7.6: Solve Applications with Rational Equations
- Solve proportions
- Solve similar figure applications
- Solve uniform motion applications
- Solve work applications
- Solve direct variation problems
- Solve inverse variation problems
Before you get started, take this readiness quiz.
- Example 2.2.13. Example 2.5.13. Example 2.2.9.
Solve Proportions
When two rational expressions are equal, the equation relating them is called a proportion .
A proportion is an equation of the form \(\dfrac{a}{b}=\dfrac{c}{d}\), where \(b \neq 0, d \neq 0\).
The proportion is read “\(a\) is to \(b\) as \(c\) is to \(d\).”
The equation \(\dfrac{1}{2}=\dfrac{4}{8}\) is a proportion because the two fractions are equal. The proportion \(\dfrac{1}{2}=\dfrac{4}{8}\) is read “1 is to 2 as 4 is to 8.”
Since a proportion is an equation with rational expressions, we will solve proportions the same way we solved rational equations. We’ll multiply both sides of the equation by the LCD to clear the fractions and then solve the resulting equation.
Solve: \(\dfrac{n}{n+14}=\dfrac{5}{7}\).
Solution
\[\dfrac{n}{n+14}=\dfrac{5}{7}, \quad n \neq-14 \nonumber \]
Multiply both sides by LCD.
\[7(n+14)\left(\dfrac{n}{n+14}\right)=7(n+14)\left(\dfrac{5}{7}\right) \nonumber \]
Remove common factors on each side.
\[7 n=5(n+14) \nonumber \]
Simplify.
\[7 n=5 n+70 \nonumber \]
Solve for \(n\).
\[\begin{aligned} 2n&=70\\ n&=35 \end{aligned} \nonumber \]
Check.
\[\dfrac{n}{n+14}=\dfrac{5}{7} \nonumber \]
Substitute \(n=35\)
\[\dfrac{35}{35+14} \overset{?}{=} \dfrac{5}{7} \nonumber \]
Simplify.
\[\dfrac{35}{49} \overset{?}{=} \dfrac{5}{7} \nonumber \]
Show common factors.
\[\dfrac{5 \cdot 7}{7 \cdot 7} \overset{?}{=} \dfrac{5}{7} \nonumber \]
Simplify.
\[\dfrac{5}{7}=\dfrac{5}{7}\; \surd \nonumber \]
Solve the proportion: \(\dfrac{y}{y+55}=\dfrac{3}{8}\).
- Answer
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\(y=33\)
Solve the proportion: \(\dfrac{z}{z-84}=-\dfrac{1}{5}\).
- Answer
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\(z=14\)
Notice in the last example that when we cleared the fractions by multiplying by the LCD, the result is the same as if we had cross-multiplied.
\[\begin{aligned} \dfrac{n}{n+14}=\dfrac{5}{7} \quad \quad \quad \dfrac{n}{n+14}=\dfrac{5}{7} \\ 7(n+14)\left(\dfrac{n}{n+14}\right)=7(n+14)\left(\dfrac{5}{7}\right) \quad \quad \quad \dfrac{n}{n+14}=\dfrac{5}{7} \\ 7n=5(n+14) \quad \quad \quad 7n=5(n+14) \end{aligned} \nonumber \]
For any proportion, \(\dfrac{a}{b}=\dfrac{c}{d}\), we get the same result when we clear the fractions by multiplying by the LCD as when we cross-multiply.
\[\begin{aligned} \dfrac{a}{b} =\dfrac{c}{d} \quad \quad \quad \dfrac{a}{b}=\dfrac{c}{d} \\ bd\left(\dfrac{a}{b}=\frac{c}{d}\right) bd \quad \quad \quad \frac{a}{b} = \frac{c}{d} \\ ad =bc \quad \quad \quad ad=bc \end{aligned} \nonumber \]
To solve applications with proportions, we will follow our usual strategy for solving applications But when we set up the proportion, we must make sure to have the units correct—the units in the numerators must match each other and the units in the denominators must also match each other.
When pediatricians prescribe acetaminophen to children, they prescribe 5 milliliters (ml) of acetaminophen for every 25 pounds of the child’s weight. If Zoe weighs 80 pounds, how many milliliters of acetaminophen will her doctor prescribe?
Solution
Identify what we are asked to find, and choose a variable to represent it.
How many ml of acetaminophen will the doctor prescribe?
Let \(a=ml\) of acetaminophen.
Write a sentence that gives the information to find it.
If 5 ml is prescribed for every 25 pounds, how much will be prescribed for 80 pounds?
Translate into a proportion—be careful of the units.
Step 1 . Write the inequality as one quotient on the left and zero on the right. Our inequality is in this form.
\[\dfrac{x-1}{x+3} \geq 0 \nonumber \]
Step 2 . Determine the critical points-the points where the rational expression will be zero or undefined.
The rational expression will be zero when the numerator is zero. Since \(x-1=0\) when \(x=1\), then 1 is a critical point. The rational expression will be undefined when the denominator is zero. Since \(x+3=0\) when \(x=-3\), then -3 is a critical point.
Step 3 . Use the critical points to divide the number line into intervals.
Step 4 . Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.
Use values in each interval to determine the value of each factor in the interval. In the interval (-3,1), zero is a good value to test. For example, when \(x=0\) then \(x-1=-1\) and \(x+3=3\) The factor \(x-1\) is marked negative and \(x+3\) marked positive. Since a negative divided by a positive is negative, the quotient is marked negative in that interval.
Step 5 . Determine the intervals where the inequality is correct. Write the solution in interval notation.
We want the quotient to be greater than or equal to zero, so the numbers in the intervals \((-\infty,-3)\) and \((1, \infty)\) are solutions. Since 3 must be excluded since it makes the rational expression 0, we cannot include it in the solution. We can include 1 in our solution.
\[(-\infty,-3) \cup[1, \infty) \nonumber \]
Multiply both sides by the LCD, 400. Remove common factors on each side. Simplify, but don’t multiply on the left. Notice what the next step will be.
\[16 \cdot 5=5 a \nonumber \]
Solve for \(a\).
\[\begin{aligned} \dfrac{16 \cdot 5}{5}&=\dfrac{5 a}{5}\\ 16&=a \end{aligned} \nonumber \]
Check. Is the answer reasonable? Write a complete sentence.
The pediatrician would prescribe 16 ml of acetaminophen to Zoe.
Pediatricians prescribe 5 milliliters (ml) of acetaminophen for every 25 pounds of a child’s weight. How many milliliters of acetaminophen will the doctor prescribe for Emilia, who weighs 60 pounds?
- Answer
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The pediatrician will prescribe 12 ml of acetaminophen to Emilia.
For every 1 kilogram (kg) of a child’s weight, pediatricians prescribe 15 milligrams (mg) of a fever reducer. If Isabella weighs 12 kg, how many milligrams of the fever reducer will the pediatrician prescribe?
- Answer
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The pediatrician will prescribe 180 mg of fever reducer to Isabella.
Solve similar figure applications
When you shrink or enlarge a photo on a phone or tablet, figure out a distance on a map, or use a pattern to build a bookcase or sew a dress, you are working with similar figures. If two figures have exactly the same shape, but different sizes, they are said to be similar. One is a scale model of the other. All their corresponding angles have the same measures and their corresponding sides have the same ratio.
Two figures are similar if the measures of their corresponding angles are equal and their corresponding sides have the same ratio.
For example, the two triangles in Figure below are similar. Each side of \(\Delta ABC\) is four times the length of the corresponding side of \(\Delta XYZ\).
This is summed up in the Property of Similar Triangles.
If \(\Delta ABC\) is similar to \(\Delta XYZ\), then their corresponding angle measure are equal and their corresponding sides have the same ratio.
To solve applications with similar figures we will follow the Problem-Solving Strategy for Geometry Applications we used earlier.
On a map, San Francisco, Las Vegas, and Los Angeles form a triangle. The distance between the cities is measured in inches. The figure on the left below represents the triangle formed by the cities on the map. If the actual distance from Los Angeles to Las Vegas is 270 miles, find the distance from Los Angeles to San Francisco.
Solution
Since the triangles are similar, the corresponding sides are proportional.
Read the problem. Draw the figures and label it with the given information. The figures are shown above.
Identify what we are looking for: the actual distance from Los Angeles to San Francisco
Name the variables: Let \(x\) = distance from Los Angeles to San Francisco.
Translate into an equation. Since the triangles are similar, the corresponding sides are proportional. We’ll make the numerators “miles” and the denominators “inches”.
\[$\dfrac{x \text { miles }}{1.3 \text { inches }}=\dfrac{270 \text { miles }}{1 \text { inch }}$ \nonumber \]
Solve the equation.
\[\begin{aligned} 1.3\left(\dfrac{x}{1.3}\right)&=1.3\left(\dfrac{270}{1}\right) \\ x&=351 \end{aligned} \nonumber \]
Check . On the map, the distance from Los Angeles to San Francisco is more than the distance from Los Angeles to Las Vegas. Since 351 is more than 270 the answer makes sense.
Check \(x=351\) in the original proportion. Use a calculator.
\[\begin {aligned} \dfrac{x \text { miles }}{1.3 \text { inches }}&=\dfrac{270 \text { miles }}{1 \text { inch }}\\ \dfrac{351 \text { miles }}{1.3 \text { inches }} &\overset{?}{=} \dfrac{270 \text { miles }}{1 \text { inch }}\\ \dfrac{270 \text { miles }}{1 \text { inch }}&=\dfrac{270 \text { miles }}{1 \text { inch }} \surd \end{aligned} \nonumber \]
Answer the question: The distance from Los Angeles to San Francisco is 351 miles.
On the map, Seattle, Portland, and Boise form a triangle. The distance between the cities is measured in inches. The figure on the left below represents the triangle formed by the cities on the map. The actual distance from Seattle to Boise is 400 miles.
Find the actual distance from Seattle to Portland.
- Answer
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The distance is 150 miles.
Find the actual distance from Portland to Boise.
- Answer
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The distance is 350 miles.
We can use similar figures to find heights that we cannot directly measure.
Tyler is 6 feet tall. Late one afternoon, his shadow was 8 feet long. At the same time, the shadow of a tree was 24 feet long. Find the height of the tree.
Solution
Read the problem and draw a figure. We are looking for \(h\), the height of the tree.
We will use similar triangles to write an equation. The small triangle is similar to the large triangle.
\[\dfrac{h}{24}=\dfrac{6}{8} \nonumber \]
Solve the proportion.
\[\begin {aligned} 24\left(\dfrac{6}{8}\right)&=24\left(\dfrac{h}{24}\right)\\ 18&=h \end{aligned} \nonumber \]
Simplify. Check.
Tyler's height is less than his shadow's length so it makes sense that the tree's height is less than the length of its shadow. Check \(h=18\) in the original proportion.
\[\begin{aligned} &\dfrac{6}{8}=\dfrac{h}{24}\\ &\dfrac{6}{8} \overset{?}{=} \dfrac{18}{24}\\ &\dfrac{3}{4}=\dfrac{3}{4} \surd \end{aligned} \nonumber \]
A telephone pole casts a shadow that is 50 feet long. Nearby, an 8 foot tall traffic sign casts a shadow that is 10 feet long. How tall is the telephone pole?
