Both leaders and advertisers inspire people to take action by choosing their words carefully and using them precisely. A good vocabulary is essential for success in any role that involves communication, and just about every role in life requires good communication skills. We include this section on vocabulary in this chapter on reading because of the connections between vocabulary building and reading. Building your vocabulary will make your reading easier, and reading is the best way to build your vocabulary.
Learning new words can be fun and does not need to involve tedious rote memorization of word lists. The first step, as in any other aspect of the learning cycle, is to prepare yourself to learn.  Work to become more aware of the words around you: the words you hear, the words you read, the words you say, and those you write.
Do you have a lazy vocabulary? Wake it up with the “lazy speech” exercise.
Activity: Lazy Speech
Recruit a friend you spend a lot of time with. Give them an index card with the following words written on it and ask them to keep a tally of the number of times you say these words sometime when you are together for an hour or more. If you have a small recorder, give it to the person and ask them to record you at a time you are not aware of it.
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Ummm or Uhh
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Like
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They
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You know
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OK
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Yeah
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Ohmigawd
Include in this list any other words, including expletives, that you may be using without thinking.
Are there words you constantly overuse? Were you surprised at how often you used some of these expressions? Now that you are aware of the frequency you use certain expressions, what strategies can you use to control or substitute more articulate and expressive words for them?
Building a stronger vocabulary should start with a strong foundation of healthy word use. Just as you can bring your overuse of certain words to your conscious awareness in the previous activity, think about the kinds of words you should be using more frequently. Some of the words you might consciously practice are actually very simple ones you already know but significantly underuse or use imprecisely. For example, many students say he or she “goes” instead of he or she “says.” If you take it a step further, you can consider more accurate choices still. Perhaps, he “claims” or she “argues.” Maybe he “insists” or “assumes.” Or it could be that she “believes” or she “suggests.” This may seem like a small matter, but it’s important from both a reader’s and a writer’s perspective to distinguish among the different meanings. And you can develop greater awareness by bringing some of these words into your speech.
These habits are easier to put into action if you have more and better material to draw upon: a stronger vocabulary. The following tips will help you gain and correctly use more words.
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Be on the lookout for new words.
Most will come to you as you read, but they may also appear in an instructor’s lecture, a class discussion, or a casual conversation with a friend. They may pop up in random places like billboards, menus, or even online ads!
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Write down the new words you encounter, along with the sentences in which they were used.
Do this in your notes with new words from a class or reading assignment. If a new word does not come from a class, you can write it on just about anything, but make sure you write it. Many word lovers carry a small notepad or a stack of index cards specifically for this purpose.
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Infer the meaning of the word.
The context in which the word is used may give you a good clue about its meaning. Do you recognize a common word root in the word? (Check Table 6.3 “Common Latin and Greek Word Roots” for common roots.) What do
you
think it means?
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Look up the word in a dictionary.
Do this as soon as possible (but only after inferring the meaning). When you are reading, you should have a dictionary at hand for this purpose. In other situations, do this within a couple hours, definitely during the same day. How does the dictionary definition compare with what you inferred?
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Write the word in a sentence, ideally one that is relevant to you.
If the word has more than one definition, write a sentence for each.
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Say the word out loud
and then say the definition and the sentence you wrote.
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Use the word.
Find occasion to use the word in speech or writing over the next two days.
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Schedule a weekly review
with yourself to go over your new words and their meanings.
Table 6.3 Common Latin and Greek Word Roots
Common Word Roots
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Root
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Meaning
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Examples
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auto
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self
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automatic, automobile
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bi
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two
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bicycle, biplane
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bio
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life
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biography, biology
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|
chrono
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time
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synchronize, chronicle
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dict
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say
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predict, dictate
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gen
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give birth
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generate, genetic
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geo
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earth
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geology, geography, geometry
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log
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thought
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biology, logic, pathology
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|
manu
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hand
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manufacture, manual
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phil
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love
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philosophy, anglophile
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port
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carry
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transport, portable
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sub
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under
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submarine, subtract
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vac
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empty
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vacuum, evacuate
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Where Have You Been All My Life?
The following are some fun ways to find new words:
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Read.
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When you look up a word in the dictionary, look at other interesting words on the same page.
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Solve crossword puzzles.
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Play word games like Scrabble, Boggle, or Pictionary.
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Watch movies.
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Listen to speeches and attend lectures.
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Go to comedy clubs.
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Have discussions (not just casual conversations) with friends.
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Read some more.
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Consider using a vocabulary application on your smartphone:
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Elevate
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Word of the Day
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FluentU
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Magoosh Vocabulary Builder
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Vocab Genius
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Word Pal
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Vocabulary.com
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Quizlet
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Vocabulary Builder
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Merriam-Webster Dictionary
Retrieved on July 18, 2024 from:
Scorebeyond.com
[scorebeyond.com]
Key Takeaways
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The best way to build your vocabulary is to read, and a stronger vocabulary makes it easier and more fun to read.
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Be aware of your own lazy vocabulary and try to avoid those words and expressions.
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Look for new words everywhere, not just in class readings.
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Before you look up a word in the dictionary, infer its meaning based on its context and roots.
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After you look up a word in the dictionary, write your own sentence using the new word. Say the word and definition out loud.
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Use the new word as soon as possible.
Checkpoint Exercises
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Which words do you habitually overuse? Do your friends overuse the same words? How can you collaborate to correct that overuse?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
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During the course of the day, find five new words in five different places. What were those words, and where did you uncover them?
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
__________________________________________________________________
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What do the words “manuscript,” “scribe,” and “scribble” have in common? Can you detect the same root in these words?
__________________________________________________________________
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What do you think the root means?
__________________________________________________________________


4: Physical Development in Infancy and Toddlerhood
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Last updated
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Learning Objectives
After this chapter, you should be able to:
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Describe the physical changes that occur during the first two years of life.
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Identify common infant reflexes.
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Discuss the sleep needs during the first two years of life.
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Summarize the sequence of both fine and gross motor skills.
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Recognize the developing sensory capacities of infants and toddlers.
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Explain how to meet the evolving nutritional needs of infants and toddlers.
Welcome to the story of development from infancy through toddlerhood; from birth until about two years of age. Researchers have given this part of the life span more attention than any other period, perhaps because changes during this time are so dramatic and so noticeable and perhaps because we have assumed that what happens during these years provides a foundation for one’s life to come. However, it has been argued that the significance of development during these years has been overstated (Bruer, 1999). Nevertheless, this is the period of life that contemporary educators, healthcare providers, and parents have focused on most heavily. We will examine growth and nutrition during infancy, as well as other prominent physical changes that take place during this time.
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4.1: Rapid Physical Changes
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The average newborn in the United States weighs about 7.5 pounds and is about 20 inches in length. After about a 5% weight loss in the first few days, there is a period of rapid growth. By the time an infant is 4 months old, it usually doubles in weight and by one year has tripled its birth weight. By age 2, the weight has quadrupled. The average length at one year is about 26-32 inches.
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4.2: Proportions of the Body
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Another dramatic physical change that takes place in the first several years of life is the change in body proportions. The head initially makes up about 50 percent of our entire length when we are developing in the womb. At birth, the head makes up about 25 percent of our length (think about how much of your length would be head if the proportions were still the same!). By age 25 it comprises about 20 percent our length.
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4.3: Reflexes
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Infants are equipped with a number of reflexes which are involuntary movements in response to stimulation.  They are referred to as gross motor skills (or large motor skills).
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4.5: Fine Motor Skills
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More exact movements of the feet, toes, hands, and fingers are fine motor skills (or small motor skills). These include the ability to reach and grasp an object in coordination with vision. Newborns cannot grasp objects voluntarily but do wave their arms toward objects of interest. At about 4 months of age, the infant is able to reach for an object, first with both arms and within a few weeks, with only one arm. Grasping an object involves the use of the fingers and palm, but no thumbs.
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4.6: Sensory Capacities
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Throughout much of history, the newborn was considered a passive, disorganized being who possessed minimal abilities. William James, an early psychologist, had described the newborn’s world as “a blooming, buzzing confusion,” (Shaffer, 1985). However, current research techniques have demonstrated just how developed the newborn is with especially organized sensory and perceptual abilities.
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4.7: Nutrition
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Nutritional needs change with age. This modules examine how caregivers should nourish children during the first years of life and some risks to nutrition that they should be aware of.
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4.8: Health
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Infants depend on the adults that care for them to promote and protect their health. The following section addresses common physical conditions that can affect infants, the danger of shaking babies, and the importance of immunizations.
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4.9: Sleep
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A newborn typically sleeps approximately 16.5 hours per 24-hour period. This is usually polyphasic sleep in that the infant is accumulating the 16.5 hours over several sleep periods throughout the day (Salkind, 2005). The infant is averaging 15 hours per 24-hour period by one month, and 14 hours by 6 months. By the time children turn two, they are averaging closer to 10 hours per 24 hours.
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4.S: Summary
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9.3: Systems of Nonlinear Equations and Inequalities - Two Variables
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Learning Objectives
- Solve a system of nonlinear equations using substitution.
- Solve a system of nonlinear equations using elimination.
- Graph a nonlinear inequality.
- Graph a system of nonlinear inequalities.
Halley’s Comet (Figure \(\PageIndex{1}\)) orbits the sun about once every \(75\) years. Its path can be considered to be a very elongated ellipse. Other comets follow similar paths in space. These orbital paths can be studied using systems of equations. These systems, however, are different from the ones we considered in the previous section because the equations are not linear.
Figure \(\PageIndex{1}\): Halley’s Comet (credit: "NASA Blueshift"/Flickr)
In this section, we will consider the intersection of a parabola and a line, a circle and a line, and a circle and an ellipse. The methods for solving systems of nonlinear equations are similar to those for linear equations.
Solving a System of Nonlinear Equations Using Substitution
A system of nonlinear equations is a system of two or more equations in two or more variables containing at least one equation that is not linear. Recall that a linear equation can take the form \(Ax+By+C=0\). Any equation that cannot be written in this form in nonlinear. The substitution method we used for linear systems is the same method we will use for nonlinear systems. We solve one equation for one variable and then substitute the result into the second equation to solve for another variable, and so on. There is, however, a variation in the possible outcomes.
Intersection of a Parabola and a Line
There are three possible types of solutions for a system of nonlinear equations involving a parabola and a line.
