 00:11, 6 August 2007 (UTC) (from a datestamped photograph)
 * --Jomegat 00:41, 15 August 2007 (UTC)

2008 Logs

 * (no archived logs yet)

Recent Logs

 * --SB_Johnny | talk 08:15, 24 June 2008 (UTC)
 * --SB_Johnny | talk 10:18, 4 July 2008 (UTC)
 * --SB_Johnny | talk 10:01, 14 July 2008 (UTC)
 * Jomegat 16:39, 23 July 2008 (UTC)
 * Jomegat 01:00, 2 August 2008 (UTC)

title: BCP/Monarda didyma/Logs


text:
User:Sjaubert

fr:Utilisateur:Sjaubert

title: User:Sjaubert


text:
BCP/Veronicastrum sibiricum

Veronicastrum sibiricum

title: BCP/Veronicastrum sibiricum


text:
BCP/Veronicastrum sibiricum/Logs

2008 Logs

 * (no archived logs yet)

Recent Logs

 * SB Johnny-LM 10:46, 25 July 2007 (UTC)
 * SB Johnny-LM 10:46, 6 August 2007 (UTC)
 * SB Johnny-LM 09:03, 14 July 2008 (UTC)

title: BCP/Veronicastrum sibiricum/Logs


text: Template:Film School Current Poster 72px right


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User talk:Sjaubert

Hello Sjaubert, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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User:Naiku2003~enwikiversity

Naiku 2003

 * B-Tech student
 * Interested in recent developments in civil engineering
 * Anxious in Learning new things

Interests

 * Structural Analysis
 * Concrete Designs
 * Geo Informatics

title: User:Naiku2003~enwikiversity


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User talk:Naiku2003~enwikiversity

Hello Naiku2003, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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Your account will be renamed
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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.

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title: User talk:Naiku2003~enwikiversity


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User:Turk~enwikiversity

Hi, I am interesting to learn english and photography. I am already working as a photographer in Turkey.

title: User:Turk~enwikiversity


text:
User talk:Turk~enwikiversity

Hello Turk, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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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.

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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) 06:24, 19 April 2015 (UTC)

title: User talk:Turk~enwikiversity


text:
Topic talk:Gastroenterology

i really want to help but im not sure on how to make "lessons" and the like. can someone teach me or email me how to make lessons in th e same wiki-style?
 * Do not worry too much about style. If you are new to wiki, try Introduction to Wikiversity and some of the other pages that introduce you to wiki editing. --JWSchmidt 02:42, 12 August 2007 (UTC)

title: Topic talk:Gastroenterology


text:
Intellectual art/Automated vehicle battery exchange systems

Purpose
See: Hydrogen economy

Electric vehicles are more efficient than hydrogen powered vehicles. If methods can be devised in order to allow for quick charging or changing of batteries, this will allow vehicles to extend their range and therefore their practicality.

Mechanisms

 * Drive car into chamber, sort of like driving into a car wash
 * The car sits on an adjustable plate that can rotate and move the car side to side and forward and backwards
 * Lasers, image processing software, rfid or other wireless communications, or manual human entrance informs the system of the make and therefore size and dimensions of the car
 * Lasers aligns the car so that the batteries can be exchanged without damage to the system or vehicle
 * Depleted battery is removed using robotic components and automatically moved to charging area
 * Newly charged battery is placed into battery


text: Category:Exams

title: Category:Exams


text:
User talk:Bdunn

Hello Bdunn, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
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Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 02:39, 12 August 2007 (UTC)

title: User talk:Bdunn


text:
User talk:Mcoa.angel

Hello Mcoa.angel, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. When leaving comments for others to read, remember to sign and date; it helps everyone keep track of who is writing messages. The signature icon in the edit window makes it simple. To help you get started:
 * Take a guided tour and learn how to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research pages; and
 * Read and help develop our community policies.

Also, don't forget to experiment with the links to your left. Be bold, and see you around Wikiversity! --JWSchmidt 02:42, 12 August 2007 (UTC)

title: User talk:Mcoa.angel


text: File:Derivative as tangent.jpg

title: File:Derivative as tangent.jpg


text:
Calculus/Derivatives

Notation
We denote the derivative of a function $$f$$ at a number $$a$$ as $$f'(a)\,\!$$.

Definition
The derivative of a function $$f$$ at a number $$a$$ a is given by the following limit (if it exists): $$f'(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$$ An analagous equation can be defined by letting $$x=(a+h)$$. Then $$h=(x-a)$$, which shows that when $$x$$ approaches $$a$$, $$h$$ approaches $$0$$: $$f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$$

As the slope of a tangent line
Given a function $$y=f(x)\,\!$$, the derivative $$f '(a)\,\!$$ can be understood as the slope of the tangent line to $$f(x)$$ at $$x=a$$:

Example
Find the equation of the tangent line to $$y=x^2$$ at $$x=1$$.

Solution
To find the slope of the tangent, we let $$y=f(x)$$ and use our first definition: $$f'(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}\Rightarrow \lim_{h\rightarrow 0}\frac{h}\Rightarrow \lim_{h\rightarrow 0}\frac{h}\Rightarrow \lim_{h\rightarrow 0}\frac{h}\Rightarrow \lim_{h\rightarrow 0}{\color{Blue}(2a+h)}$$ It can be seen that as $$h$$ approaches $$0$$, we are left with $$f'(a)={\color{Blue}2a}\,\!$$. If we plug in $$1$$ for $$a$$: $$f'({\color{Red}1})=2({\color{Red}1})\Rightarrow {\color{Red}2}$$ So our preliminary equation for the tangent line is $$y={\color{Red}2}x+b$$. By plugging in our tangent point $$(1,1)$$ to find $$b$$, we can arrive at our final equation: $${\color{Red}1}=2({\color{Red}1})+b\Rightarrow -1=b$$ So our final equation is $$y=2x-1\,\!$$.

As a rate of change
The derivative of a function $$f(x)$$ at a number $$a$$ can be understood as the instantaneous rate of change of $$f(x)$$ when $$x=a$$.

The derivative as a function
So far we have only examined the derivative of a function $$f$$ at a certain number $$a$$. If we move from the constant $$a$$ to the variable $$x$$, we can calculate the derivative of the function as a whole, and come up with an equation that represents the derivative of the function $$f$$ at any arbitrary $$x$$ value. For clarification, the derivative of $$f$$ at $$a$$ is a number, whereas the derivative of $$f$$ is a function.

Notation
Likewise to the derivative of $$f$$ at $$a$$, the derivative of the function $$f(x)$$ is denoted $$f'(x)\,\!$$.

Definition
The derivative of the function $$f$$ is defined by the following limit: $$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$ Also, $$f'(x)=\lim_{h\rightarrow x}\frac{f(x)-f(h)}{x-h}$$ or $$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}$$

d(x^n)/dx
Consider the sequences:

$$y = x^2;\ y + \Delta y = (x + h)^2 = x^2 + 2xh + h^2$$

$$y = x^3;\ y + \Delta y = (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$$

$$y = x^4;\ y + \Delta y = (x + h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4$$

$$y = x^5;\ y + \Delta y = (x + h)^5 = x^5 + 5x^4h + 10x^3h^2 + 10x^2h^3 + 5xh^4 + h^5$$

$$y = x^6;\ y + \Delta y = (x + h)^6 = x^6 + 6x^5h\ +$$ many terms containing $$h^2$$

$$y = x^n;\ y + \Delta y = (x + \Delta x)^n = x^n + nx^{n-1}\Delta x\ +$$ many terms containing $$(\Delta x)^2$$

$$\Delta y = nx^{n-1}\Delta x\ +$$ many terms containing $$(\Delta x)^2$$

Therefore:

$$\frac{\Delta y}{\Delta x} = nx^{n-1}\ +$$ many terms with each and every term containing $$\Delta x$$

$$\frac{dy}{dx} = \lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}$$

$$\frac{d(x^n)}{dx} = nx^{n-1}$$ read as "derivative with respect to $$x$$ of $$x$$ to the power $$n$$."

Later it will be shown that this is valid for all real $$n$$, positive or negative, integer or fraction.

Examples
$$\frac{d(x^4)}{dx} = 4x^{(4-1)} = 4x^3$$

$$\frac{d(7x^5)}{dx} = 7\frac{d(x^5)}{dx} = 7(5x^4) = 35x^4$$

$$\frac{d(p^3)}{dx} = \frac{d(p^3)}{dp}$$$⋅$$$\frac{dp}{dx} = 3p^2$$$⋅$$$\frac{dp}{dx}$$

$$\frac{d(2t^2 - 7T^3 + x^7)}{dx} = 4t$$$⋅$$$\frac{dt}{dx} - 7(3T^2)$$$⋅$$$\frac{dT}{dx} + 7x^6$$

$$\frac{d(x^{\frac{1}{3}})}{dx} = \frac{1}{3}x^{\frac{1}{3}-1} = \frac{1}{3}x^{-\frac{2}{3}}$$

Derivative of cubic function
In the diagram there is a stationary point at $$x = p.$$ When $$x = p,$$ there are exactly 2 values of $$x$$ that produce $$f(x) = f(p).$$

Aim of this section is to derive the condition that produces exactly 2 values of $$x.$$

See Cubic function as product of linear function and quadratic.

The associated quadratic when $$x = p:$$

$$A = a$$

$$B = Ap + b = ap+b$$

$$C = Bp + c = app + bp + c.$$

Divide $$g(x) = ax^2 + (ap+b)x + (ap^2 + bp + c)$$ by $$(x-p).$$

This division gives a quotient of $$ax + (2ap+b)$$ and remainder of $$h(p) = 3ap^2 + 2bp + c.$$

Factor $$(x-p)$$ divides $$g(x)$$ exactly. Therefore, remainder $$h(p) = 0,$$ the desired condition.

