32.2: The Domestic Mission
By the end of this section, you will be able to:
- Discuss the Bush administration’s economic theories and tax policies, and their effects on the American economy
- Explain how the federal government attempted to improve the American public education system
- Describe the federal government’s response to Hurricane Katrina
- Identify the causes of the Great Recession of 2008 and its effect on the average citizen
By the time George W. Bush became president, the concept of supply-side economics had become an article of faith within the Republican Party. The oft-repeated argument was that tax cuts for the wealthy would allow them to invest more and create jobs for everyone else. This belief in the self-regulatory powers of competition also served as the foundation of Bush’s education reform. But by the end of 2008, however, Americans’ faith in the dynamics of the free market had been badly shaken. The failure of the homeland security apparatus during Hurricane Katrina and the ongoing challenge of the Iraq War compounded the effects of the bleak economic situation.
OPENING AND CLOSING THE GAP
The Republican Party platform for the 2000 election offered the American people an opportunity to once again test the rosy expectations of supply-side economics. In 2001, Bush and the Republicans pushed through a $1.35 trillion tax cut by lowering tax rates across the board but reserving the largest cuts for those in the highest tax brackets. This was in the face of calls by Republicans for a balanced budget, which Bush insisted would happen when the so-called job creators expanded the economy by using their increased income to invest in business.
The cuts were controversial; the rich were getting richer while the middle and lower classes bore a proportionally larger share of the nation’s tax burden. Between 1966 and 2001, one-half of the nation’s income gained from increased productivity went to the top 0.01 percent of earners. By 2005, dramatic examples of income inequity were increasing; the chief executive of Wal-Mart earned $15 million that year, roughly 950 times what the company’s average associate made. The head of the construction company K. B. Homes made $150 million, or four thousand times what the average construction worker earned that same year. Even as productivity climbed, workers’ incomes stagnated; with a larger share of the wealth, the very rich further solidified their influence on public policy. Left with a smaller share of the economic pie, average workers had fewer resources to improve their lives or contribute to the nation’s prosperity by, for example, educating themselves and their children.
Another gap that had been widening for years was the education gap. Some education researchers had argued that American students were being left behind. In 1983, a commission established by Ronald Reagan had published a sobering assessment of the American educational system entitled A Nation at Risk . The report argued that American students were more poorly educated than their peers in other countries, especially in areas such as math and science, and were thus unprepared to compete in the global marketplace. Furthermore, test scores revealed serious educational achievement gaps between white students and students of color. Touting himself as the “education president,” Bush sought to introduce reforms that would close these gaps.
His administration offered two potential solutions to these problems. First, it sought to hold schools accountable for raising standards and enabling students to meet them. The No Child Left Behind Act, signed into law in January 2002, erected a system of testing to measure and ultimately improve student performance in reading and math at all schools that received federal funds (Figure \(\PageIndex{1}\)). Schools whose students performed poorly on the tests would be labeled “in need of improvement.” If poor performance continued, schools could face changes in curricula and teachers, or even the prospect of closure.
The second proposed solution was to give students the opportunity to attend schools with better performance records. Some of these might be charter schools , institutions funded by local tax monies in much the same way as public schools, but able to accept private donations and exempt from some of the rules public schools must follow. During the administration of George H. W. Bush, the development of charter schools had gathered momentum, and the American Federation of Teachers welcomed them as places to employ innovative teaching methods or offer specialized instruction in particular subjects. President George W. Bush now encouraged states to grant educational funding vouchers to parents, who could use them to pay for a private education for their children if they chose. These vouchers were funded by tax revenue that would otherwise have gone to public schools.
THE 2004 ELECTION AND BUSH’S SECOND TERM
In the wake of the 9/11 attacks, Americans had rallied around their president in a gesture of patriotic loyalty, giving Bush approval ratings of 90 percent. Even following the first few months of the Iraq war, his approval rating remained historically high at approximately 70 percent. But as the 2004 election approached, opposition to the war in Iraq began to grow. While Bush could boast of a number of achievements at home and abroad during his first term, the narrow victory he achieved in 2000 augured poorly for his chances for reelection in 2004 and a successful second term.
Reelection
As the 2004 campaign ramped up, the president was persistently dogged by rising criticism of the violence of the Iraq war and the fact that his administration’s claims of WMDs had been greatly overstated. In the end, no such weapons were ever found. These criticisms were amplified by growing international concern over the treatment of prisoners at the Guantanamo Bay detention camp and widespread disgust over the torture conducted by U.S. troops at the prison in Abu Ghraib, Iraq, which surfaced only months before the election (Figure \(\PageIndex{2}\)).
In March 2004, an ambush by Iraqi insurgents of a convoy of private military contractors from Blackwater USA in the town of Fallujah west of Baghdad, and the subsequent torture and mutilation of the four captured mercenaries, shocked the American public. But the event also highlighted the growing insurgency against U.S. occupation, the escalating sectarian conflict between the newly empowered Shia Muslims and the minority of the formerly ruling Sunni, and the escalating costs of a war involving a large number of private contractors that, by conservative estimates, approached $1.7 trillion by 2013. Just as importantly, the American campaign in Iraq had diverted resources from the war against al-Qaeda in Afghanistan, where U.S troops were no closer to capturing Osama bin Laden, the mastermind behind the 9/11 attacks.
With two hot wars overseas, one of which appeared to be spiraling out of control, the Democrats nominated a decorated Vietnam War veteran, Massachusetts senator John Kerry (Figure \(\PageIndex{3}\)), to challenge Bush for the presidency. As someone with combat experience, three Purple Hearts, and a foreign policy background, Kerry seemed like the right challenger in a time of war. But his record of support for the invasion of Iraq made his criticism of the incumbent less compelling and earned him the byname “Waffler” from Republicans. The Bush campaign also sought to characterize Kerry as an elitist out of touch with regular Americans—Kerry had studied overseas, spoke fluent French, and married a wealthy foreign-born heiress. Republican supporters also unleashed an attack on Kerry’s Vietnam War record, falsely claiming he had lied about his experience and fraudulently received his medals. Kerry’s reluctance to embrace his past leadership of Vietnam Veterans Against the War weakened the enthusiasm of antiwar Americans while opening him up to criticisms from veterans groups. This combination compromised the impact of his challenge to the incumbent in a time of war.
Urged by the Republican Party to “stay the course” with Bush, voters listened. Bush won another narrow victory, and the Republican Party did well overall, picking up four seats in the Senate and increasing its majority there to fifty-five. In the House, the Republican Party gained three seats, adding to its majority there as well. Across the nation, most governorships also went to Republicans, and Republicans dominated many state legislatures.
Despite a narrow win, the president made a bold declaration in his first news conference following the election. “I earned capital in this campaign, political capital, and now I intend to spend it.” The policies on which he chose to spend this political capital included the partial privatization of Social Security and new limits on court-awarded damages in medical malpractice lawsuits. In foreign affairs, Bush promised that the United States would work towards “ending tyranny in the world.” But at home and abroad, the president achieved few of his second-term goals. Instead, his second term in office became associated with the persistent challenge of pacifying Iraq, the failure of the homeland security apparatus during Hurricane Katrina, and the most severe economic crisis since the Great Depression.
A Failed Domestic Agenda
The Bush administration had planned a series of free-market reforms, but corruption, scandals, and Democrats in Congress made these goals hard to accomplish. Plans to convert Social Security into a private-market mechanism relied on the claim that demographic trends would eventually make the system unaffordable for the shrinking number of young workers, but critics countered that this was easily fixed. Privatization, on the other hand, threatened to derail the mission of the New Deal welfare agency and turn it into a fee generator for stock brokers and Wall Street financiers. Similarly unpopular was the attempt to abolish the estate tax. Labeled the “death tax” by its critics, its abolishment would have benefitted only the wealthiest 1 percent. As a result of the 2003 tax cuts, the growing federal deficit did not help make the case for Republicans.
The nation faced another policy crisis when the Republican-dominated House of Representatives approved a bill making the undocumented status of millions of immigrants a felony and criminalizing the act of employing or knowingly aiding illegal immigrants. In response, millions of illegal and legal immigrants, along with other critics of the bill, took to the streets in protest. What they saw as the civil rights challenge of their generation, conservatives read as a dangerous challenge to law and national security. Congress eventually agreed on a massive build-up of the U.S. Border Patrol and the construction of a seven-hundred-mile-long fence along the border with Mexico, but the deep divisions over immigration and the status of up to twelve million undocumented immigrants remained unresolved.
Hurricane Katrina
One event highlighted the nation’s economic inequality and racial divisions, as well as the Bush administration’s difficulty in addressing them effectively. On August 29, 2005, Hurricane Katrina came ashore and devastated coastal stretches of Alabama, Mississippi, and Louisiana. The city of New Orleans, no stranger to hurricanes and floods, suffered heavy damage when the levees, embankments designed to protect against flooding, failed during the storm surge, as the Army Corps of Engineers had warned they might. The flooding killed some fifteen hundred people and so overwhelmed parts of the city that tens of thousands more were trapped and unable to evacuate (Figure \(\PageIndex{4}\)). Thousands who were elderly, ill, or too poor to own a car followed the mayor’s directions and sought refuge at the Superdome, which lacked adequate food, water, and sanitation. Public services collapsed under the weight of the crisis.
CLICK AND EXPLORE
See pictures of the aftermath of Hurricane Katrina and read and view accounts of survivors of the disaster.
Although the U.S. Coast Guard managed to rescue more than thirty-five thousand people from the stricken city, the response by other federal bodies was less effective. The Federal Emergency Management Agency (FEMA), an agency charged with assisting state and local governments in times of natural disaster, proved inept at coordinating different agencies and utilizing the rescue infrastructure at its disposal. Critics argued that FEMA was to blame and that its director, Michael D. Brown, a Bush friend and appointee with no background in emergency management, was an example of cronyism at its worst. The failures of FEMA were particularly harmful for an administration that had made “homeland security” its top priority. Supporters of the president, however, argued that the scale of the disaster was such that no amount of preparedness or competence could have allowed federal agencies to cope.
While there was plenty of blame to go around—at the city, state, and national levels—FEMA and the Bush administration got the lion’s share. Even when the president attempted to demonstrate his concern with a personal appearance, the tactic largely backfired. Photographs of him looking down on a flooded New Orleans from the comfort of Air Force One only reinforced the impression of a president detached from the problems of everyday people. Despite his attempts to give an uplifting speech from Jackson Square, he was unable to shake this characterization, and it underscored the disappointments of his second term. On the eve of the 2006 midterm elections, President Bush’s popularity had reached a new low, as a result of the war in Iraq and Hurricane Katrina, and a growing number of Americans feared that his party’s economic policy benefited the wealthy first and foremost. Young voters, non-white Americans, and women favored the Democratic ticket by large margins. The elections handed Democrats control of the Senate and House for the first time since 1994, and, in January 2007, California representative Nancy Pelosi became the first female Speaker of the House in the nation’s history.