Possible Types of Solutions For Points of Intersection of a Parabola and a Line
Figure \(\PageIndex{2}\) illustrates possible solution sets for a system of equations involving a parabola and a line.
- No solution - The line will never intersect the parabola.
- One solution - The line is tangent to the parabola and intersects the parabola at exactly one point.
- Two solutions - The line crosses on the inside of the parabola and intersects the parabola at two points.
Figure \(\PageIndex{2}\)
How to: Given a system of equations containing a line and a parabola, find the solution
- Solve the linear equation for one of the variables.
- Substitute the expression obtained in step one into the parabola equation.
- Solve for the remaining variable.
- Check your solutions in both equations.
Example \(\PageIndex{1}\): Solving a System of Nonlinear Equations Representing a Parabola and a Line
Solve the system of equations.
\[\begin{align*} x−y &= −1\nonumber \\ y &= x^2+1 \nonumber \end{align*}\]
Solution
Solve the first equation for \(x\) and then substitute the resulting expression into the second equation.
\[\begin{align*} x−y &=−1\nonumber \\ x &= y−1 \;\; & \text{Solve for }x.\nonumber \\\nonumber \\ y &=x^2+1\nonumber \\ y & ={(y−1)}^2+1 \;\; & \text{Substitute expression for }x. \nonumber \end{align*}\]
Expand the equation and set it equal to zero.
\[ \begin{align*} y & ={(y−1)}^2+1\nonumber \\ &=(y^2−2y+1)+1\nonumber \\ &=y^2−2y+2\nonumber \\ 0 &= y^2−3y+2\nonumber \\ &= (y−2)(y−1) \nonumber \end{align*}\]
Solving for \(y\) gives \(y=2\) and \(y=1\). Next, substitute each value for \(y\) into the first equation to solve for \(x\). Always substitute the value into the linear equation to check for extraneous solutions.
\[\begin{align*} x−y &=−1\nonumber \\ x−(2) &= −1\nonumber \\ x &= 1\nonumber \\ x−(1) &=−1\nonumber \\ x &= 0 \nonumber \end{align*}\]
The solutions are \((1,2)\) and \((0,1)\),which can be verified by substituting these \((x,y)\) values into both of the original equations (Figure \(\PageIndex{3}\)).
Figure \(\PageIndex{3}\)
Q&A: Could we have substituted values for \(y\) into the second equation to solve for \(x\) in last example?
Yes, but because \(x\) is squared in the second equation this could give us extraneous solutions for \(x\).
For \(y=1\)
\[\begin{align*} y &= x^2+1\nonumber \\ y &= x^2+1\nonumber \\ x^2 &= 0\nonumber \\ x &= \pm \sqrt{0}=0 \nonumber \end{align*}\]
This gives us the same value as in the solution.
For \(y=2\)
\[\begin{align*} y &= x^2+1\nonumber \\ 2 &= x^2+1\nonumber \\ x^2 &= 1\nonumber \\ x &= \pm \sqrt{1}=\pm 1 \nonumber \end{align*}\]
Notice that \(−1\) is an extraneous solution.
Exercise \(\PageIndex{1}\)
Solve the given system of equations by substitution.
\[\begin{align*} 3x−y &= −2\nonumber \\ 2x^2−y &= 0 \nonumber \end{align*}\]
- Answer
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\(\left(−\dfrac{1}{2},\dfrac{1}{2}\right)\) and \((2,8)\)
Intersection of a Circle and a Line
Just as with a parabola and a line, there are three possible outcomes when solving a system of equations representing a circle and a line.
POSSIBLE TYPES OF SOLUTIONS FOR THE POINTS OF INTERSECTION OF A CIRCLE AND A LINE
Figure \(\PageIndex{4}\) illustrates possible solution sets for a system of equations involving a circle and a line.
- No solution - The line does not intersect the circle.
- One solution - The line is tangent to the circle and intersects the circle at exactly one point.
- Two solutions - The line crosses the circle and intersects it at two points.
Figure \(\PageIndex{4}\)
How to: Given a system of equations containing a line and a circle, find the solution
- Solve the linear equation for one of the variables.
- Substitute the expression obtained in step one into the equation for the circle.
- Solve for the remaining variable.
- Check your solutions in both equations.
Example \(\PageIndex{2}\): Finding the Intersection of a Circle and a Line by Substitution
Find the intersection of the given circle and the given line by substitution.
\[\begin{align*} x^2+y^2 &= 5\nonumber \\ y &= 3x−5 \nonumber \end{align*}\]
Solution
One of the equations has already been solved for \(y\). We will substitute \(y=3x−5\) into the equation for the circle.
\[\begin{align*} x^2+{(3x−5)}^2 &= 5\nonumber \\ x^2+9x^2−30x+25 &= 5\nonumber \\ 10x^2−30x+20 &= 0 \end{align*} \]
Now, we factor and solve for \(x\).
\[\begin{align*} 10(x2−3x+2) &= 0\nonumber \\ 10(x−2)(x−1) &= 0\nonumber \\ x &= 2\nonumber \\ x &= 1 \nonumber \end{align*}\]
Substitute the two \(x\)-values into the original linear equation to solve for \(y\).
\[\begin{align*} y &= 3(2)−5\nonumber \\ &= 1\nonumber \\ y &= 3(1)−5\nonumber \\ &= −2 \nonumber \end{align*}\]
The line intersects the circle at \((2,1)\) and \((1,−2)\),which can be verified by substituting these \((x,y)\) values into both of the original equations (Figure \(\PageIndex{5}\)).
Figure \(\PageIndex{5}\)
Exercise \(\PageIndex{2}\)
Solve the system of nonlinear equations.
\[\begin{align*} x^2+y^2 &= 10\nonumber \\ x−3y &= −10 \nonumber \end{align*}\]
- Answer
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\((−1,3)\)
Solving a System of Nonlinear Equations Using Elimination
We have seen that substitution is often the preferred method when a system of equations includes a linear equation and a nonlinear equation. However, when both equations in the system have like variables of the second degree, solving them using elimination by addition is often easier than substitution. Generally, elimination is a far simpler method when the system involves only two equations in two variables (a two-by-two system), rather than a three-by-three system, as there are fewer steps. As an example, we will investigate the possible types of solutions when solving a system of equations representing a circle and an ellipse.
POSSIBLE TYPES OF SOLUTIONS FOR THE POINTS OF INTERSECTION OF A CIRCLE AND AN ELLIPSE
Figure \(\PageIndex{6}\) illustrates possible solution sets for a system of equations involving a circle and an ellipse.
- No solution - The circle and ellipse do not intersect. One shape is inside the other or the circle and the ellipse are a distance away from the other.
- One solution - The circle and ellipse are tangent to each other, and intersect at exactly one point.
- Two solutions - The circle and the ellipse intersect at two points.
- Three solutions - The circle and the ellipse intersect at three points.
- Four solutions - The circle and the ellipse intersect at four points.
Figure \(\PageIndex{6}\)
Example \(\PageIndex{3}\): Solving a System of Nonlinear Equations Representing a Circle and an Ellipse
Solve the system of nonlinear equations.
\[\begin{align*} x^2+y^2 &= 26 &(1)\nonumber \\ 3x^2+25y^2 &= 100 & (2) \nonumber \end{align*}\]
Solution
Let’s begin by multiplying equation (1) by \(−3\),and adding it to equation (2).
\[\begin{align*} (−3)(x^2+y^2) = (−3)(26)&\nonumber \\ −3x^2−3y^2 = −78 &\nonumber \\ \underline{3x^2+25y^2=100}&\nonumber \\ 22y^2=22& \nonumber \end{align*}\]
After we add the two equations together, we solve for \(y\).
\[\begin{align*} y^2 &= 1\nonumber \\ y &= \pm \sqrt{1}=\pm 1 \nonumber \end{align*}\]
Substitute \(y=\pm 1\) into one of the equations and solve for \(x\).
\[\begin{align*} x^2+{(1)}^2 &= 26\nonumber \\ x^2+1 &= 26\nonumber \\ x^2 &= 25\nonumber \\ x &= \pm \sqrt{25}=\pm 5\nonumber \\ x^2+{(−1)}^2 &= 26\nonumber \\ x^2+1 &= 26\nonumber \\ x^2 &= \pm \sqrt{25}=\pm 5 \nonumber \end{align*}\]
There are four solutions: \((5,1)\), \((−5,1)\), \((5,−1)\),and \((−5,−1)\). See Figure \(\PageIndex{7}\).
Figure \(\PageIndex{7}\)
Exercise \(\PageIndex{3}\)
Find the solution set for the given system of nonlinear equations.
\[\begin{align*} 4x^2+y^2 &= 13\nonumber \\ x^2+y^2 &= 10 \nonumber \end{align*}\]
- Answer
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\({(1,3),(1,−3),(−1,3),(−1,−3)}\)
Graphing a Nonlinear Inequality
All of the equations in the systems that we have encountered so far have involved equalities, but we may also encounter systems that involve inequalities. We have already learned to graph linear inequalities by graphing the corresponding equation, and then shading the region represented by the inequality symbol. Now, we will follow similar steps to graph a nonlinear inequality so that we can learn to solve systems of nonlinear inequalities. A nonlinear inequality is an inequality containing a nonlinear expression. Graphing a nonlinear inequality is much like graphing a linear inequality.
Recall that when the inequality is greater than, \(y>a\),or less than, \(y<a\),the graph is drawn with a dashed line. When the inequality is greater than or equal to, \(y≥a\),or less than or equal to, \(y≤a\),the graph is drawn with a solid line. The graphs will create regions in the plane, and we will test each region for a solution. If one point in the region works, the whole region works. That is the region we shade (Figure \(\PageIndex{8}\)).
Figure \(\PageIndex{8}\): (a) an example of \(y>a\); (b) an example of \(y≥a\); (c) an example of \(y<a\); (d) an example of \(y≤a\)
How to: Given an inequality bounded by a parabola, sketch a graph
- Graph the parabola as if it were an equation. This is the boundary for the region that is the solution set.
- If the boundary is included in the region (the operator is \(≤\) or \(≥\)), the parabola is graphed as a solid line.
- If the boundary is not included in the region (the operator is \(<\) or \(>\)), the parabola is graphed as a dashed line.