When $$x = p,\ h(x) = 3ax^2 + 2bx + c,$$ the derivative of $$f(x).$$

Derivative of quartic function
The quartic function: $$y = f(x) = ax^4 + bx^3 + cx^2 + dx + e\ \dots\ (1)$$

In $$(1)$$ substitute $$(p + q)$$ for $$x\ \dots\ (2a).$$

In $$(1)$$ substitute $$(p - q)$$ for $$x\ \dots\ (3a).$$

$$(2a) + (3a):\ + 2apppp + 12appqq + 2aqqqq + 2bppp + 6bpqq + 2cpp + 2cqq + 2dp + 2e\ \dots\ (4)$$

$$(2a) - (3a):\ + 8apppq + 8apqqq + 6bppq + 2bqqq + 4cpq + 2dq\ \dots\ (5)$$

Reduce (4) and (5) and substitute Q for qq:

$$+ apppp + 6appQ + aQQ + bppp + 3bpQ + cpp + cQ + dp + e\ \dots\ (4a)$$

$$+ 4appp + 4apQ + 3bpp + bQ + 2cp + d\ \dots\ (5a)$$

Combine (4a) and (5a) to eliminate p and produce a function in Q:

$$C_{11}Q^{11} + C_{10}Q^{10} + C_9Q^9 +\ \cdots\ + C_2Q^2 + C_1Q + C_0$$ where:

$$C_{11} = (- 65536aaaaaaaaaabb)$$

$$C_{10} = (- 131072aaaaaaaaaabd - 229376aaaaaaaaabbc + 94208aaaaaaaabbbb)$$

$$\cdots\cdots$$

$$C_2 = (- 8192aaaaaabdeeee + 20480aaaaaacddeee \cdots - 288aabbbccccddd + 32aabbccccccdd)$$

$$C_1 = (- 4096aaaaaaddeeee - 4096aaaaabcdeeee \cdots + 64aabbcccccdde - 20aabbccccdddd)$$

Coefficient of interest is $$C_0$$ which is in fact the value $$Rs - Sr.$$ See Equal Roots of Quartic Function.

If $$C_0 == 0,\ q = Q = 0$$ is a solution and $$f(x)$$ contains at least 2 roots of format $$p+0, p-0.$$ In other words 2 equal roots where $$x = p.$$

If $$Q == 0,\ (4a)$$ and $$(5a)$$ become:

$$ap^4 + bp^3 + cp^2 + dp + e\ \dots\ (4b)$$

$$4ap^3 + 3bp^2 + 2cp + d\ \dots\ (5b)$$

$$(4b)$$ is equivalent to $$f(x)$$ and $$(5b)$$ is derivative of $$(4b).$$

Product Rule
Let $$y = u$$$⋅$$$v$$ where $$u = U(x);\ v = V(x)$$

$$y + \Delta y = (u + \Delta u)$$$⋅$$$(v + \Delta v) = u$$$⋅$$$v + u$$$⋅$$$\Delta v + v$$$⋅$$$\Delta u + \Delta u$$$⋅$$$\Delta v$$

$$\Delta y = u$$$⋅$$$\Delta v + v$$$⋅$$$\Delta u + \Delta u$$$⋅$$$\Delta v$$

$$\frac{\Delta y}{\Delta x} = u$$$⋅$$$\frac{\Delta v}{\Delta x} + v$$$⋅$$$\frac{\Delta u}{\Delta x} + \frac{(\Delta u)(\Delta v)}{\Delta x}$$

$$\frac{dy}{dx} = u$$$⋅$$$\frac{dv}{dx} + v$$$⋅$$$\frac{du}{dx}$$

Examples
Let $$y = x^{\frac{m}{n}}.$$ Calculate $$\frac{dy}{dx}$$

$$y^n = x^m$$

Differentiate both sides.

$$ny^{n-1}\cdot\frac{dy}{dx} = mx^{m-1}$$

$$\frac{dy}{dx} = \frac{mx^{m-1}}{ ny^{n-1}    }$$ $$ = \frac{mx^{m-1}}{ n(x^{\frac{m}{n}})^{n-1}    }$$ $$ = \frac{mx^{m-1}}{ nx^{m - \frac{m}{n}}    }$$ $$= \frac{m}{n}\cdot x^{m - 1 - m + \frac{m}{n} }$$ $$= \frac{m}{n}\cdot x^{\frac{m}{n} - 1}$$

This shows that $$\frac{d(x^n)}{dx}$$ as above is valid when $$n$$ is a positive fraction.

Let $$y = x^{-n} = \frac{1}{x^n}$$. Calculate $$\frac{dy}{dx}$$.

$$yx^n = 1$$

Differentiate both sides.

$$ynx^{n-1} + x^n\frac{dy}{dx} = 0$$

$$\frac{dy}{dx} = -\frac{ynx^{n-1}}{x^n} = -\frac{nx^{n-1}}{x^{n}x^n} = -nx^{n-1}x^{-2n} = -nx^{-n-1}$$

This shows that $$\frac{d(x^n)}{dx}$$ as above is valid for negative $$n$$.

Let $$y = x^{-\frac{m}{n}}$$. Calculate $$\frac{dy}{dx}$$.

$$y^n = x^{-m}$$

Differentiate both sides.

$$ny^{n-1}\frac{dy}{dx} = -mx^{-m-1}$$

$$\frac{dy}{dx} = -\frac{mx^{-m-1}}{ny^{n-1}} = -\frac{mx^{-m-1}}{n(x^{-\frac{m}{n}})^{n-1}}$$ $$ = -\frac{mx^{-m-1}}{nx^{-m+\frac{m}{n}}}$$ $$ = -\frac{m}{n}x^{-m-1+m-\frac{m}{n}}$$ $$ = -\frac{m}{n}x^{-\frac{m}{n}-1}$$

This shows that $$\frac{d(x^n)}{dx}$$ as above is valid when $$n$$ is a negative fraction.

Quotient rule
Used where $$y = \frac{U(x)}{V(x)}$$

$$y = \frac{u}{v}$$

$$yv = u$$

$$yv' + vy' = u'$$

$$vy' = u' - yv' = u' - \frac{uv'}{v}$$

$$y' = \frac{u'}{v} - \frac{uv'}{v^2} = \frac{vu'}{v^2} - \frac{uv'}{v^2} = \frac{vu' - uv'}{v^2}$$

sine(x)
$$y = \sin (x)$$

$$y + \Delta y = \sin (x + \Delta x) = \sin(x)\cos(\Delta x) + \cos(x)\sin(\Delta x)$$

$$\Delta y = \sin(x)\cos(\Delta x) + \cos(x)\sin(\Delta x) - \sin(x)$$

$$\frac{\Delta y}{\Delta x} = \sin(x)$$$⋅$$$\frac{\cos(\Delta x) - 1}{\Delta x} + \cos(x)$$$⋅$$$\frac{\sin(\Delta x)}{\Delta x} $$

The value $$\frac{\cos(\Delta x) - 1}{\Delta x}$$:

$$\lim_{\Delta x\rightarrow 0}\frac{\cos(\Delta x)-1}{\Delta x} = \lim_{\Delta x\rightarrow 0}\frac{-\sin(\Delta x)}{1} = 0$$ by L'Hôpital's rule.

The value $$\frac{\sin(\Delta x)}{\Delta x}$$:

$$\lim_{\Delta x\rightarrow 0}\frac{\sin(\Delta x)}{\Delta x} = \lim_{\Delta x\rightarrow 0}\frac{\cos(\Delta x)}{1} = 1$$ by L'Hôpital's rule.

$$\frac{dy}{dx} = \sin(x)$$$⋅$$$0 + \cos(x)$$$⋅$$$1 = \cos(x)$$

Proof of 2 limits
$$\lim_{\theta\rightarrow 0}\frac{\sin(\theta) }{\theta }:$$

In the diagram $$OA = OB = 1.$$ $$OC = OD = \cos(\theta).$$ $$CB = \sin(\theta).$$

Let $$S$$ be the area of a sector of a circle. Then $$\frac{S}{\pi r^2} = \frac{\theta}{2\pi}$$ and $$S = \frac{\theta r^2}{2}.$$