- Answer
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The telephone pole is 40 feet tall.
A pine tree casts a shadow of 80 feet next to a 30 foot tall building which casts a 40 feet shadow. How tall is the pine tree?
- Answer
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The pine tree is 60 feet tall.
Solve Uniform Motion Applications
We have solved uniform motion problems using the formula \(D=r t\) in previous chapters. We used a table like the one below to organize the information and lead us to the equation.
| Rate \(\cdot\) Time = Distance | |||
|---|---|---|---|
The formula \(D=r t\) assumes we know \(r\) and \(t\) and use them to find \(D\). If we know \(D\) and \(r\) and need to find \(t\), we would solve the equation for \(t\) and get the formula \(t=\dfrac{D}{r}\).
We have also explained how flying with or against the wind affects the speed of a plane. We will revisit that idea in the next example.
An airplane can fly 200 miles into a 30 mph headwind in the same amount of time it takes to fly 300 miles with a 30 mph tailwind. What is the speed of the airplane?
Solution
This is a uniform motion situation. A diagram will help us visualize the situation.
We fill in the chart to organize the information.
We are looking for the speed of the airplane. Let \(r\) = the speed of the airplane.
When the plane flies with the wind, the wind increases its speed and so the rate is \(r + 30\).
When the plane flies against the wind, the wind decreases its speed and the rate is \(r − 30\).
Write in the rates. Write in the distances. Since \(D=r \cdot t\), we solve for \(t\) and get \(t=\dfrac{D}{r}\). We divide the distance by the rate in each row, and place the expression in the time column.
| Rate \(\cdot\) Time = Distance | |||
|---|---|---|---|
| Headwind | \(r-30\) | \(\dfrac{200}{r-30}\) | 200 |
| Tailwind | \(r+30\) | \(\dfrac{300}{r+30}\) | 300 |
We know the times are equal and so we write our equation.
\[\dfrac{200}{r-30}=\dfrac{300}{r+30} \nonumber \]
We multiply both sides by the LCD.
\[(r+30)(r-30)\left(\frac{200}{r-30}\right)=(r+30)(r-30)\left(\frac{300}{r+30}\right) \nonumber \]
Simplify and solve.
\[\begin{aligned} (r+30)(200)&=(r-30) 300 \\ 200 r+6000&=300 r-9000 \\ 15000&=100 r \end{aligned} \nonumber \]
Check.
Is \(150 \mathrm{mph}\) a reasonable speed for an airplane? Yes. If the plane is traveling \(150 \mathrm{mph}\) and the wind is \(30 \mathrm{mph}\),
\[\text { Tailwind } \quad 150+30=180 \mathrm{mph} \quad \dfrac{300}{180}=\dfrac{5}{3} \text { hours } \nonumber \]
\[\text { Headwind } 150-30=120 \mathrm{mph} \dfrac{200}{120}=\dfrac{5}{3} \text { hours } \nonumber \]
The times are equal, so it checks. The plane was traveling \(150 \mathrm{mph}\).
Link can ride his bike 20 miles into a 3 mph headwind in the same amount of time he can ride 30 miles with a 3 mph tailwind. What is Link’s biking speed?
- Answer
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Link’s biking speed is 15 mph.
Danica can sail her boat 5 miles into a 7 mph headwind in the same amount of time she can sail 12 miles with a 7 mph tailwind. What is the speed of Danica’s boat without a wind?
- Answer
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The speed of Danica’s boat is 17 mph.
In the next example, we will know the total time resulting from traveling different distances at different speeds.
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Solution
This is a uniform motion situation. A diagram will help us visualize the situation.
We fill in the chart to organize the information. We are looking for Jazmine’s running speed. Let \(r\) = Jazmine’s running speed.
Her biking speed is 4 miles faster than her running speed. \(r + 4\) = her biking speed
The distances are given, enter them into the chart. Since \(D=r \cdot t\), we solve for \(t\) and get \(t=\dfrac{D}{r}\). We divide the distance by the rate in each row, and place the expression in the time column.
| Rate \(\cdot\) Time = Distance | |||
|---|---|---|---|
| Run | \(r\) | \(\dfrac{8}{r}\) | 8 |
| Bike | \(r+4\) | \(\dfrac{24}{r+4}\) | 24 |
| 3 |
Write a word sentence: Her time plus the time biking is 3 hours.
Translate the sentence to get the equation.
\[\dfrac{8}{r}+\dfrac{24}{r+4}=3 \nonumber \]
Solve.
\[\begin{aligned}
r(r+4)\left(\dfrac{8}{r}+\dfrac{24}{r+4}\right) &=3 \cdot r(r+4) \\
8(r+4)+24 r &=3 r(r+4) \\
8 r+32+24 r &=3 r^{2}+12 r \\
32+32 r &=3 r^{2}+12 r \\
0 &=3 r^{2}-20 r-32 \\
0 &=(3 r+4)(r-8)
\end{aligned} \nonumber \]
\[\begin{array}{lc} {(3 r+4)=0} & {(r-8)=0} \\ \cancel{r=\dfrac{4}{3}} \quad & {r=8} \end{array} \nonumber \]
Check.
A negative speed does not make sense in this problem, so \(r=8\) is the solution.
Is 8 mph a reasonable running speed? Yes.
If Jazmine’s running rate is 4, then her biking rate, \(r+4\), which is \(8+4=12\).
\[\text { Run } 8 \mathrm{mph} \quad \dfrac{8 \mathrm{miles}}{8 \mathrm{mph}}=1 \text { hour } \nonumber \]
\[\text { Bike } 12 \text { mph } \quad \dfrac{24 \text { miles }}{12 \mathrm{mph}}=2 \text { hours } \nonumber \]


22.2: Eyes
By the end of this section, you will be able to:
- Analyze the structure and function of the eyes
- Understand how to perform a nursing assessment of the eyes
- Recall common abnormalities of the eye
The nursing assessment of the eye plays a pivotal role in ensuring the overall health and well-being of individuals. The eyes are not only vital for vision but also serve as windows to the body’s general health. Through careful observation, evaluation, and documentation, nurses can uncover potential issues, monitor the progression of existing conditions, and contribute to timely interventions. Whether assessing visual acuity, examining the external and internal structures of the eye, or addressing specific concerns, the nursing assessment of the eye is a fundamental component of comprehensive patient care.
Structure and Function of the Eyes
Located within the orbit of the skill, the eye is a complex and intricate organ responsible for receiving visual information from the surrounding environment and transmitting it to the brain for interpretation. Its various structures work together to allow us to see and perceive the world around us (Figure 22.17).
External Structures
External structures of the eye consist of the following:
- Eyebrow: The eyebrow is the hair-covered ridge above the eye that helps protect the eye from sweat and direct sunlight.
- Eyelid: The eyelid is the movable fold of skin and muscle that covers and protects the front of the eye.
- Eyelashes: The eyelashes are short hairs along the eyelid edges that help prevent debris from entering the eye.
- Conjunctiva: The conjunctiva is a thin, transparent membrane covering the front surface of the eye (except the cornea) and the inner surface of the eyelids.
- Cornea: The cornea is the clear, dome-shaped front surface of the eye that focuses light onto the lens and retina.
- Sclera: The sclera is the white, tough outer covering of the eye that helps maintain its shape and protect its inner structures.
- Iris: The iris is the colored part of the eye that controls the size of the pupil and regulates the amount of light entering the eye.
- Pupil: The pupil is the black circular opening in the center of the iris that allows light to enter the eye.
- Lacrimal gland: The lacrimal gland produces tears that keep the eye moist and help with lubrication, nourishment, and protection against infections.
- Lacrimal ducts: The lacrimal duct drains tears from the eye’s surface to the nasal cavity.
Internal Structures
Internal structures of the eye consist of the following:
- Lens: The lens is a transparent, flexible structure located behind the iris. It changes shape to focus light onto the retina.
- Ciliary muscle: The ciliary muscle controls the shape of the lens to help with focusing on near and distant objects (accommodation).
- Vitreous humor: The vitreous humor is a gel-like substance that fills the larger space within the eye, helping maintain its shape and providing support to the retina.
- Retina: The retina is the innermost layer of the eye that contains specialized cells called photoreceptors (rods and cones) that detect light and transmit visual signals to the brain.
- Cones: A cone is a photoreceptor that is primarily responsible for color vision and visual acuity (sharpness).
- Rods: A rod is a photoreceptor that is responsible for vision in low-light conditions and for detecting movement.
- Macula: The macula is the small, central area of the retina responsible for central vision and color perception.
- Optic nerve: The optic nerve is a bundle of nerve fibers that carries visual information from the retina to the brain for processing.
- Optic disc: The optic disc is the area where the optic nerve exits the retina; it lacks photoreceptors, creating the “blind spot.”
Movement of the eye within the orbit occurs by the contraction of six extraocular muscles that originate from the bones of the orbit and insert into the surface of the eyeball. The extraocular muscles are innervated by the abducens nerve, trochlear nerve, and oculomotor nerve (cranial nerves III, IV, and V).
Vision
The process of vision is a complex and intricate interaction involving various structures within the eyes, brain, and neural pathways. From the visual fields that encompass our surroundings to the intricate neural pathways that relay information to the brain, and to the reflexes that adapt our visual experience to changing conditions, each facet of this process contributes to our profound sense of sight. Key elements of vision include visual fields, visual pathways , and visual reflexes.
Visual Fields and Visual Pathways
A visual field refers to the entire area that an individual can see when their gaze is fixed straight ahead. It includes all the objects and surroundings that can be perceived without moving the eyes. The visual field can be divided into two main parts for each eye: the central visual field and the peripheral visual field. The central visual field is the portion of the visual field that corresponds to the area around the point where you are directly looking. It provides the sharpest and most detailed vision and is critical for activities like reading, recognizing faces, and focusing on specific objects. The peripheral visual field encompasses the outer edges of the visual field. While it does not provide the same level of detail as the central visual field, it plays a crucial role in detecting motion, objects, and changes in the environment. It is especially important for spatial awareness and detecting potential threats from the sides.
A visual pathway is a complex network of neural connections that carries visual information from the eyes to the brain, where it is processed and interpreted to create visual perceptions. The visual pathway involves a series of structures and connections that work together to transmit visual signals and transform them into meaningful visual experiences. The process begins in the retina, where specialized cells called photoreceptors (rods and cones) detect light and convert it into electrical signals. The electrical signals generated by the photoreceptors travel along the optic nerves. These nerves carry the visual information toward the brain. Upon reaching the optic chiasm, which is located at the base of the brain, some of the nerve fibers from each optic nerve cross over to the opposite side, which allows information from both eyes to be combined and integrated. The nerve fibers that have crossed over at the optic chiasm continue as optic tracts, which carry the visual information further into the brain. The optic tracts connect to the lateral geniculate nucleus (LGN), a structure in the thalamus (a part of the brain) that processes and relays the visual signals to the visual cortex. The visual cortex, located in the occipital lobe at the back of the brain, receives and processes the visual information sent by the LGN. Different areas of the visual cortex specialize in analyzing different aspects of visual perception, such as color, shape, and motion. The visual cortex integrates and interprets the incoming visual signals to create the visual perceptions that are experienced.
