7.6: Degree of Centralization
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How can the degree of centralization/decentralization be altered to make an organization more successful?
The optimal span of control is determined by the following five factors:
- Nature of the task . The more complex the task, the narrower the span of control.
- Location of the workers . The more locations, the narrower the span of control.
- Ability of the manager to delegate responsibility . The greater the ability to delegate, the wider the span of control.
- Amount of interaction and feedback between the workers and the manager . The more feedback and interaction required, the narrower the span of control.
- Level of skill and motivation of the workers . The higher the skill level and motivation, the wider the span of control.
The final component in building an effective organizational structure is deciding at what level in the organization decisions should be made. Centralization is the degree to which formal authority is concentrated in one area or level of the organization. In a highly centralized structure, top management makes most of the key decisions in the organization, with very little input from lower-level employees. Centralization lets top managers develop a broad view of operations and exercise tight financial controls. It can also help to reduce costs by eliminating redundancy in the organization. But centralization may also mean that lower-level personnel don’t get a chance to develop their decision-making and leadership skills and that the organization is less able to respond quickly to customer demands.
Decentralization is the process of pushing decision-making authority down the organizational hierarchy, giving lower-level personnel more responsibility and power to make and implement decisions. Benefits of decentralization can include quicker decision-making, increased levels of innovation and creativity, greater organizational flexibility, faster development of lower-level managers, and increased levels of job satisfaction and employee commitment. But decentralization can also be risky. If lower-level personnel don’t have the necessary skills and training to perform effectively, they may make costly mistakes. Additionally, decentralization may increase the likelihood of inefficient lines of communication, competing objectives, and duplication of effort.
Several factors must be considered when deciding how much decision-making authority to delegate throughout the organization. These factors include the size of the organization, the speed of change in its environment, managers’ willingness to give up authority, employees’ willingness to accept more authority, and the organization’s geographic dispersion.
Decentralization is usually desirable when the following conditions are met:
- The organization is very large, like ExxonMobil, Ford, or General Electric.
- The firm is in a dynamic environment where quick, local decisions must be made, as in many high-tech industries.
- Managers are willing to share power with their subordinates.
- Employees are willing and able to take more responsibility.
- The company is spread out geographically, such as Nordstrom, Caterpillar, or Ford.
As organizations grow and change, they continually reevaluate their structure to determine whether it is helping the company to achieve its goals.
CONCEPT CHECK
- What are the characteristics of a centralized organization?
- What are the benefits of a decentralized organization?
- What factors should be considered when choosing the degree of centralization?


3: The Chemical Foundation of Life
Elements in various combinations comprise all matter, including living things. Some of the most abundant elements in living organisms include carbon, hydrogen, nitrogen, oxygen, sulfur, and phosphorus. These form the nucleic acids, proteins, carbohydrates, and lipids that are the fundamental components of living matter. Biologists must understand these important building blocks and the unique structures of the atoms that make up molecules, allowing for the formation of cells, tissues, organ systems, and entire organisms.
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- 3.1: Prelude to The Chemical Foundation of Life
- All biological processes follow the laws of physics and chemistry, so in order to understand how biological systems work, it is important to understand the underlying physics and chemistry. For example, the flow of blood within the circulatory system follows the laws of physics that regulate the modes of fluid flow. The breakdown of the large, complex molecules of food into smaller molecules is a series of chemical reactions that follow chemical laws.
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- 3.3: Atoms, Isotopes, Ions, and Molecules - The Building Blocks
- At its most fundamental level, life is made up of matter. Matter is any substance that occupies space and has mass. Elements are unique forms of matter with specific chemical and physical properties that cannot be broken down into smaller substances by ordinary chemical reactions. There are 118 elements, but only 92 occur naturally. The remaining elements are synthesized in laboratories and are unstable.
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- 3.4: Lewis Electron Dot Structures
- Ionic substances are completely held together by ionic bonds. The full charges of the ions (for example, Na+ and Cl− in sodium chloride) cause electrostatic interactions that result in a stable crystal lattice. Ionic compounds exist as extended, orderly arrangements of ions. This is quite different from the structure of molecular substances, which take the form of collections of individual molecules.
Contributors and Attributions
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Connie Rye (East Mississippi Community College), Robert Wise (University of Wisconsin, Oshkosh), Vladimir Jurukovski (Suffolk County Community College), Jean DeSaix (University of North Carolina at Chapel Hill), Jung Choi (Georgia Institute of Technology), Yael Avissar (Rhode Island College) among other contributing authors. Original content by OpenStax (CC BY 4.0; Download for free at<EMAIL_ADDRESS>).


8.15: Editing and Enhancing a PowerPoint Presentation
- Add, delete, and move slides
- Insert text boxes
- Change theme on a presentation
Now that you’ve created a presentation, let’s start adding to it! If you closed your presentation after saving, you can reopen it by starting PowerPoint and selecting your presentation (i.e., To-the-Bank Retail Trends) listed under Recents.
Adding Slides
Thumbnail pane.
From the Home Tab
- Click on New Slide (or Duplicate Slide , to copy the current slide layout)
- If you clicked New Slide , select a layout that best fits your content
- Press enter
From the Insert Tab
- Click on New Slide
- Select a layout
- Press enter
From the Thumbnail Pane
You can also add slides from the thumbnail pane, the lefthand column of your screen when you’re in the default or Slide view.
- Select (highlight) the slide you want the new slide to follow
- Right-click on the slide
- Select New Slide for the default content slide layout or Duplicate Slide to copy the slide layout and content
OR, to use a shortcut to duplicate the slide layout (not content)
- Select (highlight) the slide you want the new slide to follow
- Press Ctrl + M (PC) or Cmnd + M (Mac)
Note that your layout choice isn’t final, so don’t overthink the decision. If you’re just trying to get your thoughts down on slides, select Duplicate Slide and fine-tune later.
Deleting a Slide
As with most operations, there are multiple ways to delete a slide:
- Highlight the slide in the thumbnail pane and press delete
- Highlight the slide in the thumbnail pane and press the delete icon (Home tab)
- Highlight the slide in the thumbnail pane, right click and select Delete Slide
Organizing Slides
To move a slide, simply drag and drop in the thumbnail frame.
 One of the ways to enter text, as discussed earlier, is to click inside a placeholder box and enter your text. However, you may want to add text to a blank slide or in a place where there is no placeholder. To do so, select the Insert tab on the ribbon, and choose Text Box.
A text box will be inserted in your slide with a Click to insert text message similar to the placeholder Title and Subtitle boxes discussed earlier. The text box will expand to fit your text. If you decide you want to reposition the text box, you can click and move the text box/text around the slide. If you want to modify the width of the text box, click on a corner and drag it out as desired. The circular icon at the top of the text box activates the drawing and formatting options and allows you to rotate the text box.
Changing the Theme on a Presentation
Themes are one of the formatting options within PowerPoint. A theme is a slide design that contains pre-set colors, fonts, formatting and/or images. To view formatting options, click on the design tab and scroll through the available themes.
Clicking on a theme’s thumbnail applies the formatting to your slides, so you can evaluate potential fit. A second group of options in the ribbon provides color variations on that design. A third option, exclusive to the online version of PowerPoint, is Design Ideas, providing additional variations; this feature is still being built out. To revert to your original format, click Undo.
One caveat: given the limited number of templates and the number of people using PowerPoint, using a standard template doesn’t do anything to distinguish you or your message. Instead of selecting a standard template, use thoughtful images and complementary fonts to add design impact.
Contributors and Attributions
- Editing and Enhancing a PowerPoint Presentation. Authored by : Nina Burokas. Provided by : Lumen Learning. License : CC BY: Attribution