Area of sector $$COD = \frac{\theta \cos^2(\theta)}{2}.$$

Area of triangle $$OCB = \frac{\sin(\theta) \cos(\theta)}{2}.$$

Area of sector $$AOB = \frac{\theta \cdot 1^2}{2} = \frac{\theta}{2}.$$

Therefore $$\frac{\theta}{2}\ \ge\ \frac{\sin(\theta) \cos(\theta)}{2}\ \ge\ \frac{\theta \cos^2(\theta)}{2}.$$

$$\theta\ \ge\ \sin(\theta) \cos(\theta)\ \ge\ \theta \cos^2(\theta).$$

$$\frac{\theta}{\theta \cos(\theta)}\ \ge\ \frac{\sin(\theta) \cos(\theta)}{\theta \cos(\theta)}\ \ge\ \frac{\theta \cos^2(\theta)}{\theta \cos(\theta)}$$

$$\frac{1}{ \cos(\theta)}\ \ge\ \frac{\sin(\theta) }{\theta }\ \ge\ \cos(\theta)$$

$$\lim_{\theta\rightarrow 0}  \frac{1}{ \cos(\theta)}\ \ge\ \lim_{\theta\rightarrow 0}\frac{\sin(\theta) }{\theta }\ \ge\ \lim_{\theta\rightarrow 0}\cos(\theta)  $$

$$1 \ \ge\ \lim_{\theta\rightarrow 0}\frac{\sin(\theta) }{\theta }\ \ge\ 1 $$

Therefore $$\lim_{\theta\rightarrow 0}\frac{\sin(\theta) }{\theta } = 1.$$

$$\lim_{\theta\rightarrow 0} \frac{\cos(\theta) - 1}{\theta}:$$

$$\frac{\cos(\theta) - 1}{\theta}$$ $$= (-1)\cdot \frac{1 - \cos(\theta)}{\theta}$$ $$= (-1)\cdot \frac{1 - \cos(\theta)}{\theta} \cdot \frac{1 + \cos(\theta)}{1 + \cos(\theta)} $$ $$= (-1)\cdot \frac{ \sin(\theta)}{\theta} \cdot \frac{ \sin(\theta)}{1 + \cos(\theta)} $$

$$\lim_{\theta\rightarrow 0} \frac{\cos(\theta) - 1}{\theta}$$ $$=\lim_{\theta\rightarrow 0} (-1)\cdot \frac{ \sin(\theta)}{\theta} \cdot \frac{ \sin(\theta)}{1 + \cos(\theta)} $$ $$=(-1)\cdot \lim_{\theta\rightarrow 0} \frac{ \sin(\theta)}{\theta} \cdot \lim_{\theta\rightarrow 0} \frac{ \sin(\theta)}{1 + \cos(\theta)} $$ $$=(-1)\cdot 1 \cdot \frac{ 0}{1 + 1}  = 0.$$

cosine(x)
$$y = \cos(x) = \sqrt{1 - \sin^2(x)}$$

$$y^2 = 1 - \sin(x)$$$⋅$$$\sin(x)$$

Differentiate both sides:

$$2y$$$⋅$$$\frac{dy}{dx} = -(2\sin(x)\cos(x))$$

$$\frac{dy}{dx} = -\frac{\sin(x)\cos(x)}{\cos(x)} = -\sin(x).$$

tan(x)
$$y = \tan(x) = \frac{\sin(x)}{\cos(x)}$$

$$y\cos(x) = \sin(x)$$

Differentiate both sides:

$$y(-\sin(x)) + \cos(x)$$$⋅$$$\frac{dy}{dx} = \cos(x)$$

$$\cos(x)$$$⋅$$$\frac{dy}{dx} = \cos(x) + y\sin(x)$$

$$\frac{dy}{dx} = \frac{\cos(x) + \tan(x)\sin(x)}{\cos(x)} = 1 + \tan^2(x) = \sec^2(x)$$

arcsine(x)

$$y = \arcsin(x); \|x\| <= 1$$

$$x = \sin(y)$$

$$\frac{dx}{dy} = \cos(y)$$

$$\frac{dy}{dx} = \frac{1}{\cos(y)} = \frac{1}{\sqrt{1 - \sin ^2(y)}} = \frac{1}{\sqrt{1 - x^2}}$$

In the figure to the right you can see that the curves $$y = \arcsin(x),\ x = \sin(y)$$ are the same curve. However curve $$y = \arcsin(x)$$ is limited to $$\frac{\pi}{2} >= y >= \frac{-\pi}{2}.$$

The derivative $$\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$$ shows that the slope of $$y = \arcsin(x)$$ is $$1$$ when $$x = 0$$ and infinite when $$x = \pm1.$$

arccosine(x)

$$y = \arccos(x); \|x\| <= 1$$

$$x = \cos(y)$$

$$\frac{dx}{dy} = -\sin(y)$$

$$\frac{dy}{dx} = \frac{-1}{\sin(y)} = \frac{-1}{\sqrt{1 - \cos ^2(y)}} = \frac{-1}{\sqrt{1 - x^2}}$$

In the figure to the right you can see that the curves $$y = \arccos(x),\ x = \cos(y)$$ are the same curve. However curve $$y = \arccos(x)$$ is limited to $$\pi >= y >= 0.$$

The derivative $$\frac{dy}{dx} = \frac{-1}{\sqrt{1 - x^2}}$$ shows that the slope of $$y = \arccos(x)$$ is $$-1$$ when $$x = 0$$ and infinite when $$x = \pm1.$$

arctan(x)

$$y = \arctan(x)$$

$$x = \tan(y)$$

$$\frac{dx}{dy} = \sec^2(y)$$

$$\frac{dy}{dx} = \frac{1}{\sec^2(y)} = \cos^2(y) = \frac{1}{1 + x^2}$$

In the figure to the right you can see that the curves $$y = \arctan(x),\ x = \tan(y)$$ are the same curve. However curve $$y = \arctan(x)$$ is limited to $$\frac{\pi}{2} > y > \frac{-\pi}{2}.$$

The derivative $$\frac{dy}{dx} = \frac{1}{\sqrt{1 + x^2}}$$ shows that the slope of $$y = \arctan(x)$$ is $$1$$ when $$x = 0$$ and $$0$$ when $$x = \pm \infty$$

a^x or $$a^x$$
$$y = a^x$$

$$y + \Delta y = a^{x + \Delta x} = a^xa^{\Delta x}$$

$$\Delta y = a^xa^{\Delta x} - a^x = a^x(a^{\Delta x} - 1)$$

$$\frac{\Delta y}{\Delta x} = a^x \cdot \frac{a^{\Delta x} - 1}{\Delta x}$$

Consider the value $$\frac{a^{\Delta x} - 1}{\Delta x}$$ specifically $$\lim_{\Delta x\rightarrow 0}\frac{a^{\Delta x} - 1}{\Delta x}$$. L'Hôpital's rule cannot be used here because $$\frac{d(a^x)}{dx}$$ is what we are trying to find.

$$a^{\frac{1}{2}} = \sqrt{a}$$

$$(a^{\frac{1}{2}})^{\frac{1}{2}} = \sqrt{\sqrt{a}}$$

$$a^{\frac{1}{2^3}} = \sqrt{\sqrt{\sqrt{a}}}$$

$$a^{\frac{1}{2^n}} = \sqrt{a}$$ taken $$n$$ times.

We will look at the expression $$\frac{a^{\Delta x} - 1}{\Delta x}$$ for different values of $$a$$ with $$\Delta x = \frac{1}{2^{70}} = 8.5e-22$$ (approx) in which case the expression becomes $$(a^{1/2^{70}} - 1)*(2^{70})$$ where $$a^{1/2^{70}}$$ is $$\sqrt[70]{a}$$ or $$\sqrt{a}$$ taken $$70$$ times. This approach is used here because function  can be written so that it does not depend on logarithmic or exponential operations.

Compare the values  with  :

We know that $$8 = 2^3;\ 32 = 2^5;\ 128 = 2^7$$. The values  are behaving like logarithms. In fact $$\lim_{\Delta x\rightarrow 0}\frac{a^{\Delta x} - 1}{\Delta x}$$ is the natural logarithm of $$a$$ written as $$\ln(a).$$

Figure 5 contains graphs of $$\frac{a^x - 1}{x}$$ for $$a = 2,\ 8,\ 32,\ 128$$ with graph of $$\frac{e^x - 1}{x}$$ included for reference.

All values of $$x$$ are valid for all curves except where $$x = 0.$$

The correct value of $$\ln(2)$$ is:

Our calculation produced:

accurate to 21 places of decimals, not bad for one line of simple python code using high-school math.

This method for calculation of $$\ln(a)$$ supposes that function  is available.

Programming language python interprets expression  as $$a^b.$$ Therefore, in python, $$\ln(2)$$ can be calculated in accordance with the basic definition above:

When $$y = a^x, \frac{dy}{dx} = a^x \cdot \ln(a).$$

The base of natural logarithms is the value of $$a$$ that gives $$\lim_{\Delta x\rightarrow 0}\frac{a^{\Delta x} - 1}{\Delta x} = 1.$$

This value of $$a$$, usually called $$e, = 2.718281828459045235360287471352662497757247093699959574.....$$

When $$a = e, \frac{d(e^x)}{dx} = e^x$$$⋅$$$\ln(e) = e^x$$$⋅$$$1 = e^x.$$

ln(x)
$$y = ln(x)$$

$$x = e^y$$

$$\frac{dx}{dy} = e^y$$

$$\frac{dy}{dx} = \frac{1}{e^y} = \frac{1}{x}$$

Examples
$$y = \ln(ax)$$

$$\frac{dy}{dx} = \frac{d}{dx}\ln(ax) = \frac{d}{dx}(\ln(a) + \ln(x)) = \frac{d}{dx}\ln(a) + \frac{d}{dx}\ln(x) = \frac{1}{x}$$

$$y = \ln(x^a)$$

$$\frac{dy}{dx} = \frac{d}{dx}\ln(x^a) = \frac{d}{dx}(a \cdot \ln(x)) = a \cdot \frac{d}{dx}\ln(x) = \frac{a}{x}$$

$$y = x^{\frac{m}{n}}$$ Calculate $$\frac{dy}{dx}$$

$$y^n = x^m$$

$$n\cdot \ln(y) = m\cdot \ln(x)$$

$$n\cdot\frac{1}{y}\cdot\frac{dy}{dx} = m\cdot \frac{1}{x}$$

$$\frac{dy}{dx} = m\cdot \frac{1}{x}\cdot\frac{y}{n}$$ $$= \frac{m}{n}\cdot\frac{x^{\frac{m}{n}}}{x}$$ $$= \frac{m}{n}\cdot x^{\frac{m}{n} - 1 }$$

Careful manipulation of logarithms converts exponents into simple constants.

Chain rule
Used where $$y = G(H(I(x)))$$

Examples
$$y = \cos(2x)$$

Let $$y = \cos(u)$$ where $$u = 2x$$.

