32.8: Review Questions
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1.
Which action would the nurse perform while engaged in the mitigation phase of disaster management?
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Simulating the planned response to a hurricane event
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Placing sandbags at a river’s edge prior to flooding
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Preparing a local school for temporary housing for flood victims
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Completing a tabletop exercise to plan for flooding
2.
Using the START triage system, which victim would the nurse triage as immediate or red during a disaster response?
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A victim who is not breathing after two attempts to open the airway
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A victim who has a broken arm and can obey commands
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A victim who has multiple lacerations on extremities and chest
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A victim who has a respiratory rate greater than 30 beats/minute with flail chest
3.
The nurse is assessing the victims of a train derailment using the Sort, Assess, Life-saving Interventions, and Triage/treatment (SALT) system. Which action will the nurse take during the Sort phase of this triage system?
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Moving victims who cannot walk to a staging area
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Placing color-coded triage tags on victims
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Opening the airway of victims who are not breathing
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Asking victims to walk to a staging area
4.
Using the six steps of disaster risk assessment to determine a community’s hazard risk, which question would the nurse ask during step 3, vulnerability assessment?
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Is the hazard risk high, moderate, or low?
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Is the risk of disaster seasonal?
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What are the estimated damages that would occur with the hazard?
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What road might be impacted during the hazard?
5.
Which characteristic of a Category A biological agent will guide the nurse’s actions during an outbreak?
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They have low mortality rates.
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They are highly transmissible.
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They are not a threat to the community.
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They are difficult to disperse.
6.
Which action would the nurse perform as Logistics Section Chief following the activation of the Incident Command System after a hurricane?
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Provide boats for rescue from flooded areas
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Set objectives for hurricane response
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Create press releases regarding hurricane safety
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Conduct missing persons investigations
7.
The nurse is assessing a 6-year-old child at the scene of a school bus accident and notes that the child has a respiratory rate of 14 breaths/minute. Which color triage tag will the nurse assign to the child using the JumpSTART triage system?
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Black
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Green
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Yellow
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Red
8.
Which action will the community health nurse take during the primary prevention stage of bioterrorism preparedness?
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Participating in disaster simulations
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Recognizing signs and symptoms of a bioterrorist attack
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Immediately treating disease caused by a biologic agent
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Responding to the psychological aftermath of the bioterrorism event
9.
A community health nurse conducting a class for high school parents and teachers on mass violence would include which adolescent behavior as a warning sign for potential school violence?
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Participation in class activities
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Withdrawing from social interaction
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Crying after failing an exam
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Visiting the school counselor during lunch
10.
Which activity would the community nurse perform during the recovery phase of the disaster management cycle?
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Performing triage and emergency care
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Educating families on disaster preparedness
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Reducing the spread of communicable disease
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Performing triage and transport victims to services


6.21: Where do you go from here?
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Staying aware of how and when you communicate is important to every aspect of your life. As you learned throughout this chapter, there are many variables that come into play when you communicate. Most of the time none of us really think about all of these before we send out some kind of communication. Perhaps you would like to learn a little more about how to be mindful of your methods of communication. It will serve you well throughout your life.
Choose a topic that interests you from the list below. Then search for sources that inform this topic and create an annotated bibliography that would be useful to someone who is interested in doing further research.


26.7: Review Questions
A client is diagnosed with a hypothalamic disorder affecting the production of vasopressin. Which of the following manifestations is the nurse least likely to assess in this client?
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Hypoglycemia
A client is prescribed a drug that inhibits the action of CRH. After administration of this drug, what should the nurse monitor?
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Cortisol
A client is prescribed octreotide 150 mcg subcutaneously daily in three divided doses. The available drug comes in a concentration of 100 mcg/mL. How many milliliters should the nurse administer for each dose?
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0.5 mL
A client with adrenal insufficiency is prescribed hydrocortisone. Which action should the nurse take to detect potential adverse effects of this drug?
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Assess for hyperglycemia and hypertension
A nurse is caring for a client with acromegaly. Which explanation of this disorder will the nurse give to the client?
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“Too much growth hormone causes abnormal growth of bones and soft tissues.”
A client is prescribed fludrocortisone. Which findings indicate to the nurse that the client is having adverse effects from this medication?
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Hypokalemia and hypertension
A client requires 120 mg of methylprednisolone. The available medication comes in a 125 mg/2 mL vial. How many milliliters should the nurse administer if rounded to the nearest tenth?
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1.9 mL
A client is being discharged on a glucocorticosteroid drug for a flare-up of rheumatoid arthritis. Which instruction should the nurse give to the client?
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Taper the medication as directed.
A client with Parkinson’s disease has an order for bromocriptine. Which assessment finding would cause the nurse to not administer the drug and notify the provider?
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Blood pressure 196/108 mm Hg
A client presents with hyperaldosteronism. What assessment findings should the nurse expect?
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Increased blood pressure, decreased potassium levels


4.3: Beta-1 Antagonists
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- Ernstmeyer & Christman (Eds.)
- Chippewa Valley Technical College via OpenRN
Metoprolol is a selective Beta-1 antagonist.
Mechanism of Action: Metoprolol primarily blocks Beta-1 receptors in the heart, causing decreased heart rate and decreased blood pressure. However, higher doses can also block Beta-2 receptors in the lungs, causing bronchoconstriction.
Indications: Metoprolol is commonly used to treat high blood pressure, chest pain due to poor blood flow to the heart, as an early intervention during a myocardial infarction (MI), and in several heart conditions involving an abnormally fast heart rate.
Nursing Considerations: Don’t crush extended-release (ER) formulations. Always check patient’s apical pulse rate before giving drug. Withhold the drug and call the prescriber immediately if the heart rate is slower than 60 beats/minute, unless other parameters are provided. In diabetic patients, monitor glucose level closely because the drug masks common signs and symptoms of hypoglycemia. The most serious potential adverse effects are shortness of breath, bradycardia, and worsening heart failure. Other adverse effects include fatigue, dizziness, depression, insomnia, nightmares, gastrointestinal upset, erectile dysfunction, dyspnea, and wheezing.
Black Box Warning : When stopping therapy, the dosage should be tapered over 1 to 2 weeks because abrupt discontinuation may cause chest pain or myocardial infarction (MI).
Patient Teaching & Education: Patients should be instructed to take the medication as prescribed. They should be advised that abrupt cessation of medication therapy may result in life-threatening cardiac arrhythmias. Patients should also be taught how to self-check pulse and blood pressure to assess the effectiveness of medication therapy. Additionally, they should be cautioned against sudden changes in position due to orthostatic blood pressure changes. Patients may experience increase sensitivity to cold and should be cautioned to avoid caffeinated substances. [1]
Now let’s take a closer look at the medication grid on metoprolol in Table 4.12. [2]
|
Class/Subclass |
Prototype/Generic |
Administration Considerations |
Therapeutic Effects |
Side/Adverse Effects |
|---|---|---|---|---|
| Beta-1 antagonist | Selective B blocker: metoprolol |
Do not crush extended-release (ER) formulations
Always assess apical HR and if less than 60, do not administer and call the prescriber unless other parameters are provided Monitor blood sugar in diabetic patients because drug can mask symptoms of hypoglycemia |
Decreases blood pressure or controls rapid heart rate |
Decreased blood pressure or heart rate
Most serious: -Hypotension -Bradycardia -Worsening heart failure (HF) Other: -CNS: fatigue, dizziness, depression, insomnia, nightmares -GI upset -GU: erectile dysfunction -Respiratory: dyspnea and wheezing |
- uCentral from Unbound Medicine. https://www.unboundmedicine.com/ucentral ↵
- This work is a derivative of Daily Med by U.S. National Library of Medicine in the public domain . ↵


11.3: Absolute Motion Analysis
Absolute motion analysis is one method used to analyze bodies undergoing general planar motion . General planar motion is motion where bodies can both translate and rotate at the same time. Besides absolute motion analysis, the alternative is relative motion analysis. Either method can be used for any general planar motion problem, but one method may be significantly easier to apply for a given situation.
Absolute motion analysis will require calculus, and is generally faster for simple problems and problems where only the velocities (and not accelerations) are required. Relative motion analysis will not require calculus, but does necessitate using multiple coordinate systems; it is generally easier to use for more complex problems and problems where velocities and accelerations are being analyzed.
Utilizing Absolute Motion Analysis:
To start our discussion on absolute motion analysis, we are going to imagine a simple robotic arm such as the one below. In this arm, we have two arm sections of fixed length with motors causing rotations at joint A and joint B.
The first step in absolute motion analysis is come up with a set of equations describing the position of some point of interest. In this case we will be looking at the position of the end effector of the arm at point C, and we will write an equation for the \(x\) position and the \(y\) position of this point with respect to the fixed origin point at A. In these equations, anything that is a constant (such as the length of the arm pieces) can be put in as a number, but anything that will change, such as angles \(\theta\) and \(\phi\), will need to remain as variables in these equations even if they are known at the moment. Using the values in the diagram, we would wind up with the following two position equations.
\begin{align} x \text{-position:} \quad &\, x_C = 2 \cos (\theta) + 1.5 \cos (\phi) \\[5pt] y \text{-position:} \quad &\, y_C = 2 \sin (\theta) + 1.5 \sin (\phi) \end{align}
To find the velocity of point C in the \(x\) and \(y\) directions, we simply need to take the derivatives of the position equations. The velocity equations for our robotic arm are below.
\begin{align} x \text{-velocity:} \quad &\, v_{x \ C} = -2 \sin (\theta) \ \dot{\theta} - 1.5 \sin (\phi) \ \dot{\phi}\\[5pt] y \text{-velocity:} \quad &\, v_{y \ C} = 2 \cos (\theta) \ \dot{\theta} + 1.5 \cos (\phi) \ \dot{\phi} \end{align}
To find the acceleration of point C in the \(x\) and \(y\) directions, we simply need to take the derivatives of the velocity equations. The acceleration equations for our robotic arm in the \(x\) and \(y\) directions are shown below.
\begin{align} x \text{ acceleration:} \quad &\, a_{x \ C} = -2 \cos (\theta) \ \dot{\theta}^2 - 2 \sin (\theta) \ \ddot{\theta} - 1.5 \cos (\phi) \ \dot{\phi}^2 - 1.5 \sin (\phi) \ \ddot{\phi} \\[5pt] y \text{ acceleration:} \quad &\, a_{y \ C} = -2 \sin (\theta) \ \dot{\theta}^2 + 2 \cos (\theta) \ \ddot{\theta} - 1.5 \sin (\phi) \ \dot{\phi}^2 + 1.5 \cos (\phi) \ \ddot{\phi} \end{align}
Once we have the velocity and acceleration equations, we can start solving for any unknowns. If we have known angular velocities and accelerations (\(\dot{\theta}\), \(\ddot{\theta}\), \(\dot{\phi}\), and \(\ddot{\phi}\)) we can plug those in to find the velocity and acceleration vectors for the end effector. In other instances, we may known the desired motion of the end effector (\(v_{xC}\), \(v_{yC}\), \(a_{xC}\), and \(a_{yC}\)) and will have to plug those values into the equations to solve for unknowns such as \(\dot{\theta}\) and so on.
Example \(\PageIndex{1}\)
The robotic arm shown below has a fixed orange base at A and fixed-length members AB and BC. Motors at A and B allow for rotational motion at the joints. Based on the angular velocities and accelerations shown at each joint, determine the velocity and the acceleration of the end effector at C.
- Solution
Example \(\PageIndex{2}\)
The robotic arm from the previous problem is in the configuration shown below. Assume that \(\theta\) is currently 30 degrees and that point C currently lies along the \(x\) axis. If we want the end effector at C to travel 1 ft/s in the negative \(x\)-direction, what should the angular velocities be at joints A and B?
- Solution
Example \(\PageIndex{3}\)
A ladder is propped up against a wall as shown below. If the base of the ladder is sliding out at a speed of 2 m/s, what is the speed of the top of the ladder?
- Solution
Example \(\PageIndex{4}\)
The crank-rocker mechanism as shown below consists of a crank with a radius of 0.5 meters rotating about its fixed center at C, at a constant rate of 2 rad/s clockwise. Rocker AB is fixed at its base at A and connects to point B along the edge of the crank. The pin at point B can slide along a frictionless slot in AB. In the current state, what is the angular velocity of rocker AB?
- Solution