Visual Reflexes
A visual reflex is an automatic and involuntary response of the eyes and the body to visual stimuli. These reflexes play a crucial role in maintaining visual stability, protecting the eyes, and optimizing visual perception. Important visual reflexes include the following:
- Pupillary reflex: Pupils constrict in bright light or dilate in dark spaces to regulate the amount of light entering the eye. This reflex is controlled by the interaction between the iris and the brain’s visual centers.
- Accommodation reflex: When looking at objects at varying distances, the eyes automatically adjust the shape of the lens to focus the image clearly on the retina to ensure objects at different distances remain in focus.
- Convergence reflex: When an object moves closer to the eyes, the eyes turn inward (converge) to keep the object in focus.
- Optokinetic reflex: When presented with a moving visual stimulus, such as a rotating drum with stripes, the eyes move in the direction of the stimulus and then reset to the initial position. This reflex helps stabilize vision during continuous movement.
- Vestibulo-ocular reflex: This reflex coordinates eye movements with head movements to maintain stable vision while the head is in motion. When the head moves, the eyes move in the opposite direction to counteract the movement and keep the visual scene steady.
- Blink reflex: When an object approaches the eyes suddenly, such as a foreign object or an unexpected movement, the eyelids automatically close (blink) to protect the eyes from potential harm.
- Gaze fixation reflex: When focusing on a stationary object, the eyes make small involuntary movements that prevent visual adaptation and help maintain a clear image on the retina.
- Optical righting reflex: When the body’s orientation changes, such as when tilting the head, the eyes automatically adjust their position to align with the new gravitational reference.
- Near response reflex: When looking at a nearby object, the eyes converge to bring the object into focus and the pupils constrict. This reflex is part of the accommodation-convergence reflex loop.
Visual reflexes ensure that the eyes continuously adapt to changes in the visual environment and maintain optimal visual function. These reflexes are mediated by complex neural pathways and involve interactions between the eyes, the brain’s visual centers, and other sensory systems like the vestibular system (which controls balance and spatial orientation).
Nursing Assessment: Inspecting the Eyes
Inspecting the eyes is a fundamental component of the nursing assessment, allowing healthcare providers to gather valuable information about a person’s ocular health and overall well-being. This assessment involves careful observation and examination of the eyes and their surrounding structures. Through this process, nurses can identify potential abnormalities, monitor changes in visual health, and contribute to timely interventions. Whether assessing the external appearance of the eyes, eyelids, or conjunctiva, or observing pupil reactions and eye movements, this comprehensive examination offers crucial insights into a patient’s visual status.
Assessing the Pupils
Inspecting the pupils is an important part of a nursing assessment, as it can provide valuable information about a person’s neurological and ophthalmic health. To assess the pupils, ensure proper lighting in the room (dim lighting can help with assessing pupil size). Wash your hands and don gloves. Explain the procedure to the patient to ensure their cooperation and comfort. Position the patient comfortably sitting or lying down, facing you.
Begin by assessing the patient’s overall level of alertness and consciousness. A patient who is not fully alert may have different pupil responses. Ask the patient to look straight ahead, keeping their gaze focused on a distant object. Hold the penlight or flashlight about 12 to 15 inches away from the patient’s face and shine the light directly into one eye. Observe both pupils for their initial size and equality. They should be of equal size and react similarly to light. On average, a normal adult pupil is approximately 2 to 4 millimeters in bright lights and 4 to 8 millimeters in dark spaces. Note any differences in size between the two pupils. Test the pupil’s reaction to light by shining the light directly into one pupil and observing the response. The illuminated pupil should constrict briskly, and the opposite pupil should also constrict when light is shone into the other eye. If one or both pupils do not constrict or if the response is sluggish, it may indicate an abnormality in the neurological or ophthalmic system.
To test pupil accommodation, ask the patient to focus on a near object (e.g., your finger) and then switch their focus to a distant object. Observe the pupils for changes in size. The pupils should constrict when focusing on a near object and dilate when focusing on a distant object. If the pupils do not react as expected during accommodation, it may indicate a dysfunction in the pupil’s response to changes in focus.
Visual Acuity
Assessing visual acuity is a crucial aspect of a comprehensive eye examination. The visual acuity measures a person’s ability to see details at a specific distance. Visual acuity includes both far and near vision assessments.
To assess far-distance vision, ensure the room is well lit and the patient is wearing any corrective lenses they normally use (glasses or contact lenses). Explain to the patient they will be asked to read letters or symbols on a chart to assess their ability to see details. Far-distance acuity is measured using the Snellen chart (Figure 22.18) at a distance of 20 feet (6 meters) in the United States. The Snellen chart consists of rows of letters or symbols, with the largest and most visible letters at the top and progressively smaller ones below. Instruct the patient to cover one eye and read the letters from the top row to the bottom until they can no longer correctly identify the letters or symbols. Visual acuity is recorded as a fraction. The top number represents the distance at which the chart is viewed (20 feet), and the bottom number indicates the distance at which a person with normal vision can read the line. For example, if the patient can read the line labeled “20/40,” it means they can see at 20 feet what a person with normal vision can see at 40 feet.
A technique for how to perform a visual acuity test with a Snellen eye chart is presented in this video.
See the competency checklist for Measuring Distance Visual Acuity. You can find the checklists on the Student resources tab of your book page on openstax.org.
To assess near vision, explain to the patient that you will be assessing their ability to read small print at a close distance. A near vision chart, such as a Jaeger card (Figure 22.19), consists of paragraphs of text with varying font sizes. Ask the patient to hold the Jaeger card 14 to 16 inches from their eyes and read aloud the text on the chart starting from the largest font size. Proceed to smaller font sizes until the patient can no longer comfortably read the text. Note the font size at which the patient is able to read comfortably. For example, if they can read text labeled “J5,” it means they can read text that is typically readable by a person with normal vision at a distance of 5 inches.
Extraocular Movements
Assessing extraocular movement is an essential part of a neurological and ophthalmic examination. These movements involve the coordinated actions of the six muscles that control each eye’s movement and alignment. Ensure proper lighting in the room and position the patient comfortably sitting or lying down, facing you. Explain to the patient that you will be assessing their eye movements to ensure their eyes can move smoothly in different directions. Test the six cardinal directions of gaze by asking the patient to keep their head still and follow a target (your fingertip or a pen) with their eyes only, without moving their head. Instruct the patient to focus on the target as you move it through the six cardinal directions of gaze: right (lateral), left (lateral), up (superior), down (inferior), upper-right diagonal, and upper-left diagonal. While the patient is tracking the target with their eyes, observe their eye movements for smooth and coordinated motions in each direction. An involuntary, rhythmic, back-and-forth movement of the eyes is termed nystagmus and is considered abnormal.
To test convergence, ask the patient to focus on an object (e.g., your fingertip) held about 2 inches in front of their nose. Slowly move the object toward the patient’s nose while instructing them to keep looking at the object. Observe their eyes for convergence, where both eyes move inward. Abnormal findings, such as limited eye movements, jerky motions, or nystagmus, may indicate issues with the extraocular muscles, issues with the cranial nerves (particularly cranial nerves III, IV, and VI), or neurological problems.
Peripheral Vision
Assessing peripheral vision , also known as visual field, is important for detecting any abnormalities or defects in a person’s ability to see objects in their side or peripheral vision. Ensure adequate lighting and seat the patient comfortably facing you, at eye level. Explain to the patient that you will be testing their ability to see objects in their side vision. To conduct the confrontation test , instruct the patient to cover one eye. Instruct the patient to keep their gaze fixed on your nose or another reference point in the center. Hold your hands or an object in each of the four quadrants of the patient’s visual field: superior, inferior, temporal (side), and nasal (side). Ask the patient to tell you as soon as they see your fingers or the object you are holding in their peripheral vision. They should not look directly at the objects but should maintain their gaze on the center. Compare the patient’s responses to the objects in each quadrant of the visual field to assess whether their peripheral vision is intact. If the patient consistently misses objects in certain quadrants, it could indicate peripheral vision deficits or defects, which may warrant further evaluation by an eye care professional or a more detailed visual field test. Assessing peripheral vision is crucial for detecting conditions like glaucoma, retinal disorders, and neurological issues that can affect peripheral vision.
Validating and Documenting Data
Data should be validated if discrepancies exist between subjective and objective data , significant changes occur between visits, or highly abnormal findings are noted. To validate the data, the nurse may need to repeat the physical assessment and clarify the subjective findings with the patient. Other times, the nurse may need another nurse or healthcare professional to verify the findings.
Accurate documentation is needed to promote effective communication among the interdisciplinary team. Documenting the results of an eye assessment begins with the chief complaint or reason for the assessment, whether it is routine, due to specific symptoms, or as part of a larger medical evaluation. Document the appearance of the eyes, eyelids, eyebrows, and surrounding skin. Note any swelling, redness, lesions, or abnormalities. Describe the color, texture, and any signs of inflammation, infection, or discoloration. Note the clarity of the cornea and any irregularities. Observe the color and shape of the iris. Measure and record the size of the pupils, documenting any irregularities, such as unequal sizes. Record observations of eye movements, smoothness, coordination, and the presence of nystagmus. Record the results of visual acuity tests (Snellen chart or other methods) for both near and far vision. Document the response of the pupils to light stimuli. Note if they constrict (miosis) in bright light and dilate (mydriasis) in dim light. Describe the patient’s ability to focus on a near object and how the pupils constrict during this process. Document any additional tests that were performed, such as color vision, depth perception, or visual field tests, and also document the results.
Abnormalities of the Eye
When conducting a health assessment, it is important to be able to recognize abnormalities of the eye that impact the eyes’ structures, functions, and visual perception. From common refractive errors that blur the clarity of vision, to more complex conditions affecting the external and internal structures of the eyes, these abnormalities demand attention and timely intervention. While there are many abnormalities that may be seen during the eye assessment, this section reviews the most common abnormalities.
Abnormalities of the External Eye
Abnormalities of the external eye refer to various visual, structural, or functional deviations from the normal appearance and function of the eye’s outer structures. These abnormalities can result from various eye conditions, injuries, infections, or systemic disorders. Common abnormalities of the external eye, including conjunctivitis (pink eye), hordeolum (stye), chalazion, ptosis, ectropion, and exophthalmos can be found in Table 22.3.
| Condition | Description | Visual Example |
|---|---|---|
| Conjunctivitis (also, pink eye) | Inflammation of the conjunctiva (thin membrane covering the sclera and inner eyelids) due to infections (bacterial, viral, or allergic), causing redness, itching, discharge, and tearing | |
| Hordeolum (also, stye) | A painful, localized infection or inflammation of an eyelid gland, resulting in a red, swollen lump along the eyelid’s edge | |
| Chalazion | A noninfectious bump on the eyelid caused by a blocked meibomian gland, leading to localized swelling, tenderness, and sometimes blurred vision | |
| Ptosis | Drooping of the upper eyelid due to weakened or paralyzed levator muscles, which can partially obstruct vision | |
| Ectropion | Outward turning of the lower eyelid, leading to exposure of the inner eyelid and increased risk of dryness and irritation | |
| Exophthalmos | One or both eyeballs bulge or protrude abnormally from the eye sockets (orbits) |
The nurse is performing an assessment on a child who is being seen in the primary care office for red, itchy eyes. The nurse asks a series of questions to determine if the patient felt like they had something in their eye, experience seasonal allergies, and if there had been any discharge from the eye. The patient denied foreign bodies in the eye and allergies; however, the patient did mention the eye was crusted over with a yellow crust that morning upon waking up. The nurse analyzed the cues to hypothesize the patient was most likely experiencing conjunctivitis (pink eye).