 * $$\begin{align}

\frac{dy}{dx} =& \frac{dy}{du}\cdot\frac{du}{dx}\\ =& -\sin(u)\cdot    2\\ =& -2\sin(2x)\\ \end{align}$$

$$y = \sin^2(\ln(5x))$$

Let $$y = u^2$$ where $$u = \sin(v);\ v = \ln(w);\ w = 5x$$.

 * $$\begin{align}

\frac{dy}{dx} =& \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}\\ =& 2u \cdot \cos(v) \cdot \frac{1}{w} \cdot 5\\ =& 2\sin(v) \cdot \cos(\ln(w)) \cdot \frac{1}{5x} \cdot 5\\ =& 2\sin(\ln(w)) \cdot \cos(\ln(5x)) \cdot \frac{1}{x}\\ =& 2\sin(\ln(5x)) \cdot \cos(\ln(5x)) \cdot \frac{1}{x}\\ \end{align}$$

Shape of curves
The first derivative of $$f(x) = y'$$ or $$f'(x).$$ As shown above, $$f'(x)$$ at any point $$x_1$$ gives the slope of $$f(x)$$ at point $$x_1.$$

$$f'(x_1)$$ is the slope of $$f(x)$$ when $$x = x_1.$$

In the example to the right, $$y = f(x) = x^2 - x - 2$$ and $$y' = f'(x) = 2x - 1.$$

Of special interest is the point at which $$f'(x)$$ or slope of $$f(x) = 0.$$

When $$f'(x) = 0,\ x = 0.5$$ and $$f(x) = 0.5^2 - 0.5 - 2 = -2.25.$$

The point $$(0.5, -2.25)$$ is called a critical point or stationary point of $$f(x).$$ Because $$y'$$ has exactly one solution for $$y' = 0,\ f(x)$$ has exactly one critical point.

The value of $$y$$ at point $$(0.5, -2.25)$$ is less than $$y$$ at both $$(0, -2),\ (1,-2).$$ Therefore the critical point $$(0.5, -2.25)$$ is a minimum of $$f(x).$$

In this curve $$y = x^2 - x - 2,$$ the point $$(0.5, -2.25)$$ is both local minimum and absolute minimum.

In the example to the right, $$y = f(x) = \frac{2x^3 + 3x^2 - 12x - 8}{8}$$ and $$y' = f'(x) = \frac{6x^2 + 6x - 12}{8}.$$

Of special interest are the points at which $$f'(x)$$ or slope of $$f(x) = 0.$$

When $$f'(x) = 0,\ x = -2$$ and $$f(x) = 1.5$$ or

When $$f'(x) = 0,\ x = 1$$ and $$f(x) = -1\frac{7}{8}.$$

The points $$(-2, 1.5),\ (1, -1\frac{7}{8})$$ are critical or stationary points of $$f(x).$$ Because $$y'$$ has exactly two real solutions for $$y' = 0,\ f(x)$$ has exactly two critical points.

Slope of $$f(x)$$ to the left of $$(-2, 1.5)$$ is positive and adjacent slope of $$f(x)$$ to the right of $$(-2, 1.5)$$ is negative. Therefore point $$(-2, 1.5)$$ is local maximum. Point $$(-2, 1.5)$$ is not absolute maximum.

Adjacent slope of $$f(x)$$ to the left of $$(1, -1\frac{7}{8})$$ is negative and slope of $$f(x)$$ to the right of $$(1, -1\frac{7}{8})$$ is positive. Therefore point $$(1, -1\frac{7}{8})$$ is local minimum. Point $$(1, -1\frac{7}{8})$$ is not absolute minimum.

Electric water heater
A cylindrical water heater is standing on its base on a hard rubber pad that is a perfect thermal insulator. The vertical curved surface and the top are exposed to the free air. The design of the cylinder requires that the volume of the cylinder should be maximum for a given surface area exposed to the free air. What is the shape of the cylinder?

Let the height of the cylinder be $$h$$ and let $$h = Kr$$ where $$r$$ is the radius and $$K$$ is a constant.

Surface of cylinder $$ = S = \pi r^2 + 2\pi r h = \pi r^2 + 2\pi r (Kr) = \pi r^2 + 2\pi r^2K$$

Volume of cylinder $$ = V = \pi r^2h = \pi r^2(Kr) = \pi r^3K$$

$$K = \frac{S - \pi r^2}{2\pi r^2}$$

$$V = \pi r^3 \cdot \frac{S - \pi r^2}{2\pi r^2} = \frac{r(S - \pi r^2)}{2} = \frac{rS - \pi r^3}{2}$$

$$\frac{dV}{dr} = \frac{S - 3\pi r^2}{2}$$

For maximum volume, $$\frac{dV}{dr} = 0$$

Therefore $$S - 3\pi r^2 = 0$$

$$S = 3\pi r^2;\ \pi r^2(1 + 2K) = 3\pi r^2;\ 1 + 2K = 3;\ K =1.$$

The height of the cylinder equals the radius.

Square and triangle
A square of side $$h$$ has perimeter $$P_s = 4h$$ and area $$A_s = h^2.$$

An equilateral triangle of side $$s$$ has perimeter $$P_t = 3s$$ and area $$A_t = \frac{\sqrt{3}}{4}s^2.$$

$$P_s + P_t = 50.$$

Total area $$A = A_s + A_ t$$ and $$A$$ must be minimum. What is the value of $$s$$?

$$4h + 3s = 50$$

$$h = \frac{50 - 3s}{4}$$

$$A_s = h^2 = (\frac{50 - 3s}{4})^2$$ $$= \frac{2500 - 300s + 9s^2}{16}$$

$$A = A_s + A_t$$ $$= \frac{2500 - 300s + 9s^2}{16} + \frac{\sqrt{3}}{4}s^2$$ $$= \frac{1}{16}(2500 - 300s + 9s^2 + 4\sqrt{3}s^2)$$

For minimum $$A,\ \frac{dA}{ds} = 0.$$

$$-300 + 2(9 + 4\sqrt{3})s = 0$$

$$(9 + 4\sqrt{3})s = 150$$

$$s = \frac{150}{9 + 4\sqrt{3}}$$ $$= 9.417\dots$$

County Road

Town B is 40 miles East and 50 miles North of Town A. The county is going to construct a road from Town A to Town B. Adjacent to Town A the cost to build a road is $500k/mile. Adjacent to Town B the cost to build a road is $200k/mile. The dividing line runs East-West 30 miles North of Town A. Calculate the position of point C so that the cost of the road from Town A to Town B is minimum.

Let point $$C = (x,30).$$

Then distance from Town A to point $$C = \sqrt{x^2 + 30^2} = \sqrt{x^2 + 900}.$$

Distance from Town B to point $$C = \sqrt{(40-x)^2 + 20^2} $$ $$= \sqrt{1600 - 80x + x^2 + 400} $$ $$= \sqrt{2000 - 80x + x^2}.$$

Cost $$= 5\sqrt{x^2 + 900} + 2\sqrt{2000 - 80x + x^2}$$ in units of $100k.

For minimum cost $$f'(x) = 0.$$

$$5\cdot\frac{1}{2}\cdot\frac{2x}{\sqrt{x^2 + 900}} + 2\cdot\frac{1}{2}\cdot\frac{-80+2x}{\sqrt{2000-80x+x^2}} = 0$$

$$\frac{5x}{\sqrt{x^2+900}} + \frac{-80+2x}{\sqrt{2000-80x+x^2}} = 0$$

$$5x\sqrt{2000-80x+x^2} + (-80+2x)\sqrt{x^2+900} = 0$$

$$x = 10.52322823517\dots$$

Cardboard box

A piece of cardboard of length $$4\ ft$$ and width $$3\ ft$$ will be used to make a box with a top. Some waste will be cut out of the piece of cardboard and the remaining cardboard will be folded to make a box so that the volume of the box is maximum.

What is the height of the box?

$$2h + l = 3$$

$$2w + 2h = 4;\ w + h = 2$$

$$V = lwh = (3 - 2h)(2 - h)h = 2h^3 - 7h^2 + 6h$$

For maximum volume $$\frac{dV}{dh} = 6h^2 - 14h + 6 = 0.$$

$$3h^2 - 7h + 3 = 0$$

$$h = \frac{7 - \sqrt{13}}{6} = 6.788\dots$$ inches.

Solving ellipse at origin

An ellipse with center at origin has equation: $$Ax^2 + Bxy + Cy^2 + F = 0\ \dots\ (1)$$

Given values $$A, B, C, F$$ calculate:

 * length of major axis

 * length of minor axis.