- Test a point in one of the regions to determine whether it satisfies the inequality statement. If the statement is true, the solution set is the region including the point. If the statement is false, the solution set is the region on the other side of the boundary line.
- Shade the region representing the solution set.
Example \(\PageIndex{4}\): Graphing an Inequality for a Parabola
Graph the inequality \(y>x^2+1\).
Solution
First, graph the corresponding equation \(y=x^2+1\). Since \(y>x^2+1\) has a greater than symbol, we draw the graph with a dashed line. Then we choose points to test both inside and outside the parabola. Let’s test the points
\((0,2)\) and \((2,0)\). One point is clearly inside the parabola and the other point is clearly outside.
\[\begin{align*} y &> x^2+1\nonumber \\ 2 &> (0)^2+1\nonumber \\ 2 &>1 & \text{True}\nonumber \\\nonumber \\\nonumber \\ 0 &> (2)^2+1\nonumber \\ 0 &> 5 & \text{False} \nonumber \end{align*}\]
The graph is shown in Figure \(\PageIndex{9}\). We can see that the solution set consists of all points inside the parabola, but not on the graph itself.
Figure \(\PageIndex{9}\)
Graphing a System of Nonlinear Inequalities
Now that we have learned to graph nonlinear inequalities, we can learn how to graph systems of nonlinear inequalities. A system of nonlinear inequalities is a system of two or more inequalities in two or more variables containing at least one inequality that is not linear. Graphing a system of nonlinear inequalities is similar to graphing a system of linear inequalities. The difference is that our graph may result in more shaded regions that represent a solution than we find in a system of linear inequalities. The solution to a nonlinear system of inequalities is the region of the graph where the shaded regions of the graph of each inequality overlap, or where the regions intersect, called the feasible region .
How to: Given a system of nonlinear inequalities, sketch a graph
- Find the intersection points by solving the corresponding system of nonlinear equations.
- Graph the nonlinear equations.
- Find the shaded regions of each inequality.
- Identify the feasible region as the intersection of the shaded regions of each inequality or the set of points common to each inequality.
Example \(\PageIndex{5}\): Graphing a System of Inequalities
Graph the given system of inequalities.
\[\begin{align*} x^2−y &≤ 0\nonumber \\ 2x^2+y &≤ 12 \nonumber \end{align*}\]
Solution
These two equations are clearly parabolas. We can find the points of intersection by the elimination process: Add both equations and the variable \(y\) will be eliminated. Then we solve for \(x\).
\[\begin{align*} x^2−y = 0&\nonumber \\ \underline{2x^2+y=12}&\nonumber \\ 3x^2=12&\nonumber \\ x^2=4 &\nonumber \\ x=\pm 2 & \nonumber \end{align*}\]
Substitute the \(x\)-values into one of the equations and solve for \(y\).
\[\begin{align*} x^2−y &= 0\nonumber \\ {(2)}^2−y &= 0\nonumber \\ 4−y &= 0\nonumber \\ y &= 4\nonumber \\\nonumber \\ {(−2)}^2−y &= 0\nonumber \\ 4−y &= 0\nonumber \\ y &= 4 \nonumber \end{align*}\]
The two points of intersection are \((2,4)\) and \((−2,4)\). Notice that the equations can be rewritten as follows.
\[\begin{align*} x^2-y & ≤ 0\nonumber \\ x^2 &≤ y\nonumber \\ y &≥ x^2\nonumber \\\nonumber \\\nonumber \\ 2x^2+y &≤ 12\nonumber \\ y &≤ −2x^2+12 \nonumber \end{align*}\]
Graph each inequality. See Figure \(\PageIndex{10}\). The feasible region is the region between the two equations bounded by \(2x^2+y≤12\) on the top and \(x^2−y≤0\) on the bottom.
Figure \(\PageIndex{10}\)
Exercise \(\PageIndex{5}\)
Graph the given system of inequalities.
\[\begin{align*} y &≥ x^2−1\nonumber \\ x−y &≥ −1 \nonumber \end{align*}\]
- Answer
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Shade the area bounded by the two curves, above the quadratic and below the line.
Figure \(\PageIndex{11}\)
Media
Access these online resources for additional instruction and practice with nonlinear equations.
Key Concepts
- There are three possible types of solutions to a system of equations representing a line and a parabola: (1) no solution, the line does not intersect the parabola; (2) one solution, the line is tangent to the parabola; and (3) two solutions, the line intersects the parabola in two points. See Example \(\PageIndex{1}\).
- There are three possible types of solutions to a system of equations representing a circle and a line: (1) no solution, the line does not intersect the circle; (2) one solution, the line is tangent to the parabola; (3) two solutions, the line intersects the circle in two points. See Example \(\PageIndex{2}\).
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There are five possible types of solutions to the system of nonlinear equations representing an ellipse and a circle:
(1) no solution, the circle and the ellipse do not intersect; (2) one solution, the circle and the ellipse are tangent to each other; (3) two solutions, the circle and the ellipse intersect in two points; (4) three solutions, the circle and ellipse intersect in three places; (5) four solutions, the circle and the ellipse intersect in four points. See Example \(\PageIndex{3}\). - An inequality is graphed in much the same way as an equation, except for > or <, we draw a dashed line and shade the region containing the solution set. See Example \(\PageIndex{4}\).
- Inequalities are solved the same way as equalities, but solutions to systems of inequalities must satisfy both inequalities. See Example \(\PageIndex{5}\).
Contributors and Attributions
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Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus .


12.2: Citizenship
Substantive Citizenship
Substantive Citizenship is linked to the idea of forming an identity and cultural distinction. In this type of citizenship, minority groups are at a disadvantage as it can ascribe “imagined labor market identities to workers with different nationalities” [7] . Which ends up causing a lot of problems and unfair treatment. This concept “implies the notion of equality in that citizens are said to share a common status in respect of the rights and duties that they hold”, but instead it deals with who gets to “enjoy the rights that ensure effective membership of a national community” [8] .
An example of this seen in the world is that countries like the United States, France or England see working on a farm or the growing of crops as a lower job. While in places like Central and South America, farming is seen quite frequently and is accepted as an equal job. This means that when the people from these areas move to the United States or Europe they face discrimination as this is the only job that they know how to do.
Flexible Citizenship
Flexible citizenship [16] was defined by anthropologist Aihwa Ong as “ the strategies and effects employed by managers, technocrats, and professionals who move regularly across state boundaries who seek both to circumvent and benefit from different nation-state regimes.” This concept allows the nation-states to work together to move toward a better economy and global success. This is closely related to the concept of globalization [17]. It seemed before that in order for the nation-states to become globalized they would need to be independent, this has proven to be very contradictory due to flexible citizenship. This concept allows people such as business owners and managers to better their company and do so on a much larger scale.
This is exemplified in the Chinese culture. Chinese elite families are large components for the successes in the economy of the Pacific Rim. This all began when the Chinese moved into European empires, while doing so the Chinese were required to strengthen their bonds with their family and business partners in order to be successful. For example, a Hong Kong family business owner had their son run a part of the hotel chain in the Pacific regions, while another brother lived in San Francisco and managed the hotels that were located within Northern America and Europe.
Post-National Ethos
Within the Chinese Culture and Flexible Citizenship, there can sometimes be a few negative effects. When families are dispersed all throughout the world it can seem hard to stay happy and close together when things such as support, relationships, and parental responsibilities are neglected. It seems as though the success sometimes gets out of control and all that is seen by them is success. It is forgotten what it is like to think of things other than globalization[18], a capitalist economy, and money. The concept of Nationalism can completely lose its meaning and they seem to follow a Post-National ethos. This happens when “an attitude toward the world in which people submit to the govern mentality of the capitalist market while trying to avoid the govern mentality of nation-states.”
Pathway to and Problems with Gaining US Citizenship
For thousands of Americans, gaining citizenship to the United States was the most frustrating and time-consuming process they have ever experienced. However, it is not the civics test or the pointed questions from federal officials that make the process so hard. For most, the real block comes sooner, when prospective citizens seek to live and work in the United States by obtaining their "green card." Some 140,000 professionals, more than half working in the technology sector, are granted permanent residency out of nearly 900,000 immigrants America welcomes each year. It is this group that tends to go through an increasingly costly, risky, and tedious process. For example, most immigration lawyers charge between $5,000 to $7,500 to accompany a client through the green card process. Some cases can cost closer to $15,000 before adding on application feeds and any potential family members. The real cost, however, is harder to quantify. Many applicants can spend years marked by a feeling of lost opportunity and helplessness as they wait for the process to conclude. The process takes a while because they want to ensure that all green card members in the United States are qualified to become a citizen. They are tested for their dedication, loyalty, and trust for the country.[13]


9.3: Asset Turnover Rate
- Analyze the asset turnover rate
Asset turnover, or turnover rate, is a ratio of how many times during a selling season assets are turned over, or used. In other words, it calculates how many sales dollars are generated for each dollar invested in assets. The formula is revenue / net assets.
Think about how this might apply within a retail context. Imagine you’re the front-end manager at your local supermarket, with responsibilities including the coffee shop and floral department. Looking at the income statement for successive periods, you can see that asset turnover is slowing. You know that your buying has been steady, so this must mean that net sales are being affected.
As you walk the area, you notice that you still have mugs with hearts on them and vases that say “I love you Mom” on display—still at full retail price. This looks like stuff from Valentine’s and Mother’s Day. Remnant inventory is a concern—if they don’t sell, there will be less money to pay for planned purchases. What can you do to help the situation?
Markdown the slow selling “old merchandise” in order to stimulate sales and move the merchandise. It may not sell very well, but it will sell a lot better than leaving it fully priced. Of course, this can have a slightly negative impact on gross margin, but it will increase sales. Ultimately, this will make it possible to order new merchandise that will stimulate interest and sales. If possible, it might also be possible to work with the vendor to return the unsold, aging merchandise or to share the costs of markdowns.
Contributors and Attributions
- Asset Turnover Rate. Authored by : Patrick Williams. Provided by : Lumen Learning. License : CC BY: Attribution
- Target Store Valentine's Day. Authored by : Mike Mozart. Provided by : Flickr. Located at : https://flic.kr/p/rbK2aw . License : CC BY: Attribution


21.3: Workplace Privacy
Employees generally do not have a reasonable expectation of privacy in the workplace, especially when using company equipment or when the employer has a policy stating employees may be monitored. However, some areas such as employee restrooms and locker rooms may not be monitored. Courts have held employees do not give up all expectations of privacy by the nature of their employment. Therefore, employers should ensure that they limit monitoring activities to reasonable places where the employer has a legitimate business interest for doing so.