THE GREAT RECESSION
For most Americans, the millennium had started with economic woes. In March 2001, the U.S. stock market had taken a sharp drop, and the ensuing recession triggered the loss of millions of jobs over the next two years. In response, the Federal Reserve Board cut interest rates to historic lows to encourage consumer spending. By 2002, the economy seemed to be stabilizing somewhat, but few of the manufacturing jobs lost were restored to the national economy. Instead, the “outsourcing” of jobs to China and India became an increasing concern, along with a surge in corporate scandals. After years of reaping tremendous profits in the deregulated energy markets, Houston-based Enron imploded in 2003 over allegations of massive accounting fraud. Its top executives, Ken Lay and Jeff Skilling, received long prison sentences, but their activities were illustrative of a larger trend in the nation’s corporate culture that embroiled reputable companies like JP Morgan Chase and the accounting firm Arthur Anderson. In 2003, Bernard Ebbers, the CEO of communications giant WorldCom, was discovered to have inflated his company’s assets by as much as $11 billion, making it the largest accounting scandal in the nation’s history. Only five years later, however, Bernard Madoff’s Ponzi scheme would reveal even deeper cracks in the nation’s financial economy.
Banks Gone Wild
Notwithstanding economic growth in the 1990s and steadily increasing productivity, wages had remained largely flat relative to inflation since the end of the 1970s; despite the mild recovery, they remained so. To compensate, many consumers were buying on credit, and with interest rates low, financial institutions were eager to oblige them. By 2008, credit card debt had risen to over $1 trillion. More importantly, banks were making high-risk, high-interest mortgage loans called subprime mortgages to consumers who often misunderstood their complex terms and lacked the ability to make the required payments.
These subprime loans had a devastating impact on the larger economy. In the past, a prospective home buyer went to a local bank for a mortgage loan. Because the bank expected to make a profit in the form of interest charged on the loan, it carefully vetted buyers for their ability to repay. Changes in finance and banking laws in the 1990s and early 2000s, however, allowed lending institutions to securitize their mortgage loans and sell them as bonds, thus separating the financial interests of the lender from the ability of the borrower to repay, and making highly risky loans more attractive to lenders. In other words, banks could afford to make bad loans, because they could sell them and not suffer the financial consequences when borrowers failed to repay.
Once they had purchased the loans, larger investment banks bundled them into huge packages known as collateralized debt obligations (CDOs) and sold them to investors around the world. Even though CDOs consisted of subprime mortgages, credit card debt, and other risky investments, credit ratings agencies had a financial incentive to rate them as very safe. Making matters worse, financial institutions created instruments called credit default swaps , which were essentially a form of insurance on investments. If the investment lost money, the investors would be compensated. This system, sometimes referred to as the securitization food chain, greatly swelled the housing loan market, especially the market for subprime mortgages, because these loans carried higher interest rates. The result was a housing bubble, in which the value of homes rose year after year based on the ease with which people now could buy them.
Banks Gone Broke
When the real estate market stalled after reaching a peak in 2007, the house of cards built by the country’s largest financial institutions came tumbling down. People began to default on their loans, and more than one hundred mortgage lenders went out of business. American International Group (AIG), a multinational insurance company that had insured many of the investments, faced collapse. Other large financial institutions, which had once been prevented by federal regulations from engaging in risky investment practices, found themselves in danger, as they either were besieged by demands for payment or found their demands on their own insurers unmet. The prestigious investment firm Lehman Brothers was completely wiped out in September 2008. Some endangered companies, like Wall Street giant Merrill Lynch, sold themselves to other financial institutions to survive. A financial panic ensued that revealed other fraudulent schemes built on CDOs. The biggest among them was a pyramid scheme organized by the New York financier Bernard Madoff, who had defrauded his investors by at least $18 billion.
Realizing that the failure of major financial institutions could result in the collapse of the entire U.S. economy, the chairman of the Federal Reserve, Ben Bernanke, authorized a bailout of the Wall Street firm Bear Stearns, although months later, the financial services firm Lehman Brothers was allowed to file for the largest bankruptcy in the nation’s history. Members of Congress met with Bernanke and Secretary of the Treasury Henry Paulson in September 2008, to find a way to head off the crisis. They agreed to use $700 billion in federal funds to bail out the troubled institutions, and Congress subsequently passed the Emergency Economic Stabilization Act, creating the Troubled Asset Relief Program (TARP). One important element of this program was aid to the auto industry: The Bush administration responded to their appeal with an emergency loan of $17.4 billion—to be executed by his successor after the November election—to stave off the industry’s collapse.
The actions of the Federal Reserve, Congress, and the president prevented the complete disintegration of the nation’s financial sector and warded off a scenario like that of the Great Depression. However, the bailouts could not prevent a severe recession in the U.S. and world economy. As people lost faith in the economy, stock prices fell by 45 percent. Unable to receive credit from now-wary banks, smaller businesses found that they could not pay suppliers or employees. With houses at record prices and growing economic uncertainty, people stopped buying new homes. As the value of homes decreased, owners were unable to borrow against them to pay off other obligations, such as credit card debt or car loans. More importantly, millions of homeowners who had expected to sell their houses at a profit and pay off their adjustable-rate mortgages were now stuck in houses with values shrinking below their purchasing price and forced to make mortgage payments they could no longer afford.
Without access to credit, consumer spending declined. Some European nations had suffered similar speculation bubbles in housing, but all had bought into the mortgage securities market and suffered the losses of assets, jobs, and demand as a result. International trade slowed, hurting many American businesses. As the Great Recession of 2008 deepened, the situation of ordinary citizens became worse. During the last four months of 2008, one million American workers lost their jobs, and during 2009, another three million found themselves out of work. Under such circumstances, many resented the expensive federal bailout of banks and investment firms. It seemed as if the wealthiest were being rescued by the taxpayer from the consequences of their imprudent and even corrupt practices.


2.1: Introduction to Research Methods and Ethics in Human Sexuality
After completing this module, students have a working knowledge of the following:
- A Scientific Approach to Human Sexuality
- Some notable sex researchers and their contributions
- Populations and Samples: Representing the World of Diversity
- Methods of Observation
- Correlation
- Ethics in Sex Research
- Research Methods in Human Sexuality
How does one go about research in Human Sexuality and who came up with the idea to study such things? This section will provide you with that overview. Studying sexuality has always been difficult, mainly due to lack of funding available for this scope of research.
All research is subject to ethical considerations. When we undertake the task of studying human behavior, we must first understand that the participants are human, and deserve to be treated ethically. Because this was not always the case, and people have been hurt or killed as a result, systems have been set in place to protect research participants.


2.2: Content on LibreTexts
A textbook starts with some thoughts about content. What is it we want to teach? This part of CAP is obviously critical - but it is the least important to discuss in general for a few reasons:
- Each course within each discipline is different;
- each professor will want to cover different things in different levels of depth; and
- it is mostly the same as what you would find in a traditional hardcopy textbook.
Notice that in Figure 1, many people have studied how to approach content within the context of Architecture and Content, but the area in the Venn diagram that is completely blank is that which is only Pedagogy, as that is inherently an interaction between other areas. The only thing in the Content area is whether or not the content is authentic, and whether or not multiple representations are used.
- Content in the bookshelves is largely unchanged from a hardcopy textbook, and is thus left to the author.
- Content should never be thought of as independent from how it is taught or organized.
- How it is taught is the Pedagogy.
- How it is organized within an online textbook is the Architecture.
The biggest thing to think about is how to teach this content. How to teach is usually the definition of pedagogy. However, here we break it into two parts: Architecture and Pedagogy. Architecture is the static structure of the course materials: where are things found, how are they linked, what is the organization in general. Pedagogy, here, is how you expect people to interact with your content and what approaches, models, samples, and images within your content will be included.
- What are your learning objectives for the course?
- How many are you teaching in a given lecture? In a given exam period?
- What conceptual links between material are important for your students to understand?
- What depth does a concept need to be covered?
- How are you going to assess your content?
- How many sample problems should be included?


1.1: Conservation Biology is Still Evolving
As a distinct scientific field, conservation biology is an integrated, multidisciplinary subject that developed in response to the challenge of preserving populations, species, ecosystems, and biological interactions. The main aim of conservation biology is to ensure the long-term preservation of biodiversity. To achieve its aim, conservation biology has set three goals:
- To document Earth’s biological diversity.
- To investigate how humans influence species, evolution, and ecosystem processes.
- To investigate practical approaches to protect and restore biological communities, maintain genetic diversity, and prevent the extinction of species.
The first two goals describe typical scientific research investigating objective facts. The third goal, however, is a part of what makes conservation biology a normative discipline; that is, conservation biology incorporates human values, not just facts, to understand and achieve its value-laden goals (Lindenmayer and Hunter, 2010). In this sense, conservation biology is related to environmentalism, in which people aim to protect the natural environment for its own sake (see Section 4.3.2). However, conservation biology is at its core a scientific discipline; it is founded on scientific principles. This is not to say you must be a scientist to practice conservation biology; there are many people who are not scientists who apply the principles of conservation biology in their professional and personal lives.
The emergence of conservation biology as a distinct scientific field in the 1970s has given rise to the formation of various formal societies representing the field in a united voice. Most notable among these is the Society for Conservation Biology (SCB, Figure 1.2), which is a non-profit international professional organization with a mission to advance “the science and practice of conserving the Earth’s biological diversity”. To facilitate opportunities where like-minded people can share ideas locally, the SCB has regional branches, including an active Africa Section ( http://conbio.org/groups/sections/africa ) which hosts regular conferences. In addition to the SCB, a great number of other local, national, and regional conservation organizations also exist and act as mouthpieces for grassroots movements and as custodians of nature. Many of these groups focus on specific animals or local protected areas. Others organically adapt their missions and visions in response to a specific need or threat. For example, established in 1913 as an exchange forum between collectors of rare plants, the Botanical Society of South Africa now actively works toward protecting those rare plants in their natural habitats.
In recent years, conservation practice has evolved from just a plan to save the environment to a vision that includes sustainable development and social justice.
Conservation biology also has a history of adapting to new challenges. The very first conservation activities, in Africa and beyond, were geared towards securing the rights to valuable natural resources for people in powerful positions, such as kings and tribal chiefs, enforced through a strictly adherence to cultural norms and customary laws (Section 2.2). But as a growing human population expanded its influence on the environment, and wildlife started to decline, earliest conservation models gradually shifted towards fortress conservation approaches (Wilshusen et al., 2002) which aimed to shield wildlife from people by setting aside protected areas where human activities were strictly controlled.
