12.5B: Olfactory (I) Nerve
The olfactory nerve, or cranial nerve I, is the first of 12 cranial nerves and is responsible for the sense of smell.
- Describe the olfactory nerve (cranial nerve I)
Key Points
- The olfactory nerve consists of a collection of many sensory nerve fibers that extend from the olfactory epithelium to the olfactory bulb.
- Olfactory receptors within the olfactory mucosa in the nasal cavity receive information about smells that travel to the brain through the cranial nerve that extends from the olfactory epithelium to the olfactory bulb.
- Olfactory receptor neurons continue to emerge throughout life and extend new axons to the olfactory bulb.
- The olfactory nerve is the shortest of the 12 cranial nerves and only one of two cranial nerves (the other being the optic nerve) that do not join with the brainstem.
Key Terms
- olfactory receptors : Expressed in the cell membranes of olfactory receptor neurons, these are responsible for the detection of odor molecules. Activated olfactory receptors are the initial player in a signal transduction cascade that ultimately produces a nerve impulse that is transmitted to the brain. The olfactory receptors form a multigene family consisting of over 900 genes in humans and 1,500 genes in mice.
- cribriform plate : A sieve-like structure of the ethmoid bone that supports the olfactory bulb.
- olfactory tract : A narrow white band made of a bundle of axons connecting the olfactory bulb to several brain regions.
- ethmoid bone : An unpaired bone in the skull that separates the nasal cavity from the brain.
- olfactory mucosa : Located in the upper region of the nasal cavity, this is made up of the olfactory epithelium and the underlying lamina propria, connective tissue containing fibroblasts, blood vessels, Bowman’s glands, and bundles of fine axons from the olfactory neurons.
Olfactory bulb : Sagittal section of human head showing the olfactory bulb.
The olfactory nerve, or cranial nerve I, is the first of the 12 cranial nerves. It is instrumental in the sense of smell. The olfactory nerve is the shortest of the 12 cranial nerves and only one of two cranial nerves (the other being the optic nerve) that do not join with the brainstem.
The specialized olfactory receptor neurons of the olfactory nerve are located in the olfactory mucosa of the upper parts of the nasal cavity. The olfactory nerves consist of a collection of many sensory nerve fibers that extend from the olfactory epithelium to the olfactory bulb, passing through the many openings of the cribriform plate of the ethmoid bone.
Olfactory receptor neurons continue to emerge throughout life and extend new axons to the olfactory bulb. Olfactory-ensheathing glia wrap bundles of these axons and are thought to facilitate their passage into the central nervous system.
The sense of smell (olfaction) arises from the stimulation of olfactory (or odorant) receptors by small molecules of different spatial, chemical, and electrical properties that pass over the nasal epithelium in the nasal cavity during inhalation. These interactions are transduced into electrical activity in the olfactory bulb, which then transmits the electrical activity to other parts of the olfactory system and the rest of the central nervous system via the olfactory tract.


3.2: Factorization in Euclidean Domains
In this section, we'll seek to answer the questions:
- What is a unique factorization domain? What examples of UFDs do we possess?
- What is the ascending chain condition on ideals? What are Noetherian rings?
- What does the ascending chain condition have to do with unique factorization?
In this section, our explorations of the structural arithmetic properties that guarantee unique factorization culminate in Theorem 3.2.7 . Specifically, we'll see that all Euclidean domains possess the unique factorization property. To prove this theorem, we will rely in part on an interesting property of chains of ideals in Euclidean domains.
3.2.1 Unique Factorization Domains
We begin by describing exactly what we mean by unique factorization. The reader may find it helpful to compare Definition: Unique Factorization Domain to The Fundamental Theorem of Arithmetic.
An integral domain \(R\) is called a unique factorization domain (or UFD ) if the following conditions hold.
- Every nonzero nonunit element of \(R\) is either irreducible or can be written as a finite product of irreducibles in \(R\text{.}\)
-
Factorization into irreducibles is unique up to associates. That is, if \(s\in R\) can be written as
\begin{equation*} s = p_1 p_2 \cdots p_k \text{ and } s = q_1 q_2 \cdots q_m \end{equation*}
for some irreducibles \(p_i, q_j\in R\text{,}\) then \(k=m\) and, after reordering, \(p_i\) is an associate of \(q_i\text{.}\)
Using \(\mathbb{Z}\) as an example, illustrate the definition of UFD by factoring 20 into two sets of different irreducibles which nonetheless can be paired up as associates.
We are already familiar with several examples.
The integers \(\mathbb{Z}\) form a UFD.
Every field is a UFD.
3.2.2 The Ascending Chain Condition and Noetherian Rings
We now set our sights on a proof of Theorem 3.2.7 . In order to prove it, we will make use of an important property of ideals in Euclidean domains. First, a definition.
A commutative ring \(R\) is called Noetherian if it satisfies the ascending chain condition on ideals.
That is, \(R\) is Noetherian if whenever
\begin{equation*} I_1 \subseteq I_2\subseteq I_3\subseteq \cdots \end{equation*}
is an ascending chain of ideals in \(R\text{,}\) then there exists some \(n\) for which \(I_n = I_{n+1} = I_{n+2} = \cdots\text{.}\)
\(\subseteq\): These rings are named in honor of Emmy Noether , one of the preeminent mathematicians of the 20th century. In addition to making substantial contributions to physics, she formalized the axiomatic definition of ring that is still in use today.
Consider the ideals \(I_1 = \langle 30 \rangle\) in \(\mathbb{Z}\) and \(J_1 = \langle 32 \rangle\text{.}\) Find the longest ascending chains of ideals starting first with \(I_1\) and then with \(J_1\) that you can. When does each chain stabilize?
We next show that every PID is Noetherian.
Every principal ideal domain is Noetherian.
- Hint
-
Let \(I_1\subseteq I_2\subseteq I_3 \subseteq \cdots\) and set \(I = \cup I_j\text{.}\) Show that \(I\) is an ideal, and use your assumptions!
Every Euclidean domain is Noetherian.
3.2.3 Euclidean Domains are UFDs
We now begin collecting results to prove that every Euclidean domain is a UFD. The first condition in the UFD definition is that every nonzero nonunit factors as a product of irreducibles. We first show that every nonzero nonunit is divisible by at least one irreducible (Lemma 3.2.1 ), which we apply to show that every nonzero nonunit can be written as a finite product of irreducibles (Theorem 3.2.4 ).
Let \(R\) be a principal ideal domain, and \(r\in R\) a nonzero nonunit. Then \(r\) is divisible by an irreducible.
- Hint
-
Let \(r\in R\) be reducible and write \(r = r_1 r_2\text{.}\) Continue to factor reducibles and build an ascending chain of ideals.
Let \(R\) be a PID. Then every nonzero nonunit element of \(R\) is either irreducible or can be written as a finite product of irreducibles in \(R\text{.}\)
The second condition that must be satisfied for a domain to be a UFD is that the product of irreducibles must be unique (up to associates). In order to prove that, we will make use of Theorem 3.2.5 , which states that in PIDs, primes and irreducibles are identical concepts.
Let \(R\) be a PID and let \(p\in R\) be irreducible. Let \(a\in R\) be such that \(p\nmid a\text{.}\) Then \(1\in I = \{ax+py : x,y\in R\}\) and thus there exist \(s,t\in R\) such that \(1 = as+pt\text{.}\)
Let \(R\) be a PID and let \(p\in R\text{.}\) Then \(p\) is prime if and only if \(p\) is irreducible.
Observe that Theorem 3.2.5 implies that if \(R\) is a PID and \(p\in R\) is irreducible with \(p|ab\text{,}\) then \(p|a\) or \(p|b\text{.}\)
Our crucial final step on the road to Theorem 3.2.7 is the following.
Every PID is a UFD.
- Hint
-
For part 2 of the definition, use induction on the number of irreducible divisors of an arbitrary nonzero nonunit. Mimic the proof of Theorem 1.3.4 .
Every Euclidean domain is a unique factorization domain.
Let \(F\) be a field. Then \(F[x]\) is a UFD.
That is, if \(f(x) \in F[x]\) with \(\deg(f(x)) \ge 1\text{,}\) then \(f(x)\) is either irreducible or a product of irreducibles in \(F[x]\text{.}\) What is more, if
\begin{equation*} f(x) = p_1(x) p_2(x) \cdots p_k(x) \text{ and } f(x) = q_1(x) q_2(x) \cdots q_m(x) \end{equation*}
are two factorizations of \(f\) into irreducibles \(p_i, q_j\text{,}\) then \(m=k\) and after reordering, \(p_j\) and \(q_j\) are associates.
- Hint
-
Handle existence and uniqueness separately. For each, (strong) induction on \(\deg(f(x))\) will work. Or do something entirely different.
Thus, we see that the existence of a well-behaved division algorithm and (a lack of zero divisors) is sufficient to guarantee unique factorization. However, it is not necessary. The following theorem is included for reference, but is not intended to be proved.