1.14.22: Descriptions of Systems
Important themes in thermodynamics involve (i) properties (variables) which can be measured (e.g. volumes and/or densities) and (ii) thermodynamic variables which are rigorously defined (e.g. enthalpies). In these terms a measured property (e.g. density) is the reporter of the chemical properties or processes taking place in a system. So it is always important to ask if the “reporter” can be interrogated for the required information. In fact there is often a limit to the amount of information which a given reporter offers to the investigator. These important limitations should be borne in mind. An example makes the point.
A system is prepared by placing \(\mathrm{n}_{\mathrm{X}}^{0}\) moles of substance in a closed vessel at fixed \(\mathrm{T}\) and \(\mathrm{p}\). [The superscript ‘0’ implies at zero time.] We explore two possible descriptions of this system. Perhaps two samples were analysed by two independent laboratories.
Description A
The first laboratory reports that the system is simple and contains the single substance \(\mathrm{X}\).
Gibbs energy
\[\mathrm{G}(\mathrm{A})=\mathrm{n}_{\mathrm{X}}^{0} \, \mu_{\mathrm{X}}^{*}(\ell) \nonumber \]
and volume
\[\mathrm{V}(\mathrm{A})=\mathrm{n}_{\mathrm{X}}^{0} \, \mathrm{V}_{\mathrm{X}}^{*}(\ell) \nonumber \]
Here \(\mu_{X}^{*}(\ell)\) and \(\mathrm{V}_{\mathrm{x}}^{*}(\ell)\) are the chemical potential and molar volume of the pure chemical substance \(\mathrm{X}\).
Description B
The second laboratory identifies two substances \(\mathrm{X}\) and \(\mathrm{Y}\) in chemical equilibrium such that the equilibrium amounts of substances \(\mathrm{X}\) and \(\mathrm{Y}\) are respectively \(\mathrm{n}_{\mathrm{X}}^{\mathrm{eq}}\) and \(\mathrm{n}_{\mathrm{Y}}^{\mathrm{eq}}\).
Gibbs energy
\[\mathrm{G}(\mathrm{B})=\mathrm{n}_{\mathrm{X}}^{\mathrm{eq}} \, \mu_{\mathrm{X}}^{\mathrm{eq}}+\mathrm{n}_{\mathrm{Y}}^{\mathrm{eq}} \, \mu_{\mathrm{Y}}^{\mathrm{eq}} \nonumber \]
and volume
\[\mathrm{V}(\mathrm{B})=\mathrm{n}_{\mathrm{X}}^{\mathrm{eq}} \, \mathrm{V}_{\mathrm{X}}^{\mathrm{eq}}+\mathrm{n}_{\mathrm{Y}}^{\mathrm{eq}} \, \mathrm{V}_{\mathrm{Y}}^{\mathrm{eq}} \nonumber \]
Here \(\mu_{\mathrm{X}}^{\mathrm{eq}}\) and \(\mu_{\mathrm{Y}}^{\mathrm{eq}}\) are the equilibrium chemical potentials; \(\mathrm{V}_{\mathrm{X}}^{\mathrm{eq}}\) and \(\mathrm{V}_{\mathrm{Y}}^{\mathrm{eq}}\) are the equilibrium partial molar volumes.
Description A is “primitive” and Description B is “sophisticated”. Both Gibbs energies and volumes are functions of state so that \(\mathrm{V}(\mathrm{A})=\mathrm{V}(\mathrm{B})\) and \(\mathrm{G}(\mathrm{A})=\mathrm{G}(\mathrm{B})\). The chemical potential of substance \(\mathrm{X}\) describes the change in \(\mathrm{G}\) when \(\delta n_{X}\) moles of \(\mathrm{X}\) are added. This chemical potential is insensitive to the changes taking place in the equilibrium system;\(\mu_{X}(\mathrm{~A})=\mu_{X}(\mathrm{~B})\). Consequently, measurement of volume \(\mathrm{V}\) [and if it were possible of \(\mathrm{G}\)] would not distinguish between the two descriptions. Similarly, measurement of \(\mathrm{H}\) (if it were possible) would not distinguish between the two descriptions.
Footnotes
[1] L. P. Hammett, Physical Organic Chemistry, McGraw-Hill, New York, 2nd edn., 1970,p.16.


9.3: A.3- Definitions in Mathematics
It is difficult to overstate the importance of definitions in mathematics. Definitions play a different role in mathematics than they do in everyday life.
Suppose you give your friend a piece of paper containing the definition of the rarely-used word rodomontade .
Give your friend some time to study the definition. Then take away the paper. Ten minutes later ask her to define rodomontade. Most likely she will be able to give a reasonably accurate definition.  It is unlikely that she will have quoted the OED word-for-word. In everyday English that is fine—you would probably agree that your friend knows the meaning of the rodomontade. This is because most definitions are descriptive . They describe the common usage of a word.
Let us take a mathematical example. The OED* gives this definition of continuous .
Characterized by continuity; extending in space without interruption of substance; having no interstices or breaks; having its parts in immediate connection; connected, unbroken.
Likewise, we often hear calculus students speak of a continuous function as one whose graph can be drawn “without picking up the pencil.” This definition is descriptive. (As we learned in calculus the picking-up-the-pencil description is not a perfect description of continuous functions.) This is not a mathematical definition.
Mathematical definitions are prescriptive . The definition must prescribe the exact and correct meaning of a word. Contrast the OED’s descriptive definition of continuous with the the definition of continuous found in a real analysis textbook.
A function \(f:A\to \mathbb{R}\) is continuous at a point \(c\in A\) if, for all \(\varepsilon>0\) , there exists \(\delta>0\) such that whenever \(|x-c|<\delta\) (and \(x\in A\) ) it follows that \(|f(x)-f(c)|<\varepsilon\) . If \(f\) is continuous at every point in the domain \(A\) , then we say that \(f\) is continuous on \(A\) .*
- *
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This definition is taken from page 109 of Stephen Abbott’s Understanding Analysis , but the definition would be essentially the same in any modern real analysis textbook.
In mathematics there is very little freedom in definitions. Mathematics is a deductive theory; it is impossible to state and prove theorems without clear definitions of the mathematical terms. The definition of a term must completely, accurately, and unambiguously describe the term. Each word is chosen very carefully and the order of the words is critical. In the definition of continuity changing “there exists” to “for all,” changing the orders of quantifiers, changing \(<\) to \(\leq\) or \(>\) , or changing \(\mathbb{R}\) to \(\mathbb{Z}\) would completely change the meaning of the definition.
What does this mean for you, the student? Our recommendation is that at this stage you memorize the definitions word-for-word. It is the safest way to guarantee that you have it correct. As you gain confidence and familiarity with the subject you may be ready to modify the wording. You may want to change “for all” to “given any” or you may want to change \(|x-c|<\delta\) to \(-\delta<x-c<\delta\) or to “the distance between \(x\) and \(c\) is less than \(\delta\) .”
Of course, memorization is not enough; you must have a conceptual understanding of the term, you must see how the formal definition matches up with your conceptual understanding, and you must know how to work with the definition. It is perhaps with the first of these that descriptive definitions are useful. They are useful for building intuition and for painting the “big picture.” Only after days (weeks, months, years?) of experience does one get an intuitive feel for the \(\varepsilon,\delta\) -definition of continuity; most mathematicians have the “picking-up-the-pencil” definitions in their head. This is fine as long as we know that it is imperfect, and that when we prove theorems about continuous functions in mathematics we use the mathematical definition.
We end this discussion with an amusing real-life example in which a descriptive definition was not sufficient. In 2003 the German version of the game show Who wants to be a millionaire? contained the following question: “Every rectangle is: (a) a rhombus, (b) a trapezoid, (c) a square, (d) a parallelogram.”
The confused contestant decided to skip the question and left with €4000. Afterward the show received letters from irate viewers. Why were the contestant and the viewers upset with this problem? Clearly a rectangle is a parallelogram, so (d) is the answer. But what about (b)? Is a rectangle a trapezoid? We would describe a trapezoid as a quadrilateral with a pair of parallel sides. But this leaves open the question: can a trapezoid have two pairs of parallel sides or must there only be one pair? The viewers said two pairs is allowed, the producers of the television show said it is not. This is a case in which a clear, precise, mathematical definition is required.