Hiring Process
Employers often run background checks on prospective employees as part of their hiring process. Depending on what type of background check is done and the information used, a range of privacy issues are involved. Some states regulate the type of documents that a prospective employer may consider when making hiring decisions. Businesses need to ensure they comply with all state laws where they hire employees.
The use of artificial intelligence (AI) is a growing trend in recruiting and hiring. AI is often used to review resumes, applications, and publicly available social media. AI-powered video-interview platforms apply algorithms to video-recorded interviews to facilitate an employer’s assessment of applicants.
Illinois was the first state to pass an AI Interview Act, which requires prospective employers to notify applicants of their use of AI and to obtain their consent before using AI tools on their application materials. Although limited to its state, the Illinois law has been cited by many legal experts as a template for other federal and state laws.
Based on the Illinois law, employers who use AI during their hiring process should adopt the following best practices:
- Give notice to applicants of the use of AI-powered video-interview platforms;
- Explain what the AI is and how it works in ordinary language to applicants;
- Obtain consent of applicants to use and record their video interviews;
- Offer an alternative interview method for interviews; and
- Have a procedure in place for the destruction of recordings.
Drug and Alcohol Testing
Employers with drug and alcohol testing policies are highly regulated by the states where they operate. State requirements vary about required notice of testing, the nature and location of testing, and when testing may occur. All states protect employee privacy regarding who receives the test result and how those results are to be collected, stored, and destroyed. Employers who engage in drug and alcohol testing need to be informed about the legal consequences of enforcing their policies.
Employees frequently challenge drug and alcohol testing as a violation of their right to privacy. Employers generally win these lawsuits when:
- The employer complies with all state requirements for drug and alcohol testing;
- Conducts the test with the employee’s consent;
- Conducts the test in a manner that was not offensive; and
- The test results do not reveal information unrelated to the purpose of the test.
Employers must be careful to limit disclosure of test results to only those with a need to know. Businesses may lawfully conduct a drug or alcohol test but still be liable for privacy violations based on how they handled the results.
Health Insurance Portability and Accountability Act
The Health Insurance Portability and Accountability Act of 1996 (HIPAA) seeks to protect confidential health information and mandates standards for handling such information.
HIPAA has a Privacy Rule regulating the use and disclosure of individually identifiable health information. The Privacy Rule protects Protected Health Information (PHI) , which includes all information related to the past, present or future health status of an identified individual, of treatment received, or of payment for treatment. PHI also includes billing records, information about premium payments, and enrollment information. As a result, PHI includes medical information required by employers to carry out their obligations under the Americans with Disabilities Act, the Family Medical Leave Act, workers’ compensation, drug testing, and employer-sponsored health care plans.
HIPAA also has a Security Rule to ensure the confidentiality, integrity and availability of electronic PHI. Under the Security Rule,
- Confidentiality means PHI is not made available or disclosed to unauthorized individuals or processes;
- Integrity means PHI is not altered or destroyed in an unauthorized manner; and
- Availability means PHI is accessible and usable upon demand by an authorized individual.
The Security Rule also requires businesses to protect electronic PHI against reasonably anticipated threats and reasonably anticipated violations of the Privacy Rule.
Figure 21.1 HIPAA Security Rule
HIPAA requires businesses to designate a single person who is ultimately responsible for the security of electronic PHI. This person is also responsible for ensuring the business engages in the mandatory security management process under HIPAA. This process starts with a risk analysis of the potential vulnerabilities in the business’s system and management of PHI. The security management process is extensively regulated.
Importantly, HIPAA applies to “covered entities” rather than specific types of information. Personal fitness trackers such as Fitbit, gather what is essentially healthcare data of its consumers. However, Fitbit data can be sold as consumer information because Fitbit is not a covered entity under HIPAA with regard to its consumers. However, if Fitbit gathers PHI of its employees who request medical leaves of absence, then Fitbit is a covered entity as an employer.
Electronic Monitoring
Federal law and most state laws allow employers to monitor their employees’ electronic communications occurring over the employer’s hardware, software, and servers. If the employer provides the computer system, the employer has the right to monitor electronic communications on the system, even if those communications are not work related.
Employers may also monitor communications when employees consent to the monitoring. Therefore, many employers require employees to sign a waiver consenting to private communications sent via the employer’s equipment to be monitored. This helps defend against invasion of privacy claims better than having a policy in the employee handbook alone.
Businesses may also monitor conversations with customers in the ordinary course of business as long as they give notice. As a result, many customer service lines use a recorded message that “this call may be monitored for training purposes” before customers are connected to a customer service agent.
The most important federal law regarding monitoring of electronic communications is the Electronic Communications Privacy Act (ECPA) , which was passed by Congress in 1986. ECPA has two parts. The first part is known as the Wiretap Act and the second as the Stored Communications Act. ECPA prohibits the acquisition of the the content of a wire, oral or electronic communication using an electronic, mechanical or other device. ECPA also prohibits the use or disclosure of an unlawfully intercepted communication.
ECPA exposes businesses to multiple levels of liability within a business. For example, personnel in the IT department may be liable for unlawfully intercepting an employee’s email, and human resource personnel who use and disclose the email may be liable as well. Each unlawfully intercepted communication may give rise to liability. Therefore, a handful of communications may result in multiple individuals throughout a business repeatedly violating ECPA.
Workplace Recordings
Although recordings may be useful to capture the content of a conversation, recordings pose legal and business risks to employers. Both employers and employees may violate federal and state wiretapping laws by recording conversations without consent of the other parties. Even with consent, businesses that engage in recording employees and customers damage employee morale and risk losing customers.
Twelve states prohibit recording a conversation unless all parties consent. The majority of states allow customers and employees to hold a business liable for wiretapping violations under the respondeat superior doctrine. As a result, businesses may be liable for their employees’ unlawful recordings if done in the course and scope of employment or done to help the business.
State and federal wiretapping laws carry both civil and criminal penalties. Many state laws provide for treble damages or a statutory damage amount. Federal wiretapping laws impose fines up to one hundred dollars per day or ten thousand dollars, whichever is greater.
Another potential problem for businesses is putting confidential business information at risk. For example, employees may capture trade secrets, proprietary information, or business strategies that the business wants to protect. Recorded information can be compromised or shared against the business’s interests.
Social Media
An employer’s right to monitor electronic communications generally does not include social media. As a result, employers are not entitled to monitor social media accounts through coercion or deceit. For example, an employer cannot require employees to provide passwords to their social media accounts. Employers also cannot log onto the social media accounts of others (including employees) and pose as them to see private accounts.
However, if social media accounts are public, then employers are entitled to review them to the same extent as other members of the public.
Videotaping and Surveillance Cameras
ECPA only protects electronic communications. As a result, ECPA does not apply to video or camera surveillance without an audio component. To avoid violating ECPA, businesses should ensure their security and surveillance cameras do not capture human voices.
Security cameras cannot be used in areas in which employees and customers have a reasonable expectation of privacy. For example, retailers cannot use cameras in changing rooms, restrooms, and locker rooms. Businesses need to place cameras so that private activity cannot, and is not, monitored and recorded.
Businesses engaged in surveillance must use the most limited means available to conduct the surveillance. Companies should have a legitimate business reason to use security cameras, and they need to ensure the surveillance is targeted and limited in duration and scope.
Retailers who use cameras to prevent theft at entryways and cash registers should place cameras in positions that are open and obvious to act as notice to customers. Signs giving express notice are also a best practice to avoid legal liability.
Biometrics and Wearable Technology
Biometrics is the automated identification of people using their physical characteristics. While many metrics can be used, fingerprints and facial recognition are the most common. It is helpful to think of biometrics as measurements of some aspect of a person.
There is a growing trend among businesses to move from traditional time clocks to biometric time clocks that scan fingerprints, retinas, or irises to verify an employee’s identity and clock the employee in and out of work. Biometric time clocks prevent time clock fraud, increase timekeeping efficiency, and increase accuracy of wages.
The type of biometric technology used impacts the privacy rights involved. Technology storing biometric data directly impacts privacy rights more than technology creating a “template” through an algorithm to create a representation of a fingerprint. Whether the technology captures and uses existing personal information or creates a replica has legal consequences.
There have been a series of class action lawsuits against employers that have not notified employees when their biometric identifiers and data were being shared with third party timekeeping vendors. Consent is only a defense for employers if they give notice and obtain consent for all uses of the information.
Another technology trend is the use of wearable technology. Wearable technology is a category of electronic devices that can be worn as accessories, embedded in clothing, implanted in the user’s body, or even tattooed on skin. The devices are intended to be hands-free, are powered by microprocessors and connect with the Internet. Wearable technology includes smartwatches, fitness trackers, and medical devices. It is helpful to think of wearable technology as something that an employee has.
Wearable technology is often used to track employee locations and grant access to areas. Concerned about private companies coercing employees to be microchipped, states are passing laws prohibiting employers to require, coerce, or compel an individual to receive a microchip implant or use wearable technology as a condition of employment.


5.8: Stokes' Theorem
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Learning Objectives
- Explain the meaning of Stokes’ theorem.
- Use Stokes’ theorem to evaluate a line integral.
- Use Stokes’ theorem to calculate a surface integral.
- Use Stokes’ theorem to calculate a curl.
In this section, we study Stokes’ theorem, a higher-dimensional generalization of Green’s theorem. This theorem, like the Fundamental Theorem for Line Integrals and Green’s theorem, is a generalization of the Fundamental Theorem of Calculus to higher dimensions. Stokes’ theorem relates a vector surface integral over surface \(S\) in space to a line integral around the boundary of \(S\). Therefore, just as the theorems before it, Stokes’ theorem can be used to reduce an integral over a geometric object \(S\) to an integral over the boundary of \(S\). In addition to allowing us to translate between line integrals and surface integrals, Stokes’ theorem connects the concepts of curl and circulation. Furthermore, the theorem has applications in fluid mechanics and electromagnetism. We use Stokes’ theorem to derive Faraday’s law , an important result involving electric fields.