If \(R\) is a UFD, then \(R[x]\) is a UFD.
Thus, \(\mathbb{Z}[x]\) is a UFD. That is, every nonconstant polynomial in \(\mathbb{Z}[x]\) is either irreducible or can be factored uniquely into a product of irreducibles. However, as we will see later, \(\mathbb{Z}[x]\) is not a PID.


3.6: Reading- Utility Software
Introduction
Utility software is system software designed to help analyze, configure, optimize or maintain a computer.
Utility software usually focuses on how the computer infrastructure (including the computer hardware, operating system, software and data storage) operates. Utility software, along with operating system software, is a type of system software, distinguishing it from application software.
Utility software
- Anti-virus utilities scan for computer viruses.
- Archivers output a stream or a single file when provided with a directory or a set of files. Archive utilities, unlike archive suites, usually do not include compression or encryption capabilities. Some archive utilities may even have a separate un-archive utility for the reverse operation.
- Backup software can make copies of all information stored on a disk and restore either the entire disk (e.g. in an event of disk failure) or selected files (e.g. in an event of accidental deletion).
- Clipboard managers expand the clipboard functionality of an operating system .
- Cryptographic utilities encrypt and decrypt streams and files.
- Data compression utilities output a shorter stream or a smaller file when provided with a stream or file.
-
Data synchronization
utilities establish consistency among data from a source to a target data storage and vice versa. There are several branches of this type of utility:
- File synchronization utilities maintain consistency between two sources. They may be used to create redundancy or backup copies but are also used to help users carry their digital music, photos and video in their mobile devices.
- Revision control utilities are intended to deal with situations where more than one user attempts to simultaneously modify the same file.
- Debuggers are used to test and “debug” other programs, mainly to solve programming errors. Also utilized for reverse engineering of software or systems.
- Disk checkers can scan operating hard drive.
- Disk cleaners can find files that are unnecessary to computer operation, or take up considerable amounts of space. Disk cleaner helps the user to decide what to delete when their hard disk is full.
- Disk compression utilities can transparently compress/uncompress the contents of a disk, increasing the capacity of the disk.
- Disk defragmenters can detect computer files whose contents are scattered across several locations on the hard disk, and move the fragments to one location to increase efficiency.
- Disk partitions can divide an individual drive into multiple logical drives, each with its own file system which can be mounted by the operating system and treated as an individual drive.
- Disk space analyzers for the visualization of disk space usage by getting the size for each folder (including sub folders) & files in folder or drive. showing the distribution of the used space.
- Disk storage utilities
- File managers provide a convenient method of performing routine data management tasks, such as deleting, renaming, cataloging, uncataloging, moving, copying, merging, generating and modifying data sets.
- Hex editors directly modify the text or data of a file. These files could be data or an actual program.
- Memory testers check for memory failures.
- Network utilities analyze the computer’s network connectivity, configure network settings, check data transfer or log events.
- Package managers are used to configure, install or keep up to date other software on a computer.
- Registry cleaners clean and optimize the Windows Registry by removing old registry keys that are no longer in use.
- Screensavers were desired to prevent phosphor burn-in on CRT and plasma computer monitors by blanking the screen or filling it with moving images or patterns when the computer is not in use. Contemporary screensavers are used primarily for entertainment or security.
- System monitors for monitoring resources and performance in a computer system.
- System profilers provide detailed information about the software installed and hardware attached to the computer.
CC licensed content, Shared previously
- Utility Software. Provided by : Wikipedia. Located at : https://en.wikipedia.org/wiki/Utility_software . License : CC BY-SA: Attribution-ShareAlike
- Blue Wrench Emulsion. Authored by : Bill Gracey. Located at : https://www.flickr.com/photos/9422878@N08/5396568649/ . License : CC BY-NC-ND: Attribution-NonCommercial-NoDerivatives
- Screensaver. Authored by : Russ Morris. Located at : https://www.flickr.com/photos/russmorris/939663869/ . License : CC BY-NC-ND: Attribution-NonCommercial-NoDerivatives


1.3: Integumentary System
- Apply the rules of medical language to build, analyze, spell, pronounce, abbreviate, and define terms as they relate to the integumentary system
- Identify meanings of key word components of the integumentary system
- Categorize diagnostic, therapeutic, procedural or anatomic terms related to the integumentary system
- Use terms related to the integumentary system
- Use terms related to the diseases and disorders of the integumentary system
Integumentary System Word Parts
Click on prefixes, combining forms, and suffixes to reveal a list of word parts to memorize for the Integumentary System.
Introduction to the Integumentary System
The integumentary system refers to the skin and its accessory structures. In the adult human body, the skin makes up about 16 percent of body weight and covers an area of 16-21 ft 2 .
In fact, the skin and accessory structures are the largest organ system in the human body. The skin protects your inner organs and it is in need of daily care and protection to maintain its health.
Practice integumentary system medical terms.
Anatomy (Structures) of the Integumentary System
The skin and its accessory structures make up the integumentary system, which provides the body with overall protection. The skin is made of multiple layers of cells and tissues, which are held to underlying structures by connective tissue. The deeper layer of skin is well vascularized. It also has numerous sensory, and autonomic and sympathetic nerve fibers ensuring communication to and from the brain.
The skin is composed of two main layers:
- The epidermis
-
The dermis
- Beneath the dermis lies the hypodermis
Concept Check
- On the diagram above find the two layers of the skin; epidermis and dermis .
- The literal breakdown for hypodermis is below the dermis. On the diagram above where can you locate it?
- Can you find a hair follicle , hair root and hair shaft ?
- Keep reading to find out what the arrector pilli muscle does when you are frightened.
Epidermis
The epidermis is composed of keratinized, stratified squamous epithelium. It is made of four or five layers of epithelial cells, depending on its location in the body. It is avascular.
The cells in all of the layers except the stratum basale are called keratinocytes. Keratin is an intracellular fibrous protein that gives hair, nails, and skin their hardness and water-resistant properties. The keratinocytes in the stratum corneum are dead and regularly slough away, being replaced by cells from the deeper layers.
Dermis
The dermis contains blood and lymph vessels, nerves, and other structures, such as hair follicles and sweat glands. The dermis is made of two layers (papillary layer and reticular layer) of connective tissue that compose an interconnected mesh of elastin and collagenous fibers, produced by fibroblasts (see Figure 3.2).Papillary Layer
The papillary layer is made of loose, areolar connective tissue, which means the collagen and elastin fibers of this layer form a loose mesh. This superficial layer of the dermis projects into the stratum basale of the epidermis to form finger-like dermal papillae (see Figure 3.2).  In addition, the papillary layer contains phagocytes, that help fight bacteria or other infections that have breached the skin. This layer also contains lymphatic capillaries, nerve fibers, and Meissner corpuscles.
Reticular Layer
Underlying the papillary layer is the much thicker reticular layer, composed of dense, irregular connective tissue. This layer is well vascularized and has a rich sensory and sympathetic nerve supply. The reticular layer appears reticulated due to a tight meshwork of fibers. Elastin fibers provide some elasticity to the skin, enabling movement. Collagen fibers provide structure and tensile strength, with strands of collagen extending into both the papillary layer and the hypodermis. In addition, collagen binds water to keep the skin hydrated. Collagen injections and Retin-A creams help restore skin turgor by either introducing collagen externally or stimulating blood flow and repair of the dermis, respectively.
Hypodermis
The hypodermis serves to connect the skin to the underlying fascia of the bones and muscles. It is not strictly a part of the skin, although the border between the hypodermis and dermis can be difficult to distinguish. The hypodermis consists of well-vascularized, loose, areolar connective tissue and [pb_glossary id="1697"]adipose[/pb_glossary] tissue, which functions as a mode of fat storage and provides insulation and cushioning for the integument.
Practice labeling the layers of the skin.
Physiology (Function) of the Integumentary System
The skin and accessory structures perform a variety of essential functions, such as protecting the body from invasion by microorganisms, chemicals, and other environmental factors; preventing dehydration; acting as a sensory organ; modulating body temperature and electrolyte balance; and synthesizing vitamin D. The underlying hypodermis has important roles in storing fats, forming a “cushion” over underlying structures, and providing insulation from cold temperatures.
Protection
The skin protects the body from wind, water, and UV sunlight. It acts as a protective barrier against water loss and it also is the first line of defense against abrasive activity such as grit, microbes, or harmful chemicals. Sweat excreted from sweat glands deters microbes from over-colonizing the skin surface.