1.2: Defining the Opportunity
Defining the Opportunity
Finding the intersection of technology and market
So where is the next million-dollar idea hiding, just waiting to be discovered? The answer: within you. Truly. In fact, you might believe you've got a million-dollar idea in your head already. But a great idea alone doesn't guarantee success: you also need to be able to recognize when a concept has the potential to become something tangible and hit it big. It all comes down to opportunity recognition (there's that word again).
How do you recognize an opportunity? What exactly does one look like? It has to do with analyzing both the product and the market and seeing if they intersect. What you're looking for is a product that fulfills (or creates) a need at just the right time, in just the right place. Opportunities are built using a combination of your ideas and entrepreneurial creativity mixed with market need and good situational timing.
Think about your idea as you read the rest of this chapter—or, if you don't have an idea yet (which is perfectly fine), let it inform your idea generation process—and ask yourself these questions:
- Is my idea realistic? Is it technically and financially feasible?
- Is there a strong need for my potential product?
- Is this a need I passionately want to address?
- Does pursuing this opportunity meet my goals as an entrepreneur?
Where's the pain?
There's an old metaphor about meeting customer needs:.Come up with a product that scratches a customer's itch, and maybe they'll buy it. Come up with a product that relieves a customer's pain, and you've got a winner. "Pain," of course, is a relative term and could describe just about anything-a toothache, a slow computer, a boring job-but the idea behind the metaphor is that it's the entrepreneur's responsibility to find out what the customer needs most, then fulfill that need. It goes without saying that if your product isn't grounded in reality and serves no need at all-doesn't even scratch an itch-it has no chance. No one buys useless products.
On the cutting edge
Quite a few products and services out there simply fulfill a pre-existing need. Returning to the previous example, if you have a toothache, you go to the dentist and she fulfills your (painful, immediate) need. Simple. But when you're talking about high-tech products, things get a lot more complex. You may very well come up with an idea so new and innovative that it addresses a need that doesn't even exist yet. This is a difficult spot to be in as your product may have a hard time gaining traction.
When you're on the cutting edge:
- Get into your customers' heads as much as possible
- Develop cheap, easy-to-test prototypes
- Put the prototypes into people's hands and listen to what they say
- Be open to your product perhaps being useful in a market you never originally imagined
The last point is particularly instructive. When 3-M chemist Spence Silver invented a new adhesive that would not dry or permanently bond to things, he had no idea what to do with it. When another 3-M chemist, Arthur Fry, needed a bookmark for his choir book the idea for applying the glue to small pieces of paper emerged, and Post-It Notes were born.
What you know
Understandably, the majority of entrepreneurs found businesses based on ideas gathered from jobs and activities and lifestyles they're familiar with. For one thing, the entrepreneur understands what target customers want because she's part of the group. Two, because the entrepreneur is familiar with the products currently on the market, she can usually introduce a product that trumps the competition. And finally, when selling the product to customers, the entrepreneur appears not as a salesperson, but rather as "one of us." It's a powerful situation that doesn't guarantee success, but definitely gives you a good chance.
Levels of need
But now let's reverse the situation and talk about what you don't know, what you aren't familiar with. It can be a very good thing to actually go beyond what you know-to keep an open mind about opportunity and think broadly about the needs of the world beyond your comfort zone. Undoubtedly a new mp3 player or a new snowboard boot addresses the needs of a certain population and has a chance at success. But UV water purification systems for villages in the developing world address the needs of an entirely different population, and have just as good a chance at success, whether measured in terms of profits or lives saved, or both. What we're talking about are levels of need. If you haven't already, what you need to do is figure out your priorities as an entrepreneur. Whose needs to do you want meet? Those of the US market, or the basic human needs of the developing world, people with disabilities, the elderly? How about products designed with the environment in mind? Your answers to these questions will shape your business from the very beginning. Weigh them carefully.
When is an idea an opportunity?
An idea is an opportunity when it is attractive, durable and timely and is anchored in a product or service that creates or adds value for its end buyer or user. The most successful entrepreneurs are opportunity-focused; that is, they start with what customers and the marketplace want and do not lose sight of this.
You have a chance
Don't think you can't possibly compete with large, multibillion dollar players. The high-tech industry welcomes small business like no other. The performance of smaller firms in technological innovation is remarkable: 95% of the radical innovations since WWII have come from new small firms. A National Science Foundation study found that smaller firms generated 24 times more innovations per research and development dollar than firms with 10,000 or more employees.
New venture realities
- Success is highly situational
- Starting a company is a lot harder than it looks, or you think it will be.
- Most new ventures are works in process.
- Speed, adroitness of reflex, and adaptability are crucial.
- The key to succeeding is failing quickly and recouping quickly.
- The best entrepreneurs specialize in making "new mistakes" only.
Quotes from "New Venture Creation," 2004, by Timmons and Spinelli
Newton’s recipe for failure: Ignoring customer needs
For a high-tech product to succeed it must be polished and refined until its technological capabilities and features meet the needs of the market. Two examples from the dawn of the Personal Digital Assistant (PDA) era in the late 90s illustrate how—and how not—to bring a high-tech product to market.
Apple’s Newton belongs in the “how not” category. Despite a high-quality handwriting recognition system, the Newton was large and cumbersome, expensive, and prone to problems. Larry Tesler, a former member of the Newton development group, says that although Apple had learned about key customer needs during market research, it ignored these needs when it came to developing the product, opting instead to push ahead and beat its competitors to launch. Tesler says, “In the end, we cut corners and ignored problems to try to meet a price range and a ship date that we had prematurely announced to gain an edge in a reckless public relations battle.” (from “Why the Apple Newton Failed,” by Larry Tesler, CEO and chief scientist, Stagecast Software) With promotional materials boasting that “Newton can read your handwriting,” Apple launched its product on a platform of unsubstantiated claims and inflated expectations, dooming the product to consumer disappointment, ridicule, and elimination from the Apple product line in 1998.
The other side of the story is the PalmPilot. Introduced in 1996, the PalmPilot 1000 won PC Computing’s MVP Usability Achievement of the Year Award and led to a revolution in handheld computing. Jeff Hawkins, one of the original creators of the PalmPilot, suggests that the device’s commercial success has a lot to do with responding to consumers’ needs, since the major market for handhelds was and is the individual purchase. He maintains that it’s important to make tradeoffs on “what to put in and what not to put in,” so that the product maintains the correct balance of techno-logical features, usability, and affordability.


6.7: Brainstorming and Freewriting
- Explain how brainstorming and freewriting can help you start writing
As you begin thinking about a topic, before you begin your official draft, you’ll want to write down ideas and concepts associated with your assignment to develop your ideas. This is a critical step in helping to shape and organize your paper. Brainstorming and freewriting are two great ways to get started.
Brainstorming
Brainstorming allows you to quickly generate a large number of ideas. You can brainstorm with others or you can brainstorm by yourself, which sometimes turns into freewriting. To effectively brainstorm, write down whatever ideas come to mind.
To brainstorm, let your thoughts about a specific topic flow, and list those thoughts.
Example: squirrels
- How to get them out of the garden
- How to get rid of them ethically (without killing)
- Squirrel traps
- Repellents for squirrels
- Types of squirrels
- Brown vs. black vs. red squirrels
- Flying squirrels
- What they eat
- Different types of play
- Training squirrels
- Hunting squirrels
- Squirrels and cats
- How they nest
- Build nests in the same place each year
So, what happens once you’ve brainstormed a topic? Look over the list. Are there items that group together? Are there items that catch your interest as a thinker, researcher, and writer—items you want to know more about? Are there items that seem unrelated or not useful? Use your list as a starting place; it creates ideas for you, as a writer, to work with.
Sometimes it works better to write down each idea on a separate piece of paper. It also helps to ask yourself some questions:
- What do I care about or what am I interested in?
- What do I know that I could teach others?
- What irritates or annoys me about this issue?
In order to capture more of your thoughts, you may want to brainstorm a few times until you have enough ideas to start writing.
Imagine you are in a class. Your instructor says you will have to write a paper on your favorite freetime activity , and that you must also persuade your reader to try it.
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First ask yourself, What do I care about? or What am I interested in? It is easiest to write about a topic that you are interested in. This could be anything from gardening to ice skating, or from writing poetry to playing the piano. Your list, in this example, would then read:
- gardening
- ice skating
- writing poetry
- playing the piano
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Second, ask yourself, What do I know that I could teach others?
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You may be able to teach someone else something that you really enjoy. Good for you. If you cannot, don’t worry; you are still just brainstorming. Perhaps you teach swimming lessons or t-ball, or maybe you bake really well and are able to offer some of your insights. Your list, in this example, would then read:
- swimming lessons
- t-ball
- baking
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You may be able to teach someone else something that you really enjoy. Good for you. If you cannot, don’t worry; you are still just brainstorming. Perhaps you teach swimming lessons or t-ball, or maybe you bake really well and are able to offer some of your insights. Your list, in this example, would then read:
Let’s think of another example. How about the common situation in which the instructor wants you to write about “something you care about” or an “issue you have” ?
-
Again start by asking yourself a question. Ask yourself, What irritates me? Everyone has things that irritate them, some small and others large.
- An example of something small that’s irritating could be people in your dorm who leave trails of toothpaste by the sink and never clean up after themselves. A personal example can be useful as a bridge to a larger issue that will be your topic—in this case it could be community living and personal responsibility.
- In academic writing with a less personal slant, the source of irritation is often another writer/theorist with whom you disagree. Your “irritation” then would lead to an effective piece about why you have a better conception of what’s really going on.
- A less direct version of this would be a writer/theorist who makes some good points but lacks something in his/her argument that you can add to the “conversation.”
This video demonstrates that writers of all levels and experiences value the process of brainstorming. This clip shows how writers brainstorm ideas for the show, Bernie Mac, which aired in the early 2000s.
Freewriting
Freewriting is just what it says—writing freely, whatever comes into your mind, without caring about spelling, punctuation, etc. It’s a way to free up your thoughts, help you know where your interests lie, and get your fingers moving on the keyboard (and this physical act can be a way to get your thoughts flowing).
Try a series of timed freewritings. Set a timer for five minutes. The object is to keep your fingers moving constantly and write down whatever thoughts come into your head during that time. If you can’t think of anything to say, keep writing I don’t know or this is silly until your thoughts move on. Stop when the timer rings. Shake out your hands, wait awhile, and then do more timed freewriting. After you have a set of five or so freewritings, review them to see if you’ve come back to certain topics, or whether you recorded some ideas that might be the basis for a piece of writing.
Here’s a sample freewrite that could yield a number of topics for writing:
I don’t think this is useful or helpful in any way.  I’m looking out of my window and it’s the end of may and I can see that white cotton stuff flying around in the air, from the trees. One of my aunts was always allergic to that stuff when it started flying around in the spring. Don’t know offhand what type of tree that comes from. That aunt is now 94 years old and is in a nursing home for a while after she had a bad episode. She seems to have one now every spring. It’s like that old tree cotton triggers something in her body. Allergies. Spring. Trying to get the flowers to grow but one of the neighbors who is also in his 90s keeps feeding the squirrels and they come and dig up everyone’s flowerbed to store their peanuts. Plant the flowers and within thirty minutes there’s a peanut there. Wonder if anyone has grown peanut bushes yet?
Don’t know . . . know . . .
Possible topics from this freewrite:
- Allergy causes
- Allergies on the rise in the U.S.
- Consequences of humanizing wild animals
- Squirrel behavior patterns and feeding habits
- Growing your own food
Contributors and Attributions
- Video: Writers' Room. Authored by : TV411. Located at : http://www.tv411.org/writing/creative-personal-writing/video-writers-room . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike
- Brainstorming. Authored by : Marianne Botos, Lynn McClelland, Stephanie Polliard, Pamela Osback . Located at : https://pvccenglish.files.wordpress.com/2010/09/eng-101-inside-pages-proof2-no-pro.pdf . Project : Horse of a Different Color: English Composition and Rhetoric . License : CC BY-SA: Attribution-ShareAlike
- Brainstorming. Provided by : Excelsior College OWL. Located at : https://owl.excelsior.edu/writing-process/prewriting-strategies/prewriting-strategies-brainstorming/ . License : CC BY: Attribution
- Freewriting. Provided by : Excelsior OWL. Located at : https://owl.excelsior.edu/writing-process/prewriting-strategies/prewriting-strategies-freewriting/ . License : CC BY: Attribution