Stokes’ Theorem
Stokes’ theorem says we can calculate the flux of \( curl \,\vecs{F}\) across surface \(S\) by knowing information only about the values of \(\vecs{F}\) along the boundary of \(S\). Conversely, we can calculate the line integral of vector field \(\vecs{F}\) along the boundary of surface \(S\) by translating to a double integral of the curl of \(\vecs{F}\) over \(S\).
Let \(S\) be an oriented smooth surface with unit normal vector \(\vecs{N}\). Furthermore, suppose the boundary of \(S\) is a simple closed curve \(C\). The orientation of \(S\) induces the positive orientation of \(C\) if, as you walk in the positive direction around \(C\) with your head pointing in the direction of \(\vecs{N}\), the surface is always on your left. With this definition in place, we can state Stokes’ theorem .
 If \(\vecs{F}\) is a vector field with component functions that have continuous partial derivatives on an open region containing \(S\), then
\[\int_C \vecs{F} \cdot d \vecs{r} = \iint_S curl \, \vecs{F} \cdot d\vecs S. \label{Stokes1} \]
Suppose surface \(S\) is a flat region in the \(xy\)-plane with upward orientation. Then the unit normal vector is \(\vecs{k}\) and surface integral
\[\iint_S curl \, \vecs{F} \cdot d\vecs{S} \nonumber \]
is actually the double integral
\[\iint_S curl \, \vecs{F} \cdot \vecs{k} \, dA. \nonumber \]
In this special case, Stokes’ theorem gives
\[\int_C \vecs{F} \cdot d\vecs{r} = \iint_S curl \, \vecs{F} \cdot \vecs{k} \, dA. \nonumber \]
However, this is the flux form of Green’s theorem, which shows us that Green’s theorem is a special case of Stokes’ theorem. Green’s theorem can only handle surfaces in a plane, but Stokes’ theorem can handle surfaces in a plane or in space.
The complete proof of Stokes’ theorem is beyond the scope of this text. We look at an intuitive explanation for the truth of the theorem and then see proof of the theorem in the special case that surface \(S\) is a portion of a graph of a function, and \(S\), the boundary of \(S\), and \(\vecs{F}\) are all fairly tame.
Proof
First, we look at an informal proof of the theorem. This proof is not rigorous, but it is meant to give a general feeling for why the theorem is true. Let \(S\) be a surface and let \(D\) be a small piece of the surface so that \(D\) does not share any points with the boundary of \(S\). We choose \(D\) to be small enough so that it can be approximated by an oriented square \(E\). Let \(D\) inherit its orientation from \(S\), and give \(E\) the same orientation. This square has four sides; denote them \(E_l, \, E_r, \, E_u\), and \(E_d\) for the left, right, up, and down sides, respectively. On the square, we can use the flux form of Green’s theorem:
\[\int_{E_l+E_d+E_r+E_u} \vecs{F} \cdot d \vecs{r} = \iint_E curl \, \vecs{F} \cdot \vecs{N} \, d \vecs{S} = \iint_E curl \, \vecs{F} \cdot d\vecs{S}. \nonumber \]
To approximate the flux over the entire surface, we add the values of the flux on the small squares approximating small pieces of the surface (Figure \(\PageIndex{2}\)).
By Green’s theorem, the flux across each approximating square is a line integral over its boundary. Let \(F\) be an approximating square with an orientation inherited from \(S\) and with a right side \(E_l\) (so \(F\) is to the left of \(E\)). Let \(F_r\) denote the right side of \(F\); then, \(E_l = - F_r\). In other words, the right side of \(F\) is the same curve as the left side of \(E\), just oriented in the opposite direction. Therefore,
\[\int_{E_l} \vecs F \cdot d\vecs r = - \int_{F_r} \vecs F \cdot d\vecs r. \nonumber \]
As we add up all the fluxes over all the squares approximating surface \(S\), line integrals
\[\int_{E_l} \vecs{F} \cdot d \vecs{r} \nonumber \]
and
\[ \int_{F_r} \vecs{F} \cdot d\vecs{r} \nonumber \]
cancel each other out. The same goes for the line integrals over the other three sides of \(E\). These three line integrals cancel out with the line integral of the lower side of the square above \(E\), the line integral over the left side of the square to the right of \(E\), and the line integral over the upper side of the square below \(E\) (Figure \(\PageIndex{3}\)). After all this cancelation occurs over all the approximating squares, the only line integrals that survive are the line integrals over sides approximating the boundary of \(S\). Therefore, the sum of all the fluxes (which, by Green’s theorem, is the sum of all the line integrals around the boundaries of approximating squares) can be approximated by a line integral over the boundary of \(S\). In the limit, as the areas of the approximating squares go to zero, this approximation gets arbitrarily close to the flux.
Let’s now look at a rigorous proof of the theorem in the special case that \(S\) is the graph of function \(z = f(x,y)\), where \(x\) and \(y\) vary over a bounded, simply connected region \(D\) of finite area (Figure \(\PageIndex{4}\)). Furthermore, assume that \(f\) has continuous second-order partial derivatives. Let \(C\) denote the boundary of \(S\) and let \(C'\) denote the boundary of \(D\). Then, \(D\) is the “shadow” of \(S\) in the plane and \(C'\) is the “shadow” of \(C\). Suppose that \(S\) is oriented upward. The counterclockwise orientation of \(C\) is positive, as is the counterclockwise orientation of \(C'\). Let \(\vecs F(x,y,z) = \langle P,Q,R \rangle\) be a vector field with component functions that have continuous partial derivatives.
We take the standard parameterization of \(S \, : \, x = x, \, y = y, \, z = g(x,y)\). The tangent vectors are \(\vecs t_x = \langle 1,0,g_x \rangle\) and \(\vecs t_y = \langle 0,1,g_y \rangle\), and therefore \(\vecs t_x \times \vecs t_y = \langle -g_x, \, -g_y, \, 1 \rangle\).
\[\iint_S curl \, \vecs{F} \cdot d\vecs{S} = \iint_D [- (R_y - Q_z)z_x - (P_z - R_x)z_y + (Q_x - P_y)] \, dA, \nonumber \]
where the partial derivatives are all evaluated at \((x,y,g(x,y))\), making the integrand depend on \(x\) and \(y\) only. Suppose \(\langle x (t), \, y(t) \rangle, \, a \leq t \leq b\) is a parameterization of \(C'\). Then, a parameterization of \(C\) is \(\langle x (t), \, y(t), \, g(x(t), \, y(t))\rangle, \, a \leq t \leq b\). Armed with these parameterizations, the Chain rule, and Green’s theorem, and keeping in mind that \(P\), \(Q\) and \(R\) are all functions of \(x\) and \(y\), we can evaluate line integral
\[ \begin{align*} \int_C \vecs{F} \cdot d \vecs{r} &= \int_a^b (Px'(t) + Qy'(t) + Rz'(t)) \, dt \\[4pt] &= \int_a^b \left[Px'(t) + Qy'(t) + R\left(\dfrac{\partial z}{\partial x} \dfrac{dx}{dt} + \dfrac{\partial z}{\partial y} \dfrac{dy}{dt}\right) \right] dt \\[4pt] &= \int_a^b \left[ \left(P + R \dfrac{\partial z}{\partial x} \right) x' (t) + \left(Q + R \dfrac{\partial z}{\partial y} \right) y'(t) \right] dt \\[4pt] &= \int_{C'} \left(P + R \dfrac{\partial z}{\partial x} \right)\, dx + \left(Q + R \dfrac{\partial z}{\partial y} \right) \, dy \\[4pt] &= \iint_D \left[ \dfrac{\partial}{\partial x} \left( Q + R \dfrac{\partial z}{\partial y} \right) - \dfrac{\partial}{\partial y} \left(P + R \dfrac{\partial z}{\partial x} \right) \right] \, dA \\[4pt] &=\iint_D \left(\dfrac{\partial Q}{\partial x} + \dfrac{\partial Q}{\partial z} \dfrac{\partial z}{\partial x} + \dfrac{\partial R}{\partial x} \dfrac{\partial z}{\partial y} + \dfrac{\partial R}{\partial z}\dfrac{\partial z}{\partial x} \dfrac{\partial z}{\partial y} + R \dfrac{\partial^2 z}{\partial x \partial y} \right) - \left(\dfrac{\partial P}{\partial y} + \dfrac{\partial P}{\partial z} \dfrac{\partial z}{\partial y} + \dfrac{\partial R}{\partial z} \dfrac{\partial z}{\partial y} \dfrac{\partial z}{\partial x} + R \dfrac{\partial^2 z}{\partial y \partial x} \right) \end{align*} \nonumber \]
By Clairaut’s theorem,
\[\dfrac{\partial^2 z}{\partial x \partial y} = \dfrac{\partial^2 z}{\partial y \partial x} \nonumber \]
Therefore, four of the terms disappear from this double integral, and we are left with
\[\iint_D [- (R_y - Q_z)Z_x - (P_z - R_x) z_y + (Q_x - P_y)] \, dA, \nonumber \]
which equals
\[\iint_S curl \, \vecs{F} \cdot d\vecs{S}. \nonumber \]
\(\Box\)
We have shown that Stokes’ theorem is true in the case of a function with a domain that is a simply connected region of finite area. We can quickly confirm this theorem for another important case: when vector field \(\vecs{F}\) is a conservative field. If \(\vecs{F}\) is conservative, the curl of \(\vecs{F}\) is zero, so
\[\iint_S curl \, \vecs{F} \cdot d\vecs{S} = 0. \nonumber \]
Since the boundary of \(S\) is a closed curve, the integral
\[\int_C \vecs{F} \cdot d\vecs{r}. \nonumber \]
is also zero.
Example \(\PageIndex{1}\): Verifying Stokes’ Theorem for a Specific Case
Verify that Stokes’ theorem is true for vector field \(\vecs{F}(x,y) = \langle -z,x,0 \rangle\) and surface \(S\), where \(S\) is the hemisphere, oriented outward, with parameterization \(\vecs r(\phi, \theta) = \langle \sin \phi \, \cos \theta, \, \sin \phi \, \sin \theta, \, \cos \phi \rangle, \, 0 \leq \theta \leq \pi, \, 0 \leq \phi \leq \pi\) as shown in Figure \(\PageIndex{5}\).