Page 3.5: Energy Flow through Ecosystems
An ecosystem is a community of organisms and their abiotic (non-living) environment. Ecosystems can be small, such as the tide pools found near the rocky shores of many oceans, or large, such as those found in the tropical rainforest of the Amazon in Brazil (Figure \(\PageIndex{1}\)).
There are three broad categories of ecosystems based on their general environment: freshwater, marine, and terrestrial. Within these three categories are individual ecosystem types based on the environmental habitat and organisms present.
Freshwater ecosystems are the least common, occurring on only 1.8 percent of Earth’s surface. These systems comprise lakes, rivers, streams, and springs; they are quite diverse and support a variety of animals, plants, fungi, protists and prokaryotes.
Marine ecosystems are the most common, comprising 75 percent of Earth’s surface and consisting of three basic types: shallow ocean, deep ocean water, and deep ocean bottom. Shallow ocean ecosystems include extremely biodiverse coral reef ecosystems. Small photosynthetic organisms suspended in ocean waters, collectively known as phytoplankton, perform 40 percent of all photosynthesis on Earth. Deep ocean bottom ecosystems contain a wide variety of marine organisms. These ecosystems are so deep that light is unable to reach them.
Terrestrial ecosystems , also known for their diversity, are grouped into large categories called biomes. A biome is a large-scale community of organisms, primarily defined on land by the dominant plant types that exist in geographic regions of the planet with similar climatic conditions. Examples of biomes include tropical rainforests, savannas, deserts, grasslands, temperate forests, and tundras. Grouping these ecosystems into just a few biome categories obscures the great diversity of the individual ecosystems within them. For example, the saguaro cacti ( Carnegiea gigantean ) and other plant life in the Sonoran Desert, in the United States, are relatively diverse compared with the desolate rocky desert of Boa Vista, an island off the coast of Western Africa (Figure \(\PageIndex{2}\)).
Food Chains and Food Webs
Figure \(\PageIndex{2}\). Desert ecosystems, like all ecosystems, can vary greatly. The desert in (a) Saguaro National Park, Arizona, has abundant plant life, while the rocky desert of (b) Boa Vista island, Cape Verde, Africa, is devoid of plant life. (credit a: modification of work by Jay Galvin; credit b: modification of work by Ingo Wölbern)A food chain is a linear sequence of organisms through which nutrients and energy pass as one organism eats another. The levels in the food chain are producers, primary consumers, higher-level consumers, and finally decomposers. These levels are used to describe ecosystem structure and dynamics. There is a single path through a food chain. Each organism in a food chain occupies a specific trophic level (energy level), its position in the food chain or food web.
Figure \(\PageIndex{3}\). These are the trophic levels of a food chain in Lake Ontario at the United States–Canada border. Energy and nutrients flow from photosynthetic green algae at the base to the top of the food chain: the Chinook salmon. (credit: modification of work by National Oceanic and Atmospheric Administration/NOAA)In many ecosystems, the base, or foundation, of the food chain consists of photosynthetic organisms (plants or phytoplankton), which are called producers . The organisms that consume the producers are herbivores called primary consumers . Secondary consumers are usually carnivores that eat the primary consumers. Tertiary consumers are carnivores that eat other carnivores. Higher-level consumers feed on the next lower trophic levels, and so on, up to the organisms at the top of the food chain. In the Lake Ontario food chain, shown in Figure \(\PageIndex{3}\), the Chinook salmon is the apex consumer at the top of this food chain.
One major factor that limits the number of steps in a food chain is energy. Energy is lost at each trophic level and between trophic levels as heat and in the transfer to decomposers (Figure \(\PageIndex{4}\) below). Thus, after a limited number of trophic energy transfers, the amount of energy remaining in the food chain may not be great enough to support viable populations at higher trophic levels.
There is a one problem when using food chains to describe most ecosystems. Even when all organisms are grouped into appropriate trophic levels, some of these organisms can feed at more than one trophic level. In addition, species feed on and are eaten by more than one species. In other words, the linear model of ecosystems, the food chain, is a hypothetical and overly simplistic representation of ecosystem structure. A holistic model—which includes all the interactions between different species and their complex interconnected relationships with each other and with the environment—is a more accurate and descriptive model for ecosystems. A food web is a concept that accounts for the multiple trophic (feeding) interactions between each species (Figure \(\PageIndex{5}\) below).
Figure \(\PageIndex{4}\). The relative energy in trophic levels in a Silver Springs, Florida, ecosystem is shown. Each trophic level has less energy available, and usually, but not always, supports a smaller mass of organisms at the next level.Two general types of food webs are often shown interacting within a single ecosystem. A grazing food web has plants or other photosynthetic organisms at its base, followed by herbivores and various carnivores. A detrital food web consists of a base of organisms that feed on decaying organic matter (dead organisms), including decomposers (which break down dead and decaying organisms) and detritivores (which consume organic detritus). These organisms are usually bacteria, fungi, and invertebrate animals that recycle organic material back into the biotic part of the ecosystem as they themselves are consumed by other organisms.
Figure \(\PageIndex{5}\). This food web shows the interactions between organisms across trophic levels. Arrows point from an organism that is consumed to the organism that consumes it. All the producers and consumers eventually become nourishment for the decomposers (fungi, mold, earthworms, and bacteria in the soil). (credit “fox”: modification of work by Kevin Bacher, NPS; credit “owl”: modification of work by John and Karen Hollingsworth, USFWS; credit “snake”: modification of work by Steve Jurvetson; credit “robin”: modification of work by Alan Vernon; credit “frog”: modification of work by Alessandro Catenazzi; credit “spider”: modification of work by “Sanba38″/Wikimedia Commons; credit “centipede”: modification of work by “Bauerph”/Wikimedia Commons; credit “squirrel”: modification of work by Dawn Huczek; credit “mouse”: modification of work by NIGMS, NIH; credit “sparrow”: modification of work by David Friel; credit “beetle”: modification of work by Scott Bauer, USDA Agricultural Research Service; credit “mushrooms”: modification of work by Chris Wee; credit “mold”: modification of work by Dr. Lucille Georg, CDC; credit “earthworm”: modification of work by Rob Hille; credit “bacteria”: modification of work by Don Stalons, CDC)How Organisms Acquire Energy in a Food Web
All living things require energy in one form or another. At the cellular level, energy is used in most metabolic pathways (usually in the form of ATP), especially those responsible for building large molecules from smaller compounds. Living organisms would not be able to assemble complex organic molecules (proteins, lipids, nucleic acids, and carbohydrates) without a constant energy input.
Food-web diagrams illustrate how energy flows directionally through ecosystems. They can also indicate how efficiently organisms acquire energy, use it, and how much remains for use by other organisms of the food web. Energy is acquired by living things in two ways: autotrophs harness light or chemical energy and heterotrophs acquire energy through the consumption and digestion of other living or previously living organisms.
Photosynthetic and chemosynthetic organisms are autotrophs, which are organisms capable of synthesizing their own food (more specifically, capable of using inorganic carbon as a carbon source). Photosynthetic autotrophs ( photoautotrophs ) use sunlight as an energy source, and chemosynthetic autotrophs ( chemoautotrophs ) use inorganic molecules as an energy source. Autotrophs are critical for ecosystems because they occupy the trophic level containing producers. Without these organisms, energy would not be available to other living organisms, and life would not be possible.
Photoautotrophs, such as plants, algae, and photosynthetic bacteria, are the energy source for a majority of the world’s ecosystems. Photoautotrophs harness the Sun’s solar energy by converting it to chemical energy. The rate at which photosynthetic producers incorporate energy from the Sun is called gross primary productivity . However, not all of the energy incorporated by producers is available to the other organisms in the food web because producers must also grow and reproduce, which consumes energy. Net primary productivity is the energy that remains in the producers after accounting for these organisms’ metabolism and heat loss. The net productivity is then available to the primary consumers at the next trophic level.
Figure \(\PageIndex{6}\). Swimming shrimp, a few squat lobsters, and hundreds of vent mussels are seen at a hydrothermal vent at the bottom of the ocean. As no sunlight penetrates to this depth, the ecosystem is supported by chemoautotrophic bacteria and organic material that sinks from the ocean’s surface. This picture was taken in 2006 at the submerged NW Eifuku volcano off the coast of Japan by the National Oceanic and Atmospheric Administration (NOAA). The summit of this highly active volcano lies 1535 m below the surface.
Chemoautotrophs are primarily bacteria and archaea that are found in rare ecosystems where sunlight is not available, such as those associated with dark caves or hydrothermal vents at the bottom of the ocean (Figure \(\PageIndex{6}\)). Many chemoautotrophs in hydrothermal vents use hydrogen sulfide (H 2 S), which is released from the vents, as a source of chemical energy. This allows them to synthesize complex organic molecules, such as glucose, for their own energy and, in turn, supplies energy to the rest of the ecosystem.
One of the most important consequences of ecosystem dynamics in terms of human impact is biomagnification. Biomagnification is the increasing concentration of persistent, toxic substances in organisms at each successive trophic level. These are substances that are lipid soluble and are stored in the fat reserves of each organism. Many substances have been shown to biomagnify, including classical studies with the pesticide dichlorodiphenyltrichloroethane (DDT), which were described in the 1960s bestseller Silent Spring by Rachel Carson. DDT was a commonly used pesticide before its dangers to apex consumers, such as the bald eagle, became known. DDT and other toxins are taken in by producers and passed on to successive levels of consumers at increasingly higher rates. As bald eagles feed on contaminated fish, their DDT levels rise. It was discovered that DDT caused the eggshells of birds to become fragile, which contributed to the bald eagle being listed as an endangered species under U.S. law. The use of DDT was banned in the United States in the 1970s.
Figure \(\PageIndex{7}\). This chart shows the PCB concentrations found at the various trophic levels in the Saginaw Bay ecosystem of Lake Huron. Notice that the fish in the higher trophic levels accumulate more PCBs than those in lower trophic levels. (credit: Patricia Van Hoof, NOAA)Another substances that biomagnifies is polychlorinated biphenyl (PCB), which was used as coolant liquids in the United States until its use was banned in 1979. PCB was best studied in aquatic ecosystems where predatory fish species accumulated very high concentrations of the toxin that is otherwise exists at low concentrations in the environment. As illustrated in a study performed by the NOAA in the Saginaw Bay of Lake Huron of the North American Great Lakes (Figure \(\PageIndex{7}\) below), PCB concentrations increased from the producers of the ecosystem (phytoplankton) through the different trophic levels of fish species. The apex consumer, the walleye, has more than four times the amount of PCBs compared to phytoplankton. Also, research found that birds that eat these fish may have PCB levels that are at least ten times higher than those found in the lake fish.
Other concerns have been raised by the biomagnification of heavy metals, such as mercury and cadmium, in certain types of seafood. The United States Environmental Protection Agency recommends that pregnant women and young children should not consume any swordfish, shark, king mackerel, or tilefish because of their high mercury content. These individuals are advised to eat fish low in mercury: salmon, shrimp, pollock, and catfish. Biomagnification is a good example of how ecosystem dynamics can affect our everyday lives, even influencing the food we eat.
Suggested Supplementary Reading
Canales, M. et al. 2018. 6 Things that Make Like on Earth Possible [Infographic]. National Geographic . March.
Contributors and Attributions
- Energy Flow through Ecosystems by OpenStax is licensed under CC BY 3.0. Modified from the original by Matthew R. Fisher.


13.10: Chapter Review - Practice Exercises
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Practice Exercises
Build a simple database and create a query that joins information. You are currently working with information about the sales and marketing team. You want to keep track of the team members and the regions they work with. Set up a database with the following tables (remember, you would separate this data into at least two tables to make the database more efficient and easy to work with):
Table:TeamMemberTMID
TMFirstName
TMLastName
TMPhone
TMEmail
Table:SalesRegion
SalesRegionName
SalesRegionNotes
Table:SalesRecord
TMID
SalesRegion
Once you have set up the tables, create a query that returns the team member’s first and last names and the sales region they are assigned to. Add any fields you think are necessary to create a list of team members’ names with their region. You may want to try sorting to improve the query output.
Table: EMPLInfo
EmplID
EmplFirstName
EmplLastName
EmplAddress
EmplCity
EmplState
EmplZip
Table: EMPLLearning
EmplID
TrainingName
TrainingDate
Importing records is an important function when working with a database like Access. Complete the following steps to create a database from scratch:
- Create a blank database using Access and then import the colors.csv file from the LEGO Database found on Kaggle.
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Then use your navigation buttons to do the following:
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Add a record with the values:
ID: 2000
Name: Vivid burgundy
RGBN: 990033
is_trans: f - Modify the record to say “dragon-scale burgundy.”
- Delete the record.
- Undo the change.
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Add a record with the values:
The WorldCorp Human Resources Department has asked you to create reports to support the following inquiries using the Human Resources Data Set from the Kaggle website. Before creating the reports, look at the type of data available and decide what data needs to be included or omitted. Which fields should be considered to answer these inquiries?
- Which employees have high performance scores? Low performance scores?
- Which departments are currently the most diverse?
- How has the company acquired its most diverse employees to date?
- Is pay equitable across all divisions and positions in the company?
- Evaluating employee performance (including attendance and lateness)
- Understanding employee roles (including position, department, location)
- Reviewing employee hiring information (including recruitment and position)


