6.8: Fitting Exponential Models to Data
- Build an exponential model from data.
- Build a logarithmic model from data.
- Build a logistic model from data.
In previous sections of this chapter, we were either given a function explicitly to graph or evaluate, or we were given a set of points that were guaranteed to lie on the curve. Then we used algebra to find the equation that fit the points exactly. In this section, we use a modeling technique called regression analysis to find a curve that models data collected from real-world observations. With regression analysis , we don’t expect all the points to lie perfectly on the curve. The idea is to find a model that best fits the data. Then we use the model to make predictions about future events.
Do not be confused by the word model . In mathematics, we often use the terms function , equation , and model interchangeably, even though they each have their own formal definition. The term model is typically used to indicate that the equation or function approximates a real-world situation.
We will concentrate on three types of regression models in this section: exponential, logarithmic, and logistic. Having already worked with each of these functions gives us an advantage. Knowing their formal definitions, the behavior of their graphs, and some of their real-world applications gives us the opportunity to deepen our understanding. As each regression model is presented, key features and definitions of its associated function are included for review. Take a moment to rethink each of these functions, reflect on the work we’ve done so far, and then explore the ways regression is used to model real-world phenomena.
 What do these phenomena have in common? For one thing, all the models either increase or decrease as time moves forward. But that’s not the whole story. It’s the way data increase or decrease that helps us determine whether it is best modeled by an exponential equation. Knowing the behavior of exponential functions in general allows us to recognize when to use exponential regression, so let’s review exponential growth and decay.
Recall that exponential functions have the form \(y=ab^x\) or \(y=A_0e^{kx}\). When performing regression analysis, we use the form most commonly used on graphing utilities, \(y=ab^x\). Take a moment to reflect on the characteristics we’ve already learned about the exponential function \(y=ab^x\) (assume \(a>0\)):
- \(b\) must be greater than zero and not equal to one.
-
The initial value of the model is \(y=a\).
- If \(b>1\), the function models exponential growth. As \(x\) increases, the outputs of the model increase slowly at first, but then increase more and more rapidly, without bound.
- If \(0<b<1\), the function models exponential decay . As \(x\) increases, the outputs for the model decrease rapidly at first and then level off to become asymptotic to the x -axis. In other words, the outputs never become equal to or less than zero.
As part of the results, your calculator will display a number known as the correlation coefficient , labeled by the variable \(r\), or \(r^2\). (You may have to change the calculator’s settings for these to be shown.) The values are an indication of the “goodness of fit” of the regression equation to the data. We more commonly use the value of \(r^2\) instead of \(r\), but the closer either value is to \(1\), the better the regression equation approximates the data.
Exponential regression is used to model situations in which growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero. We use the command “ExpReg” on a graphing utility to fit an exponential function to a set of data points. This returns an equation of the form,
\[y=ab^x\]
Note that:
- \(b\) must be non-negative.
- when \(b>1\), we have an exponential growth model.
- when \(0<b<1\), we have an exponential decay model.
-
Use the STAT then EDIT menu to enter given data.
- Clear any existing data from the lists.
- List the input values in the L1 column.
- List the output values in the L2 column.
-
Graph and observe a scatter plot of the data using the STATPLOT feature.
- Use ZOOM [9] to adjust axes to fit the data.
- Verify the data follow an exponential pattern.
-
Find the equation that models the data.
- Select “ExpReg” from the STAT then CALC menu.
- Use the values returned for a and b to record the model, \(y=ab^x\).
- Graph the model in the same window as the scatterplot to verify it is a good fit for the data.
Example \(\PageIndex{1}\): Using Exponential Regression to Fit a Model to Data
In 2007, a university study was published investigating the crash risk of alcohol impaired driving. Data from \(2,871\) crashes were used to measure the association of a person’s blood alcohol level (BAC) with the risk of being in an accident. Table \(\PageIndex{1}\) shows results from the study. The relative risk is a measure of how many times more likely a person is to crash. So, for example, a person with a BAC of \(0.09\) is \(3.54\) times as likely to crash as a person who has not been drinking alcohol.
| BAC | 0 | 0.01 | 0.03 | 0.05 | 0.07 | 0.09 |
|---|---|---|---|---|---|---|
| Relative Risk of Crashing | 1 | 1.03 | 1.06 | 1.38 | 2.09 | 3.54 |
| BAC | 0.11 | 0.13 | 0.15 | 0.17 | 0.19 | 0.21 |
| Relative Risk of Crashing | 6.41 | 12.6 | 22.1 | 39.05 | 65.32 | 99.78 |
- Let \(x\) represent the BAC level, and let \(y\) represent the corresponding relative risk. Use exponential regression to fit a model to these data.