Solution
Let \(C\) be the boundary of \(S\). Note that \(C\) is a circle of radius 1, centered at the origin, sitting in plane \(y = 0\). This circle has parameterization \(\langle \cos t, \, 0, \, \sin t \rangle, \, 0 \leq t \leq 2\pi\). the equation for scalar surface integrals
\[ \begin{align*} \int_C \vecs{F} \cdot d \vecs{r} &= \int_0^{2\pi} \langle -\sin t, \, \cos t, \, 0 \rangle \cdot \langle - \sin t, \, 0, \, \cos t \rangle \, dt \\[4pt] &= \int_0^{2\pi} \sin^2 t \, dt \\[4pt] &= \pi. \end{align*}\]
By the equation for vector line integrals,
\[ \begin{align*} \iint_S \, curl \, \vecs{F} \cdot d\vecs S &= \iint_D curl \, \vecs{F} (\vecs r (\phi,\theta)) \cdot ( \vecs t_{\phi} \times \vecs t_{\theta}) \, dA \\[4pt] &= \iint_D \langle 0, -1, 1 \rangle \cdot \langle \cos \theta \, \sin^2 \phi, \, \sin \theta \, \sin^2 \phi, \, \sin \phi \, \cos \phi \rangle \, dA \\[4pt] &= \int_0^{\pi} \int_0^{\pi} (\sin \phi \, \cos \phi - \sin \theta \, \sin^2 \phi ) \, d\phi d\theta \\[4pt] &= \dfrac{\pi}{2} \int_0^{\pi} \sin \theta \, d\theta \\[4pt] &= \pi.\end{align*}\]
Therefore, we have verified Stokes’ theorem for this example.
Exercise \(\PageIndex{1}\)
Verify that Stokes’ theorem is true for vector field \(\vecs{F}(x,y,z) = \langle y,x,-z \rangle \) and surface \(S\), where \(S\) is the upwardly oriented portion of the graph of \(f(x,y) = x^2 y\) over a triangle in the \(xy\)-plane with vertices \((0,0), \, (2,0)\), and \((0,2)\).
- Hint
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Calculate the double integral and line integral separately.
- Answer
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Both integrals give \(-\dfrac{136}{45}\):
Example \(\PageIndex{3}\): Calculating a Line Integral
Calculate the line integral
\[\int_C \vecs{F} \cdot d\vecs{r}, \nonumber \]
where \(\vecs{F} = \langle xy, \, x^2 + y^2 + z^2, \, yz \rangle\) and \(C\) is the boundary of the parallelogram with vertices \((0,0,1), \, (0,1,0), \, (2,0,-1)\), and \((2,1,-2)\).
Solution
To calculate the line integral directly, we need to parameterize each side of the parallelogram separately, calculate four separate line integrals, and add the result. This is not overly complicated, but it is time-consuming.
By contrast, let’s calculate the line integral using Stokes’ theorem. Let \(S\) denote the surface of the parallelogram. Note that \(S\) is the portion of the graph of \(z = 1 - x - y\) for \((x,y)\) varying over the rectangular region with vertices \((0,0), \, (0,1), \, (2,0)\), and \((2,1)\) in the \(xy\)-plane. Therefore, a parameterization of \(S\) is \(\langle x,y, \, 1 - x - y \rangle, \, 0 \leq x \leq 2, \, 0 \leq y \leq 1\). The curl of \(\vecs{F}\) is \( \langle -z, \, 0, \, x \rangle\),and Stokes’ theorem and the equation for scalar surface integrals
\[ \begin{align*} \int_C \vecs{F} \cdot d\vecs{r} &= \iint_S curl \, \vecs{F} \cdot d\vecs{S} \\[4pt] &= \int_0^2 \int_0^1 curl \, \vecs{F} (x,y) \cdot (\vecs t_x \times \vecs t_y) \, dy\, dx \\[4pt] &= \int_0^2 \int_0^1 \langle - (1 - x - y), \, 0, \, x \rangle \cdot ( \langle 1, \, 0, -1 \rangle \times \langle 0, \, 1, \, -1 \rangle ) \, dy \,dx \\[4pt] &= \int_0^2 \int_0^1 \langle x + y - 1, \, 0, \, x \rangle \cdot \langle 1, 1, 1 \rangle \, dy \, dx \\[4pt] &= \int_0^2 \int_0^1 2x + y - 1 \, dy \, dx \\[4pt] &= 3.\end{align*} \nonumber \]
Exercise \(\PageIndex{3}\)
Use Stokes’ theorem to calculate line integral
\[\int_C \vecs{F} \cdot d\vecs{r}, \nonumber \]
where \(\vecs{F} = \langle z,x,y \rangle \) and \(C\) is the boundary of a triangle with vertices \((0,0,1), \, (3,0,-2)\), and \((0,1,2)\).
- Hint
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This triangle lies in plane \(z = 1 - x + y\).
- Answer
-
\(\dfrac{3}{2}\)
Interpretation of Curl
In addition to translating between line integrals and flux integrals, Stokes’ theorem can be used to justify the physical interpretation of curl that we have learned. Here we investigate the relationship between curl and circulation, and we use Stokes’ theorem to state Faraday’s law—an important law in electricity and magnetism that relates the curl of an electric field to the rate of change of a magnetic field.
Recall that if \(C\) is a closed curve and \(\vecs{F}\) is a vector field defined on \(C\), then the circulation of \(\vecs{F}\) around \(C\) is line integral
\[\int_C \vecs{F} \cdot d\vecs{r}. \nonumber \]
If \(\vecs{F}\) represents the velocity field of a fluid in space, then the circulation measures the tendency of the fluid to move in the direction of \(C\).
Let \(\vecs{F}\) be a continuous vector field and let \(D_{\tau}\) be a small disk of radius \(r\) with center \(P_0\) (Figure \(\PageIndex{7}\)). If \(D_{\tau}\) is small enough, then \((curl \, \vecs{F})(P) \approx (curl \, \vecs F)(P_0)\) for all points \(P\) in \(D_{\tau}\) because the curl is continuous. Let \(C_{\tau}\) be the boundary circle of \(D_{\tau}\): By Stokes’ theorem,
\[\int_{C_{\tau}} \vecs{F} \cdot d\vecs{r} = \iint_{D_{\tau}} curl \, \vecs{F} \cdot \vecs{N} \, d\vecs S \approx \iint_{D_{\tau}} (curl \, \vecs{F})(P_0) \cdot \vecs{N} (P_0) \, d\vecs S. \nonumber \]
The quantity \( (curl \, \vecs F)(P_0) \cdot \vecs N (P_0) \) is constant, and therefore
\[\iint_{D_{\tau}} (curl \, \vecs F)(P_0) \cdot \vecs N (P_0) \, d\vecs S = \pi r^2 [(curl \, \vecs F)(P_0) \cdot \vecs N (P_0)]. \nonumber \]
Thus
\[\int_{C_{\tau}} \vecs F \cdot d\vecs r \approx \pi r^2 [ (curl \, \vecs F)(P_0) \cdot \vecs N (P_0)], \nonumber \]
and the approximation gets arbitrarily close as the radius shrinks to zero. Therefore Stokes’ theorem implies that
\[(curl \, \vecs F)(P_0) \cdot \vecs N (P_0) = \lim_{r\rightarrow 0^+} \dfrac{1}{\pi r^2} \int_{C_{\tau}} \vecs F \cdot d\vecs r. \nonumber \]
This equation relates the curl of a vector field to the circulation. Since the area of the disk is \(\pi r^2\), this equation says we can view the curl (in the limit) as the circulation per unit area. Recall that if \(\vecs F\) is the velocity field of a fluid, then circulation \[\oint_{C_{\tau}} \vecs F \cdot d\vecs r = \oint_{C_{\tau}} \vecs F \cdot \vecs T \, ds \nonumber \] is a measure of the tendency of the fluid to move around \(C_{\tau}\): The reason for this is that \(\vecs F \cdot \vecs T\) is a component of \(\vecs F\) in the direction of \(\vecs T\), and the closer the direction of \(\vecs F\) is to \(\vecs T\), the larger the value of \(\vecs F \cdot \vecs T\) (remember that if \(\vecs a\) and \(\vecs b\) are vectors and \(\vecs b\) is fixed, then the dot product \(\vecs a \cdot \vecs b\) is maximal when \(\vecs a\) points in the same direction as \(\vecs b\)). Therefore, if \(\vecs F\) is the velocity field of a fluid, then \(curl \, \vecs F \cdot \vecs N\) is a measure of how the fluid rotates about axis \(\vecs N\). The effect of the curl is largest about the axis that points in the direction of \(\vecs N\), because in this case \(curl \, \vecs F \cdot \vecs N\) is as large as possible.
To see this effect in a more concrete fashion, imagine placing a tiny paddlewheel at point \(P_0\) (Figure \(\PageIndex{8}\)). The paddlewheel achieves its maximum speed when the axis of the wheel points in the direction of curl \(\vecs F\). This justifies the interpretation of the curl we have learned: curl is a measure of the rotation in the vector field about the axis that points in the direction of the normal vector \(\vecs N\), and Stokes’ theorem justifies this interpretation.
Now that we have learned about Stokes’ theorem, we can discuss applications in the area of electromagnetism. In particular, we examine how we can use Stokes’ theorem to translate between two equivalent forms of Faraday’s law . Before stating the two forms of Faraday’s law, we need some background terminology.