Today, however, as human populations are exploding, and consumption is increasing, even protected areas are increasingly unable to withstand the multitude of threats to biodiversity that ignore property boundaries and political borders. In response, fortress conservation approaches are beginning to make way for large-scale integrated activities that highlight the social and economic benefits of biodiversity conservation. To do this, new alliances are being formed and new agendas are being established, such as those that directly link human health with environmental health (Box 1.1). These integrated conservation philosophies that pursue strategies that benefit both humans and biodiversity show much promise because they focus on fundamental extinction drivers, and advocate for more inclusive sustainable development. In this way, the practice of conservation has evolved from just a plan to save the environment to a vision that accomplishes its goals through sustainable development and social justice.
Yet, as we consider how to best invest limited conservation resources, some difficult questions arise. With seemingly more work to be done than can be accomplished, should we let some species go extinct (Bottrill et al., 2008)? Which species? Who decides? How can we even dare to think that we can play god? Such questions predictably bring about strongly opinionated and emotional debate (Soulé, 2013 vs. Marvier, 2014; Tallis and Lubchenco, 2014). Given the successful track record of fortress conservation initiatives in preventing extinctions despite limited budgets (Young et al., 2014), as well as the promising progress of more complex people-centred initiatives (Pooley et al., 2014), it seems clear that conservation relies on some balance between these two conservation philosophies (Sodhi et al., 2011). Conservation biologists of tomorrow will be able to fine-tune the balance between these strategies by closely inspecting the successes and failures of our actions today.
Gladys Kalema-Zikusoka
Conservation Through Public Health,
Kampala, Uganda.
Conservation Through Public Health (CTPH) is a grassroots non-governmental organization (NGO) and non-profit that promotes biodiversity conservation by enabling people, wildlife, and livestock to coexist. The organization was founded in 2003 after fatal scabies skin disease outbreaks in mountain gorillas (Gorilla beringei beringei, EN) were traced to people living around Bwindi Impenetrable National Park, Uganda, who had limited access to basic health services (Kalema-Zikusoka et al., 2002). Since then, CTPH has contributed to conservation and sustainable development in Africa by improving human and animal health and welfare in and around protected areas.
One of the main goals of CTPH is to reduce disease transfer between humans and gorillas. We accomplish this through an integrated population, health, and environment (PHE) program that was established in 2007 with funding from the US Agency for International Development (USAID). As a first step, piloted around Bwindi, CTPH held consultative meetings with local leaders, during which at least one Village Health and Conservation Team (VHCT) volunteer was selected from each village and two from each parish (consisting of 11 villages) to oversee distribution of family planning supplies. This initiative rapidly expanded into a sustainable social service delivery network that promotes family planning, hygiene, and sanitation. The network resulted in a 20% to 60% (national average is 30%) increase in new users to modern family planning, and a 10% to 60% increase in adoption of hand washing facilities at homes visited by VHCTs. VHCT volunteers also refer people suffering from infectious diseases and malnutrition to local health centres and promote more sustainable alternative livelihoods. Another group of community volunteers, the “Human and Gorilla Conflict Resolution” (HUGO) team, in turn collect gorilla faecal samples left on communal land to monitor their health (Figure 1.A), and visually monitor gorillas for clinical signs of disease inside and outside protected areas (Gaffikin and Kalema-Zikusoka, 2010). In the process, we have seen reduced disease incidences in the gorillas, reduced conflict between people and gorillas, and improved attitudes toward conservation. One unintended outcome has been increased gender equality: men are now more involved in family planning, and women are more involved in natural resource management.
Our experience in initiating and managing PHE programs for the past 10 years has taught us several lessons. One of the most important lessons to ensure project sustainability is to regularly engage with local leaders and the government. The Uganda Wildlife Authority, Uganda’s Ministry of Health, and local health centres all attend CTPH meetings with VHCTs. Attendance by and representation of these groups not only informs them of our activities, but also provides a platform to inform or train the VHCTs in what they would like them to disseminate to the local communities.
We have also learnt that PHE-implementing partners and projects need to be well-suited to each other and each site; this remains true even though health needs are often the same, regardless of the location. For example, at Mount Elgon National Park in Uganda, we found that training VHCTs in reducing conflict with park management played a key role in changing community attitudes toward conservation. In contrast, at Virunga National Park, Democratic Republic of the Congo (DRC), we found that VHCTs needed to work more closely with local health centres to prevent disease transmission between people and gorillas, and to promote family planning in a largely Catholic country.
Lastly, we found that establishing income-generating projects for groups rather than individuals was key to sustaining VHCT networks and program goals beyond donor funding cycles where we have had no volunteer dropouts in the first 10 years of initiating the PHE program. These key components were accomplished by initiating livestock group enterprises and by encouraging VHCT volunteers to invest generated income into Village Saving and Loan Associations (see http://www.care.org/vsla ).


5.4: Conservative Vector Fields
-
- Last updated
- Save as PDF
Learning Objectives
- Describe simple and closed curves; define connected and simply connected regions.
- Explain how to find a potential function for a conservative vector field.
- Use the Fundamental Theorem for Line Integrals to evaluate a line integral in a vector field.
- Explain how to test a vector field to determine whether it is conservative.
In this section, we continue the study of conservative vector fields. We examine the Fundamental Theorem for Line Integrals, which is a useful generalization of the Fundamental Theorem of Calculus to line integrals of conservative vector fields. We also discover show how to test whether a given vector field is conservative, and determine how to build a potential function for a vector field known to be conservative.
Curves and Regions
Before continuing our study of conservative vector fields, we need some geometric definitions. The theorems in the subsequent sections all rely on integrating over certain kinds of curves and regions, so we develop the definitions of those curves and regions here. We first define two special kinds of curves: closed curves and simple curves. As we have learned, a closed curve is one that begins and ends at the same point. A simple curve is one that does not cross itself. A curve that is both closed and simple is a simple closed curve (Figure \(\PageIndex{1}\)).
DEFINITION: Closed Curves
Curve \(C\) is a closed curve if there is a parameterization \(\vecs r(t)\), \(a≤t≤b\) of \(C\) such that the parameterization traverses the curve exactly once and \(\vecs r(a)=\vecs r(b)\). Curve \(C\) is a simple curve if \(C\) does not cross itself. That is, \(C\) is simple if there exists a parameterization \(\vecs r(t)\), \(a≤t≤b\) of \(C\) such that \(\vecs r\) is one-to-one over \((a,b)\). It is possible for \(\vecs r(a)=\vecs r(b)\), meaning that the simple curve is also closed.
Example \(\PageIndex{1}\): Determining Whether a Curve Is Simple and Closed
Is the curve with parameterization \(\vecs{r}(t)=\left\langle\cos t,\frac{\sin(2t)}{2}\right\rangle\), \(0≤t≤2\pi\) a simple closed curve?
Solution
Note that \(\vecs{r}(0)=⟨1,0⟩=\vecs r(2\pi)\); therefore, the curve is closed. The curve is not simple, however. To see this, note that \(\vecs{r}\left(\frac{\pi}{2}\right)=⟨0,0⟩=\vecs{r}\left(\frac{3\pi}{2}\right)\), and therefore the curve crosses itself at the origin (Figure \(\PageIndex{2}\)).
Exercise \(\PageIndex{1}\)
Is the curve given by parameterization \(\vecs{r}(t)=⟨2\cos t,3\sin t⟩\), \(0≤t≤6\pi\), a simple closed curve?
- Hint
-
Sketch the curve.
- Answer
-
Yes
Many of the theorems in this chapter relate an integral over a region to an integral over the boundary of the region, where the region’s boundary is a simple closed curve or a union of simple closed curves. To develop these theorems, we need two geometric definitions for regions: that of a connected region and that of a simply connected region. A connected region is one in which there is a path in the region that connects any two points that lie within that region. A simply connected region is a connected region that does not have any holes in it. These two notions, along with the notion of a simple closed curve, allow us to state several generalizations of the Fundamental Theorem of Calculus later in the chapter. These two definitions are valid for regions in any number of dimensions, but we are only concerned with regions in two or three dimensions.
DEFINITION: connected regions
A region D is a connected region if, for any two points \(P_1\) and \(P_2\), there is a path from \(P_1\) to \(P_2\) with a trace contained entirely inside D . A region D is a simply connected region if D is connected for any simple closed curve C that lies inside D , and curve C can be shrunk continuously to a point while staying entirely inside D . In two dimensions, a region is simply connected if it is connected and has no holes.
All simply connected regions are connected, but not all connected regions are simply connected (Figure \(\PageIndex{3}\)).
Exercise \(\PageIndex{2}\)
Is the region in the below image connected? Is the region simply connected?
- Hint
-
Consider the definitions.
- Answer
-
The region in the figure is connected. The region in the figure is not simply connected.
Fundamental Theorem for Line Integrals
Now that we understand some basic curves and regions, let’s generalize the Fundamental Theorem of Calculus to line integrals. Recall that the Fundamental Theorem of Calculus says that if a function \(f\) has an antiderivative \(F\), then the integral of \(f\) from \(a\) to \(b\) depends only on the values of \(F\) at \(a\) and at \(b\)—that is,
\[\int_a^bf(x)\,dx=F(b)−F(a). \nonumber \]
If we think of the gradient as a derivative, then the same theorem holds for vector line integrals. We show how this works using a motivational example.
Example \(\PageIndex{2}\): Evaluating a Line Integral and the Antiderivatives of the Endpoints
Let \(\vecs{F}(x,y)=⟨2x,4y⟩\). Calculate \(\displaystyle \int_C \vecs{F} \cdot d\vecs{r}\), where C is the line segment from \((0,0)\) to \((2,2)\) (Figure \(\PageIndex{4}\)).
Solution
We use the method from the previous section to calculate \(\int_C \vecs{F} \cdot d\vecs{r}\). Curve C can be parameterized by \(\vecs{r}(t)=⟨2t,2t⟩\), \(0≤t≤1\). Then, \(\vecs{F}(\vecs r(t))=⟨4t,8t⟩\) and \(\vecs r′(t)=⟨2,2⟩\), which implies that
\[\begin{align*} \int_C \vecs{F}·d\vecs{r} &=\int_0^1⟨4t,8t⟩·⟨2,2⟩dt \\[4pt] &=\int_0^1(8t+16t)dt=\int_0^1 24tdt\\[4pt] &={\big[12t^2\big]}_0^1=12. \end{align*}\]
Notice that \(\vecs{F}=\vecs \nabla f\), where \(f(x,y)=x^2+2y^2\). If we think of the gradient as a derivative, then \(f\) is an “antiderivative” of \(\vecs{F}\). In the case of single-variable integrals, the integral of derivative \(g′(x)\) is \(g(b)−g(a)\), where a is the start point of the interval of integration and b is the endpoint. If vector line integrals work like single-variable integrals, then we would expect integral \(\vecs{F}\) to be \(f(P_1)−f(P_0)\), where \(P_1\) is the endpoint of the curve of integration and \(P_0\) is the start point. Notice that this is the case for this example:
\[\int_C \vecs{F} \cdot d\vecs{r}=\int_C \vecs \nabla f \cdot d\vecs{r}=12 \nonumber \]
and
\[f(2,2)−f(0,0)=4+8−0=12. \nonumber \]
In other words, the integral of a “derivative” can be calculated by evaluating an “antiderivative” at the endpoints of the curve and subtracting, just as for single-variable integrals.