In Figure 1 $$t$$ is any line from origin to ellipse and $$\theta$$ is angle between $$X$$ axis and $$t.$$

Aim of this section is to calculate $$\theta$$ so that length of $$t$$ is maximum, in which case length of major axis = $$2t.$$

Let $$c = \cos(\theta)$$ and $$s = \sin(\theta).$$

Then $$x = t \cos(\theta) = tc,$$ and $$y = t \sin(\theta) = ts.$$

Substitute these values in $$(1):$$

$$Attcc + Bttcs + Cttss + F = 0\ \dots\ (2)$$

Calculate $$t' = \frac{dt}{d\theta}$$

$$A(tt(-2cs) + cc2tt') + B(tt(cc-ss) + cs2tt') + C(tt(2sc) + ss2tt') = 0$$

$$Att(-2cs) + Acc2tt' + Btt(cc-ss) + Bcs2tt' + Ctt(2sc) + Css2tt' = 0$$

$$-2Attcs + Acc2tt' + Bttcc-Bttss + Bcs2tt' + 2Cttsc + 2Csstt' = 0$$

$$ + Acc2tt' + 2Bcstt' + 2Csstt' - 2Attcs + Bttcc - Bttss + 2Cttsc = 0$$

$$ + t'(Acc2t + 2Bcst + 2Csst) = + 2Attcs - Bttcc + Bttss - 2Cttsc$$

$$t' = \frac{+ 2Attcs - Bttcc + Bttss - 2Cttsc}{(Acc2t + 2Bcst + 2Csst)}$$

For maximum or minimum $$t:$$

$$2Attcs - Bttcc + Bttss - 2Cttsc = 0$$

$$2Acs - Bcc + Bss - 2Csc = 0$$

$$2Acs - 2Csc = Bcc - Bss $$

Square both sides, substitute $$1 - ss$$ for $$cc$$ and result is:

$$aS^2 + bS + c = 0\ \dots\ (3)$$ where:

$$S = \sin^2(\theta)$$

$$a = (+ 4AA - 8AC + 4BB + 4CC)$$

$$b = -a$$

$$c = BB$$

An example
Let equation of ellipse be: $$55x^2 - 24xy + 48y^2 - 2496 = 0$$ The solutions of quadratic equation $$(3)$$ are $$.36$$ or $$.64$$.

Therefore $$\sin(\theta) = \pm 0.6$$ or $$\sin(\theta) = \pm 0.8$$.

From $$(2)$$ above: $$t = \sqrt{\frac{-F}{Acc + Bcs + Css}}$$

Minimum value of $$t = 6.244997998398398.$$ Length of minor axis $$= 2 * 6.244997998398398$$

Maximum value of $$t = 7.999999999999999.$$ Length of major axis $$= 8 * 2$$

The car jack

In triangle $$ABC$$ to the right:

 * length $$AB = 10$$ inches,

 * length $$BC = x$$ inches and is horizontal,

 * length $$CA = y$$ inches and is vertical,

 * $$\angle ABC = \theta.$$

Point $$B$$ is moving towards point $$C$$ at the rate of $$5$$ inches$$/$$minute.

Vertical motion
[[File:0216_0901CarJack3.png|thumb|400px| Figure 3: Curves and values associated with car jack.

When $$x = 1,\ \frac{dy}{dx} = 0.1\dots$$ When $$\theta = 45^\circ,\ \frac{dy}{dx} = 1$$ When $$x = 9,\ \frac{dy}{dx} = 2.06\dots$$ ]] At what rate is point $$A$$ moving upwards:

(a) when $$x = 9$$?

(b) when $$x = 1$$?

(c) when $$\theta = 45^\circ$$?

We have to calculate $$\frac{dy}{dt}$$ when $$\frac{dx}{dt}$$ is given.

$$\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}$$

$$x^2 + y^2 = 10^2$$ (equation of circle)

$$y = \sqrt{100 - x^2}$$

$$y^2 = 100 - x^2$$

$$2y\cdot\frac{dy}{dx} = -2x$$

$$\frac{dy}{dx} = \frac{-2x}{2y} = \frac{-x}{\sqrt{100 - x^2}}$$

For convenience we'll use the negative value of the square root and say that $$\frac{dy}{dx} = \frac{x}{\sqrt{100 - x^2}}.$$

Relative to line $$BC:$$

When $$x = 9,\ \frac{dy}{dx} = 2.06,\ \frac{dy}{dt} = 2.06\cdot5 = 10.3$$ inches$$/$$minute.

When $$x = 1,\ \frac{dy}{dx} = 0.1,\ \frac{dy}{dt} = 0.1\cdot5 = 0.5$$ inches$$/$$minute.

When $$\theta = 45^\circ, x = 10\cdot\cos 45^\circ = 7.071$$ and $$\frac{dy}{dt} = 1\cdot 5 = 5$$ inches$$/$$minute.

This example highlights the mechanical advantage of this simple but effective tool. When the top of the jack is low, it moves quickly. As the jack takes more and more weight, the top of the jack moves more slowly.

Change of area

At what rate is the area of $$\Delta ABC$$ changing when

(i) $$x = 9$$?

(ii) $$x = 1$$?

(iii) $$\theta = 45^{\circ}$$?

$$\frac{da}{dt} = \frac{da}{dx}\cdot\frac{dx}{dt}$$ where $$a$$ is area of $$\Delta ABC$$ and $$\frac{dx}{dt} = 5$$ inches $$/$$ minute.

$$a = \frac{b\cdot h}{2} = \frac{x \cdot y}{2} = \frac{x\sqrt{100 - x^2}}{2}$$

Calculate $$\frac{da}{dx}:$$

$$2a = x\sqrt{100 - x^2}$$

$$4a^2 = x^2(100 - x^2) = 100x^2 - x^4$$

$$8a\cdot\frac{da}{dx} = 200x - 4x^3$$

$$\frac{da}{dx} = \frac{200x - 4x^3}{8a} = \frac{50x - x^3}{2a} = \frac{50x - x^3}{ x\sqrt{100 - x^2} }$$ $$ = \frac{50 - x^2}{ \sqrt{100 - x^2} }$$

When $$x = 9,\ \frac{da}{dx} = 7.11\dots$$ and area of $$\Delta ABC$$ is increasing at rate of $$(5\cdot 7.11\dots)$$ square inches/minute.

When $$x = 1,\ \frac{da}{dx} = -4.92\dots$$ and area of $$\Delta ABC$$ is decreasing at rate of $$(5\cdot 4.92\dots)$$ square inches/minute.

When $$\theta = 45^{\circ},\ h = b = x = y = 10\cos(45^{\circ}) = 10\cdot\frac{\sqrt{2}}{2} = 5\sqrt{2},\ $$ $$\frac{da}{dx} = 0,\ a$$ is a maximum of $$25$$ square inches and $$\frac{da}{dt} = 0.$$ $$$$

On the clock
An old fashion analog clock with American style face (12 hours) keeps accurate time. The length of the minute hand is $$4$$ inches and the length of the hour hand is $$3$$ inches.

At what rate is the tip of the minute hand approaching the tip of the hour hand at 3 o'clock?

Let $$a$$ be distance between the two tips.

The task is to calculate $$\frac{da}{dt}.$$

$$\frac{da}{dt} = \frac{da}{d\theta}\cdot \frac{d\theta}{dt}$$ where $$\theta$$ is angle subtended by side $$a$$ at center of clock.

Calculating $$\frac{d\theta}{dt}:$$

Angular velocity of minute hand $$= \frac{2\pi}{1}$$ radians/hour.

Angular velocity of hour hand $$= \frac{2\pi}{12} = \frac{\pi}{6}$$ radians/hour.

$$\frac{d\theta}{dt} =$$ angular velocity of minute hand relative to hour hand $$= 2\pi - \frac{\pi}{6} = \frac{11\pi}{6}$$ radians/hour.

Calculating $$\frac{da}{d\theta}:$$

$$a^2 = 3^2 + 4^2 - 2\cdot 3\cdot 4\cdot\cos \theta = 25 - 24\cos \theta.$$

$$2a\cdot\frac{da}{d\theta} = 24\sin \theta.$$

$$\frac{da}{d\theta} = \frac{24\sin \theta}{2a} = \frac{24\sin 90^\circ}{2\cdot 5} = \frac{24}{10} = 2.4.$$

$$\frac{da}{dt} = 2.4\cdot\frac{11\pi}{6} = 4.4\pi$$ inches/hour.

Radar Speed Trap

A multi-lane highway is oriented East-West. A vehicle is moving in the inside lane from West to East. A law-enforcement officer with a radar gun is in position 50 feet South of the center of the inside lane. When the vehicle is 200 feet from the radar gun, it shows the vehicle's speed to be 60 mph. What is the actual speed of the vehicle?

Let length $$AC = b.$$

$$\frac{db}{dt} = \frac{db}{dc}\cdot\frac{dc}{dt}$$ where $$\frac{dc}{dt} = 60$$ mph.

The derivative $$\frac{db}{dc}:$$
 * $$\begin{align}

\ \ \ \ &b^2 = c^2 - a^2;\ b = \sqrt{c^2 - a^2}\\ &2b\cdot \frac{db}{dc} = 2c\\ &\frac{db}{dc} = \frac{2c}{2b} = \frac{c}{\sqrt{c^2 - a^2}} \end{align}$$

$$\frac{db}{dt} = \frac{200}{\sqrt{200^2 - 50^2}}\cdot 60 = 61.967\dots$$ mph.

On the Water
A ship is at sea $$15$$ nautical miles East and $$8$$ nautical miles South of a lighthouse, and the ship is steaming South-West at $$10$$ nautical miles/hour or $$10$$ knots.

At what rate is ship approaching lighthouse?

$$\frac{da}{dt} = 10$$ knots.

$$\frac{dc}{dt} = \frac{dc}{da}\cdot\frac{da}{dt}$$ knots.

Calculating $$\frac{dc}{da}:$$

$$c^2 = a^2 + b^2 - 2ab\cos\theta$$

$$2c\cdot\frac{dc}{da} = 2a - 2b\cos\theta$$

$$\frac{dc}{da} = \frac{a - b\cos\theta}{c} = -0.291161615782696$$

The $$-$$ sign indicates that, when $$a$$ is increasing, $$c$$ is decreasing.

$$\frac{dc}{dt} = \frac{dc}{da}\cdot10$$ $$= 2.91161615782696$$ knots towards the lighthouse.

Reciprocating engine
The diagram illustrates the piston, connecting rod and crankshaft of an internal combustion reciprocating engine.

The connecting rod $$a$$ is connected to piston on $$y$$ axis, and to crankshaft at end of $$c$$ on $$x$$ axis.

This section analyzes the motion of the piston as the crankshaft rotates through angle $$x$$ and the piston moves up and down on the $$y$$ axis.