Let \(C\) be a closed curve that models a thin wire. In the context of electric fields, the wire may be moving over time, so we write \(C(t)\) to represent the wire. At a given time \(t\), curve \(C(t)\) may be different from original curve \(C\) because of the movement of the wire, but we assume that \(C(t)\) is a closed curve for all times \(t\). Let \(D(t)\) be a surface with \(C(t)\) as its boundary, and orient \(C(t)\) so that \(D(t)\) has positive orientation. Suppose that \(C(t)\)is in a magnetic field \(\vecs B(t)\) that can also change over time. In other words, \(\vecs{B}\) has the form
\[\vecs B(x,y,z) = \langle P(x,y,z), \, Q(x,y,z), \, R(x,y,z) \rangle, \nonumber \]
where \(P\), \(Q\), and \(R\) can all vary continuously over time. We can produce current along the wire by changing field \(\vecs B(t)\) (this is a consequence of Ampere’s law). Flux \(\displaystyle \phi (t) = \iint_{D(t)} \vecs B(t) \cdot d\vecs S\) creates electric field \(\vecs E(t)\) that does work. The integral form of Faraday’s law states that
\[Work = \int_{C(t)} \vecs E(t) \cdot d\vecs r = - \dfrac{\partial \phi}{\partial t}. \nonumber \]
In other words, the work done by \(\vecs{E}\) is the line integral around the boundary, which is also equal to the rate of change of the flux with respect to time. The differential form of Faraday’s law states that
\[curl \, \vecs{E} = - \dfrac{\partial \vecs B}{\partial t}. \nonumber \]
Using Stokes’ theorem, we can show that the differential form of Faraday’s law is a consequence of the integral form. By Stokes’ theorem, we can convert the line integral in the integral form into surface integral
\[-\dfrac{\partial \phi}{\partial t} = \int_{C(t)} \vecs E(t) \cdot d\vecs r = \iint_{D(t)} curl \,\vecs E(t) \cdot d\vecs S. \nonumber \]
Since \[\phi (t) = \iint_{D(t)} B(t) \cdot d\vecs S, \nonumber \] then as long as the integration of the surface does not vary with time we also have
\[- \dfrac{\partial \phi}{\partial t} = \iint_{D(t)} - \dfrac{\partial \vecs B}{\partial t} \cdot d\vecs S. \nonumber \]
Therefore,
\[\iint_{D(t)} - \dfrac{\partial \vecs B}{\partial t} \cdot d\vecs S = \iint_{D(t)} curl \,\vecs E \cdot d\vecs S. \nonumber \]
To derive the differential form of Faraday’s law, we would like to conclude that \(curl \,\vecs E = -\dfrac{\partial \vecs B}{\partial t}\): In general, the equation
\[\iint_{D(t)} - \dfrac{\partial \vecs B}{\partial t} \cdot d\vecs S = \iint_{D(t)} curl \,\vecs E \cdot d\vecs S \nonumber \]
is not enough to conclude that \(curl \, \vecs E = -\dfrac{\partial \vecs B}{\partial t}\): The integral symbols do not simply “cancel out,” leaving equality of the integrands. To see why the integral symbol does not just cancel out in general, consider the two single-variable integrals \(\displaystyle \int_0^1 x \, dx\) and \(\displaystyle \int_0^1 f(x)\, dx\), where
\[f(x) = \begin{cases}1, &\text{if } 0 \leq x \leq 1/2 \\ 0, & \text{if } 1/2 \leq x \leq 1. \end{cases} \nonumber \]
Both of these integrals equal \(\dfrac{1}{2}\), so \(\displaystyle \int_0^1 x \, dx = \int_0^1 f(x) \, dx\).
However, \(x \neq f(x)\). Analogously, with our equation \[\iint_{D(t)} - \dfrac{\partial \vecs B}{\partial t} \cdot d\vecs S = \iint_{D(t)} curl \, \vecs E \cdot d\vecs S, \nonumber \] we cannot simply conclude that \(curl \, \vecs E = -\dfrac{\partial \vecs B}{\partial t}\) just because their integrals are equal. However, in our context, equation
\[\iint_{D(t)} - \dfrac{\partial \vecs B}{\partial t} \cdot d\vecs S = \iint_{D(t)} curl \, \vecs E \cdot d\vecs S \nonumber \]
is true for any region, however small (this is in contrast to the single-variable integrals just discussed). If \(\vecs F\) and \(\vecs G\) are three-dimensional vector fields such that
\[\iint_S \vecs F \cdot d\vecs S = \iint_S \vecs G \cdot d\vecs S \nonumber \]
for any surface \(S\), then it is possible to show that \(\vecs F = \vecs G\) by shrinking the area of \(S\) to zero by taking a limit (the smaller the area of \(S\), the closer the value of \(\displaystyle \iint_S \vecs F \cdot d\vecs S\) to the value of \(\vecs F\) at a point inside \(S\)). Therefore, we can let area \(D(t)\) shrink to zero by taking a limit and obtain the differential form of Faraday’s law:
\[curl \,\vecs E = - \dfrac{\partial \vecs B}{\partial t}. \nonumber \]
In the context of electric fields, the curl of the electric field can be interpreted as the negative of the rate of change of the corresponding magnetic field with respect to time.
Example \(\PageIndex{4}\): Using Faraday’s Law
Calculate the curl of electric field \(\vecs{E}\) if the corresponding magnetic field is constant field \(\vecs B(t) = \langle 1, -4, 2 \rangle\).
Solution
Since the magnetic field does not change with respect to time, \(-\dfrac{\partial \vecs B}{\partial t} = \vecs 0\). By Faraday’s law, the curl of the electric field is therefore also zero.
AnalysisA consequence of Faraday’s law is that the curl of the electric field corresponding to a constant magnetic field is always zero.
Exercise \(\PageIndex{4}\)
Calculate the curl of electric field \(\vecs{E}\) if the corresponding magnetic field is \(\vecs B(t) = \langle tx, \, ty, \, -2tz \rangle, \, 0 \leq t < \infty.\)
- Hint
-
- Use the differential form of Faraday’s law.
- Notice that the curl of the electric field does not change over time, although the magnetic field does change over time.
- Answer
-
\(curl \, \vecs{E} = \langle x, \, y, \, -2z \rangle\)
Key Concepts
- Stokes’ theorem relates a flux integral over a surface to a line integral around the boundary of the surface. Stokes’ theorem is a higher dimensional version of Green’s theorem, and therefore is another version of the Fundamental Theorem of Calculus in higher dimensions.
- Stokes’ theorem can be used to transform a difficult surface integral into an easier line integral, or a difficult line integral into an easier surface integral.
- Through Stokes’ theorem, line integrals can be evaluated using the simplest surface with boundary \(C\).
- Faraday’s law relates the curl of an electric field to the rate of change of the corresponding magnetic field. Stokes’ theorem can be used to derive Faraday’s law.
Key Equations
- Stokes’ theorem
\[\int_C \vecs{F} \cdot d\vecs{r} = \iint_S curl \, \vecs{F} \cdot d\vecs{S} \nonumber \]
Glossary
- Stokes’ theorem
- relates the flux integral over a surface \(S\) to a line integral around the boundary \(C\) of the surface \(S\)
- surface independent
- flux integrals of curl vector fields are surface independent if their evaluation does not depend on the surface but only on the boundary of the surface
Applying Stokes’ Theorem
Stokes’ theorem translates between the flux integral of surface \(S\) to a line integral around the boundary of \(S\). Therefore, the theorem allows us to compute surface integrals or line integrals that would ordinarily be quite difficult by translating the line integral into a surface integral or vice versa. We now study some examples of each kind of translation.
Example \(\PageIndex{2}\): Calculating a Surface Integral
Calculate surface integral
\[\iint_S curl \, \vecs{F} \cdot d\vecs S, \nonumber \]
where \(S\) is the surface, oriented outward, in Figure \(\PageIndex{6}\) and \(\vecs{F} = \langle z,\, 2xy, \, x + y \rangle\).
Solution
Note that to calculate
\[ \iint_S curl \, \vecs F \cdot d\vecs S \nonumber \]
without using Stokes’ theorem, we would need the equation for scalar surface integrals. Use of this equation requires a parameterization of \(S\). Surface \(S\) is complicated enough that it would be extremely difficult to find a parameterization. Therefore, the methods we have learned in previous sections are not useful for this problem. Instead, we use Stokes’ theorem, noting that the boundary \(C\) of the surface is merely a single circle with radius 1.
The curl of \(\vecs{F}\) is \(\langle 1,1,2y \rangle\). By Stokes’ theorem,
\[\iint_S curl \, \vecs F \cdot d\vecs S = \int_C \vecs F \cdot d\vecs r, \nonumber \]
where \(C\) has parameterization \(\langle \cos t, \, \sin t, \, 1 \rangle, 0 \leq t \leq 2\pi\). By the equation for vector line integrals,
\[ \begin{align*} \iint_S curl \, F \cdot d\vecs S &= \int_C \vecs{F} \cdot d \vecs{r} \\[4pt] &= \int_0^2 \langle 1, \, \sin t \, \cos t, \, \cos t + \sin t \rangle \cdot \langle - \sin t, \, \cos t, \, 0 \rangle \, dt \\[4pt] &= \int_0^{2\pi} ( - \sin t + 2 \, \sin t \, \cos^2 t ) \, dt \\[4pt] &= \left[ \cos t - \dfrac{2 \, \cos^3 t}{3} \right]_0^{2\pi} \\[4pt] &= \cos (2\pi) - \dfrac{2 \, \cos^3 (2\pi)}{3} - \left(\cos (0) - \dfrac{2 \, \cos^3 (0)}{3} \right) \\[4pt] &= 0. \end{align*} \nonumber \]
An amazing consequence of Stokes’ theorem is that if \(S'\) is any other smooth surface with boundary \(C\) and the same orientation as \(S\), then \[\iint_S curl \, \vecs F \cdot d\vecs S = \int_C \vecs F \cdot d\vecs r = 0 \nonumber \] because Stokes’ theorem says the surface integral depends on the line integral around the boundary only.
In Example \(\PageIndex{2}\), we calculated a surface integral simply by using information about the boundary of the surface. In general, let \(S_1\) and \(S_2\) be smooth surfaces with the same boundary \(C\) and the same orientation. By Stokes’ theorem,
\[\iint_{S_1} curl \, \vecs{F} \cdot d\vecs{S} = \int_C \vecs{F} \cdot d\vecs{r} = \iint_{S_2} curl \, \vecs{F} \cdot d\vecs{S}. \label{20} \]
Therefore, if
\[\iint_{S_1} curl \, \vecs{F} \cdot d\vecs{S} \nonumber \]
is difficult to calculate but
\[\iint_{S_2} curl \, \vecs{F} \cdot d\vecs S \nonumber \]
is easy to calculate, Stokes’ theorem allows us to calculate the easier surface integral. In Example \(\PageIndex{2}\), we could have calculated
\[\iint_S curl \, \vecs{F} \cdot d \vecs{S} \nonumber \]
by calculating
\[\iint_{S'} curl \, \vecs{F} \cdot d\vecs{S}, \nonumber \]
where \(\vecs{S}'\) is the disk enclosed by boundary curve \(C\) (a much more simple surface with which to work).