The following theorem says that, under certain conditions, what happened in the previous example holds for any gradient field. The same theorem holds for vector line integrals, which we call the Fundamental Theorem for Line Integrals .
Theorem: THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS
Let C be a piecewise smooth curve with parameterization \(\vecs r(t)\), \(a≤t≤b\). Let \(f\) be a function of two or three variables with first-order partial derivatives that exist and are continuous on C .  Therefore
\[\int_C \vecs F·d\vecs r=\int_C\vecs \nabla f·d\vecs{r}=f(\vecs r(b))−f(\vecs r(a)). \nonumber \]
In other words, just as with the Fundamental Theorem of Calculus, computing the line integral \(\int_C \vecs F·d\vecs{r}\), where \(\vecs{F}\) is conservative, is a two-step process:
- Find a potential function (“antiderivative”) \(f\) for \(\vecs{F}\) and
- Compute the value of \(f\) at the endpoints of \(C\) and calculate their difference \(f(\vecs r(b))−f(\vecs r(a))\).
Keep in mind, however, there is one major difference between the Fundamental Theorem of Calculus and the Fundamental Theorem for Line Integrals:
A function of one variable that is continuous must have an antiderivative. However, a vector field, even if it is continuous, does not need to have a potential function.
Example \(\PageIndex{3}\): Applying the Fundamental Theorem
Calculate integral \(\int_C \vecs{F} \cdot d\vecs{r}\), where \(\vecs{F}(x,y,z)=⟨2x\ln y,\dfrac{x^2}{y}+z^2,2yz⟩\) and \(C\) is a curve with parameterization \(\vecs{r}(t)=⟨t^2,t,t⟩\), \(1≤t≤e\)
- without using the Fundamental Theorem of Line Integrals and
- using the Fundamental Theorem of Line Integrals.
Solution
1. First, let’s calculate the integral without the Fundamental Theorem for Line Integrals and instead use the method we learned in the previous section:
\[\begin{align*} \int_C \vecs{F} \cdot dr &=\int_1^e\vecs F(\vecs r(t)) \cdot \vecs r′(t)\,dt\\[4pt] &=\int_1^e⟨2t^2\ln t,\dfrac{t^4}{t}+t^2,2t^2⟩ \cdot ⟨2t,1,1⟩\,dt\\[4pt] &=\int_1^e(4t^3\ln t+t^3+3t^2)\,dt \\[4pt] &=\int_1^e 4t^3\ln t \,dt+\int_1^e(t^3+3t^2)\,dt \\[4pt] &=\int_1^e 4t^3\ln t\,dt+{\Big[\dfrac{t^4}{4}+t^3\Big]}_1^e \\[4pt] &=\int_1^e 4t^3\ln t\,dt+\dfrac{e^4}{4}+e^3 −\dfrac{1}{4} −1 \\[4pt] &= \int_1^e 4t^3\ln t\,dt+\dfrac{e^4}{4}+e^3 −\dfrac{5}{4}\end{align*}\]
Integral \(\displaystyle \int_1^e t^3\ln t\,dt\) requires integration by parts. Let \(u=\ln t\) and \(dv=t^3\). Then \(u=\ln t\), \(dv=t^3\)
and
\[du=\dfrac{1}{t}\,dt, \;\;v=\dfrac{t^4}{4}.\nonumber \]
Therefore,
\[\begin{align*} \int_1^e t^3\ln t\,dt &={\Big[\dfrac{t^4}{4}\ln t\Big]}_1^e−\dfrac{1}{4}\int_1^e t^3\,dt \\[4pt] &=\dfrac{e^4}{4}−\dfrac{1}{4}\left(\dfrac{e^4}{4}−\dfrac{1}{4}\right). \end{align*}\]
Thus,
\[\begin{align*} \int_C \vecs F \cdot d\vecs{r} &= 4\int_1^e t^3\ln t\, dt\quad +\quad \dfrac{e^4}{4}+e^3 − \dfrac{5}{4} \\[4pt] &=4\left(\dfrac{e^4}{4}−\dfrac{1}{4}\left(\dfrac{e^4}{4}−\dfrac{1}{4}\right)\right)+\dfrac{e^4}{4}+e^3−\dfrac{5}{4}\\[4pt] &=e^4−\dfrac{e^4}{4}+\dfrac{1}{4}+\dfrac{e^4}{4}+e^3−\dfrac{5}{4} \\[4pt] &=e^4+e^3−1. \end{align*}\]
2. Given that \(f(x,y,z)=x^2\ln y+yz^2\) is a potential function for \(\vecs F\), let’s use the Fundamental Theorem for Line Integrals to calculate the integral. Note that
\[\begin{align*} \int_C \vecs F \cdot d\vecs{r} &=\int_C \vecs \nabla f \cdot d\vecs{r} \\[4pt] &=f(\vecs r(e))−f(\vecs r(1)) \\[4pt] &=f(e^2,e,e)−f(1,1,1)\\[4pt] &=e^4+e^3−1. \end{align*}\]
This calculation is much more straightforward than the calculation we did in (a). As long as we have a potential function, calculating a line integral using the Fundamental Theorem for Line Integrals is much easier than calculating without the theorem.
Example \(\PageIndex{3}\) illustrates a nice feature of the Fundamental Theorem of Line Integrals: it allows us to calculate more easily many vector line integrals. As long as we have a potential function, calculating the line integral is only a matter of evaluating the potential function at the endpoints and subtracting.
Exercise \(\PageIndex{3}\)
Given that \(f(x,y)={(x−1)}^2y+{(y+1)}^2x\) is a potential function for \(\vecs F(x,y)=⟨2xy−2y+{(y+1)}^2,{(x−1)}^2+2yx+2x⟩\), calculate integral \(\int_C \vecs F·d\vecs r\), where \(C\) is the lower half of the unit circle oriented counterclockwise.
- Hint
-
The Fundamental Theorem for Line Intervals says this integral depends only on the value of \(f\) at the endpoints of \(C\).
- Answer
-
2
The Fundamental Theorem for Line Integrals has two important consequences. The first consequence is that if \(\vecs{F}\) is conservative and \(C\) is a closed curve, then the circulation of \(\vecs{F}\) along \(C\) is zero—that is, \(\int_C \vecs F·d\vecs r=0\). To see why this is true, let \(f\) be a potential function for \(\vecs{F}\). Since \(C\) is a closed curve, the terminal point \(\vecs r(b)\) of \(C\) is the same as the initial \(\vecs r(a)\) of \(C\)—that is, \(\vecs r(a)=\vecs r(b)\). Therefore, by the Fundamental Theorem for Line Integrals,
\[\begin{align} \oint_C \vecs F·d\vecs r &=\oint_C \vecs \nabla f·d\vecs r\\[4pt] &=f(\vecs r(b))−f(\vecs r(a)) \\[4pt] &=f(\vecs r(b))−f(\vecs r(b)) \\[4pt] &=0. \end{align} \nonumber \]
Recall that the reason a conservative vector field \(\vecs{F}\) is called “conservative” is because such vector fields model forces in which energy is conserved. We have shown gravity to be an example of such a force. If we think of vector field \(\vecs{F}\) in integral \(\oint_C \vecs F·d\vecs r\) as a gravitational field, then the equation \(\oint_C \vecs{F}·d\vecs{r}=0\) follows. If a particle travels along a path that starts and ends at the same place, then the work done by gravity on the particle is zero.
The second important consequence of the Fundamental Theorem for Line Integrals (Equation \ref{FunTheLine}) is that line integrals of conservative vector fields are independent of path—meaning, they depend only on the endpoints of the given curve, and do not depend on the path between the endpoints.
DEFINITION: Path Independence
Let \(\vecs{F}\) be a vector field with domain \(D\); it is independent of path (or path independent) if
\[\int_{C_1} \vecs{F}·d\vecs{r}=\int_{C_2} \vecs{F}·d\vecs{r} \nonumber \]
for any paths \(C_1\) and \(C_2\) in \(D\) with the same initial and terminal points.
The second consequence is stated formally in the following theorem.
Theorem: CONSERVATIVE FIELDS
If \(\vecs{F}\) is a conservative vector field , then \(\vecs{F}\) is independent of path.
Proof
Let \(D\) denote the domain of \(\vecs{F}\) and let \(C_1\) and \(C_2\) be two paths in \(D\) with the same initial and terminal points (Figure \(\PageIndex{5}\)). Call the initial point \(P_1\) and the terminal point \(P_2\). Since \(\vecs{F}\) is conservative, there is a potential function \(f\) for \(\vecs{F}\). By the Fundamental Theorem for Line Integrals,
\[\int_{C_1} \vecs{F}·d\vecs{r}=f(P_2)−f(P_1)=\int_{C_2} \vecs{F}·d\vecs{r}. \nonumber \]
Therefore, \(\int_{C_1}\vecs F·d\vecs r=\int_{C_2}\vecs F·d\vecs r\) and \(\vecs{F}\) is independent of path.
\(\square\)
To visualize what independence of path means, imagine three hikers climbing from base camp to the top of a mountain. Hiker 1 takes a steep route directly from camp to the top. Hiker 2 takes a winding route that is not steep from camp to the top. Hiker 3 starts by taking the steep route but halfway to the top decides it is too difficult for him. Therefore he returns to camp and takes the non-steep path to the top. All three hikers are traveling along paths in a gravitational field. Since gravity is a force in which energy is conserved, the gravitational field is conservative. By independence of path, the total amount of work done by gravity on each of the hikers is the same because they all started in the same place and ended in the same place. The work done by the hikers includes other factors such as friction and muscle movement, so the total amount of energy each one expended is not the same, but the net energy expended against gravity is the same for all three hikers.
We have shown that if \(\vecs{F}\) is conservative, then \(\vecs{F}\) is independent of path. It turns out that if the domain of \(\vecs{F}\) is open and connected, then the converse is also true. That is, if \(\vecs{F}\) is independent of path and the domain of \(\vecs{F}\) is open and connected, then \(\vecs{F}\) is conservative. Therefore, the set of conservative vector fields on open and connected domains is precisely the set of vector fields independent of path.
