23.5: Using Aldol Reactions in Synthesis
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Objectives
After completing this section, you should be able to identify the aldehyde or ketone and other necessary reagents that should be used to prepare a given enone by an aldol condensation.
Study Notes
This section stresses the importance of being able to think logically.
The experience that you have already gained through designing multi‑step syntheses and solving road‑map problems should help you to recognize when an aldol reaction may have been one of the steps in the synthesis of a given compound.
It is important that you recognize that the aldol condensation is an important part of a synthetic chemist’s repertoire, both because it involves the formation of a new carbon‑carbon bond, and also because it yields a product containing two functional groups.
Aldol reactions are excellent methods for the synthesis of many enones or beta-hydroxy carbonyls. Because of this, being able to predict when an aldol reaction might be used in a synthesis in an important skill. This can be accomplished by identifying these combinations of atoms and bonds and then, working backwards, theoretically breaking the target molecule apart into possible reactants.
Fragments which are easily formed by an aldol reaction
Determining the Reactants for an Aldol Condensation
During an aldol condensation a C-C sigma and a C-C pi bond are formed. This makes the key bond cleavage in the target molecule the C=C bond between the carbons alpha and beta away from the carbonyl. After the cleavage, the carbon that was in the alpha position (on the fragment with the carbonyl) gains two hydrogens. The carbon that was in the beta gains a =O to form a carbonyl.
Determining the Reactants for an Aldol Reaction
During an aldol condensation a C-C sigma bond is formed. This makes the key bond cleavage in the target molecule the C-C bond between the carbons alpha and beta away from the carbonyl. After the cleavage, the carbon that was in the alpha position (the fragment with the carbonyl) gains one hydrogen. The fragment loses a hydrogen from the OH and then forms a C=O carbonyl bond.
Worked Example
Show how the following molecule could be made using an aldol condensation?
- Answer
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Analysis: The C=C bond in the target molecule is cleaved to form two fragments. The fragment with the carbonyl gains two alpha hydrogens. The other fragment gains a =O to form a carbonyl. Both fragments end up producing the same reactant which is typical for an aldol condensation.
Solution
Additional Synthetic Considerations
The enone product of an aldol condensation is versatile because it contains two functional groups (alkene & carbonyl) which can be subject to further reactions. Among many possible reactions, an enone can undergo hydrogenation to produce an aldehyde or ketone. Also, the carbonyl group can undergo hydride reduction to produce a beta-hydroxyalkene.
These additional reactions can be applied with the consideration of using an aldol reaction in the synthesis of a target molecule. A similar analysis can be extrapolated to the other reactions possible with the alkene and carbonyl present in an enone.
Analysis for Hydrogenation
To consider a hydrogenation, remove a hydrogen from a carbon in both the alpha and beta positions relative to the carbonyl. Then connect these two carbons with a C=C double bond. This create a possible enone which can be broken apart further using the analysis described above. Target molecules can have often have multiple alkyl chains which can be used to form a double bond.
Analysis for Hydride Reduction
To consider a hydride reduction remove a hydrogen from the alcohol oxygen then a hydrogen from the adjacent carbon. Connect the oxygen and carbon with a double bond to form a C=O carbonyl. This provides an enone which can undergo further analysis.
Exercise \(\PageIndex{1}\)
Please devise a synthesis pathway for the following molecule using an aldol reaction:
- Answer
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Analysis: Because the target molecule has two beta-carbons with hydrogens there are two possible synthesis pathways. Both should be considered for their effectiveness. Because a carbonyl is already present fragmentations will first be performed to create two possible enone intermediates. These enones will both be broken into their aldol reactants. When looking a possible aldol reactants produced by each pathway it is clear that pathway 2 is preferred. Because both fragments are the same they will react to form on major product.
Solution
Exercises \(\PageIndex{1}\)
1) Which of the following molecules could be made using an aldol reaction or condensation. Please show the starting material for the reactions that are possible.
2) Please show the reaction steps required to make 4-methyl-2-pentanol from an aldol condensation.
3) Please design a synthesis for the following molecule: Red = Oxygen, Grey = Carbon, White = Hydrogen
- Answers
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1) Molecules A and C are possible.
Solution for molecule A
Solution for molecule C
2)
3)


15.11: Reference Material
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This chapter was edited by Michael Hammond, with content contributions by Eladio Bobadilla, Andrew Chadwick, Zach Fredman, Leif Fredrickson, Michael Hammond, Richara Hayward, Joseph Locke, Mark Kukis, Shaul Mitelpunkt, Michelle Reeves, Elizabeth Skilton, Bill Speer, and Ben Wright.
Recommended citation: Eladio Bobadilla et al., “The Recent Past,” Michael Hammond, ed., in The American Yawp , eds. Joseph Locke and Ben Wright (Stanford, CA: Stanford University Press, 2018).
Recommended Reading
- Alexander, Michelle. The New Jim Crow: Mass Incarceration in the Age of Colorblindness. New York: New Press, 2012.
- Canaday, Margot. The Straight State: Sexuality and Citizenship in Twentieth-Century America. Princeton, NJ: Princeton University Press, 2011.
- Carter, Dan T. From George Wallace to Newt Gingrich: Race in the Conservative Counterrevolution, 1963–1994. Baton Rouge: LSU Press, 1996.
- Cowie, Jefferson. Capital Moves: RCA’s 70-Year Quest for Cheap Labor. New York: New Press, 2001.
- Ehrenreich, Barbara. Nickel and Dimed: On (Not) Getting By in America. New York: Metropolitan, 2001.
- Evans, Sara. Tidal Wave: How Women Changed America at Century’s End. New York: Free Press, 2003.
- Gardner, Lloyd C. The Long Road to Baghdad: A History of U.S. Foreign Policy from the 1970s to the Present. New York: Free Press, 2008.
- Hinton, Elizabeth. From the War on Poverty to the War on Crime. Cambridge, MA: Harvard University Press, 2016.
- Hollinger, David. Postethnic America: Beyond Multiculturalism. New York: Basic Books, 1995.
- Hunter, James D. Culture Wars: The Struggle to Define America. New York: Basic Books, 1992.
- Meyerowitz, Joanne. How Sex Changed: A History of Transsexuality in the United States. Cambridge, MA: Harvard University Press, 2004.
- Mittelstadt, Jennifer. The Rise of the Military Welfare State. Cambridge, MA: Harvard University Press, 2015.
- Moreton, Bethany. To Serve God and Walmart: The Making of Christian Free Enterprise. Cambridge, MA: Harvard University Press, 2009.
- Nadasen, Premilla. Welfare Warriors: The Welfare Rights Movement in the United States. New York: Routledge, 2005.
- Osnos, Evan. Age of Ambition: Chasing Fortune, Truth and Faith in the New China. New York: Farrar, Straus and Giroux, 2014.
- Packer, George. The Unwinding: An Inner History of the New America. New York: Farrar, Straus and Giroux, 2013.
- Patterson, James T. Restless Giant: The United States from Watergate to Bush v. Gore. New York: Oxford University Press, 2005.
- Piketty, Thomas. Capital in the Twenty-First Century. Translated from the French by Arthur Goldhammer. Cambridge, MA: Belknap Press, 2013.
- Ricks, Thomas E. Fiasco: The American Military Adventure in Iraq. New York: Penguin, 2006.
- Schlosser, Eric. Fast Food Nation: The Dark Side of the All-American Meal. New York: Houghton Mifflin Harcourt, 2001.
- Stiglitz, Joseph. Freefall: America, Free Markets, and the Sinking of the World Economy. New York: Norton, 2010.
- Taylor, Paul. The Next America: Boomers, Millennials, and the Looming Generational Showdown. New York: Public Affairs, 2014.
- Wilentz, Sean. The Age of Reagan: A History, 1974–2008. New York: HarperCollins, 2008.
- Williams, Daniel K. God’s Own Party: The Making of the Christian Right. New York: Oxford University Press, 2007.
- Wright, Lawrence. The Looming Tower: Al Qaeda and the Road to 9/11. New York: Knopf, 2006.
Notes
- William Faulker, Requiem for a Nun (New York: Random House, 1954), 73.
- Bill Minutaglio, First Son: George W. Bush and the Bush Family Dynasty (New York: Random House, 1999), 210–224.
- Roger Simon, “How a Murderer and Rapist Became the Bush Campaign’s Most Valuable Player,” Baltimore Sun , November 11, 1990.
- See especially Dan T. Carter, From George Wallace to Newt Gingrich: Race in the Conservative Counterrevolution, 1963–1994 (Baton Rouge: LSU Press, 1996), 72–80.
- Sean Wilentz, The Age of Reagan: A History, 1974–2008 (New York: HarperCollins, 2008).
- James F. Clarity, “End of the Soviet Union,” New York Times , December 26, 1991.
- Francis Fukuyama, “The End of History?” National Interest (Summer 1989)
- William Thomas Allison, The Gulf War, 1990–91 (New York: Palgrave Macmillan, 2012), 145, 165.
- Charles W. Dunn, The Presidency in the Twenty-first Century (Lexington: University Press of Kentucky, 2011), 152.
- Robert M. Collins, Transforming America: Politics and Culture During the Reagan Years (New York: Columbia University Press, 2009), 171, 172
- For Clinton’s presidency and the broader politics of the 1990s, see James T. Patterson, Restless Giant: The United States from Watergate to Bush v. Gore (New York: Oxford University Press, 2005); and Wilentz, Age of Reagan .
- Patterson, Restless Giant , 298–299.
- United Nations International Organization for Migration , “Migrant Deaths Remain High Despite Sharp Fall in US-Mexico Border Crossings in 2017,” press release, February 6, 2018. https://news.un.org/en/story/2018/02/1002101 .
- Carolyn Skorneck, “Final Welfare Bill Written,” Washington Post , July 30, 1996, A1
- Frank Newport, “ClinDexter Filkins, The Forever War (New York: Vintage Books, 2009).ton Receives Record High Job Approval Rating,” Gallup, December 24, 1998. http://news.gallup.com/poll/4111/cli...hment-vot.aspx ).
- Dexter Filkins, The Forever War (New York: Vintage Books, 2009).
- See, for instance, Lawrence Wright, The Looming Tower: Al Qaeda and the Road to 9/11 (New York: Knopf, 2006).
- Thomas R. Mockaitis, The Iraq War: A Documentary and Reference Guide (Santa Barbara, CA: ABC-Clio, 2012), 26.
- Judy Keen, “Bush to Troops: Mission Accomplished,” USA Today , June 5, 2003 .
- Bruce Nolan, “Katrina: The Storm We’ve Always Feared,” New Orleans Times-Picayune , August 30, 2005.
- Douglas Brinkley, The Great Deluge: Hurricane Katrina, New Orleans, and the Mississippi Gulf Coast (New York: HarperCollins, 2006).
- On the Great Recession, see Joseph Stiglitz, Freefall: America, Free Markets, and the Sinking of the World Economy (New York: Norton, 2010); and Michael Lewis, The Big Short: Inside the Doomsday Machine (New York: Norton: 2010)
- Harold Meyerson, “Corporate America’s Chokehold on Wages,” Washington Post , July 19, 2011.
- Thomas J. Sugrue, Not Even Past: Barack Obama and the Burden of Race (Princeton, NJ: Princeton University Press, 2012).
- Kate Zernike and Megan Thee-Brenan, “Poll Finds Tea Party Backers Wealthier and More Educated,” New York Times , April 14, 2010; Jill Lepore, The Whites of Their Eyes: The Tea Party’s Revolution and the Battle over American History (Princeton, NJ: Princeton University Press, 2011).
- Kerry Close, “The 1% Pocketed 85% of Post-Recession Income Growth,” Time , June 16, 2016. http://time.com/money/4371332/income...ity-recession/ . See also Justin Wolfers, “The Gains from the Economic Recovery Are Still Limited to the Top One Percent,” New York Times , January 27, 2015. http://www.nytimes.com/2015/01/28/up...e-percent.html .