Sensory Function
The skin acts as a sense organ because the epidermis, dermis, and the hypodermis contain specialized sensory nerve structures that detect touch, surface temperature, and pain. These receptors are more concentrated on the tips of the fingers, which are most sensitive to touch, especially the Meissner corpuscle, which responds to light touch, and the Pacinian corpuscle , which responds to vibration. Merkel cells, seen scattered in the stratum basale, are also touch receptors. In addition to these specialized receptors, there are sensory nerves connected to each hair follicle, pain and temperature receptors scattered throughout the skin, and motor nerves innervate the arrector pili muscles and glands. This rich innervation helps us sense our environment and react accordingly,
Thermoregulation
The integumentary system helps regulate body temperature through its tight association with the sympathetic nervous system. The sympathetic nervous system is continuously monitoring body temperature and initiating appropriate motor responses.
-
When the
body becomes warm
sweat glands, accessory structures to the skin, secrete water, salt, and other substances to cool the body.
- Even when the body does not appear to be noticeably sweating, approximately 500 mL of sweat are secreted a day.
-
If the
body becomes excessively warm
due to high temperatures, vigorous activity, or a combination of the two, sweat glands will be stimulated by the sympathetic nervous system to produce large amounts of sweat.
- When the sweat evaporates from the skin surface, the body is cooled as body heat is dissipated.
- In addition to sweating, arterioles in the dermis dilate so that excess heat carried by the blood can dissipate through the skin and into the surrounding environment .
- This accounts for the skin redness that many people experience when exercising.
-
When
body temperatures drop
, the arterioles constrict to minimize heat loss, particularly in the ends of the digits and tip of the nose.
- This reduced circulation can result in the skin taking on a whitish hue.
- Although the temperature of the skin drops as a result, passive heat loss is prevented, and internal organs and structures remain warm.
- If the temperature of the skin drops too much (such as environmental temperatures below freezing), the conservation of body core heat can result frostbite .
Concept Check
Can you describe the thermoregulation process between the integumentary system and the sympathetic system?
- When body temperature is too warm.
- When body temperature is too cold.
Vitamin D Synthesis
The epidermal layer of human skin synthesizes Vitamin D when exposed to UV radiation. In the presence of sunlight, a form of Vitamin D 3 called cholecalciferol is synthesized from a derivative of the steroid cholesterol in the skin. The liver converts cholecalciferol to calcidiol, which is then converted to calcitriol (the active chemical form of the vitamin) in the kidneys.
- Vitamin D is essential for normal absorption of calcium and phosphorous, which are required for healthy bones.
- The absence of sun exposure can lead to a lack of vitamin D in the body, in children this can cause rickets. Vitamin D deficiency in elderly individuals may lead to osteomalacia.
- In present day society, Vitamin D is added as a supplement to many foods, including milk and orange juice, compensating for the need for sun exposure. In addition to its essential role in bone health, Vitamin D is essential for general immunity against bacterial, viral, and fungal infections.
Watch this video:
Media 3.2. The Integumentary System, Part 2 – Skin Deeper: Crash Course A&P #7 [Online video]. Copyright 2015 by CrashCourse .Media 3.2. The Integumentary System, Part 2 – Skin Deeper: Crash Course A&P #7 [Online video]. Copyright 2015 by CrashCourse .
Accessory Structures
Accessory structures of the skin include hair, nails, sweat glands, and sebaceous glands. These structures embryologically originate from the epidermis and can extend down through the dermis into the hypodermis.
Hair
Hair is a keratinous filament growing out of the epidermis. It is primarily made of dead, keratinized cells. Strands of hair originate in an epidermal penetration of the dermis called the hair follicle. The hair shaft is the part of the hair not anchored to the follicle, and much of this is exposed at the skin’s surface. The rest of the hair, which is anchored in the follicle, lies below the surface of the skin and is referred to as the hair root. The hair root ends deep in the dermis at the hair bulb, and includes a layer of mitotically active basal cells called the hair matrix. The hair bulb surrounds the hair papilla, which is made of connective tissue and contains blood capillaries and nerve endings from the dermis (see Figure 3.4).
Hair Function
Hair serves a variety of functions, including protection, sensory input, thermoregulation, and communication. For example:
- Hair on the head protects the skull from the sun.
- Hair in the nose and ears, and around the eyes (eyelashes) defends the body by trapping and excluding dust particles that may contain allergens and microbes.
- Hair of the eyebrows prevents sweat and other particles from dripping into and bothering the eyes.
Hair also has a sensory function due to sensory innervation by a hair root plexus surrounding the base of each hair follicle. Hair is extremely sensitive to air movement or other disturbances in the environment, much more so than the skin surface. This feature is also useful for the detection of the presence of insects or other potentially damaging substances on the skin surface.
Each hair root is connected to a smooth muscle called the arrector pili that contracts in response to nerve signals from the sympathetic nervous system, making the external hair shaft “stand up.” The primary purpose for this is to trap a layer of air to add insulation. This is visible in humans as goose bumps and even more obvious in animals, such as when a frightened cat raises its fur. Of course, this is much more obvious in organisms with a heavier coat than most humans, such as dogs and cats.
Hair Growth, Loss and Color
Hair grows and is eventually shed and replaced by new hair. Hair typically grows at the rate of 0.3 mm per day. On average, 50 hairs are lost and replaced per day. Hair loss occurs if there is more hair shed than what is replaced and can happen due to hormonal or dietary changes. Hair loss can also result from the aging process, or the influence of hormones. Similar to the skin, hair gets its color from the pigment melanin, produced by melanocytes in the hair papilla. Different hair color results from differences in the type of melanin. As a person ages, the melanin production decreases, and hair tends to lose its color and becomes gray and/or white.
Nails
The nail bed is a specialized structure of the epidermis that is found at the tips of our fingers and toes. The nail body is formed on the nail bed, and protects the tips of our fingers and toes as they are the farthest extremities and the parts of the body that experience the maximum mechanical stress (see Figure 3.5). The nail body forms a back-support for picking up small objects with the fingers. The nail body is composed of densely packed dead keratinocytes.
The epidermis in this part of the body has evolved a specialized structure upon which nails can form. The nail body forms at the nail root, which has a matrix of proliferating cells from the stratum basale that enables the nail to grow continuously. The lateral nail fold overlaps the nail on the sides, helping to anchor the nail body. The nail fold that meets the proximal end of the nail body forms the nail cuticle, also called the eponychium.
The nail bed is rich in blood vessels, making it appear pink, except at the base, where a thick layer of epithelium over the nail matrix forms a crescent-shaped region called the lunula (the “little moon”). The area beneath the free edge of the nail, furthest from the cuticle, is called the hyponychium. It consists of a thickened layer of stratum corneum.
Sweat Glands
Sudoriferous Glands
When the body becomes warm, sudoriferous glands produce sweat to cool the body. Sweat glands develop from epidermal projections into the dermis and are classified as merocrine glands; that is, the secretions are excreted by exocytosis through a duct without affecting the cells of the gland. There are two types of sweat glands, each secreting slightly different products.
An eccrine sweat gland is type of gland that produces a hypotonic sweat for thermoregulation as described previously. These glands are found all over the skin’s surface, but are especially abundant on the palms of the hand, the soles of the feet, and the forehead (Figure 3.6). They are coiled glands lying deep in the dermis, with the duct rising up to a pore on the skin surface, where the sweat is released. This type of sweat, released by exocytosis, is hypotonic and composed mostly of water, with some salt, antibodies, traces of metabolic waste, and dermicidin, an antimicrobial peptide. Eccrine glands are a primary component of thermoregulation in humans and thus help to maintain homeostasis .
An apocrine sweat gland is usually associated with hair follicles in densely hairy areas, such as armpits and genital regions. Apocrine sweat glands are larger than eccrine sweat glands and lie deeper in the dermis, sometimes even reaching the hypodermis, with the duct normally emptying into the hair follicle. In addition to water and salts, apocrine sweat includes organic compounds that make the sweat thicker and subject to bacterial decomposition and subsequent smell. The release of this sweat is under both nervous and hormonal control, and plays a role in the poorly understood human pheromone response. Most commercial antiperspirants use an aluminum-based compound as their primary active ingredient to stop sweat. When the antiperspirant enters the sweat gland duct, the aluminum-based compounds precipitate due to a change in pH and form a physical block in the duct, which prevents sweat from coming out of the pore.
Sebaceous Glands
A sebaceous gland is a type of oil gland that is found all over the body and helps to lubricate and waterproof the skin and hair. Most sebaceous glands are associated with hair follicles. They generate and excrete sebum, a mixture of lipids, onto the skin surface, thereby naturally lubricating the dry and dead layer of keratinized cells of the stratum corneum, keeping it pliable. The fatty acids of sebum also have antibacterial properties, and prevent water loss from the skin in low-humidity environments. The secretion of sebum is stimulated by hormones, many of which do not become active until puberty. Thus, sebaceous glands are relatively inactive during childhood.
Words not Easily Broken into Word Parts
Common Integumentary System Abbreviations
Many terms and phrases related to the integumentary system are abbreviated. Learn these common abbreviations by expanding the list below.