Theorem: THE PATH INDEPENDENCE TEST FOR CONSERVATIVE FIELDS
If \(\vecs{F}\) is a continuous vector field that is independent of path and the domain \(D\) of \(\vecs{F}\) is open and connected, then \(\vecs{F}\) is conservative.
Proof
We prove the theorem for vector fields in \(ℝ^2\). The proof for vector fields in \(ℝ^3\) is similar. To show that \(\vecs F=⟨P,Q⟩\) is conservative, we must find a potential function \(f\) for \(\vecs{F}\). To that end, let \(X\) be a fixed point in \(D\). For any point \((x,y)\) in \(D\), let \(C\) be a path from \(X\) to \((x,y)\). Define \(f(x,y)\) by \(f(x,y)=\int_C \vecs F·d\vecs r\). (Note that this definition of \(f\) makes sense only because \(\vecs{F}\) is independent of path. If \(\vecs{F}\) was not independent of path, then it might be possible to find another path \(C′\) from \(X\) to \((x,y)\) such that \(\int_C \vecs F·d\vecs r≠\int_{C'} \vecs F·d\vecs r\), and in such a case \(f(x,y)\) would not be a function.) We want to show that \(f\) has the property \(\vecs \nabla f=\vecs F\).
Since domain \(D\) is open, it is possible to find a disk centered at \((x,y)\) such that the disk is contained entirely inside \(D\). Let \((a,y)\) with \(a<x\) be a point in that disk. Let \(C\) be a path from \(X\) to \((x,y)\) that consists of two pieces: \(C_1\) and \(C_2\). The first piece, \(C_1\), is any path from \(C\) to \((a,y)\) that stays inside \(D\); \(C_2\) is the horizontal line segment from \((a,y)\) to \((x,y)\) (Figure \(\PageIndex{6}\)). Then
\[f(x,y)=\int_{C_1} \vecs F·d\vecs r+\int_{C_2}\vecs F \cdot d\vecs r.\nonumber \]
The first integral does not depend on \(x\), so
\[f_x(x,y)=\dfrac{∂}{∂x}\int_{C_2} \vecs F \cdot d\vecs r. \nonumber \]
If we parameterize \(C_2\) by \(\vecs r(t)=⟨t,y⟩\), \(a≤t≤x\), then
\[\begin{align*} f_x(x,y) &=\dfrac{∂}{∂x}\int_{C_2} \vecs F \cdot d\vecs r \\[4pt] &=\dfrac{∂}{∂x}\int_a^x \vecs F(\vecs r(t)) \cdot \vecs r′(t)\,dt \\[4pt] &=\dfrac{∂}{∂x}\int_a^x \vecs F(\vecs r(t)) \cdot \dfrac{d}{dt}(⟨t,y⟩)\,dt \\[4pt] &=\dfrac{∂}{∂x}\int_a^x \vecs F(\vecs r(t)) \cdot ⟨1,0⟩\,dt \\[4pt] &=\dfrac{∂}{∂x}\int_a^x P(t,y)\,dt.\\[4pt] \end{align*}\]
By the Fundamental Theorem of Calculus (part 1),
\[f_x(x,y)=\dfrac{∂}{∂x}\int_a^x P(t,y)\,dt=P(x,y).\nonumber \]
A similar argument using a vertical line segment rather than a horizontal line segment shows that \(f_y(x,y)=Q(x,y)\).
Therefore \(\vecs \nabla f=\vecs F\) and \(\vecs{F}\) is conservative.
\(\square\)
We have spent a lot of time discussing and proving the theorems above, but we can summarize them simply: a vector field \(\vecs F\) on an open and connected domain is conservative if and only if it is independent of path. This is important to know because conservative vector fields are extremely important in applications, and these theorems give us a different way of viewing what it means to be conservative using path independence.
Example \(\PageIndex{4}\): Showing That a Vector Field Is Not Conservative
Use path independence to show that vector field \(\vecs F(x,y)=⟨x^2y,y+5⟩\) is not conservative.
Solution
We can indicate that \(\vecs{F}\) is not conservative by showing that \(\vecs{F}\) is not path independent. We do so by giving two different paths, \(C_1\) and \(C_2\), that both start at \((0,0)\) and end at \((1,1)\), and yet \(\int_{C_1} \vecs F \cdot d\vecs r≠\int_{C_2} \vecs F \cdot d\vecs r\).
Let \(C_1\) be the curve with parameterization \(\vecs r_1(t)=⟨t,\,t⟩\), \(0≤t≤1\) and let \(C_2\) be the curve with parameterization \(\vecs r_2(t)=⟨t,\,t^2⟩\), \(0≤t≤1\) (Figure \(\PageIndex{7}\).). Then
\[\begin{align*} \int_{C_1} \vecs{F}·d\vecs r &=\int_0^1 \vecs F(\vecs r_1(t))·\vecs r_1′(t)\,dt \\[4pt] &=\int_0^1⟨t^3,t+5⟩·⟨1,1⟩\,dt=\int_0^1(t^3+t+5)\,dt\\[4pt] &={\Big[\dfrac{t^4}{4}+\dfrac{t^2}{2}+5t\Big]}_0^1=\dfrac{23}{4} \end{align*}\]
and
\[\begin{align*} \int_{C_2}\vecs F·d\vecs r &=\int_0^1 \vecs F(\vecs r_2(t))·\vecs r_2′(t)\,dt \\[4pt] &=\int_0^1⟨t^4,t^2+5⟩·⟨1,2t⟩\,dt=\int_0^1(t^4+2t^3+10t)\,dt \\[4pt] &={\Big[\dfrac{t^5}{5}+\dfrac{t^4}{2}+5t^2\Big]}_0^1=\dfrac{57}{10}. \end{align*}\]
Since \(\int_{C_1} \vecs F \cdot d\vecs r≠\int_{C_2} \vecs F \cdot d\vecs r\), the value of a line integral of \(\vecs{F}\) depends on the path between two given points. Therefore, \(\vecs{F}\) is not independent of path, and \(\vecs{F}\) is not conservative.
Exercise \(\PageIndex{4}\)
Show that \(\vecs{F}(x,y)=⟨xy,\,x^2y^2⟩\) is not path independent by considering the line segment from \((0,0)\) to \((0,2)\) and the piece of the graph of \(y=\dfrac{x^2}{2}\) that goes from \((0,0)\) to \((0,2)\).
- Hint
-
Calculate the corresponding line integrals.
- Answer
-
If \(C_1\) and \(C_2\) represent the two curves, then \[\int_{C_1} \vecs F \cdot d\vecs r≠\int_{C_2} \vecs F \cdot d\vecs r. \nonumber \]
Conservative Vector Fields and Potential Functions
As we have learned, the Fundamental Theorem for Line Integrals says that if \(\vecs{F}\) is conservative, then calculating \(\int_C \vecs F·d\vecs r\) has two steps: first, find a potential function \(f\) for \(\vecs{F}\) and, second, calculate \(f(P_1)−f(P_0)\), where \(P_1\) is the endpoint of \(C\) and \(P_0\) is the starting point. To use this theorem for a conservative field \(\vecs{F}\), we must be able to find a potential function \(f\) for \(\vecs{F}\). Therefore, we must answer the following question: Given a conservative vector field \(\vecs{F}\), how do we find a function \(f\) such that \(\vecs \nabla f=\vecs{F}\)? Before giving a general method for finding a potential function, let’s motivate the method with an example.
Example \(\PageIndex{5}\): Finding a Potential Function
Find a potential function for \(\vecs F(x,y)=⟨2xy^3,3x^2y^2+\cos(y)⟩\), thereby showing that \(\vecs{F}\) is conservative.
Solution
Suppose that \(f(x,y)\) is a potential function for \(\vecs{F}\). Then, \(\vecs \nabla f=\vecs F\), and therefore
\[f_x(x,y)=2xy^3 \; \; \text{and} \;\; f_y(x,y)=3x^2y^2+\cos y. \nonumber \]
Integrating the equation \(f_x(x,y)=2xy^3\) with respect to \(x\) yields the equation
\[f(x,y)=x^2y^3+h(y). \nonumber \]
Notice that since we are integrating a two-variable function with respect to \(x\), we must add a constant of integration that is a constant with respect to \(x\), but may still be a function of \(y\). The equation \(f(x,y)=x^2y^3+h(y)\) can be confirmed by taking the partial derivative with respect to \(x\):
\[\dfrac{∂f}{∂x}=\dfrac{∂}{∂x}(x^2y^3)+\dfrac{∂}{∂x}(h(y))=2xy^3+0=2xy^3. \nonumber \]
Since \(f\) is a potential function for \(\vecs{F}\),
\[f_y(x,y)=3x^2y^2+\cos(y), \nonumber \]
and therefore
\[3x^2y^2+g′(y)=3x^2y^2+\cos(y). \nonumber \]
This implies that \(h′(y)=\cos y\), so \(h(y)=\sin y+C\). Therefore, any function of the form \(f(x,y)=x^2y^3+\sin(y)+C\) is a potential function. Taking, in particular, \(C=0\) gives the potential function \(f(x,y)=x^2y^3+\sin(y)\).
To verify that \(f\) is a potential function, note that \(\vecs \nabla f(x,y)=⟨2xy^3,3x^2y^2+\cos y⟩=\vecs F\).
Exercise \(\PageIndex{5}\)
Find a potential function for \(\vecs{F}(x,y)=⟨e^xy^3+y,3e^xy^2+x⟩\).
- Hint
-
Follow the steps in Example \(\PageIndex{5}\).
- Answer
-
\(f(x,y)=e^xy^3+xy\)
The logic of the previous example extends to finding the potential function for any conservative vector field in \(ℝ^2\). Thus, we have the following problem-solving strategy for finding potential functions:
PROBLEM-SOLVING STRATEGY: FINDING A POTENTIAL FUNCTION FOR A CONSERVATIVE VECTOR FIELD \(\vecs{F}(x,y)=⟨P(x,y),Q(x,y)⟩\)
- Integrate \(P\) with respect to \(x\). This results in a function of the form \(g(x,y)+h(y)\), where \(h(y)\) is unknown.
- Take the partial derivative of \(g(x,y)+h(y)\) with respect to \(y\), which results in the function \(gy(x,y)+h′(y)\).
- Use the equation \(gy(x,y)+h′(y)=Q(x,y)\) to find \(h′(y)\).
- Integrate \(h′(y)\) to find \(h(y)\).
- Any function of the form \(f(x,y)=g(x,y)+h(y)+C\), where \(C\) is a constant, is a potential function for \(\vecs{F}\).
We can adapt this strategy to find potential functions for vector fields in \(ℝ^3\), as shown in the next example.
