1.16: Skeletal System
- Identify the anatomy of the skeletal system
- Describe the main functions of the skeletal system
- Spell the medical terms of the skeletal system and use correct abbreviations
- Explore common diseases, disorders, and procedures related to the skeletal system
- Identify the medical specialties associated with the skeletal system
Skeletal System Word Parts
Click on prefixes, combining forms, and suffixes to reveal a list of word parts to memorize for the Musculoskeletal System.
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Introduction to the Skeletal System
The skeletal system forms the framework of the body. It is the body system composed of bones, cartilage and ligaments. Each bone serves a particular function and varies in size, shape and strength. Bones are weight-bearing structures in your body and can therefore change in thickness as you gain or lose weight. The skeletal system performs the following critical functions for the human body:
- supports the body
- facilitates movement
- protects internal organs
- produces blood cells
- stores and releases minerals and fat
Watch this video:
Media 16.1. The Skeletal System: Crash Course A&P #19 [Online video]. Copyright 2015 by CrashCourse .
Skeletal System Medical Terms
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Anatomy (Structures) of the Skeletal System
The skeletal system includes all of the bones, cartilages, and ligaments of the body that support and give shape to the body and body structures. The skeleton consists of the bones of the body. For adults, there are 206 bones in the skeleton. Younger individuals have higher numbers of bones because some bones fuse together during childhood and adolescence to form an adult bone. The primary functions of the skeleton are to provide a rigid, internal structure that can support the weight of the body against the force of gravity, and to provide a structure upon which muscles can act to produce movements of the body.
In addition to providing for support and movements of the body, the skeleton has protective and storage functions. It protects the internal organs, including the brain, spinal cord, heart, lungs, and pelvic organs. The bones of the skeleton serve as the primary storage site for important minerals such as calcium and phosphate. The bone marrow found within bones stores fat and houses the blood-cell producing tissue of the body.
The skeleton is subdivided into two major divisions: the axial and appendicular .
The Axial Skeleton
The axial skeleton forms the vertical, central axis of the body and includes all bones of the head, neck, chest, and back (see Figure 16.1 ). It serves to protect the brain, spinal cord, heart, and lungs. It also serves as the attachment site for muscles that move the head, neck, and back, and for muscles that act across the shoulder and hip joints to move their corresponding limbs.
The axial skeleton of the adult consists of 80 bones including the skull , the vertebral column , and the thoracic cage . The skull is formed by 22 bones. Also associated with the head are an additional seven bones, including the hyoid bone and the ear ossicles (three small bones found in each middle ear). The vertebral column consists of 24 bones, each called a vertebra , plus the sacrum and coccyx . The thoracic cage includes the 12 pairs of ribs , and the sternum , the flattened bone of the anterior chest.
The axial skeleton has 80 bones and includes bones of the skull (and face), vertebral column, and thoracic cage.
The cranium or skull supports the face and protects the brain. It is subdivided into the bones of the skull and the bones of the face.
Bones of the Skull
- Frontal – forms the forehead
- Parietal – the upper lateral sides of the cranium
- Occipital – the posterior skull and base of the cranial cavity
- Temporal – the lower lateral sides of the cranium
- Sphenoid -the ‘keystone’ bone that forms part of the base of skull and eye sockets
- Ethmoid – forms part of the nose and orbit and base of cranium
- Auditory ossicles – the small bones of the middle ear
- External auditory meatus – the external opening of ear and temporal bone
Bones of the Face
- Zygomatic – the cheekbone
- Maxillary – the upper jaw and hard palate
- Palatine – the lateral walls of the nose
- Lacrimal – the walls of the orbit
- Inferior conchae – the lower lateral wall of the nasal cavity
- Vomer – the separates the left and right nasal cavity
- Mandible – the lower jaw bone (The only movable bone of the skull)
- Hyoid – the bone located between the mandible and larynx, not connected to other bones
Bones of the Vertebral Column
The vertebral column is also known as the spinal column or spine (see Figure 16.2 ). It consists of a sequence of vertebrae (singular = vertebra), each of which is separated and united by an intervertebral disc . Together, the vertebrae and intervertebral discs form the vertebral column. It is a flexible column that supports the head, neck, and body and allows for their movements. It also protects the spinal cord, which passes down the back through openings in the vertebrae.
Types of Vertebrae
- Cervical – C1 to C7 – the first 7 vertebrae in the neck region
- Thoracic – T1 to T12 – the next 12 vertebrae that forms the outward curvature of the spine
- Lumbar – L1 to L5 – the next 5 vertebrae that forms the inner curvature of spine
- Sacrum – the triangular-shaped bone at the base of the spine
- Coccyx – the tailbone
Bones of the Thoracic Cavity
The thoracic cage (rib cage) forms the thorax (chest) portion of the body. It consists of the 12 pairs of ribs with their costal cartilages and the sternum (see Figure 16.3 ). The ribs are anchored posteriorly to the 12 thoracic vertebrae (T1–T12). The thoracic cage protects the heart and lungs.
Ribs
There are 12 sets of ribs and can be divided as such:
- 7 true ribs as they are attached to the front of the sternum
- 3 false ribs as they are attached to the cartilage that joins the sternum
- 2 floating ribs as they are not attached to the front of the sternum
Sternum
The sternum , also known as the breast bone, is divided into 3 parts:
- manubrium – the upper portion of the breast bone
- body – the middle portion of the breast bone
- xiphoid process – the lower portion of the breast bone and is made up of cartilage
Answer the following questions:
- What is the medical term for the upper jaw bone and for the lower jaw bone?
- What medical term is used for the bones of the inner ear?
- How many bones make up the cervical region of the vertebral column?
The Appendicular Skeleton
The appendicular skeleton has 126 bones. It is divided into the bones of the u pper limbs and l ower limb s that attach each limb to skeleton. (Betts, et al., 2013)
Bones of the Pectoral Girdle
- Scapula – the shoulder blades
- Clavicle – the collar bones. It connects the sternum to the scapula
- Acromion – the extension that forms the bony point of the shoulder
Bones of the Upper Limbs
The bones of the upper limbs include the bones of the arms, wrists, and hands.
Bones of the Arm
- Humerus – the bone in upper arm
- Radius – the bone that runs thumb-side of the forearm
- Ulna – the bone that runs on the side of the little finger of the forearm
Bones of the Wrist and Hand
- Carpals – the wrist bones
- Metacarpals – the bones in the palm of hand
- Phalanges – the finger and toe bones
Each phalanx has three bones: the distal, medial, and proximal. The exception is the thumb and big toe which has two bones: distal and proximal. See Fig 16.5 below. There are 30 bones in each upper limb. Can you count them on your limb?
Bones of the Pelvic Region
The bones of the pelvic region protect the reproductive, urinary, and excretory organs.
- Pelvic girdle – the hip or coxal bone. It is formed by the fusion of three bones during adolescence
- Illium – the largest part of the hip bone
- Ischium – the lower portion of pelvic girdle
- Pubis – the anterior portion of pelvic girdle
- Pelvis – consists of four bones: the left and right hip bones as well as the sacrum and coccyx
- Acetabulum – the large socket in the pelvic bones that holds the head of the femur
The shape of the pelvic girdle is different for males than females. In the male, it is a funnel shape. In the female it is shaped like a basin to accommodate for the fetus during pregnancy.
The femur is the longest and strongest bone of the body, and accounts for approximately one-quarter of a person’s total height (Betts, et al., 2013) .
Bones of the Lower Limbs
The bones of the lower limb include bones of the leg and the feet.
Bones of the Leg
- Femur – the thigh bone and is also referred to the upper leg bone. It is the longest and strongest bone in the human body
- Patella – the knee cap
- Tibia – the shin bone. It is a medial bone and the main weight-bearing bone of the lower leg
- Fibula – the smaller of the lower leg bone (see Figure 16.6 )
Bones of the Ankles and Feet
- Tarsals – the ankle bones (7 total)
- Malleous – the bony protrusions of the ankle bones
- Talus – the superior ankle bones
- Calcaneous – the heel bones
- Metatarsals – the foot bones
- Phalanges – the bones of the toes (see Figure 16.7 )
Answer the following questions:
- Is the humerus the same as the funny bone?
- What is the medical term for the kneecap?
Anatomy Labeling Activity
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Physiology (Function) of the Skeletal System
The bones of the skeletal system is comprised of an inner spongy tissue referred to as bone marrow. There are two types of bone marrow, red and yellow. The red bone marrow produces the red blood cells and it does so by a process called hematopoiesis . The yellow bone marrow contains adipose tissues which can be a source of energy. The bones of the skeletal system also store minerals such as calcium and phosphate. These minerals are important for the physiological processes in the body and are released into the bloodstream when levels are low in the body.
Joints
Watch this video:
Media 16.2. Joints: Crash Course A&P #20 [Online video]. Copyright 2015 by CrashCourse .
Most bones connect to at least one other bone in the body. The area where bones meet bones or where bones meet cartilage are called articulations . Joints can be classified based on their ability to move. At movable joints, the articulating surfaces of the adjacent bones can move smoothly against each other. However, other joints may be connected to each other by connective tissue or cartilage. These joints are designed for stability and provide for little or no movement. Importantly, joint stability and movement are related to each other. This means that stable joints allow for little or no mobility between the adjacent bones. Conversely, joints that provide the most movement between bones are the least stable.
Based on the function of joints , there are 3 types of joints:
The left and right hip bones are connected by an amphiarthrosis joint.
-
Synarthrosis
joints which allow no movement
- example: joints of the skull
-
Amphiarthrosis
joints which allow some movement
- example: joints of the pubic symphysis
-
Diarthrosis
joints which allow for free movement
- example: joints of the knee
Structures associated with joints are:
- Cartilage – the elastic connective tissue that is found at the ends of bones, nose tip, etc.
- Synovial membrane – the lining or covering of synovial joints
- Synovial fluid – the lubricating fluid found between synovial joints
- Ligaments – the tough, elastic connective tissue that connects bone to bone
- Tendons – the fibrous connective tissue that attaches muscle to bone
- Bursa – the closed, fluid-filled sacs that works as a cushion
- Meniscus – C_shaped cartilage that act as shock absorbers between bones
Body Movements
Synovial joints are movable joints and provide most of the body movements. Body movement occurs when the bones, joints and muscles work together.