Changes Due to Aging
All systems in the body accumulate subtle and some not-so-subtle changes as a person ages. Among these changes are reductions in cell division, metabolic activity, blood circulation, hormonal levels, and muscle strength (see Figure 3.7). In the skin, these changes are reflected in decreased mitosis in the stratum basale, leading to a thinner epidermis. The dermis, which is responsible for the elasticity and resilience of the skin, exhibits a reduced ability to regenerate, which leads to slower wound healing. The hypodermis, with its fat stores, loses structure due to the reduction and redistribution of fat, which in turn contributes to the thinning and sagging of skin.
The accessory structures also have lowered activity, generating thinner hair and nails, and reduced amounts of sebum and sweat. A reduced sweating ability can cause some elderly to be intolerant to extreme heat. Other cells in the skin, such as melanocytes and dendritic cells, also become less active, leading to a paler skin tone and lowered immunity. Wrinkling of the skin occurs due to breakdown of its structure, which results from decreased collagen and elastin production in the dermis, weakening of muscles lying under the skin, and the inability of the skin to retain adequate moisture.
Disease and Disorders
The integumentary system is susceptible to a variety of diseases, disorders, and injuries. These range from annoying but relatively benign bacterial or fungal infections that are categorized as disorders, to skin cancer and severe burns, which can be fatal. In this section, you will learn several of the most common skin conditions.
One of the most talked about diseases is skin cancer. Most cancers are identified by the organ or tissue in which the cancer originates. One common form of cancer is skin cancer.
In general, cancers result from an accumulation of DNA mutations. These mutations can result in cell populations that do not die when they should and uncontrolled cell proliferation that leads to tumors. Although many tumors are benign, some metastasize. Cancers are characterized by their ability to metastasize.
Sun Damage
It requires about 10 days after initial sun exposure for melanin synthesis to peak, which is why pale-skinned individuals tend to suffer sunburns of the epidermis initially. Dark-skinned individuals can also get sunburns, but are more protected than are pale-skinned individuals. Too much sun exposure can eventually lead to wrinkling due to the destruction of the cellular structure of the skin, and in severe cases, can cause sufficient DNA damage to result in skin cancer. When there is an irregular accumulation of melanocytes in the skin, freckles appear. Moles are larger masses of melanocytes, and although most are benign, they should be monitored for changes that might indicate the presence of cancer (see Figure 3.8).
Basal Cell Carcinoma (BCC)
Basal cell carcinoma is a form of cancer that affects the mitotically active stem cells in the stratum basale of the epidermis. It is the most common of all cancers that occur in the United States and is frequently found on the head, neck, arms, and back, which are are as that are most susceptible to long-term sun exposure. Although UV rays are the main culprit, exposure to other agents, such as radiation and arsenic, can also lead to this type of cancer. Wounds on the skin due to open sores, tattoos, burns, etc. may be predisposing factors. Basal cell carcinomas start in the stratum basale and usually spread along this boundary. At some point, they begin to grow toward the surface and become an uneven patch, bump, growth, or scar on the skin surface (see Figure 3.9). Like most cancers, basal cell carcinomas respond best to treatment when caught early. Treatment options include surgery, freezing (cryosurgery), and topical ointments.
Squamous Cell Carcinoma (SCC)
Squamous cell carcinoma is a cancer that affects the keratinocytes of the stratum spinosum and presents as lesions commonly found on the scalp, ears, and hands (see Figure 3.10). It is the second most common skin cancer. The American Cancer Society reports that two of 10 skin cancers are squamous cell carcinomas, and it is more aggressive than basal cell carcinoma. If not removed, these carcinomas can metastasize. Surgery and radiation are used to cure squamous cell carcinoma .
Melanoma
A melanoma is a cancer characterized by the uncontrolled growth of melanocytes, the pigment-producing cells in the epidermis. Typically, a melanoma develops from a mole. It is the most fatal of all skin cancers, as it is highly metastatic and can be difficult to detect before it has spread to other organs. Melanomas usually appear as asymmetrical brown and black patches with uneven borders and a raised surface (see Figure 3.11). Treatment typically involves surgical excision and immunotherapy.
ABCDE for Early Diagnosis
Doctors often give their patients the following ABCDE mnemonic to help with the diagnosis of early-stage melanoma (see Fig 3.12). If you observe a mole on your body displaying these signs, consult a doctor.
- A symmetry – the two sides are not symmetrical
- B orders – the edges are irregular in shape
- C olor – the color is varied shades of brown or black
- D iameter – it is larger than a pencil eraser
- E volving – its sh ape has change d
Some specialists cite the following additional signs for the most serious form, nodular melanoma:
Elevated – it is raised on the skin surface
Firm – it feels hard to the touch
Growing – it is getting larger
Albinism
Albinism is a genetic disorder that affects (completely or partially) the coloring of skin, hair, and eyes. This is primarily due to the inability of melanocytes to produce melanin. Individuals with albinism tend to appear white or very pale due to the lack of melanin in their skin and hair. Recall that melanin helps protect the skin from the harmful effects of UV radiation. Individuals with albinism tend to need more protection from UV radiation, as they are more prone to sunburns and skin cancer. They also tend to be more sensitive to light and have vision problems due to the lack of pigmentation on the retinal wall (Betts, et al., 2021)
Treatment of this disorder usually involves addressing the symptoms, such as limiting UV light exposure to the skin and eyes. In vitiligo , the melanocytes in certain areas lose their ability to produce melanin, possibly due to an autoimmune reaction. This leads to a loss of color in patches (see Figure 3.13 ). Neither albinism nor vitiligo directly affects the lifespan of an individual (Betts, et al., 2021)
Changes in Skin Coloration
Other changes in the appearance of skin coloration can be indicative of diseases associated with other body systems.
- Liver disease or liver cancer can cause the accumulation of bile and the yellow pigment bilirubin, leading to xanthoderma or jaundice.
- Tumors of the pituitary gland can result in the secretion of large amounts of melanocyte-stimulating hormone (MSH), which results in a darkening of the skin.
- Addison’s disease can stimulate the release of excess amounts of adrenocorticotropic hormone (ACTH), which can give the skin a deep bronze color
- A sudden drop in oxygenation can affect skin color, causing the skin to initially turn ashen (white).
- A prolonged reduction in oxygen levels, dark red deoxyhemoglobin becomes dominant in the blood, making the skin appear blue , a condition referred to as cyanosis. This happens when the oxygen supply is restricted, as when someone is experiencing difficulty in breathing because of asthma or a heart attack. However, in these cases the effect on skin color has nothing do with the skin’s pigmentation (Betts, et al., 2021)
Skin Disorders
Two common skin disorders are eczema and acne . Eczema is an inflammatory condition and occurs in individuals of all ages. Acne involves the clogging of pores, which can lead to infection and inflammation, and is often seen in adolescents. Other disorders, include seborrheic dermatitis (on the scalp), psoriasis, fungal infections, cold sores, impetigo, scabies, hives, and warts (Betts, et al., 2021).
Eczema
Eczema is an allergic reaction that manifests as dry, itchy patches of skin that resemble rashes (see Figure 3.14). It may be accompanied by swelling of the skin, flaking, and in severe cases, bleeding. Symptoms are usually managed with moisturizers, corticosteroid creams, and immunosuppressants (Betts, et al., 2021).
Acne
Acne is a skin disturbance that typically occurs on areas of the skin that are rich in sebaceous glands (face and back). It is most common along with the onset of puberty due to associated hormonal changes, but can also occur in infants and continue into adulthood. Hormones, such as androgens, stimulate the release of sebum. An overproduction and accumulation of sebum along with keratin can block hair follicles. This plug is initially white. The sebum, when oxidized by exposure to air, turns black. Acne results from infection by acne-causing bacteria ( Propionibacterium and Staphylococcus ), which can lead to redness and potential scarring due to the natural wound healing process (see Figure 3.15) (Betts, et al., 2021).
Ringworm
Tinea or dermatophytosis is often referred to as ringworm. Ringworm presents as a circular rash that is itchy and red and can be found on various parts of the body. It is referred to by the location that it is found:
-
- Tinea Pedis – feet or commonly referred to as athlete’s feet
- Tinea Capitis – scalp
- Tinea barbae – beard
- Tinea manuum – hands
- Tinea unguium – Toenails and fingernails also called onychomycosis
- Tinea corporis – Body parts such as arms and legs (Center for Disease Control and Prevention, 2018a)
To learn more about ringworm, visit the Center for Disease Control and Prevention’s web page on fungal infections.
Psoriasis
Psoriasis is a chronic autoimmune disorder that results in patches of thick red skin with the appearance of silvery scales.  Psoriasis can be confused with other skin disease so a dermatologist is the best physician to diagnosis psoriasis. Treatments may include creams, ointments, ultraviolet light therapy and medication (Center for Disease Control and Prevention, 2018). To learn more, visit the Center for Disease Control and Prevention’s web page on psoriasis .