Equation \ref{20} shows that flux integrals of curl vector fields are surface independent in the same way that line integrals of gradient fields are path independent. Recall that if \(\vecs{F}\) is a two-dimensional conservative vector field defined on a simply connected domain, \(f\) is a potential function for \(\vecs{F}\), and \(C\) is a curve in the domain of \(\vecs{F}\), then
\[\int_C \vecs{F} \cdot d\vecs{r} \nonumber \]
depends only on the endpoints of \(C\). Therefore if \(C'\) is any other curve with the same starting point and endpoint as \(C\) (that is, \(C'\) has the same orientation as \(C\)), then
\[\int_C \vecs{F} \cdot d\vecs{r} = \int_{C'} \vecs{F} \cdot d\vecs{r} \nonumber \]
In other words, the value of the integral depends on the boundary of the path only; it does not really depend on the path itself.
Analogously, suppose that \(S\) and \(S'\) are surfaces with the same boundary and same orientation, and suppose that \(\vecs{G}\) is a three-dimensional vector field that can be written as the curl of another vector field \(\vecs{F}\) (so that \(\vecs{F}\) is like a “potential field” of \(\vecs{G}\)). By Equation \ref{20},
\[ \begin{align*} \iint_S \vecs G \cdot d\vecs S = \iint_S curl \, \vecs F \cdot d\vecs S = \int_C \vecs F \cdot d\vecs r = \iint_{S'} curl \, \vecs F \cdot d\vecs S = \iint_{S'} \vecs G \cdot d\vecs S.\end{align*} \nonumber \]
Therefore, the flux integral of \(\vecs{G}\) does not depend on the surface, only on the boundary of the surface. Flux integrals of vector fields that can be written as the curl of a vector field are surface independent in the same way that line integrals of vector fields that can be written as the gradient of a scalar function are path independent.
Exercise \(\PageIndex{1}\)
Use Stokes’ theorem to calculate surface integral \[\iint_S curl \, \vecs{F} \cdot d\vecs{S}, \nonumber \] where \(\vecs{F} = \langle x,y,z \rangle\) and \(S\) is the surface as shown in the following figure.
Parameterize the boundary of \(S\) and translate to a line integral.
\(-\pi\)


28.1: Nucleotides and Nucleic Acids
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- Steven Farmer & Dietmar Kennepohl
- University of Illinois Springfield
Objectives
After completing this section, you should be able to
- outline the relationship between nucleic acids, nucleotides and nucleosides.
- identify, in general terms, the enzymatic hydrolysis products of nucleosides.
- explain the structural difference between the sugar components of DNA and RNA.
- identify by name the four heterocyclic amine bases found in deoxyribonucleotides.
- identify by name the four heterocyclic amine bases found in ribonucleotides.
- draw the general structure of a nucleotide and a nucleoside.
- indicate the nitrogen atom by which a given purine or pyrimidine base attaches to the sugar component in nucleotides and nucleosides.
- sketch a section of nucleic acid to show how the nucleotide units are joined together.
Key Terms
Make certain that you can define, and use in context, the key terms below.
- deoxyribonucleic acid (DNA)
- nucleosides nucleotides
- ribonucleic acid (RNA)
Study Notes
The five bases that are found in nucleotides are often represented by their initial letter: adenine, A; guanine, G; cytosine, C; thymine, T; and uracil, U. Note that A, G, C and T occur in DNA; A, G, C and U occur in RNA. You are not required to memorize the structures of these bases, but you must know how each one bonds to the sugar unit in a nucleotide.
To fulfill Objective 6, you should be able to reproduce the figure below.
The Learning Objective of this Module is to identify the different molecules that combine to form nucleotides.
The repeating, or monomer, units that are linked together to form nucleic acids are known as nucleotides . The deoxyribonucleic acid (DNA) of a typical mammalian cell contains about 3 × 10 9 nucleotides. Nucleotides can be further broken down to a phosphate group (PO 4 -3 ) and a nucleoside which is composed of a aldopentose sugar (a sugar with five carbon atoms), and a heterocyclic purine or pyrimidine base (a base containing nitrogen atoms).
\[\mathrm{nucleic\: acids \underset{down\: into}{\xrightarrow{can\: be\: broken}} nucleotides \underset{down\: into}{\xrightarrow{can\: be\: broken}} H_3PO_4 + nitrogen\: base + pentose\: sugar} \nonumber \]
If the pentose sugar is ribose, the nucleotide is more specifically referred to as a ribonucleotide , and the resulting nucleosides are used in ribonucleic acid (RNA). If the sugar is 2-deoxyribose, the nucleotide is a deoxyribonucleotide , and the nucleoside are used in deoxyribonucleic acid (DNA). The prefix -deoxy implies that there is an oxygen missing from the 2' position of ribose.
The nitrogenous bases found in nucleotides are classified as pyrimidines or purines. Pyrimidines are heterocyclic amines with two nitrogen atoms in a six-member ring and include uracil, thymine, and cytosine. Purines are heterocyclic amines consisting of a pyrimidine ring fused to a five-member ring with two nitrogen atoms. Adenine and guanine are the major purines found in nucleic acids (Figure \(\PageIndex{1}\)). The numbering convention is that primed numbers designate the atoms of the pentose ring, and unprimed numbers designate the atoms of the purine or pyrimidine ring.
Nucleosides are formed by a bond between the anomeric C1′ of the pentose sugar and N1 position of the pyrimidine base or the N9 position of the purine base. The addition of a phospate groups at the 5' position of a nucleoside creates a corresponding nucleotide. DNA is made from four deoxyribonucleotides (Cytosine, Thymine, Adenine, Guanine) and RNA is made from four ribonucleotides Cytosine, Uracil, Adenine, Guanine. The names and structures of the major ribonucleotides and deoxyribonucleotides are given in Figure \(\PageIndex{2}\).
Apart from being the monomer units of DNA and RNA, the nucleotides and some of their derivatives have other functions as well. Adenosine diphosphate (ADP) and adenosine triphosphate (ATP), shown in Figure \(\PageIndex{3}\), have a role in cell metabolism. Moreover, a number of coenzymes, including flavin adenine dinucleotide (FAD), nicotinamide adenine dinucleotide (NAD + ), and coenzyme A, contain adenine nucleotides as structural components.
Primary Structure of Nucleic Acids
Nucleotides are joined together through the phosphate group of one nucleotide connecting in an ester linkage to the OH group on the 3' carbon atom of the sugar unit of a second nucleotide. This unit joins to a third nucleotide, and the process is repeated to produce a long nucleic acid chain (Figure 28.1.4). The backbone of the chain consists of alternating phosphate and sugar units (2-deoxyribose in DNA and ribose in RNA). The purine and pyrimidine bases branch off this backbone.
Like proteins, nucleic acids have a primary structure that is defined as the sequence of their nucleotides. Unlike proteins, which have 20 different kinds of amino acids, there are only 4 different kinds of nucleotides in nucleic acids. For amino acid sequences in proteins, the convention is to write the amino acids in order starting with the N-terminal amino acid. In writing nucleotide sequences for nucleic acids, the convention is to write the nucleotides (usually using the one-letter abbreviations for the bases, shown in Figure 28.1.4) starting with the nucleotide having a free phosphate group, which is known as the 5′ end, and indicate the nucleotides in order. For DNA, a lowercase d is often written in front of the sequence to indicate that the monomers are deoxyribonucleotides. The final nucleotide has a free OH group on the 3′ carbon atom and is called the 3′ end . The sequence of nucleotides in the DNA segment shown in Figure 28.1.4 would be written 5′-dG-dT-dA-dC-3′, which is often further abbreviated to dGTAC or just GTAC.
Note
Each phosphate group has one acidic hydrogen atom that is ionized at physiological pH. This is why these compounds are known as nucleic acids.
Exercise \(\PageIndex{1}\)
Classify each compound as a pentose sugar, a purine, or a pyrimidine.
- adenine
- guanine
- deoxyribose
- thymine
- ribose
- cytosine
- Answer
-
- purine
- purine
- pentose sugar
- pyrimidine
- pentose sugar
- pyrimidine
Exercise \(\PageIndex{2}\)
Identify the three molecules needed to form the nucleotides in each nucleic acid.
- DNA
- RNA
- Answer
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- nitrogenous base (adenine, guanine, cytosine, and thymine), 2-deoxyribose, and H 3 PO 4
- nitrogenous base (adenine, guanine, cytosine, and uracil), ribose, and H 3 PO 4
Exercise \(\PageIndex{3}\)
For each structure, circle the sugar unit and identify the nucleotide as a ribonucleotide or a deoxyribonucleotide.
- Answer
-
a.
b.
Exercise \(\PageIndex{4}\)
For each structure, circle the nitrogenous base and identify it as a purine or pyrimidine.
- Answer
-
-


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14.2: Cognitive Changes in the Brain
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Early in adolescence, changes in Dopamine , a chemical in the brain that is a neurotransmitter and produces feelings of pleasure, can contribute to increases in adolescents’ sensation-seeking and reward motivation. During adolescence, people tend to do whatever activities produce the most dopamine without fully considering the consequences of such actions. Later in adolescence, the prefrontal cortex , the area of the brain responsible for outcomes, forming judgments, controlling impulses and emotions, also continues to develop (Goldberg, 2001). The difference in timing of the development of these different regions of the brain contributes to more risk taking during middle adolescence because adolescents are motivated to seek thrills (Steinberg, 2008). One of the world’s leading experts on adolescent development, Laurence Steinberg, likens this to engaging a powerful engine before the braking system is in place. The result is that adolescents are prone to risky behaviors more often than children or adults.
Although the most rapid cognitive changes occur during childhood, the brain continues to develop throughout adolescence, and even into the 20s (Weinberger, Elvevåg, & Giedd, 2005). The brain continues to form new neural connections and becomes faster and more efficient because it prunes , or casts off unused neurons and connections (Blakemore, 2008), and produces myelin , the fatty tissue that forms around axons and neurons, which helps speed transmissions between different regions of the brain (Rapoport et al., 1999). This time of rapid cognitive growth for teens, making them more aware of their potential and capabilities, causes a great amount of disequilibrium for them. Theorists have researched cognitive changes and functions and have formed theories based on this developmental period. 3
Contributors and Attributions
3. Content by Dawn Rymond is licensed under CC BY 4.0
Adolescent Development by Jennifer Lansford is licensed under CC BY-NC-SA 4.0