Flexion and Extension
Flexion and extension are movements that take place within the sagittal plane and involve anterior or posterior movements of the body or limbs. For the vertebral column, flexion (anterior flexion) is an anterior (forward) bending of the neck or body, while extension involves a posterior-directed motion, such as straightening from a flexed position or bending backward. Lateral flexion is the bending of the neck or body toward the right or left side. These movements of the vertebral column involve both the joints as well as the associated intervertebral disc.
In the limbs, flexion decreases the angle between the bones (bending of the joint), while extension increases the angle and straightens the joint (see Figure 16.8(a-d) ). You will discover in the muscular system chapter that the associated muscles to these movements are flexor and extensor.
Abduction and Adduction
Abduction and adduction motions occur within the coronal plane and involve medial-lateral motions of the limbs, fingers, toes, or thumb. For example, abduction is raising the arm at the shoulder joint, moving it laterally away from the body, while adduction brings the arm down to the side of the body (see Figure 16.8(e) ). In the muscular system chapter you will discover that the associated muscles to these movements are abductor and adductor.
Circumduction
Circumduction is the movement of a body region in a circular manner, in which one end of the body region being moved stays relatively stationary while the other end describes a circle. It involves the sequential combination of flexion, adduction, extension, and abduction at a joint (see Figure 16.8(e) ).
Rotation
Rotation can occur within the vertebral column, at a pivot joint , or at a ball-and-socket joint . Rotation of the neck or body is the twisting movement produced by the summation of the small rotational movements available between adjacent vertebrae. At a pivot joint, one bone rotates in relation to another bone.
Rotation can also occur at the ball-and-socket joints of the shoulder and hip. Here, the humerus and femur rotate around their long axis, which moves the anterior surface of the arm or thigh either toward or away from the midline of the body (see see Figure 16.8(f) ).
Supination and Pronation
Supination and pronation are movements of the forearm. In the anatomical position, the upper limb is held next to the body with the palm facing forward . This is the supinated position of the forearm. In this position, the radius and ulna are parallel to each other. When the palm of the hand faces backward , the forearm is in the pronated position , and the radius and ulna form an X-shape.
Pronation is the movement that allows the palm of the hand to face backward while in supination the palm of the hand faces forward . It helps to remember that supination is the motion you use when scooping up soup with a spoon (see Figure 16.9(g) ).
Dorsiflexion and Plantar Flexion
Dorsiflexion and plantar flexion are movements at the ankle joint, which is a hinge joint. Lifting the front of the foot, so that the top of the foot moves (upward) toward the anterior leg is dorsiflexion, while lifting the heel of the foot from the ground or pointing the toes downward is plantar flexion. These are the only movements available at the ankle joint (see Figure 16.9(h) ).
Inversion and Eversion
Inversion and eversion are complex movements that involve the multiple plane joints among the tarsal bones of the posterior foot (intertarsal joints) and thus are not motions that take place at the ankle joint. Inversion is the turning of the foot to angle the bottom of the foot toward the midline , while eversion turns the bottom of the foot away from the midline. The foot has a greater range of inversion than eversion motion. These are important motions that help to stabilize the foot when walking or running on an uneven surface and aid in the quick side-to-side changes in direction used during active sports such as basketball, racquetball, or soccer (see Figure 16.9(i) ).
Protraction and Retraction
Protraction and retraction are anterior-posterior movements of the scapula or mandible. Protraction of the scapula occurs when the shoulder is moved forward , as when pushing against something or throwing a ball. Retraction is the opposite motion, with the scapula being pulled posteriorly and medially, toward the vertebral column. For the mandible, protraction occurs when the lower jaw is pushed forward, to stick out the chin, while retraction pulls the lower jaw backward (see Figure 16.9(j) ).
Depression and Elevation
Depression and elevation are downward and upward movements of the scapula or mandible. The upward movement of the scapula and shoulder is elevation, while a downward movement is depression. These movements are used to shrug your shoulders. Similarly, elevation of the mandible is the upward movement of the lower jaw used to close the mouth or bite on something, and depression is the downward movement that produces opening of the mouth (see Figure 16.9(k) ).
- Discuss the joints involved and movements required for you to cross your arms together in front of your chest.
- Differentiate between pronation and supination.
Musculoskeletal System Movement Terms
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Medical Terms in Context
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Common Diseases and Disorders
Osteoporosis
Health Canada (2018) describes osteoporosis as bone loss that causes bones to become weak and thin over time. This weakness can lead to fractures from simple movements and occur often in the wrist, shoulder, spine, and hip. To learn more about osteoporosis please visit the Osteoporosis Health Canada website.
Arthritis
Arthritis often presents as edema , arthralgia , and ankylosis (Centers for Disease Control and Prevention, 2019). Common types of arthritis are osteoarthritis (OA), rheumatoid arthritis (RA), fibromyalgia , Gout and lupus. To learn more about arthritis visit the CDC’s Arthritis Basics web page.
Osteoarthritis
Osteoarthritis is the most common form of arthritis and according to the Arthritis Society affects Canadians more than the combination of all other types of arthritis. The breakdown of cartilage and bone occurs over time when joints are exposed to heavy workloads either through occupation, obesity and/or prior injury to a joint. Common symptoms are pain, stiffness and aching that worsens over time. While there is no cure, symptoms can be managed through exercise, medications and in severe cases joint replacements (Arthritis Society, 2020).
Rheumatoid arthritis
The CDC describes rheumatoid arthritis (RA) as an autoimmune and inflammatory disease. Autoimmune diseases are disorders in which the immune system over reacts and begins to attack itself. In the case of RA inflammation to the joint tissues of the hands, wrists and knees is painful and debilitating. Treatments may include immunosuppressive drugs and anti-inflammatory drugs (Betts, et al., 2013). RA can also affect other tissues throughout the body and cause problems in organs such as the lungs, heart, and eyes. RA can affect children and in this case it is referred to as juvenile rheumatoid arthritis (Centers for Disease Control and Prevention, 2020a).
Gout
According to the Arthritis Society, gout is an inflammatory arthritis caused when the immune system attacks the crystals that form when uric acid builds up in a joint. Gout has periods of exacerbation and remission and is commonly treated through lifestyle changes and medication. While any joint can be effected it is common in the lower extremities and most often in the big toe (Choy & MacMullan, 2017). To learn more about the causes and treatments please visit the Arthritis Society’s web page about gout.
Myasthenia Gravis
The National Institute of Neurological Disorders and Strokes describes myasthenia gravis as a "chronic autoimmune neuromuscular disorder that causes weakness in the skeletal muscle” (Office of Communications and Public Liaison, 2020). To learn more, visit the National Institute of Neurological Disorders and Stroke’s Myasthenia Gravis fact sheet.
Fibromyalgia
Fibromyalgia is a challenging disease to diagnose since symptoms manifest differently and are similar to other diseases. Symptoms may include chronic fatigue, gastrointestinal problems, headaches and increased pain sensitivity. Historically, fibromyalgia was often misdiagnosed or dismissed as not real. According to The Canadian Women’s Health Network, there is agreement on the definition and treatment for fibromyalgia but it is recommended to find a specialist who is familiar with fibromyalgia (Canadian Women’s Health Network, 2012). To learn more about the diagnosis and treatment for fibromyalgia please visit the Canadian Women’s Health Network’s fibromyalgia web page.
Osteomyelitis
Osteomyelitis is a bone infection caused when staphylococcus bacteria travels through the blood stream from an infection in one part of the body to the bone. Staphylococcus bacteria is found on the skin and it can transfer to the bone through a wound and/or surgical contamination. The risk increases as people age or if their immune system is compromised (Mayo Clinic Staff, 2018). To learn more about the causes, symptoms and treatments for osteomyelitis please visit the Mayo Clinic’s osteomyelitis web page.
Disorders of the Curvature of the Spine
Developmental anomalies, pathological changes, or obesity can enhance the normal vertebral column curves, resulting in the development of abnormal or excessive curvatures (see Figure 16.10 ). Disorders associated with the curvature of the spine include:
- Kyphosis: Also referred to as humpback, is an excessive posterior curvature of the thoracic region. This can develop when osteoporosis causes weakening and erosion of the anterior portions of the upper thoracic vertebrae, resulting in their gradual collapse (see Figure 16.11 ).
- Lordosis: Also referred to as swayback, is an excessive anterior curvature of the lumbar region and is most commonly associated with obesity or late pregnancy. The accumulation of body weight in the abdominal region results an anterior shift in the line of gravity that carries the weight of the body. This causes in an anterior tilt of the pelvis and a pronounced enhancement of the lumbar curve.
- Scoliosis: An abnormal, lateral curvature, accompanied by twisting of the vertebral column. Scoliosis is the most common vertebral abnormality among girls. The cause is usually unknown, but it may result from weakness of the back muscles, defects such as differential growth rates in the right and left sides of the vertebral column, or differences in the length of the lower limbs. When present, scoliosis tends to get worse during adolescent growth spurts. Although most individuals do not require treatment, a back brace may be recommended for growing children. In extreme cases, surgery may be required (Betts, et al., 2013).
Fractures
A fracture is a broken bone. It will heal whether or not a physician resets it in its anatomical position. If the bone is not reset correctly, the healing process will keep the bone in its deformed position. Crepitation or crepitus is the creaking or popping sound that is heard when fractured bones move against each other Fractures are classified by their complexity, location, and other features (see Figure 16.12 ). Some fractures may be described using more than one term because it may have the features of more than one type (e.g., an open transverse fracture) (Betts, et al., 2013; Canadian Orthopaedics Foundation, n.d.).
Types of fractures include:
- Closed or simple – bones are broken but does not protrude the skin
- Open or compound – bones are broken and pierce through the skin
- Transverse – bone is broken straight across
- Spiral – bone has twisted apart
- Comminuted – bones are broken and crushed into pieces
- Greenstick – bones are partially broken; occurs mainly in children
- Oblique – bones are broken at an angle
- Coles – bones are broken and occurs at the wrist or distal radius
- Stress – small crack in bone
Bone Cancer
There are three types of primary bone cancers: osteosarcoma, Ewing Sarcomas and chondrosarcoma. These are considered primary cancers because they originate in the bones. Osteosarcoma and Ewing Sarcomas are cancers found in children, teenagers, and young adults. Ewing Sarcomas is considered to be the more aggressive of the two cancers since it tends to metastasize quickly. Osteosarcoma is the most common type of bone cancer and it begins in the tissues of growing bones. Chondrosarcoma is a slow-growing bone cancer that affects adults and rarely metastasizes (Government of Canada, 2013). To learn more, visit the Public Health Agency of Canada’s web page on bone cancer.