Position of piston
$$a^2 = y^2 + c^2 - 2yc\cos(x)$$

$$y^2 - 2c\cos(x)y + c^2 - a^2 = 0$$

$$y^2 - 2(5)\cos(x)y + (5)^2 - (13)^2 = 0$$

$$y^2 - 10\cos(x)y - 144 = 0$$

$$y = \frac{10\cos(x) + \sqrt{ (10\cos(x))^2 -4(-144) } }{2}$$

$$y = \frac{10\cos(x) + \sqrt{ 100(\cos(x))^2 + 576 } }{2}$$

$$y = 5\cos(x) + \sqrt{ 25(\cos(x))^2 + 144 }$$

Code supplied to grapher (without white space) is: When expressed in this way, it's easy to convert the code to python code:

Positions of interest:

Velocity of piston
Calculation of $$\frac{dy}{dx}$$ by implicit differentiation.

$$y^2 - 10 \cos(x)y - 144 = 0$$

$$2y\cdot y' - 10(\cos(x)y' + y(-\sin(x)) ) = 0$$

$$2y\cdot y' - 10\cos(x)y' + 10\sin(x)y = 0$$

$$y\cdot y' - 5\cos(x)y' = -5\sin(x)y$$

$$y'(y - 5\cos(x)) = -5\sin(x)y$$

$$y' = \frac{5\sin(x)y}{5\cos(x) - y}$$

Acceleration of piston
Acceleration introduces the second derivative. While velocity was the first derivative of position with respect to time, acceleration is the first derivative of velocity or the second derivative of position.

From velocity above $$yy' - 5\cos(x)y' + 5\sin(x)y = 0$$

By implicit differentation:

$$yy + y'y' - 5(\cos(x)y + y'(-\sin(x))) + 5(\sin(x)y' + y\cos(x)) = 0$$

$$yy + y'y' - 5\cos(x)y + 5y'\sin(x) + 5\sin(x)y' + 5y\cos(x) = 0$$

$$yy - 5\cos(x)y = -( y'y' + 5y'\sin(x) + 5\sin(x)y' + 5y\cos(x) )$$

$$y''(y - 5\cos(x)) = -( y'y' + 10\sin(x)y' + 5y\cos(x) )$$

$$y'' = \frac{y'y' + 10\sin(x)y' + 5y\cos(x) }{5\cos(x) - y}$$

Substitute for $$y$$ and $$y'$$ as defined above, and you see the code input to grapher at top of diagram to right.

"Kinks" in the curve:

It is not obvious by looking at the curve of velocity that there are slight irregularities in the curve when velocity is increasing.

However, the irregularities are obvious in the curve of acceleration.

During one revolution of the crankshaft there is less time allocated for negative acceleration than for positive acceleration. Therefore, the maximum absolute value of negative acceleration is greater than the maximum value of positive acceleration.

Minimum and maximum velocity
Velocity is rate of change of position. See also Figure 3 above.

Minimum velocity:

Velocity is zero when slope of curve of position is zero. This occurs at top dead center and at bottom dead center, ie, when $$x = 0$$ and $$x = \pi.$$

Maximum velocity:

Intuition suggests that the position of maximum velocity might be the point at which the connecting rod is tangent to the circle of the crankshaft. In other words:

$$x = \arctan(\frac{13}{5}) = 68.96\dots^{\circ}$$

or, that the position of maximum velocity might be the point at which the piston is half-way between top dead center and bottom dead center. In other words:

$$x = \arccos(\frac{2.5}{13}) = 78.9\dots^{\circ}$$

However, velocity is maximum when acceleration is $$0,$$ which occurs when $$x = \pm 71.26\dots^{\circ}.$$

Suppose that the engine is rotating at $$100$$ radians/second or approx. $$955$$ RPM.

$$v_{max} = y' \frac{inches}{radian} \cdot 100\frac{radians}{second}$$

abs$$(v_{max}) = 5.36 (100)$$ inches/second or approx. $$30.5$$ mph.

Minimum and maximum acceleration
Acceleration is rate of change of velocity. See also Figures 4a and 4b above.

Minimum acceleration:

Acceleration is zero when slope of curve of velocity is zero. This occurs at maximum velocity or when $$\angle x$$ is approx. $$\pm 71.3^{\circ}.$$

Maximum acceleration:

Acceleration is maximum when slope of curve of velocity is maximum.

Maximum negative acceleration occurs when slope of curve of velocity is maximum negative. This happens at top dead center when $$\angle x = 0.$$

Maximum positive acceleration occurs when slope of curve of velocity is maximum positive. This happens before and after bottom dead center when $$\angle x$$ is approx. $$\pm 127^{\circ}.$$

Let the engine continue to rotate at $$100$$ radians/second.

$$acc_{max} = y''\frac{inches}{radian^2}\cdot 100\frac{radians}{second} \cdot 100\frac{radians}{second}$$

abs$$(acc_{max}) = 6.92*100*100$$ inches/second$$^2$$ or approx. $$180$$ times the acceleration due to terrestrial gravity.

This maximum value of acceleration is maximum negative when $$\angle x = 0.$$

According to Newtonian physics $$f = ma$$, force = mass*acceleration, and $$w = fs$$, work = force*distance. In this engine energy expended in just accelerating piston to maximum velocity is proportional to rpm$$^2$$.

Perhaps this helps to explain why a big marine diesel engine rotating at low RPM can achieve efficiency of $$55 \%.$$

Simple laws of motion
Let a body move in accordance with the following function of $$t,\ f(t)$$ where $$t$$ means time:

$$p_t = f(t) = at^2 + bt + c$$ where $$p_t$$ is position at time $$t.$$

$$p_t$$ has the dimension of length. Therefore, each component of $$f(t)$$ must have the dimension of length.

For $$bt$$ to have the dimension of length, $$b$$ must have the dimensions of $$\frac{length}{time}$$ or velocity.

For $$at^2$$ to have the dimension of length, $$a$$ must have the dimensions of $$\frac{length}{time^2}$$ or acceleration.

If $$t ==0,\ p_0 = c$$ and $$p_t = f(t) = at^2 + bt + p_0.$$

The derivatives enable us to assign specific values to $$a,\ b.$$

$$\frac{d(p_t)}{dt} = f'(t) = v_t = 2at + b$$ where $$v_t$$ is velocity at time $$t.$$

If $$t ==0,\ v_0 = b$$ and $$p_t = f(t) = at^2 + v_0t + p_0.$$

$$\frac{d(v_t)}{dt} = f''(t) = 2a,$$ a constant equal to the acceleration to which the body is subjected.

For convenience let us say that $$p_t = \frac{1}{2}at^2 + v_0t + p_0$$ where $$a$$ is the (constant) acceleration to which the body is subjected.

Then $$v_t = \frac{1}{2}\cdot 2at + v_0 = at + v_0$$ and $$\frac{d(v_t)}{dt} = a.$$

title: Calculus/Derivatives


text:
Zeroth Law of Thermodynamics

 * If two thermodynamic systems are in thermal equilibrium with a third, they are also in thermal equilibrium with each other.

This is named zeroth law because it is basic law that is formulated after the three laws are formulated. The above is the actual statement for Zeroth law of Thermodynamics. This can be understood as follows

 * When two bodies, of which one is at a higher temperature than the other, are kept in contact, heat energy flows from the one at the higher temperature to the one at lower temperature.

title: Zeroth Law of Thermodynamics


text:
MES College

MYSORE EDUCATION SOCIETY

HISTORY

A group of enlightened citizens of Malleshwaram felt a need to open a college in Malleshwaram particularly to provide facilities to girl students, who had to go as far as Maharanis College for their courses in higher learning. The founders Sri.G.A.Acharya, Prof.B.R. Subba Rao, Sri. M.Chinnaswamy, Sri M.R.Narasimha Iyengar, Prof.M.P.L.Sastry and Smt Vimala Rangachar responded to this need by starting an intermediate college in 15th Cross, Malleshwaram, were introduced in 1956-57 with a student strength of 68.

In 1962 courses in Science were introduced. In the course of its historical development MES College has acquired great prominence not only in Karnataka but also in the country. The growth is both quantitative and qualitative. The college is now a four-floor stone structure with an annexe to accommodate the growing strength of students. Spacious, well-lit and airy classrooms are most suited for effective learning and teaching. Anybody who enters this college will feel and experience the silken touch of care, compassion and concern, which embody the philosophy of learning pursued here. Consequently character, confidence and credibility evolve.

In its infant stages the college was housed in a small dwelling Anandashram, which recalled the ancient Guru Kula with a good rapport between the small number of students and their teachers. From this humble beginning the college has grown gradually into a full-fledged college.

MES has grown with its ideal of spreading academic culture in several ways. Its aim has always been to provide value based education at affordable cost and to strive for the cultural growth of the student. The college celebrated its Silver Jubilee in 1982 and its 40th year of meaningful functioning in 1997.

The main college was bifurcated according to the Karnataka Government order DCE|ROB|134|134|internal|2001-02 dated 7.8.2001 into MES Degree College of Arts, Commerce and Science and MES Pre-University College of Arts, Commerce and Science.

The two colleges have been working in the same campus.

MES Degree college has been accredited by NAAC with 'A' grade in 2015-16

title: MES College


text:
BCP/Platycodon grandiflorus/Logs

2008 Logs

 * (no archived logs yet)

Recent Logs

 * --SB_Johnny | talk 11:22, 4 July 2007 (UTC)
 * --SB_Johnny | talk 10:15, 30 June 2008 (UTC)
 * --SB_Johnny | talk 10:24, 14 July 2008 (UTC)
 * --SB_Johnny talk 20:52, 9 August 2008 (UTC)
 * --SB_Johnny talk 09:13, 5 September 2008 (UTC)
 * --SB_Johnny talk 13:37, 5 October 2008 (UTC)

title: BCP/Platycodon grandiflorus/Logs


text:
BCP/Physostegia virginiana/Logs

This page contains recent bloom logs for the plant Physostegia virginiana (Obedient Plant).