7.6.2: Methicillin-Resistant Staphylococcus Aureus (MRSA)
Methicillin-Resistant Staphylococcus Aureus is a type of bacterial infection commonly referred to as “staph”. Although most staph bacteria are harmless, some can cause severe staph infections. Not all staph infections are resistant to treatment, but one particular strain of this bacteria (MRSA) has developed a resistance to the antibiotics commonly used to treat it. Healthcare settings are common sources of staph infections due to the depressed immunity of patients and the invasive procedures being done, which allows many opportunities for staph bacteria to be transmitted to susceptible people via surfaces or medical equipment. Community spread of staph infections, particularly MRSA, can happen in athletic facilities, in the military or childcare centers or schools as well - particularly in places where people do physical activity. Staph bacteria can only get through if there is a break in the skin, so keeping cuts and abrasions covered as well as cleaning surfaces and equipment are the best prevention strategies. MRSA most commonly causes infections in wounds or surgical sites, and in rare cases it can get into the bloodstream or lungs (causing pneumonia). In some cases this infection can cause sepsis - which is the body’s own amplified immune response to the infection, and can be deadly (CDC, 2019).


7: Liquid Activity Coefficients
Distillation Science (a blend of Chemistry and Chemical Engineering)
This is Part VII, Liquid Activity Coefficients of a ten-part series of technical articles on Distillation Science, as is currently practiced on an industrial level. See also Part I, Overview for introductory comments, the scope of the article series, and nomenclature.
Part VII, Liquid Activity Coefficients builds on Part VI, Fugacity regarding the departure of vapor-liquid equilibria (VLE) of binary mixtures from ideal behavior, as is commonly found in practical application of distillation science. In conjunction with previous articles, the goal of this article to to complete the explanation of equilibrium behavior of binary systems; such that Part IX can illustrate an example of a distillation process design.
In Part VI the combination of Raoult’s Law and Dalton’s Law was introduced governing the ideal vapor-liquid behavior of volatile fluid mixtures, with Equation 6-1 repeated for convenience.
\( Y_{1}=(X_{1} \times VP_{1})/(X_{1} \times VP_{1} + X_{2} \times VP_{2}) \nonumber\) \( Y_{2}=(X_{2} \times VP_{2})/(X_{1} \times VP_{1} + X_{2} \times VP_{2}) \nonumber\) repeat Equation 6-1
Also in Part VI the real-world departure from ideal behavior was discussed as far as the fugacity coefficients, but stopped short of Liquid Activity Coefficients. Equation 6-4 repeated below gave the complete relationship between the partial pressure of the ith component, PP i , and the pure-component vapor pressure VP i , for liquid and vapor mole fractions X i and Y i , respectively. The total system pressure (sum of the partial pressures) is denoted as \(\pi \nonumber\); (\(\phi_{i}^V\nonumber\)& \(\phi_{i}^L\nonumber\)) are the liquid and vapor fugacity coefficients. \(\gamma_{i} \nonumber\) is the ith component’s Liquid Activity Coefficient for that specific set of components.
\(Y_{i} \times \pi =PP_{i} =(\phi_{i}^L/ \phi_{i}^V) \times \gamma_{i} \times VP_{i} \times X_{i} \nonumber\) repeat Equation 6-4
An example was given in Part VI of a vaporizing TCS-STC mixture, contrasting the vapor-liquid equilibria (VLE) expected from Raoult/Dalton; from what actually occurs in the real world as a result of the ideality departures. Summarizing the previous article, the more volatile fluid exerts an effect on the less volatile fluid, to increase its apparent vapor pressure; and the less volatile fluid reduces the more volatile fluid’s vapor pressure. The net effect is to make the component separations more difficult via distillation (e.g., more theoretical trays or increased amount of reflux than expected). For detailed discussion of fugacity coefficients, see Part VI .
As a “house-keeping” note, it must be emphasized that Parts VI and VII treat fugacity coefficients and Liquid Activity Coefficients separately. In some texts, the two topics are compressed together using specific models of mixing and Equations of State (EOS). However, the two ideality departure factors are differently based, and their combination often gives erroneous or nonsense results when used with polar fluids, or those that have a high degree of hydrogen bonding (such as the chlorosilanes that are used as continuing examples). Fugacity coefficients naturally result from differences in fluid volatility. Liquid Activity Coefficients result from differences in fluid properties, including critical temperature (T c ), critical pressure (P c ), critical volume (V c ) and acentric factor (ω), as well as the entropy & enthalpy effects of mixing (referred to in thermodynamic texts as excess Gibbs Free Energy).
For a given temperature and combination of fluids, the value of \(\gamma \nonumber\) is a function of the liquid mole fraction, rising asymptotically from unity at the pure component condition. Typically the values of \(\gamma \nonumber\) for a system are plotted as a function of the more volatile component’s mole fraction. An example of such a plot is given in Figure 7-1, for the TCS-STC binary at 73.9°C (the temperature of the continuing example of previous articles).
There are several models for Liquid Activity Coefficients that are thermodynamically consistent (i.e., follow the Gibbs-Duhem Rule of binary system thermodynamics). The two-constant Van Laar model is the easiest to manipulate, but is limited to binary systems. The Margules model is not that different from the Van Laar model, and also limited to binary systems. The Wilson model is more complex mathematically, but can be applied to ternary or greater-numbered component systems. Fortunately the Wilson model parameters can be calculated from the Van Laar model. Even more complex Liquid Activity Coefficient models are known (e.g., NRTL and UNIQUAC), but these require more data points to effectively evaluate. In the case of the above plot, the Van Laar model is shown. In Part VIII , the techniques are discussed to evaluate experimental data on Liquid Activity Coefficients and fit the data to a model. In most cases with chlorosilanes (and their impurities) the data quality is not that great, so the simpler Van Laar is typically used to correlate data and for evaluating binary systems, and the Wilson model is used for ternary systems and beyond. For a more in-depth discussion of Liquid Activity Coefficients, the reader is referred to the text " The Properties of Gases and Liquids", by Reid, Prausnitz and Sherwood, (McGraw-Hill). The third edition is more readable, but the fifth edition is more updated.
The Van Laar model is:
\[Ln(\gamma_{1}) =A_{1} \times [1+(A_{1} \times X_{1})/(A_{2} \times X_{2})]^{-2} \label{7-1} \]
\(Ln(\gamma_{2}) =A_{2} \times [1+(A_{2} \times X_{2})/(A_{1} \times X_{1})]^{-2} \nonumber\)
with A 1 and A 2 being the Van Laar constants for the more volatile and less volatile component, respectively.
Conveniently, \[EXP(A_{1}) = \gamma_{1} @ X_{2} \rightarrow 0 \label{7-2} \]
\(EXP(A_{2}) = \gamma_{2} @ X_{1} \rightarrow 0 \nonumber\)
which simplifies evaluation. Note from Figure 7-1 that the Liquid Activity Coefficients plot is not always symmetrical (i.e., A 1 ≠ A 2 ). In the plot of Figure 7-1, A 1 (TCS) = 0.1752, so the TCS curve asymptotes at \(\gamma \nonumber\)= 1.191; and A 2 (STC) = 0.2086, so the TCS curve asymptotes at \(\gamma \nonumber\)= 1.232 .
Completing the on-going example from Part VI , for a TCS-STC liquid mixture, with 40% molar TCS, at 73.9°C, to determine the vapor mole fraction in equilibrium (but now including the Liquid Activity Coefficient effects):
In Part VI , the pure-component vapor pressures of TCS and STC respectively were calculated from the VP equation of Part IV to be 3.500 and 1.651 atmospheres, respectively at 73.9°C. Also from Part V , the liquid and vapor fugacities were calculated as per Table 6-1, repeated below, using the Peng-Robinson EOS ( Part V ).
|
TCS |
STC |
|
|---|---|---|
|
\(\phi^L \nonumber\) |
0.9976 |
1.0039 |
|
\(\phi^V \nonumber\) |
1.0488 |
0.9562 |
|
\(\phi^L / \phi^V \nonumber\) |
0.9512 |
1.0498 |
The values for \(\gamma \nonumber\) are calculated for TCS and STC @ 0.40 mf TCS, per E quation ( \ref{7-1} ) using the values for A 1 and A 2 above (or just read from Figure 7-1), to be 1.075 and 1.027 respectively. Plugging all these values into Equation 6-4 (repeated above), the partial pressures of TCS and STC @ 73.9°C are:
\(PP_{TCS} =0.9512 \times 1.075 \times 3.500 \times 0.40 =1.432 \nonumber\) atmospheres, and
\(PP_{STC} =1.0498 \times 1.027 \times 1.651 \times 0.60 =1.068 \nonumber\) atmospheres.
So the value of \(Y_{TCS} = 1.432/(1.432+1.068) =.573 \nonumber\), which was given in Part VI, with the STC mole fraction being 0.427.
The careful reader will see that not only are the partial pressures of TCS and STC different from Raoult/Dalton, but so is the total pressure. In the ideal behavior of Raoult/Dalton, the partial pressures were given in Part VI as 1.400 and 0.991 atmospheres, for a total pressure of 2.39 atmospheres. But in reality, at 73.9°C and 0.40 mf TCS, the total pressure would instead be 2.50 atmospheres. So while the ideality departures of TCS made the vapor mole fraction smaller (0.573 vs 0.586), both the TCS and STC exerted slightly more partial pressure than Raoult would indicate. But the TCS mole fraction still went down, because the ideality departure of STC was greater than that of TCS.
The take-away from this example is that while Raoult/Dalton will give you a “quick & dirty” value for Y/X, it is easy to get the overall wrong answer for determining the requirements of a distillation column design. In almost every instance of industrial practice, the use of Raoult/Dalton will undersize the number of trays needed to make a separation (or for the same number of trays, undersize the reflux required). For this reason, it is important to know how to work out the better answer.
For many binary systems, VLE data exists – but typically not at the temperature/pressure needed for industrial design. Most industrial designs use higher pressures for economy of size, as well as to accommodate available energy sources for driving the column reboiler and condenser. Yet the practicalities of data collection in the laboratory usually require pressures slightly above ambient, or at slight vacuums. This is especially true of chlorosilanes, since the normal materials of construction for higher pressures can catalyze slow side reactions, such as disproportionation and dimerization.
Lab data is normally collected based on small amounts of fluid that are used repeatedly at somewhat different combinations of temperature and composition. In the case of electronic impurities, some of these compounds are not very stable outside of a chlorosilane matrix. So a correlating method must be used that allows a certain degree of extrapolation.
In 1981 Chung-Ton Lin and Thomas Daubert from Penn State University developed such a Liquid Activity Coefficient model and published same in Industrial & Engineering Chemistry Process Design and Development, basing their work on non-polar hydrocarbon mixtures. I have used their general method, but modified two of the constants so as to better fit available VLE data on chlorosilanes. Using these re-evaluated constants, the revised “Lin & Daubert” method (detailed below as E quation ( \ref{7-3} ) was used to check against industrially obtained data on the distillation purification of TCS and silane, with good results. Obviously more data would be preferable, especially on ternary mixtures and low-concentration electronic impurities in chlorosilanes. However until such becomes available, E quation ( \ref{7-3} ) below is recommended.
To develop a generic expression for the Van Laar parameters (\(A_{i} \nonumber\) & \(A_{j} \nonumber\) ), Lin & Daubert go back to Van Laar’s original two-term expression regarding the excess Gibbs energy of mixing, using the SRK Equation of State for partial molar compressibility and simple mixing rules; and expand it into the following:
\[A_{i} = c \times F_{i} +d \times k_{ij} \times G_{i} \label{7-3} \]
\(A_{j} = c \times F_{j} +d \times k_{ij} \times G_{j} \nonumber\)
- \(F_{i} =1/[T_{ri}P_{ci}] \times (R_{i} \times (1+M_{j})^{0.5} \times P_{ci}^{0.5} - R_{j} \times (1+M_{j})^{0.5} \times P_{cj}^{0.5})^2 \nonumber\)
- \(F_{j} =1/[T_{rj}P_{cj}] \times (R_{j} \times (1+M_{i})^{0.5} \times P_{cj}^{0.5} - R_{i} \times (1+M_{i})^{0.5} \times P_{ci}^{0.5})^2 \nonumber\)
- \(G_{i} = T_{ri}^{-1} (P_{cj}/P_{ci})^{0.5} \times R_{i} \times (1+M_{i})^{0.5} \times R_{j} (1+M_{i})^{0.5} \nonumber\)
- \(G_{j} = T_{rj}^{-1} (P_{ci}/P_{cj})^{0.5} \times R_{j} \times (1+M_{j})^{0.5} \times R_{i} (1+M_{j})^{0.5} \nonumber\)
- \(M_{i} = 0.480 + 1.574\omega_{i} -.176\omega_{i}^2 \nonumber\) \(M_{j} = 0.480 + 1.574\omega_{j} -.176\omega_{j}^2 \nonumber\)
- \(R_{i} =[1+M_{i} \times (1-T_{ri}^{0/5})]^{0.5} \nonumber\) \(R_{j} =[1+M_{j} \times (1-T_{rj}^{0/5})]^{0.5} \nonumber\)
- \(k_{ij} =1 - 2 \times [V_{ci}^{1/3} \times V_{cj}^{1/3}]^{0.5} / [V_{ci}^{1/3} + V_{cj}^{1/3}] \nonumber \)
Using an improved set of curve-fits for “c”,
- \(0< \Delta \omega <0.03 ⇒ \nonumber\)\( Ln(c)= -8.0637E+04(\Delta \omega)^3 +8.2649E+03(\Delta \omega)^ 2 - 3.2891E+02 (\Delta \omega) +5.5051 \nonumber\)
- \(0.03 < \Delta \omega <0.30 ⇒ \nonumber\)\( Ln(c)=6.4645E+02(\Delta \omega)^4 -7.9857E+02(\Delta \omega)^3 +3.4768E+02(\Delta \omega)^2 -6.5099E+01(\Delta \omega) +2.5667 \nonumber\)
Reducing available VLE data to Van Laar parameters, and adjusting for vapor and liquid fugacity coefficients, the best value of “d” is found to be 112.1 for chlorosilanes and similar fluids.
Lin & Daubert suggested that the “c” function be capped at \(0.005 < \Delta \omega=|\omega_{i} - \omega_{j}| <0.15 \nonumber\), based on their data; but using the above improved set of polynomials for “c” avoids singularities during iterative calculations, with a minimum “c” of 0.135 (as “Δω” → ∞ )
Note that the "c" variable in the first term of Equation 7-3 is based on the \(F_{i} \nonumber\) values and the differences between the two fluid's acentric factor, ω ( i.e., their molecular complexity); whereas the second term of Equation 7-3 is based on the \(G_{i} \nonumber\) values and the differences between the two fluid's molecular volumes at critical (V c ).
Returning to the continuing previous TCS-STC example, and using the two components’ critical temperature (T c ), critical pressure (P c ), critical volume (V c ) and acentric factor (ω), as well as basing the reduced temperature (T r ) on 73.9°C = 347.05°K, E quation ( \ref{7-3} ) gives the values of A 1 and A 2 used earlier in this article, of 0.1752 and 0.2086 respectively.
Frequently in evaluating different distillation designs for a process, the designer is faced with the need to make process temperature changes and determine the effect on Y/X using Equation 6-4 , repeated above. That causes no problem in the evaluation of the fugacity ratio,\(\phi_{i}^L / \phi_{i}^V \nonumber\), since those factors are derived from the EOS ( see Part VI ). However, if using experimentally obtained values for \(\gamma \nonumber\) that were obtained at a different temperature, there needs to be further adjustment. The general “rule-of-thumb” seen in some texts is that Ln(\(\gamma \nonumber\)) is proportional to 1/T (absolute temperature) for small temperature changes. In reality such a “rule-of-thumb” is tantamount to making an assumption that the mixing effects of the multi-component fluid are closer to an isothermal model than an isenthalpic one. In most cases, the temperature dependency on activity coefficient is a blend of two theoretical models, which leads to the relationship:
\[Ln(\gamma_{i})=a + \frac{b}{T} \label{7-4} \]
where “a” and “b” are the constants of a linear relationship (“a” = 0 leading to the isenthalpic model, or “b” = 0 leading to the isothermal model). The use of E quation ( \ref{7-3} ) solves the problem, since it predicts such temperature changes along the lines of E quation ( \ref{7-4} ) , and so can be used in a relative manner. For example, if E quation ( \ref{7-3} ) is found to over-predict the value of \(Ln(\gamma) \nonumber\) by 10% of an experimentally obtained activity coefficient value, then calculate how much E quation ( \ref{7-3} ) would predict for the different temperature, and apply that 10% on the \(Ln(\gamma) \nonumber\).
Note that to all intents, the Liquid Activity Coefficient is not a direct function of pressure, other than increasing the boiling temperature increases pressure. However, fugacities are a function of both temperature and pressure, and that is why they should be included in Y/X calculations (many texts advocate setting fugacities to unity, which is only close to being valid at very low pressures).
To demonstrate how E quation ( \ref{7-3} ) performs with changing temperature (from 330°K to 380°K), Figure 7-2 shows the dependency for the asymptotic BIP’s of the TCS-STC mixture in the example above. Note that the slope of the STC Liquid Activity Coefficient line is about 20% steeper than that of the TCS Liquid Activity Coefficient line (i.e., in the TCS-STC binary system, increasing the temperature has a greater effect on the STC's \(\gamma \nonumber\) than it does on the TCS's \(\gamma \nonumber\)).