4.5: Force
Learning Objectives
- Explain force and relativity
Force is a concept that is seldom needed in relativity, and that’s why this section is optional.
Four-force
By analogy with Newtonian mechanics, we define a relativistic force vector
\[F = m\cdot a\]
where \(a\) is the acceleration four-vector (section 3.5) and \(m\) is the mass of a particle that has that acceleration as a result of the force \(F\). This is equivalent to
\[F = \frac{\mathrm{d} p}{\mathrm{d} \tau }\]
where \(p\) is the mass of the particle and \(τ\) its proper time. Since the timelike part of \(p\) is the particle’s mass-energy, the timelike component of the force is related to the power expended by the force. These definitions only work for massive particles, since for a massless particle we can’t define \(a\) or \(τ\). \(F\) has been defined in terms of Lorentz invariants and four-vectors, and therefore it transforms as a god-fearing four-vector itself.
The force measured by an observer
The trouble with all this is that \(F\) isn’t what we actually measure when we measure a force, except if we happen to be in a frame of reference that momentarily coincides with the rest frame of the particle. As with velocity and acceleration (
section 3.7), we have a four-vector that has simple, standard transformation properties, but a different \(F_o\), which is what is actually measured by the observer \(o\). It’s defined as\[F_o = \frac{\mathrm{d} p}{\mathrm{d} t }\]
with a \(dt\) in the denominator rather than a \(dτ\). In other words, it measures the rate of transfer of momentum according to the observer, whose time coordinate is \(t\), not \(τ\) — unless the observer happens to be moving along with the particle. Unlike the three-vectors \(v_o\) and \(a_o\), whose timelike components are zero by definition according to observer \(o\), \(F_o\) usually has a nonvanishing timelike component, which is the rate of change of the particle’s mass-energy, i.e., the power. We can refer to the spacelike part of \(F_o\) as the three-force.
The following two examples show that an object moving at relativistic speeds has less inertia in the transverse direction than in the longitudinal one. A corollary is that the three-acceleration need not be parallel to the three-force.
Example \(\PageIndex{1}\): Circular motion
For a particle in uniform circular motion, \(γ\) is constant, and we have
\[F_o = \frac{\mathrm{d} }{\mathrm{d} t }(m\gamma v) = m\gamma \frac{\mathrm{d} v}{\mathrm{d} t}\]
The particle’s mass-energy is constant, so the timelike component of \(F_o\) does happen to be zero in this example. In terms of the three-vectors \(v_o\) and \(a_o\) defined in
section 3.7, we have\[F_o = m\gamma \frac{\mathrm{d} v_o}{\mathrm{d} t} = m\gamma a_o\]
which is greater than the Newtonian value by the factor \(γ\). As a practical example, in a cathode ray tube(CRT) such as the tube in an old-fashioned oscilloscope or television, a beam of electrons is accelerated upto relativistic speed. To paint a picture on the screen, the beam has to be steered by transverse forces, and since the deﬂection angles are small, the world-line of the beam is approximately that of uniform circular motion. The force required to deﬂect the beam is greater by a factor of \(γ\) than would have been expected according to Newton’s laws.
Example \(\PageIndex{2}\): Linear motion
For accelerated linear motion in the x direction, ignoring \(y\) and \(z\), we have a velocity vector
\[v = \frac{\mathrm{d} r}{\mathrm{d} \tau }\]
whose \(x\) component is \(γv\). Then
\[\begin{align*} F_{o,x} &= m\frac{\mathrm{d} (\gamma v)}{\mathrm{d} t}\\ &= m\frac{\mathrm{d} (\gamma )}{\mathrm{d} t}v + m\gamma \frac{\mathrm{d} v}{\mathrm{d} t}\\ &= m\frac{\mathrm{d} \gamma }{\mathrm{d} v}\frac{\mathrm{d} v}{\mathrm{d} t} + m\gamma a\\ &= m\left ( v^2 \gamma ^3 a + \gamma a \right )\\ &= ma\gamma ^3 \end{align*}\]
The particle’s apparent inertia is increased by a factor of \(γ^3\) due to relativity.
The results of above two examples can be combined as follows:
\[F_o = m\gamma a_{o,\perp } + m\gamma ^3 a_{o,\parallel }\]
where the subscripts \(\perp\) and \(\parallel\) refer to the parts of \(a_o\) perpendicular and parallel to \(v_o\).
Transformation of the force measured by an observer
Define a frame of reference \(o\) for the inertial frame of reference of an observer who does happen to be moving along with the particle at a particular instant in time. Then \(t\) is the same as \(τ\), and \(F_o\) the same as \(F\). In this frame, the particle is momentarily at rest, so the work being done on it vanishes, and the timelike components of \(F_o\) and \(F\) are both zero.
Suppose we do a Lorentz transformation from o to a new frame \(o'\), and suppose the boost is parallel to \(F_o\) and \(F\) (which are both purely spatial in frame \(o\)). Call this direction \(x\). Then \(dp = (dp_t,dp_x) = (0,dp_x)\) transforms to \(dp' = (-γv dp_x,γ dp_x)\), so that \(F_{o',x} = dp'_x/dt' = (γ dp_x)/(γ dt) = F_{o,x}\). The two factors of \(γ\) cancel, and we find that \(F_{o',x} = F_{o,x}\).
Now let’s do the case where the boost is in the \(y\) direction, perpendicular to the force. The Lorentz transformation doesn’t change \(dp_y\), so
\[\begin{align*} F_{o',y} &= \frac{\mathrm{d} p'_y}{\mathrm{d} t'}\\ &= \frac{\mathrm{d} p_y}{(\gamma \mathrm{d} t)}\\ &= \frac{F_{o',y}}{\gamma } \end{align*}\]
The summary of our results is as follows. Let \(F_o\) be the force acting on a particle, as measured in a frame instantaneously comoving with the particle. Then in a frame of reference moving relative to this one, we have
\[F_{o',\parallel } = F_{o,\parallel }\]
and
\[F_{o',\perp } = \frac{F_{o,\perp }}{\gamma }\]
where \(\parallel\) indicates the direction parallel to the relative velocity of the two frames, and \(\perp\) a direction perpendicular to it.
Work
Consider the one-dimensional version of the three-force, \(F = dp/dt\). An advantantage of this quantity is that it allows us to use the Newtonian form of the (one-dimensional) work-kinetic energy relation \(dE/dx = F\) without correction. Proof:
\[\frac{\mathrm{d} E}{\mathrm{d} x} = \frac{\mathrm{d} E}{\mathrm{d} p} \frac{\mathrm{d} p}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} x} = \frac{\mathrm{d} E}{\mathrm{d} p} \frac{F}{v}\]
By implicit differentiation of the definition of mass, we find that \(dE/dp = p/E\), and this in turn equals \(v\) by the identity proved in Example 4.3.2. This leads to the claimed result, which is valid for both massless and material particles.


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6.1: What is an Acid and a Base?
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General properties of acids and bases are described. Arrhenius's definitions and nomenclature of Arrhenius's acids and bases are also presented.
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6.2: Brønsted–Lowry acids and bases
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Brønsted–Lowry's definition, i.e., acids are proton donors, and bases are proton acceptors are described. The related terms: acid/conjugate bases pair, mono- di-, tri-protic acids, and amphoteric substances are also described.
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6.3: Strength of acids and bases
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The strength of acids and bases, i.e., the extent of dissociation of the dissolved acid or base into ions in water is described. The relative strength of the acid/conjugate base pair is also explained.
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6.4: Acid-base equilibrium
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The establishment of an equilibrium between an acids-base mixture and its conjugates is described. Ways to move the equilibrium forward or reverse, by varying concentrations, pressure in the cases involving gases, and temperature is explained based on Le Châtelier’s principle.
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6.5: Dissociation of water
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About one in half a billion water molecules act as an acid by donating a proton to another water molecule that acts as a base, producing hydroxide ion and hydronium ion, respectively. The product of hydroxide ion and hydronium ion concentration is constant, but the acidic solution has more hydronium ions, the basic solution has more hydroxide ions, and the neutral solution has equal.
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6.6: The pH
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pH, i.e., log of reciprocal of hydronium ion concentration, its and measurement using pH paper and pH indicator is described. The importance of and pH of body fluid, acid rain, and its effects are introduced.
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6.7: Acid-base reactions
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Reactions of acids with metals, carbonate and bicarbonate salts, and Arrhenius bases are introduced. Ways of representing acid-base reaction by molecular equation, complete ionic equations, net ionic equations, and determination of acid or base concentration by titration method are described.
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6.8: pH Buffers
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pH Buffer, i.e., a mixture of a weak acid and its conjugate base or vice versa, is used to regulate the pH within a narrow range. Mechanisms of buffer action and buffer action in the blood are explained.