- Julia Queen and Christian Hilland, “2008 Presidential Campaign Financial Activity Summarized: Receipts Nearly Double 2004 Total,” Federal Election Commission, June 8, 2009. www.fec.gov/press/press2009/2...PresStat.shtml; Andre Tartar and Eric Benson, “The Forever Campaign,” New York Magazine (October 14, 2012). http://nymag.com/news/politics/elect...eline-2012-10/ .
- Intergovernmental Panel on Climate Change, Climate Change 2013: The Physical Science Basis (Cambridge, UK: Cambridge University Press, 2014).
- Intergovernmental Panel on Climate Change, Climate Change 2014: Impacts, Adaptation and Vulnerability: Global and Sectoral Aspects (Cambridge, UK: Cambridge University Press, 2014)
- Philip Bump, “A Quarter of Americans Dislike Both Major-Party Presidential Candidates,” Washington Post , July 14, 2016. https://www.washingtonpost.com/news/...tes/?tid=a_inl ; Aaron Zitner and Julia Wolfe, “Drumpf and Clinton’s Popularity Problem,” Wall Street Journal , May 24, 2016. http://graphics.wsj.com/elections/20...arity-problem/ .
- Paul Taylor, The Next America: Boomers, Millennials, and the Looming Generational Showdown (New York: Public Affairs, 2014).
- “Gay and Lesbian Rights,” Gallup, December 5–7, 2003. http://www.gallup.com/poll/1651/gay-...an-rights.aspx .
- Janet Hook, “Support for Gay Marriage Hits All-Time High,” Wall Street Journal , March 9, 2015.
- Ibid.
- Monica Anders, “Technology Device Ownership: 2015,” Pew Research Center, October 29, 2015. http://www.pewglobal.org/2016/02/22/...ing-economies/
- Monica Anderson and Andrea Caumont, “How Social Media Is Reshaping News,” Pew Research Center, September 24, 2014. http://www.pewresearch.org/fact-tank...eshaping-news/
- See, for instance, Nicholas G. Carr’s 2010 The Shallows: What the Internet Is Doing to Our Brains , a 2011 Pulitzer Prize finalist.
- Bijan Stephen, “Social Media Helps Black Lives Matter Fight the Power,” Wired (November 2015). www.wired.com/2015/10/how-bla...ght-the-power/.
- U.S. Census Bureau, 2016 American Community Survey 1-Year Estimates. https://factfinder.census.gov/bkmk/t...0500000US48215 .


5.9: Naming Acids
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A spot test for gold has been in use for decades. The sample is first treated with nitric acid. Other metals may react or dissolve in this acid, but gold will not. Then the sample is added to a mixture of nitric acid and hydrochloric acid. Gold will only dissolve in this mixture. The term "acid test" arose from the California gold rush in the late 1840's when this combination was used to test for the presence of real gold. It has since come to mean, "tested and approved" in a number of fields.
Acids
An acid can be defined in several ways. The most straightforward definition is that an acid is a molecular compound that contains one or more hydrogen atoms and produces hydrogen ions \(\left( \ce{H^+} \right)\) when dissolved in water.
This is a different type of compound than the others we have seen so far. Acids are molecular, which means that in their pure state they are individual molecules and do not adopt the extended three-dimensional structures of ionic compounds like \(\ce{NaCl}\). However, when these molecules are dissolved in water, the chemical bond between the hydrogen atom and the rest of the molecule breaks, leaving a positively-charged hydrogen ion and an anion. This can be symbolized in a chemical equation:
\[\ce{HCl} \rightarrow \ce{H^+} + \ce{Cl^-} \nonumber \]
Since acids produce \(\ce{H^+}\) cations upon dissolving in water, the \(\ce{H}\) of an acid is written first in the formula of an inorganic acid. The remainder of the acid (other than the \(\ce{H}\)) is the anion after the acid dissolves. Organic acids are also an important class of compounds, but will not be discussed here.
Naming Acids
Since all acids contain hydrogen, the name of an acid is based on the anion that goes with it. These anions can either be monatomic or polyatomic.
Naming Binary acids (in aqueous form)
A binary acid is an acid that consists of hydrogen and one other element. The most common binary acids contain a halogen. The acid name begins with the prefix hydro- . followed by the base name of the anion, followed by the suffix -ic .
Naming Oxyacids
An oxyacid is an acid that consists of hydrogen, oxygen, and a third element. The third element is usually a nonmetal.
a. Oxyanions with -ite ending.
The name of the acid is the root of the anion followed by the suffix -ous . There is no prefix.
b. Oxyanions with -ate ending.
The name of the acid is the root of the anion followed by the suffix -ic . There is no prefix.
Note
The base name for sulfur containing oxyacid is sulfur- instead of just sulf- . The same is true for a phosphorus containing oxyacid. The base name is phosphor- instead of simply phosph- .
Writing Formulas for Acids
Like other compounds that we have studied, acids are electrically neutral. Therefore, the charge of the anion part of the formula must be exactly balanced out by the \(\ce{H^+}\) ions. Another way to think about writing the correct formula is to utilize the crisscross method, shown below for sulfuric acid.
Formula: H 2 SO 4
Figure \(\PageIndex{2}\): Crisscross approach to writing formula for sulfuric acid.
Summary
- Acids are molecular compounds that release hydrogen ions.
- A binary acid consists of hydrogen and one other element.
- Oxyacids contain hydrogen, oxygen, and one other element.
- The name of the acid is based on the anion attached to the hydrogen.


7.4: Lesson 3- Levels
Levels create visual interest and provide opportunities for dynamic staging. Levels are necessary to staging and should be introduced as much as possible. One strong principal to keep in mind is to create a visual triangle (Figure \(\PageIndex{1}\)) whenever possible. Triangles are visually pleasing and create a focal point. Anytime you have 3 or more actors on stage try to position them into either a triangle or a shape with a visual apex. When looking at staging you need to view levels in terms of depth and height in regards to the stage.
When actors stand in a straight line (Figure \(\PageIndex{2}\)) the image for the audience is flat and uninteresting, yet when one member of the group moves slightly upstage or downstage to create a triangle (Figure \(\PageIndex{3}\)), the image gains visual interest.
You can also create levels through height. If two actors are sitting and one is standing the person standing becomes the apex of the triangle and receives the focus and power. The same relationship applies in reverse. If two characters are standing the character sitting is again the apex of the triangle and draws focus.
Utilizing levels built into the set is another way to enhance visual interest and create beautiful stage pictures (Figure \(\PageIndex{4}\) and Figure \(\PageIndex{5}\)). As the director step back from the stage and try to create large visual levels when possible. Use the entire stage. When training use acting blocks, tables, and chairs to create your set and provide levels.
Staging Exercise 1
- Create a silent scene with a beginning, middle, and end in which a character enters a dangerous environment, encounters obstacles, and is transformed into something new.
- Utilize levels in at least 3 moments in the scene.
- Rehearse for 1 hour and then present
- Discuss and critique each scene
- Remember to always focus on successful elements first. People need to know what they are doing well more than they need to know what to improve. Encourage progress and try to give 5 positive comments to each suggestion for improvement.
- Rehearse for an additional 15 minutes in order to incorporate notes.
- Present scenes a final time without notes or critique.
Staging Exercise 2
- Create a scene where someone in the scene is sitting, standing, kneeling, and laying down at all times. You have to justify each movement and as someone moves to a new position, another character must shift to keep the balance.
- Rehearse for 1 hour and then present
- Discuss and critique each scene
- Remember to always focus on successful elements first. People need to know what they are doing well more than they need to know what to improve. Encourage progress and try to give 5 positive comments to each suggestion for improvement.
- Rehearse for an additional 15 minutes in order to incorporate notes.
- Present scenes a final time without notes or critique.
Staging Exercise 3
- Create a scene where a character goes on a silent journey and something familiar become new and abstract. Use levels in at least 3 seperate moments within the scene.
- Rehearse for 1 hour and then present
- Discuss and critique each scene
- Remember to always focus on successful elements first. People need to know what they are doing well more than they need to know what to improve. Encourage progress and try to give 5 positive comments to each suggestion for improvement.
- Rehearse for an additional 15 minutes in order to incorporate notes.
- Present scenes a final time without notes or critique.


13: Trigonometric Functions
The trigonometric functions are functions of an angle. and relate the angles of a triangle to the lengths of its sides. They are important in the study of triangles and modeling periodic phenomena, among many other applications.
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- 13.0: Prelude to Trigonometric Functions
- A function that repeats its values in regular intervals is known as a periodic function. The graphs of such functions show a general shape reflective of a pattern that keeps repeating. This means the graph of the function has the same output at exactly the same place in every cycle. And this translates to all the cycles of the function having exactly the same length.
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- 13.1: Angles
- An angle is formed from the union of two rays, by keeping the initial side fixed and rotating the terminal side. The amount of rotation determines the measure of the angle. An angle is in standard position if its vertex is at the origin and its initial side lies along the positive x-axis. A positive angle is measured counterclockwise from the initial side and a negative angle is measured clockwise.
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- 13.2: Unit Circle - Sine and Cosine Functions
- In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.
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- 13.3: The Other Trigonometric Functions
- Trigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. We have already defined the sine and cosine functions of an angle. Though sine and cosine are the trigonometric functions most often used, there are four others. Together they make up the set of six trigonometric functions. In this section, we will investigate the remaining functions.
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- 13.4: Right Triangle Trigonometry
- We have previously defined the sine and cosine of an angle in terms of the coordinates of a point on the unit circle intersected by the terminal side of the angle. In this section, we will see another way to define trigonometric functions using properties of right triangles.
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- 13.R: Trigonometric Functions (Review)
- We have previously defined the sine and cosine of an angle in terms of the coordinates of a point on the unit circle intersected by the terminal side of the angle. In this section, we will see another way to define trigonometric functions using properties of right triangles.


4.1: Introduction
Mammals come in a wide variety of shapes and sizes. Some species are incredibly tiny. For example, the bumblebee bat, weighing in at 2 g, competes for the title of smallest mammal with the slightly lighter (but also slightly longer) Etruscan shrew (Hill 1974) . Other species are huge, as anyone who has encountered a blue whale knows. Body size is important as a biological variable because it predicts so many other aspect of an animal’s life, from the physiology of heat exchange to the biomechanics of locomotion. Thus, the rate at which body size evolves is of great interest among mammalian biologists. Throughout this chapter, I will discuss the evolution of body size across different species of mammals. The data I will analyze is taken from Garland (1992) .
Sometimes one might be interested in calculating the rate of evolution of a particular character like body size in a certain clade, say, mammals. You have a phylogenetic tree with branch lengths that are proportional to time, and data on the phenotypes of species on the tips of that tree. It is usually a good idea to log-transform your data if they involve a measurement from a living thing (see Box 4.1, below). If we assume that the character has been evolving under a Brownian motion model, we have two parameters to estimate: \(\bar{z}(0)\), the starting value for the Brownian motion model – equivalent to the ancestral state of the character at the root of the tree – and σ 2 , the diffusion rate of the character. It is this latter parameter that is commonly considered as the rate of evolution for comparative approaches 1 .
Box 4.1: Biology under the log
One general rule for continuous traits in biology is to carry out a log-transformation (usually natural log, base e , denoted l n ) of your data before undertaking any analysis. This also applies to comparative data. There are two main reasons for this, one statistical and the other biological. The statistical reason is that many methods assume that variables follow normal distributions. One can observe that, in general, measurements of species' traits have a distribution that is skewed to the right. A log-transformation will often result in trait distributions that are closer to normal. But why is this the case? The answer is related to the biological reason for log-transformation. When you log transform a variable, the new scale for that variable is a ratio scale, so that a certain differences between points reflects a constant ratio of the two numbers represented by the points. So, for example, if any two numbers are separated by 0.693 units on a natural log scale, one will be exactly two times the other. Ratio scales make sense for living things because it is usually percentage changes rather than absolute changes that matter. For example, a change in body size of 1 mm might matter a lot for a termite, but be irrelevant for an elephant; whereas a change in body size of 50% might be expected to matter for them both.