21: Respiratory System Last updated Save as PDF Page ID 7918 21.10: Respiration Control 21.10A: Neural Mechanisms (Respiratory Center) 21.10B: Neural Mechanisms (Cortex) 21.10C: Chemoreceptor Regulation of Breathing 21.10D: Proprioceptor Regulation of Breathing 21.11: Respiratory Adjustments 21.11A: Adjustments During Exercise 21.11B: Adjustments at High Altitude 21.1: Overview of the Respiratory System 21.1A: The Reason for Breathing 21.1B: Functional Anatomy of the Respiratory System 21.2: Conducting Zone 21.2A: Nose and Paranasal Sinuses 21.2B: Pharynx 21.2C: Larynx 21.2D: Structures Used in Voice Production 21.2E: Trachea 21.3: Respiratory Zone 21.3A: Bronchi and Subdivisions 21.3B: Alveoli 21.4: The Lungs and Pleurae 21.4A: Lungs 21.4B: Lobes, Fissures, and Lobules 21.4C: Blood Supply to the Lungs 21.5: Mechanics of Breathing 21.5A: Pressure Changes During Pulmonary Ventilation 21.5B: Inspiration 21.5C: Expiration 21.5D: Breathing Patterns 21.6: Factors Affecting Pulmonary Ventilation 21.6A: Factors Affecting Pulmonary Ventilation: Surface Tension of Alveolar Fluid 21.6B: Factors Affecting Pulmonary Ventilation: Compliance of the Lungs 21.6C: Factors Affecting Pulmonary Ventilation: Airway Resistance 21.7: Nonrespiratory Lung Functions 21.7A: Lung Capacity and Volume 21.7B: Nonrespiratory Air Movements 21.8: Gas Laws 21.8A: Dalton’s Law of Partial Pressure 21.8B: Henry’s Law 21.9: Gas Exchange 21.1A: External Respiration 21.9B: Internal Respiration 21.9C: Oxygen Transport 21.9D: Carbon Dioxide Transport