Abnormalities of the Internal Eye
Abnormalities of the internal eye refer to various disorders and conditions that affect the structures within the eye, including the retina, optic nerve, vitreous humor, and other components. These abnormalities can lead to changes in vision, visual disturbances, and potential vision loss. Common abnormalities of the internal eye include the following:
- Pterygium: A pterygium is a growth of tissue on the conjunctiva that can extend onto the cornea, often caused by excessive ultraviolet (UV) light exposure and leading to redness, irritation, and potential visual disturbances.
- Corneal abrasions or scars: A corneal abrasion is a scratch or injury to the cornea’s surface, while a corneal scar is tissue that forms after the healing of a deeper injury. Both can cause pain, discomfort, and changes in vision.
- Cataracts: A cataract is a clouding of the eye’s natural lens, leading to gradual vision loss, glare sensitivity, and decreased color perception (Figure 22.20).
- Irregularly shaped iris: An irregularly shaped iris is observed when the colored part of the eye that surrounds the pupil has irregular contours, which can be congenital or associated with certain eye conditions.
- Mydriasis: In mydriasis , there is abnormal dilation of the pupil, often due to medication, neurological issues, or eye trauma (Figure 22.21).
- Anisocoria: In anisocoria , there are unequal pupil sizes in both eyes, which can be normal or a sign of underlying neurological or eye problems.
- Miosis: In miosis , there is abnormal constriction of the pupil, often due to medication, bright light, or neurological conditions.
- Papilledema: In papilledema , there is swelling of the optic disc at the back of the eye due to increased intracranial pressure, which can be a sign of serious conditions like intracranial tumors or head injuries.
- Glaucoma: The term glaucoma is used for a group of eye disorders characterized by increased intraocular pressure that damages the optic nerve, causing gradual peripheral vision loss and potential blindness (Figure 22.22).
- Optic atrophy: In optic atrophy , there is degeneration of the optic nerve fibers, leading to vision loss and pale optic discs on examination.
- Macular degeneration: In macular degeneration , there is gradual and progressive damage to the macula, which can lead to significant vision loss and impairment of daily activities that require clear central vision, such as reading and recognizing faces (Figure 22.23).
Visual Abnormalities
Common visual abnormalities that individuals might experience include the following:
- Myopia (nearsightedness): In myopia (nearsightedness) , distant objects appear blurry, while close objects are clear. It occurs when light focuses in front of the retina instead of directly on it.
- Hyperopia (farsightedness): In hyperopia (farsightedness) , distant objects are clearer than close ones. Light focuses behind the retina instead of directly on it.
- Astigmatism: With an astigmatism , blurred or distorted vision is due to an irregularly shaped cornea or lens, causing light to focus on multiple points rather than a single point.
- Presbyopia: In the age-related condition presbyopia , an individual has difficulty focusing on close objects due to a gradual loss of the eye’s ability to change focus.
- Color blindness: Individuals with color blindness have difficulty distinguishing certain colors, often red and green or blue and yellow, due to a genetic deficiency in color-sensitive cone cells.
- Amblyopia (lazy eye): In amblyopia (also, lazy eye) , there is reduced vision in one eye due to improper development during childhood, often caused by unequal refractive errors or strabismus (misaligned eyes).
- Strabismus: Individuals with strabismus have misalignment of the eyes, causing one eye to turn inward, outward, upward, or downward while the other eye remains focused.
- Diplopia (double vision): When an individual has diplopia (double vision), they see two images of a single object. This is often caused by misalignment of the eyes, neurological issues, or eye muscle problems.
- Floater: A floater is a small, semitransparent spot or shape that appears to “float” in the field of vision due to debris in the vitreous humor.
- Flash: A flash is a brief, flickering light or sensation that occurs in the visual field, often caused by traction on the retina or vitreous humor.
- Photophobia: An individual who has photophobia has sensitivity to light, leading to discomfort and squinting in bright environments.
- Night blindness: An individual who has night blindness has difficulty seeing in low-light conditions due to reduced function of the rod cells in the retina.
- Halo: A halo is a glowing circle or ring around a light source, often seen by individuals with cataracts or after refractive surgery.


3.3: Conservative and Non-Conservative Forces
Path-Dependence for Work
As we saw in our introduction to the work-energy theorem, the calculation of the work done by a force requires that we perform a line integral. In such an integral, the actual path that the particle takes from the start to the end is, in general, an important factor. There are some special cases, however, where the line integral calculation of the work done on a particle by a force as it moves from point A to point B results in a single, unique, value, no matter what path is taken between those points. A force that satisfies such a property is called a conservative force . Conversely, a force for which the work done on a particle depends upon the path it takes between two endpoints is called a non-conservative force .
Another way to characterize these two types of forces is to look at the work they do to a particle that follows a closed path – a journey that begins and ends at the same position in space. Suppose a particle starts at a point A , and is moved around for awhile while acted upon by a force until it finally returns to point A . If the force is conservative, then every path that brings it back around to its starting point results in the same work done by that force. One can imagine choosing paths that are shorter and shorter until finally the "path" we try is just not moving it at all. Clearly there is no work done in this case, and if it is true for one path (however trivial), it is true for all paths. The therefore conclude:
The work done by a conservative force around any closed path is equal to zero.
This fact gives us a nice mathematical trick for recognizing conservative forces. To see the source of this trick, we will (as usual) use a simple case, which we will then be able to generalize. Consider a rectangular closed path in the \(x\)-\(y\) plane with the sides of the rectangle parallel to the \(x\) and \(y\) axes:
Figure 3.3.1 – A Simple Closed Path
The work done by a force \(\overrightarrow F\left(x,y\right)=F_x\left(x,y\right)\;\hat i+F_y\left(x,y\right)\;\hat j\) for the part of the journey from \(A\) to \(B\) only includes the \(y\)-component of the force, evaluated at \(x\)-position \(x+\Delta x\):
\[W\left(A\rightarrow B\right) = \overrightarrow F\cdot \overrightarrow{\Delta l} = \left(F_x\;\hat i+F_y\;\hat j\right)\cdot\left(\Delta y \;\hat j\right)=F_y\left(x+\Delta x,y\right)\Delta y\]
The part of the journey from point \(C\) to \(D\) looks similar, except that the direction is negative of the \(A\rightarrow B\) path, and the force is evaluated at \(x\), rather than \(x+\Delta x\):
\[W\left(C\rightarrow D\right) = \overrightarrow F\cdot \overrightarrow{\Delta l} = \left(F_x\;\hat i+F_y\;\hat j\right)\cdot\left(-\Delta y \;\hat j\right)=-F_y\left(x,y\right)\Delta y\]
The work done over the \(B\rightarrow C\) and \(D\rightarrow A\) paths are found similarly, with the \(x\)'s and \(y\)'s swapped:
\[W\left(B\rightarrow C\right) = -F_x\left(x+\Delta x,y\right)\Delta x\;\;\;\;\; W\left(D\rightarrow A\right) = F_x\left(x,y\right)\Delta x\]
Now we add all of these contributions together to get the work done around the closed loop, and if the force is conservative, we set this equal to zero, giving:
\[0=\left[F_y\left(x+\Delta x,y\right)-F_y\left(x,y\right)\right]\Delta y - \left[F_x\left(x,y+\Delta y\right)-F_x\left(x,y\right)\right]\Delta x\]
Dividing both sides by \(\Delta x\Delta y\) gives:
\[0=\dfrac{F_y\left(x+\Delta x,y\right)-F_y\left(x,y\right)}{\Delta x}-\dfrac{F_x\left(x,y+\Delta y\right)-F_x\left(x,y\right)}{\Delta y}\]We have taken a bit of license here, when we did not concern ourselves with (for example) how the \(y\)-component of the force might change during the path \(A\rightarrow B.\) But we can remedy this by allowing the dimensions to shrink to zero, \(\Delta x\rightarrow 0\), \(\Delta y\rightarrow 0\). This turns the two terms into derivatives. More specifically, it turns them into partial derivatives . A partial derivative is a derivative that acts on a function of multiple variables (in this case, \(x\) and \(y\)), but only measures the rate of change of the function with respect to one of the variables, while holding the other one fixed. The symbols used for partial derivatives are slightly different than those for regular derivatives:
\[\dfrac{\partial f}{\partial x}\equiv\lim\limits_{\Delta x\rightarrow 0}\dfrac{f\left(x+\Delta x,y\right)-f\left(x,y\right)}{\Delta x}\]
Here's an example:
\[f\left(x,y\right)=x^2y^3 + 3y + 6x^3\;\;\;\Rightarrow\;\;\;\dfrac{\partial f}{\partial x} = 2xy^3+0+18x^2\;,\;\;\;\;\dfrac{\partial f}{\partial y} = 3x^2y^2+3+0\]
The condition we have for a force \(\overrightarrow F\left(x,y\right)=F_x\left(x,y\right)\;\hat i+F_y\left(x,y\right)\;\hat j\) to be conservative is therefore most compactly written as:
\[0=\dfrac{\partial F_y}{\partial x}-\dfrac{\partial F_x}{\partial y}\]
If the force has components in three dimensions, then of course this mathematical condition has to also apply to the \(y\)-\(z\) and \(x\)-\(z\) planes as well.