9.3: Activity 3 - Sorting Algorithms by Merge and Partition methods
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Introduction
This activity will present two sorting algorithms. The first algorithm called merge sort is based on merge method, and the second algorithm called quick sort is based on partition method.
Activity Details
Merge Sort Algorithm
The Merge sort algorithm is based on the divide-and-conquer strategy. Its worst-case running time has a lower order of growth than insertion sort. Since the algorithm deals with sub-problems, we state each sub-problem as sorting a sub-array A[p .. r]. Initially, p = 1 and r = n, but these values change as recursion proceeds through sub-problems.
To sort A[p .. r], the merge sort algorithm proceeds as follows:
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Divide step - If a given array A has zero or one element, simply return as it is already sorted. Otherwise, split A[p .. r] into two sub-arrays A[p .. q] and A[q + 1 .. r], each containing about half of the elements of A[p .. r].
That is, q is the halfway point of A[p .. r].
- Conquer step - Conquer by recursively sorting the two sub-arrays A[p .. q] and A[q + 1 .. r].
- Combine step - Combine the elements back in A[p .. r] by merging the two sorted sub-arrays A[p .. q] and A[q + 1 .. r] into a sorted sequence. This step will be accomplished by procedure MERGE (A, p, q, r).
Note that the recursion ends when the sub-array has just one element, so that it is trivially sorted (base case). The following are the description of the merge sort algorithm: To sort the entire sequence A[1 .. n], make the initial call to the procedure MERGE-SORT (A, 1, n).
MERGE-SORT (A, p, r)
- IF p < r // Check for base case
- THEN q = FLOOR[(p + r)/2] // Divide step
- MERGE (A, p, q) // Conquer step
- MERGE (A, q + 1, r) // Conquer step
- MERGE (A, p, q, r) // Conquer step
Figure 3.3.1 depicts a bottom-up view of the merge sort procedure for n = 8.
What remains is the MERGE procedure. The following is the input and output of the MERGE procedure.
INPUT - Array A and indices p, q, r such that p ≤ q ≤ r and sub-array A[p .. q] is sorted and sub-array A[q + 1 .. r] is sorted. By restrictions on p, q, r, neither sub-array is empty.
OUTPUT - The two sub-arrays are merged into a single sorted sub-array in A[p .. r].
The merge procedure takes Θ(n) time, where n = r − p + 1, which is the number of elements being merged. The idea behind linear time merging can be easily understood by thinking of two piles of cards; each pile is sorted and placed face-up on a table with the smallest cards on top. Merge these into a single sorted pile, face-down on the table, by completing the following basic steps:
- Choose the smaller of the two top cards.
- Remove it from its pile, thereby exposing a new top card.
- Place the chosen card face-down onto the output pile.
- Repeat steps 1 - 3 until one input pile is empty.
- Eventually, take the remaining input pile and place it face-down onto the output pile.
Each basic step takes constant time, i.e., O(1), since it involves checking the two top cards. There are at most n basic steps, since each basic step removes one card from the input piles, and we started with n cards in the input piles. Therefore, this procedure should take Θ(n) time.
Now the question is do we actually need to check whether a pile is empty before each basic step?
The answer is no, we do not. Put on the bottom of each input pile a special guard card. It contains a special value that we use to simplify the code. We use ∞, since that’s guaranteed to drop to any other value. The only way that ∞ cannot drop is when both piles have ∞ exposed as their top cards. But when that happens, all the non-guard cards have already been placed into the output pile. We know in advance that there are exactly r − p + 1 non-guard cards so stop once we have performed r − p + 1 basic steps. There is no need to check for guards, since they will always drop. Rather than even counting basic steps, just fill up the output array from index p up through and including index r.
The pseudocode of the MERGE procedure is as follows:
MERGE (A, p, q, r )
1. n1 ← q − p + 1 2. n2 ← r − q 3. Create arrays L[1 . . n1 + 1] and R[1 . . n2 + 1] 4. FOR i ← 1 TO n1 5. DO L[i] ← A[p + i − 1] 6. FOR j ← 1 TO n2 7. DO R[j] ← A[q + j ] 8. L[n1 + 1] ← ∞ 9. R[n2 + 1] ← ∞ 10. i ← 1 11. j ← 1 12. FOR k ← p TO r 13. DO IF L[i ] ≤ R[ j] 14. THEN A[k] ← L[i] 15. i ← i + 1 16. ELSE A[k] ← R[j] 17. j ← j + 1
The time complexity of the merge procedure is as follows: The first two for loops (that is, the loop in line 4 and the loop in line 6) take Θ(n1 + n2) = Θ(n) time. The last for loop (that is, the loop in line 12) makes n iterations, each taking constant time, for Θ(n) times. Therefore, the total running time is Θ(n).
The analysis of the merge sort algorithm is as follows: For convenience assume that n is a power of 2 so that each divide step yields two sub-problems, both of size exactly n/2. The base case occurs when n = 1.
When n ≥ 2, time for merge sort steps are:
- Divide - Just compute q as the average of p and r, which takes constant time, i.e., Θ(1).
- Conquer- Recursively solve 2 sub-problems, each of size n/2, which is 2T(n/2).
- Combine- MERGE on an n-element sub-array takes Θ(n) time.
Summed together they give a function that is linear in n, which is Θ(n). Therefore, the recurrence for merge sort running time is
\[T(n)=
\begin{cases}
\Theta(1) & \text{ if } n=1, \\
2T(n/2)+\Theta(n) & \text{ if } n>1.
\end{cases}\nonumber\]
Solving the merge sort recurrence we can show that T(n) = Θ(n lg n).
The implementation of the merge sort algorithm proceeds as follows:
void mergeSort(int numbers[], int temp[], int array_size) { m_sort(numbers, temp, 0, array_size - 1); } void m_sort(int numbers[], int temp[], int left, int right) { int mid; if (right > left) { mid = (right + left) / 2; m_sort(numbers, temp, left, mid); m_sort(numbers, temp, mid+1, right); merge(numbers, temp, left, mid+1, right); } void merge(int numbers[], int temp[], int left, int mid, int right) { int i, left_end, num_elements, tmp_pos; left_end = mid - 1; tmp_pos = left; num_elements = right - left + 1; while ((left <= left_end) && (mid <= right)) { if (numbers[left] <= numbers[mid]) { temp[tmp_pos] = numbers[left]; tmp_pos = tmp_pos + 1; left = left +1; } else { temp[tmp_pos] = numbers[mid]; tmp_pos = tmp_pos + 1; mid = mid + 1; } while (left <= left_end) { temp[tmp_pos] = numbers[left]; left = left + 1; tmp_pos = tmp_pos + 1; } while (mid <= right) { temp[tmp_pos] = numbers[mid]; mid = mid + 1; tmp_pos = tmp_pos + 1;} for (i = 0; i <= num_elements; i++) { numbers[right] = temp[right]; right = right - 1;}
Quick Sort Algorithm
The quick sort algorithm operates on the principle of divide-and-conquer. Quick sort is an algorithm of choice in many situations because it is not difficult to implement, it is a good “general purpose” sort and it consumes relatively fewer resources during execution. The following are some pros and cons of the quick sort algorithm:
Pros
- well established and uses only a small auxiliary stack
- requires only n log(n) time to sort n items.
- has an extremely short inner loop
Cons
- is recursive. Especially if recursion is not available, the implementation is extremely complicated.
- is fragile i.e., a simple mistake in the implementation can go unnoticed and cause it to perform badly.
The quick sort algorithm works by partitioning a given array A[p . . r] into two non-empty sub array A[p . . q] and A[q+1 . . r] such that every key in A[p . . q] is less than or equal to every key in A[q+1 . . r]. Then the two sub-arrays are sorted by recursive calls to quick sort. The exact position of the partition depends on the given array and index q is computed as a part of the partitioning procedure. In general, the quick sort algorithm performs the following steps:
Quick_Sort
- If p < r then
- q Partition (A, p, r)
- Recursive call to Quick Sort (A, p, q)
- Recursive call to Quick Sort (A, q + r, r)
Note that to sort the entire array, initially call Quick_Sort(A, 1, length[A]).
As a first step, the quick sort algorithm would choose as a pivot an element from one of the items in the array to be sorted. Subsequently, the array is partitioned on either side of the pivot. Elements that are less than or equal to pivot will move toward the left and elements that are greater than or equal to pivot will move toward the right.
The partition procedure is defined below:
PARTITION (A, p, r)
1. x ← A[p] 2. i ← p-1 3. j ← r+1 4. while TRUE do 5. Repeat j ← j-1 6. until A[j] ≤ x 7. Repeat i ← i+1 8. until A[i] ≥ x 9. if i < j 10. then exchange A[i] ← A[j] 11. else return j
The partition procedure selects the first key, A[p] as a pivot key about which the array will be partitioned:
- Keys ≤ A[p] will be moved towards the left
- Keys ≥ A[p] will be moved towards the right
The time complexity of the partition procedure is Θ(n) where n = r - p +1 which is the number of keys in the array.
The time complexity of quick sort algorithm depends on whether partition is balanced or unbalanced, which in turn depends on which elements of an array to be sorted are used for partitioning. A very good partition splits an array up into two equal sized arrays. A bad partition, on other hand, splits an array up into two arrays of very different sizes. The worst partition puts only one element in one array and all other elements in the other array. If the partitioning is balanced, the quick sort runs asymptotically as fast as merge sort. On the other hand, if partitioning is unbalanced, the quick sort runs asymptotically as slow as insertion sort.
In the best case each partitioning stage divides the array exactly in half. In other words, the best to be a median of the keys in A[p . . r] every time procedure PARTITION is called. The procedure PARTITION always split the array to be sorted into two equal sized arrays. If the procedure PARTITION produces two regions of size n/2, the recurrence relation is then:
\[\begin{align}
\mathrm{T(n)}
&= \mathrm{T(n/2) + T(n/2) + \Theta(n)}\nonumber\\
&= \mathrm{2T(n/2) + \Theta(n)}\nonumber
\end{align}\nonumber\]
And
\[\mathrm{T(n) = \Theta(\textrm{n lg n}).}\nonumber\]
The worst-case occurs if the input array A[1 . . n] is already sorted. The call to PARTITION (A, p, r) always returns p, so successive calls to partition will split arrays of length n, n-1, n-2, . . . , 2. Hence, the running time is proportional to n + (n-1) + (n-2) + . . . + 2 = [(n+2)(n-1)]/2 = Θ(n2). The worst-case also occurs if A[1 . . n] starts out in reverse order.
The implementation of the quick sort algorithm is presented below:
void quickSort(int numbers[], int array_size) {q_sort(numbers, 0, array_size - 1); } void q_sort(int numbers[], int left, int right) { int pivot, l_hold, r_hold; l_hold = left; r_hold = right; pivot = numbers[left]; while (left < right) { while ((numbers[right] >= pivot) && (left < right)) right--; if (left != right) { numbers[left] = numbers[right]; left++; } while ((numbers[left] <= pivot) && (left < right)) left++; if (left != right) { numbers[right] = numbers[left]; right--; } } numbers[left] = pivot; pivot = left; left = l_hold; right = r_hold; if (left < pivot) q_sort(numbers, left, pivot-1); if (right > pivot) q_sort(numbers, pivot+1, right);
Conclusion
In this activity we presented two sorting algorithms, the merge sort and quick sort algorithm based on merge and partition approach respectively. Both algorithms use the divide-and-conquer strategy and have comparable run time complexity on average. Quick sort is a well established sorting algorithm whose worst-case running time is Θ(n2) and expected running time is Θ(n lg n) where the constants hidden in Θ(n lg n) are small.
Assessment
- Why would in practice one use quick sort algorithm than merge sort algorithm even though, merge sort promises a run time complexity of (n lg n) for both worst and average cases?
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Show the steps taken when sorting the following list of integers [80, 30, 50, 70, 60, 90, 20, 30, 40] using
- merge sort
- quick sort


2.2.5: The combined gas law
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The laws relating to pressure \(P\), volume \(V\), and temperature \(T\) for a constant amount \(n\) of a gas are the following:
- If \(n\) and \(T\) are constant: \(P_{1} V_{1}=P_{2} V_{2}\), that is Boyle's law.
- If \(P\) and \(n\) are constant: \(\dfrac{V_{1}}{T_{1}}=\dfrac{V_{2}}{T_{2}}\), that is Charles's law.
- If \(\mathrm{V}\) and \(\mathrm{n}\) are constant: \(\dfrac{P_{1}}{T_{1}}=\dfrac{P_{2}}{T_{2}}\), that is Gay Lussac's law.
All three relationships are combined in the following law.
Combined gas law
If \(\mathrm{n}\) is constant: \[\dfrac{P_{1} V_{1}}{T_{1}}=\dfrac{P_{2} V_{2}}{T_{2}}\nonumber\], that is the combined gas law.
The combined gas law allows calculating the effect of varying two parameters on the third.
Example \(\PageIndex{1}\)
A weather balloon contains \(212 \mathrm{~L}\) of helium at \(25^{\circ} \mathrm{C}\) and \(750 \mathrm{~mm} \mathrm{Hg}\). What is the volume of the balloon when it ascends to an altitude where the temperature is \(-40{ }^{\circ} \mathrm{C}\) and \(540 \mathrm{~mm} \mathrm{Hg}\), assuming the quantity of gas remains the same?
Solution
Given and desired parameters (temperatures must be converted to Kelvin scale):
\[\begin{array}{lll}
\mathrm{P}_{1}=750 \mathrm{~mm} \mathrm{Hg}, & \mathrm{V}_{1}=212 \mathrm{~L}, & \mathrm{~T}_{1}=25^{\circ} \mathrm{C}+273.15=298.15 \mathrm{~K} \\
\mathrm{P}_{2}=540 \mathrm{~mm} \mathrm{Hg}, & \mathrm{V}_{2}=? & \mathrm{~T}_{2}=-40^{\circ} \mathrm{C}+273.15=233.15 \mathrm{~K}
\end{array}\nonumber\]
Formula:
\[\dfrac{P_{1} V_{1}}{T_{1}}=\dfrac{P_{2} V_{2}}{T_{2}}, \nonumber\]
rearrange the formula to isolate the desired parameter:
\[V_{2}=\dfrac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}. \nonumber\]
Calculations:
\[V_{2}=\dfrac{750 \cancel{\mathrm{~mm} \mathrm{Hg}} \times 212 \mathrm{~L} \times 233.15 \cancel{\mathrm{~K}}}{298.15 \cancel{\mathrm{~K}} \times 540 \cancel{\mathrm{~mm} \mathrm{Hg}}}=230 \mathrm{~L}. \nonumber\]