Example \(\PageIndex{6}\): Finding a Potential Function in \(ℝ^3\)
Find a potential function for \(F(x,y,z)=⟨2xy,x^2+2yz^3,3y^2z^2+2z⟩\), thereby showing that \(\vecs{F}\) is conservative.
Solution
Suppose that \(f\) is a potential function. Then, \(\vecs \nabla f= \vecs{F}\) and therefore \(f_x(x,y,z)=2xy\). Integrating this equation with respect to \(x\) yields the equation \(f(x,y,z)=x^2y+g(y,z)\) for some function \(g\). Notice that, in this case, the constant of integration with respect to \(x\) is a function of \(y\) and \(z\).
Since \(f\) is a potential function,
\[x^2+2yz^3=f_y(x,y,z)=x^2+g_y(y,z). \nonumber \]
Therefore,
\[g_y(y,z)=2yz^3. \nonumber \]
Integrating this function with respect to \(y\) yields
\[g(y,z)=y^2z^3+h(z) \nonumber \]
for some function \(h(z)\) of \(z\) alone. (Notice that, because we know that \(g\) is a function of only \(y\) and \(z\), we do not need to write \(g(y,z)=y^2z^3+h(x,z)\).) Therefore,
\[f(x,y,z)=x^2y+g(y,z)=x^2y+y^2z^3+h(z). \nonumber \]
To find \(f\), we now must only find \(h\). Since \(f\) is a potential function,
\[3y^2z^2+2z=g_z(y,z)=3y^2z^2+h′(z). \nonumber \]
This implies that \(h′(z)=2z\), so \(h(z)=z^2+C\). Letting \(C=0\) gives the potential function
\[f(x,y,z)=x^2y+y^2z^3+z^2. \nonumber \]
To verify that \(f\) is a potential function, note that \(\vecs \nabla f(x,y,z)=⟨2xy,x^2+2yz^3,3y^2z^2+2z⟩=\vecs F(x,y,z)\).
Exercise \(\PageIndex{6}\)
Find a potential function for \(\vecs{F}(x,y,z)=⟨12x^2,\cos y\cos z,1−\sin y\sin z⟩\).
- Hint
-
Following Example \(\PageIndex{6}\), begin by integrating with respect to \(x\) .
- Answer
-
\(f(x,y,z)=4x^3+\sin y\cos z+z\)
We can apply the process of finding a potential function to a gravitational force. Recall that, if an object has unit mass and is located at the origin, then the gravitational force in \(ℝ^2\) that the object exerts on another object of unit mass at the point \((x,y)\) is given by vector field
\(\vecs F(x,y)=−G\left\langle\dfrac{x}{ {(x^2+y^2)}^{3/2} },\dfrac{y}{ {(x^2+y^2)}^{3/2} }\right\rangle\),
where \(G\) is the universal gravitational constant. In the next example, we build a potential function for \(\vecs{F}\), thus confirming what we already know: that gravity is conservative.
Example \(\PageIndex{7}\): Finding a Potential Function
Find a potential function \(f\) for \(\vecs{F}(x,y)=−G\left\langle\dfrac{x}{ {(x^2+y^2)}^{3/2} },\dfrac{y}{ {(x^2+y^2)}^{3/2} }\right\rangle\).
Solution
Suppose that \(f\) is a potential function. Then, \(\vecs \nabla f= \vecs{F}\) and therefore
\[f_x(x,y)=\dfrac{−Gx}{ {(x^2+y^2)}^{3/2} }.\nonumber \]
To integrate this function with respect to \(x\) , we can use \(u\)-substitution. If \(u=x^2+y^2\), then \(\dfrac{du}{2}=x\,dx\), so
\[\begin{align*} \int \dfrac{−Gx}{ {(x^2+y^2)}^{3/2} }\,dx &=\int \dfrac{−G}{2u^{3/2}} \,du \\[4pt] &=\dfrac{G}{\sqrt{u}}+h(y) \\[4pt] &=\dfrac{G}{\sqrt{x^2+y^2}}+h(y) \end{align*}\]
for some function \(h(y)\). Therefore,
\[f(x,y)=\dfrac{G}{ \sqrt{x^2+y^2}}+h(y).\nonumber \]
Since \(f\) is a potential function for \(\vecs{F}\),
\[f_y(x,y)=\dfrac{−Gy}{ {(x^2+y^2)}^{3/2} }\nonumber \].
Since \(f(x,y)=\dfrac{G}{ \sqrt{x^2+y^2}}+h(y)\), \(f_y(x,y)\) also equals \(\dfrac{−Gy}{ {(x^2+y^2)}^{3/2} }+h′(y)\).
Therefore,
\[\dfrac{−Gy}{ {(x^2+y^2)}^{3/2} }+h′(y)=\dfrac{−Gy}{ {(x^2+y^2)}^{3/2} }, \nonumber \]
which implies that \(h′(y)=0\). Thus, we can take \(h(y)\) to be any constant; in particular, we can let \(h(y)=0\). The function
\[f(x,y)=\dfrac{G}{ \sqrt{x^2+y^2} } \nonumber \]
is a potential function for the gravitational field \(\vecs{F}\). To confirm that \(f\) is a potential function, note that
\[\begin{align*} \vecs\nabla f(x,y) &=⟨−\dfrac{1}{2} \dfrac{G}{ {(x^2+y^2)}^{3/2} } (2x),−\dfrac{1}{2} \dfrac{G}{ {(x^2+y^2)}^{3/2} }(2y)⟩ \\[4pt] &=⟨\dfrac{−Gx}{ {(x^2+y^2)}^{3/2} },\dfrac{−Gy}{ {(x^2+y^2)}^{3/2} }⟩\\[4pt] &=\vecs F(x,y). \end{align*}\]
Exercise \(\PageIndex{7}\)
Find a potential function \(f\) for the three-dimensional gravitational force \(\vecs{F}(x,y,z)=\left\langle\dfrac{−Gx}{ {(x^2+y^2+z^2)}^{3/2} },\dfrac{−Gy}{ {(x^2+y^2+z^2)}^{3/2} },\dfrac{−Gz}{ {(x^2+y^2+z^2)}^{3/2} }\right\rangle\).
- Hint
-
Follow the Problem-Solving Strategy.
- Answer
-
\(f(x,y,z)=\dfrac{G}{\sqrt{x^2+y^2+z^2}}\)
Testing a Vector Field
Until now, we have worked with vector fields that we know are conservative, but if we are not told that a vector field is conservative, we need to be able to test whether it is conservative. Recall that, if \(\vecs{F}\) is conservative, then \(\vecs{F}\) has the cross-partial property (see The Cross-Partial Property of Conservative Vector Fields). That is, if \(\vecs F=⟨P,Q,R⟩\) is conservative, then \(P_y=Q_x\), \(P_z=R_x\), and \(Q_z=R_y\). So, if \(\vecs{F}\) has the cross-partial property, then is \(\vecs{F}\) conservative? If the domain of \(\vecs{F}\) is open and simply connected, then the answer is yes.
Theorem: THE CROSS-PARTIAL TEST FOR CONSERVATIVE FIELDS
If \(\vecs{F}=⟨P,Q,R⟩\) is a vector field on an open, simply connected region \(D\) and \(P_y=Q_x\), \(P_z=R_x\), and \(Q_z=R_y\) throughout \(D\), then \(\vecs{F}\) is conservative.
Although a proof of this theorem is beyond the scope of the text, we can discover its power with some examples. Later, we see why it is necessary for the region to be simply connected.
Combining this theorem with the cross-partial property, we can determine whether a given vector field is conservative:
Theorem: CROSS-PARTIAL PROPERTY OF CONSERVATIVE FIELDS
Let \(\vecs{F}=⟨P,Q,R⟩\) be a vector field on an open, simply connected region \(D\) . Then \(P_y=Q_x\), \(P_z=R_x\), and \(Q_z=R_y\) throughout \(D\) if and only if \(\vecs{F}\) is conservative.
The version of this theorem in \(ℝ^2\) is also true. If \(\vecs F(x,y)=⟨P,Q⟩\) is a vector field on an open, simply connected domain in \(ℝ^2\), then \(\vecs F\) is conservative if and only if \(P_y=Q_x\).
Example \(\PageIndex{8}\): Determining Whether a Vector Field Is Conservative
Determine whether vector field \(\vecs F(x,y,z)=⟨xy^2z,x^2yz,z^2⟩\) is conservative.
Solution
Note that the domain of \(\vecs{F}\) is all of \(ℝ^2\) and \(ℝ^3\) is simply connected. Therefore, we can use The Cross-Partial Property of Conservative Vector Fields to determine whether \(\vecs{F}\) is conservative. Let
\[P(x,y,z)=xy^2z \nonumber \]
\[Q(x,y,z)=x^2yz \nonumber \]
and
\[R(x,y,z)=z^2.\nonumber \]
Since \(Q_z(x,y,z)=x^2y\) and \(R_y(x,y,z)=0\), the vector field is not conservative.
Example \(\PageIndex{9}\): Determining Whether a Vector Field Is Conservative
Determine vector field \(\vecs{F}(x,y)=⟨x\ln (y), \,\dfrac{x^2}{2y}⟩\) is conservative.
Solution
Note that the domain of \(\vecs{F}\) is the part of \(ℝ^2\) in which \(y>0\). Thus, the domain of \(\vecs{F}\) is part of a plane above the \(x\)-axis, and this domain is simply connected (there are no holes in this region and this region is connected). Therefore, we can use The Cross-Partial Property of Conservative Vector Fields to determine whether \(\vecs{F}\) is conservative. Let
\[P(x,y)=x\ln (y) \;\; \text{and} \;\;\ Q(x,y)=\dfrac{x^2}{2y}. \nonumber \]
Then \(P_y(x,y)=\dfrac{x}{y}=Q_x(x,y)\) and thus \(\vecs{F}\) is conservative.
Exercise \(\PageIndex{8}\)
Determine whether \(\vecs{F}(x,y)=⟨\sin x\cos y,\,\cos x\sin y⟩\) is conservative.
- Hint
-
Use The Cross-Partial Property of Conservative Vector Fields from the previous section.
- Answer
-
It is conservative.
When using The Cross-Partial Property of Conservative Vector Fields , it is important to remember that a theorem is a tool, and like any tool, it can be applied only under the right conditions. In the case of The Cross-Partial Property of Conservative Vector Fields , the theorem can be applied only if the domain of the vector field is simply connected.