Digression: Curl
While it is beyond the scope of this course, a reader with more mathematical background than required for this class may recognize this condition for a conservative force as being the zero "curl" of the force vector field:
\[\text{curl of }\overrightarrow F\left(x,y,z\right)=\overrightarrow\nabla\times\overrightarrow F\equiv\left| \begin{array}{c}\hat i && \hat j && \hat k \\ \dfrac{\partial}{\partial x} && \dfrac{\partial}{\partial y} && \dfrac{\partial}{\partial z} \\ F_x && F_y && Fz \end{array}\right| = \left(\dfrac{\partial F_z}{\partial y}-\dfrac{\partial F_y}{\partial z}\right)\hat i + \left(\dfrac{\partial F_x}{\partial z}-\dfrac{\partial F_z}{\partial x}\right)\hat y + \left(\dfrac{\partial F_x}{\partial y}-\dfrac{\partial F_y}{\partial x}\right)\hat k=0\nonumber\]
Consider the following forces that act on a particle as it moves in the \(\left(x,y\right)\) plane (\(\alpha\) is a constant):
- \(\overrightarrow F\left(x,y\right)=\alpha\left(x \;\hat i-y \;\hat j\right)\)
- \(\overrightarrow F\left(x,y\right)=\alpha\left(y \;\hat i-x \;\hat j\right)\)
- Analysis
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There are a few ways we can check whether this force is conservative or not. The simplest involves taking the partial derivatives of the components of the force:
\(\overrightarrow F\left(x,y\right)=\alpha\left(x \;\hat i-y \;\hat j\right)\) :
\[\dfrac{\partial}{\partial x}F_y=\dfrac{\partial}{\partial x}\left(-\alpha y\right)=0\nonumber\]
\[\dfrac{\partial}{\partial y}F_x=\dfrac{\partial}{\partial y}\left(\alpha x\right)=0\nonumber\]
With both of these derivatives vanishing, our check confirms that this force is conservative:
\[\dfrac{\partial F_y}{\partial x}-\dfrac{\partial F_x}{\partial y}=0-0=0\nonumber\]
\(\overrightarrow F\left(x,y\right)=\alpha\left(y \;\hat i-x \;\hat j\right)\) :
\[\dfrac{\partial}{\partial x}F_y=\dfrac{\partial}{\partial x}\left(-\alpha x\right)=-\alpha\nonumber\]
\[\dfrac{\partial}{\partial y}F_x=\dfrac{\partial}{\partial y}\left(\alpha y\right)=\alpha \nonumber\]
For this case, our check shows that the force is non-conservative:
\[\dfrac{\partial F_y}{\partial x}-\dfrac{\partial F_x}{\partial y}=-\alpha-\alpha=-2\alpha\ne 0\nonumber\]
Another method would be to perform a work integral between two points over different paths. Unfortunately, this method will only work as a proof that the force is non-conservative if the two integrals come out different. If the two integrals come out the same, it does not prove that the force is conservative, since those two paths could coincidentally result in the same work for a non-conservative force. As we cannot perform the work integral for all paths, we can't use that method to conclusively prove that the force is conservative.
Examples of Conservative Forces
Let's look at a few important examples of conservative forces.
central forces
The first is very fundamental to all of physics, and it goes like this:
Any force on a particle that originates from a single point is conservative.
This can be proven mathematically with the tools we have, but we will refrain from doing so here. The way this is usually expressed by physicists is, " central forces are conservative ," where a central force is one that originates from a single point (the "center"). Reminding ourselves that forces are interactions between two objects, a force that "originates" from a single point is one that is exerted by an object located at that single point. Put another way, this is a force from another particle! We labeled forces that act between individual particles as "fundamental forces," so we therefore conclude that all fundamental forces are conservative .
Digression: Spherical Sources
While all forces that have point particles as a source are certainly central forces, the converse is not true. It is possible to have a central force whose source is a collection of particles, if those particles are placed very symmetrically – in the shape of a sphere. This fact will become important later when we get to gravitation, because stars and planets very closely approximate spheres, which means we can treat the gravitational forces they exert as conservative to a very good approximation.
This means that to the extent that we see non-conservative forces in nature, the source of its non-conservative nature must be that the forces are exerted by objects that are collections of particles rather than by individual particles. This is not to say that the composite forces we have discussed (gravity, elastic, drag, etc.) are inherently non-conservative. It just means that they require just the right conditions to be "considered" conservative to a good approximation.
Alert
Resist the temptation to label a certain type of force as either conservative or non-conservative. One cannot tell whether a type (gravity, tension, etc.) is conservative or not without more details of how this type of force is acting. It is perhaps better not to say that a given force is conservative, and instead say that the force is "being applied in a conservative manner." But this is not the language you will find elsewhere, so we will not use it here.
gravity
As a composite force, we have approximated gravity as a constant force at all points in space – we assume that the region involved is small compared to the size of the Earth. Subject to the limits of this approximation, we can declare gravity to be conservative:
\[\overrightarrow F_{grav}=-mg\;\hat j\;\;\;\Rightarrow\;\;\; \dfrac{\partial F_y}{\partial x}-\dfrac{\partial F_x}{\partial y}=\dfrac{\partial}{\partial x}\left(mg\right)-\dfrac{\partial}{\partial y}\left(0\right)=0\]
It is enlightening to look at how much work is actually done by gravity as a particle moves around. The work done by gravity on a particle that moves from point \(A\) to point \(B\) (near the surface of the Earth, of course) is given by:
\[W\left(A\rightarrow B\right) = \int\limits_{A\rightarrow B}\overrightarrow F_{grav}\cdot\overrightarrow{dl}\]
As usual, we have that \(\overrightarrow{dl}=dx\;\hat i+dy\;\hat j\), but as we will see, we will be able to determine the work done without specifying a path from \(A\) to \(B\) by expressing \(y\) as a function of \(x\):
The work done by gravity depends only upon the change in height of the particle. It is important to note a few things about this result:
- It doesn't depend upon the path taken from the starting height to the ending height – it could go straight up, or take many loop-de-loops that go above and below the starting and ending points, and the result is the same.
- The actual value of the starting and ending heights are not relevant, it is only the difference of the starting and ending heights that matters.
- According to the work-energy theorem, if gravity is the only force that does work on a particle (or if other forces that might be acting happen to cancel out), then the change of kinetic energy of a particle (and therefore its speed) can be calculated using nothing more than the change in the particle's height.
Exercise
A particle follows projectile motion free of air resistance. According to the work-energy theorem, its change in kinetic energy equals the work done on it by the net force, which in this case is only gravity, so:
\[W_{grav}=\Delta KE\;\;\;\Rightarrow\;\;\;-mg\Delta y=\Delta\left(\frac{1}{2}mv^2\right)\nonumber\]
Confirm that this is true using tools we have from kinematics.
- Solution
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We will assume that the particle travels from point A to point B. For projectile motion, the \(x\)-component of velocity doesn't change, which means that:
\[v_{Ax}=v_{Bx}\nonumber\]
The \(y\)-component does change, and since we know nothing about how long the particle is in the air, we will use the " no time equation " for vertical motion accelerated by gravity (\(a_y=-g\)):
\[v_{By}^2-v_{Ay}^2=2\left(-g\right)\left(y_B-y_A\right)\nonumber\]
Using the result of the first equation, we can add \(0=v_{Ax}^2-v_{Bx}^2\) to the left side of this equation without changing it, giving
\[\left(v_{Bx}^2+v_{By}^2\right)-\left(v_{Ax}^2+v_{Ay}^2\right)=-2g\left(y_B-y_A\right)\nonumber\]
The quantities in parentheses are just the squares of the speeds at points \(B\) and \(A\), respectively, so:
\[v_B^2-v_A^2=2g\left(y_B-y_A\right)\;\;\;\Rightarrow\;\;\;\Delta \left(v^2\right) = -2g\Delta y\nonumber\]
Now multiplying both sides by the mass of the particle and dividing both sides by 2 reconstructs the work-energy theorem result given above.
A small block slides along a frictionless, horizontal surface into a frictionless vertical half-circle track, and it remains in contact with the track, until at least the \(\theta=90^o\) point (with \(\theta\) defined in the diagram).
- Analysis
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With no friction present, the only forces acting on the block are gravity and the normal force. We can ask how much work each of these forces do on the block. We assume the gravitational force is a constant \(mg\) downward, so from our efforts above, we conclude that the work done by gravity is:
\[W_{grav}=-mg\Delta y\nonumber\]
If we call the radius of the circle \(R\), then we can express the change in the height in terms of \(R\) and \(\theta\):
\[W_{grav}=-mgR\left[1+\sin\left(\theta-90^o\right)\right]=-mgR\left[1-\cos\theta\right]\nonumber\]
The work done by the normal force might seem complicated at first, since on the curved track it is constantly changing direction, but every infinitesimal displacement by the block \(\overrightarrow {dl}\) is parallel to the track, while the normal force is always perpendicular to the track. Each contribution to the work done is \(dW=\overrightarrow N\cdot\overrightarrow {dl}\), and since the normal force and displacement are perpendicular to each other at all times the work done by this force is zero.
Now that we know the total work done by all forces, we can apply the work-energy theorem to get the speed of the block after it climbs the angle \(\theta\):
\[\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2=-mgR\left(1-\cos\theta\right)\;\;\;\Rightarrow\;\;\;v_f=\sqrt{v_o^2-2gR\left(1-\cos\theta\right)}\nonumber\]
There is one other thing that we can extract from this analysis. We know that the block is still in contact with the track when it reaches the angle \(\theta\). It has been slowing down as it gets higher, so the centripetal acceleration \(\dfrac{v^2}{R}\) that keeps it going in a circle has gotten smaller, but it is not zero. The normal force combined with a component of the gravity force is what is maintaining this acceleration, and as it goes higher from its current position, the component of the gravity force acting toward the center of the circle will get larger. So this means that the decaying centripetal acceleration is due to a decreasing normal force. The normal force can only go to zero (it can't become negative), and if it does, then the block will start to lose contact with the track. We can therefore determine the minimum speed the block must have at angle \(\theta\) to maintain contact with the track (anything less, and it falls off). The gravity force can be broken into components radial (toward the center of the circle) and tangential to the track (use a FBD and geometry to determine these components), and setting the normal force equal to zero gives:
\[N-mg\cos\theta=m\dfrac{v^2}{R}\;\;\;\Rightarrow\;\;\;v_{min}=\sqrt{-gR\cos\theta}\nonumber\]
[Note that the minus sign is present because angle being used (the cosine of \(\theta>90^o\) is negative).]
elastic
In one way, the elastic force is an even easier case than that of gravity – the particle's displacement is entirely confined to one dimension. But it does include the twist that unlike gravity, it does not maintain a constant magnitude. Let's define the one dimension we are working in to be the \(x\)-axis. Let's further assume that the particle that is subject to the elastic force experiences zero force when it is located at \(x=x_o\). That is, when the particle is here, then the spring is at its equilibrium length. Then according to Hooke's law, the force on the particle is:
\[\overrightarrow F_{elastic} =-k\left(x-x_o\right)\hat i\]
When the particle is located in the region \(x>x_o\) it experiences a force in the \(-x\)-direction (back toward \(x_o\)), and when it is in the region \(x<x_o\), it experiences a force in the \(+x\)-direction (also back toward \(x_o\)). So this is the restoring force we expect for the elastic force. The partial derivative check once again demonstrates that this is a conservative force. As with the case of gravity, we can compute the work done by the spring on the particle. All the displacements in this case are confined to the \(x\)-axis, so we have simply that \(\overrightarrow{dl}=dx\;\hat i\), giving:
\[W\left(A\rightarrow B\right) = \int\limits_{A\rightarrow B}\overrightarrow F_{elastic}\cdot\overrightarrow{dl}= \int\limits_A^B\left[-k\left(x-x_o\right)\hat i\right]\cdot\left(dx\;\hat i \right)=\left[-\frac{1}{2}kx^2+kx_ox\right]_{x_A}^{x_B}=-\frac{1}{2}k\Delta\left(x^2\right)+kx_o\Delta x\]
We can simplify the look of this result a bit by defining the equilibrium point of the spring to be the origin \(x_o=0\), giving:
\[W\left(A\rightarrow B\right) = -\frac{1}{2}k\Delta \left(x^2\right)\]
We see that this result differs from that of gravity, in that the work done by gravity is proportional to how much the (vertical) position changes, while the work done by the elastic force is proportional to how much the square of the position changes. So if one particle changes its height by twice as much as another particle, gravity does twice as much work on it. But if a particle changes its position from the equilibrium point twice as much as another particle, the spring does four times as much work on it.