Diagnostic Procedures
Common diagnostic procedures related specifically to the skeletal system include x-rays, bone mineral density testing, and arthroscopy.
- X-rays are common diagnostic tests used to confirm or rule out fractures and broken bones. The radiation dose is low so it is considered a safe diagnostic test (Ontario Association of Radiologist, 2020).
- Dual x-ray absorptionmetry (BMD) , also called a bone mineral density test, is a test to determine osteoporosis by measuring the amount of bone mineral in a particular amount of bone (National Cancer Insitute, n.d.).
- Arthroscopy is a common procedure performed by orthopedic surgeons to view the inside of a joint to diagnose and/or to repair joint problems. The patient is given a local anesthetic and the surgeon inserts an arthroscope through an incision in the skin. Depending on what the surgeons finds, a repair of the joint may take place during the procedure (Mayo Clinic Staff, 2018a).
Medical Specialties Related to the Skeletal System
Orthopedic Surgeon
Orthopedic Surgeons are medical doctors who complete an additional 5-years of specialized training in the prevention, diagnosis, treatment and surgery of disorders and diseases related to the musculoskeletal systems (Canadian Medical Association, 2018). For more details please visit the Canadian Medical Association’s page on Orthopedic Surgery (PDF file).
Rheumatologist
Rheumatologists are medical doctors who have additional training as internists with a sub-specialty in rheumatology. Many rheumatology disorders have an underlying autoimmune disorders. Subsequently, rheumatologists are interested in autoimmune disorders and their impact on multiple body systems including the musculoskeletal systems (Canadian Medical Association, 2018a). For more details please follow the link to the Canadian Medical Association’s page on Rheumatology (PDF file).
Doctor of Chiropractic (DC)/Chiropractor
A Doctor of Chiropractic (DC) is regulated and licensed by each province in Canada. Chiropractors have seven years of University Education, a supervised internship, and national examinations. Chiropractors are trained in the prevention, assessment and treatment of the spine, muscular system and nervous system. Chiropractors focus on spinal adjustments, nutrition, and preventing injury without the use of pharmaceuticals or surgical procedures (Canadian Chiropractic Association, 2020; 2020a).. To learn more visit the Canadian Chiropractic Association website.
Physiotherapist
A physiotherapist in Canada has a Master’s degree in physiotherapy and has successfully completed a national Physiotherapy Competency Examination (PCE). Physiotherapists use an evidenced-based approach when assessing and designing treatment plans for their clients. Treatments may include exercises, massage, joint manipulation, and occupational retraining (Canadian Physiotherapy Association, 2020). To learn more please visit the Canadian Physiotherapy Association website.
Skeletal System Vocabulary
Amphiarthrosis
Joints with some movement.
Ankylosis
Abnormal condition of stiffness.
Appendicular Skeleton
Consists of all the bones in the upper and lower limbs.
Arthralgia
Painful joint(s).
Arthritis
Inflammation of the joints.
Articulations
Also known as joints. It is where bones meet bones or bones meet bones.
Autoimmune Diseases/Disorders
Autoimmune diseases are disorders in which the immune system over reacts and begins to attack itself.
Axial Skeleton
Forms the vertical, central axis of the body and includes all bones of the head, neck, chest, and back.
Chronic
A condition that lasts a long time with periods of remission and exacerbation.
Diarthrosis
Freely moveable joints.
Edema
Swelling.
Hematopoiesis
The process in which the body produces blood.
Osteoarthritis
Inflammation of bones and joints.
Osteoporosis
Abnormal condition of bones that are porous.
Synarthrosis
Joints with no movements.
Test Yourself
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References
Arthritis Society. (2020). Osteoarthritis . https://arthritis.ca/about-arthritis/arthritis-types-(a-z)/types/osteoarthritis
Canadian Chiropractic Association. (2020). Regulation and education. https://www.chiropractic.ca/about-chiropractic/canadian-chiropractors/regulation-and-education/
Canadian Chiropractic Association. (2020a). How our experts can help. https://www.chiropractic.ca/about-chiropractic/how-our-experts-can-help/
Canadian Medical Association. (2018, August). Orthopedic surgery profile. CMA Specialty Profiles. https://www.cma.ca/sites/default/files/2019-01/orthopedic-surgery-e.pdf
Canadian Medical Association. (2018a, August). Rheumatology profile. CMA Specialty Profiles. https://www.cma.ca/sites/default/files/2019-01/rheumatology-e.pdf
Canadian Orthopaedic Foundation. (n.d.). Long bone fractures. https://whenithurtstomove.org/about-orthopaedics/conditions-and-ailments/long-bone-fractures/
Canadian Physiotherapy Association. (2020). Physiotherapy in Canada. https://physiotherapy.ca/physiotherapy-canada
Canadian Women’s Health Network. (2012). Fibromyalgia . http://www.cwhn.ca/node/40784
Centers for Disease Control and Prevention. (2019). Arthritis basics. CDC Arthritis Program. https://www.cdc.gov/arthritis/basics/index.html
Centers for Disease Control and Prevention. (2020). Rheumatoid arthritis. CDC Arthritis Program. https://www.cdc.gov/arthritis/basics/rheumatoid-arthritis.html
Choy, G., & MacMullan, P. (2017). Gout . Arthritis Society. https://arthritis.ca/about-arthritis/arthritis-types-(a-z)/types/gout
Government of Canada. (2013). Bone cancer. Public Health Agency of Canada. https://www.canada.ca/en/public-health/services/chronic-diseases/cancer/bone-cancer.html
Government of Canada. (2018). Osteoporosis. Public Health Agency of Canada. https://www.canada.ca/en/public-health/services/chronic-diseases/osteoporosis.html
Mayo Clinic Staff. (2018). Osteomyelitis. Mayo Clinic Patient Care and Health Information. https://www.mayoclinic.org/diseases-conditions/osteomyelitis/symptoms-causes/syc-20375913
Mayo Clinic Staff. (2018a). Arthroscopy. Mayo Clinic Patient Care and Health Information. https://www.mayoclinic.org/tests-procedures/arthroscopy/about/pac-20392974
Office of Communications and Public Liaison. (2020). Myasthenia gravis fact sheet. National Institute of Neurological Disorders and Stroke. https://www.ninds.nih.gov/Disorders/Patient-Caregiver-Education/Fact-Sheets/Myasthenia-Gravis-Fact-Sheet
Ontario Association of Radiologists. (2020). X-ray . https://oarinfo.ca/patient-info/radiology-medical-services/x-ray
National Cancer Institute. Dual x-ray absorptiometry . National Institutes of Health. https://www.cancer.gov/publications/dictionaries/cancer-terms/def/dual-x-ray-absorptiometry
Image Descriptions
Figure 16.1 image description: This diagram shows the human skeleton and identifies the major bones. The left panel shows the anterior view (from the front) and the right panel shows the posterior view (from the back). Labels read (from the top of skull): skull (cranial portion, facial portion), pectoral shoulder girdle, clavicle, scapula, thoracic cage (sternum, ribs), upper limb (humerus, ulna, radius, carpals, meta carpals, phalanges), vertebral column, pelvic girdle (hip bones), lower limb (femur, patella, tibia, fibula, tarsals, metatarsals, phalanges). [Return to Figure 16.1].
Figure 16.2 image description: This image shows the structure of the vertebral column. The left panel shows the front view of the vertebral column. Labels and the right panel shows the side view of the vertebral column. labels read (from top): 7 cervical vertebrae (C1-C7) form cervical curve, 12 thoracic vertebrae (T1-T12) form thoracic curve, intervertebral disc, 5 lumbar vertebrae (L1-L5) form lumbar curve, Fused vertebrae of sacrum and coccyx form sacrococcygeal curve, sacrum, coccyx. [Return to Figure 16.2].
Figure 16.3 image description: This figure shows the skeletal structure of the rib cage. The left panel shows the anterior view of the sternum. Labels read (from top): clavicular notch, jugular notch, manubrium, sternal angle, body, xiphoid process. The right panel shows the anterior panel of the sternum including the entire rib cage. Labels read (from top): jugular notch, clavicular notch, clavicle, sternum (manubrium, body, xyphoid process), scapula, sternal angle, costal cartilages, intercostal space. Ribs are numbered 1-12 from the top. [Return to Figure 16.3].
Figure 16.4 image description: This diagram labels the bones of the lower arm (excluding the hands). Labels read (from top): olecranon process, head of radius, radial notch of the ulna, trochlear notch, coronoid process, radial tuberosity, proximal radioulnar joint, neck of radius, radius, interosseous membrane, ulna, ulnar notch of the radius, head of the ulna, distal radioulnar joint, styloid process of ulna, styloid process of radius. [Return to Figure 16.4].
Figure 16.5 image description: This diagram shows an anterior and posterior view of the hands with corresponding labels. Anterior view labels read (from top): middle finger, ring finger, index finger, little finger, thumb, phalanges (distal, proximal), metacarpals, carpals, ulna, radius. Posterior view lables read (frop top): Phalanges (distal, middle, proximal), head shaft and base of proximal phalange, head shaft and base of metatarsal, metatarsals 1-5, carpals, ulna, radius. [Return to Figure 16.5].
Figure 16.6 image description: This image shows the structure of the tibia and the fibula. The left panel shows the anterior view. Labels read (from top): lateral condyle, medial condyle, tibial tuberosity, anterior border, interosseous membrane, fibula, tibia, medial malleolus, lateral malleolus, articular surface.The right panel shows the posterior view. Labels read (from top): articular surface of medial and lateral condyles, medial condyle, head of fibula, soleal line, interosseous membrane, tibla, fibula, medial malleolus, lateral malleolus, articular surface. [Return to Figure 16.6].
Figure 16.7 image description: This figure shows the bones of the foot. The left panel shows the superior view. Labels read (from toes): distal, proximal phalanges, distal phalage, middle phalange, proximal phalange, medial cuneiform, intermediate and lateral cuneiforms, navicular, cuboid, talus, trochlea of talus, calcaneus. The top right panel shows the medial view. Labels read (from left to right starting at toe): first metatarsal, medial cuneiform, intermediate cuneiform, navicular, talus, calcaneus, facet for medial malleolus, sustentaculum tali (talar shelf), calcaneal tuberosity. The bottom right panel shows the lateral view. Labels read (from left at the heel, to right): calcaneus, talus, facelt for lateral malleolus, cuboid, navicular, intermediate and lateral cuneiforms, fifth metatarsal. [Return to Figure 16.7].