Injuries
Because the skin is the part of our bodies that meets the world most directly, it is especially vulnerable to injury. Injuries include burns, wounds , as well as scars and calluses . They can be caused by sharp objects, heat, or excessive pressure or friction to the skin (Betts, et al., 2021).Skin injuries set off a healing process that occurs in several overlapping stages.
- The first step to repairing damaged skin is the formation of a blood clot that helps stop the flow of blood and scabs over with time. Many different types of cells are involved in wound repair, especially if the surface area that needs repair is extensive.
- Before the basal stem cells of the stratum basale can recreate the epidermis, fibroblasts mobilize and divide rapidly to repair the damaged tissue by collagen deposition, forming granulation tissue.
- Blood capillaries follow the fibroblasts and help increase blood circulation and oxygen supply to the area.
- Immune cells, such as macrophages, roam the area and engulf any foreign matter to reduce the chance of infection (Betts, et al., 2021).
Burns
A burn results when the skin is damaged by intense heat, radiation, electricity, or chemicals. The damage results in the death of skin cells, which can lead to a massive loss of fluid. Dehydration, electrolyte imbalance, and renal and circulatory failure follow, which can be fatal. Burn patients are treated with intravenous fluids to offset dehydration, as well as intravenous nutrients that enable the body to repair tissues and replace lost proteins. Another serious threat to the lives of burn patients is infection. Burned skin is extremely susceptible to bacteria and other pathogens, due to the loss of protection by intact layers of skin (Betts, et al., 2021).
Burn Classification
Burns are sometimes measured in terms of the size of the total surface area affected. This is referred to as the rule of nines , which associates specific anatomical areas with a percentage that is a factor of nine (see Figure 3.16) (Betts, et al., 2021).
Burns are also classified by the degree of their severity .
- A first-degree burn is a superficial burn that affects only the epidermis. Although the skin may be painful and swollen, these burns typically heal on their own within a few days. Mild sunburn fits into the category of a first-degree burn.
- A second-degree burn goes deeper and affects both the epidermis and a portion of the dermis. These burns result in swelling and a painful blistering of the skin. It is important to keep the burn site clean and sterile to prevent infection. If this is done, the burn will heal within several weeks.
- A third-degree burn fully extends into the epidermis and dermis, destroying the tissue and affecting the nerve endings and sensory function. These are serious burns that may appear white, red, or black; they require medical attention and will heal slowly without it.
- A fourth-degree burn is even more severe, affecting the underlying muscle and bone.
Oddly, third and fourth-degree burns are usually not as painful because the nerve endings themselves are damaged. Full-thickness burns cannot be repaired by the body, because the local tissues used for repair are damaged and require debridement, or amputation in severe cases, followed by grafting of the skin from an unaffected part of the body, or from skin grown in tissue culture for grafting purposes. Skin grafts are required when the damage from trauma or infection cannot be closed with sutures or staples (Betts et al., 2021).
Scars and Keloids
Most cuts or wounds, with the exception of ones that only scratch the epidermis, lead to scar formation. Scarring occurs in cases in which there is repair of skin damage, but the skin fails to regenerate the original skin structure. Fibroblasts generate scar tissue in the form of collagen, and the bulk of repair is due to the basket-weave pattern generated by collagen fibers and does not result in regeneration of the typical cellular structure of skin. Instead, the tissue is fibrous in nature and does not allow for the regeneration of accessory structures, such as hair follicles, sweat glands, or sebaceous glands (Betts, et al., 2021).
Sometimes, there is an overproduction of scar tissue, because the process of collagen formation does not stop when the wound is healed; this results in a keloid (see Figure 3.17). In contrast, scars that result from acne and chickenpox have a sunken appearance and are called atrophic scars (Betts, et al., 2021).
Scarring of skin after wound healing is a natural process and does not need to be treated further. Application of mineral oil and lotions may reduce the formation of scar tissue. However, modern cosmetic procedures, such as dermabrasion, laser treatments, and filler injections have been invented as remedies for severe scarring. All of these procedures try to reorganize the structure of the epidermis and underlying collagen tissue to make it look more natural (Betts, et al., 2021).
Bedsores and Stretch Marks
Skin and its underlying tissue can be affected by excessive pressure. One example of this is called a bedsore. Bedsores, also called decubitis ulcers, are caused by constant, long-term, unrelieved pressure on certain body parts that are bony, reducing blood flow to the area and leading to necrosis . Bedsores are most common in elderly patients who have debilitating conditions that cause them to be immobile. Most hospitals and long-term care facilities have the practice of turning the patients every few hours to prevent the incidence of bedsores. If left untreated bedsores can be fatal if they become infected (Betts, et al., 2021)
The skin can also be affected by pressure associated with rapid growth. A stretch mark results when the dermis is stretched beyond its limits of elasticity, as the skin stretches to accommodate the excess pressure. Stretch marks usually accompany rapid weight gain during puberty and pregnancy. They initially have a reddish hue, but lighten over time. Other than for cosmetic reasons, treatment of stretch marks is not required. They occur most commonly over the hips and abdomen (Betts, et al., 2021). For detailed images of types of bedsores, see Minnesota Hospital Association’s Pressure Ulcer Staging Cards .
Calluses
When you wear shoes that do not fit well and are a constant source of abrasion on your toes, you tend to form a callus at the point of contact. This occurs because the basal stem cells in the stratum basale are triggered to divide more often to increase the thickness of the skin at the point of abrasion to protect the rest of the body from further damage. This is an example of a minor or local injury, and the skin manages to react and treat the problem independent of the rest of the body. Calluses can also form on your fingers if they are subject to constant mechanical stress, such as long periods of writing, playing string instruments, or video games. A corn is a specialized form of callus. Corns form from abrasions on the skin that result from an elliptical-type motion (Betts, et al., 2021).
Medical Terms in Context
Medical Specialties and Procedures Related to the Integumentary System
A dermatologist is a medical doctor with specialized training in treating diseases, disorders and injuries related to the integumentary system and its accessory structures. There are many dermatologic subspecialties such as cosmetic dermatology, dermatopathology and pediatric dermatology. To learn more visit the ‘What is a Dermatologist?’ page of the American Academy of Dermatology Association website.
Dermatologists can be specially trained to perform a procedure called Mohs surgery. Mohs surgery excises skin cancers in thin layers until all cancer is removed from the tissue (Mayo Clinic Staff, 2017). For more details about each step of the process visit the Skin Cancer Foundation’s site on Mohs Surgery .
Test Yourself
References
Centers for Disease Control and Prevention. (2018, October 25). Psoriasis . Centers for Disease Control and Prevention: Fungal Diseases. https://www.cdc.gov/psoriasis/
Centers for Disease Control and Prevention. (2018a, August 6). Ringworm . Centers for Disease Control and Prevention: Fungal Diseases. https://www.cdc.gov/fungal/diseases/ringworm/definition.html
CrashCourse. (2015, February 16). The Integumentary system, part 2 – skin deeper: Crash Course A&P #7 [Video]. YouTube. https://youtu.be/EN-x-zXXVwQ
Mayo Clinic Staff. (2017, September 6). Mohs surgery . Mayo Clinic. https://www.mayoclinic.org/tests-procedures/mohs-surgery/about/pac-20385222#:~:text=Mohs%20surgery%20is%20a%20precise,known%20as%20Mohs%20micrographic%20surgery .
Ted-Ed. (2018). The science of skin- Emma Bryce [Video]. https://www.youtube.com/watch?v=OxPlCkTKhzY&t=1s
Unless otherwise indicated, this chapter contains material adapted from Anatomy and Physiology (on OpenStax ), by Betts, et al. and is used under a a CC BY 4.0 international license . Download and access this book for free at https://openstax.org/books/anatomy-and-physiology/pages/1-introduction .


1.5: Intermolecular Forces
- To describe the intermolecular forces in liquids.
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to intra molecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, inter molecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances without breaking covalent bonds .
The properties of liquids are intermediate between those of gases and solids, but are more similar to solids.
Intermolecular forces determine bulk properties, such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid.
Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures.
In this section, we explicitly consider three kinds of intermolecular interactions. There are two additional types of electrostatic interaction that you are already familiar with: the ion–ion interactions that are responsible for ionic bonding, and the ion–dipole interactions that occur when ionic substances dissolve in a polar substance such as water. The first two are often described collectively as van der Waals forces.
Dipole–Dipole Interactions
Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a dipole ). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in Figure \(\PageIndex{1a}\).
These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\PageIndex{1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\PageIndex{1b}\), are attractive intermolecular interactions , whereas those in Figure \(\PageIndex{1d}\) are repulsive intermolecular interactions . Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(\PageIndex{2}\). On average, however, the attractive interactions dominate.
Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/ r , where r is the distance between the ions. Doubling the distance ( r → 2 r ) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/ r 3 , so doubling the distance between the dipoles decreases the strength of the interaction by 2 3 , or 8-fold. Thus a substance such as \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure. Conversely, \(\ce{NaCl}\), which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(\PageIndex{1}\).
| Compound | Molar Mass (g/mol) | Dipole Moment (D) | Boiling Point (K) |
|---|---|---|---|
| C 3 H 6 (cyclopropane) | 42 | 0 | 240 |
| CH 3 OCH 3 (dimethyl ether) | 46 | 1.30 | 248 |
| CH 3 CN (acetonitrile) | 41 | 3.9 | 355 |
The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6.