26.2: Australia
Aboriginal Australian Art
Aboriginal Australian art can be traced back at least 30,000 years and is one of the longest continuously practiced artistic traditions in the world.
Differentiate between the periods and regional styles of Aboriginal Australian rock paintings
Key Points
- The history of visual arts in Australia ranges from ancient Aboriginal rock paintings to the western-influenced contemporary works of today.
- Rock paintings are divided into three periods based on the styles and content of the art: the Pre-Estuarine (c. 40,000–6000 BCE), Estuarine (c. 6000 BCE–500 CE), and Fresh Water (c. 500 CE–present) periods.
- There are three regional styles of cave art: the geometric style (known for its concentric circles, arcs, and dots), the simple figurative style, and the complex figurative style.
- Rock paintings are believed to have served several functions, from ceremonial to merely decorative. Today, they are preserved and protected in national parks throughout the continent.
- The influence of Aboriginal artwork in Australia carries over to the 19th and 20th centuries in the works of many artists, as well as in the resurgence of Aboriginal art today.
Key Terms
- advent : Coming; coming to; approach; arrival.
- indigenous : Born or engendered in, native to a land or region, especially before an intrusion.
- aboriginal : First according to historical or scientific records; original; indigenous.
Aboriginal and Indigenous Art
The visual arts have a long history in Australia, from ancient Aboriginal rock paintings to colonial landscapes to contemporary movements of today. Aboriginal art in Australia can be traced back at least 30,000 years; the rock art of Australian Aborigines is one of the longest continuously practiced artistic traditions in the world.
Rock paintings are divided into three periods based on the styles and content of the art: the Pre-Estuarine Period (c. 40,000–6000 BCE), the Estuarine Period (c. 6000 BCE–500 CE), and the Fresh Water Period (c. 500 CE–present). There are also three regional styles of cave art:
- The geometric style (known for its concentric circles, arcs, and dots) found in Central Australia, Tasmania, the Kimberly, and Victoria;
- The simple figurative style found in Queensland; and
- The complex figurative style found in Arnhem Land.
Rock paintings are believed to have served several functions, from ceremonial to merely decorative. Today, they are preserved in national parks throughout the continent and protected through organizations such as the Friends of Australian Rock Art.
Influence of Aboriginal Art Today
The influence of Aboriginal artwork in Australia carries over to the 19th and 20th centuries in the works of William Barak, who recorded traditional aboriginal ways for the education of Westerners: Margaret Preston, a non-indigenous painter incorporating Aboriginal influences in her works; Albert Namatjira, an Arrernte artist whose landscapes inspired the Hermannsburg School of art; and Elizabeth Durack, notable for her fusion of Western and indigenous influences.
In 1971, Geoffrey Bardon encouraged the Aboriginal people of Papunya to paint their Dreamtime stories about creation, people, animals, and customs on canvas. This led to the development of the Papunya Tula School, or dot art, now possibly Australia’s most recognizable style of art worldwide. Clifford Possum Tjapaltjarri, Long Jack Phillipus Tjakamarra, and William Sandy are some of the best known Papunya artists. Other Aboriginal artists have incorporated western mediums into their work, such as Emily Kngwarreye, Rover Thomas, and Freddy Timms.
Colonial Australian Art
European colonization had a profound and permanent affect on the artwork of Australia.
Discuss the impact colonialism had on Australian art
Key Points
- Europeans first began depicting the natural landscape and wildlife of the Australian continent during initial voyages in the late 1700s.
- The colonial art market primarily desired landscape paintings, which are often described as making a gradual shift from a European sense of light to a more Australian one, as the lighting in Australia is notably different from that of Europe.
- The first artistic representations of the Australia scene by European artists were mainly natural history illustrations, depicting the distinctive flora and fauna of the land for scientific purposes and the topography of the coast.
- From 1851, the Victorian Gold Rush resulted in a huge influx of settlers and new wealth, and wealthy landowners and merchants recorded their material success by commissioning landscape paintings.
- The origin of distinctly Australian painting is associated with the Heidelberg School of the 1880s–1890s, which integrated an impressionistic plein air approach and idealized the rural landscape.
Key Terms
- plein air : A french term referring to the practice of painting outdoors, where a painter reproduces the actual visual conditions seen at the time of the painting.
Colonial Art (1770–1900)
The visual arts have a long history in Australia, from ancient Aboriginal rock paintings to colonial landscapes to contemporary movements of today. Europeans depicted the natural landscape, plant life, and wildlife of the Australian continent during initial voyages in the late 1700s. John Lewin and Harriet and Helena Scott were among the first professional natural history illustrators, while artists such as Augustus Earle focused on ethnographic portraiture of Aboriginal Australians.
The colonial art market primarily desired landscape paintings, which are often described as making a gradual shift from a European sense of light to a more Australian one. The lighting in Australia is notably different from that of Europe, and early attempts at landscapes attempted to reflect this. The art of this era is also characterized by transformation, where artistic ideas originating from beyond (primarily Europe) gained new meaning and purpose when transplanted into the new continent and the emerging colonial society.
Early Colonial Art (1770–1850)
The first artistic representations of the Australia scene by European artists were mainly natural history illustrations, depicting the distinctive flora and fauna of the land for scientific purposes and the topography of the coast. Sydney Parkinson, the Botanical illustrator on James Cook’s 1770 voyage that first charted the eastern coastline of Australia, made a large number of such drawings under the direction of naturalist Joseph Banks. Until the turn of the century, all drawings made in the colony were crafted by soldiers, including British naval officers George Raper and John Hunter, and convict artists, including Thomas Watling. Most are in the style of naval draughtsmanship and cover natural history topics, specifically birds, while a few depict the infant colony itself.
Later Colonial Art (1850–1895)
From 1851, the Victorian Gold Rush resulted in a huge influx of settlers and new wealth. S. T. Gill (1818–1880) documented life on the Australian gold fields; however the colonial art market primarily desired landscape paintings, which were commissioned by wealthy landowners or merchants wanting to record their material success. Some of the artists of note included Eugene von Guerard, Nicholas Chevalier, William Strutt, John Skinner Prout, and Knut Bull. Louis Buvelot was a key figure in landscape painting, illustrating a more domesticated and settled view of the land. Among the first Australian artists to gain a reputation overseas were the impressionist John Peter Russell and landscape painter Rupert Bunny.
The Heidelberg School (1885–1910)
The origin of distinctly Australian painting is associated with the Heidelberg School of the 1880s–1890s. Like European Impressionists, artists such as Arthur Streeton, Frederick McCubbin, Tom Roberts, and Charles Conder applied themselves to recreating a truer sense of light and color as seen in the Australian landscape. They began an impressionistic plein air approach to the Australian landscape that remains embedded in Australia’s popular consciousness, both inside and outside the art world.
Their most recognized paintings involve scenes of pastoral and outback Australia. Central themes of their art are considered those of work, conquering the land, and an idealization of the rural pioneer. By the 1890s, most Australians were city-dwellers, and a romantic view of pioneer life gave great power and popularity to images of the rural landscape.
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14.1: Introduction
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Objective
After completing this section, you should be able to determine whether or not a molecule contains a conjugated system, given its Kekulé, condensed or shorthand formula.
Key Terms
Make certain that you can define, and use in context, the key terms below.
- conjugated diene
- conjugated double bonds
- diene
- enone
- polyene
Study Notes
Conjugated double bonds are double bonds which are separated by one carbon-carbon single bond. Thus the double bonds in butadiene, $\ce{\sf{CH2=CH-CH=CH2}}$, are conjugated, and this compound is an example of a conjugated diene .
Just as the term diene indicates the presence of two carbon-carbon double bonds in a compound, so the term polyene is used to describe compounds containing many carbon-carbon double bonds.
An enone is a compound containing a carbon-carbon double bond (ene) and a carbonyl group (one). A conjugated enone contains the structural unit:
Conjugated Dienes
A diene is a hydrocarbon chain that has two double bonds that may or may not be adjacent to each other. This section focuses on the delocalization of \(\pi\) systems by comparing two neighboring double bonds. The arrangements of these double bonds can have varying affects on the compounds reactivity and stability.
Naming Dienes
First identify the longest chain containing both carbons with double bonds in the compound. Then give the lowest possible number for the location of the carbons with double bonds and any other functional groups present (remember when naming alkenes that some groups take priority such as alcohols). Do not forget stereochemistry or any other orientation of the double bond such as (E/Z,cis or trans).
Examples:
Conjugated vs. Nonconjugated vs. Cumulated Dienes
Conjugated dienes are two double bonds separated by a single bond
Nonconjugated (Isolated) dienes are two double bonds are separated by more than one single bond.
Cumulated dienes (allenes) are two double bond connected to a carbon atom.
The reactivity of these molecules is substantially different from that of alkenes which have isolated C=C. These molecules are thus considered a different class of organic molecule. Conjugated dienes, especially butadiene, are very important materials in the production of rubber, and thus for the tires of our cars.


3.4: Numerical Approximation of Multiple Integrals
As you have seen, calculating multiple integrals is tricky even for simple functions and regions. For complicated functions, it may not be possible to evaluate one of the iterated integrals in a simple closed form. Luckily there are numerical methods for approximating the value of a multiple integral. The method we will discuss is called the Monte Carlo method . The idea behind it is based on the concept of the average value of a function, which you learned in single-variable calculus. Recall that for a continuous function \(f (x)\), the average value \(\bar f \text{ of }f\) over an interval \([a,b]\) is defined as
\[\bar f = \dfrac{1}{b-a}\int_a^b f (x)\,dx \label{Eq3.11}\]
The quantity \(b − a\) is the length of the interval \([a,b]\), which can be thought of as the “volume” of the interval. Applying the same reasoning to functions of two or three variables, we define the average value of \(f (x, y)\) over a region \(R\) to be
\[\bar f = \dfrac{1}{A(R)} \iint\limits_R f (x, y)\,d A \label{Eq3.12}\]
where \(A(R)\) is the area of the region \(R\), and we define the average value of \(f (x, y, z)\) over a solid \(S\) to be
\[\bar f = \dfrac{1}{V(S)} \iiint\limits_S f (x, y, z)\,dV \label{Eq3.13}\]
where \(V(S)\) is the volume of the solid \(S\). Thus, for example, we have
\[\iint\limits_R f (x, y)\,d A = A(R) \bar f \label{Eq3.14}\]
The average value of \(f (x, y)\) over \(R\) can be thought of as representing the sum of all the values of \(f\) divided by the number of points in \(R\). Unfortunately there are an infinite number (in fact, uncountably many) points in any region, i.e. they can not be listed in a discrete sequence. But what if we took a very large number \(N\) of random points in the region \(R\) (which can be generated by a computer) and then took the average of the values of \(f\) for those points, and used that average as the value of \(\bar f\) ? This is exactly what the Monte Carlo method does. So in Formula \ref{Eq3.14} the approximation we get is
\[\iint\limits_R f (x, y)\,d A \approx A(R) \bar f \pm A(R) \sqrt{\dfrac{\bar {f^2}-(\bar f)^2}{N}}\label{Eq3.15}\]
where
\[\bar f = \dfrac{\sum_{i=1}^{N}f(x_i,y_i)}{N}\text{ and } \bar {f^2} = \dfrac{\sum_{i=1}^{N}(f(x_i,y_i))^2}{N}\label{Eq3.16}\]
with the sums taken over the \(N\) random points \((x_1 , y_1), ..., (x_N , y_N )\). The \(\pm\) “error term” in Formula \ref{Eq3.15} does not really provide hard bounds on the approximation. It represents a single standard deviation from the expected value of the integral. That is, it provides a likely bound on the error. Due to its use of random points, the Monte Carlo method is an example of a probabilistic method (as opposed to deterministic methods such as Newton’s method, which use a specific formula for generating points).
For example, we can use Formula \ref{Eq3.15} to approximate the volume \(V\) under the plane \(z = 8x + 6y\) over the rectangle \(R = [0,1] \times [0,2]\). In Example 3.1 in Section 3.1, we showed that the actual volume is 20. Below is a code listing (montecarlo.java) for a Java program that calculates the volume, using a number of points \(N\) that is passed on the command line as a parameter.
Listing 3.1 Program listing for montecarlo.java
The results of running this program with various numbers of random points (e.g. java montecarlo 100) are shown below:
As you can see, the approximation is fairly good. As \(N \to \infty\), it can be shown that the Monte Carlo approximation converges to the actual volume (on the order of \(O( \sqrt{N})\), in computational complexity terminology).
In the above example the region \(R\) was a rectangle. To use the Monte Carlo method for a nonrectangular (bounded) region \(R\), only a slight modification is needed. Pick a rectangle \(\tilde R \text{ that encloses }R\), and generate random points in that rectangle as before. Then use those points in the calculation of \(\bar f\) only if they are inside \(R\). There is no need to calculate the area of \(R\) for Equation \ref{Eq3.15} in this case, since the exclusion of points not inside \(R\) allows you to use the area of the rectangle \(\tilde R\) instead, similar to before.
For instance, in Example 3.4 we showed that the volume under the surface \(z = 8x + 6y\) over the nonrectangular region \(R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2x^2 }\) is 6.4. Since the rectangle \(\tilde R = [0,1] \times [0,2] \text{ contains }R\), we can use the same program as before, with the only change being a check to see if \(y < 2x^2\) for a random point \((x, y) \text{ in }[0,1] \times [0,2]\). Listing 3.2 below contains the code (montecarlo2.java):
Listing 3.2 Program listing for montecarlo2.java
The results of running the program with various numbers of random points (e.g. java montecarlo2 1000) are shown below:
To use the Monte Carlo method to evaluate triple integrals, you will need to generate random triples \((x, y, z)\) in a parallelepiped, instead of random pairs \((x, y)\) in a rectangle, and use the volume of the parallelepiped instead of the area of a rectangle in Equation \ref{Eq3.15} (see Exercise 2). For a more detailed discussion of numerical integration methods, see PRESS et al.