8.8: Chapter Review - Review Questions
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Review Questions
1 .
Once you learn the context of data, it is considered ________.
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analyzed data
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knowledge
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information
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content
2 .
A content management system manages ________.
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a blog
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e-commerce
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site visitors
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web content
3 .
________ organize internal documents and the information employees may need to do their jobs.
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Enterprise content management systems
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Mobile content management systems
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Social media content management systems
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Web content management systems
4 .
A CMS allows editing, publishing, modifying content, and site administration. If you are managing a team of developers, designers, writers, and editors, which feature are you most likely to use and manage?
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editing
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publishing
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modifying content
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site administration
5 .
Multiple people are working on updating the website to match the theme of the new launch. An employee was supposed to edit images on the site, but when you log in, you find that some files are missing. The system allows you to see the time, date, and user, so you can identify when any changes were made and who made them. This feature is known as ________.
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authoring
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content versioning
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multiple users
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site administration
6 .
________ allows user engagement to actually create content on a company website.
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Enterprise content
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Web 1.0
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Web 2.0
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Open-source
7 .
A(n) ________ CMS involves no initial costs and allows the user to download the software without a contract.
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open-source
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proprietary
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enterprise-enabled
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CMS
8 .
WordPress uses ________ to control user access to different features in the site.
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cookies
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usernames
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roles
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code
9 .
Where do you find the metrics in WordPress that are used to measure the effectiveness of your website?
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dashboard
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stats
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edit
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posts
10 .
Google uses ________ to generate search results.
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an algorithm
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crawling
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indexing
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a URL
11 .
The purpose of SEO is to ________.
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categorize blogs
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search for keywords
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target paid traffic to a website
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improve the quality of a website
12 .
When using multiple words in the URL, you should use a(n) ________ to separate words.
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underscore
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hyphen
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space
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period
13 .
What is one reason having a social media presence would help a small business?
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Guarantee sales.
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Increase brand awareness.
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Reduce product costs.
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Automatically respond to complaints.
14 .
What is one key advantage of engagement management (EM)?
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creating content
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encouraging single site visits
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not having a website
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scheduling posts
15 .
What is a social media calendar?
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data collected from different social channels to measure performance
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a tool that allows organizations to engage with members and involves a team of client relations to satisfy clients
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a spreadsheet or app used to schedule social posts in advance
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hyperlinks in a web page that take you to another website or another location on your site


15.3A: Pharyngitis
- List the symptoms and bacterial causes associated with pharyngitis
Pharyngitis
Pharyngitis is an inflammation of the throat. In most cases, it is quite painful and is the most common cause of a sore throat. Like many types of inflammation, pharyngitis can be acute or chronic. Acute cases are characterized by a rapid onset and, typically, a relatively short course of inflammation. Pharyngitis can result in very large tonsils. This can make swallowing and breathing difficult. It can be accompanied by a cough or fever, for example, if it is caused by a systemic infection. Most acute cases are caused by viral infections (40–80%). The remainder are caused by bacterial infections, fungal infections, or irritants such as pollutants or chemical substances. The treatment of viral causes is mainly symptomatic. Bacterial or fungal causes are often amenable to antibiotics and anti-fungal treatments, respectively.
Bacterial Causes of Pharyngitis
A number of different bacteria can infect the human throat. The most common is Group A streptococcus, but others include Corynebacterium diphtheriae, Neisseria gonorrhoeae, Chlamydophila pneumoniae, and Mycoplasma pneumoniae.
Streptococcal pharyngitis, more commonly known as strep throat, is caused by group A beta-hemolytic streptococcus (GAS). This is the most common bacterial cause of pharyngitis (15–30%). Common symptoms of strep throat include fever, sore throat, and large lymph nodes. It is a contagious infection, spread by close contact with an infected individual. A throat culture is the gold standard for the diagnosis of streptococcal pharyngitis, with a sensitivity of 90–95%. A rapid strep test (also called rapid antigen detection testing, or RADT) is also occasionally used as a diagnostic. While the rapid strep test is quicker, it has a lower sensitivity (70%) and a statistically equal specificity (98%) as a throat culture. For step throat, antibiotics are useful in preventing complications and expediting recovery.
Fusobacterium necrophorum are normal inhabitants of the oropharyngeal flora. Occasionally, however, these bacteria can create a peritonsillar abscess. In 1 out of 400 untreated cases, Lemierre’s syndrome can occur as a result of these abscesses.
Diphtheria is a potentially life threatening upper respiratory infection caused by Corynebacterium diphtheriae. As a result of childhood vaccination programs, diphtheria has has been largely eradicated in developed nations, but it is still reported in the Third World, and, increasingly, in some areas in Eastern Europe. Antibiotics are effective in the early stages, but recovery is generally slow.
Key Points
- In most cases, pharyngitis is caused by a systemic viral infection and is typified by a painful swelling inflammation of the throat. This can cause difficulty swallowing or breathing.
- The most common bacterial cause of pharyngitis is streptococcus.
- Several types of bacteria can cause pharyngitis. The most common and effective treatment for these infections are antibiotics.
Key Terms
- hemolytic : producing hemolysis; destroying red blood cells
- peritonsillar abscess : Peritonsillar abscess (PTA), also called a quinsy, or abbreviated as a PTA, is a recognized complication of tonsillitis. It consists of a collection of pus beside the tonsil in what is referred to as Peritonsilar space (Peri – meaning surrounding).


2.1.2: Electronegativity and the Bonding Continuum
Electronegativity
Electronegativity is the measure of the ability of an atom to attract electrons within a chemical bond. It is related to the electron affinity in the sense that both period properties measure the force by which an electron is attracted to a neutral atom. However, the electron affinity is associated with isolated atoms, that do not make bonds to neighbored atoms, while the electronegativity refers to atoms that are bonded to neighbored atoms. The importance of the electronegativity stems largely from its ability to predict and understand the nature of the bonding between atoms, in particular, covalent, ionic, and metallic bonding.
Pauling Electronegativity Scale
Electronegativity was a concept first developed by Linus Pauling to describe the relative polarity of bonds and molecules. He argued that the electronegativity of an atom could be derived from bond energy differences between homoleptic and heteroleptic bonds (Eq. 1.3.1).
Equation 1.3.1 Electronegativity according to Pauling, eV = electron volts.
For example, the energy required to break the bond in H 2 is 432 kJ mol -1 , and for F 2 it is 159 kJ mol -1 . However the energy required to break a bond in HF is 565 kJ mol -1 , which is much higher than expected just by averaging the energy of the two homonuclear bonds (296 kJ mol -1 ). Pauling argued that the difference could be assigned to electrostatic attractions between the F and H “ends” of the molecule whereby the F end has more electron density and the H end has less electron density. The greater the difference between the average energy of the homoleptic bonds and the heteroleptic bonds, the greater the electronegativity difference between the atoms, and the greater the polarity of the heteroleptic bond.
First, Pauling used the arithmetic means of the heteroleptic bond energies, later he used the geometric means, because he found empirically that it worked better (Equation 1.3.1). The geometric mean of the homoleptic bond energies is the square root of the product of the homoleptic bond energies. This method only provided electronegativity differences. He therefore needed to define a reference atom with an arbitrarily defined electronegativity value, and then determine the electronegativity values of all other atoms relative to that value. He chose the most electronegative atom, the fluorine atom, as the standard and assigned a value of 4.0. The values of all other atoms vary between 0.7 (Fr) and 3.44 (O), and are shown in the periodic table on the below (Fig. 1.3.6).
What are the periodic trends? One can see that within a period the electronegativity values of the main group elements strictly increase with the group number. Note though that the noble gases do not have a Pauling electronegativity because, with few exceptions, noble gases do not make compounds and thus no bond energies are available for them. We can also see that for main group elements the electronegativity strictly decreases down a group. This makes the fluorine the most electronegative atom, and the cesium the least electronegative atom. Cesium has an electronegativity value of 0.7. We ignore the radioactive francium here. For the d-block, the trends are less strict, but there is a tendency of electronegativity increase from group 3 to group 11. All group 12 elements have smaller electronegativities compared to their neighbored group 11 elements. What about trends within a group? In group 3 to group 5 there is a small decrease in electronegativity down a group. For group 6 and 9, the period 5 elements have higher values than their neighbored elements in period 4 and 6. For group 10 to 12 the electronegativity increases down a group. The d-block element with the highest electronegativity is the gold, it has a value of 2.4. Note that this is similar to the electronegativity of non-metals such as I, S, and P. For these reasons, sometimes gold can behave like a non-metal in compounds with very low electronegativity. For example, there is the compound cesium auride (CsAu) which is a transparent, ionic crystalline compound due to the high electronegativity difference between cesium and gold. The f-block elements generally have low electronegativity.
Allred-Rochow Electronegativity Scale
The Pauling electronegativity scale is derived from empirical data on bond energies. It works very well in practice and it is up to date the most used electronegativity scale. However, it does not to relate electronegativity to other periodic properties, and the quantum-mechanical model of the atoms. One electronegativity scale that addresses this short-coming the Allred-Rochow scale. It relates the electronegativity to the Coulomb force that acts on an electron on the surface of an atom. The Coulomb force is proportional to the effective nuclear charge Z* and inverse proportional to the atomic radius square. Z*/r 2 is multiplied with a factor of 3590 and a value of 0.744 is added to this term. These numbers are empirically chosen so that the values of the Allred-Rochow scale become comparable to the Pauling scale.
Equation 1.3.2 Allred-Rochow definition of electronegativity
Mulliken Electronegativity Scale
Another frequently used scale is the Mulliken electronegativity scale. The Mulliken scale relates the electronegativity to the sum of the first ionization energy IE and the first electron affinity EA. We would intuitively agree that the ability of an atom to attract electrons within a chemical bond is the higher, the harder it is to remove an electron from an isolated atom, and the easier it is to add an electron to an isolated atom. The numbers 0.118 and -0.207 are empirically chosen to make comparisons to the Pauling scale possible. The Mulliken scale is related to the Allred-Rochow scale in the sense that you can explain ionization energies and electron affinities of atoms with the concept of the effective nuclear charge.
Equation 1.3.3 Mulliken definition of electronegativity
Allen Electronegativity Scale
Another electronegativity scale has been developed by Leland C. Allen. It designed for main group elements in particular. It argues that the electronegativity is proportional to the average energy of the s and p valence electrons in the atom. This electronegativity concept also has a relationship to that of Allred-Rochow, because orbital energies can be calculated from the effective nuclear charge.
Equation 1.3.4 Leland C. Allen definition of electronegativity
Overall, you can see that the electronegativity scales are inter-related, they say essentially the same, but present electronegativity from a somewhat different perspective.
Electronegativity of the Elements and Chemical Bonding
One of the most powerful attributes of electronegativity is that it can predict what type of chemical bond is to expect in elements and compounds. Let us first look at the bonding between elements.
In the depicted periodic table (Figure 1.3.7) you can see the type of chemical bonding between atoms indicated by different colors. There is metallic bonding in metals, indicated by the color blue. In metallic bonds the electrons are shared between atoms but are delocalized over many atoms in the metal. The green color indicates covalent bonding seen in non-metals. In covalent bonding the electrons are shared, but localized in between typically two atoms. Metalloids have chemical bonding with a mix of covalent and metallic character shown in orange. There is electron-sharing with a moderate degree of delocalization. The Pauling electronegativity values of the elements are written underneath the element symbols.
Can we relate the electronegativity values to the chemical bonding in the elements? We can clearly see that elements with high electronegativity values significantly above 2.0, the non-metals, tend to make covalent bonds in between them. All metalloids with a hybrid covalent-metallic character, have intermediate electronegativity around 2.0. Most metals have low electronegativity values below 2.0 except noble metals like Pt and Au. Overall, we can say that there is a clear relationship between the bonding type in an element and its electronegativity. Now let us see if electronegativity can also predict chemical bonding in compounds.