Figure 16.8 image description: This multi-part image shows different types of movements that are possible by different joints in the body. Labels read (from top, left): a and b angular movements: flexion and extension at the shoulders and knees, c) angular movements: flexion and extension of the neck (arrows pointing left and right to indicate movement). Labels (from bottom, left) read: d) angular movements: flexion and extension of the vertical column, e) angular movements abduction, adduction, and cicumduction of the upper limb at the shoulder, f) rotation of the head, neck, and lower limb. [Return to Figure 16.8].
Figure 16.10 image description: This image shows the changes to the abnormal curves of the vertebral columns in different diseases. The left panel shows the change in the curve of the vertebral column in scoliosis, the middle panel shows the change in the curve of the vertebral column in kyphosis, and the right panel shows the change in the curve of the vertebral column in lordosis. [Return to Figure 16.10].
Figure 16.11 image description: This figure shows the changes to the spine in osteoporosis. The left panel shows the structure of normal vertebrae and the right panel shows the curved vertebrae in osteoporosis. [Return to Figure 16.11].
Figure 16.12 image description: In this illustration, each type of fracture is shown on the right femur from an anterior view. In the closed fracture, the femur is broken in the middle of the shaft with the upper and lower halves of the bone completely separated. However, the two halves of the bones are still aligned in that the broken edges are still facing each other. In an open fracture, the femur is broken in the middle of the shaft with the upper and lower halves of the bone completely separated. Unlike the closed fracture, in the open fracture, the two bone halves are misaligned. The lower half is turned laterally and it has protruded through the skin of the thigh. The broken ends no longer line up with each other. In a transverse fracture, the bone has a crack entirely through its width, however, the broken ends are not separated. The crack is perpendicular to the long axis of the bone. Arrows indicate that this is usually caused by compression of the bone in a superior-inferior direction. A spiral fracture travels diagonally through the diameter of the bone. In a comminuted fracture, the bone has several connecting cracks at its middle. It is possible that the bone could splinter into several small pieces at the site of the comminuted fracture. In an impacted fracture, the crack zig zags throughout the width of the bone like a lightning bolt. An arrow indicates that these are usually caused by an impact that pushes the femur up into the body. A greenstick fracture is a small crack that does not extend through the entire width of the bone. The oblique fracture shown here is travelling diagonally through the shaft of the femur at about a thirty degree angle. [Return to Figure 16.12].
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 .


11.7: Refraction
Learning Objectives
- Determine the index of refraction, given the speed of light in a medium.
It is easy to notice some odd things when looking into a fish tank. For example, you may see the same fish appearing to be in two different places. (See Figure \(\PageIndex{1}\).) This is because light coming from the fish to us changes direction when it leaves the tank, and in this case, it can travel two different paths to get to our eyes. The changing of a light ray’s direction (loosely called bending) when it passes through variations in matter is called refraction . Refraction is responsible for a tremendous range of optical phenomena, from the action of lenses to voice transmission through optical fibers.
Definition: REFRACTION
The changing of a light ray’s direction (loosely called bending) when it passes through variations in matter is called refraction.
Why does light change direction when passing from one material (medium) to another? It is because light changes speed when going from one material to another. So before we study the law of refraction, it is useful to discuss the speed of light and how it varies in different media.
The Speed of Light
Early attempts to measure the speed of light, such as those made by Galileo, determined that light moved extremely fast, perhaps instantaneously. The first real evidence that light traveled at a finite speed came from the Danish astronomer Ole Roemer in the late 17th century. Roemer had noted that the average orbital period of one of Jupiter’s moons, as measured from Earth, varied depending on whether Earth was moving toward or away from Jupiter. He correctly concluded that the apparent change in period was due to the change in distance between Earth and Jupiter and the time it took light to travel this distance. From his 1676 data, a value of the speed of light was calculated to be \(2.26 \times 10^{8} \mathrm{~m} / \mathrm{s}\) (only 25% different from today’s accepted value). In more recent times, physicists have measured the speed of light in numerous ways and with increasing accuracy. One particularly direct method, used in 1887 by the American physicist Albert Michelson (1852–1931), is illustrated in Figure \(\PageIndex{2}\). Light reflected from a rotating set of mirrors was reflected from a stationary mirror 35 km away and returned to the rotating mirrors. The time for the light to travel can be determined by how fast the mirrors must rotate for the light to be returned to the observer’s eye.
The speed of light is now known to great precision. In fact, the speed of light in a vacuum \(c\) is so important that it is accepted as one of the basic physical quantities and has the fixed value
\[c=2.9972458 \times 10^{8} \mathrm{~m} / \mathrm{s} \approx 3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}, \nonumber \]
where the approximate value of \(3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\) is used whenever three-digit accuracy is sufficient. The speed of light through matter is less than it is in a vacuum, because light interacts with atoms in a material. The speed of light depends strongly on the type of material, since its interaction with different atoms, crystal lattices, and other substructures varies. We define the index of refraction \(n\) of a material to be
\[n=\frac{c}{v}, \nonumber \]
where \(v\) is the observed speed of light in the material. Since the speed of light is always less than \(c\) in matter and equals \(c\) only in a vacuum, the index of refraction is always greater than or equal to one.
VALUE OF THE SPEED OF LIGHT
\[c=2.9972458 \times 10^{8} \mathrm{~m} / \mathrm{s} \approx 3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} \nonumber\]
Definition: INDEX OF REFRACTION
\[n=\frac{c}{v} \nonumber\]
That is, \(n \geq 1\). Table \(\PageIndex{1}\) gives the indices of refraction for some representative substances. The values are listed for a particular wavelength of light, because they vary slightly with wavelength. (This can have important effects, such as colors produced by a prism.) Note that for gases, \(n\) is close to 1.0. This seems reasonable, since atoms in gases are widely separated and light travels at \(c\) in the vacuum between atoms. It is common to take \(n=1\) for gases unless great precision is needed. Although the speed of light \(v\) in a medium varies considerably from its value \(c\) in a vacuum, it is still a large speed.
Example \(\PageIndex{1}\): Speed of Light in Matter
Calculate the speed of light in zircon, a material used in jewelry to imitate diamond.
Strategy
The speed of light in a material, \(v\), can be calculated from the index of refraction \(n\) of the material using the equation \(n=c / v\).
Solution
The equation for index of refraction states that \(n=c / v\). Rearranging this to determine \(v\) gives
\[v=\frac{c}{n}. \nonumber\]
The index of refraction for zircon is given as 1.923 in Table \(\PageIndex{1}\), and \(c\) is given in the equation for speed of light. Entering these values in the last expression gives
\[\begin{aligned}
v &=\frac{3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}}{1.923} \\
&=1.56 \times 10^{8} \mathrm{~m} / \mathrm{s}.
\end{aligned} \nonumber\]
Discussion
This speed is slightly larger than half the speed of light in a vacuum and is still high compared with speeds we normally experience. The only substance listed in Table \(\PageIndex{1}\) that has a greater index of refraction than zircon is diamond. We shall see later that the large index of refraction for zircon makes it sparkle more than glass, but less than diamond.
Law of Refraction
Figure \(\PageIndex{3}\) shows how a ray of light changes direction when it passes from one medium to another. As before, the angles are measured relative to a perpendicular to the surface at the point where the light ray crosses it. (Some of the incident light will be reflected from the surface, but for now we will concentrate on the light that is transmitted.) The change in direction of the light ray depends on how the speed of light changes. The change in the speed of light is related to the indices of refraction of the media involved. In the situations shown in Figure \(\PageIndex{3}\), medium 2 has a greater index of refraction than medium 1. This means that the speed of light is less in medium 2 than in medium 1. Note that as shown in Figure \(\PageIndex{3}\)(a), the direction of the ray moves closer to the perpendicular when it slows down. Conversely, as shown in Figure \(\PageIndex{3}\)(b), the direction of the ray moves away from the perpendicular when it speeds up. The path is exactly reversible. In both cases, you can imagine what happens by thinking about pushing a lawn mower from a footpath onto grass, and vice versa. Going from the footpath to grass, the front wheels are slowed and pulled to the side as shown. This is the same change in direction as for light when it goes from a fast medium to a slow one. When going from the grass to the footpath, the front wheels can move faster and the mower changes direction as shown. This, too, is the same change in direction as for light going from slow to fast.
The amount that a light ray changes its direction depends both on the incident angle and the amount that the speed changes. For a ray at a given incident angle, a large change in speed causes a large change in direction, and thus a large change in angle.
 Do this and observe the shape of the pencil when you look at the pencil sideways, that is, through air, glass, water. Explain your observations. Draw ray diagrams for the situation.
Section Summary
- The changing of a light ray’s direction when it passes through variations in matter is called refraction.
- The speed of light in vacuum \(c=2.9972458 \times 10^{8} \mathrm{~m} / \mathrm{s} \approx 3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} \)
- Index of refraction \(n=\frac{c}{v}\), where \(v\) is the speed of light in the material, \(c\) is the speed of light in vacuum, and \(n\) is the index of refraction.
Glossary
- refraction
- changing of a light ray’s direction when it passes through variations in matter
- index of refraction
- for a material, the ratio of the speed of light in vacuum to that in the material


4.1: Exponential Functions
- Refresher Topics
-
-
Systems
- Solving 2x2 Systems of Linear Equations - Substitution Method
-
Systems
In this section, you will:
- Evaluate exponential functions.
- Find the equation of an exponential function.
- Use compound interest formulas.
- Evaluate exponential functions with base \(e\).
When the original iteration of this textbook was written (2013), India was the second most populous country in the world with a population of about \(1.25\) billion people. At that time, the population was growing at a rate of about \(1.2\%\) each year. If that rate continued, the population of India would have exceeded China's population by the year \(2031\). Interestingly, the rate of growth for India's population actually sped up! In 2025, India's population was \( 1.42 \) billion while China's was \( 1.41 \) billion. When populations grow rapidly, we often say that the growth is "exponential," meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions , which model this kind of rapid growth.
Identifying Exponential Functions
When exploring linear growth, we observed a constant rate of change - a constant number by which the output increased for each unit increase in input. For example, in the equation \(f(x)=3x+4\), the slope tells us the output increases by \( 3 \) each time the input increases by \( 1 \). We call this the Slope Addition Property and it is unique to linear functions. The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people.