Recent Logs

 * SB Johnny-LM 11:49, 7 July 2007 (UTC)
 * SB Johnny-LM 11:49, 6 August 2007 (UTC) (from photograph datestamp)

=====Archived Logs=====

(If there are more than 5 signatures under the recent logs heading, move the oldest logs to under this heading (generally leave the 3 most recent).
 * (no archived logs yet)

title: BCP/Physostegia virginiana/Logs


text:
Electrical Power Transmission

Welcome to the Home page for Transmission of Electrical Power

In this course we will discuss transmission and distribution systems of electrical power. At the end of this course, you should be familiar with the principles of electrical power transmission systems as applied in real life.

You are advised to read carefully through the material and attempt all quizzes in this course. Don't forget to follow the safety code outlined in this course.

Course Outcomes
At the end of this course the student is expected to have understood the following:
 * 1) Basic principles of supply systems
 * 2) Mechanical & electrical design of overhead lines
 * 3) Performance of transmission lines

Prerequisites
It is important that the student be at ease with the following courses since the concepts taught will be employed here:
 * 1) Electrical Engineering Fundamentals
 * 2) Electric Circuit Analysis

Organization
This course consists of:
 * 4 Lessons → [ 4 weeks ]
 * 1 Quiz Test → [ 1 Day ]

All this should take about 4 weeks. Plan your time wisely; start a journal and track your progress. If you work consistently and keep to your planned study time, you will well.

Here are some of the resources to help you understand and make better use of your personal learning experience:
 * The history of free learning
 * Learning by doing
 * How to use wiki technology as a free learner

Lessons

 * Introduction
 * OverHead Lines (Mechanical)
 * OverHead Lines (Electrical)
 * Line Performance

Resources
Wikiversity has a respectable body of content related to this course, you are advised to use the resources wisely to achieve your success. This course includes a formula sheet that can help you with formulae and principles. However, do not rely on the formula sheet entirely without understanding essential course material.

title: Electrical Power Transmission


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BCP/Lotus corniculatus

Lotus corniculatus

title: BCP/Lotus corniculatus


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BCP/Lotus corniculatus/Logs

2007 Logs

 * --Luai lashire 15:36, 26 August 2007 (UTC)
 * --Jomegat 19:37, 30 August 2007 (UTC)
 * --Luai lashire 21:58, 1 September 2007 (UTC)
 * --Jomegat 17:07, 17 September 2007 (UTC)
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 * --Jomegat 19:08, 14 August 2007 (UTC)

2008 Logs

 * --Jomegat 20:53, 31 May 2008 (UTC)
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 * Jomegat 23:56, 9 June 2008 (UTC)
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2009 Logs

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Recent Logs

 * Jomegat 23:00, 1 June 2010 (UTC)

title: BCP/Lotus corniculatus/Logs


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User:Haresh kumar~enwikiversity

SYED MESBAH UDDIN AHMED GP-CIC (ENTERPENEUR) KHORDHID PLAZA,SHOP NO-3&4 T.A ROAD,BRAHMAANBARIA.

title: User:Haresh kumar~enwikiversity


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Wikiversity:Colloquium/archives/August 2006/Project Logo

Project Logo
Right now we are using the English Wikipedia logo, and that needs to change ASAP to something else. If there are any suggestions for a interim logo for Wikiversity, that would be appreciated (and no, we are not going to use the U.S. Flag like was done with Wikipedia). --Robert Horning 14:08, 15 August 2006 (UTC)

 * I've uploaded the one on the right Wikiversity-logo-en.png as a temporary logo (I know it's the wrong message and it's crap, but it's better than the Wikipedia one), but I'm trying to get a developer to add it as the logo. Cormaggio 14:11, 15 August 2006 (UTC)

 * As you may have noticed, I've uploaded the logo at the wrong size - but I can't now re-upload the re-sized image, as apparently you need to have a four-day-old account to do so. As soon as we have an admin, can s/he delete the image, and i'll re-upload the new one? Thanks. Cormaggio 14:46, 15 August 2006 (UTC)

 * Yes. Is somebody getting bureaucrat-status? -- sebmol ? 15:03, 15 August 2006 (UTC)

 * Trying.. :-) Cormaggio 15:05, 15 August 2006 (UTC)

I have uploaded that logo to commons; Image:Wikiversity beta.png and created commons:Category:Wikiversity --Walter 22:11, 15 August 2006 (UTC)

To get Wikiversity on the main portal this template needs to be updated; Www.wikimedia.org template. But maybe it is better to wait a bit for that. --Walter 22:11, 15 August 2006 (UTC)

 * For a logo how about a large magnifying glass on a piece of a computer monitor showing some cryptic math equations or pictographs? Mirwin 11:28, 16 August 2006 (UTC)

meta:User:Nightstallion (away for the moment) is good at arranging logo design competitions. I suggest you center it at Meta, similar to the current Wikibooks/Wiktionary discussions there. I don't like the current one, but it's OK for the moment. Dbmag9 11:56, 16 August 2006 (UTC)

 * The French, German, and Swedish Wikiversities all have the same logo at present -- which is this nice amalgamated image combining the middle ages, the renaissance, and a modern guy of today. we can either affirm our solidarity with the EU (although Italy doesn't have a logo yet), and use theirs or come up with something equally complex and rich.  I like suggesting our affiliation --  albeit distant -- with the medieval university which was a bastion of learning, a refuge and repository of knowledge and an international  center for the exchange of ideas in a very chaotic and war-torn world. --Smithgrrl 00:42, 22 August 2006 (UTC)

 * A discussion about Wikiversity logo has started here: Wikiversity/logo. Feel free to share your ideas. guillom 16:22, 23 August 2006 (UTC)


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Example for the WikiU Film School Lesson on Lesson:3D Storyboard

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Used with the WikiU Film School Lesson on how to create 3D storyboards with FrameForge 3D Studio Demo Version.

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Used with the WikiU Film School Lesson on how to create 3D storyboards with FrameForge 3D Studio Demo Version.

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Nonlinear finite elements/Matrices

Much of finite elements revolves around forming matrices and solving systems of linear equations using matrices. This learning resource gives you a brief review of matrices.

/Matrices/
Suppose that you have a linear system of equations

\begin{align} a_{11} x_1 + a_{12} x_2 + a_{13} x_3 + a_{14} x_4 &= b_1 \\ a_{21} x_1 + a_{22} x_2 + a_{23} x_3 + a_{24} x_4 &= b_2 \\ a_{31} x_1 + a_{32} x_2 + a_{33} x_3 + a_{34} x_4 &= b_3 \\ a_{41} x_1 + a_{42} x_2 + a_{43} x_3 + a_{44} x_4 &= b_4 \end{align} ~. $$

Matrices provide a simple way of expressing these equations. Thus, we can instead write

\begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} =    \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix} ~. $$

An even more compact notation is

\left[\mathsf{A}\right] \left[\mathsf{x}\right] = \left[\mathsf{b}\right] ~\text{or}~ \mathbf{A} \mathbf{x} = \mathbf{b} ~. $$

Here $$\mathbf{A}$$ is a $$4\times 4$$ matrix while $$\mathbf{x}$$ and $$\mathbf{b}$$ are $$4\times 1$$ matrices. In general, an $$m \times n$$ matrix $$\mathbf{A}$$ is a set of numbers arranged in $$m$$ rows and $$n$$ columns.

\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \dots & a_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & a_{m3} & \dots & a_{mn} \end{bmatrix}~. $$

Practice Exercises
Practice: Expressing Linear Equations As Matrices

Types of Matrices
Common types of matrices that we encounter in finite elements are:

 * a  row vector that has one row and $$n$$ columns.

\mathbf{v} = \begin{bmatrix} v_1 & v_2 & v_3 & \dots & v_n \end{bmatrix} $$
 * a  column vector that has $$n$$ rows and one column.

\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \\ \vdots \\ v_n \end{bmatrix} $$ diagonal elements ($$a_{ii}$$) nonzero.
 * a  square matrix that has an equal number of rows and columns.
 * a  diagonal matrix which is a square matrix with only the

\mathbf{A} = \begin{bmatrix} a_{11} & 0 & 0 & \dots & 0 \\ 0 & a_{22} & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & a_{nn} \end{bmatrix}~. $$ with each of its nonzero elements ($$a_{ii}$$) equal to 1.
 * the  identity matrix ($$\mathbf{I}$$) which is a diagonal matrix and

\mathbf{A} = \begin{bmatrix} 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \end{bmatrix}~. $$ such that $$a_{ij} = a_{ji}$$.
 * a  symmetric matrix which is a square matrix with elements

\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ a_{12} & a_{22} & a_{23} & \dots & a_{2n} \\ a_{13} & a_{23} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & a_{3n} & \dots & a_{nn} \end{bmatrix}~. $$ such that $$a_{ij} = -a_{ji}$$.
 * a  skew-symmetric matrix which is a square matrix with elements

\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ -a_{12} & a_{22} & a_{23} & \dots & a_{2n} \\ -a_{13} & -a_{23} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ -a_{1n} & -a_{2n} & -a_{3n} & \dots & a_{nn} \end{bmatrix}~. $$ Note that the diagonal elements of a skew-symmetric matrix have to be zero: $$a_{ii} = -a_{ii} \Rightarrow a_{ii} = 0$$.