7.1: Energy in Living Systems
Skills to Develop
- Discuss the importance of electrons in the transfer of energy in living systems
- Explain how ATP is used by the cell as an energy source
Energy production within a cell involves many coordinated chemical pathways. Most of these pathways are combinations of oxidation and reduction reactions. Oxidation and reduction occur in tandem. An oxidation reaction strips an electron from an atom in a compound, and the addition of this electron to another compound is a reduction reaction. Because oxidation and reduction usually occur together, these pairs of reactions are called oxidation reduction reactions, or redox reactions .
Electrons and Energy
The removal of an electron from a molecule, oxidizing it, results in a decrease in potential energy in the oxidized compound. The electron (sometimes as part of a hydrogen atom), does not remain unbonded, however, in the cytoplasm of a cell. Rather, the electron is shifted to a second compound, reducing the second compound. The shift of an electron from one compound to another removes some potential energy from the first compound (the oxidized compound) and increases the potential energy of the second compound (the reduced compound). The transfer of electrons between molecules is important because most of the energy stored in atoms and used to fuel cell functions is in the form of high-energy electrons. The transfer of energy in the form of electrons allows the cell to transfer and use energy in an incremental fashion—in small packages rather than in a single, destructive burst. This chapter focuses on the extraction of energy from food; you will see that as you track the path of the transfers, you are tracking the path of electrons moving through metabolic pathways.
Electron Carriers
In living systems, a small class of compounds functions as electron shuttles: They bind and carry high-energy electrons between compounds in pathways. The principal electron carriers we will consider are derived from the B vitamin group and are derivatives of nucleotides. These compounds can be easily reduced (that is, they accept electrons) or oxidized (they lose electrons). Nicotinamide adenine dinucleotide (NAD) (Figure \(\PageIndex{1}\)) is derived from vitamin \(B_3\), niacin. NAD + is the oxidized form of the molecule; NADH is the reduced form of the molecule after it has accepted two electrons and a proton (which together are the equivalent of a hydrogen atom with an extra electron).
NAD + can accept electrons from an organic molecule according to the general equation:
\[ \underset{\text{reducing agent}}{\ce{RH}} + \underset{\text{oxidizing agent}}{\ce{NAD^+}} \rightarrow \underset{\text{reduced}}{\ce{NADH}} + \underset{\text{oxidized}}{\ce{R^+}} \nonumber\]
When electrons are added to a compound, they are reduced. A compound that reduces another is called a reducing agent. In the above equation, RH is a reducing agent, and NAD + is reduced to NADH. When electrons are removed from compound, it oxidized. A compound that oxidizes another is called an oxidizing agent. In the above equation, NAD + is an oxidizing agent, and RH is oxidized to R.
Similarly, flavin adenine dinucleotide (FAD + ) is derived from vitamin \(B_2\), also called riboflavin. Its reduced form is FADH 2 . A second variation of NAD, NADP, contains an extra phosphate group. Both NAD + and FAD + are extensively used in energy extraction from sugars, and NADP plays an important role in anabolic reactions and photosynthesis.
ATP in Living Systems
A living cell cannot store significant amounts of free energy. Excess free energy would result in an increase of heat in the cell, which would result in excessive thermal motion that could damage and then destroy the cell. Rather, a cell must be able to handle that energy in a way that enables the cell to store energy safely and release it for use only as needed. Living cells accomplish this by using the compound adenosine triphosphate (ATP). ATP is often called the “energy currency” of the cell, and, like currency, this versatile compound can be used to fill any energy need of the cell. How? It functions similarly to a rechargeable battery.
When ATP is broken down, usually by the removal of its terminal phosphate group, energy is released. The energy is used to do work by the cell, usually by the released phosphate binding to another molecule, activating it. For example, in the mechanical work of muscle contraction, ATP supplies the energy to move the contractile muscle proteins. Recall the active transport work of the sodium-potassium pump in cell membranes. ATP alters the structure of the integral protein that functions as the pump, changing its affinity for sodium and potassium. In this way, the cell performs work, pumping ions against their electrochemical gradients.
ATP Structure and Function
At the heart of ATP is a molecule of adenosine monophosphate (AMP), which is composed of an adenine molecule bonded to a ribose molecule and to a single phosphate group (Figure \(\PageIndex{2}\)). Ribose is a five-carbon sugar found in RNA, and AMP is one of the nucleotides in RNA. The addition of a second phosphate group to this core molecule results in the formation of adenosine di phosphate (ADP); the addition of a third phosphate group forms adenosine tri phosphate (ATP).
The addition of a phosphate group to a molecule requires energy. Phosphate groups are negatively charged and thus repel one another when they are arranged in series, as they are in ADP and ATP. This repulsion makes the ADP and ATP molecules inherently unstable. The release of one or two phosphate groups from ATP, a process called dephosphorylation , releases energy.
Energy from ATP
Hydrolysis is the process of breaking complex macromolecules apart. During hydrolysis, water is split, or lysed, and the resulting hydrogen atom (H + ) and a hydroxyl group (OH - ) are added to the larger molecule. The hydrolysis of ATP produces ADP, together with an inorganic phosphate ion (P i ), and the release of free energy. To carry out life processes, ATP is continuously broken down into ADP, and like a rechargeable battery, ADP is continuously regenerated into ATP by the reattachment of a third phosphate group. Water, which was broken down into its hydrogen atom and hydroxyl group during ATP hydrolysis, is regenerated when a third phosphate is added to the ADP molecule, reforming ATP.
Obviously, energy must be infused into the system to regenerate ATP. Where does this energy come from? In nearly every living thing on earth, the energy comes from the metabolism of glucose. In this way, ATP is a direct link between the limited set of exergonic pathways of glucose catabolism and the multitude of endergonic pathways that power living cells.
Phosphorylation
Recall that, in some chemical reactions, enzymes may bind to several substrates that react with each other on the enzyme, forming an intermediate complex. An intermediate complex is a temporary structure, and it allows one of the substrates (such as ATP) and reactants to more readily react with each other; in reactions involving ATP, ATP is one of the substrates and ADP is a product. During an endergonic chemical reaction, ATP forms an intermediate complex with the substrate and enzyme in the reaction. This intermediate complex allows the ATP to transfer its third phosphate group, with its energy, to the substrate, a process called phosphorylation. Phosphorylation refers to the addition of the phosphate (~P). This is illustrated by the following generic reaction:
\[\text{A} + \text{enzyme} + \text{ATP} \rightarrow \text{[A − enzyme − ∼P]} \rightarrow \text{B} + \text{enzyme} + \text{ADP} + \text{phosphate ion} \nonumber\]
Substrate Phosphorylation
ATP is generated through two mechanisms during the breakdown of glucose. A few ATP molecules are generated (that is, regenerated from ADP) as a direct result of the chemical reactions that occur in the catabolic pathways. A phosphate group is removed from an intermediate reactant in the pathway, and the free energy of the reaction is used to add the third phosphate to an available ADP molecule, producing ATP (Figure \(\PageIndex{3}\)). This very direct method of phosphorylation is called substrate-level phosphorylation .
Oxidative Phosphorylation
Most of the ATP generated during glucose catabolism, however, is derived from a much more complex process, chemiosmosis, which takes place in mitochondria (Figure \(\PageIndex{4}\)) within a eukaryotic cell or the plasma membrane of a prokaryotic cell. Chemiosmosis , a process of ATP production in cellular metabolism, is used to generate 90 percent of the ATP made during glucose catabolism and is also the method used in the light reactions of photosynthesis to harness the energy of sunlight. The production of ATP using the process of chemiosmosis is called oxidative phosphorylation because of the involvement of oxygen in the process.
Career Connections: Mitochondrial Disease Physician
What happens when the critical reactions of cellular respiration do not proceed correctly? Mitochondrial diseases are genetic disorders of metabolism. Mitochondrial disorders can arise from mutations in nuclear or mitochondrial DNA, and they result in the production of less energy than is normal in body cells. In type 2 diabetes, for instance, the oxidation efficiency of NADH is reduced, impacting oxidative phosphorylation but not the other steps of respiration. Symptoms of mitochondrial diseases can include muscle weakness, lack of coordination, stroke-like episodes, and loss of vision and hearing. Most affected people are diagnosed in childhood, although there are some adult-onset diseases. Identifying and treating mitochondrial disorders is a specialized medical field. The educational preparation for this profession requires a college education, followed by medical school with a specialization in medical genetics. Medical geneticists can be board certified by the American Board of Medical Genetics and go on to become associated with professional organizations devoted to the study of mitochondrial diseases, such as the Mitochondrial Medicine Society and the Society for Inherited Metabolic Disease.
Summary
ATP functions as the energy currency for cells. It allows the cell to store energy briefly and transport it within the cell to support endergonic chemical reactions. The structure of ATP is that of an RNA nucleotide with three phosphates attached. As ATP is used for energy, a phosphate group or two are detached, and either ADP or AMP is produced. Energy derived from glucose catabolism is used to convert ADP into ATP. When ATP is used in a reaction, the third phosphate is temporarily attached to a substrate in a process called phosphorylation. The two processes of ATP regeneration that are used in conjunction with glucose catabolism are substrate-level phosphorylation and oxidative phosphorylation through the process of chemiosmosis.
Glossary
- chemiosmosis
- process in which there is a production of adenosine triphosphate (ATP) in cellular metabolism by the involvement of a proton gradient across a membrane
- dephosphorylation
- removal of a phosphate group from a molecule
- oxidative phosphorylation
- production of ATP using the process of chemiosmosis and oxygen
- phosphorylation
- addition of a high-energy phosphate to a compound, usually a metabolic intermediate, a protein, or ADP
- redox reaction
- chemical reaction that consists of the coupling of an oxidation reaction and a reduction reaction
- substrate-level phosphorylation
- production of ATP from ADP using the excess energy from a chemical reaction and a phosphate group from a reactant


2.3: The Multicore and Multiprocessor Segments
Introduction
This section introduces the learner to the multicore and multiprocessor. It also highlights why computer architecture is moving towards multiprocessor architecture
Activity details
Central Processing Unit is what is typically referred to as a processor. A processor contains many discrete parts within it, such as one or more memory caches for instructions and data, instruction decoders, and various types of execution units for performing arithmetic or logical operations.
Multicore: is a type of architecture where a single physical processor contains the core logic of two or more processors.
A multicore CPU has multiple execution cores on one CPU. This can mean different things depending on the exact architecture, but it basically means that a certain subset of the CPU’s components is duplicated, so that multiple “cores” can work in parallel on separate operations.
This is called CMP, Chip-level Multiprocessing.
A multiprocessor system contains more than one such CPU, allowing them to work in parallel. This is called SMP, or Simultaneous Multiprocessing. That is Multi-processing simply means
putting multiple processors in one system.
For example, a multicore processor may have a separate L1 cache and execution unit for each core, while it has a shared L2 cache for the entire processor. That means that while the processor has one big pool of slower cache, it has separate fast memory and arithmetic/logic units for each of several cores. This would allow each core to perform operations at the same time as the others.
Single-core CPU Chip
Figure 1 below illustrates the single-core CPU chip. it has the components register files, ALU, bus interface and the system bus. The figure also shows the single core clearly marked out
Figure 1
Multi-core architectures
The cores fit on a single processor socket, also called CMP (Chip Multi-Processor)
Multicore System:
A Multicore system usually refers to a multiprocessor system that has all its processors on the same chip. It could also refer to a system where the processors are on different chips but use the same package (i.e., a multichip module). Multicore systems were developed primarily to enhance the system performance while limiting its power consumption. It consists of
1. General - purpose programmable cores,
2. Special - purpose accelerator cores,
3. Shared memory modules,
4. NoC (interconnection network), and
5. I/O interface.
Figure 2 below illustrates the Multi-core CPU chip
Figure 2
In Figure 2 all processors are on the same chip. The Multi-core processors are MIMD, i.e. different cores execute different threads (Multiple Instructions), operating on different parts of memory (Multiple Data). Also Multi-core is a shared memory multiprocessor, i.e. all cores share the same memory
NB
The main reason for which computer architecture is moving towards multicore systems is scalability. That is, as we increase the number of processors to enhance performance, multicore systems allow limiting power consumption and interprocessor communication overhead. A Multicore system can be scaled by adding more CPU cores and adjusting the interconnection network. More system programming work has to be done to be able to utilize the increased resources. It is one thing to increase the number of CPU resources. It is another to be able to schedule all of them to do useful tasks.
Multiprocessor
Multiprocessor. is the use of two or more central processing units (CPUs) within a single computer system.
Multiprocessor System
A multiprocessor system has its processors residing in separate chips and processors are interconnected by a backplane bus. Multiprocessor systems were developed to enhance the system performance with little regard to power consumption. A multiprocessor system has good performance and its constituent processors are high - performing processors.
Main Differences between Multicore Systems and Multiprocessor Systems:
The differences are as expressed in Table 4
Table 4
| Multiprocessor system | Multicore system | |
| Integration level | Each processor in a chip | All processors on the same chip |
| Processor performance | High | Low |
| System performance | Very high | High |
| Processor power consumption | High | Low |
| Total power consumption | Relatively high | Relatively low |
Conclusion
This section introduced the main reason why computer architecture is moving towards multicore systems which is scalability. It enables an increase in the number of processors.
This has enhanced performance, multicore systems allow limiting power consumption and interprocessor communication overhead
Assessment
Distinguish the multicore and multiprocessor architectures
A CPU, or Central Processing Unit, is what is typically referred to as a processor. A processor contains many discrete parts within it, such as one or more memory caches for instructions and data, instruction decoders, and various types of execution units for performing arithmetic or logical operations.
A multiprocessor system contains more than one such CPU, allowing them to work in parallel. This is called SMP, or Simultaneous Multiprocessing.
A multicore CPU has multiple execution cores on one CPU. Now, this can mean different things depending on the exact architecture, but it basically means that a certain subset of the CPU’s components is duplicated, so that multiple “cores” can work in parallel on separate operations.
This is called CMP, Chip-level Multiprocessing.