We have been careful to define all of these conservative forces in terms of the displacement of a particle, rather than an "object." But just as we saw in earlier sections, we can extend these results to non-rotating objects whose constituent particles remain rigidly in place within the object. If they do not, then an object that follows a closed path may end up with a different internal energy, which can have an effect on the amount of kinetic energy it has. And if the kinetic energy changes around a closed path, then the work done around that closed path is not zero, and the force on that object is not conservative.
A block is attached to a vertical spring, the other end of which is attached to the ceiling. The block is held stationary at a height where the spring is at its equilibrium length. The block is then released.
- Analysis
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There are two forces acting on the block when it is released. One is the force of gravity downward, and the other is the elastic force of the spring upward. Both of these conservative forces do work at the same time on the block, and since the block moves downward, the gravity force (which acts in the direction of motion) does positive work, while the elastic force (which acts opposite to the direction of motion does negative work.
One thing we can determine, if given all the numbers, is how far the block will fall before stopping and bouncing back up. If we are careless, we might conclude too quickly that this must be where the two forces cancel each other out – but this is wrong! As the block is falling, when it reaches the point where the forces cancel, then the net force is zero, which means it stops accelerating . This does not mean it stops moving.
The way we find the distance it falls before stopping is to use the work-energy theorem. The total work done by all forces equals the change in kinetic energy. The kinetic energy starts at zero, and we are interested in where the block once again has zero kinetic energy – when it stops falling. As stated above, gravity does positive work as the block falls, and the spring does negative work, so we are looking for the distance where these cancel out to give zero total work. Calling the stretch of the spring from equilibrium at the point where the block stops \(+\Delta y\), then the change of the block's height is \(-\Delta y\), and we can use the work results for the gravity and elastic forces above to get:
\[0=W_{tot}=W_{spring}+W_{grav}=\left[-\frac{1}{2}k\left(\Delta y\right)^2\right]+\left[-mg\left(-\Delta y\right)\right]\;\;\;\Rightarrow\;\;\;\Delta y = \frac{2mg}{k}\nonumber\]
The solution also included the possibility that \(\Delta y=0\) – naturally no total work is done if the block never moves, or (since the forces are conservative) if it comes back to the position where it started.
Also notice that if the block was hung stationary from the spring, then the forces would balance, giving \(k\Delta y = mg\). The value of \(\Delta y\) in this case is half as much as that found above. So the zero-net-force position is exactly halfway between the maximum and minimum heights of the block.
Now for Some Non-Conservative Forces
Next we will discuss a couple of non-conservative forces. The first is fairly obvious, but the second may be surprising.
kinetic friction
The kinetic friction force is quite unlike the others we have discussed. It's magnitude is simple enough – if the normal force and the surfaces remain unchanged, then the kinetic friction is constant – but the direction is quite another matter. The kinetic friction force on one object always acts in the direction opposite to the direction of motion of that object relative to the surface it is rubbing against. Consider a hockey puck sliding on a horizontal table. If it is sliding north when it is located at point \(\left(x,y\right)\), then the friction force points south. If, at a later time, it is sliding east at that same point \(\left(x,y\right)\), then the direction of the friction force points west. What this tells us is that the direction of the force cannot be determined from the location alone – one must know the direction of motion.
If we now look back at our partial derivative formula for determining whether a force is conservative, it requires that we know the \(x\) and \(y\) components of the force as a function of the position \(\left(x,y\right)\). We cannot get this information in the case of kinetic friction. The partial derivative equation is both a necessary and sufficient condition for the force to be conservative, so when it is unusable for a given force, that force is necessarily non-conservative.
Perhaps this is an unsatisfying explanation. Fair enough, let's do a more rigorous analysis of the case of kinetic friction for a specific case:
A puck is slid along a horizontal rough surface in a straight line along the diameter of a circle (the gravity, contact and friction forces are the only forces on the puck). The same puck is then slid on the same surface starting at the same speed along the circle defined by the diameter indicated in the first experiment (it slides around the inside surface of a frictionless circular wall). The figure shown depicts a top view of these two paths.
- Solution
-
The kinetic friction force always acts in the direction directly opposite to the motion. So for every small displacement \(\overrightarrow{dl}\) along the path taken, the direction of the friction force is always the same as the direction of \(-\overrightarrow{dl}\). When we compute the work done by friction, we therefore have:
\[dW=\overrightarrow f_k\cdot\overrightarrow{dl}=\left|\overrightarrow f_k\right|\left|\overrightarrow{dl}\right|\cos180^o=-f_kdl\nonumber\]
The magnitude of the kinetic friction force is constant, as it is proportional to the unchanging normal force (the surface remains horizontal). So the line integral can pull the constant \(f_k\) out of the integral, leaving:
\[W\left(A\rightarrow B\right)=-\int\limits_A^B f_kdl=f_k\int\limits_A^B dl = -f_k\left(\text{length of path}\right)\nonumber\]
The length of the path that follows a half-circle path (#2) is longer than the straight-line path (#1) by a factor of \(\frac{\pi}{2}\), so the work done on the puck by friction is greater (more negative) for path #2 than for path #1 by that factor.
Notice that both paths begin and end at the same points, but the work done is different for the two paths, demonstrating that kinetic friction is non-conservative.
One might ask, "The frictional force between the particles on the surfaces is ultimately electric, which is a fundamental force, and therefore conservative, so what is the physical source of this force being non-conservative?" The answer is that by only interacting at their surfaces, some particles in the rubbing objects are accelerated while others are not. This results in changes in the internal energies of the rubbing objects. In the analysis example above, path #1 results in less energy being transferred into internal energy than path #2 (more surface particles are affected in the longer path). With the internal energies changed by different amounts, the kinetic energies change by different amounts. This can only occur if different amounts of work are performed, and since the two paths begin and end at the same place, and by definition this means that the force is non-conservative.
gravity
Wait... gravity? Didn't we just learn that gravity is a conservative force? Gravitation as a force between individual particles is conservative as all fundamental forces are, and gravity near the surface of the Earth is approximately constant, making it approximately conservative, but here we are going to look at an example where this force does not behave as a conservative force.
The Voyager probes were launched in 1977. Neither was given sufficient kinetic energy to escape the Sun's gravitational pull without "help." If we approximate a probe and the sun as individual particles (when they are far apart, this is reasonable), then when the gravitational pull of the Sun brings the probe back to the point where it began, no net work will be done, and the probe's kinetic energy goes back to its starting value. Now let's introduce a third particle – Jupiter. We will declare the Sun and Jupiter to be a two-particle object (we'll call it the "solar system"), and examine what can happen to the individual Voyager "particle" when it interacts with this system.
Some very bright minds at NASA launched the Voyager probes so that they would get gravitational "kicks" from flying close to planets, the result of which is increased speed for the Voyager probes that is sufficient to escape the Sun's gravitational pull. Now suppose NASA made a small miscalculation, and sent the probe (with its extra gravitational kick) back toward the Sun, instead of away from it. Then when it returns to the point where it started, it is moving faster than when it was launched. This means that the starting and ending kinetic energies at the same point in space are not equal, and the force on the probe by the solar system is not conservative. [ If you are curious about how this gravitational "kick" is accomplished, see the simplified explanation at the end of Section 7.3 . ]
As with the case of friction, this result comes about because the force on the particle comes from different parts of the system acting independently, and those different parts moving relative to each other (as Jupiter moves around the Sun). The Sun and Jupiter "particles" each exert their own conservative gravitational force on the probe, and the aggregate of these forces comes out to be non-conservative.


7.4: Reading- Strain
As we’ve just learned, the earth’s crust is constantly subjected to forces that push, pull, or twist it. These forces are called stress. In response to stress, the rocks of the earth undergo strain , also known as deformation.
Strain is any change in volume or shape.There are four general types of stress. One type of stress is uniform, which means the force applies equally on all sides of a body of rock. The other three types of stress, tension, compression and shear, are non-uniform, or directed, stresses.All rocks in the earth experience a uniform stress at all times. This uniform stress is called lithostatic pressure and it comes from the weight of rock above a given point in the earth. Lithostatic pressure is also called hydrostatic pressure. (Included in lithostatic pressure are the weight of the atmosphere and, if beneath an ocean or lake, the weight of the column of water above that point in the earth. However, compared to the pressure caused by the weight of rocks above, the amount of pressure due to the weight of water and air above a rock is negligible, except at the earth’s surface.) The only way for lithostatic pressure on a rock to change is for the rock’s depth within the earth to change.Because lithostatic pressure is a uniform stress, a change in lithostatic pressure does not cause fracturing and slippage along faults. Nevertheless, it may be the cause of certain types of earthquakes. In subducting tectonic plates, the increased pressure of greater depth within the earth may cause the minerals in the plate to metamorphose spontaneously into a new set of denser minerals that are stable at the higher pressure. This is thought to be the likely cause of certain types of deep earthquakes in subduction zones, including the deepest earthquakes ever recorded.
Rocks are also subjected to the three types of directed (non-uniform) stress – tension, compression, and shear.
- Tension is a directed (non-uniform) stress that pulls rock apart in opposite directions. The tensional (also called extensional) forces pull away from each other.
- Compression is a directed (non-uniform) stress that pushes rocks together. The compressional forces push towards each other.
- Shear is a directed (non-uniform) stress that pushes one side of a body of rock in one direction, and the opposite side of the body of rock in the opposite direction. The shear forces are pushing in opposite ways.
In response to stress, rock may undergo three different types of strain – elastic strain, ductile strain, or fracture.
- Elastic strain is reversible. Rock that has undergone only elastic strain will go back to its original shape if the stress is released.
- Ductile strain is irreversible. A rock that has undergone ductile strain will remain deformed even if the stress stops. Another term for ductile strain is plastic deformation.
- Fracture is also called rupture. A rock that has ruptured has abruptly broken into distinct pieces. If the pieces are offset—shifted in opposite directions from each other—the fracture is a fault.
Ductile and Brittle Strain
Earth’s rocks are composed of a variety of minerals and exist in a variety of conditions. In different situations, rocks may act either as ductile materials that are able to undergo an extensive amount of ductile strain in response to stress, or as brittle materials, which will only undergo a little or no ductile strain before they fracture. The factors that determine whether a rock is ductile or brittle include:
- Composition—Some minerals, such as quartz, tend to be brittle and are thus more likely to break under stress. Other minerals, such as calcite, clay, and mica, tend to be ductile and can undergo much plastic deformation. In addition, the presence of water in rock tends to make it more ductile and less brittle.
- Temperature—Rocks become softer (more ductile) at higher temperature. Rocks at mantle and core temperatures are ductile and will not fracture under the stresses that occur deep within the earth. The crust, and to some extent the lithosphere, are cold enough to fracture if the stress is high enough.
- Lithostatic pressure—The deeper in the earth a rock is, the higher the lithostatic pressure it is subjected to. High lithostatic pressure reduces the possibility of fracture because the high pressure closes fractures before they can form or spread. The high lithostatic pressures of the earth’s sub-lithospheric mantle and solid inner core, along with the high temperatures, are why there are no earthquakes deep in the earth.