Video Discussing Dipole Intermolecular Forces. Source: Dipole Intermolecular Force, YouTube(opens in new window) [youtu.be]
Arrange ethyl methyl ether (CH 3 OCH 2 CH 3 ), 2-methylpropane [isobutane, (CH 3 ) 2 CHCH 3 ], and acetone (CH 3 COCH 3 ) in order of increasing boiling points. Their structures are as follows:
Given: compounds.
Asked for: order of increasing boiling points.
Strategy:
Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points.
Solution:
The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds.
The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point.
Ethyl methyl ether has a structure similar to H 2 O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point.
Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point.
Thus we predict the following order of boiling points:
2-methylpropane < ethyl methyl ether < acetone
This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D.
Arrange carbon tetrafluoride (CF 4 ), ethyl methyl sulfide (CH 3 SC 2 H 5 ), dimethyl sulfoxide [(CH 3 ) 2 S=O], and 2-methylbutane [isopentane, (CH 3 ) 2 CHCH 2 CH 3 ] in order of decreasing boiling points.
- Answer
-
dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C)
London Dispersion Forces
Thus far, we have considered only interactions between polar molecules. Other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature; why others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(\PageIndex{2}\)).
What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances.
| Substance | Molar Mass (g/mol) | Melting Point (°C) | Boiling Point (°C) |
|---|---|---|---|
| Ar | 40 | −189.4 | −185.9 |
| Xe | 131 | −111.8 | −108.1 |
| N 2 | 28 | −210 | −195.8 |
| O 2 | 32 | −218.8 | −183.0 |
| F 2 | 38 | −219.7 | −188.1 |
| I 2 | 254 | 113.7 | 184.4 |
| CH 4 | 16 | −182.5 | −161.5 |
Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(\PageIndex{3}\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/ r 6 . Doubling the distance therefore decreases the attractive energy by 2 6 , or 64-fold.
Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H 2 molecules in part (b) in Figure \(\PageIndex{3}\), tends to become more pronounced as atomic and molecular masses increase (Table \(\PageIndex{2}\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1 s electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more polarizable than lighter ones.
For similar substances, London dispersion forces get stronger with increasing molecular size.
The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus, London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(\PageIndex{4}\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(\PageIndex{4}\) shows 2,2-dimethylpropane (neopentane) and n -pentane, both of which have the empirical formula C 5 H 12 . Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas n -pentane has an extended conformation that enables it to come into close contact with other n -pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of n -pentane (36.1°C).
All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate.
Video Discussing London/Dispersion Intermolecular Forces. Source: Dispersion Intermolecular Force, YouTube(opens in new window) [youtu.be]
Arrange n -butane, propane, 2-methylpropane [isobutene, (CH 3 ) 2 CHCH 3 ], and n -pentane in order of increasing boiling points.
Given: compounds
Asked for: order of increasing boiling points
Strategy:
Determine the intermolecular forces in the compounds, and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution:
The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and n -pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and n -butane has the more extended shape. Consequently, we expect intermolecular interactions for n -butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < n -butane (−0.5°C) < n -pentane (36.1°C).
Arrange GeH 4 , SiCl 4 , SiH 4 , CH 4 , and GeCl 4 in order of decreasing boiling points.
- Answer
-
GeCl 4 (87°C) > SiCl 4 (57.6°C) > GeH 4 (−88.5°C) > SiH 4 (−111.8°C) > CH 4 (−161°C)
Hydrogen Bonds
Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent, Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(\PageIndex{5}\). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H 2 Te and H 2 Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C.
Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure \(\PageIndex{6}\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the hydrogen bond donor ) and the atom that has the lone pair of electrons (the hydrogen bond acceptor ). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are not equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water, rather than sinks.
Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cage like structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion.
Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor.
Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze.
Video Discussing Hydrogen Bonding Intermolecular Forces. Source: Hydrogen Bonding Intermolecular Force, YouTube(opens in new window) [youtu.be]
Considering CH 3 OH , C 2 H 6 , Xe, and (CH 3 ) 3 N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
Given: compounds
Asked for: formation of hydrogen bonds and structure
Strategy:
- Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors.
- Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding.
Solution:
A. Of the species listed, xenon (Xe), ethane (C 2 H 6 ), and trimethylamine [(CH 3 ) 3 N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors.
B. The one compound that can act as a hydrogen bond donor, methanol (CH 3 OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows:
Considering CH 3 CO 2 H, (CH 3 ) 3 N, NH 3 , and CH 3 F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures.
- Answer
-
CH 3 CO 2 H and NH 3 ;
Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH 3 . Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms four hydrogen bonds at a time.
Arrange C 60 (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N 2 O in order of increasing boiling points.
Given: compounds.
Asked for: order of increasing boiling points.
Strategy:
Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point.
Solution
Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N 2 O have very similar molar masses (40 and 44 g/mol, respectively), but N 2 O is polar while Ar is not. Consequently, N 2 O should have a higher boiling point. A C 60 molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N 2 O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C 60 should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses:
He (−269°C) < Ar (−185.7°C) < N 2 O (−88.5°C) < C 60 (>280°C) < NaCl (1465°C).
Arrange 2,4-dimethylheptane, Ne, CS 2 , Cl 2 , and KBr in order of decreasing boiling points.
- Answer
-
KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS 2 (46.6°C) > Cl 2 (−34.6°C) > Ne (−246°C)
Identify the most significant intermolecular force in each substance.
- C 3 H 8
- CH 3 OH
- H 2 S
Solution
- Although C–H bonds are polar, they are only minimally polar. The most significant intermolecular force for this substance would be dispersion forces.
- This molecule has an H atom bonded to an O atom, so it will experience hydrogen bonding.
- Although this molecule does not experience hydrogen bonding, the Lewis electron dot diagram and VSEPR indicate that it is bent, so it has a permanent dipole. The most significant force in this substance is dipole-dipole interaction.
Identify the most significant intermolecular force in each substance.
- HF
- HCl
- Answer a
-
hydrogen bonding
- Answer b
-
dipole-dipole interactions
Summary
Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid, or the liquid and gas phases, are due to changes in intermolecular interactions, but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as van der Waals forces ), and hydrogen bonds. Dipole–dipole interactions arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/ r 3 , where r is the distance between dipoles. London dispersion forces are due to the formation of instantaneous dipole moments in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an induced dipole in adjacent molecules; their energy falls off as 1/ r 6 . Larger atoms tend to be more polarizable than smaller ones, because their outer electrons are less tightly bound and are therefore more easily perturbed. Hydrogen bonds are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the hydrogen bond donor ) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the hydrogen bond acceptor ). Because of strong O⋅⋅⋅H hydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cage like structure that is less dense than liquid water.


5.4: Rational Functions
- What is a rational function?
- How can we determine key information about a rational function from its algebraic structure?
- Why are rational functions important?
The average rate of change of a function on an interval always involves a ratio. Indeed, for a given function \(f\) that interests us near \(t = 2\text{,}\) we can investigate its average rate of change on intervals near this value by considering
Suppose, for instance, that \(f\) meausures the height of a falling ball at time \(t\) and is given by \(f(t) = -16t^2 + 32t + 48\text{,}\) which happens to be a polynomial function of degree \(2\text{.}\) For this particular function, its average rate of change on \([1,1+h]\) is
Structurally, we observe that \(AV_{[2,2+h]}\) is a ratio of the two functions \(-64h - 16h^2\) and \(h\text{.}\) Moreover, both the numerator and the denominator of the expression are themselves polynomial functions of the variable \(h\text{.}\) Note that we may be especially interested in what occurs as \(h \to 0\text{,}\) as these values will tell us the average velocity of the moving ball on shorter and shorter time intervals starting at \(t = 2\text{.}\) At the same time, \(AV_{[2,2+h]}\) is not defined for \(h = 0\text{.}\)
Ratios of polynomial functions arise in several different important circumstances. Sometimes we are interested in what happens when the denominator approaches \(0\text{,}\) which makes the overall ratio undefined. In other situations, we may want to know what happens in the long term and thus consider what happens when the input variable increases without bound.
A drug company 1 estimates that to produce a new drug, it will cost $5 million in startup resources, and that once they reach production, each gram of the drug will cost $2500 to make.
- Determine a formula for a function \(C(q)\) that models the cost of producing \(q\) grams of the drug. What familiar kind of function is \(C\text{?}\)
- The drug company needs to sell the drug at a price of more than $2500 per gram in order to at least break even. To investigate how they might set prices, they first consider what their average cost per gram is. What is the total cost of producing \(1000\) grams? What is the average cost per gram to produce \(1000\) grams?