4.8: Homework- Derivatives in Space
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We’ve talked a lot about how derivatives measure slope, but literal slope on a topographic map is a good example of derivatives in space. Consider this picture:
src: usgs.gov’s topoview
The thinnest contours on the map represent a height gain of \(20\) feet, and the slightly thicker contours are a height gain of \(100\) feet. Each thick red line has a length of roughly \(500\) feet. What is the slope of the mountain at the thick red lines?
-
Consider the following thermal image of wolves howling:
src: usgs.gov
- Why is the eyes and mouth much hotter than other parts of the wolves, like the neck?
- According to the heat equation, things get colder or warmer depending on the second derivative with respect to space. Assuming these wolves are a constant temperature, that means the second derivative of temperature with respect to space is zero. What does it mean for the first derivative if the second derivative is zero? What does that mean for the temperature of the wolves at different places on its body where the fur is thinner or thicker?
-
Consider a cylinder filled with particles of some important nutrient (say oxygen). Suppose the concentration isn’t constant, but instead looks something like this:
The oxygen particles will tend to move from areas of high concentration to areas of low concentration, a process called diffusion. Diffusion is extremely important in biology, since that is how many cells get their nutrients.
- Based on the diagram of the cylinder, draw a rough sketch of the concentration graph. On the \(y\)-axis should be concentration \(C\) (molecules per cm\(^3\)), and on the \(x\)-axis should be distance \(x\) from the back of the cylinder measured in cm.
- Based on your graph, estimate the slope at two points of your choosing. The notation for your answers would be \(\frac{dC}{dx}\).
-
The rate at which the particles move in diffusion is called flux. It follows Fick’s law:
\(\text{Flux} = -D \cdot \frac{dC}{dx}\)
Here, flux is how fast the particles are moving, \(D\) is the diffusion coefficient (a constant), and \(\frac{dC}{dx}\) is what we found in the previous part. Why is there a negative in the equation? Explain it in words, and talk about your graph and your answers for \(\frac{dC}{dx}\) from the previous part.


14.1: Guiding Principles in Supporting Children’s Health, Safety, and Nutrition
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An integrated and comprehensive approach is most effective when preschool children are taught about health. Health education does not stand alone in the preschool curriculum. It is integrated with the other domains. Health is comprehensive. Health education involves ideas directly relevant to the child, such as “How do I grow?” Preschool teachers work with children who are naturally curious and eager to learn about their bodies and how each part works. A developmentally appropriate curriculum promotes overall health (e.g., wellness, safety, oral health, nutrition) and integrates topic areas. For example, a discussion about safety rules might include nutrition and sanitation.
Teachers address ideas and concepts that children can grasp at their developmental level and then progressively build on what children already know and understand. This approach applies to all children, including children with various abilities, disabilities, or other special needs (such as delays in language, cognition, or physical ability).
[1] Image by the California Department of Education is used with permission
[2] The California Preschool Curriculum Framework, Volume 2 by the California Department of Education is used with permission


II: Ratio Analysis and Forecasting Modeling
-
- 5: Financial Ratios and Forecasting; Liquidity and Solvency Ratios
-
- 5.1: Chapter Five- Learning Outcomes
- 5.2: Financial Ratios and Forecasting
- 5.3: Financial Ratios
- 5.4: Longitudinal vs. Cross-sectional Analysis (Example)
- 5.5: Liquidity and Liquidity Ratios
- 5.6: The Income Statement versus the Balance Sheet
- 5.7: Accounts Receivable Aging Schedule
- 5.8: Solvency Ratios
-
- 7: Market Ratios
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- 7.1: Chapter Seven- Learning Outcomes
- 7.2: Market Ratios
- 7.3: Earnings Retention and Growth
- 7.4: The P/BV and P/E Ratios
- 7.5: Ratio Analysis Exercise
- 7.6: Solution Template for Ratio Analysis Problem
- 7.7: Solution for Ratio Analysis Problem
- 7.8: Adjustments to Basic Financial Ratios for Companies That Have Preferred Stock
- 7.9: Exhibit of Effect of Preferred Stock on Earnings Retention
- 7.10: Industry Data Benchmarks
- 7.11: Some Limitations of Financial Ratios
- 7.12: Chapters 5 - 7- Review Questions
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- 9: Corporate Forecasting Models
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- 9.1: Chapter Nine- Learning Outcomes
- 9.2: Free Cash Flow
- 9.3: Free Cash Flow Exercises
- 9.4: External Funds Needed Formula (EFN)
- 9.5: Internal and External Funds (Summary)
- 9.6: The EFN Formula Explained
- 9.7: EFN Application
- 9.8: EFN Solution
- 9.9: Summary- The Fundamentals of Accounting and Financial Analysis
- 9.10: Chapters Eight and Nine- Review Questions