Ketelaar's Triangle
The ability of electronegativity to predict the bonding type in compounds can be understood by Ketelaar’s triangle, named after J.A.A. Ketelaar.
You can see that Ketelaar’s triangle has three corners (Figure 1.3.8). Two of the corners are occupied by the elements F and Cs, and the third corner is occupied by CsF. All elements are located on the horizontal edge of the triangle, and there many compound indicated by dots either on the two other edges or within the triangle. There are no compounds or elements outside of the triangle. How can we understand this triangle and what does it say about the relationship between bonding character and electronegativity? To understand this, look at the two axes. The horizontal x-axis represents the average electronegativity of the atoms in an element or a compound. The vertical y-axis represents the electronegativity difference between the atoms in the elements or the compound. For elements the electronegativity difference between the atoms is zero, because in an element all atoms are of the same type. Therefore, all the elements lie on the horizontally oriented edge of the triangle. Because Cs is the element with the lowest electronegativity, it lies furthest to the left on this edge. Fluorine is the element with the highest electronegativity, and thus lies the furthest on the right side. The other elements are in located in between the cesium and the fluorine on the edge. The higher the electronegativity of the elements, the further right their position on this edge of the triangle.
For compounds, the electronegativity difference between atoms is never exactly zero, therefore all compounds are located above the horizontal edge of the triangle. The compound with the highest electronegativity difference is the CsF. Its average electronegativity is the sum the electronegativity of Cs and F divided by 2: Therefore, CsF defines the third corner of the triangle. All other compounds must lie on the edges or within the triangle. All cesium compounds are located on the edge between the Cs and CsF, and all the fluorine compounds are located on the edge between CsF and F2. The compounds of all other elements are inside the triangle. The position of the compound on or in the triangle defines its bonding character. The closer the position of the compound toward the Cs corner, the more metallic, the closer the position is to the F2 corner, the more covalent, and the closer it is to the CsF corner, the more ionic. For example Li 2 O is located close to CsF, and the bonding would be predominantly ionic, although there is also a small degree of covalent and metallic bonding. In contrast to that, SrMg has mostly metallic bonding, a little bit of covalent bonding, and even less ionic bonding. The bonding situation in LiH is in between metallic and ionic, with a little bit of covalent bonding character mixed in. Note that a 100% ionic bond is not possible because the electronegativity difference is finite, and thus there must always be a certain degree of electron sharing. In contrast to that, a 100% covalent bond is possible because electrons can be equally shared between two atoms when the electronegativity difference is zero.
The take home message is that in compounds and elements there is usually a mix of bonding types, and not a single bonding type, even through one bonding type may strongly dominate. The concept of electronegativity, and Ketelaar’s triangle in particular, is extremely helpful to predict to which degree the three bonding types are present in a substance.
Dr. Kai Landskron ( Lehigh University ). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate .


1.6: Measurements and Units
- Express quantities properly, using a number and a unit.
- State the different measurement systems used in chemistry.
Measurements
A coffee maker’s instructions tell you to fill the coffeepot with 4 cups of water and to use 3 scoops of coffee. When you follow these instructions, you are measuring. When the nurse checks your temperature, height, weight, and blood pressure at the doctor’s office, (Figure 1.6.1), the nurse is also measuring.
Chemists measure the properties of matter and express these measurements as quantities. A quantity is an amount of something and consists of a number and a unit. The number tells us how many (or how much), and the unit tells us what the scale of measurement is. For example, when a distance is reported as “5 kilometers,” we know that the quantity has been expressed in units of kilometers and that the number of kilometers is 5. If you ask a friend how far they walk from home to school, and the friend answers “12” without specifying a unit, you do not know whether your friend walks 12 kilometers, 12 miles, 12 furlongs, or 12 yards. Both a number and a unit must be included to express a quantity properly.
Identify the number and the unit in each quantity.
- one dozen eggs
- 2.54 centimeters
- a box of pencils
- 88 meters per second
Solution
- The number is one, and the unit is dozen eggs.
- The number is 2.54, and the unit is centimeter.
- The number 1 is implied because the quantity is only a box. The unit is box of pencils.
- The number is 88, and the unit is meters per second. Note that in this case the unit is actually a combination of two units: meters and seconds.
Units of Measurements
To understand chemistry, we need a clear understanding of the units chemists work with and the rules they follow for expressing numbers. How long is a yard? It depends on whom you ask and when you asked the question. Today we have a standard definition of the yard, which you can see marked on every football field. If you move the ball ten yards, you get a first down and it does not matter whether you are playing in Los Angeles, Dallas, or Green Bay. But at one time that yard was arbitrarily defined as the distance from the tip of the king's nose to the end of his outstretched hand. Of course, the problem there is simple: new king, new distance (and then you have to re-mark all of those football fields).
SI Base Units
All measurements depend on the use of units that are well known and understood. The English system of measurement units (inches, feet, ounces, etc.) are not used in science. The metric system is used because all metric units are based on multiples of 10, making conversions very simple. The metric system was originally established in France in 1795. The International System of Units is a system of measurement based on the metric system. The acronym SI is commonly used to refer to this system and stands for the French term, Le Système International d'Unités . The SI was adopted by international agreement in 1960. Five of the base units are shown in Table \(\PageIndex{1}\).
| Quantity | Base Unit | Symbol |
|---|---|---|
| Length | meter | \(\text{m}\) |
| Mass | kilogram | \(\text{kg}\) |
| Temperature | kelvin | \(\text{K}\) |
| Time | second | \(\text{s}\) |
| Amount of a Substance | mole | \(\text{mol}\) |
These units are frequently encountered in chemistry. All other measurement quantities, such as volume, force, and energy, can be derived from these base units.
The map below shows the adoption of the SI units in countries around the world. The United States has legally adopted the metric system for measurements, but does not use it in everyday practice. Great Britain and much of Canada use a combination of metric and imperial units.
Summary
- Identify a quantity properly with a number and a unit.
- The basic units in the SI system such as mass, length, temperature, time, and amount of substance are commonly used in chemistry.
Contributions & Attributions
This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality:
-
Henry Agnew (UC Davis)


3: Types of Data
To process data it is not enough just to obtain them. You need to convert it to the appropriate format, typically to numbers. Since Galileo Galilei, who urged to “ measure what can be measured, and make measurable what cannot be measured ”, European science aggregated tremendous experience in transferring surrounding events into numbers. Most of our instruments are devices which translate environment features (e.g., temperature, distance) to the numerical language.
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- 3.3: Colors, Names and Sexes - Nominal Data
- Nominal, or categorical, data, unlike ranked, are impossible to order or align. They are even farther away from numbers. For example, if we assign numerical values to males and females (say, “1” and “2”), it would not imply that one sex is somehow “larger” then the other. An intermediate value (like “1.5”) is also hard to imagine. Consequently, nominal indices may be labeled with any letters, words or special characters—it does not matter.


2.7: Directional Derivatives and the Gradient
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In Partial Derivatives , we introduced the partial derivative. A function \(z=f(x,y)\) has two partial derivatives: \(∂z/∂x\) and \(∂z/∂y\). These derivatives correspond to each of the independent variables and can be interpreted as instantaneous rates of change (that is, as slopes of a tangent line). For example, \(∂z/∂x\) represents the slope of a tangent line passing through a given point on the surface defined by \(z=f(x,y),\) assuming the tangent line is parallel to the \(x\)-axis . Similarly, \(∂z/∂y\) represents the slope of the tangent line parallel to the \(y\)-axis. Now we consider the possibility of a tangent line parallel to neither axis.
Directional Derivatives
We start with the graph of a surface defined by the equation \(z=f(x,y)\). Given a point \((a,b)\) in the domain of \(f\), we choose a direction to travel from that point. We measure the direction using an angle \(θ\), which is measured counterclockwise in the \(xy\)-plane, starting at zero from the positive \(x\)-axis (Figure \(\PageIndex{1}\)). The distance we travel is \(h\) and the direction we travel is given by the unit vector \(\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}.\) Therefore, the \(z\)-coordinate of the second point on the graph is given by \(z=f(a+h\cos θ,b+h\sin θ).\)
We can calculate the slope of the secant line by dividing the difference in \(z\)-values by the length of the line segment connecting the two points in the domain. The length of the line segment is \(h\). Therefore, the slope of the secant line is
\[m_{sec}=\dfrac{f(a+h\cos θ,b+h\sin θ)−f(a,b)}{h}\]
To find the slope of the tangent line in the same direction, we take the limit as \(h\) approaches zero.
Definition: Directional Derivatives
Suppose \(z=f(x,y)\) is a function of two variables with a domain of \(D\). Let \((a,b)∈D\) and define \(\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}\). Then the directional derivative of \(f\) in the direction of \(\vecs u\) is given by
\[D_{\vecs u}f(a,b)=\lim_{h→0}\dfrac{f(a+h \cos θ,b+h\sin θ)−f(a,b)}{h} \label{DD}\]
provided the limit exists.
Equation \ref{DD} provides a formal definition of the directional derivative that can be used in many cases to calculate a directional derivative.
Note that since the point \((a, b)\) is chosen randomly from the domain \(D\) of the function \(f\), we can use this definition to find the directional derivative as a function of \(x\) and \(y\).
That is,
\[D_{\vecs u}f(x,y)=\lim_{h→0}\dfrac{f(x+h \cos θ,y+h\sin θ)−f(x,y)}{h} \label{DDxy}\]
Example \(\PageIndex{1}\): Finding a Directional Derivative from the Definition
Let \(θ=\arccos(3/5).\) Find the directional derivative \(D_{\vecs u}f(x,y)\) of \(f(x,y)=x^2−xy+3y^2\) in the direction of \(\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}\).
Then determine \(D_{\vecs u}f(−1,2)\).
Solution
First of all, since \(\cos θ=3/5\) and \(θ\) is acute, this implies
\[\sin θ=\sqrt{1−\left(\dfrac{3}{5}\right)^2}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}. \nonumber\]
Using \(f(x,y)=x^2−xy+3y^2,\) we first calculate \(f(x+h\cos θ,y+h\sin θ)\):
\[\begin{align*} f(x+h\cos θ,y+h\sin θ)&=(x+h\cos θ)^2−(x+h\cos θ)(y+h\sin θ)+3(y+h\sin θ)^2 \\&=x^2+2xh\cos θ+h^2\cos^2 θ−xy−xh\sin θ−yh\cos θ−h^2\sin θ\cos θ+3y^2+6yh\sin θ+3h^2\sin^2 θ \\ &=x^2+2xh(\dfrac{3}{5})+\dfrac{9h^2}{25}−xy−\dfrac{4xh}{5}−\dfrac{3yh}{5}−\dfrac{12h^2}{25}+3y^2+6yh(\dfrac{4}{5})+3h^2(\dfrac{16}{25})\\&=x^2−xy+3y^2+\dfrac{2xh}{5}+\dfrac{9h^2}{5}+\dfrac{21yh}{5}. \end{align*}\]
We substitute this expression into Equation \ref{DD} with \(a = x\) and \(b = y\):
\[\begin{align*} D_{\vecs u}f(x,y)&=\lim_{h→0}\dfrac{f(x+h\cos θ,y+h\sin θ)−f(x,y)}{h}\\ &=\lim_{h→0}\dfrac{(x^2−xy+3y^2+\dfrac{2xh}{5}+\dfrac{9h^2}{5}+\dfrac{21yh}{5})−(x^2−xy+3y^2)}{h}\\ &=\lim_{h→0}\dfrac{\dfrac{2xh}{5}+\dfrac{9h^2}{5}+\dfrac{21yh}{5}}{h}\\ &=\lim_{h→0}\dfrac{2x}{5}+\dfrac{9h}{5}+\dfrac{21y}{5}\\&=\dfrac{2x+21y}{5}. \end{align*}\]
To calculate \(D_{\vecs u}f(−1,2),\) we substitute \(x=−1\) and \(y=2\) into this answer (Figure \(\PageIndex{2}\)):
\[ D_{\vecs u}f(−1,2)=\dfrac{2(−1)+21(2)}{5}=\dfrac{−2+42}{5}=8. \nonumber\]
An easier approach to calculating directional derivatives that involves partial derivatives is outlined in the following theorem.