To see what can go wrong when misapplying the theorem, consider the vector field from Example \(\PageIndex{4}\):
\[\vecs F(x,y)=\dfrac{y}{x^2+y^2}\,\hat{\mathbf i}+\dfrac{−x}{x^2+y^2}\,\hat{\mathbf j}. \nonumber \]
This vector field satisfies the cross-partial property, since
\[\dfrac{∂}{∂y}\left(\dfrac{y}{x^2+y^2}\right)=\dfrac{(x^2+y^2)−y(2y)}{ {(x^2+y^2)}^2}=\dfrac{x^2−y^2}{ {(x^2+y^2)}^2} \nonumber \]
and
\[\dfrac{∂}{∂x}\left(\dfrac{−x}{x^2+y^2}\right)=\dfrac{−(x^2+y^2)+x(2x)}{ {(x^2+y^2)}^2}=\dfrac{x^2−y^2}{ {(x^2+y^2)}^2}. \nonumber \]
Since \(\vecs{F}\) satisfies the cross-partial property, we might be tempted to conclude that \(\vecs{F}\) is conservative. However, \(\vecs{F}\) is not conservative. To see this, let
\[\vecs r(t)=⟨\cos t,\sin t⟩,\;\; 0≤t≤\pi \nonumber \]
be a parameterization of the upper half of a unit circle oriented counterclockwise (denote this \(C_1\)) and let
\[\vecs s(t)=⟨\cos t,−\sin t⟩,\;\; 0≤t≤\pi \nonumber \]
be a parameterization of the lower half of a unit circle oriented clockwise (denote this \(C_2\)). Notice that \(C_1\) and \(C_2\) have the same starting point and endpoint. Since \({\sin}^2 t+{\cos}^2 t=1\),
\[\vecs F(\vecs r(t)) \cdot \vecs r′(t)=⟨\sin(t),−\cos(t)⟩ \cdot ⟨−\sin(t), \cos(t)⟩=−1 \nonumber \]
and
\[\vecs F(\vecs s(t))·\vecs s′(t)=⟨−\sin t,−\cos t⟩·⟨−\sin t,−\cos t⟩={\sin}^2 t+{\cos}^2t=1. \nonumber \]
Therefore,
\[\int_{C_1} \vecs F·d\vecs r=\int_0^{\pi}−1\,dt=−\pi \nonumber \]
and
\[\int_{C_2}\vecs F·d\vecs r=\int_0^{\pi} 1\,dt=\pi. \nonumber \]
Thus, \(C_1\) and \(C_2\) have the same starting point and endpoint, but \(\int_{C_1} \vecs F·d\vecs r≠\int_{C_2} \vecs F·d\vecs r\). Therefore, \(\vecs{F}\) is not independent of path and \(\vecs{F}\) is not conservative.
To summarize: \(\vecs{F}\) satisfies the cross-partial property and yet \(\vecs{F}\) is not conservative. What went wrong? Does this contradict The Cross-Partial Property of Conservative Vector Fields ? The issue is that the domain of \(\vecs{F}\) is all of \(ℝ^2\) except for the origin. In other words, the domain of \(\vecs{F}\) has a hole at the origin, and therefore the domain is not simply connected. Since the domain is not simply connected, The Cross-Partial Property of Conservative Vector Fields does not apply to \(\vecs{F}\).
We close this section by looking at an example of the usefulness of the Fundamental Theorem for Line Integrals. Now that we can test whether a vector field is conservative, we can always decide whether the Fundamental Theorem for Line Integrals can be used to calculate a vector line integral. If we are asked to calculate an integral of the form \(\int_C \vecs F·d\vecs r\), then our first question should be: Is \(\vecs{F}\) conservative? If the answer is yes, then we should find a potential function and use the Fundamental Theorem for Line Integrals to calculate the integral. If the answer is no, then the Fundamental Theorem for Line Integrals cannot help us and we have to use other methods, such as using the method from the previous section (using \(\vecs F(\vecs r(t))\) and \(\vecs r'(t)\)).
Example \(\PageIndex{10}\): Using the Fundamental Theorem for Line Integrals
Calculate line integral \(\int_C \vecs F·d\vecs r\), where \(\vecs F(x,y,z)=⟨2xe^yz+e^xz,\,x^2e^yz,\,x^2e^y+e^x⟩\) and \(C\) is any smooth curve that goes from the origin to \((1,1,1)\).
Solution
Before trying to compute the integral, we need to determine whether \(\vecs{F}\) is conservative and whether the domain of \(\vecs{F}\) is simply connected. The domain of \(\vecs{F}\) is all of \(ℝ^3\), which is connected and has no holes. Therefore, the domain of \(\vecs{F}\) is simply connected. Let
\[P(x,y,z)=2xe^yz+e^xz, \;\; Q(x,y,z)=x^2e^yz, \;\; \text{and} \;\; R(x,y,z)=x^2e^y+e^x \nonumber \]
so that \(\vecs{F}(x,y,z)=⟨P,Q,R⟩\). Since the domain of \(\vecs{F}\) is simply connected, we can check the cross partials to determine whether \(\vecs{F}\) is conservative. Note that
\[\begin{align*} P_y(x,y,z) &=2xe^yz=Q_x(x,y,z) \\[4pt]P_z(x,y,z) &=2xe^y+e^x=R_x(x,y,z) \\[4pt] Q_z(x,y,z) &=x^2e^y=R_y(x,y,z).\end{align*}\]
Therefore, \(\vecs{F}\) is conservative.
To evaluate \(\int_C \vecs F·d\vecs r\) using the Fundamental Theorem for Line Integrals, we need to find a potential function \(f\) for \(\vecs{F}\). Let \(f\) be a potential function for \(\vecs{F}\). Then, \(\vecs \nabla f=\vecs F\), and therefore \(f_x(x,y,z)=2xe^yz+e^xz\). Integrating this equation with respect to \(x\) gives \(f(x,y,z)=x^2e^yz+e^xz+h(y,z)\) for some function \(h\). Differentiating this equation with respect to \(y\) gives \(x^2e^yz+h_y(y,z)=Q(x,y,z)=x^2e^yz\), which implies that \(h_y(y,z)=0\). Therefore, \(h\) is a function of \(z\) only, and \(f(x,y,z)=x^2e^yz+e^xz+h(z)\). To find \(h\), note that \(f_z=x^2e^y+e^x+h′(z)=R=x^2e^y+e^x\). Therefore, \(h′(z)=0\) and we can take \(h(z)=0\). A potential function for \(\vecs{F}\) is \(f(x,y,z)=x^2e^yz+e^xz\).
Now that we have a potential function, we can use the Fundamental Theorem for Line Integrals to evaluate the integral. By the theorem,
\[\begin{align*} \int_C \vecs F·d\vecs r &=\int_C \vecs \nabla f·d\vecs r\\[4pt] &=f(1,1,1)−f(0,0,0)\\[4pt] &=2e. \end{align*}\]
Analysis
Notice that if we hadn’t recognized that \(\vecs{F}\) is conservative, we would have had to parameterize \(C\) and use the method from the previous section. Since curve \(C\) is unknown, using the Fundamental Theorem for Line Integrals is much simpler.
Exercise \(\PageIndex{9}\)
Calculate integral \(\int_C \vecs F·d\vecs r\), where \(\vecs{F}(x,y)=⟨\sin x\sin y, 5−\cos x\cos y⟩\) and \(C\) is a semicircle with starting point \((0,\pi)\) and endpoint \((0,−\pi)\).
- Hint
-
Use the Fundamental Theorem for Line Integrals.
- Answer
-
\(−10\pi\)
Example \(\PageIndex{11}\): Work Done on a Particle
Let \(\vecs F(x,y)=⟨2xy^2,2x^2y⟩\) be a force field. Suppose that a particle begins its motion at the origin and ends its movement at any point in a plane that is not on the \(x\)-axis or the \(y\)-axis. Furthermore, the particle’s motion can be modeled with a smooth parameterization. Show that \(\vecs{F}\) does positive work on the particle.
Solution
We show that \(\vecs{F}\) does positive work on the particle by showing that \(\vecs{F}\) is conservative and then by using the Fundamental Theorem for Line Integrals.
To show that \(\vecs{F}\) is conservative, suppose \(f(x,y)\) were a potential function for \(\vecs{F}\). Then, \(\vecs \nabla f(x,y)=\vecs F(x,y)=⟨2xy^2,2x^2y⟩\) and therefore \(f_x(x,y)=2xy^2\) and \(f_y(x,y)=2x^2y\). The equation \(fx(x,y)=2xy^2\) implies that \(f(x,y)=x^2y^2+h(y)\). Deriving both sides with respect to \(y\) yields \(f_y(x,y)=2x^2y+h′(y)\). Therefore, \(h′(y)=0\) and we can take \(h(y)=0\).
If \(f(x,y)=x^2y^2\), then note that \(\vecs \nabla f(x,y)=⟨2xy^2,2x^2y⟩=\vecs F\), and therefore \(f\) is a potential function for \(\vecs{F}\).
Let \((a,b)\) be the point at which the particle stops is motion, and let \(C\) denote the curve that models the particle’s motion. The work done by \(\vecs{F}\) on the particle is \(\int_C \vecs{F}·d\vecs{r}\). By the Fundamental Theorem for Line Integrals,
\[\begin{align*} \int_C \vecs F·d\vecs r &=\int_C \nabla f·d\vecs r \\[4pt] &=f(a,b)−f(0,0)\\[4pt] &=a^2b^2. \end{align*}\]
Since \(a≠0\) and \(b≠0\), by assumption, \(a^2b^2>0\). Therefore, \(\int_C \vecs F·d\vecs r>0\), and \(\vecs{F}\) does positive work on the particle.
Analysis
Notice that this problem would be much more difficult without using the Fundamental Theorem for Line Integrals. To apply the tools we have learned, we would need to give a curve parameterization and use the method from the previous section. Since the path of motion \(C\) can be as exotic as we wish (as long as it is smooth), it can be very difficult to parameterize the motion of the particle.
Exercise \(\PageIndex{10}\)
Let \(\vecs{F}(x,y)=⟨4x^3y^4,4x^4y^3⟩\), and suppose that a particle moves from point \((4,4)\) to \((1,1)\) along any smooth curve. Is the work done by \(\vecs{F}\) on the particle positive, negative, or zero?
- Hint
-
Use the Fundamental Theorem for Line Integrals.
- Answer
-
Negative
Key Concepts
- The theorems in this section require curves that are closed, simple, or both, and regions that are connected or simply connected.
- The line integral of a conservative vector field can be calculated using the Fundamental Theorem for Line Integrals. This theorem is a generalization of the Fundamental Theorem of Calculus in higher dimensions. Using this theorem usually makes the calculation of the line integral easier.
- Conservative fields are independent of path. The line integral of a conservative field depends only on the value of the potential function at the endpoints of the domain curve.
- Given vector field \(\vecs{F}\), we can test whether \(\vecs{F}\) is conservative by using the cross-partial property. If \(\vecs{F}\) has the cross-partial property and the domain is simply connected, then \(\vecs{F}\) is conservative (and thus has a potential function). If \(\vecs{F}\) is conservative, we can find a potential function by using the Problem-Solving Strategy.