- Strain rate—The faster a rock is being strained, the greater its chance of fracturing. Even brittle rocks and minerals, such as quartz, or a layer of cold basalt at the earth’s surface, can undergo ductile deformation if the strain rate is slow enough.
Most earthquakes occur in the earth’s crust. A smaller number of earthquakes occur in the uppermost mantle (to about 700 km deep) where subduction is taking place. Rocks in the deeper parts of the earth do not undergo fracturing and do not produce earthquakes because the temperatures and pressures there are high enough to make all strain ductile. No earthquakes originate from below the the earth’s upper mantle.
Stress and Fault Types
The following correlations can be made between types of stress in the earth, and the type of fault that is likely to result:
- Tension leads to normal faults.
- Compression leads to reverse or thrust faults.
- Horizontal shear leads to strike-slip faults.
Correlations between type of stress and type of fault can have exceptions. For example, zones of horizontal stress will likely have strike-slip faults as the predominant fault type. However there may be active normal and thrust faults in such zones as well, particularly where there are bends or gaps in the major strike-slip faults.
To give another example, in a region of compression stress in the crust, where sheets of rock are stacked on active thrust faults, strike-slip faults commonly connect some of the thrust faults together.
Contributors and Attributions
- Basics -- Earthquakes. Authored by : Ralph L. Dawes and Cheryl D. Dawes. Provided by : Wenatchee Valley College. Located at : http://commons.wvc.edu/rdawes/G101OCL/Basics/earthquakes.html . Project : Geology 101 - Introduction to Physical Geology. License : CC BY: Attribution


2.4.9: Derivability and Consistency
We will now establish a number of properties of the derivability relation. They are independently interesting, but each will play a role in the proof of the completeness theorem.
If \(\Gamma \Proves A\) and \(\Gamma \cup \{A\}\) is inconsistent, then \(\Gamma\) is inconsistent.
Proof. There are finite \(\Gamma_0\) and \(\Gamma_1 \subseteq \Gamma\) such that \(\Log{LK}\) derives \(\Gamma_0 \Sequent A\) and \(A, \Gamma_1 \Sequent \quad\) . Let the \(\Log{LK}\) -derivation of \(\Gamma_0 \Sequent A\) be \(\pi_0\) and the \(\Log{LK}\) -derivation of \(\Gamma_1, A \Sequent \quad\) be \(\pi_1\) . We can then derive
Since \(\Gamma_0 \subseteq \Gamma\) and \(\Gamma_1 \subseteq \Gamma\) , \(\Gamma_0 \cup \Gamma_1 \subseteq \Gamma\) , hence \(\Gamma\) is inconsistent. ◻
\(\Gamma \Proves A\) iff \(\Gamma \cup \{\lnot A\}\) is inconsistent.
Proof. First suppose \(\Gamma \Proves A\) , i.e., there is a derivation \(\pi_0\) of \(\Gamma \Sequent A\) . By adding a \(\LeftR{\lnot}\) rule, we obtain a derivation of \(\lnot A, \Gamma \Sequent \quad\) , i.e., \(\Gamma \cup \{\lnot A\}\) is inconsistent.
If \(\Gamma \cup \{\lnot A\}\) is inconsistent, there is a derivation \(\pi_1\) of \(\lnot A, \Gamma \Sequent \quad\) . The following is a derivation of \(\Gamma \Sequent A\) :
◻
Prove that \(\Gamma \Proves \lnot A\) iff \(\Gamma \cup \{A\}\) is inconsistent.
If \(\Gamma \Proves A\) and \(\lnot A \in \Gamma\) , then \(\Gamma\) is inconsistent.
Proof. Suppose \(\Gamma \Proves A\) and \(\lnot A \in \Gamma\) . Then there is a derivation \(\pi\) of a sequent \(\Gamma_0 \Sequent A\) . The sequent \(\lnot A, \Gamma_0 \Sequent \quad\) is also derivable:
Since \(\lnot A \in \Gamma\) and \(\Gamma_0 \subseteq \Gamma\) , this shows that \(\Gamma\) is inconsistent. ◻
If \(\Gamma \cup \{A\}\) and \(\Gamma \cup \{\lnot A\}\) are both inconsistent, then \(\Gamma\) is inconsistent.
Proof. There are finite sets \(\Gamma_0 \subseteq \Gamma\) and \(\Gamma_1 \subseteq \Gamma\) and \(\Log{LK}\) -derivations \(\pi_0\) and \(\pi_1\) of \(A, \Gamma_0 \Sequent \quad\) and \(\lnot A, \Gamma_1 \Sequent \quad\) , respectively. We can then derive
Since \(\Gamma_0 \subseteq \Gamma\) and \(\Gamma_1 \subseteq \Gamma\) , \(\Gamma_0 \cup \Gamma_1 \subseteq \Gamma\) . Hence \(\Gamma\) is inconsistent. ◻


2: Conjugation, Resonance, and Aromaticity
After Chapter 2, you should be able to
- identify conjugated and non-conjugated bonding in molecules
- use the information presented in this chapter, along with material from earlier chapters, to solve problems, particularly road-map problems and those requiring an understanding of spectroscopy.
- explain the concept of aromaticity and the stability of aromatic compounds.
- define, and use in context, the key terms introduced.
In Chapter 2, we describe conjugation within molecules and how it relates to electronic communication. Lewis' method for drawing organic molecules does not always accurately predict some chemical and/or physical phenomenon. Resonance theory is a way to represent molecules as a combination of multiple Lewis structures - the hybrid of which best describes the chemical and physical properties of the molecule. Molecules that are considered aromatic 1) are conjugated, 2) can be represented using a combination of Lewis structures, and 3) have chemical properties that make them uniquely stable.
-
- 2.3: Resonance
- Resonance structures are a set of two or more Lewis Structures that collectively describe the electronic bonding a single polyatomic species including fractional bonds and fractional charges. Resonance structure are capable of describing delocalized electrons that cannot be expressed by a single Lewis formula with an integer number of covalent bonds.
-
- 2.4: Rules for Resonance Forms
- The above resonance structures show that the electrons are delocalized within the molecule and through this process the molecule gains extra stability. Ozone with both of its opposite formal charges creates a neutral molecule and through resonance it is a stable molecule. The extra electron that created the negative charge on one terminal oxygen can be delocalized by resonance through the other terminal oxygen.
-
- 2.5: Drawing Resonance Forms
- Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms. The net sum of valid resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. A molecule that has several resonance structures is more stable than one with fewer.
-
- 2.7: Aromaticity and the Hückel 4n + 2 Rule
- In 1931, German chemist and physicist Erich Hückel proposed a theory to help determine if a planar ring molecule would have aromatic properties. His rule states that if a cyclic, planar molecule has 4n+2 π electrons, it is considered aromatic. This rule would come to be known as Hückel's Rule.


3.1: Prelude to Linear Regression and Correlation
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CHAPTER OBJECTIVES
By the end of this chapter, the student should be able to:
Professionals often want to know how two or more numeric variables are related. For example, is there a relationship between the grade on the second math exam a student takes and the grade on the final exam? If there is a relationship, what is the relationship and how strong is it? In another example, your income may be determined by your education, your profession, your years of experience, and your ability. The amount you pay a repair person for labor is often determined by an initial amount plus an hourly fee.
Figure \(\PageIndex{1}\): Linear regression and correlation can help you determine if an auto mechanic’s salary is related to his work experience. (credit: Joshua Rothhaas)The type of data described in the examples is bivariate data — "bi" for two variables. In reality, statisticians use multivariate data, meaning many variables. In this chapter, you will be studying the simplest form of regression, "linear regression" with one independent variable (\(x\)). This involves data that fits a line in two dimensions. You will also study correlation which measures how strong the relationship is.
Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>.


8.2: Another look at distributing apples or folders
A recurring problem so far in this book has been to consider problems that ask about distributing indistinguishable objects (say apples) to distinct entities (say children). We started in Chapter 2 by asking how many ways there were to distribute 40 apples to 5 children so that each child is guaranteed to get at least one apple and saw that the answer was \(C(39,4)\). We even saw how to restrict the situation so that one of the children was limited and could receive at most 10 apples. In Chapter 7, we learned how to extend the restrictions so that more than one child had restrictions on the number of apples allowed by taking advantage of the Principle of Inclusion-Exclusion . Before moving on to see how generating functions can allow us to get even more creative with our restrictions, let's take a moment to see how generating functions would allow us to solve the most basic problem at hand.
We already know that the number of ways to distribute \(n\) apples to 5 children so that each child gets at least one apple is \(C(n−1,4)\), but it will be instructive to see how we can derive this result using generating functions. Let's start with an even simpler problem: how many ways are there to distribute n apples to one child so that each child receives at least one apple? Well, this isn't too hard, there's only one way to do it—give all the apples to the lucky kid! Thus the sequence that enumerates the number of ways to do this is \(\{a_n:n \geq 1\}\) with \(a_n=1\) for all \(n \geq 1\). Then the generating function for this sequence is
\(x+x^2+x^3+ \cdot \cdot \cdot = x(1+x+x^2+x^3+ \cdot \cdot \cdot) = \dfrac{x}{1-x}\)
How can we get from this fact to the question of five children? Notice what happens when we multiply
\((x+x^2+\cdot \cdot \cdot)(x+x^2+ \cdot \cdot \cdot)(x+x^2+ \cdot \cdot \cdot)(x+x^2+\cdot \cdot \cdot)(x+x^2+\cdot \cdot \cdot)\).
To see what this product represents, first consider how many ways can we get an \(x^6\)? We could use the \(x^2\) from the first factor and \(x\) from each of the other four, or \(x^2\) from the second factor and \(x\) from each of the other four, etc., meaning that the coefficient on \(x^6\) is \(5=C(5,4)\). More generally, what's the coefficient on \(x^n\) in the product? In the expansion, we get an \(x^n\) for every product of the form \(x^{k_1}x^{k_2}x^{k_3}x^{k_4}x^{k_5}\) where \(k_1+k_2+k_3+k_4+k_5=n\). Returning to the general question here, we're really dealing with distributing \(n\) apples to 5 children, and since \(k_i>0\) for \(i=1,2,…,5\), we also have the guarantee that each child receives at least one apple, so the product of the generating function for one child gives the generating function for five children.
Let's pretend for a minute that we didn't know that the coefficients must be \(C(n−1,4)\). How could we figure out the coefficients just from the generating function? The generating function we're interested in is \(x^5/(1−x)^5\), which you should be able to pretty quickly see satisfies
\(\dfrac{x^5}{(1-x)^5} = \dfrac{x^5}{4!} \dfrac{d^4}{dx^4}(\dfrac{1}{1-x}) = \dfrac{x^5}{4!} \displaystyle \sum_{n=0}^{\infty} n(n-1)(n-2)(n-3)x^{n-4}\)
\(= \displaystyle \sum_{n=0}^{\infty} \dfrac{n(n-1)(n-2)(n-3)}{4!} x^{n+1} = \sum_{n=0}^{\infty} \dbinom{n}{4} x^{n+1}\)
The coefficient on \(x^n\) in this series \(C(n−1,4)\), just as we expected.