Theorem \(\PageIndex{1}\): Directional Derivative of a Function of Two Variables
Let \(z=f(x,y)\) be a function of two variables \(x\) and \(y\), and assume that \(f_x\) and \(f_y\) exist. Then the directional derivative of \(f\) in the direction of \(\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}\) is given by
\[D_{\vecs u}f(x,y)=f_x(x,y)\cos θ+f_y(x,y)\sin θ. \label{DD2v}\]
Proof
Applying the definition of a directional derivative stated above in Equation \ref{DD}, the directional derivative of \(f\) in the direction of \(\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}\) at a point \((x_0, y_0)\) in the domain of \(f\) can be written
\[D_{\vecs u}f((x_0, y_0))=\lim_{t→0}\dfrac{f(x_0+t \cos θ,y_0+t\sin θ)−f(x_0,y_0)}{t}.\]
Let \(x=x_0+t\cos θ\) and \(y=y_0+t\sin θ,\) and define \(g(t)=f(x,y)\). Since \(f_x\) and \(f_y\) both exist, we can use the chain rule for functions of two variables to calculate \(g′(t)\):
\[g′(t)=\dfrac{∂f}{∂x}\dfrac{dx}{dt}+\dfrac{∂f}{∂y}\dfrac{dy}{dt}=f_x(x,y)\cos θ+f_y(x,y)\sin θ.\]
If \(t=0,\) then \(x=x_0\) and \(y=y_0,\) so
\[g′(0)=f_x(x_0,y_0)\cos θ+f_y(x_0,y_0)\sin θ\]
By the definition of \(g′(t),\) it is also true that
\[g′(0)=\lim_{t→0}\dfrac{g(t)−g(0)}{t}=\lim_{t→0}\dfrac{f(x_0+t\cos θ,y_0+t\sin θ)−f(x_0,y_0)}{t}.\]
Therefore, \(D_{\vecs u}f(x_0,y_0)=f_x(x_0,y_0)\cos θ+f_y(x_0,y_0)\sin θ\).
Since the point \( (x_0,y_0) \) is an arbitrary point from the domain of \(f\), this result holds for all points in the domain of \(f\) for which the partials \(f_x\) and \(f_y\) exist.
Therefore, \[D_{\vecs u}f(x,y)=f_x(x,y)\cos θ+f_y(x,y)\sin θ.\]
□
Example \(\PageIndex{2}\): Finding a Directional Derivative: Alternative Method
Let \(θ=\arccos (3/5).\) Find the directional derivative \(D_{\vecs u}f(x,y)\) of \(f(x,y)=x^2−xy+3y^2\) in the direction of \(\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}\).
Then determine \(D_{\vecs u}f(−1,2)\).
Solution
First, we must calculate the partial derivatives of \(f\):
\[\begin{align*}f_x(x,y)&=2x−y \\ f_y(x,y)&=−x+6y, \end{align*}\]
Then we use Equation \ref{DD2v} with \(θ=\arccos (3/5)\):
\[\begin{align*} D_{\vecs u}f(x,y)&=f_x(x,y)\cos θ+f_y(x,y)\sin θ \\&=(2x−y)\dfrac{3}{5}+(−x+6y)\dfrac{4}{5} \\ &=\dfrac{6x}{5}−\dfrac{3y}{5}−\dfrac{4x}{5}+\dfrac{24y}{5}\\&=\dfrac{2x+21y}{5}. \end{align*}\]
To calculate \(D_{\vecs u}f(−1,2),\) let \(x=−1\) and \(y=2\):
\[D_{\vecs u}f(−1,2)=\dfrac{2(−1)+21(2)}{5}=\dfrac{−2+42}{5}=8.\]
This is the same answer obtained in Example \(\PageIndex{1}\).
Exercise \(\PageIndex{1}\):
Find the directional derivative \(D_{\vecs u}f(x,y)\) of \(f(x,y)=3x^2y−4xy^3+3y^2−4x\) in the direction of \(\vecs u=(\cos \dfrac{π}{3})\,\hat{\mathbf i}+(\sin \dfrac{π}{3})\,\hat{\mathbf j}\) using Equation \ref{DD2v}.
What is \(D_{\vecs u} f(3,4)\)?
- Hint
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Calculate the partial derivatives and determine the value of \(θ\).
- Answer
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\(D_{\vecs u}f(x,y)=\dfrac{(6xy−4y^3−4)(1)}{2}+\dfrac{(3x^2−12xy^2+6y)\sqrt{3}}{2}\)
\(D_{\vecs u}f(3,4)=\dfrac{72−256−4}{2}+\dfrac{(27−576+24)\sqrt{3}}{2}=−94−\dfrac{525\sqrt{3}}{2}\)
If the vector that is given for the direction of the derivative is not a unit vector, then it is only necessary to divide by the norm of the vector. For example, if we wished to find the directional derivative of the function in Example \(\PageIndex{2}\) in the direction of the vector \(⟨−5,12⟩\), we would first divide by its magnitude to get \(\vecs u\). This gives us \(\vecs u=⟨−\frac{5}{13},\frac{12}{13}⟩\).
Then
\[ \begin{align*} D_{\vecs u}f(x,y)&=f_x(x,y)\cos θ+f_y(x,y)\sin θ \\ &=−\dfrac{5}{13}(2x−y)+\dfrac{12}{13}(−x+6y) \\ &=−\dfrac{22}{13}x+\dfrac{17}{13}y \end{align*}\]
Gradient
The right-hand side of Equation \ref{DD2v} is equal to \(f_x(x,y)\cos θ+f_y(x,y)\sin θ,\) which can be written as the dot product of two vectors. Define the first vector as \(\vecs ∇f(x,y)=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j}\) and the second vector as \(\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}\). Then the right-hand side of the equation can be written as the dot product of these two vectors:
\[D_{\vecs u}f(x,y)=\vecs ∇f(x,y)⋅\vecs u. \label{gradDirDer}\]
The first vector in Equation \ref{gradDirDer} has a special name: the gradient of the function \(f\). The symbol \(∇\) is called nabla and the vector \(\vecs ∇f\) is read “ del \(f\).”
Definition: The Gradient
Let \(z=f(x,y)\) be a function of \(x\) and \(y\) such that \(f_x\) and \(f_y\) exist. The vector \(\vecs ∇f(x,y)\) is called the gradient of \(f\) and is defined as
\[\vecs ∇f(x,y)=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j}. \label{grad}\]
The vector \(\vecs ∇f(x,y)\) is also written as “ grad \(f\).”
Example \(\PageIndex{3}\): Finding Gradients
Find the gradient \(\vecs ∇f(x,y)\) of each of the following functions:
- \(f(x,y)=x^2−xy+3y^2\)
- \(f(x,y)=\sin 3 x \cos 3y\)
Solution
For both parts a. and b., we first calculate the partial derivatives \(f_x\) and \(f_y\), then use Equation \ref{grad}.
a. \( f_x(x,y)=2x−y\) and \(f_y(x,y)=−x+6y\), so
\[\begin{align*} \vecs ∇f(x,y)&=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j}\\&=(2x−y)\,\hat{\mathbf i}+(−x+6y)\,\hat{\mathbf j}.\end{align*}\]
b. \( f_x(x,y)=3\cos 3x \cos 3y\) and \(f_y(x,y)=−3\sin 3x \sin 3y\), so
\[\begin{align*} \vecs ∇f(x,y)&=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j} \\ &=(3\cos 3x \cos 3y)\,\hat{\mathbf i}−(3\sin 3x \sin 3y)\,\hat{\mathbf j}. \end{align*}\]
Exercise \(\PageIndex{2}\)
Find the gradient \(\vecs ∇f(x,y)\) of \(f(x,y)=\dfrac{x^2−3y^2}{2x+y}\).
- Hint
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Calculate the partial derivatives, then use Equation \ref{grad}.
- Answer
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\(\vecs ∇f(x,y)=\dfrac{2x^2+2xy+6y^2}{(2x+y)^2}\,\hat{\mathbf i}−\dfrac{x^2+12xy+3y^2}{(2x+y)^2}\,\hat{\mathbf j}\)
The gradient has some important properties. We have already seen one formula that uses the gradient: the formula for the directional derivative. Recall from The Dot Product that if the angle between two vectors \(\vecs a\) and \(\vecs b\) is \(φ\), then \(\vecs a⋅\vecs b=‖\vecs a‖‖\vecs b‖\cos φ.\) Therefore, if the angle between \(\vecs ∇f(x_0,y_0)\) and \(\vecs u=(cosθ)\,\hat{\mathbf i}+(sinθ)\,\hat{\mathbf j}\) is \(φ\), we have
\[D_{\vecs u}f(x_0,y_0)=\vecs ∇f(x_0,y_0)⋅\vecs u=\|\vecs ∇f(x_0,y_0)\|‖\vecs u‖\cos φ=\|\vecs ∇f(x_0,y_0)\|\cos φ.\]
The \(‖\vecs u‖\) disappears because \(\vecs u\) is a unit vector. Therefore, the directional derivative is equal to the magnitude of the gradient evaluated at \((x_0,y_0)\) multiplied by \(\cos φ\). Recall that \(\cos φ\) ranges from \(−1\) to \(1\).
If \(φ=0,\) then \(\cos φ=1\) and \(\vecs ∇f(x_0,y_0)\) and \(\vecs u\) both point in the same direction.
If \(φ=π\), then \(\cos φ=−1\) and \(\vecs ∇f(x_0,y_0)\) and \(\vecs u\) point in opposite directions.
In the first case, the value of \(D_{\vecs u}f(x_0,y_0)\) is maximized and in the second case, the value of \(D_{\vecs u}f(x_0,y_0)\) is minimized.
We can also see that if \(\vecs ∇f(x_0,y_0)=\vecs 0\), then
\[ D_{\vecs u}f(x_0,y_0)=\vecs ∇f(x_0,y_0)⋅\vecs u=0\]
for any vector \(\vecs u\). These three cases are outlined in the following theorem.
Theorem \(\PageIndex{2}\): Properties of the Gradient
Suppose the function \(z=f(x,y)\) is differentiable at \((x_0,y_0)\) (Figure \(\PageIndex{3}\)).
- If \(\vecs ∇f(x_0,y_0)=\vecs 0\), then \(D_{\vecs u}f(x_0,y_0)=0\) for any unit vector \(\vecs u\).
- If \(\vecs ∇f(x_0,y_0)≠\vecs 0\), then \(D_{\vecs u}f(x_0,y_0)\) is maximized when \(\vecs u\) points in the same direction as \(\vecs ∇f(x_0,y_0)\). The maximum value of \(D_{\vecs u}f(x_0,y_0)\) is \(\|\vecs ∇f(x_0,y_0)\|\).
- If \(\vecs ∇f(x_0,y_0)≠\vecs 0\), then \(D_{\vecs u}f(x_0,y_0)\) is minimized when \(\vecs u\) points in the opposite direction from \(\vecs ∇f(x_0,y_0)\). The minimum value of \(D_{\vecs u}f(x_0,y_0)\) is \(−\|\vecs ∇f(x_0,y_0)\|\).
Note: Gradient indicates direction of steepest ascent
Since the gradient vector points in the direction within the domain of \(f\) that corresponds to the maximum value of the directional derivative, \(D_{\vecs u}f(x_0,y_0)\), we say that the gradient vector points in the direction of steepest ascent or most rapid increase in \(f\) , that is, at any given point, the gradient points in the direction with the steepest uphill slope.
Example \(\PageIndex{4}\): Finding a Maximum Directional Derivative
Find the direction for which the directional derivative of \(f(x,y)=3x^2−4xy+2y^2\) at \((−2,3)\) is a maximum. What is the maximum value?