Defining an Exponential Function
A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, \( 2.5\% \) of the population was vegan, adhering to a diet that does not include any animal products - no meat, poultry, fish, dairy, or eggs. If this rate continued, vegans would have made up \( 10\% \) of the U.S. population in 2015, \( 40\% \) in 2019, and \( 80\% \) in 2021.
What exactly does it mean to grow exponentially ? What does the word double have in common with percent increase ? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media.
- Percent change refers to a change based on a percent of the original amount.
- Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time.
- Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time.
For us to gain a clear understanding of exponential growth, let us contrast exponential growth with linear growth. We will construct two functions. The first function is exponential. We will start with an input of \( 0 \), and increase each input by \( 1 \). We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of \( 0 \), and increase each input by \( 1 \). We will add \( 2 \) to the corresponding consecutive outputs. See Table \( \PageIndex{ 1 } \).
| \(x\) | \(f(x)= 2^x\) | \(g(x)=2x\) |
|---|---|---|
| 0 | 1 | 0 |
| 1 | 2 | 2 |
| 2 | 4 | 4 |
| 3 | 8 | 6 |
| 4 | 16 | 8 |
| 5 | 32 | 10 |
| 6 | 64 | 12 |
| 7 | 128 | 14 |
| 8 | 256 | 16 |
| 9 | 512 | 18 |
| 10 | 1024 | 20 |
From Table \( \PageIndex{ 1 } \) we can infer that for these two functions, exponential growth dwarfs linear growth.
- Exponential growth means the original value from the range increases by the same percentage over equal increments in the domain.
- Linear growth means the original value from the range increases by the same amount over equal increments in the domain.
Apparently, the difference between "the same percentage" and "the same amount" is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. We call this the Base Multiplier Property , and it is unique to exponential functions. For linear growth, the constant additive rate of change over equal increments resulted in adding 2 to the output whenever the input was increased by one. As mentioned previously, we call this the Slope Addition Property .
The general form of the exponential function is \(f(x) = a b^x \), where \(a\) is any nonzero number, \(b\) is a positive real number not equal to \( 1 \).
- If \(b > 1\), the function grows at a rate proportional to its size.
- If \(0 < b < 1\), the function decays at a rate proportional to its size.
Let's look at the function \(f(x)= 2^x\) from our example. We will create a table (Table \( \PageIndex{ 2 } \)) to determine the corresponding outputs over an interval in the domain from \(−3\) to \(3\).
| \(x\) | \(−3\) | \(−2\) | \(−1\) | \(0\) | \(1\) | \(2\) | \(3\) |
| \(f(x)= 2^x\) | \(2^{−3} = \dfrac{1}{8}\) | \(2^{−2} = \dfrac{1}{4}\) | \(2^{−1} = \dfrac{1}{2}\) | \(2^0 = 1\) | \(2^1 = 2\) | \(2^2 = 4\) | \(2^3 = 8\) |
Let us examine the graph of \(f\) by plotting the ordered pairs we observe on the table in Figure \( \PageIndex{ 1 } \), and then make a few observations.
Figure \( \PageIndex{ 1 } \)
Let's define the behavior of the graph of the exponential function \(f(x)= 2^x\) and highlight some its key characteristics.
- the domain is \(( − \infty , \infty )\),
- the range is \(( 0, \infty )\),
- as \(x \to \infty , f(x) \to \infty \),
- as \(x \to − \infty , f(x) \to 0\),
- \(f(x)\) is always increasing,
- the graph of \(f(x)\) will never touch the \( x \)-axis because base two raised to any exponent never has the result of zero.
- \(y=0\) is the horizontal asymptote.
- the \( y \)-intercept is \( 1 \).
For any real number \(x\), an exponential function is a function with the form\[f(x)=a b^x, \nonumber \]where
- \(a\) is a non-zero real number called the initial value .
- \(b\) is any positive real number such that \(b \neq 1\).
- The domain of \(f\) is all real numbers.
- The range of \(f\) is all positive real numbers if \(a > 0\).
- The range of \(f\) is all negative real numbers if \(a < 0\).
- The \( y \)-intercept is \(( 0,a )\), and the horizontal asymptote is \(y=0\).
Which of the following equations are not exponential functions?
- \(f(x)= 4^{3( x−2 )}\)
- \(g(x)= x^3\)
- \(h(x)= \left( \frac{1}{3} \right)^x\)
- \(j(x)= ( −2 )^x\)
- Solution
-
By definition, an exponential function has a constant as a base and an independent variable as an exponent. Thus, \(g(x)= x^3\) does not represent an exponential function because the base is an independent variable. In fact, \(g(x)= x^3\) is a power function.
Recall that the base \( b \) of an exponential function is always a positive constant, and \(b \neq 1\). Thus, \(j(x)= ( −2 )^x\) does not represent an exponential function because the base, \(−2\), is less than \(0\).
Which of the following equations represent exponential functions?
- \(f(x)=2 x^2 −3x+1\)
- \(g(x)= 0.875^x\)
- \(h(x)=1.75x+2\)
- \(j(x)= 1095.6^{-2x}\)
Evaluating Exponential Functions
Recall that the base of an exponential function must be a positive real number other than \(1\). Why do we limit the base \(b\) to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive:
- Let \(b=−9\) and \(x= \frac{1}{2} \). Then \(f(x)=f\left( \frac{1}{2} \right)= ( −9 )^{1/2} =\sqrt{−9} \), which is not a real number.
Why do we limit the base to positive values other than \(1\)? Because base \(1\) results in the constant function. Observe what happens if the base is \(1\):
- Let \(b=1\). Then \(f(x)= 1^x =1\) for any value of \(x\).
To evaluate an exponential function with the form \(f(x)= b^x \), we simply substitute \(x\) with the given value, and calculate the resulting power. For example, let \(f(x)= 2^x \). What is \(f(3)\)?\[ \begin{array}{rrclcl} & f( x ) & = & 2^x & & \\[6pt] \implies & f( 3 ) & = & 2^3 & \quad & \left( \text{substituting} \right) \\[6pt] & & = & 8 & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]To evaluate an exponential function with a form other than the basic form, it is important to follow the Order of Operations. For example, let \(f(x)=30( 2 )^x \). What is \(f(3)\)?\[ \begin{array}{rrclcl} & f( x ) & = & 30(2)^x & & \\[6pt] \implies & f( 3 ) & = & 30(2)^3 & \quad & \left( \text{substituting} \right) \\[6pt] & & = & 30(8) & \quad & \left( \text{exponents before multiplication} \right) \\[6pt] & & = & 240 & \quad & \left( \text{multiplication} \right) \\[6pt] \end{array} \nonumber \]Note that if the Order of Operations were not followed, the result would be incorrect:\[f(3) = 30( 2 )^3 \neq 60^3 = 216,000. \nonumber \]
Let \(f( x )=5 ( 3 )^{x+1} \). Evaluate \(f( 2 )\) without using a calculator.
- Solution
-
Follow the Order of Operations. Be sure to pay attention to the parentheses.\[ \begin{array}{rrclcl} & f(x) & = & 5(3)^{x + 1} & & \\[6pt] \implies & f(2) & = & 5(3)^{2 + 1} & \quad & \left( \text{substituting} \right) \\[6pt] & & = & 5(3)^{3} & \quad & \left( \text{simplifying} \right) \\[6pt] & & = & 5(27) & \quad & \left( \text{simplifying} \right) \\[6pt] & & = & 135 & \quad & \left( \text{simplifying} \right) \\[6pt] \end{array} \nonumber \]
Let \(f( x )=8 ( 1.2 )^{x−5} \). Evaluate \(f( 3 )\) using a calculator. Round to four decimal places.
Defining Exponential Growth
Because the output of exponential functions increases very rapidly, the term "exponential growth" is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth.
A function that models exponential growth grows by a rate proportional to the amount present. For any real number \(x\) and any positive real numbers \(a\) and \(b\) such that \(b \neq 1\), an exponential growth function has the form\[f(x)=a \, b^x, \nonumber \]where
- \(a\) is the initial or starting value of the function and
- \(b\) is the growth factor or growth multiplier per unit \(x\) .
In more general terms, we have an exponential function , in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let's consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function \(A( x )=100+50x\). Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function \(B(x)=100 ( 1+0.5 )^{x} \).
A few years of growth for these companies are illustrated in Table \( \PageIndex{ 3 } \).
| Year, \(x\) | Stores, Company A | Stores, Company B |
|---|---|---|
| \(0\) | \(100+50( 0 )=100\) | \(100 ( 1+0.5 )^0 =100\) |
| \(1\) | \(100+50( 1 )=150\) | \(100 ( 1+0.5 )^1 =150\) |
| \(2\) | \(100+50( 2 )=200\) | \(100 ( 1+0.5 )^2 =225\) |
| \(3\) | \(100+50( 3 )=250\) | \(100 ( 1+0.5 )^3 =337.5\) |
| \(x\) | \(A( x )=100+50x\) | \(B(x)=100 ( 1+0.5 )^x\) |
The graphs comparing the number of stores for each company over a five-year period are shown in Figure \( \PageIndex{ 2 } \). We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth.
Figure \( \PageIndex{ 2 } \): The graph shows the numbers of stores Companies A and B opened over a five-year period.
Notice that the domain for both functions is \([0, \infty )\), and the range for both functions is \([100, \infty )\). After year 1, Company B always has more stores than Company A.
Now we will turn our attention to the function representing the number of stores for Company B, \(B(x)=100 ( 1+0.5 )^x \). In this exponential function, \( 100 \) represents the initial number of stores, \( 0.50 \) represents the growth rate , and \(1+0.5=1.5\) represents the growth factor . Generalizing further, we can write this function as \(B(x)=100 ( 1.5 )^x \), where \( 100 \) is the initial value, \(1.5\) is called the base , and \(x\) is called the exponent .
At the beginning of this section, we learned that the population of India was about \(1.25\) billion in the year 2013, with an annual growth rate of about \(1.2\%\). This situation is represented by the growth function \(P(t)=1.25 ( 1.012 )^t \), where \(t\) is the number of years since 2013. To the nearest thousandth, and using this model, what will the population of India be in 2031?
- Solution
-
To estimate the population in 2031, we evaluate the model for \(t=18\), because 2031 is \(18\) years after 2013. Rounding to the nearest thousandth,\[P(18)=1.25 ( 1.012 )^{18} \approx 1.549. \nonumber \]There will be about 1.549 billion people in India in the year 2031.
The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about \(0.6\%\). This situation is represented by the growth function \(P(t)=1.39 ( 1.006 )^t \), where \(t\) is the number of years since 2013. To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example \( \PageIndex{ 3 } \)?