9.4: Manufacturing Overhead Budget
Learning Outcomes
- Create a manufacturing overhead budget
Well, we are working through these budgets, but now we got to an interesting one. There is more that goes in to the production of our shoes that just the raw materials and the people working. We have equipment, and small supplies, as well as repairs and utilities. These things can get costly, so we need to make sure we are on top of them as we work through this budget!
Manufacturing overhead includes all the costs of production other than labor and raw materials. This can include some variable and some fixed components.
Variable manufacturing overhead is based on direct labor hours. This can include things like electricity, production supplies (perhaps needles for the machines that sew together the shoe components) and other miscellaneous items needed to produce the shoes.
Fixed manufacturing overhead includes depreciation on the equipment, rent or mortgage on the facility and costs to process purchase orders, customer calls and such. There may come a point where the current facility, equipment or customer reps can’t handle the volume. If this occurs, the fixed costs may change. For our purposes, let’s assume that the current facility and equipment can handle the budgeted output!
So let’s assume our variable manufacturing overhead to be $3 per labor hour. Let’s further assume our monthly fixed manufacturing overhead is $2050 per month. So, included in our fixed overhead is $500 of depreciation. Remember depreciation is not a cash outlay, so we can deduct it from our total manufacturing overhead for cash purposes!! We will talk more about that when we get to our cash budget in a bit.
So plugging the information above into our manufacturing overhead budget, we can come up with a predetermined overhead rate for the year. We also have figured out the cash outlay, as well as the total manufacturing overhead.
| Hupana Running Company Manufacturing Overhead Budget | |||||
|---|---|---|---|---|---|
| Quarter | Q1 | Q2 | Q3 | Q4 | Total |
| Budgeted direct labor hours | 225 | 250 | 250 | 300 | 1025 |
| Variable manufacturing overhead rate | $3 | $3 | $3 | $3 | $3 |
| Variable manufacturing overhead | $675 | $750 | $750 | $900 | $3,075 |
| Fixed manufacturing overhead | $2,050 | $2,050 | $2,050 | $2,050 | $8,200 |
| Total manufacturing overhead | $2,725 | $2,800 | $2,800 | $2,950 | $11,275 |
| Minus depreciation | $500 | $500 | $500 | $500 | $2,000 |
| Cash disbursements for manufacturing overhead | $2,225 | $2,300 | $2,300 | $2,450 | $9,275 |
| A. Total manufacturing overhead | $11,275 | ||||
| B. Budgeted direct labor hours | $1,052 | ||||
| Predetermined overhead rate for the year A/B | $11 |
So if you look at our previous budget for direct labor, the hours shown here on our manufacturing overhead budget come directly from that budget! The variable manufacturing overhead total is simply the hours multiplied by the rate. We now have the information needed to continue moving forward in our budgeting process! We also calculated our predetermined overhead rate for the year. Since manufacturing overhead i s an indirect cost, we can’t assign it to a particular product or job. Because we are using our direct labor hours as our allocation base, dividing the entire manufacturing overhead by the direct hours, gives us a dollar amount we can use, as an addition to the direct labor cost, for each hour worked.
So, as an example, our direct labor cost per hour is $20 (as you remember from our direct labor budget). We an now add the $11 per hour manufacturing overhead to each hour worked as we price our products. This is a helpful calculation to spread out those costs that we cannot directly tie to a given product. This comes in very handy when more than one product line is manufactured. So if Hupana Running Company had 12 lines of shoes with varying labor and material costs, we could use this standard amount per hour of labor to allocate all of the manufacturing overhead costs!
Okay, so now we can move on!! We have all kinds of information now to complete our budgeting tasks!
Practice Questions