We could revisit an example from Chapter 7 to see that if we wanted to limit a child to receive at most 4 apples, we would use \((x+x^2+x^3+x^4)\) as its generating function instead of \(x/(1−x)\), but rather than belabor that here, let's try something a bit more exotic.
A grocery store is preparing holiday fruit baskets for sale. Each fruit basket will have 20 pieces of fruit in it, chosen from apples, pears, oranges, and grapefruit. How many different ways can such a basket be prepared if there must be at least one apple in a basket, a basket cannot contain more than three pears, and the number of oranges must be a multiple of four?
- Solution
-
In order to get at the number of baskets consisting of 20 pieces of fruit, let's solve the more general problem where each basket has \(n\) pieces of fruit. Our method is simple: find the generating function for how to do this with each type of fruit individually and then multiply them. As in the previous example, the product will contain the term \(x^n\) for every way of assembling a basket of \(n\) pieces of fruit subject to our restrictions. The apple generating function is \(x/(1−x)\), since we only want positive powers of \(x\) (corresponding to ensuring at least one apple). The generating function for pears is \((1+x+x^2+x^3)\), since we can have only zero, one, two, or three pears in basket. For oranges we have \(1/(1−x^4)=1+x^4+x^8+ \cdot \cdot \cdot \), and the unrestricted grapefruit give us a factor of \(1/(1−x)\). Multiplying, we have
\(\dfrac{x}{1-x}(1+x+x^2+x^3) \dfrac{1}{1-x^4} \dfrac{1}{1-x} = \dfrac{x}{(1-x)^2(1-x^4)}(1+x+x^2+x^3)\)
Now we want to make use of the fact that \((1+x+x^2+x^3)=(1−x^4)/(1−x)\) (by (8.1.1) ) to see that our generating function is
\(\dfrac{x}{(1-x)^3} = \dfrac{x}{2} \displaystyle \sum_{n=0}^{\infty} n(n-1)x^{n-2} = \sum_{n=0}^{\infty} \dfrac{n(n-1)}{2}x^{n-1}\)
\(= \displaystyle \sum_{n=0}^{\infty} \dbinom{n}{2}x^{n-1} = \sum_{n=0}^{\infty} \dbinom{n+1}{2}x^n\).
Thus, there are \(C(n+1,2)\) possible fruit baskets containing \(n\) pieces of fruit, meaning that the answer to the question we originally asked is \(C(21,2)=210\).
The compact form of the solution to Example 8.5 suggests that perhaps there is a way to come up with this answer without the use of generating functions. Thinking about such an approach would be a good way to solidify your understanding of a variety of the enumerative topics we have already covered.
Find the number of integer solutions to the equation
\(x_1+x_2+x_3=n\)
(\(n \geq 0\) an integer) with \(x_1 \geq 0\) even, \(x_2 \geq 0\), and \(0 \leq x_3 \leq 2\).
- Solution
-
Again, we want to look at the generating function we would have if each variable existed individually and take their product. For \(x_1\), we get a factor of \(1/(1−x^2)\); for \(x_2\), we have \(1/(1−x)\); and for \(x_3\) our factor is \((1+x+x^2)\). Therefore, the generating function for the number of solutions to the equation above is
\(\dfrac{1+x+x^2}{(1-x)(1-x^2)} = \dfrac{1+x+x^2}{(1+x)(1-x)^2}\).
In calculus, when we wanted to integrate a rational function of this form, we would use the method of partial fractions to write it as a sum of “simpler” rational functions whose antiderivatives we recognized. Here, our technique is the same, as we can readily recognize the formal power series for many rational functions. Our goal is to write
\(\dfrac{1+x+x^2}{(1+x)(1-x)^2} = \dfrac{A}{1+x} + \dfrac{B}{1-x} + \dfrac{C}{(1-x)^2}\)
for appropriate constants, \(A, B\), and \(C\). To find the constants, we clear the denominators, giving
\(1+x+x^2 = A(1-x)^2 + B(1-x^2) + C(1+x)\).
Equating coefficients on terms of equal degree, we have:
\(1=A+B+C\)
\(1=-2A+C\)
\(1=A-B\)
Solving the system, we find \(A=1/4\), \(B=−3/4\), and \(C=3/2\). Therefore, our generating function is
\(\dfrac{1}{4} \dfrac{1}{1+x} - \dfrac{3}{4} \dfrac{1}{1-x} + \dfrac{3}{2} \dfrac{1}{(1-x)^2}\)
\(=\dfrac{1}{4} \displaystyle \sum_{n=0}^{\infty} (-1)^n x^n - \dfrac{3}{4} \sum_{n=0}^{\infty} x^n + \dfrac{3}{2} \sum_{n=0}^{\infty} nx^{n-1}\)
The solution to our question is thus the coefficient on xn in the above generating function, which is
\(\dfrac{(-1)^n}{4} - \dfrac{3}{4} + \dfrac{3(n+1)}{2}\),
a surprising answer that would not be too easy to come up with via other methods!
The invocation of partial fractions in
Example 8.6
is powerful, but solving the necessary system of equations and then hoping that the resulting formal power series have expansions we immediately recognize can be a challenge. If
Example 8.6
had not asked about the general case with n on the right-hand side of the equation but instead asked specifically about ,n=30, you might be wondering if it would just be faster to write some Python code to generate all the solutions or more interesting to huddle up and devise some clever strategy to count them. Fortunately, technology can help us out when working with generating functions. In SageMath, we can use the
series()
method to get the power series expansion of a given function. The two arguments to
series
are the variable and the degree of the terms you want to truncate. In the cell below, we ask SageMath to expand the generating function from
Example 8.6
by giving us all the terms of degree at most 30 and then collapsing the rest of the series into its form of big-Oh notation, which we discard by storing the output from
series()
in a polynomial
f(x)
.
\\ SageMath Code
If all we really want is the coefficient on a specific term, we can use the
list()
method to turn the polyomial into a list of its coefficients and then index into that list using standard SageMath or Python syntax:
\\Code
Let's see that the answer agrees with what our formula in the solution to Example 8.6 gives us for \(n=30\):
\\Code
That's a relief, and so long as we only need a single coefficient, we're now in good shape. But what if we really need a formula for the coefficient on \(x^n\) in general? Let's see how we can use SageMath to help us with some of the other steps in
Example 8.6
. The first thing we'll want is the
partial_fraction()
method:
\\Code
If you don't like the way that looks, the
pretty_print()
function can make it easier to read:
\\Code
Up to the location of a minus sign, this is what we got by hand, but we get it much faster! From this stage, it's frequently possible to use our knowledge of certain fundamental power series that appear when doing the partial fractions expansion to come up with the general form for the coefficient on an arbitrary term of the power series. To facilitate this, we close this section with an example that illustrates how we can use solutions to counting problems we have already studied in order to figure out the coefficients on generating functions.
Let \(n\) be a positive integer. What is the coefficient on \(x^k\) in the generating function
\(\dfrac{1}{(1-x)^n}\)?
- Solution
-
We have already encountered the case n=5 in the midst of working on Example 8.4 , but there we appealed to calculus. Let's take a look at this from the perspective of just counting. The generating function \(1/(1−x)=1+x+x^2+ \cdot \cdot \cdot \) encodes the sequence for the number of ways to distribute \(n\) apples to one child. There's only one way to do that task: give the lucky kid all the apples. Multiplying together a bunch of copies of \(1/(1−x)\) then serves to increase the number of children to whom the apples are being distributed, and since each power series being multiplied starts with 1, we are in the situation where the number of apples each child receives must be nonnegative . This is therefore a problem from Section 2.5. We have \(n\) children and the coefficient on \(x^k\) is the number of ways of distributing \(k\) apples to them. This requires \(n\) artificial apples, so we distribute \(k+n\) apples, which determine \(k+n−1\) gaps and we must choose \(n−1\) of them as the locations for dividers. Therefore, we can conclude that
\(\dfrac{1}{(1-x)^n} = \displaystyle \sum_{k=0}^{\infty} \dbinom{k+n-1}{n-1}x^k = \sum_{k=0}^{\infty} \dbinom{k+n-1}{n}x^k\).
It's possible to arrive at this conclusion using techniques from calculus, but there are a lot of factorials and −1s to monitor, so this combinatorial approach may be less error prone!


1: Reactions using Chiral Lewis Acids and Brønsted Acid
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This module presents the recent developments in chiral Lewis acid and Brønsted acid catalysis, especially the systems having the combination of Lewis acids and Brønsted acids. This combined catalytic system has been useful in asymmetric synthesis over the past 20 years.
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1.1: Brønsted Acid-Assisted Lewis Acid (BLA)
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Chiral Brønsted acid-assisted Lewis acids (BLAs) are efficient and versatile chiral Lewis acids for a wide range of catalytic asymmetric cycloaddition reactions.
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1.2: Lewis Acid-Assisted Lewis Acid (LLA)
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In Lewis acid assisted chiral Lewis acids (LLAs), achiral Lewis acid is added to activate chiral Lewis acid via complex formation. The reactivity of LLA is much greater compared to that of achiral Lewis acid, and thus, the latter's presence does not affect the selectivity of the reaction.
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1.3: LBA Catalysts
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The combination of Lewis acids and chiral Brønsted acids affords LBA catalysts. In this system, the coordination of the Lewis acids to the heteroatom of the chiral Brønsted acid results in increase the acidity of the latter.
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1.4: Problems + Reference
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1.5: Chiral Phosphoric Acids (PAs)
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The combination of Lewis acids and chiral Brønsted acids affords LBA catalysts. In this system, the coordination of the Lewis acids to the heteroatom of the chiral Brønsted acid results in increase the acidity of the latter.


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anchoring strategies
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in the ethical context, the process of making ethics a foundation (anchor) of your business so that decision makers act keeping in mind key ethical considerations
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antitrust laws
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prevent companies from abusing position or power in the market through exclusion or limitation of competitor access to the market
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business ethics
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conduct by which companies and their agents abide by the law and respect the rights of their stakeholders, particularly their customers, clients, employees, and the surrounding community and environment
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conflict of interest
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when an individual or organization has interests in multiple areas that come into conflict with each other
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copyright
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grants the creator of a work the exclusive right to reproduce the work, typically for a specified period of time
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corporate social responsibility (CSR)
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practice in which a business views itself within a broader context, as a member of society with certain implicit social obligations and responsibility for its own effects on environmental and social well-being
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design patent
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protects product design
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sharing the needed facts and details about a subject in a transparent and truthful way
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diversity
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environmental entrepreneurship
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entrepreneurial ventures focused on identifying an environmental problem and creating a product or process to solve that problem
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ethics
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standards of behavior to which we hold ourselves in our personal and professional lives
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fairness
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impartial, unbiased compliance with rules and standards of what is right, just, and equitable
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integrity
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adherence to a code of moral values implying trustworthiness and incorruptibility because there is unity between what we say and what we do
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intellectual property
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prejudice
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way of thinking about or treating employees in a manner that favors or hinders one person or group over another, usually in a way considered to be unfair
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social entrepreneurship
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entrepreneurial ventures focused on identifying a societal problem and creating a product or process to solve that problem
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sustainability
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practice of preserving resources and operating in a way that is ecologically responsible in the long term
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