Finding Equations of Exponential Functions
In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants \(a\) and \(b\), and evaluate the function.
Given two data points, write an exponential model.
- If one of the data points has the form \(( 0,a )\), then \(a\) is the initial value. Using \(a\), substitute the second point into the equation \(f(x)=a ( b )^x \), and solve for \(b\).
- If neither of the data points have the form \(( 0,a )\), substitute both points into two equations with the form \(f(x)=a ( b )^x \). Solve the resulting system of two equations in two unknowns to find \(a\) and \(b\).
- Using the \(a\) and \(b\) found in the steps above, write the exponential function in the form \(f(x)=a ( b )^x \).
In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an exponential function \(N(t)\) representing the population \(( N )\) of deer over time \(t\).
- Solution
-
We let our independent variable \(t\) be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: \( (0, 80) \) and \( (6, 180) \). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, \(a=80\). We can now substitute the second point into the equation \(N(t)=80 b^t\) to find \(b\):\[ \begin{array}{rrclcl} & N(t) & = & 80b^t & & \\[6pt] \implies & 180 & = & 80b^6 & \quad & \left( \text{substituting} \right) \\[6pt] \implies & \dfrac{9}{4} & = & b^6 & \quad & \left( \text{dividing both sides by }80 \right) \\[6pt] \implies & \left(\dfrac{9}{4}\right)^{1/6} & = & b & \quad & \left( \text{raising both sides to the }\frac{1}{6}\text{ power} \right) \\[6pt] \implies & b & \approx & 1.1447 & \quad & \left( \text{rounding to four decimal places} \right) \\[6pt] \end{array} \nonumber \] NOTES: When we took the sixth root of both sides, we did not bother writing \( b = \pm \left( \frac{9}{4} \right)^{1/6} \). This is because the base of an exponential function is never negative. Moreover, do not round any intermediate calculations. Doing so leads to error propagation.
The exponential model for the population of deer is \(N(t) \approx 80 ( 1.1447 )^t \). (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)
We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph in Figure \( \PageIndex{ 3 } \) passes through the initial points given in the problem, \(( 0,80 )\) and \(( 6,180 )\). We can also see that the domain for the function is \([0, \infty )\), and the range for the function is \([80, \infty )\).
Figure \( \PageIndex{ 3 } \): Graph showing the population of deer over time, \(N(t) \approx =80 ( 1.1447 )^t \), \(t\) years after 2006
A wolf population is growing exponentially. In 2011, \(129\) wolves were counted. By 2013, the population had reached \( 236 \) wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population \(N\) of wolves over time \(t\).
Find an exponential function that passes through the points \(( −2,6 )\) and \(( 2,1 )\).
- Solution
-
Because we don't have the initial value, we substitute both points into an equation of the form \(f(x)=a \, b^x \), and then solve the system for \(a\) and \(b\).
- Substituting \(( −2,6 )\) gives \(6=a \, b^{−2}\)
- Substituting \(( 2,1 )\) gives \(1=a \, b^2\)
Use the first equation to solve for \(a\) in terms of \(b\):\[ \begin{array}{rrclcl} & 6 & = & a \, b^{-2} & & \\[6pt] \implies & 6b^2 & = & a & \quad & \left( \text{multiplying both sides by }b^2 \right) \\[6pt] \end{array} \nonumber \]Substitute \(a\) in the second equation, and solve for \(b\):\[ \begin{array}{rrclcl} & 1 & = & a \, b^{2} & & \\[6pt] \implies & 1 & = & 6 b^2 b^2 & \quad & \left( \text{substituting }a = 6 b^2 \right) \\[6pt] \implies & 1 & = & 6 b^4 & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] \implies & \dfrac{1}{6} & = & b^4 & \quad & \left( \text{dividing both sides by }6 \right) \\[6pt] \implies & \left(\dfrac{1}{6}\right)^{1/4} & = & b & \quad & \left( \text{raise both sides to the }\frac{1}{4}\text{ power} \right) \\[6pt] \end{array} \nonumber \]Now that we have \( b \), we go back to the first equation to find \( a \):\[ a = 6 b^2 = 6 \left(\left( \dfrac{1}{6} \right)^{1/4}\right)^2 = 6 \left( \dfrac{1}{6} \right)^{2/4} = 6 \left( \dfrac{1}{6} \right)^{1/2} = \dfrac{6}{6^{1/2}} = 6^{1 - 1/2} = 6^{1/2} = \sqrt{6}. \nonumber \]Thus, the equation is\[ f(x) = \sqrt{6} \left( \left( \dfrac{1}{6} \right)^{1/4} \right)^x = \sqrt{6} \left( \dfrac{1}{6} \right)^{x/4}.\nonumber \]We can graph our model to check our work. Notice that the graph in Figure \( \PageIndex{ 4 } \) passes through the initial points given in the problem, \(( −2, 6 )\) and \(( 2, 1 )\). The graph is an example of an exponential decay function.
Figure \( \PageIndex{ 4 } \): The graph of \(f(x)=\sqrt{6} \left( \frac{1}{6} \right)^{x/4}\) models exponential decay.
Given the two points \(( 1,3 )\) and \(( 2,4.5 )\), find the equation of the exponential function that passes through these two points.
Do two points always determine a unique exponential function?
Yes, provided the two points are either both above the \( x \)-axis or both below the \( x \)-axis and have different \( x \)-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in \(x\), which in many real world cases involves time.
Find an equation for the exponential function graphed in Figure \( \PageIndex{ 5 } \).
Figure \( \PageIndex{ 5 } \)
- Solution
-
We can choose the \( y \)-intercept of the graph, \(( 0,3 )\), as our first point. This gives us the initial value, \(a=3\). Next, choose a point on the curve some distance away from \(( 0,3 )\) that has integer coordinates. One such point is \((2,12)\).\[ \begin{array}{rrclcl} & y & = & a b^x & \quad & \left( \text{writing the general form} \right) \\[6pt] \implies & y & = & 3 b^x & \quad & \left( \text{substituting }a = 3 \right) \\[6pt] \implies & 12 & = & 3 b^2 & \quad & \left( \text{substituting }x = 2\text{ and }y = 12 \right) \\[6pt] \implies & 4 & = & b^2 & \quad & \left( \text{dividing both sides by }3 \right) \\[6pt] \implies & \pm 2 & = & b & \quad & \left( \text{Extraction of Roots} \right) \\[6pt] \implies & 2 & = & b & \quad & \left( \text{exponential bases are positive} \right) \\[6pt] \end{array} \nonumber \]Substitute \(a\) and \(b\) into the standard form to yield the equation \(f(x)=3 (2)^x \).
Find an equation for the exponential function graphed in Figure \( \PageIndex{ 6 } \).
Figure \( \PageIndex{ 6 } \)
Applying the Compound-Interest Formula
Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account.
The annual percentage rate (APR) of an account, also called the nominal rate , is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing.
We can calculate the compound interest using the compound interest formula , which is an exponential function of the variables time \(t\), principal \(P\), APR \(r\), and number of compounding periods in a year \(n\):\[A(t)=P \left( 1 + \dfrac{r}{n} \right)^{nt}. \nonumber \]For example, observe Table \( \PageIndex{ 4 } \), which shows the result of investing $1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases.
| Frequency | Value after 1 year |
|---|---|
| Annually | $1100 |
| Semiannually | $1102.50 |
| Quarterly | $1103.81 |
| Monthly | $1104.71 |
| Daily | $1105.16 |
Compound interest can be calculated using the formula\[A(t)=P \left( 1 + \dfrac{r}{n} \right)^{nt}, \nonumber \]where
- \(A(t)\) is the account value,
- \(t\) is measured in years,
- \(P\) is the starting amount of the account, often called the principal , or more generally present value ,
- \(r\) is the annual percentage rate (APR) expressed as a decimal, and
- \(n\) is the number of compounding periods in one year.
If we invest $3,000 in an investment account paying 3% interest compounded quarterly, how much will the account be worth in 10 years?
- Solution
-
Because we are starting with $3,000, \(P=3000\). Our interest rate is 3%, so \(r=0.03\). Because we are compounding quarterly, we are compounding 4 times per year, so \(n=4\). We want to know the value of the account in 10 years, so we are looking for \(A( 10 )\), the value when \(t=10\).\[ \begin{array}{rrclcl} & A(t) & = & P\left( 1 + \dfrac{r}{n} \right)^{nt} & & \\[6pt] \implies & A(10) & = & 3000 \left( 1 + \dfrac{0.03}{4} \right)^{4 \cdot 10} & \quad & \left( \text{substituting known values} \right) \\[6pt] & & \approx & \$4,045.05 & \quad & \left( \text{rounding to two decimal places} \right) \\[6pt] \end{array} \nonumber \]The account will be worth about $4,045.05 in 10 years.
An initial investment of $100,000 at 12% interest is compounded weekly (use 52 weeks in a year). What will the investment be worth in 30 years?
A 529 Plan is a college-savings plan that allows relatives to invest money to pay for a child's future college tuition; the account grows tax-free. Lily wants to set up a 529 account for her new granddaughter and wants the account to grow to $40,000 over 18 years. She believes the account will earn 6% compounded semi-annually (twice a year). To the nearest dollar, how much will Lily need to invest in the account now?
- Solution
-
The nominal interest rate is 6%, so \(r=0.06\). Interest is compounded twice a year, so \(k=2\).
We want to find the initial investment, \(P\), needed so that the value of the account will be worth $40,000 in \(18\) years. Substitute the given values into the compound interest formula, and solve for \(P\).\[ \begin{array}{rrclcl} & A(t) & = & P\left( 1 + \dfrac{r}{n} \right)^{nt} & & \\[6pt] \implies & 40,000 & = & P \left( 1 + \dfrac{0.06}{2} \right)^{2 \cdot 18} & \quad & \left( \text{substituting known values} \right) \\[6pt] \implies & 40,000 & = & P \left( 1.03 \right)^{36} & \quad & \left( \text{simplifying} \right) \\[6pt] \implies & \dfrac{40,000}{1.03^{36}} & = & P & \quad & \left( \text{isolating }P \right) \\[6pt] \implies & P & \approx & \$13,801 & \quad & \left( \text{rounding to the nearest dollar} \right) \\[6pt] \end{array} \nonumber \]Lily will need to invest $13,801 to have $40,000 in 18 years.
Refer to Example \( \PageIndex{ 8 } \). To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?