5.8: Rectifiable Arcs. Absolute Continuity
This page is a draft and is under active development.
If a function \(f : E^{1} \rightarrow E\) is of bounded variation (§7) on an interval \(I=[a, b],\) we can define a real function \(v_{f}\) on \(I\) by
\[v_{f}(x)=V_{f}[a, x](=\text { total variation of } f \text { on }[a, x]) \text { for } x \in I;\]
\(v_{f}\) is called the total variation function, or length function, generated by \(f\) on \(I\). Note that \(v_{f} \uparrow\) on \(I.\) (Why?) We now consider the case where \(f\) is also relatively continuous on \(I,\) so that the set \(A=f[I]\) is a rectifiable arc (see §7, Note 1 and Definition 2).
A function \(f : E^{1} \rightarrow E\) is (weakly) absolutely continuous on \(I=[a, b]\) iff \(V_{f}[I]<+\infty\) and \(f\) is relatively continuous on \(I\).
The following are equivalent:
(i) \(f\) is (weakly) absolutely continuous on \(I=[a, b]\);
(ii) \(v_{f}\) is finite and relatively continuous on \(I ;\) and
(iii) \((\forall \varepsilon>0) \text{ } (\exists \delta>0) \text{ } (\forall x, y \in I | 0 \leq y-x<\delta) \text{ } V_{f}[x, y]<\varepsilon\).
- Proof
-
We shall show that (ii) \(\Rightarrow\) (iii) \(\Rightarrow\) (i) \(\Rightarrow\) (ii).
(ii) \(\Rightarrow\) (iii). As \(I=[a, b]\) is compact, (ii) implies that \(v_{f}\) is uniformly continuous on \(I\) (Theorem 4 of Chapter 4, §8). Thus
\[(\forall \varepsilon>0) \text{ } (\exists \delta>0) \text{ } (\forall x, y \in I | 0 \leq y-x<\delta) \quad v_{f}(y)-v_{f}(x)<\varepsilon.\]
However,
\[v_{f}(y)-v_{f}(x)=V_{f}[a, y]-V_{f}[a, x]=V_{f}[x, y]\]
by additivity (Theorem 1 in §7). Thus (iii) follows.
(iii) \(\Rightarrow\) (i). By Corollary 3 of §7, \(|f(x)-f(y)| \leq V_{f}[x, y].\) Therefore, (iii) implies that
\[(\forall \varepsilon>0) \text{ } (\exists \delta>0) \text{ } (\forall x, y \in I| | x-y |<\delta) \quad|f(x)-f(y)|<\varepsilon,\]
and so \(f\) is relatively (even uniformly) continuous on \(I\).
Now with \(\varepsilon\) and \(\delta\) as in (iii), take a partition \(P=\left\{t_{0}, \ldots, t_{m}\right\}\) of \(I\) so fine that
\[t_{i}-t_{i-1}<\delta, \quad i=1,2, \ldots,\text{ } m.\]
Then \((\forall i) V_{f}\left[t_{i-1}, t_{i}\right]<\varepsilon.\) Adding up these \(m\) inequalities and using the additivity of \(V_{f},\) we obtain
\[V_{f}[I]=\sum_{i=1}^{m} V_{f}\left[t_{i-1}, t_{i}\right]<m \varepsilon<+\infty.\]
Thus (i) follows, by definition.
That (i) \(\Rightarrow\) (ii) is given as the next theorem. \(\quad \square\)
If \(V_{f}[I]<+\infty\) and if \(f\) is relatively continuous at some \(p \in I\) (over \(I=[a, b]),\) then the same applies to the length function \(v_{f}\).
- Proof
-
We consider left continuity first, with \(a<p \leq b\).
Let \(\varepsilon>0.\) By assumption, there is \(\delta>0\) such that
\[|f(x)-f(p)|<\frac{\varepsilon}{2} \text { when }|x-p|<\delta \text { and } x \in[a, p].\]
Fix any such \(x.\) Also, \(V_{f}[a, p]=\sup _{P} S(f, P)\) over \([a, p].\) Thus
\[V_{f}[a, p]-\frac{\varepsilon}{2}<\sum_{i=1}^{k}\left|\Delta_{i} f\right|\]
for some partition
\[P=\left\{t_{0}=a, \ldots, t_{k-1}, t_{k}=p\right\} \text { of }[a, p]. \text { (Why?)}\]
We may assume \(t_{k-1}=x, x\) as above. (If \(t_{k-1} \neq x,\) add \(x\) to \(P. )\) Then
\[\left|\Delta_{k} f\right|=|f(p)-f(x)|<\frac{\varepsilon}{2},\]
and hence
\[V_{f}[a, p]-\frac{\varepsilon}{2}<\sum_{i=1}^{k-1}\left|\Delta_{i} f\right|+\left|\Delta_{k} f\right|<\sum_{i=1}^{k-1}\left|\Delta_{i} f\right|+\frac{\varepsilon}{2} \leq V_{f}\left[a, t_{k-1}\right]+\frac{\varepsilon}{2}.\]
However,
\[V_{f}[a, p]=v_{f}(p)\]
and
\[V_{f}\left[a, t_{k-1}\right]=V_{f}[a, x]=v_{f}(x).\]
Thus (1) yields
\[\left|v_{f}(p)-v_{f}(x)\right|=V_{f}[a, p]-V_{f}[a, x]<\varepsilon \text { for } x \in[a, p] \text { with }|x-p|<\delta.\]
This shows that \(v_{f}\) is left continuous at \(p\).
Right continuity is proved similarly on noting that
\[v_{f}(x)-v_{f}(p)=V_{f}[p, b]-V_{f}[x, b] \text { for } p \leq x<b. \text { (Why?)}\]
Thus \(v_{f}\) is, indeed, relatively continuous at \(p.\) Observe that \(v_{f}\) is also of bounded variation on \(I,\) being monotone and finite (see Theorem 3(ii) of §7).
This completes the proof of both Theorem 2 and Theorem 1. \(\quad \square\)
We also have the following.
If \(f\) is real and absolutely continuous on \(I=[a, b]\) (weakly), so are the nondecreasing functions \(g\) and \(h(f=g-h)\) defined in Theorem 3 of §7.
Indeed, the function \(g\) as defined there is simply \(v_{f}.\) Thus it is relatively continuous and finite on \(I\) by Theorem 1. Hence so also is \(h=f-g.\) Both are of bounded variation (monotone!) and hence absolutely continuous (weakly).
Note 1. The proof of Theorem 1 shows that (weak) absolute continuity implies uniform continuity. The converse fails, however (see Problem 1(iv) in §7).
We now apply our theory to antiderivatives (integrals).
If \(F=\int f\) on \(I=[a, b]\) and if \(f\) is bounded \(\left(|f| \leq K \in E^{1}\right)\) on \(I-Q\) (\(Q\) countable), then \(F\) is weakly absolutely continuous on \(I.\)
(Actually, even the stronger variety of absolute continuity follows. See Chapter 7, §11, Problem 17).
- Proof
-
By definition, \(F=\int f\) is finite and relatively continuous on \(I,\) so we only have to show that \(V_{F}[I]<+\infty.\) This, however, easily follows by Problem 3 of §7 on noting that \(F^{\prime}=f\) on \(I-S\) (\(S\) countable). Details are left to the reader. \(\quad \square\)
Our next theorem expresses arc length in the form of an integral.
If \(f : E^{1} \rightarrow E\) is continuously differentiable on \(I=[a, b]\) (§6), then \(v_{f}=\int\left|f^{\prime}\right|\) on \(I\) and
\[V_{f}[a, b]=\int_{a}^{b}\left|f^{\prime}\right|.\]
- Proof
-
Let \(a<p<x \leq b, \Delta x=x-p,\) and
\[\Delta v_{f}=v_{f}(x)-v_{f}(p)=V_{f}[p, x] . \quad \text {(Why?)}\]
As a first step, we shall show that
\[\frac{\Delta v_{f}}{\Delta x} \leq \sup _{[p, x]}\left|f^{\prime}\right|.\]
For any partition \(P=\left\{p=t_{0}, \ldots, t_{m}=x\right\}\) of \([p, x],\) we have
\[S(f, P)=\sum_{i=1}^{m}\left|\Delta_{i} f\right| \leq \sum_{i=1}^{m} \sup _{\left[t_{i-1}, t_{i}\right]}\left|f^{\prime}\right|\left(t_{i}-t_{i-1}\right) \leq \sup _{[p, x]}\left|f^{\prime}\right| \Delta x.\]
Since this holds for any partition \(P,\) we have
\[V_{f}[p, x] \leq \sup _{[p, x]}\left|f^{\prime}\right| \Delta x,\]
which implies (2).
On the other hand,
\[\Delta v_{f}=V_{f}[p, x] \geq|f(x)-f(p)|=|\Delta f|.\]
Combining, we get
\[\left|\frac{\Delta f}{\Delta x}\right| \leq \frac{\Delta v_{f}}{\Delta x} \leq \sup _{[p, x]}\left|f^{\prime}\right|<+\infty\]
since \(f^{\prime}\) is relatively continuous on \([a, b],\) hence also uniformly continuous and bounded. (Here we assumed \(a<p<x \leq b\). However, (3) holds also if \(a \leq x<p<b,\) with \(\Delta v_{f}=-V[x, p]\) and \(\Delta x<0.\) Verify!)
Now
\[| | f^{\prime}(p)|-| f^{\prime}(x)| | \leq\left|f^{\prime}(p)-f^{\prime}(x)\right| \rightarrow 0 \quad \text { as } x \rightarrow p,\]
so, taking limits as \(x \rightarrow p,\) we obtain
\[\lim _{x \rightarrow p} \frac{\Delta v_{f}}{\Delta x}=\left|f^{\prime}(p)\right|.\]
Thus \(v_{f}\) is differentiable at each \(p\) in \((a, b),\) with \(v_{f}^{\prime}(p)=\left|f^{\prime}(p)\right|.\) Also, \(v_{f}\) is relatively continuous and finite on \([a, b]\) (by Theorem 1). Hence \(v_{f}=\int\left|f^{\prime}\right|\) on \([a, b],\) and we obtain
\[\int_{a}^{b}\left|f^{\prime}\right|=v_{f}(b)-v_{f}(a)=V_{f}[a, b], \text { as asserted.} \quad \square\]
Note 2. If the range space \(E\) is \(E^{n}\) (*or \(C^{n}\)), \(f\) has \(n\) components
\[f_{1}, f_{2}, \ldots, f_{n}.\]
By Theorem 5 in §1, \(f^{\prime}=\left(f_{1}^{\prime}, f_{2}^{\prime}, \ldots, f_{n}^{\prime}\right),\) so
\[\left|f^{\prime}\right|=\sqrt{\sum_{k=1}^{n}\left|f_{k}^{\prime}\right|^{2}},\]
and we get
\[V_{f}[a, b]=\int_{a}^{b} \sqrt{\sum_{k=1}^{n}\left|f_{k}^{\prime}\right|^{2}}=\int_{a}^{b} \sqrt{\sum_{k=1}^{n}\left|f_{k}^{\prime}(t)\right|^{2}} d t \quad\text {(classical notation).}\]
In particular, for complex functions, we have (see Chapter 4, §3, Note 5)
\[V_{f}[a, b]=\int_{a}^{b} \sqrt{f_{\mathrm{re}}^{\prime}(t)^{2}+f_{\mathrm{im}}^{\prime}(t)^{2}} d t.\]
In practice, formula (5) is used when a curve is given parametrically by
\[x_{k}=f_{k}(t), \quad k=1,2, \ldots, \text{ }n,\]
with the \(f_{k}\) differentiable on \([a, b].\) Curves in \(E^{2}\) are often given in nonparametric form as
\[y=F(x), \quad F : E^{1} \rightarrow E^{1}.\]
Here \(F[I]\) is \(n o t\) the desired curve but simply a set in \(E^{1}.\) To apply (5) here, we first replace "\(y=F(x)\)" by suitable parametric equations,
\[x=f_{1}(t) \text { and } y=f_{2}(t);\]
i.e., we introduce a function \(f : E^{1} \rightarrow E,\) with \(f=\left(f_{1}, f_{2}\right).\) An obvious (but not the only) way of achieving it is to set
\[x=f_{1}(t)=t \text { and } y=f_{2}(t)=F(t)\]
so that \(f_{1}^{\prime}=1\) and \(f_{2}^{\prime}=F^{\prime}.\) Then formula (5) may be written as
\[V_{f}[a, b]=\int_{a}^{b} \sqrt{1+F^{\prime}(x)^{2}} d x, \quad f(x)=(x, F(x)).\]
Find the length of the circle
\[x^{2}+y^{2}=r^{2}.\]
Here it is convenient to use the parametric equations
\[x=r \cos t \text { and } y=r \sin t,\]
i.e., to define \(f : E^{1} \rightarrow E^{2}\) by
\[f(t)=(r \cos t, r \sin t),\]
or, in complex notation,
\[f(t)=r e^{t i}.\]
Then the circle is obtained by letting \(t\) vary through \([0,2 \pi].\) Thus (5) yields
\[V_{f}[0,2 \pi]=\int_{a}^{b} r \sqrt{\cos ^{2} t+\sin ^{2} t} d t=r \int_{a}^{b} 1 d t=r\left.t\right|_{0} ^{2 \pi}=2 r \pi.\]
Note that \(f\) describes the same circle \(A=f[I]\) over \(I=[0,4 \pi].\) More generally, we could let \(t\) vary through any interval \([a, b]\) with \(b-a \geq 2 \pi.\) However, the length, \(V_{f}[a, b],\) would change (depending on \(b-a)\). This is because the circle \(A=f[I]\) is not a simple arc (see §7, Note 1), so \(\ell A\) depends on \(f\) and \(I,\) and one must be careful in selecting both appropriately.