10.5: Solving Quadratic Equations Using the Method of Completing the Square
The Logic Behind The Method
Suppose we wish to solve the quadratic equation \(x^2 - 3x - 1 = 0\). Since the equation is not of the form \(x^2 = K\), we cannot use extraction of roots. Next, we try factoring, but after a few trials we see that \(x^2 - 3x - 1\) is not factorable. We need another method for solving quadratic equations.
The method we shall study is based on perfect square trinomials and extraction of roots. The method is called solving quadratic equations by completing the square. Consider the equation
\[x^2 + 6x + 5 = 0.\]
This quadratic equation could be solved by factoring, but we'll use the method of completing the square. We will explain the method in detail after we look at this example. First we'll rewrite the equation as
\[x^2 + 6x = -5\]
Then, we'll add \(9\) to each side. We get
\[x^2 + 6x + 9 = -5 + 9\]
The left side factors as a perfect square trinomial.
\[(x + 3)^2 = 4\]
We can solve this by the extraction of roots.
\[\begin{array}{flushleft}
x + 3 &= \pm \sqrt{4}\\
x + 3 &= \pm 2\\
x &= \pm 2 - 3\\
x &= + 2 - 3 & \text{ and } & x = -2 - 3\\
x &= -1 & \text{ and } -5
\end{array}\]
Notice that when the roots are rational numbers, the equation is factorable.
The big question is, "How did we know to add 9 to each side of the equation?" We can convert any quadratic trinomial appearing in an equation into a perfect square trinomial if we know what number to add to both sides. We can determine that particular number by observing the following situation:
Consider the square of the binomial and the resulting perfect square trinomial
\[(x + p)^2 = x^2 + 2px + p^2\]
Notice that the constant term (the number we are looking for) can be obtained from the linear term \(2px\). If we take one half the coefficient of \(x\), \(\dfrac{20}{2} = p\), and square it, we get the constant term \(p^2\). This is true for every perfect square trinomial with a leading coefficient \(1\).
In a perfect square trinomial with leading coefficient \(1\), the constant term is the square of one-half the coefficient of the linear term.
Study these examples to see what constant term will make the given binomial into a perfect square trinomial.
\(x^2 + 6x\). The constant must be the square of one half the coefficient of \(x\). Since the coefficient of \(x\) is \(6\), we have
\(\dfrac{6}{2} = 3\) and \(3^2 = 9\)
The constant is \(9\).
\[x^2 + 6x + 9 = (x + 3)^2\]
This is a perfect square trinomial.
\(a^2 + 10a\). The constant must be the square of one half the coefficient of \(a\). Since the coefficient of \(a\) is \(10\), we have
\(\dfrac{10}{2} = 5\) and \(5^2 = 25\).
The constant is \(25\)
\[a^2 + 10a + 25 = (a + 5)^2\]
\(y^2 + 3y\). The constant must be the square of one-half the coefficient of \(y\). Since the coefficient of \(y\) is \(3\), we have
\(\dfrac{3}{2}\) and \((\dfrac{3}{2})^2 = \dfrac{9}{4}\)
The constant is \(\dfrac{9}{4}\).
\[y^2 + 3y + \dfrac{9}{4} = (y + \dfrac{3}{2})^2\]
The Method Of Completing The Square
Now, with these observations, we can describe the method of completing the square.
- Write the equation so that the constant term appears on the right side of the equation.
- If the leading coefficient is different from 1, divide each term of the equation by that coefficient.
- Take one half of the coefficient of the linear term, square it, then add it to both sides of the equation.
- The trinomial on the left is now a perfect square trinomial and can be factored as \(()^2\). The first term in the parentheses is the square root of the quadratic term. The last term in the parentheses is one-half the coefficient of the linear term.
- Solve this equation by extraction of roots.
Sample Set A
Solve the following equations.
\(\begin{array}{flushleft}
x^2 + 8x - 9 &= 0 & \text{ Add } 9 \text{ to both sides.}\\
x^2 + 8x &= 9 & \text{ One half the coefficient of } x \text{ is } 4, \text{ and } 4^2 \text{ is } 16. \text{ Add } 16 \text{ to both sides. }\\
x^2 + 8x + 16 &= 9 + 16\\
x^2 + 8x + 16 &= 25 & \text{ Factor. }\\
(x+4)^2 &= 25 & \text{ Take square roots. }\\
x + 4 &= \pm 5\\
x &= \pm 5 - 4 & +5 - 4 = 1, -5 - 4 = -9\\
x &= 1, -9
\end{array}\)
\(\begin{array}{flushleft}
x^2 - 3x - 1 &= 0 & \text{ Add } 1 \text{ to both sides.}\\
&& \text{ One half the coefficient of } x \text{ is } \dfrac{-3}{2}.\\
x^2 - 3x &= 1 & \text{ Square it: } (\dfrac{-3}{2})^2 = \dfrac{9}{4}. \text{ Add } \dfrac{9}{4} \text{to each side.}\\
x^2 - 3x + \dfrac{9}{4} &= 1 + \dfrac{9}{4}\\
x^2 - 3x + \dfrac{9}{4} &= \dfrac{13}{4} & \text{Factor. Notice that since the sign of the middle term of the trinomial is "-", its factored form has a "-" sign.}\\
(x - \dfrac{3}{2})^2 &= \dfrac{13}{4} & \text{ Now take square roots }\\
x - \dfrac{3}{2} &= \pm \sqrt{\dfrac{13}{4}}\\
x - \dfrac{3}{2} &= \pm \sqrt{\dfrac{13}{2}}\\
x &= \pm \dfrac{\sqrt{13}}{2} + \dfrac{3}{2}\\
x &= \dfrac{\pm \sqrt{13} + 3}{2}\\
x &= \dfrac{3 \pm \sqrt{13}}{2}
\end{array}\)
\(\begin{array}{flushleft}
3a^2 - 36a - 39 &= 0 & \text{ Add } 39 \text{ to both sides.}\\
3a^2 - 36a &= 39 & \text{ The leading coefficient is } 3 \text{ and we need it to be } 1. \text{ Divide each term by } 3 \text{. One half of the coefficient of } a \text{ is } -6\\
a^2 - 12a &= 13 & \text{ Square it: } (-6)^2 = 36 \text{ Add } 36 \text{ to each side. }\\
a^2 - 12a + 36 &= 13 + 36\\
a^2 - 12a + 36 &= 49\\
(a-6)^2 &= 49 & \text{ Factor. }\\
a - 6 &= \pm 7\\
a &= \pm 7 + 6 & +7 + 6 = 13, -7 + 6 = -1
\end{array}\)
\(a = 13, -1\).
\(\begin{array}{flushleft}
2x^2 + x + 4 &= 0\\
2x^2 + x &= -4\\
x^2 + \dfrac{1}{2}x &= -2\\
x^2 + \dfrac{1}{2}x + (\dfrac{1}{4})^2 &= -2 + (\dfrac{1}{4})^2
\end{array}\)
\((x + \dfrac{1}{4})^2 = -2 + \dfrac{1}{16} = \dfrac{-32}{16} + \dfrac{1}{16} = \dfrac{-31}{16}\)
Since we know that the square of any number is positive, this equation has no real number solution.
Calculator Problem:
Solve \(7a^2 - 5a - 1 = 0\). Round each solution to the nearest tenth.
\(\begin{array}{flushleft}
7a^2 - 5a - 1 &= 0\\
7a^2 - 5a &= 1\\
a^2 - \dfrac{5}{7}a &= \dfrac{1}{7}\\
a^2 - \dfrac{5}{7}a + (\dfrac{5}{14})^2 &= \dfrac{1}{7} + (\dfrac{5}{14})^2\\
(a - \dfrac{5}{14})^2 &= \dfrac{1}{7} + \dfrac{25}{196} &= \dfrac{28}{196} + \dfrac{25}{196} = \dfrac{53}{196}\\
a - \dfrac{5}{14} &= \pm \sqrt{\dfrac{53}{196}} &= \pm \dfrac{\sqrt{53}}{14}
\end{array}\)
\(a = \dfrac{5}{14} \pm \dfrac{\sqrt{53}}{14} = \dfrac{5 \pm \sqrt{53}}{14}\)
Rounding to tenths, we get \(a \approx -0.2\). Thus, \(a \approx 0.9\) and \(-0.2\) to the nearest tenth.
Practice Set A
Solve each of the following quadratic equations using the method of completing the square.
\(x^2 - 2x - 48 = 0\)
- Answer
-
\(x=−6, 8\)
\(x^2 + 3x - 5 = 0\)
- Answer
-
\(x = \dfrac{-3 \pm \sqrt{29}}{2}\)
\(4m^2 + 5m = -1\)
- Answer
-
\(m = \dfrac{-1}{4}, -1\)
\(5y^2 - 2y - 4 = 0\)
- Answer
-
\(y = \dfrac{1 \pm \sqrt{21}}{5}\)
Calculator problem:
Solve \(3x^2 - x - 1= 0\). Round each solution to the nearest tenth.
- Answer
-
\(x = 0.8, -0.4\)
Exercises
For the following problems, solve the equations by completing the square.
\(x^2 +2x - 8 = 0\)
- Answer
-
\(x=−4, 2\)
\(y^2 - 5y - 6 = 0\)
\(a^2 + 7a + 12 = 0\)
- Answer
-
\(a=−3, −4\)
\(x^2 - 10x + 16 = 0\)
\(y^2 - 2y - 24 = 0\)
- Answer
-
\(y = -4, 6\)
\(a^2 + 2a - 35 = 0\)
\(x^2 + 2x + 5 = 0\)
- Answer
-
No real number solution.
\(x^2 - 6x + 1 = 0\)
\(x^2 + 4x + 4 = 0\)
- Answer
-
\(x=−2\)
\(a^2 + 4a + 7 = 0\)
\(b^2 + 5b - 3 = 0\)
- Answer
-
\(b = \dfrac{-5 \pm \sqrt{37}}{2}\)
\(b^2 - 6b = 72\)
\(a^2 + 10a - 9 = 0\)
- Answer
-
\(a = -5 \pm \sqrt{34}\)
\(a^2 - 2a - 3 = 0\)
\(x^2 - 10x = 0\)
- Answer
-
\(x=10, 0\)
\(y^2 - 8y = 0\)
\(a^2 - 6a = 0\)
- Answer
-
\(a = 6, 0\)
\(b^2 + 6b = 0\)
\(-x^2 - 14x = 13\)
- Answer
-
\(x=−13, −1\)
\(-x^2 +8x = -84\)
\(2a^2 + 2a - 1 = 0\)
- Answer
-
\(a = \dfrac{-1 \pm \sqrt{3}}{2}\)
\(4b^2 - 8b = 16\)
\(9x^2 + 12x - 5 = 0\)
- Answer
-
\(x = \dfrac{1}{3}, -\dfrac{5}{3}\)
\(16y^2 - 8y - 3 = 0\)
\(2x^2 + 5x - 4 = 0\)
- Answer
-
\(x = \dfrac{-5 \pm \sqrt{57}}{4}\)
\(3a^2 + 2a - 24 = 0\)
\(x^2 + 2x + 8 = 0\)
- Answer
-
No real number solution.
\(y^2 - 3y + 10 = 0\)
\(7a^2 + 3a - 1 = 0\)
- Answer
-
\(a = \dfrac{-3 \pm \sqrt{37}}{14}\)
Calculator Problems
For the following problems, round each solution to the nearest hundredth.
\(5m^2 - 2m - 6 = 0\)
\(3y^2 + 5y = 7\)
- Answer
-
\(y=0.91, −2.57\)
\(1.8x^2 + 2.3x - 4.1 = 0\)
\(0.04a^2 - 0.03a + 0.02 = 0\)
- Answer
-
No real number solution.
Exercises For Review
Factor \(12ax - 6bx + 20ay - 10by\) by grouping.
Graph the compound inequality \(-6 \le -2x + 2 < -4\)
- Answer
-
Find the equation of the line that passes through the points \((1, -2)\) and \((0, 4)\)
Find the product: \(\dfrac{x^2 - 4x - 12}{x^2 - 2x - 8} \cdot \dfrac{x^2 - 3x - 4}{x^2 - 3x - 18}\)
- Answer
-
\(\dfrac{x+1}{x+3}\)
Use the method of extraction of roots to solve \((x-2)^2 = 25\)


20.2: Thinking about Life's Origins- A Short Summary of a Long History
By all accounts, the earth must have been a very unpleasant place soon after its formation! For that reason, the period from 4.8 to 4.0 billion years ago is called the Hadean Eon, after Hades, the hell of the ancient Greeks! Until recently, geological, geochemical and fossil evidence suggested that life arose between 3.8 and 4.1 billion years ago. The 2017 discovery of 3.95 billion year-old sedimentary rocks in Labrador with evidence of life, points to an even earlier origin of life, (see From Canada Comes the Oldest Evidence of Life on Earth ). In fact, questions about life’s origins are probably “as old as the hills…” or at least as old as the ancient Greeks! We only have records of human notions of life’s origins dating from biblical accounts and, just a bit later, from Aristotle’s musings. While Aristotle did not suggest that life began in hell, he and other ancient Greeks did speculate about life’s origins by spontaneous generation , in the sense of abiogenesis ( life originating from non-life ). He further speculated that the origins of life were gradual.
Later, the dominant theological accounts of creation in Europe in the middle ages muted any notions of origins and evolution. While a few mediaeval voices ran counter to strict biblical readings of the creation stories, it was not until the Renaissance in the 14th-17th century that an appreciation of ancient Greek humanism was reawakened, and with it, scientific curiosity and the ability to engage in rational questioning and research.
Many will recall that Louis Pasteur in the mid-19th century put to rest any lingering notions of life forming from dead (e.g., rotten, or fecal) matter. He showed that life would not form in sterilized nutrient solutions unless the broth was exposed to the air. Fewer know that much earlier, Anton Van Leeuwenhoek, the 17th century microscopist who first described bacteria and animalcules , mostly protozoa in pond water, had already tested the notion of spontaneous generation. By observing open and sealed containers of meat over time, he became convinced that ‘large’ animals like fleas and frogs do not arise de novo from putrid meat or slime. He also declared that insects come from other insects, and not from the flowers that they visited.
No lesser light than Charles Darwin suggested in 1859 that life might have begun in a " warm little pond, with all sorts of ammonia and phosphoric salts, light, heat, electricity, &c., present, that a proteine compound was chemically formed ready to undergo still more complex changes. " He even realized that these chemical constituents would not have survived in the atmosphere and waters of his day, but must have done so in a prebiotic world. In On the Origin of Species , he referred to life having been ‘created’. There, Darwin was not referring to a biblical basis of creation; he clearly meant that life originated “ by some wholly unknown process " at a time before which there was no life. Finally, Pasteur’s 1861 contribution was the irrefutable, definitive proof that ‘invisible’ microbial life likewise did not arise by spontaneous generation. Thus for creatures already on earth, they could only arise by biogenesis ( life - from - life ), the opposite of abiogenesis , a term that now applies to only the first origins of life!
Among Darwin’s friends and contemporaries were Charles Lyell and Roderick Murchison, both geologists who understood much about the slow geological changes that shaped the earth. Darwin was therefore familiar with the concept of extended periods of geological time, amounts of time he believed was necessary for the natural selection of traits leading to species divergence.
Fast-forward to the 1920s when J.H.B.S. Haldane and A. Oparin offered an hypothesis about the life’s origins based on notions of the chemistry and physical conditions that might have existed on a prebiotic earth . Their proposal assumed that the earth’s atmosphere was hot, hellish and reducing (i.e., filled with inorganic molecules able to give up electrons and hydrogens). There are more than a few hypotheses for which chemicals were already present on earth, or that formed when the planet formed about 4.8 billion years ago. We’ll start our exploration with Oparin and Haldane’s reducing atmosphere . Then we look at the possibility that life began under non-reducing conditions (with passing reference to a few other ideas).