Evaluating Functions with Base \( e \)
As we saw earlier, the amount earned on an account increases as the compounding frequency increases. Table \( \PageIndex{ 5 } \) shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.
Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies, listed in Table \( \PageIndex{ 5 } \).
| Frequency | \(A(n)= \left( 1+ \frac{1}{n}\right)^n\) | Value |
|---|---|---|
| Annually | \(\left( 1 + \frac{1}{1} \right)^{1}\) | $2 |
| Semiannually | \(\left( 1 + \frac{1}{2} \right)^{2}\) | $2.25 |
| Quarterly | \(\left( 1 + \frac{1}{4} \right)^{4}\) | $2.441406 |
| Monthly | \(\left( 1 + \frac{1}{12} \right)^{12}\) | $2.613035 |
| Daily | \(\left( 1 + \frac{1}{365} \right)^{365}\) | $2.714567 |
| Hourly | \(\left( 1 + \frac{1}{8760} \right)^{8760}\) | $2.718127 |
| Once per minute | \(\left( 1 + \frac{1}{525600} \right)^{525600}\) | $2.718279 |
| Once per second | \(\left( 1 + \frac{1}{31536000} \right)^{31536000}\) | $2.718282 |
These values appear to be approaching a limit as \(n\) increases without bound. In fact, as \(n\) gets larger and larger, the expression \(\left( 1 + \frac{1}{n} \right)^n\) approaches a number used so frequently in Mathematics that it has its own name: the letter \(e\). This value is an irrational number, which means that its decimal expansion goes on forever without repeating.
The letter \( e \) represents the irrational number\[\left( 1 + \dfrac{1}{n} \right)^n, \text{ as } n \text{ increases without bound.} \nonumber \]
The letter \( e \) is used as a base for many real-world exponential models. To work with base \( e \), we use the approximation, \(e \approx 2.718282\). The constant was named by the Swiss mathematician Leonhard Euler (1707–1783) who first investigated and discovered many of its properties.
Calculate \(e^{3.14} \). Round to five decimal places.
- Solution
-
On a calculator, press the button labeled \( \mathrm{[e^x]}\). The window shows \(\mathrm{e^(}\). Type \(3.14\) and then close parenthesis, \(\mathrm{)}\). Press \(\mathrm{[ENTER]}\). Rounding to \(5\) decimal places, \(e^{3.14} \approx 23.10387\).
Caution: Many scientific calculators have an \(\mathrm{Exp}\) button, which is used to enter numbers in scientific notation. It is not used to find powers of \(e\).
Use a calculator to find \(e^{−0.5} \). Round to five decimal places.
Investigating Continuous Growth
So far we have worked with rational bases for exponential functions. For most real-world phenomena, however, \( e \) is used as the base for exponential functions. Exponential models that use \(e\) as the base are called continuous growth or decay models . We see these models in finance, computer science, and most of the sciences, such as physics, toxicology, and fluid dynamics.
For all real numbers \(t\), and all positive numbers \(a\) and \(r\), continuous growth or decay is represented by the formula\[A(t)=a e^{rt}, \nonumber \]where
- \(a\) is the initial value,
- \(r\) is the continuous growth rate per unit time,
- and \(t\) is the elapsed time.
If \(r > 0\) , then the formula represents continuous growth. If \(r < 0\) , then the formula represents continuous decay.
For business applications, the continuous growth formula is called the continuous compounding formula and takes the form\[A(t)=P e^{rt}, \nonumber \]where
- \(P\) is the principal or the initial invested,
- \(r\) is the growth or interest rate per unit time,
- and \(t\) is the period or term of the investment.
A person invested $1,000 in an account earning a nominal 10% per year compounded continuously. How much was in the account at the end of one year?
- Solution
-
Since the account is growing in value, this is a continuous compounding problem with growth rate \(r=0.10\). The initial investment was $1,000, so \(P=1000\). We use the continuous compounding formula to find the value after \(t=1\) year:\[ \begin{array}{rclcl} A(t) & = & Pe^{rt} & & \\[6pt] & = & 1000 e^{0.1} & \quad & \left( \text{substituting known values} \right) \\[6pt] & \approx & 1105.17 & \quad & \left( \text{rounding to the nearest hundredth} \right) \\[6pt] \end{array} \nonumber \]The account is worth $1,105.17 after one year.
A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years?
Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days?
- Solution
-
Since the substance is decaying, the rate, \(17.3\%\), is negative. So, \(r=−0.173\). The initial amount of radon-222 was \(100\) mg, so \(a=100\). We use the continuous decay formula to find the value after \(t=3\) days:\[ \begin{array}{rclcl} A(t) & = & a e^{rt} & & \\[6pt] & = & 100 e^{-0.173(3)} & \quad & \left( \text{substituting known values} \right) \\[6pt] & \approx & 59.5115 & \quad & \left( \text{rounding to four decimal places} \right) \\[6pt] \end{array} \nonumber \]So approximately 59.5115 mg of radon-222 will remain.
Using the data in Example \( \PageIndex{ 11 } \), how much radon-222 will remain after one year?
Access these online resources for additional instruction and practice with exponential functions.


6.7: Stimulants
Stimulant medications are prescribed to treat children, adolescents, or adults diagnosed with attention deficit hyperactivity disorder (ADHD). Stimulants block the reuptake of norepinephrine and dopamine in the synapse and increase the overall level of these substances in the brain, but they have a paradoxical calming effect and improve the ability to focus and concentrate for individuals diagnosed with ADHD. Common stimulants used to treat ADHD include methylphenidate, amphetamine, dextroamphetamine, and lisdexamfetamine dimesylate. Stimulant medications are safe when prescribed with close supervision, but they are a Schedule II controlled substance because they have a high potential for misuse and dependence.
Read more information about medications used to treat attention deficit hyperactivity disorder (ADHD), adverse side effects, and patient education in the “ Common Disorders and Disabilities in Children and Adolescents ” section of the “Childhood and Adolescent Disorders” chapter.


4.1: Carl Jung
Carl Jung brought an almost mystical approach to psychodynamic theory. An early associate and follower of Freud, Jung eventually disagreed with Freud on too many aspects of personality theory to remain within a strictly Freudian perspective. Subsequently, Jung developed his own theory, which applied concepts from natural laws (primarily in physics) to psychological functioning. Jung also introduced the concept of personality types, and began to address personality development throughout the lifespan. In his most unique contribution, at least from a Western perspective, Jung proposed that the human psyche contains within itself psychological constructs developed throughout the evolution of the human species.
-
- 4.1.1: Prelude to Carl Jung
- Jung has always been controversial and confusing. His blending of psychology and religion, as well as his openness to different religious and spiritual philosophies, was not easy to accept for many psychiatrists and psychologists trying to pursue a purely scientific explanation of personality and mental illness. Perhaps no one was more upset than Freud, whose attitude toward Jung changed dramatically over just a few years.
-
- 4.1.2: A Brief Biography of Carl Jung
- Jung believed that our personality begins with a collective unconscious, developed within our species throughout time, and that we have only limited ability to control the psychic process that is our own personality. Thus, our true personality arises from within as our collective unconscious comes forth into our personal unconscious and then our consciousness.


5.4: Bulk Transport
Skills to Develop
- Describe endocytosis, including phagocytosis, pinocytosis, and receptor-mediated endocytosis
- Understand the process of exocytosis
In addition to moving small ions and molecules through the membrane, cells also need to remove and take in larger molecules and particles (see Table \(\PageIndex{1}\) for examples). Some cells are even capable of engulfing entire unicellular microorganisms. You might have correctly hypothesized that the uptake and release of large particles by the cell requires energy. A large particle, however, cannot pass through the membrane, even with energy supplied by the cell.
Endocytosis
Endocytosis is a type of active transport that moves particles, such as large molecules, parts of cells, and even whole cells, into a cell. There are different variations of endocytosis, but all share a common characteristic: The plasma membrane of the cell invaginates, forming a pocket around the target particle. The pocket pinches off, resulting in the particle being contained in a newly created intracellular vesicle formed from the plasma membrane.
Phagocytosis
Phagocytosis (the condition of “cell eating”) is the process by which large particles, such as cells or relatively large particles, are taken in by a cell. For example, when microorganisms invade the human body, a type of white blood cell called a neutrophil will remove the invaders through this process, surrounding and engulfing the microorganism, which is then destroyed by the neutrophil (Figure \(\PageIndex{1}\)).
In preparation for phagocytosis, a portion of the inward-facing surface of the plasma membrane becomes coated with a protein called clathrin , which stabilizes this section of the membrane. The coated portion of the membrane then extends from the body of the cell and surrounds the particle, eventually enclosing it. Once the vesicle containing the particle is enclosed within the cell, the clathrin disengages from the membrane and the vesicle merges with a lysosome for the breakdown of the material in the newly formed compartment (endosome). When accessible nutrients from the degradation of the vesicular contents have been extracted, the newly formed endosome merges with the plasma membrane and releases its contents into the extracellular fluid. The endosomal membrane again becomes part of the plasma membrane.
Pinocytosis
A variation of endocytosis is called pinocytosis . This literally means “cell drinking” and was named at a time when the assumption was that the cell was purposefully taking in extracellular fluid. In reality, this is a process that takes in molecules, including water, which the cell needs from the extracellular fluid. Pinocytosis results in a much smaller vesicle than does phagocytosis, and the vesicle does not need to merge with a lysosome (Figure \(\PageIndex{2}\)).
A variation of pinocytosis is called potocytosis . This process uses a coating protein, called caveolin , on the cytoplasmic side of the plasma membrane, which performs a similar function to clathrin. The cavities in the plasma membrane that form the vacuoles have membrane receptors and lipid rafts in addition to caveolin. The vacuoles or vesicles formed in caveolae (singular caveola) are smaller than those in pinocytosis. Potocytosis is used to bring small molecules into the cell and to transport these molecules through the cell for their release on the other side of the cell, a process called transcytosis.
Receptor-mediated Endocytosis
A targeted variation of endocytosis employs receptor proteins in the plasma membrane that have a specific binding affinity for certain substances (Figure \(\PageIndex{3}\)).