1.7: Falling Objects
Learning Objectives
- Describe the effects of gravity on objects in motion.
- Describe the motion of objects that are in free fall.
- Calculate the position and velocity of objects in free fall.
Falling objects form an interesting class of motion problems. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. By applying the kinematics developed so far to falling objects, we can examine some interesting situations and learn much about gravity in the process.
Gravity
The most remarkable and unexpected fact about falling objects is that, if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration , independent of their mass . This experimentally determined fact is unexpected, because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones.
In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball will reach the ground after a hard baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, while friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them. For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall .
The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called the acceleration due to gravity . The acceleration due to gravity is constant , which means we can apply the kinematics equations to any falling object where air resistance and friction are negligible. This opens a broad class of interesting situations to us. The acceleration due to gravity is so important that its magnitude is given its own symbol, \(g\). It is constant at any given location on Earth and has the average value
\[g=9.80 \mathrm{~m} / \mathrm{s}^{2}. \nonumber \]
Although \(g\) varies from \(9.78 \mathrm{~m} / \mathrm{s}^{2}\) to \(9.83 \mathrm{~m} / \mathrm{s}^{2}\), depending on latitude, altitude, underlying geological formations, and local topography, the average value of \(9.80 \mathrm{~m} / \mathrm{s}^{2}\) will be used in this text unless otherwise specified. The direction of the acceleration due to gravity is downward (towards the center of Earth) . In fact, its direction defines what we call vertical. Note that whether the acceleration \(a\) in the kinematic equations has the value \(+g\) or \(−g\) depends on how we define our coordinate system. If we define the upward direction as positive, then \(a=-g=-9.80 \mathrm{~m} / \mathrm{s}^{2}\), and if we define the downward direction as positive, then \(a=g=9.80 \mathrm{~m} / \mathrm{s}^{2}\). In certain examples and homework exercises, slightly inaccurate but more convenient value of \(g \simeq 10 \mathrm{~m} / \mathrm{s}^{2}\) will be used with appropriate specifications.
One-Dimensional Motion Involving Gravity
The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So we start by considering straight up and down motion with no air resistance or friction. These assumptions mean that the velocity (if there is any) is vertical. If the object is dropped, we know the initial velocity is zero. Once the object has left contact with whatever held or threw it, the object is in free-fall. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude \(g\). We will also represent vertical displacement with the symbol \(y\) and use \(x\) for horizontal displacement.
KINEMATIC EQUATIONS FOR OBJECTS IN FREE-FALL
\[v=v_{0}-g t \nonumber \]
\[y=y_{0}+v_{0} t-\frac{1}{2} g t^{2} \nonumber \]
\[v^{2}=v_{0}^{2}-2 g\left(y-y_{0}\right) \nonumber \]
Example \(\PageIndex{1}\): Calculating Position and Velocity of a Falling Object: A Rock Thrown Upward
A person standing on the edge of a high cliff throws a rock straight up with an initial velocity of 13.0 m/s . The rock misses the edge of the cliff as it falls back to earth. Calculate the position and velocity of the rock 1.00 s, 2.00 s, and 3.00 s after it is thrown, neglecting the effects of air resistance.
Strategy
Draw a sketch.
We are asked to determine the position \(y\) at various times. It is reasonable to take the initial position \(y_{0}\) to be zero. This problem involves one-dimensional motion in the vertical direction. We use plus and minus signs to indicate direction, with up being positive and down negative. Since up is positive, and the rock is thrown upward, the initial velocity must be positive too. The acceleration due to gravity is downward, so \(a\) is negative. It is crucial that the initial velocity and the acceleration due to gravity have opposite signs. Opposite signs indicate that the acceleration due to gravity opposes the initial motion and will slow and eventually reverse it.
Since we are asked for values of position and velocity at three times, we will refer to these as \(y_{1}\) and \(v_{1}\); \(y_{2}\) and \(v_{2}\); and \(y_{3}\) and \(v_{3}\).
Solution for Position \(y_{1}\)
1. Identify the knowns. We know that \(y_{0}=0 ; \ v_{0}=13.0 \mathrm{~m} / \mathrm{s} ; \ a=-g=-9.80 \mathrm{~m} / \mathrm{s}^{2}\); and \(t=1.00\) s.
2. Identify the best equation to use. We will use \(y=y_{0}+v_{0} t+\frac{1}{2} a t^{2}\) because it includes only one unknown, \(y\) (or \(y_{1}\), here), which is the value we want to find.
3. Plug in the known values and solve for \(y_{1}\).
\[y_{1}=0+(13.0 \mathrm{~m} / \mathrm{s})(1.00 \mathrm{~s})+\frac{1}{2}\left(-9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(1.00 \mathrm{~s})^{2}=8.10 \mathrm{~m} \nonumber\]
Discussion
The rock is 8.10 m above its starting point at \(t=1.00\) s, since \(y_{1}>y_{0}\). It could be moving up or down; the only way to tell is to calculate \(v_{1}\) and find out if it is positive or negative.
Solution for Velocity \(v_{1}\)
1. Identify the knowns. We know that \(y_{0}=0 ; \ v_{0}=13.0 \mathrm{~m} / \mathrm{s} ; \ a=-g=-9.80 \mathrm{~m} / \mathrm{s}^{2}\); and \(t=1.00 s\) . We also know from the solution above that \(y_{1}=8.10 \mathrm{~m}\).
2. Identify the best equation to use. The most straightforward is \(v=v_{0}-g t\) (from \(v=v_{0}+a t\), where \(a=\text { gravitational acceleration }=-g\)).
3. Plug in the knowns and solve.
\[v_{1}=v_{0}-g t=13.0 \mathrm{~m} / \mathrm{s}-\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(1.00 \mathrm{~s})=3.20 \mathrm{~m} / \mathrm{s}\nonumber\]
Discussion
The positive value for \(v_{1}\) means that the rock is still heading upward at \(t\)=1.00s . However, it has slowed from its original 13.0 m/s, as expected.
Solution for Remaining Times
The procedures for calculating the position and velocity at \(t\) = 2.00 s and 3.00 s are the same as those above. The results are summarized in Table \(\PageIndex{1}\) and illustrated in Figure \(\PageIndex{3}\).
| Time, t | Position, y | Velocity, v | Acceleration, a |
|---|---|---|---|
| 1.00 s | 8.10 m | 3.20 m | \(-9.80 \mathrm{~m} / \mathrm{s}^{2}\) |
| 2.00 s | 6.40 m | −6.60 m | \(-9.80 \mathrm{~m} / \mathrm{s}^{2}\) |
| 3.00 s | −5.10 m | −16.4 m | \(-9.80 \mathrm{~m} / \mathrm{s}^{2}\) |
Graphing the data helps us understand it more clearly.
Discussion
The interpretation of these results is important. At 1.00 s the rock is above its starting point and heading upward, since \(y_{1}\) and \(v_{1}\) are both positive. At 2.00 s, the rock is still above its starting point, but the negative velocity means it is moving downward. At 3.00 s, both \(y_{3}\) and \(v_{3}\) are negative, meaning the rock is below its starting point and continuing to move downward. Notice that when the rock is at its highest point (at 1.5 s), its velocity is zero, but its acceleration is still \(-9.80 \mathrm{~m} / \mathrm{s}^{2}\). Its acceleration is \(-9.80 \mathrm{~m} / \mathrm{s}^{2}\) for the whole trip—while it is moving up and while it is moving down. Note that the values for \(y\) are the positions (or displacements) of the rock, not the total distances traveled. Finally, note that free-fall applies to upward motion as well as downward. Both have the same acceleration—the acceleration due to gravity, which remains constant the entire time. Astronauts training in the famous Vomit Comet, for example, experience free-fall while arcing up as well as down, as we will discuss in more detail later.
 Have a friend hold a ruler between your thumb and index finger, separated by about 1 cm. Note the mark on the ruler that is right between your fingers. Have your friend drop the ruler unexpectedly, and try to catch it between your two fingers. Note the new reading on the ruler. Assuming acceleration is that due to gravity, calculate your reaction time. How far would you travel in a car (moving at 30 m/s) if the time it took your foot to go from the gas pedal to the brake was twice this reaction time?
Example \(\PageIndex{2}\): Calculating Velocity of a Falling Object: A Rock Thrown Down
What happens if the person on the cliff throws the rock straight down, instead of straight up? To explore this question, calculate the velocity of the rock when it is 5.10 m below the starting point, and has been thrown downward with an initial speed of 13.0 m/s.
Strategy
Draw a sketch.
Since up is positive, the final position of the rock will be negative because it finishes below the starting point at \(y_{0}=0\). Similarly, the initial velocity is downward and therefore negative, as is the acceleration due to gravity. We expect the final velocity to be negative since the rock will continue to move downward.
Solution
1. Identify the knowns. \(y_{0}=0 ; \ y_{1}=-5.10 \mathrm{~m} ; \ v_{0}=-13.0 \mathrm{~m} / \mathrm{s} ; \ a=-g=-9.80 \mathrm{~m} / \mathrm{s}^{2}\).
2. Choose the kinematic equation that makes it easiest to solve the problem. The equation \(v^{2}=v_{0}^{2}+2 a\left(y-y_{0}\right)\) works well because the only unknown in it is \(v\). (We will plug \(y_{1}\) in for \(y\).)
3. Enter the known values
\[v^{2}=(-13.0 \mathrm{~m} / \mathrm{s})^{2}+2\left(-9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(-5.10 \mathrm{~m}-0 \mathrm{~m})=268.96 \mathrm{~m}^{2} / \mathrm{s}^{2}, \nonumber\]
where we have retained extra significant figures because this is an intermediate result.
Taking the square root, and noting that a square root can be positive or negative, gives
\[v=\pm 16.4 \mathrm{~m} / \mathrm{s}. \nonumber\]
The negative root is chosen to indicate that the rock is still heading down. Thus,
\[v=-16.4 \mathrm{~m} / \mathrm{s} \nonumber.\]
Discussion
Note that this is exactly the same velocity the rock had at this position when it was thrown straight upward with the same initial speed . (See Example \(\PageIndex{1}\) and Figure \(\PageIndex{5}\) (a).) This is not a coincidental result. Because we only consider the acceleration due to gravity in this problem, the speed of a falling object depends only on its initial speed and its vertical position relative to the starting point. For example, if the velocity of the rock is calculated at a height of 8.10 m above the starting point (using the method from Example \(\PageIndex{1}\)) when the initial velocity is 13.0 m/s straight up, a result of \(\pm 3.20 \mathrm{~m} / \mathrm{s}\) is obtained. Here both signs are meaningful; the positive value occurs when the rock is at 8.10 m and heading up, and the negative value occurs when the rock is at 8.10 m and heading back down. It has the same speed but the opposite direction.
Another way to look at it is this: In Example \(\PageIndex{1}\), the rock is thrown up with an initial velocity of 13.0 m/s . It rises and then falls back down. When its position is \(y=0\) on its way back down, its velocity is −13.0 m/s . That is, it has the same speed on its way down as on its way up. We would then expect its velocity at a position of \(y=−5.10 \mathrm{m}\) to be the same whether we have thrown it upwards at +13.0 m/s or thrown it downwards at −13.0 m/s . The velocity of the rock on its way down from \(y=0\) is the same whether we have thrown it up or down to start with, as long as the speed with which it was initially thrown is the same.
Example \(\PageIndex{3}\): Find \(g\) from Data on a Falling Object
The acceleration due to gravity on Earth differs slightly from place to place, depending on topography (e.g., whether you are on a hill or in a valley) and subsurface geology (whether there is dense rock like iron ore as opposed to light rock like salt beneath you.) The precise acceleration due to gravity can be calculated from data taken in an introductory physics laboratory course. An object, usually a metal ball for which air resistance is negligible, is dropped and the time it takes to fall a known distance is measured. See, for example, Figure \(\PageIndex{6}\). Very precise results can be produced with this method if sufficient care is taken in measuring the distance fallen and the elapsed time.
Suppose the ball falls 1.0000 m in 0.45173 s. Assuming the ball is not affected by air resistance, what is the precise acceleration due to gravity at this location?
Strategy
Draw a sketch.
We need to solve for acceleration \(a\). Note that in this case, displacement is downward and therefore negative, as is acceleration.
Solution
1. Identify the knowns. \(y_{0}=0 ; \ y=-1.0000 \mathrm{~m} ; \ t=0.45173 ; \ v_{0}=0 .\).
2. Choose the equation that allows you to solve for \(a\) using the known values.
\[y=y_{0}+v_{0} t+\frac{1}{2} a t^{2} \nonumber\]
3. Substitute 0 for \(v_{0}\) and rearrange the equation to solve for \(a\). Substituting 0 for \(v_{0}\) yields
\[y=y_{0}+\frac{1}{2} a t^{2}. \nonumber\]
Solving for \(a\) gives
\[a=\frac{2\left(y-y_{0}\right)}{t^{2}}\nonumber.\]
4. Substitute known values yields
\[a=\frac{2(-1.0000 \mathrm{~m}-0)}{(0.45173 \mathrm{~s})^{2}}=-9.8010 \mathrm{~m} / \mathrm{s}^{2} \nonumber,\]
so, because \(a=−g\) with the directions we have chosen,
\[g=9.8010 \mathrm{~m} / \mathrm{s}^{2}. \nonumber\]
Discussion
The negative value for \(a\) indicates that the gravitational acceleration is downward, as expected. We expect the value to be somewhere around the average value of \(9.80 \mathrm{~m} / \mathrm{s}^{2}\), so \(9.8010 \mathrm{~m} / \mathrm{s}^{2}\) makes sense. Since the data going into the calculation are relatively precise, this value for \(g\) is more precise than the average value of \(9.80 \mathrm{~m} / \mathrm{s}^{2}\); it represents the local value for the acceleration due to gravity.
Exercise \(\PageIndex{1}\)
A chunk of ice breaks off a glacier and falls 30.0 meters before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water?
- Answer
-
We know that initial position \(y_{0}=0\), final position \(y=-30.0 \mathrm{~m}\), and \(a=-g=-9.80 \mathrm{~m} / \mathrm{s}^{2}\). We can then use the equation \(y=y_{0}+v_{0} t+\frac{1}{2} a t^{2}\) to solve for \(t\). Inserting \(a=-g\), we obtain
\[\begin{aligned}
&y=0+0-\frac{1}{2} g t^{2} \\
&t^{2}=\frac{2 y}{-g} \\
&t \quad=\pm \sqrt{\frac{2 y}{-g}}=\pm \sqrt{\frac{2(-30.0 \mathrm{~m})}{-9.80 \mathrm{~m} / \mathrm{s}^{2}}}=\pm \sqrt{6.12 \mathrm{~s}^{2}}=2.47 \mathrm{~s} \approx 2.5 \mathrm{~s}
\end{aligned} \nonumber\]where we take the positive value as the physically relevant answer. Thus, it takes about 2.5 seconds for the piece of ice to hit the water.
Section Summary
- An object in free-fall experiences constant acceleration if air resistance is negligible.
-
On Earth, all free-falling objects have an acceleration due to gravity \(g\), which averages
\[g=9.80 \mathrm{~m} / \mathrm{s}^{2} \nonumber.\]
- Whether the acceleration \(a\) should be taken as \(+g\) or \(−g\) is determined by your choice of coordinate system. If you choose the upward direction as positive,\(a=-g=-9.80 \mathrm{~m} / \mathrm{s}^{2}\) is negative. In the opposite case, \(a=+\mathrm{g}=9.80 \mathrm{~m} / \mathrm{s}^{2}\) is positive. Since acceleration is constant, the kinematic equations above can be applied with the appropriate \(+g\) or \(−g\) substituted for \(a\).
- For objects in free-fall, up is normally taken as positive for displacement, velocity, and acceleration.
Glossary
- free-fall
- the state of movement that results from gravitational force only
- acceleration due to gravity
- acceleration an object undergoes while in free-fall