Solution:
The maximum value of the directional derivative occurs when \(\vecs ∇f\) and the unit vector point in the same direction. Therefore, we start by calculating \(\vecs ∇f(x,y\)):
\[f_x(x,y)=6x−4y \; \text{and}\; f_y(x,y)=−4x+4y \nonumber\]
so
\[\vecs ∇f(x,y)=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j}=(6x−4y)\,\hat{\mathbf i}+(−4x+4y)\,\hat{\mathbf j}. \nonumber\]
Next, we evaluate the gradient at \((−2,3)\):
\[\vecs ∇f(−2,3)=(6(−2)−4(3))\,\hat{\mathbf i}+(−4(−2)+4(3))\,\hat{\mathbf j}=−24\,\hat{\mathbf i}+20\,\hat{\mathbf j}. \nonumber\]
The gradient vector gives the direction of the maximum value of the directional derivative.
The maximum value of the directional derivative at \((−2,3)\) is \(\|\vecs ∇f(−2,3)\|=4\sqrt{61}\) (see the Figure \(\PageIndex{4}\)).
Exercise \(\PageIndex{3}\):
Find the direction for which the directional derivative of \(g(x,y)=4x−xy+2y^2\) at \((−2,3)\) is a maximum. What is the maximum value?
- Hint
-
Evaluate the gradient of \(g\) at point \((−2,3)\).
- Answer
-
The gradient of \(g\) at \((−2,3)\) is \(\vecs ∇g(−2,3)=\,\hat{\mathbf i}+14\,\hat{\mathbf j}\). This gives the direction of the maximum value of the directional derivative at the point \((−2,3)\).
The maximum value of the directional derivative is \(\|\vecs ∇g(−2,3)\|=\sqrt{197}\).
Figure \(\PageIndex{5}\) shows a portion of the graph of the function \(f(x,y)=3+\sin x \sin y\). Given a point \((a,b)\) in the domain of \(f\), the maximum value of the directional derivative at that point is given by \(\|\vecs ∇f(a,b)\|\). This would equal the rate of greatest ascent if the surface represented a topographical map. If we went in the opposite direction, it would be the rate of greatest descent.
When using a topographical map, the steepest slope is always in the direction where the contour lines are closest together (Figure \(\PageIndex{6}\)). This is analogous to the contour map of a function, assuming the level curves are obtained for equally spaced values throughout the range of that function.
Gradients and Level Curves
Recall that if a curve is defined parametrically by the function pair \((x(t),y(t)),\) then the vector \(x′(t)\,\hat{\mathbf i}+y′(t)\,\hat{\mathbf j}\) is tangent to the curve for every value of \(t\) in the domain. Now let’s assume \(z=f(x,y)\) is a differentiable function of \(x\) and \(y\), and \((x_0,y_0)\) is in its domain. Let’s suppose further that \(x_0=x(t_0)\) and \(y_0=y(t_0)\) for some value of \(t\), and consider the level curve \(f(x,y)=k\). Define \(g(t)=f(x(t),y(t))\) and calculate \(g′(t)\) on the level curve. By the chain Rule,
\[g′(t)=f_x(x(t),y(t))x′(t)+f_y(x(t),y(t))y′(t).\]
But \(g′(t)=0\) because \(g(t)=k\) for all \(t\). Therefore, on the one hand,
\[f_x(x(t),y(t))x′(t)+f_y(x(t),y(t))y′(t)=0;\]
on the other hand,
\[f_x(x(t),y(t))x′(t)+f_y(x(t),y(t))y′(t)=\vecs ∇f(x,y)⋅⟨x′(t),y′(t)⟩.\]
Therefore,
\[\vecs ∇f(x,y)⋅⟨x′(t),y′(t)⟩=0.\]
Thus, the dot product of these vectors is equal to zero, which implies they are orthogonal. However, the second vector is tangent to the level curve, which implies the gradient must be normal to the level curve, which gives rise to the following theorem.
Theorem \(\PageIndex{3}\): Gradient Is Normal to the Level Curve
Suppose the function \(z=f(x,y)\) has continuous first-order partial derivatives in an open disk centered at a point \((x_0,y_0)\). If \(\vecs ∇f(x_0,y_0)≠\vecs 0\), then \(\vecs ∇f(x_0,y_0)\) is normal to the level curve of \(f\) at \((x_0,y_0)\).
We can use this theorem to find tangent and normal vectors to level curves of a function.
Example \(\PageIndex{5}\): Finding Tangents to Level Curves
For the function \(f(x,y)=2x^2−3xy+8y^2+2x−4y+4,\) find a tangent vector to the level curve at point \((−2,1)\). Graph the level curve corresponding to \(f(x,y)=18\) and draw in \(\vecs ∇f(−2,1)\) and a tangent vector.
Solution:
First, we must calculate \(\vecs ∇f(x,y):\)
\[f_x(x,y)=4x−3y+2 \;\text{and}\; f_y=−3x+16y−4 \;\text{so}\; \vecs ∇f(x,y)=(4x−3y+2)\,\hat{\mathbf i}+(−3x+16y−4)\,\hat{\mathbf j}.\]
Next, we evaluate \(\vecs ∇f(x,y)\) at \((−2,1):\)
\[\vecs ∇f(−2,1)=(4(−2)−3(1)+2)\,\hat{\mathbf i}+(−3(−2)+16(1)−4)\,\hat{\mathbf j}=−9\,\hat{\mathbf i}+18\,\hat{\mathbf j}.\]
This vector is orthogonal to the curve at point \((−2,1)\). We can obtain a tangent vector by reversing the components and multiplying either one by \(−1\). Thus, for example, \(−18\,\hat{\mathbf i}−9\,\hat{\mathbf j}\) is a tangent vector (see the following graph).
Exercise \(\PageIndex{4}\):
For the function \(f(x,y)=x^2−2xy+5y^2+3x−2y+3\), find the tangent to the level curve at point \((1,1)\). Draw the graph of the level curve corresponding to \(f(x,y)=8\) and draw \(\vecs ∇f(1,1)\) and a tangent vector.
- Hint
-
Calculate the gradient at point \((1,1)\).
- Answer
-
\(\vecs ∇f(x,y)=(2x−2y+3)\,\hat{\mathbf i}+(−2x+10y−2)\,\hat{\mathbf j}\)
\(\vecs ∇f(1,1)=3\,\hat{\mathbf i}+6\,\hat{\mathbf j}\)
Tangent vector: \(6\,\hat{\mathbf i}−3\,\hat{\mathbf j}\) or \(−6\,\hat{\mathbf i}+3\,\hat{\mathbf j}\)
The fact that the gradient of a surface always points in the direction of steepest increase/decrease is very useful, as illustrated in the following example.
Example \(\PageIndex{6}\): The flow of water downhill
Consider the surface given by \(f(x,y)= 20-x^2-2y^2\). Water is poured on the surface at \((1,\frac{1}{4})\). What path does it take as it flows downhill?
SOLUTION
Let \(\vecs r (t) = \langle x(t), y(t)\rangle\) be the vector--valued function describing the path of the water in the \(xy\)-plane; we seek \(x(t)\) and \(y(t)\). We know that water will always flow downhill in the steepest direction; therefore, at any point on its path, it will be moving in the direction of \(-\vecs\nabla f\). (We ignore the physical effects of momentum on the water.) Thus \(\vecs r '(t)\) will be parallel to \(\vecs\nabla f\), and there is some constant \(c\) such that \(c\vecs\nabla f(t) = \vec r '(t) = \langle x'(t), y'(t)\rangle\).
We find \(\vecs\nabla f(x,y) = \langle -2x, -4y\rangle\) and write \(x'(t)\) as \(\frac{dx}{dt}\) and \(y'(t)\) as \(\frac{dy}{dt}\). Then
\[\begin{align*}
c\vecs\nabla f(t) &= \langle x'(t), y'(t)\rangle \\
\langle -2cx, -4cy \rangle & = \langle \frac{dx}{dt}, \frac{dy}{dt}\rangle.
\end{align*}\]
This implies
\[-2cx = \frac{dx}{dt} \quad \text{and} \quad -4cy =\frac{dy}{dt}, \text{ i.e.,}\]
\[c = -\frac{1}{2x}\frac{dx}{dt} \quad \text{and} \quad c =-\frac{1}{4y}\frac{dy}{dt}.\]
As \(c\) equals both expressions, we have
\[\frac{1}{4y}\frac{dy}{dt} = \frac{1}{2x}\frac{dx}{dt}.\]
To find an explicit relationship between \(x\) and \(y\), we can integrate both sides with respect to \(t\). Recall from our study of differentials that \( \frac{dx}{dt}dt = dx\). Thus:
\[\begin{align*}
\int \frac{1}{4y}\frac{dy}{dt}\ dt &= \int \frac{1}{2x}\frac{dx}{dt}\ dt \\
\int\frac{1}{4y}\ dy &= \int \frac{1}{2x}\ dx \\
\frac14\ln|y| &= \frac 12\ln|x| +C_1\\
\ln|y| &= 2\ln|x| +C_1\\
\ln|y| &= \ln |x^2|+C_1 \end{align*}\]
Now raise both sides as a power of \(e\):
\[\begin{align*}
|y| &= e^{\ln |x^2|+C_1}\\
|y| &= e^{\ln x^2}e^{C_1}\\
y &= \pm e^{C_1}x^2
\end{align*}\]
Then \[ y = Cx^2, \quad \text{where}\; C = \pm e^{C_1} \; \text{or} \; C = 0.\]
As the water started at the point \((1,\frac{1}{4})\), we can now solve for \(C\):
\[C(1)^2 = \frac14 \quad \Rightarrow \quad C = \frac14.\]
Figure \(\PageIndex{8}\):
A graph of the surface along with the path in the \(xy\)-plane with the level curves.
Thus the water follows the curve \(y=\frac{x^2}{4}\) in the \(xy\)-plane. The surface and the path of the water is graphed in Figure \(\PageIndex{8}\). In part (b) of the figure, the level curves of the surface are plotted in the \(xy\)-plane, along with the curve \(y=\frac{x^2}{4}\). Notice how the path intersects the level curves at right angles. As the path follows the gradient downhill, this reinforces the fact that the gradient is orthogonal to level curves.
Three-Dimensional Gradients and Directional Derivatives
The definition of a gradient can be extended to functions of more than two variables.
Definition: Gradients in 3D
Let \(w=f(x, y, z)\) be a function of three variables such that \(f_x, \, f_y\), and \(f_z\) exist. The vector \(\vecs ∇ f(x,y,z)\) is called the gradient of \(f\) and is defined as
\[\vecs ∇f(x,y,z)=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k}.\label{grad3d}\]
\(\vecs ∇f(x,y,z)\) can also be written as grad \(f(x,y,z).\)
Calculating the gradient of a function in three variables is very similar to calculating the gradient of a function in two variables. First, we calculate the partial derivatives \(f_x, \, f_y,\) and \(f_z\), and then we use Equation \ref{grad3d}.