Matrix addition
Let $$\mathbf{A}$$ and $$\mathbf{B}$$ be two $$m \times n$$ matrices with components $$a_{ij}$$ and $$b_{ij}$$, respectively. Then

\mathbf{C} = \mathbf{A} + \mathbf{B} \implies c_{ij} = a_{ij} + b_{ij} $$

Multiplication by a scalar
Let $$\mathbf{A}$$ be a $$m \times n$$ matrix with components $$a_{ij}$$ and let $$\lambda$$ be a scalar quantity. Then,

\mathbf{C} = \lambda\mathbf{A} \implies c_{ij} = \lambda a_{ij} $$

Multiplication of matrices
Let $$\mathbf{A}$$ be a $$m \times n$$ matrix with components $$a_{ij}$$. Let $$\mathbf{B}$$ be a $$p \times q$$ matrix with components $$b_{ij}$$.

The product $$\mathbf{C} = \mathbf{A} \mathbf{B}$$ is defined only if $$n = p$$. The matrix $$\mathbf{C}$$ is a $$m \times q$$ matrix with components $$c_{ij}$$. Thus,

\mathbf{C} = \mathbf{A} \mathbf{B} \implies c_{ij} = \sum^n_{k=1} a_{ik} b_{kj} $$ Similarly, the product $$\mathbf{D} = \mathbf{B} \mathbf{A}$$ is defined only if $$q = m$$. The matrix $$\mathbf{D}$$ is a $$p \times n$$ matrix with components $$d_{ij}$$. We have

\mathbf{D} = \mathbf{B} \mathbf{A} \implies d_{ij} = \sum^m_{k=1} b_{ik} a_{kj} $$

Clearly, $$\mathbf{C} \ne \mathbf{D}$$ in general, i.e., the matrix product  is not commutative.

However, matrix multiplication is  distributive. That means

\mathbf{A} (\mathbf{B} + \mathbf{C}) = \mathbf{A} \mathbf{B} + \mathbf{A} \mathbf{C} ~. $$

The product is also  associative. That means

\mathbf{A} (\mathbf{B} \mathbf{C}) = (\mathbf{A} \mathbf{B}) \mathbf{C} ~. $$

Transpose of a matrix
Let $$\mathbf{A}$$ be a $$m \times n$$ matrix with components $$a_{ij}$$. Then the  transpose of the matrix is defined as the $$n \times m$$ matrix $$\mathbf{B} = \mathbf{A}^T$$ with components $$b_{ij} = a_{ji}$$. That is,

\mathbf{B} = \mathbf{A}^T = \begin{bmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \dots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & a_{m3} & \dots & a_{mn} \end{bmatrix}^T =    \begin{bmatrix} a_{11} & a_{21} & a_{31} & \dots & a_{m1} \\ a_{12} & a_{22} & a_{32} & \dots & a_{m2} \\ a_{13} & a_{23} & a_{33} & \dots & a_{m3} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{1n} & a_{2n} & a_{3n} & \dots & a_{mn} \end{bmatrix} $$

An important identity involving the transpose of matrices is

{    (\mathbf{A} \mathbf{B})^T = \mathbf{B}^T \mathbf{A}^T }~. $$

Determinant of a matrix
The  determinant of a matrix is defined only for square matrices.

For a $$2 \times 2$$ matrix $$\mathbf{A}$$, we have

\mathbf{A} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \implies \det(\mathbf{A}) = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}\end{vmatrix} = a_{11} a_{22} - a_{12} a_{21} ~. $$

For a $$n \times n$$ matrix, the determinant is calculated by expanding into  minors as
 * $$\begin{align}

&\det(\mathbf{A}) = \begin{vmatrix} a_{11} & a_{12} & a_{13} & \dots & a_{1n} \\ a_{21} & a_{22} & a_{23} & \dots & a_{2n} \\ a_{31} & a_{32} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & a_{n3} & \dots & a_{nn} \end{vmatrix} \\ &= a_{11} \begin{vmatrix} a_{22} & a_{23} & \dots & a_{2n} \\ a_{32} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n2} & a_{n3} & \dots & a_{nn} \end{vmatrix} - a_{12} \begin{vmatrix} a_{21} & a_{23} & \dots & a_{2n} \\ a_{31} & a_{33} & \dots & a_{3n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n3} & \dots & a_{nn} \end{vmatrix} + \dots \pm a_{1n} \begin{vmatrix} a_{21} & a_{22} & \dots & a_{2(n-1)} \\ a_{31} & a_{32} & \dots & a_{3(n-1)} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{n(n-1)} \end{vmatrix} \end{align}$$ In short, the determinant of a matrix $$\mathbf{A}$$ has the value

{   \det(\mathbf{A}) = \sum^n_{j=1} (-1)^{1+j} a_{1j} M_{1j} } $$ where $$M_{ij}$$ is the determinant of the submatrix of $$\mathbf{A}$$ formed by eliminating row $$i$$ and column $$j$$ from $$\mathbf{A}$$.

Some useful identities involving the determinant are given below.

 * If $$\mathbf{A}$$ is a $$n \times n$$ matrix, then

\det(\mathbf{A}) = \det(\mathbf{A}^T)~. $$
 * If $$\lambda$$ is a constant and $$\mathbf{A}$$ is a $$n \times n$$ matrix, then

\det(\lambda\mathbf{A}) = \lambda^n\det(\mathbf{A}) \implies \det(-\mathbf{A}) = (-1)^n\det(\mathbf{A}) ~. $$
 * If $$\mathbf{A}$$ and $$\mathbf{B}$$ are two $$n \times n$$ matrices, then

\det(\mathbf{A}\mathbf{B}) = \det(\mathbf{A})\det(\mathbf{B})~. $$

If you think you understand determinants, take the quiz.

Inverse of a matrix
Let $$\mathbf{A}$$ be a $$n \times n$$ matrix. The  inverse of $$\mathbf{A}$$ is denoted by $$\mathbf{A}^{-1}$$ and is defined such that

{   \mathbf{A} \mathbf{A}^{-1} = \mathbf{I} } $$ where $$\mathbf{I}$$ is the $$n \times n$$ identity matrix.

The inverse exists only if $$\det(\mathbf{A}) \ne 0$$. A  singular matrix does not have an inverse.

An important identity involving the inverse is

{   (\mathbf{A}\mathbf{B})^{-1} = \mathbf{B}^{-1} \mathbf{A}^{-1}, } $$ since this leads to: $$   {    (\mathbf{A} \mathbf{B})^{-1} (\mathbf{A} \mathbf{B}) = (\mathbf{B}^{-1} \mathbf{A}^{-1}) (\mathbf{A} \mathbf{B} ) = \mathbf{B}^{-1} \mathbf{A}^{-1} \mathbf{A} \mathbf{B} = \mathbf{B}^{-1} (\mathbf{A}^{-1} \mathbf{A}) \mathbf{B} = \mathbf{B}^{-1} \mathbf{I} \mathbf{B} = \mathbf{B}^{-1} \mathbf{B} = \mathbf{I}. }  $$

Some other identities involving the inverse of a matrix are given below.

determinant of its inverse.
 * The determinant of a matrix is equal to the multiplicative inverse of the

\det(\mathbf{A}) = \cfrac{1}{\det(\mathbf{A}^{-1})}~. $$ is equal to the original matrix.
 * The determinant of a  similarity transformation of a matrix

\det(\mathbf{B} \mathbf{A} \mathbf{B}^{-1}) = \det(\mathbf{A}) ~. $$ We usually use numerical methods such as Gaussian elimination to compute the inverse of a matrix.

Eigenvalues and eigenvectors
A thorough explanation of this material can be found at Eigenvalue, eigenvector and eigenspace. However, for further study, let us consider the following examples:

\mathbf{A} = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix}, \mathbf{v} = \begin{bmatrix} 6 \\ -5   \end{bmatrix},  \mathbf{t} = \begin{bmatrix} 7 \\ 4   \end{bmatrix}~. $$ Which vector is an eigenvector for $$ \mathbf{A} $$ ?
 * Let :$$

We have $$   \mathbf{A}\mathbf{v} = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix}\begin{bmatrix} 6 \\ -5   \end{bmatrix} = \begin{bmatrix} -24 \\ 20   \end{bmatrix} = -4\begin{bmatrix} 6 \\ -5   \end{bmatrix} $$, and $$ \mathbf{A}\mathbf{t} = \begin{bmatrix} 1 & 6 \\ 5 & 2 \end{bmatrix}\begin{bmatrix} 7 \\ 4   \end{bmatrix} = \begin{bmatrix} 31 \\ 43   \end{bmatrix}~.$$

Thus, $$ \mathbf{v}$$ is an eigenvector.

1 \\ 4   \end{bmatrix} $$   an eigenvector for $$ \mathbf{A} = \begin{bmatrix} -3 & -3 \\ 1 & 8 \end{bmatrix}$$ ?
 * Is $$ \mathbf{u} = \begin{bmatrix}

We have that since $$ \mathbf{A}\mathbf{u} = \begin{bmatrix} -3 & -3 \\ 1 & 8 \end{bmatrix}\begin{bmatrix} 1 \\ 4   \end{bmatrix} = \begin{bmatrix} -15 \\ 33   \end{bmatrix}$$, $$ \mathbf{u} = \begin{bmatrix} 1 \\ 4   \end{bmatrix} $$   is not an eigenvector for $$ \mathbf{A} = \begin{bmatrix} -3 & -3 \\ 1 & 8 \end{bmatrix}~.$$

title: Nonlinear finite elements/Matrices


text: Nonlinear finite elements/Partial differential equations


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