- What is the total cost of producing \(10000\) grams? What is the average cost per gram to produce \(10000\) grams?
- Our computations in (b) and (c) naturally lead us to define the “average cost per gram” function, \(A(q)\text{,}\) whose output is the average cost of producing \(q\) grams of the drug. What is a formula for \(A(q)\text{?}\)
- Explain why another formula for \(A\) is \(A(q) = 2500 + \frac{5000000}{q}\text{.}\)
- What can you say about the long-range behavior of \(A\text{?}\) What does this behavior mean in the context of the problem?
Long-range behavior of rational functions
The functions \(AV_{[2,2+h]} = \frac{-64h - 16h^2}{h}\) and \(A(q) = \frac{5000000 + 2500q}{q}\) are both examples of rational functions, since each is a ratio of polynomial functions. Formally, we have the following definition.
A function \(r\) is rational provided that it is possible to write \(r\) as the ratio of two polynomials, \(p\) and \(q\text{.}\) That is, \(r\) is rational provided that for some polynomial functions \(p\) and \(q\text{,}\) we have
\begin{equation*} r(x) = \frac{p(x)}{q(x)}\text{.} \end{equation*}
Like with polynomial functions, we are interested in such natural questions as
- What is the long range behavior of a given rational function?
- What is the domain of a given rational function?
- How can we determine where a given rational function's value is \(0\text{?}\)
We begin by focusing on the long-range behavior of rational functions. It's important first to recall our earlier work with power functions of the form \(p(x) = x^{-n}\) where \(n = 1, 2, \ldots\text{.}\) For such functions, we know that \(p(x) = \frac{1}{x^n}\) where \(n \gt 0\) and that
since \(x^n\) increases without bound as \(x \to \infty\text{.}\) The same is true when \(x \to -\infty\text{:}\) \(\lim_{x \to -\infty} \frac{1}{x^n} = 0\text{.}\) Thus, any time we encounter a quantity such as \(\frac{1}{x^3}\text{,}\) this quantity will approach \(0\) as \(x\) increases without bound, and this will also occur for any constant numerator. For instance,
since \(2500\) times a quantity approaching \(0\) will still approach \(0\) as \(x\) increases.
Consider the rational function \(r(x) = \frac{3x^2 - 5x + 1}{7x^2 + 2x - 11}\text{.}\)
Observe that the largest power of \(x\) that's present in \(r(x)\) is \(x^2\text{.}\) In addition, because of the dominant terms of \(3x^2\) in the numerator and \(7x^2\) in the denominator, both the numerator and denominator of \(r\) increase without bound as \(x\) increases without bound. In order to understand the long-range behavior of \(r\text{,}\) we choose to write the function in a different algebraic form.
- Note that we can multiply the formula for \(r\) by the form of \(1\) given by \(1 = \frac{\frac{1}{x^2}}{\frac{1}{x^2}}\text{.}\) Do so, and distribute and simplify as much as possible in both the numerator and denominator to write \(r\) in a different algebraic form.
-
Having rewritten \(r\text{,}\) we are in a better position to evaluate \(\lim_{x \to \infty} r(x)\text{.}\) Using our work from (a), we have
\begin{equation*} \lim_{x \to \infty} r(x) = \lim_{x \to \infty} \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{7 + \frac{2}{x} - \frac{11}{x^2}}\text{.} \end{equation*}
What is the exact value of this limit and why?
-
Next, determine
\begin{equation*} \lim_{x \to -\infty} r(x) = \lim_{x \to -\infty} \frac{3 - \frac{5}{x} + \frac{1}{x^2}}{7 + \frac{2}{x} - \frac{11}{x^2}}\text{.} \end{equation*}
- Use Desmos to plot \(r\) on the interval \([-10,10]\text{.}\) In addition, plot the horizontal line \(y = \frac{3}{7}\text{.}\) What is the meaning of the limits you found in (b) and (c)?
Let \(s(x) = \frac{3x - 5}{7x^2 + 2x - 11}\) and \(u(x) = \frac{3x^2 - 5x + 1}{7x + 2}\text{.}\) Note that both the numerator and denominator of each of these rational functions increases without bound as \(x \to \infty\text{,}\) and in addition that \(x^2\) is the highest order term present in each of \(s\) and \(u\text{.}\)
-
Using a similar algebraic approach to our work in Activity 5.4.2, multiply \(s(x)\) by \(1 = \frac{\frac{1}{x^2}}{\frac{1}{x^2}}\) and hence evaluate
\begin{equation*} \lim_{x \to \infty} \frac{3x - 5}{7x^2 + 2x - 11}\text{.} \end{equation*}
What value do you find?
- Plot the function \(y = s(x)\) on the interval \([-10,10]\text{.}\) What is the graphical meaning of the limit you found in (a)?
-
Next, use appropriate algebraic work to consider \(u(x)\) and evaluate
\begin{equation*} \lim_{x \to \infty} \frac{3x^2 - 5x + 1}{7x + 2}\text{.} \end{equation*}
What do you find?
- Plot the function \(y = u(x)\) on the interval \([-10,10]\text{.}\) What is the graphical meaning of the limit you computed in (c)?
We summarize and generalize the results of Activity \(\PageIndex{2}\) and Activity \(\PageIndex{3}\) as follows.
Let \(p\) and \(q\) be polynomial functions so that \(r(x) = \frac{p(x)}{q(x)}\) is a rational function. Suppose that \(p\) has degree \(n\) with leading term \(a_n x^n\) and \(q\) has degree \(m\) with leading term \(b_m x^m\) for some nonzero constants \(a_n\) and \(b_m\text{.}\) There are three possibilities (\(n \lt m\text{,}\) \(n = m\text{,}\) and \(n \gt m\)) that result in three different behaviors of \(r\text{:}\)
-
if \(n \lt m\text{,}\) then the degree of the numerator is less than the degree of the denominator, and thus
\begin{equation*} \lim_{n \to \infty} r(x) = \lim_{n \to \infty} \frac{a_n x^n + \cdots + a_0}{b_m x^m + \cdots + b_0} = 0\text{,} \end{equation*}
which tells us that \(y = 0\) is a horizontal asymptote of \(r\text{;}\)
-
if \(n = m\text{,}\) then the degree of the numerator equals the degree of the denominator, and thus
\begin{equation*} \lim_{n \to \infty} r(x) = \lim_{n \to \infty} \frac{a_n x^n + \cdots + a_0}{b_n x^n + \cdots + b_0} = \frac{a_n}{b_n}\text{,} \end{equation*}
which tells us that \(y = \frac{a_n}{b_n}\) (the ratio of the coefficients of the highest order terms in \(p\) and \(q\)) is a horizontal asymptote of \(r\text{;}\)
-
if \(n \gt m\text{,}\) then the degree of the numerator is greater than the degree of the denominator, and thus
\begin{equation*} \lim_{n \to \infty} r(x) = \lim_{n \to \infty} \frac{a_n x^n + \cdots + a_0}{b_m x^m + \cdots + b_0} = \pm \infty\text{,} \end{equation*}
(where the sign of the limit depends on the signs of \(a_n\) and \(b_m\)) which tells us that \(r\) is does not have a horizontal asymptote.
In both situations (a) and (b), the value of \(\lim_{x \to -\infty} r(x)\) is identical to \(\lim_{x \to \infty} r(x)\text{.}\)
The domain of a rational function
Because a rational function can be written in the form \(r(x) = \frac{p(x)}{q(x)}\) for some polynomial functions \(p\) and \(q\text{,}\) we have to be concerned about the possibility that a rational function's denominator is zero. Since polynomial functions always have their domain as the set of all real numbers, it follows that any rational function is only undefined at points where its denominator is zero.
Let \(p\) and \(q\) be polynomial functions so that \(r(x) = \frac{p(x)}{q(x)}\) is a rational function. The domain of \(r\) is the set of all real numbers except those for which \(q(x) = 0\text{.}\)
Determine the domain of the function \(r(x) = \frac{5x^3 + 17x^2 - 9x + 4}{2x^3 - 6x^2 - 8x}\text{.}\)
- Solution
-
To find the domain of any rational function, we need to determine where the denominator is zero. The best way to find these values exactly is to factor the denominator. Thus, we observe that
\begin{equation*} 2x^3 - 6x^2 - 8x = 2x(x^2 - 3x - 4) = 2x(x+1)(x-4)\text{.} \end{equation*}
By the Zero Product Property, it follows that the denominator of \(r\) is zero at \(x = 0\text{,}\) \(x = -1\text{,}\) and \(x = 4\text{.}\) Hence, the domain of \(r\) is the set of all real numbers except \(-1\text{,}\) \(0\text{,}\) and \(4\text{.}\)
We note that when it comes to determining the domain of a rational function, the numerator is irrelevant: all that matters is where the denominator is \(0\text{.}\)
Determine the domain of each of the following functions. In each case, write a sentence to accurately describe the domain.