In receptor-mediated endocytosis , as in phagocytosis, clathrin is attached to the cytoplasmic side of the plasma membrane. If uptake of a compound is dependent on receptor-mediated endocytosis and the process is ineffective, the material will not be removed from the tissue fluids or blood. Instead, it will stay in those fluids and increase in concentration. Some human diseases are caused by the failure of receptor-mediated endocytosis. For example, the form of cholesterol termed low-density lipoprotein or LDL (also referred to as “bad” cholesterol) is removed from the blood by receptor-mediated endocytosis. In the human genetic disease familial hypercholesterolemia, the LDL receptors are defective or missing entirely. People with this condition have life-threatening levels of cholesterol in their blood, because their cells cannot clear LDL particles from their blood.
Although receptor-mediated endocytosis is designed to bring specific substances that are normally found in the extracellular fluid into the cell, other substances may gain entry into the cell at the same site. Flu viruses, diphtheria, and cholera toxin all have sites that cross-react with normal receptor-binding sites and gain entry into cells.
Link to Learning
Video \(\PageIndex{1}\): See receptor-mediated endocytosis in action, and click on different parts for a focused animation.
Exocytosis
The reverse process of moving material into a cell is the process of exocytosis. Exocytosis is the opposite of the processes discussed above in that its purpose is to expel material from the cell into the extracellular fluid. Waste material is enveloped in a membrane and fuses with the interior of the plasma membrane. This fusion opens the membranous envelope on the exterior of the cell, and the waste material is expelled into the extracellular space (Figure \(\PageIndex{4}\)). Other examples of cells releasing molecules via exocytosis include the secretion of proteins of the extracellular matrix and secretion of neurotransmitters into the synaptic cleft by synaptic vesicles.
| Transport Method | Active/Passive | Material Transported |
|---|---|---|
| Diffusion | Passive | Small-molecular weight material |
| Osmosis | Passive | Water |
| Facilitated transport/diffusion | Passive | Sodium, potassium, calcium, glucose |
| Primary active transport | Active | Sodium, potassium, calcium |
| Secondary active transport | Active | Amino acids, lactose |
| Phagocytosis | Active | Large macromolecules, whole cells, or cellular structures |
| Pinocytosis and potocytosis | Active | Small molecules (liquids/water) |
| Receptor-mediated endocytosis | Active | Large quantities of macromolecules |
Summary
Active transport methods require the direct use of ATP to fuel the transport. Large particles, such as macromolecules, parts of cells, or whole cells, can be engulfed by other cells in a process called phagocytosis. In phagocytosis, a portion of the membrane invaginates and flows around the particle, eventually pinching off and leaving the particle entirely enclosed by an envelope of plasma membrane. Vesicle contents are broken down by the cell, with the particles either used as food or dispatched. Pinocytosis is a similar process on a smaller scale. The plasma membrane invaginates and pinches off, producing a small envelope of fluid from outside the cell. Pinocytosis imports substances that the cell needs from the extracellular fluid. The cell expels waste in a similar but reverse manner: it pushes a membranous vacuole to the plasma membrane, allowing the vacuole to fuse with the membrane and incorporate itself into the membrane structure, releasing its contents to the exterior.
Glossary
- caveolin
- protein that coats the cytoplasmic side of the plasma membrane and participates in the process of liquid update by potocytosis
- clathrin
- protein that coats the inward-facing surface of the plasma membrane and assists in the formation of specialized structures, like coated pits, for phagocytosis
- endocytosis
- type of active transport that moves substances, including fluids and particles, into a cell
- exocytosis
- process of passing bulk material out of a cell
- pinocytosis
- a variation of endocytosis that imports macromolecules that the cell needs from the extracellular fluid
- potocytosis
- variation of pinocytosis that uses a different coating protein (caveolin) on the cytoplasmic side of the plasma membrane
- receptor-mediated endocytosis
- variation of endocytosis that involves the use of specific binding proteins in the plasma membrane for specific molecules or particles, and clathrin-coated pits that become clathrin-coated vesicles


2.2.3: Active Listening
“Most people do not listen with the intent to understand; they listen with the intent to reply”–Stephen R. Covey
Active listening means what it sounds like.It means that you are fully concentrating on what is being said, rather than passively listening or focusing on your reply. You are actively paying attention or listening to the other person with whom you are speaking. Active listening is an important skill for a HHA/PCA to develop. It demonstrates to their patients that they care and are interested in them. It also allows them to obtain important information about their patient’s health care needs.
Active listening means giving the speaker, or sender of the message, your full attention. It means using all your senses while listening. Active listening means paying attention to the words the speaker is saying and the nonverbal messages that are also being expressed. Active listening means conveying (showing) the other person that you are interested in what they are saying. We demonstrate active listening by not only our spoken words. Unspoken words or our nonverbal messages also demonstrate that we are actively listening. Nonverbal messages, such as use of eye contact, posture, nodding of the head, touch, and facial expressions like smiling convey the powerful message that we are actively listening to the speaker.
How to Be a Good Listener:
Allow the person the opportunity to express his/her ideas completely. Concentrate on what they are saying. Do not interrupt. Avoid finishing their sentences, even if you are in a hurry, if you think you know what they are going to say, if they have told you the same thing before, or if they are having trouble communicating and are taking a long time to respond. If you find it particularly difficult to not interrupt, practice counting to ten when you feel the urge to interrupt.
Restate the message you heard the other person saying to ensure you heard them correctly. This feedback allows the other person to correct any misunderstandings and demonstrates you were listening to and heard them. For instance, you can something like, “I heard you saying that you are not happy with your new diet.”
Provide feedback while you are listening to encourage the person to continue and show you are interested. Leading statements such as, “Go on,” “Mmm–hmm,” “Yes,” and “Tell me more,” encourages the person to continue and demonstrates your interest in what they have to say. Use of body language such as nodding your head helps to encourage the person to continue.
Allow for silence. Silence can be a powerful tool. Silence allows the person a chance to fully explore and express their thoughts. Often, we try to avoid silence because it makes us uncomfortable. Silence is especially important when discussing sensitive or difficult topics. Don’t rush to fill in the silence with meaningless conversation or clichés.
Avoid changing the subject even if the topic makes you uncomfortable. This conveys to the patient that you are not interested in what they are saying. For example, a patient might say, “I am really afraid to die,” do not say something like, “Don‘t think like that. Come on, let‘s do something to cheer you up.”  Tell me what‘s on your mind.”
Use your body language to demonstrate your interest in the other person.  Stand, or better yet, sit, facing them. If appropriate for the culture, make eye contact. If a patient enjoys touch as a means of communication, a HHA/PCA can touch their hand or arm while they are talking. Use facial expressions to convey your interest and attention. Smile as appropriate and appear concerned when a difficult topic is being discussed. Lean forward to convey your interest and avoid crossing your arms over your chest, which closes you off from the other person.
Be aware of cultural differences. Get to know a patient’s culture and what is acceptable. Learn the words and expressions of a patient’s culture to fully understand the message they are trying to convey. When in doubt, ask a patient to explain what a phrase or expression means. Do not assume you understand.
Do not talk to an elderly or disabled person as if they are a child. Talk to them as if you would anyone else. Do not call them names such as “Honey,” “Sweetheart,” or “Dear.” Call the person by the name they wish to be called. Always ask how the person would like to be referred. Some people may find it inappropriate if you call them by their first name. They may prefer to be addressed as “Mr.” or “Mrs.”
Avoid fidgeting during conversation. Do not look at the clock while people are speaking as this conveys that you are in a hurry or bored. Avoid picking at your fingernails or clothing, playing with your hair, doodling, or using your cell phone while the person is speaking. Home Health Aides/Personal Care Aides should give their patient their full attention.
Use empathy. Try to put yourself in another person’s shoes to understand what they are going through. Don’t state, “I understand how you feel” as you cannot know exactly how they feel.
Ensure communication aides such as hearing aids, eyeglasses, and dentures are clean and in good working order. Home Health Aides/Personal Care Aides should provide them to their patient to help them communicate more effectively.
Which of these two scenarios demonstrates active listening?
1. Mrs. Pompi is in her bed and is telling her HHA that she is scared she will never walk again. The HHA is cleaning the room as Mrs. Pompi talks and periodically says, “Mhmmm” and “Okay” during the conversation. The HHA is worried that she will not get all her tasks completed before she has to leave for the day. She looks at the clock on the wall every now and then as she works and Mrs. Pompi talks. Mrs. Pompi says, “I just don’t want to live the rest of my life in this bed. I’m so scared.” The HHA says to her, “Don’t worry about that, honey.I understand how you feel. Everything will be okay. Now, let’s get you washed and dressed.”
2. Mrs. Williams is in her wheelchair doing a crossword puzzle while the HHA makes her bed. Suddenly she looks up from her crossword and says, “I am so frightened that I am going to die.  The HHA stops what she is doing, pulls a chair next to Mrs. Williams and touches her hand. The HHA says, “You sound scared and worried about dying. Would you like to talk about it?”
- Answer
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Scenario 2 demonstrates active listening.
Feedback :
The healthcare worker stopped completion of her tasks, positioned herself so she could focus on her patient and so that her patient would know she cared about what she had to say. She also used reflection and an open-ended question, inviting Mrs. Williams to share her thoughts and feelings.


8.7: Spotlight Application
Emma is assigned to care for a 64-year-old patient readmitted with heart failure (HF). Emma recognizes the patient’s name during shift report as the reporting nurse comments how this patient was a “frequent flyer,” and he was just discharged last week. The reporting nurse spends little time describing the patient’s actual condition and focuses comments on how the patient “might as well put his name plaque on the door” because he is here so frequently. Emma does agree that the patient has been readmitted very frequently with HF exacerbations but is concerned by the dismissive comments by her peer. She is determined to spend time with this patient on this admission and find out more about his home environment, support system, and self-care.
Reflective Questions
- How do the frequent readmissions affect the client’s health care costs and the hospital’s reimbursement levels by Medicare and private insurance companies?
- What additional nursing interventions might help prevent future frequent readmissions and reduce costs?
Emma engages in a conversation with the client about his dietary habits at home. He reports that his wife passed away one year ago from breast cancer, and he has been preparing his own meals since that time by using easy-to-prepare and cheap foods. Canned soups, boxed meals, and chips are common staples in his diet. Through Emma’s conversation with the patient, she can see that the patient needs dietary instruction and further education about monitoring his fluid retention with daily weights. He seems engaged in exploring different food preparations and was unaware that foods could exacerbate his heart failure. When Emma asked if he had been taught about this on previous admissions, he stated that no one took the time to talk with him about his home situation. He reports that he received pamphlets but never really took the time to read them because the print was so small. Emma reflects on how preventative education opportunities were missed and increased costs were incurred by the patient and the health care organization as a result.