Example \(\PageIndex{7}\): Finding Gradients in Three Dimensions
Find the gradient \(\vecs ∇f(x,y,z)\) of each of the following functions:
- \(f(x,y,z)=5x^2−2xy+y^2−4yz+z^2+3xz\)
- \(f(x,y,z)=e^{−2z}\sin 2x \cos 2y\)
Solution:
For both parts a. and b., we first calculate the partial derivatives \(f_x,f_y,\) and \(f_z\), then use Equation \ref{grad3d}.
a. \(f_x(x,y,z)=10x−2y+3z\), \(f_y(x,y,z)=−2x+2y−4z\), and \( f_z(x,y,z)=3x−4y+2z\), so
\[\begin{align*} \vecs ∇f(x,y,z)&=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k} \\ &=(10x−2y+3z)\,\hat{\mathbf i}+(−2x+2y−4z)\,\hat{\mathbf j}+(3x-4y+2z)\,\hat{\mathbf k}. \end{align*}\]
b. \(f_x(x,y,z) =2e^{−2z}\cos 2x \cos 2y\), \( f_y(x,y,z)=−2e^{−2z} \sin 2x \sin 2y\), and \(f_z(x,y,z)=−2e^{−2z}\sin 2x \cos 2y\), so
\[\begin{align*} \vecs ∇f(x,y,z)&=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k} \\&=(2e^{−2z}\cos 2x \cos 2y)\,\hat{\mathbf i}+(−2e^{−2z}\sin 2x \sin 2y)\,\hat{\mathbf j}+(−2e^{−2z}\sin 2x \cos 2y)\,\hat{\mathbf k} \\ & =2e^{−2z}(\cos 2x \cos 2y \,\hat{\mathbf i}−\sin 2x \sin 2y\,\hat{\mathbf j}−\sin 2x \cos 2y\,\hat{\mathbf k}). \end{align*}\]
Exercise \(\PageIndex{5}\):
Find the gradient \(\vecs ∇f(x,y,z)\) of \(f(x,y,z)=\dfrac{x^2−3y^2+z^2}{2}\) and find its gradient vector at the point \( (-1, 2, 3) \).
- Answer
-
\(\vecs ∇f(x,y,z)=x\,\hat{\mathbf i}−3y\,\hat{\mathbf j}+z\,\hat{\mathbf k}\)
\(\vecs ∇f(-1,2,3)=-\hat{\mathbf i}−6\,\hat{\mathbf j}+3\,\hat{\mathbf k}\)
Theorem \(\PageIndex{4}\): The Gradient of a Function of Three Variables is Normal to the Level Surface
Suppose the function \(z=f(x,y,z)\) has continuous first-order partial derivatives in an open sphere centered at a point \((x_0,y_0,z_0)\). If \(\vecs ∇f(x_0,y_0,z_0)≠\vecs 0\), then \(\vecs ∇f(x_0,y_0,z_0)\) is normal to the level surface of \(f\) at \((x_0,y_0,z_0)\).
Figure \(\PageIndex{9}\) shows the gradient vectors at various points on a level surface of the function in Exercise \(\PageIndex{5}\). The points shown on the level surface are: \( (-1, 2, 3) \), \( (3, -2, -1) \), \( (0, \sqrt{\frac{2}{3}}, 0) \), \( (2, \sqrt{\frac{10}{3}}, 2) \), and \( (2, \sqrt{\frac{10}{3}}, -2) \).
Figure \(\PageIndex{9}\): A level surface of the function \(f(x,y,z)=\dfrac{x^2−3y^2+z^2}{2}\) for \(C = -1\) along with various points on this level surface and the corresponding gradient vectors. Note how these gradient vectors are normal to this level surface .
The directional derivative can also be generalized to functions of three variables. To determine a direction in three dimensions, a vector with three components is needed. This vector is a unit vector, and the components of the unit vector are called directional cosines . Given a three-dimensional unit vector \(\vecs u\) in standard form (i.e., the initial point is at the origin), this vector forms three different angles with the positive \(x\)-, \(y\)-, and \(z\)-axes. Let’s call these angles \(α,β,\) and \(γ\). Then the directional cosines are given by \(\cos α,\cos β,\) and \(\cos γ\). These are the components of the unit vector \(\vecs u\); since \(\vecs u\) is a unit vector, it is true that \(\cos^2 α+\cos^2 β+\cos^2 γ=1.\)
Definition: Directional Derivative of a Function of Three variables
Suppose \(w=f(x,y,z)\) is a function of three variables with a domain of \(D\). Let \((x_0,y_0,z_0)∈D\) and let \(\vecs u=\cos α\,\hat{\mathbf i}+\cos β\,\hat{\mathbf j}+\cos γ\,\hat{\mathbf k}\) be a unit vector. Then, the directional derivative of \(f\) in the direction of \(u\) is given by
\[D_{\vecs u}f(x_0,y_0,z_0)=\lim_{t→0}\dfrac{f(x_0+t \cos α,y_0+t\cos β,z_0+t\cos γ)−f(x_0,y_0,z_0)}{t}\]
provided the limit exists.
We can calculate the directional derivative of a function of three variables by using the gradient, leading to a formula that is analogous to Equation \ref{DD2v}.
Theorem \(\PageIndex{5}): Directional Derivative of a Function of Three Variables
Let \(f(x,y,z)\) be a differentiable function of three variables and let \(\vecs u=\cos α\,\hat{\mathbf i}+\cos β\,\hat{\mathbf j}+\cos γ\,\hat{\mathbf k}\) be a unit vector. Then, the directional derivative of \(f\) in the direction of \(\vecs u\) is given by
\[D_{\vecs u}f(x,y,z)=\vecs ∇f(x,y,z)⋅\vecs u=f_x(x,y,z)\cos α+f_y(x,y,z)\cos β+f_z(x,y,z)\cos γ. \label{DDv3}\]
The three angles \(α,β,\) and \(γ\) determine the unit vector \(\vecs u\). In practice, we can use an arbitrary (nonunit) vector, then divide by its magnitude to obtain a unit vector in the desired direction.
Example \(\PageIndex{8}\): Finding a Directional Derivative in Three Dimensions
Calculate \(D_{\vecs v}f(1,−2,3)\) in the direction of \(\vecs v=−\,\hat{\mathbf i}+2\,\hat{\mathbf j}+2\,\hat{\mathbf k}\) for the function
\[ f(x,y,z)=5x^2−2xy+y^2−4yz+z^2+3xz. \nonumber\]
Solution:
First, we find the magnitude of \(v\):
\[‖\vecs v‖=\sqrt{(−1)^2+(2)^2}=3. \nonumber\]
Therefore, \(\dfrac{\vecs v}{‖\vecs v‖}=\dfrac{−\hat{\mathbf i}+2\,\hat{\mathbf j}+2\,\hat{\mathbf k}}{3}=−\dfrac{1}{3}\,\hat{\mathbf i}+\dfrac{2}{3}\,\hat{\mathbf j}+\dfrac{2}{3}\,\hat{\mathbf k}\) is a unit vector in the direction of \(\vecs v\), so \(\cos α=−\dfrac{1}{3},\cos β=\dfrac{2}{3},\) and \(\cos γ=\dfrac{2}{3}\). Next, we calculate the partial derivatives of \(f\):
\[\begin{align*} f_x(x,y,z)& =10x−2y+3z \\ f_y(x,y,z)&=−2x+2y−4z \\ f_z(x,y,z)&=−4y+2z+3x, \end{align*} \]
then substitute them into Equation \ref{DDv3}:
\[\begin{align*} D_{\vecs v}f(x,y,z)&=f_x(x,y,z)\cos α+f_y(x,y,z)\cos β+f_z(x,y,z)\cos γ \\ &=(10x−2y+3z)(−\dfrac{1}{3})+(−2x+2y−4z)(\dfrac{2}{3})+(−4y+2z+3x)(\dfrac{2}{3}) \\ &=−\dfrac{10x}{3}+\dfrac{2y}{3}−\dfrac{3z}{3}−\dfrac{4x}{3}+\dfrac{4y}{3}−\dfrac{8z}{3}−\dfrac{8y}{3}+\dfrac{4z}{3}+\dfrac{6x}{3} \\ &=−\dfrac{8x}{3}−\dfrac{2y}{3}−\dfrac{7z}{3}. \end{align*}\]
Last, to find \(D_{\vecs v}f(1,−2,3),\) we substitute \(x=1,\, y=−2\), and \(z=3:\)
\[\begin{align*} D_{\vecs v}f(1,−2,3)&=−\dfrac{8(1)}{3}−\dfrac{2(−2)}{3}−\dfrac{7(3)}{3} \\ &=−\dfrac{8}{3}+\dfrac{4}{3}−\dfrac{21}{3} \\&=−\dfrac{25}{3}. \end{align*}\]
Exercise \(\PageIndex{6}\):
Calculate \(D_{\vecs v}f(x,y,z)\) and \(D_{\vecs v}f(0,−2,5)\) in the direction of \(\vecs v=−3\,\hat{\mathbf i}+12\,\hat{\mathbf j}−4\,\hat{\mathbf k}\) for the function
\[f(x,y,z)=3x^2+xy−2y^2+4yz−z^2+2xz.\nonumber \]
- Hint
-
First, divide \(\vecs v\) by its magnitude, calculate the partial derivatives of \(f\), then use Equation \ref{DDv3}.
- Answer
-
\(D_{\vecs v}f(x,y,z)=−\dfrac{3}{13}(6x+y+2z)+\dfrac{12}{13}(x−4y+4z)−\dfrac{4}{13}(2x+4y−2z)\)
\(D_{\vecs v}f(0,−2,5)=\dfrac{384}{13}\)
Summary
- A directional derivative represents a rate of change of a function in any given direction.
- The gradient can be used in a formula to calculate the directional derivative.
- The gradient indicates the direction of greatest change of a function of more than one variable.
Key Equations
- directional derivative (two dimensions)
\(D_{\vecs u}f(a,b)=\lim_{h→0}\dfrac{f(a+h\cos θ,b+h\sin θ)−f(a,b)}{h}\)
or
\(D_{\vecs u}f(x,y)=f_x(x,y)\cos θ+f_y(x,y)\sin θ\)
or
\(D_{\vecs u}f(x,y)=\vecs ∇f(x,y) \cdot \vecs u\), where \(\vecs u\) is a unit vector in the \(xy\)-plane
- gradient (two dimensions)
\(\vecs ∇f(x,y)=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j}\)
- gradient (three dimensions)
\(\vecs ∇f(x,y,z)=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k}\)
- directional derivative (three dimensions)
\(D_{\vecs u}f(x,y,z)=\vecs ∇f(x,y,z)⋅\vecs u=f_x(x,y,z)\cos α+f_y(x,y,z)\cos β+f_x(x,y,z)\cos γ\)
Glossary
- directional derivative
-
the derivative of a function in the direction of a given unit vector
- gradient
-
the gradient of the function \(f(x,y)\) is defined to be \(\vecs ∇f(x,y)=(∂f/∂x)\,\hat{\mathbf i}+(∂f/∂y)\,\hat{\mathbf j},\) which can be generalized to a function of any number of independent variables
Contributors
-
Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .
- Example \(\PageIndex{6}\) is adapted from Apex Calculus by Gregory Hartman (Virginia Military Institute)
- Edited and expanded by Paul Seeburger (Monroe Community College)


4.3.5: Barriers to Effective Decision-Making
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- Jamie Hammond and Western Technical College
- OpenStax
Learning Objectives
- Understand the barriers that exist that make effective decision making difficult.
There are a number of barriers to effective decision-making. Effective managers are aware of these potential barriers and try to overcome them as much as possible.
Bounded Rationality
While we might like to think that we can make completely rational decisions, this is often unrealistic given the complex issues faced by managers. Nonrational decision-making is common, especially with nonprogrammed decisions. Since we haven’t faced a particular situation previously, we don’t always know what questions to ask or what information to gather. Even when we have gathered all the possible information, we may not be able to make rational sense of all of it, or to accurately forecast or predict the outcomes of our choice. Bounded rationality is the idea that for complex issues we cannot be completely rational because we cannot fully grasp all the possible alternatives, nor can we understand all the implications of every possible alternative. Our brains have limitations in terms of the amount of information they can process. Similarly, as was alluded to earlier in the chapter, even when managers have the cognitive ability to process all the relevant information, they often must make decisions without first having time to collect all the relevant data—their information is incomplete.