- \(\displaystyle \displaystyle f(x) = \frac{x^2-1}{x^2 + 1}\)
- \(\displaystyle \displaystyle g(x) = \frac{x^2 - 1}{x^2 + 3x - 4}\)
- \(\displaystyle \displaystyle h(x) = \frac{1}{x} + \frac{1}{x-1} + \frac{1}{x-2}\)
- \(\displaystyle \displaystyle j(x) = \frac{(x+5)(x-3)(x+1)(x-4)}{(x+1)(x+3)(x-5)}\)
- \(\displaystyle \displaystyle k(x) = \frac{2x^2 + 7}{3x^3 - 12x}\)
- \(\displaystyle \displaystyle m(x) = \frac{5x^2 - 45}{7(x-2)(x-3)^2(x^2 + 9)(x+1)}\)
Applications of rational functions
Rational functions arise naturally in the study of the average rate of change of a polynomial function, leading to expressions such as
We will study several subtle issues that correspond to such functions further in Section 5.5. For now, we will focus on a different setting in which rational functions play a key role.
In Section 5.3 , we encountered a class of problems where a key quantity was modeled by a polynomial function. We found that if we considered a container such as a cylinder with fixed surface area, then the volume of the container could be written as a polynomial of a single variable. For instance, if we consider a circular cylinder with surface area \(10\) square feet, then we know that
and therefore \(h = \frac{10 - 2\pi r^2}{2 \pi r}\text{.}\) Since the cylinder's volume is \(V = \pi r^2 h\text{,}\) it follows that
which is a polynomial function of \(r\text{.}\)
What happens if we instead fix the volume of the container and ask about how surface area can be written as a function of a single variable?
Suppose we want to construct a circular cylinder that holds \(20\) cubic feet of volume. How much material does it take to build the container? How can we state the amount of material as a function of a single variable?
- Answer
-
Neglecting any scrap, the amount of material it takes to construct the container is its surface area, which we know to be
\begin{equation*} S = 2\pi r^2 + 2\pi r h\text{.} \end{equation*}Because we want the volume to be fixed, this results in a constraint equation that enables us to relate \(r\) and \(h\text{.}\) In particular, since
\begin{equation*} V = 20 = \pi r^2 h\text{,} \end{equation*}it follows that we can solve for \(h\) and get \(h = \frac{20}{\pi r^2}\text{.}\) Substituting this expression for \(h\) in the equation for surface area, we find that
\begin{equation*} S = 2\pi r^2 + 2\pi r \cdot \frac{20}{\pi r^2} = 2 \pi r^2 + \frac{40}{r}\text{.} \end{equation*}Getting a common denominator, we can also write \(S\) in the form
\begin{equation*} S(r) = \frac{2 \pi r^3 + 40}{r} \end{equation*}and thus we see that \(S\) is a rational function of \(r\text{.}\) Because of the physical context of the problem and the fact that the denominator of \(S\) is \(r\text{,}\) the domain of \(S\) is the set of all positive real numbers.
Suppose that we want to build an open rectangular box (that is, without a top) that holds \(15\) cubic feet of volume. If we want one side of the base to be twice as long as the other, how does the amount of material required depend on the shorter side of the base? We investigate this question through the following sequence of prompts.
- Draw a labeled picture of the box. Let \(x\) represent the shorter side of the base and \(h\) the height of the box. What is the length of the longer side of the base in terms of \(x\text{?}\)
- Use the given volume constraint to write an equation that relates \(x\) and \(h\text{,}\) and solve the equation for \(h\) in terms of \(x\text{.}\)
- Determine a formula for the surface area, \(S\text{,}\) of the box in terms of \(x\) and \(h\text{.}\)
- Using the constraint equation from (b) together with your work in (c), write surface area, \(S\text{,}\) as a function of the single variable \(x\text{.}\)
- What type of function is \(S\text{?}\) What is its domain?
- Plot the function \(S\) using Desmos . What appears to be the least amount of material that can be used to construct the desired box that holds \(15\) cubic feet of volume?
Summary
- A rational function is a function whose formula can be written as the ratio of two polynomial functions. For instance, \(r(x) = \frac{7x^3 - 5x + 16}{-4x^4 + 2x^3 - 11x + 3}\) is a rational function.
-
Two aspects of rational functions are straightforward to determine for any rational function. Given \(r(x) = \frac{p(x)}{q(x)}\) where \(p\) and \(q\) are polynomials, the domain of \(r\) is the set of all real numbers except any values of \(x\) for which \(q(x) = 0\text{.}\) In addition, we can determine the long-range behavior of \(r\) by examining the highest order terms in \(p\) and \(q\text{:}\)
- if the degree of \(p\) is less than the degree of \(q\text{,}\) then \(r\) has a horizontal asymptote at \(y = 0\text{;}\)
- if the degree of \(p\) equals the degree of \(q\text{,}\) then \(r\) has a horizontal asymptote at \(y = \frac{a_n}{b_n}\text{,}\) where \(a_n\) and \(b_n\) are the leading coefficients of \(p\) and \(q\) respectively;
- and if the degree of \(p\) is greater than the degree of \(q\text{,}\) then \(r\) does not have a horizontal asymptote.
- Two reasons that rational functions are important are that they arise naturally when we consider the average rate of change on an interval whose length varies and when we consider problems that relate the volume and surface area of three-dimensional containers when one of those two quantities is constrained.
Exercises
For each rational function below, determine the function's domain as well as the exact value of any horizontal asymptote.
- \(\displaystyle \displaystyle f(x) = \frac{17x^2 + 34}{19x^2 - 76}\)
- \(\displaystyle \displaystyle g(x) = \frac{29}{53} + \frac{1}{x-2}\)
- \(\displaystyle \displaystyle h(x) = \frac{4-31x}{11x - 7}\)
- \(\displaystyle \displaystyle r(x) = \frac{151(x-4)(x+5)^2(x-2)}{537(x+5)(x+1)(x^2+1)(x-15)}\)
A rectangular box is being constructed so that its base is \(1.5\) times as long as it is wide. In addition, suppose that material for the base and top of the box costs $\(3.75\) per square foot, while material for the sides costs $\(2.50\) per square foot. Finally, we want the box to hold \(8\) cubic feet of volume.
- Draw a labeled picture of the box with \(x\) as the length of the shorter side of the box's base and \(h\) as its height.
- Determine a formula involving \(x\) and \(h\) for the total surface area, \(S\text{,}\) of the box.
- Use your work from (b) along with the given information about cost to determine a formula for the total cost, \(C\text{,}\) oif the box in terms of \(x\) and \(h\text{.}\)
- Use the volume constraint given in the problem to write an equation that relates \(x\) and \(h\text{,}\) and solve that equation for \(h\) in terms of \(x\text{.}\)
- Combine your work in (c) and (d) to write the cost, \(C\text{,}\) of the box as a function solely of \(x\text{.}\)
- What is the domain of the cost function? How does a graph of the cost function appear? What does this suggest about the ideal box for the given constraints?
A cylindrical can is being constructed so that its volume is \(16\) cubic inches. Suppose that material for the lids (the top and bottom) cost $\(0.11\) per square inch and material for the “side” of the can costs $\(0.07\) per square inch. Determine a formula for the total cost of the can as a function of the can's radius. What is the domain of the function and why?
- Hint
-
You may find it helpful to ask yourself a sequence of questions like those stated in Exercise 10).
<1> This activity is based on p. 457ff in Functions Modeling Change , by Connally et al.


8.3: Sparklines
Learning Outcomes
- Create sparklines.
Now that you have seen the value of data visualization in charts, we can address a much simpler and compact visualization: a sparkline. Sparklines are small graphs that are created to offer simple visual clues about the data graphed.
There are three types of sparklines:
- Line: visualizes the data in a line graph form
- Column: visualizes the data in column form, similar to the clustered column chart
- Win/Loss: visualizes the data as positive or negative based on color
The best use of a sparkline is to place it close to the data you are visualizing. You may be wonder what use a sparkline is since you know how to create a chart. Well, the great thing about sparklines is they automatically print when you print a worksheet. Charts do not print when you print a worksheet; they must be separately printed or moved to another program like Word or PowerPoint.
Sparklines are also great in that they automatically update if you change the data they display.
To create your own sparklines, follow the steps below.
- Click on the cell where you want your sparkline to appear.
- Click on the Insert tab and find the Sparklines group. Select the type of sparkline you want to create.
- A new data box will appear. This is where you will enter the range of cells to be graphed by your sparkline.
- Hit Enter.
Note that like any other chart, you can change the display options of a sparkline, including colors and style, after you have created it. To change a sparkline, you will need to access the Design tab that appears under Sparkline Tools once you have created a sparkline. This tab will only appear if you have selected the cell with your sparkline in it.
Contributors and Attributions
- Sparklines. Authored by : Shelli Carter. Provided by : Lumen Learning. License : CC BY: Attribution