- The circulation of a conservative vector field on a simply connected domain over a closed curve is zero.
Key Equations
-
Fundamental Theorem for Line Integrals
\(\displaystyle \int_C \vecs \nabla f·d\vecs r=f(\vecs r(b))−f(\vecs r(a))\) -
Circulation of a conservative field over curve
C
that encloses a simply connected region
\(\displaystyle \oint_C \vecs \nabla f·d\vecs r=0\)
Glossary
- closed curve
- a curve that begins and ends at the same point
- connected region
- a region in which any two points can be connected by a path with a trace contained entirely inside the region
- Fundamental Theorem for Line Integrals
- the value of line integral \(\displaystyle \int_C\vecs ∇f⋅d\vecs r\) depends only on the value of \(f\) at the endpoints of \(C: \displaystyle \int_C \vecs ∇f⋅d\vecs r=f(\vecs r(b))−f(\vecs r(a))\)
- independence of path
- a vector field \(\vecs{F}\) has path independence if \(\displaystyle \int_{C_1} \vecs F⋅d\vecs r=\displaystyle \int_{C_2} \vecs F⋅d\vecs r\) for any curves \(C_1\) and \(C_2\) in the domain of \(\vecs{F}\) with the same initial points and terminal points
- simple curve
- a curve that does not cross itself
- simply connected region
- a region that is connected and has the property that any closed curve that lies entirely inside the region encompasses points that are entirely inside the region


5.3: Placement, Style, and Coloring
Learning Outcome
- Describe the impact of placement, style, and coloring when incorporating graphics into a message
Once you’ve decided what graph or chart to use, we need to make sure it fits with our visual media usage standards. It needs to make your message more accessible by being:
- Clean and simple
- Uniform
- Persuasive
- On brand
Let’s build a pie chart and apply each of these standards to it as we go along.
Say you want to give a quarter bonus to the best member of each department, and you have settled on giving a $50 gift card for a movie theater. You have surveyed the company to determine which local cinemas are used by your employees to see movies. You have decided to make a pie chart from the results, because you’re looking to show the composition of your employees and their movie-going preferences.
At the moment, your pie chart looks like this:
Figure 1 is definitely not clean and simple. There is so much to look at here that it’s hard to see anything. No one will find this useful. In fact, a rule of thumb for pie charts is that if you have more than ten categories, you should present the information differently. (There’s something called an exploding pie chart if you want to check that out.)
So how do we simplify the data? To begin with, we need to narrow down the categories. In this case, you could display theaters by company, rather than location, since all theaters of the same company will take the same gift card. Let’s take a look at how the revised chart would look in Figure 2:
Now we’re getting somewhere! This is a much easier chart to read. We can see at a glance that nearly half of our employees go to a movie theater owned by AAA, and smaller portions see films at the BBB and CCC theaters. This information has become way more accessible for the reader.
We’ve also chosen some bold, pleasing colors here. We can clearly tell what part of the pie belongs to BBB and what part belongs to the others. This is clean and simple!
Once we’ve achieved that, we must make it uniform. Do the rest of your charts, tables and graphs use the same bold, pleasing colors we use here? Have we chosen the same font size for our chart key? If the answer is yes, we can move on. We know that we are sharing information without distracting the reader.
Now we ask ourselves, is this data persuasive? Well, that depends on the story you’re looking to tell. In this case, your best option is to purchase gift cards for AAA theaters because that is what your data is telling, since most of your employees enjoy seeing movies there. So let’s give this a title:
Figure 3 a bit more persuasive. This immediately tells the reader “AAA Complexes are our employees’ favorite movie theater.”
Now, is your chart on brand? If your company’s palette of colors includes blue, red, yellow and green, then yes! We are on brand. If your company uses different colors, go ahead and change them. A lot of companies have very specific color requirements, even for internal projects, so be sure to look out for your company’s style requirements! For this project, it’s as easy as that.
You’ve made the point of this communication very easy and accessible by making these changes and following our standards of visual media communication!
- Misleading Graphs: Real Life Examples
- Good Data, Bad Graphs


5.3: Mean and Standard Deviation of Binomial Distribution
If you list all possible values of \(x\) in a Binomial distribution, you get the Binomial Probability Distribution (pdf). You can draw a histogram of the pdf and find the mean, variance, and standard deviation of it.
For a general discrete probability distribution, you can find the mean, the variance, and the standard deviation for a pdf using the general formulas
\(\mu=\sum x P(x), \sigma^{2}=\sum(x-\mu)^{2} P(x), \text { and } \sigma=\sqrt{\sum(x-\mu)^{2} P(x)}\)
These formulas are useful, but if you know the type of distribution, like Binomial, then you can find the mean and standard deviation using easier formulas. They are derived from the general formulas.
Note
For a Binomial distribution, \(\mu\), the expected number of successes, \(\sigma^{2}\), the variance, and \(\sigma\), the standard deviation for the number of success are given by the formulas:
\(\mu=n p \quad \sigma^{2}=n p q \quad \sigma=\sqrt{n p q}\)
Where p is the probability of success and q = 1 - p .
Example \(\PageIndex{1}\) Finding the Probability Distribution, Mean, Variance, and Standard Deviation of a Binomial Distribution
When looking at a person’s eye color, it turns out that 1% of people in the world has green eyes ("What percentage of," 2013). Consider a group of 20 people.
- State the random variable.
- Write the probability distribution.
- Draw a histogram.
- Find the mean.
- Find the variance.
- Find the standard deviation.
Solution
a. x = number of people who have green eyes
b. In this case you need to write each value of x and its corresponding probability. It is easiest to do this by using the binompdf command, but don’t put in the r value. You may want to set your calculator to only three decimal places, so it is easier to see the values and you don’t need much more precision than that. The command would look like \(\text{binompdf}(20, .01)\). This produces the information in Example \(\PageIndex{1}\).
| x | P (x=r) |
|---|---|
| 0 | 0.818 |
| 1 | 0.165 |
| 2 | 0.016 |
| 3 | 0.001 |
| 4 | 0.000 |
| 5 | 0.000 |
| 6 | 0.000 |
| 7 | 0.000 |
| 8 | 0.000 |
| 9 | 0.000 |
| 10 | 0.000 |
| \(\vdots\) | \(\vdots\) |
| 20 | 0.000 |
Notice that after x = 4, the probability values are all 0.000. This just means they are really small numbers.
c. You can draw the histogram on the TI-83/84 or other technology. The graph would look like in Figure \(\PageIndex{1}\) .
Figure \(\PageIndex{1}\): Histogram Created on TI-83/84This graph is very skewed to the right.
d. Since this is a binomial, then you can use the formula \(\mu=n p\). So \(\mu=20(0.01)=0.2\) people.
You expect on average that out of 20 people, less than 1 would have green eyes.
e. Since this is a binomial, then you can use the formula \(\sigma^{2}=n p q\).
\(q=1-0.01=0.99\)
\(\sigma^{2}=20(0.01)(0.99)=0.198 \text { people }^{2}\)
f. Once you have the variance, you just take the square root of the variance to find the standard deviation.
\(\sigma=\sqrt{0.198} \approx 0.445\)
Homework
Exercise \(\PageIndex{1}\)
-
Suppose a random variable,
x
, arises from a binomial experiment. Suppose
n
= 6, and
p
= 0.13.
- Write the probability distribution.
- Draw a histogram.
- Describe the shape of the histogram.
- Find the mean.
- Find the variance.
- Find the standard deviation.
-
Suppose a random variable,
x
, arises from a binomial experiment. Suppose
n
= 10, and
p
= 0.81.
- Write the probability distribution.
- Draw a histogram.
- Describe the shape of the histogram.
- Find the mean.
- Find the variance.
- Find the standard deviation.
-
Suppose a random variable,
x
, arises from a binomial experiment. Suppose
n
= 7, and
p
= 0.50.
- Write the probability distribution.
- Draw a histogram.
- Describe the shape of the histogram.
- Find the mean.
- Find the variance.
- Find the standard deviation.
-
Approximately 10% of all people are left-handed. Consider a grouping of fifteen people.
- State the random variable.
- Write the probability distribution.
- Draw a histogram.
- Describe the shape of the histogram.
- Find the mean.
- Find the variance.
- Find the standard deviation.
-
According to an article in the American Heart Association’s publication Circulation, 24% of patients who had been hospitalized for an acute myocardial infarction did not fill their cardiac medication by the seventh day of being discharged (Ho, Bryson & Rumsfeld, 2009). Suppose there are twelve people who have been hospitalized for an acute myocardial infarction.
- State the random variable.
- Write the probability distribution.
- Draw a histogram.
- Describe the shape of the histogram.
- Find the mean.
- Find the variance.
- Find the standard deviation.
-
Eyeglassomatic manufactures eyeglasses for different retailers. In March 2010, they tested to see how many defective lenses they made, and there were 16.9% defective lenses due to scratches. Suppose Eyeglassomatic examined twenty eyeglasses.
- State the random variable.
- Write the probability distribution.
- Draw a histogram.
- Describe the shape of the histogram.
- Find the mean.
- Find the variance.
- Find the standard deviation.
-
The proportion of brown M&M’s in a milk chocolate packet is approximately 14% (Madison, 2013). Suppose a package of M&M’s typically contains 52 M&M’s.
- State the random variable.
- Find the mean.
- Find the variance.
- Find the standard deviation.
- Answer
-
1. a. See solutions, b. See solutions, c. Skewed right, d. 0.78, e. 0.6786, f. 0.8238
3. a. See solutions, b. See solutions, c. Symmetric, d. 3.5, e. 1.75, f. 1.3229
5. a. See solutions, b. See solutions, c. See solutions, d. Skewed right, e. 2.88, f. 2.1888, g. 1.479
7. a. See solutions, b. 7.28, c. 6.2608, d. 2.502
Data Sources: 11 little-known facts about left-handers . (2013, October 21). Retrieved from www.huffingtonpost.com/2012/1...n_2005864.html
CDC-data and statistics, autism spectrum disorders - ncbdd . (2013, October 21). Retrieved from http://www.cdc.gov/ncbddd/autism/data.html
Ho, P. M., Bryson, C. L., & Rumsfeld, J. S. (2009). Medication adherence. Circulation , 119 (23), 3028-3035. Retrieved from http://circ.ahajournals.org/content/119/23/3028
Households by age of householder and size of household: 1990 to 2010 . (2013, October 19). Retrieved from www.census.gov/compendia/stat...es/12s0062.pdf
Madison, J. (2013, October 15). M&M's color distribution analysis. Retrieved from http://joshmadison.com/2007/12/02/mm...tion-analysis/
What percentage of people have green eyes?. (2013, October 21). Retrieved from www.ask.com/question/what-per...ave-green-eyes