Escalation of Commitment
Given the lack of complete information, managers don’t always make the right decision initially, and it may not be clear that a decision was a bad one until after some time has passed. For example, consider a manager who had to choose between two competing software packages that her organization will use on a daily basis to enhance efficiency. She initially chooses the product that was developed by the larger, more well-established company, reasoning that they will have greater financial resources to invest in ensuring that the technology is good. However, after some time it becomes clear that the competing software package is going to be far superior. While the smaller company’s product could be integrated into the organization’s existing systems at little additional expense, the larger company’s product will require a much greater initial investment, as well as substantial ongoing costs for maintaining it. At this point, however, let’s assume that the manager has already paid for the larger company’s (inferior) software. Will she abandon the path that she’s on, accept the loss on the money that’s been invested so far, and switch to the better software? Or will she continue to invest time and money into trying to make the first product work? Escalation of commitment is the tendency of decision makers to remain committed to poor decision, even when doing so leads to increasingly negative outcomes. Once we commit to a decision, we may find it difficult to reevaluate that decision rationally. It can seem easier to “stay the course” than to admit (or to recognize) that a decision was poor. It’s important to acknowledge that not all decisions are going to be good ones, in spite of our best efforts. Effective managers recognize that progress down the wrong path isn’t really progress, and they are willing to reevaluate decisions and change direction when appropriate.
Time Constraints
Managers often face time constraints that can make effective decision-making a challenge. When there is little time available to collect information and to rationally process it, we are much less likely to make a good nonprogrammed decision. Time pressures can cause us to rely on heuristics rather than engage in deep processing. While heuristics save time, however, they don’t necessarily lead to the best possible solution. The best managers are constantly assessing the risks associated with acting too quickly against those associated with not acting quickly enough.
Uncertainty
In addition, managers frequently make decisions under conditions of uncertainty—they cannot know the outcome of each alternative until they’ve actually chosen that alternative. Consider, for example, a manager who is trying to decide between one of two possible marketing campaigns. The first is more conservative but is consistent with what the organization has done in the past. The second is more modern and edgier, and might bring much better results . . . or it might be a spectacular failure. The manager making the decision will ultimately have to choose one campaign and see what happens, without ever knowing what the results would have been with the alternate campaign. That uncertainty can make it difficult for some managers to make decisions, because committing to one option means forgoing other options.
Personal Biases
Our decision-making is also limited by our own biases. We tend to be more comfortable with ideas, concepts, things, and people that are familiar to us or similar to us. We tend to be less comfortable with that which is unfamiliar, new, and different. One of the most common biases that we have, as humans, is the tendency to like other people who we think are similar to us (because we like ourselves). 7 While these similarities can be observable (based on demographic characteristics such as race, gender, and age), they can also be a result of shared experiences (such as attending the same university) or shared interests (such as being in a book club together). This “similar to me” bias and preference for the familiar can lead to a variety of problems for managers: hiring less-qualified applicants because they are similar to the manager in some way, paying more attention to some employees’ opinions and ignoring or discounting others, choosing a familiar technology over a new one that is superior, sticking with a supplier that is known over one that has better quality, and so on.
It can be incredibly difficult to overcome our biases because of the way our brains work. The brain excels at organizing information into categories, and it doesn’t like to expend the effort to re-arrange once the categories are established. As a result, we tend to pay more attention to information that confirms our existing beliefs and less attention to information that is contrary to our beliefs, a shortcoming that is referred to as confirmation bias . 8
In fact, we don’t like our existing beliefs to be challenged. Such challenges feel like a threat, which tends to push our brains towards the reactive system and prevent us from being able to logically process the new information via the reflective system. It is hard to change people’s minds about something if they are already confident in their convictions. So, for example, when a manager hires a new employee who she really likes and is convinced is going to be excellent, she will tend to pay attention to examples of excellent performance and ignore examples of poor performance (or attribute those events to things outside the employee’s control). The manager will also tend to trust that employee and therefore accept their explanations for poor performance without verifying the truth or accuracy of those statements. The opposite is also true; if we dislike someone, we will pay attention to their negatives and ignore or discount their positives. We are less likely to trust them or believe what they say at face value. This is why politics tend to become very polarized and antagonistic within a two-party system. It can be very difficult to have accurate perceptions of those we like and those we dislike. The effective manager will try to evaluate situations from multiple perspectives and gather multiple opinions to offset this bias when making decisions.
Conflict
Finally, effective decision-making can be difficult because of conflict. Most individuals dislike conflict and will avoid it when possible. However, the best decision might be one that is going to involve some conflict. Consider a manager who has a subordinate who is often late to work, causing others to have to step away from their responsibilities in order to cover for the late employee. The manager needs to have a conversation with that employee to correct the behavior, but the employee is not going to like the conversation and may react in a negative way. Both of them are going to be uncomfortable. The situation is likely to involve conflict, which most people find stressful. Yet, the correct decision is still to have the conversation even if (or especially if) the employee otherwise is an asset to the department.
If the bad behavior is not corrected, it will continue, which is going to cause more problems in the workplace in the long run. Other employees may recognize that this behavior is allowed, and they may also start coming to work late or engaging in other negative behaviors. Eventually, some employees may become sufficiently frustrated that they look for another place to work. It’s worth noting that in this situation, the best employees will find new jobs the most quickly. It’s important for managers to recognize that while conflict can be uncomfortable (especially in the short-term), there are times when it is necessary for the group, department, or organization to function effectively in the long run.
It is also helpful to think about conflict in terms of process conflict or relationship conflict. 9 Process conflict , conflict about the best way to do something, can actually lead to improved performance, as individuals explore various options together in order to identify superior solutions. Relationship conflict is conflict between individuals that is more personal and involves attacks on a person rather than an idea. This kind of conflict is generally harmful and should be quelled when possible. The harm from relationship conflict arises at least in part because feeling personally attacked will cause an individual to revert to the reactive system of the brain.
Effective managers should be particularly aware of the possibility of relationship conflict when giving feedback and should keep feedback focused on behaviors and activities (how things are done) rather than on the individual. Being aware of and dealing with relationship conflict points to why emotional intelligence and empathy are beneficial in organizational leaders. Such leaders are more likely to be attentive to the harmful consequences of relationship conflict. The “Managerial Leadership” segment shows how one CEO encourages empathetic collaboration and how that effort is proving beneficial.
Managerial Leadership
Satya Nadella’s Transformation of Microsoft
When Satya Nadella became the CEO of Microsoft in 2014, he set in motion a major transformation of the organization’s culture. He wanted it to shift from a culture that valued “know-it-alls” to one that values “learn-it-all.” Instead of employees feeling the need to prove that they were the smartest person in the room, he wanted them to become curious and effective listeners, learners, and communicators. Only through continual learning and collaboration with one another, and with customers, would Microsoft remain able to develop and provide great technology solutions.
One of Nadella’s first mandates as CEO was to ask all the members of the top management team to read the book Nonviolent Communication by Marshall Rosenberg. The primary focus of the book is on empathetic communication—a kinder, gentler approach than Microsoft employees were accustomed to. Nadella believes that developing empathy leads to a heightened understanding of consumer needs and wants and an enhanced ability to develop better products and services through collaboration.
Nadella has also embraced diversity and inclusion initiatives, though he readily acknowledges that there is more to be done. This is, in part, an extension of his focus on empathy. However, it’s also good business, because increasing the diversity of perspectives can help to drive innovation.
This cultural shift is reflected in Microsoft’s new mission statement: “To empower every person and every organization on the planet to achieve more.” Empowering every person includes Microsoft’s own employees. Achieving diversity is particularly a challenge in an industry that is male dominated, and Nadella admits that he has made mistakes based on his own biases. At a Women in Computing conference early in his tenure as CEO, Nadella suggested that women did not need to ask for raises when they deserved them; the system, he said, would work it out. He later admitted that he was wrong and used the mistake as a platform for making greater strides in this arena.
Senior management team meetings at Microsoft have apparently changed dramatically as a result of the culture change driven by Nadella. Previously, members felt the need to constantly prove that they knew all the right answers at team meetings. Nadella has established different norms; he seeks out honest opinions from team members and gives positive feedback on a regular basis. By moving the focus away from always being right and toward a focus of continual learning, the culture at Microsoft has become more collaborative, and employees are more willing to take risks to create something amazing. The culture shift seems to be paying off: Microsoft’s products are being described as “cool” and “exciting,” its cloud-computing platform is outperforming the competition, and its financial performance has improved dramatically. Transforming the culture of an organization is a massive undertaking, but Nadella’s leadership of Microsoft clearly shows that it’s a decision that can pay off.
Discussion Questions
- Do you think a culture focused on learning makes sense for Microsoft? Why or why not?
- What are the advantages of a culture that emphasizes empathetic communication? Can you think of any disadvantages?
- The job of CEO means making big decisions that impact the entire organization—like deciding to change the culture. How do you think you prepare for that job?
Sources: Kendall Baker, “Confirmed: Microsoft is a legit threat to Apple,”The Hustle, March 16, 2017. Bob Evans, “10 Powerful examples of Microsoft CEO Satya Nadella’s Transformative Vision,” Forbes, July 26, 2017. Harry McCraken, “Satya Nadella Rewrites Microsoft’s Code,”Fast Company, September 18, 2017, www.fastcompany.com/40457458...icrosofts-code. Annie Palmer, “Microsoft has been reborn under CEO Satya Nadella,”The Street, September 20, 2017.
Concept Check
- Explain the concept of confirmation bias.
- List and describe at least three barriers to effective decision-making.
- When is conflict beneficial, and when is it harmful? Why?
References
7. Christopher L. Aberson, Michael Healy, & Victoria Romero. 2000. Ingroup Bias and Self-Esteem: A Meta-Analysis. Personality and Social Psychology Review , 4: 157-173.
8. Elizabeth Kolbert. 2017. Why Facts Don’t Change our Minds. The New Yorker , February 27, 2017.
9. Karen A. Jehn & Elizabeth A. Mannix. 2001. The Dynamic Nature of Conflict: A Longitudinal Study of Intragroup Conflict and Group Performance. Academy of Management Journal , 44: 238-251.


7.1: Introduction
Parenteral medications are medications administered directly into body tissue or the circulatory system (according to Merriam-Webster, “parenteral” is a term taken from the Greeks meaning “to avoid the intestines”). They are synonymous with “injectables,” as syringes and needles are used to administer these medications by subcutaneous, intradermal, intramuscular, and intravenous routes. Injections are a direct and reliable way to deliver medication for fast absorption. However, parenteral medications pose a greater risk of harm and adverse reactions than nonparenteral medications. Parenteral medications require special equipment and a specific skill set to ensure that the medication is prepared correctly to have the right therapeutic effect, and to avoid complications (Perry, Potter, & Ostendorf, 2014).
Learning Objectives
- Describe the advantages and disadvantages of injecting medications by each parenteral route
- Explain how to safely administer parenteral medication, prevent needle-stick injuries, prevent infection, and minimize patient discomfort during an injection
- Discuss factors related to equipment and selection of injection site for each parenteral route
- Describe how to correctly administer intradermal, subcutaneous, and intramuscular injections
- Review how to administer an intravenous medication via the direct IV route (bolus) and by a piggyback infusion through a continuous IV line and a saline lock
- Describe how to manage adverse reactions to IV medications
- Explain the complications associated with intravenous medications
- Summarize how to manage and report medication errors in the health care setting


2.2: Introduction to Inventory Management
The Role of Inventory Management
Inventory management plays a crucial role in industrial operations by overseeing the material resources that generate future revenue for an organization. This responsibility falls under the purview of the Operations Manager. Consider a retail store selling various products like packaged foods, groceries, apparel, and electronics. Typically, not all merchandise is stored within the store premises. A portion of the stock is kept in a warehouse facility. The combined inventory, comprising goods in the store and warehouse at any given time, constitutes the total inventory
This chapter will explore inventory management models that deal with certain or known demands. These models aim to address two fundamental questions:
- How many units to order
- When to place the order for those units
The primary objective of inventory control or inventory management is to achieve satisfactory levels of customer service while keeping inventory costs reasonably low. Therefore, there is a trade-off between the amount of inventory held and the level of customer service provided.
Source
" 8.1 Introduction to Inventory Management " from Fundamentals of Operations Management Copyright © 2024 by Azim Abbas and Seyed Goosheh is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.