- After \(6\) drinks, a person weighing \(160\) pounds will have a BAC of about \(0.16\). How many times more likely is a person with this weight to crash if they drive after having a \(6\)-pack of beer? Round to the nearest hundredth.
Solution
- Using the STAT then EDIT menu on a graphing utility, list the BAC values in L1 and the relative risk values in L2. Then use the STATPLOT feature to verify that the scatterplot follows the exponential pattern shown in Figure \(\PageIndex{1}\):
Figure \(\PageIndex{1}\)
Use the “ExpReg” command from the STAT then CALC menu to obtain the exponential model,
\(y=0.58304829{(2.20720213E10)}^x\)
Converting from scientific notation, we have:
\(y=0.58304829{(22,072,021,300)}^x\)
Notice that \(r^2≈0.97\) which indicates the model is a good fit to the data. To see this, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure \(\PageIndex{2}\):
Figure \(\PageIndex{2}\)
- Use the model to estimate the risk associated with a BAC of \(0.16\). Substitute \(0.16\) for \(x\) in the model and solve for \(y\).
\[\begin{align*} y&= 0.58304829{(22,072,021,300)}^x \qquad \text{Use the regression model found in part } (a)\\ &= 0.58304829{(22,072,021,300)}^{0.16} \qquad \text{Substitute 0.16 for x}\\ &\approx 26.35 \qquad \text{Round to the nearest hundredth} \end{align*}\]
If a \(160\)-pound person drives after having \(6\) drinks, he or she is about \(26.35\) times more likely to crash than if driving while sober.
Table \(\PageIndex{2}\) shows a recent graduate’s credit card balance each month after graduation.
| Month | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|
| Debt ($) | 620.00 | 761.88 | 899.80 | 1039.93 | 1270.63 | 1589.04 | 1851.31 | 2154.92 |
- Use exponential regression to fit a model to these data.
- If spending continues at this rate, what will the graduate’s credit card debt be one year after graduating?
- Answer a
-
The exponential regression model that fits these data is \(y=522.88585984{(1.19645256)}^x\).
- Answer b
-
If spending continues at this rate, the graduate’s credit card debt will be \($4,499.38\) after one year.
No. Remember that models are formed by real-world data gathered for regression. It is usually reasonable to make estimates within the interval of original observation (interpolation). However, when a model is used to make predictions, it is important to use reasoning skills to determine whether the model makes sense for inputs far beyond the original observation interval (extrapolation).
Building a Logarithmic Model from Data
Just as with exponential functions, there are many real-world applications for logarithmic functions: intensity of sound, pH levels of solutions, yields of chemical reactions, production of goods, and growth of infants. As with exponential models, data modeled by logarithmic functions are either always increasing or always decreasing as time moves forward. Again, it is the way they increase or decrease that helps us determine whether a logarithmic model is best.
Recall that logarithmic functions increase or decrease rapidly at first, but then steadily slow as time moves on. By reflecting on the characteristics we’ve already learned about this function, we can better analyze real world situations that reflect this type of growth or decay. When performing logarithmic regression analysis , we use the form of the logarithmic function most commonly used on graphing utilities, \(y=a+b\ln(x)\). For this function
- All input values, \(x\),must be greater than zero.
- The point \((1,a)\) is on the graph of the model.
- If \(b>0\),the model is increasing. Growth increases rapidly at first and then steadily slows over time.
- If \(b<0\),the model is decreasing. Decay occurs rapidly at first and then steadily slows over time.
Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time. We use the command “LnReg” on a graphing utility to fit a logarithmic function to a set of data points. This returns an equation of the form,
\[y=a+b\ln(x)\]
Note that
- all input values, \(x\),must be non-negative.
- when \(b>0\), the model is increasing.
- when \(b<0\), the model is decreasing.
-
Use the STAT then EDIT menu to enter given data.
- Clear any existing data from the lists.
- List the input values in the L1 column.
- List the output values in the L2 column.
-
Graph and observe a scatter plot of the data using the STATPLOT feature.
- Use ZOOM [9] to adjust axes to fit the data.
- Verify the data follow a logarithmic pattern.
-
Find the equation that models the data.
- Select “LnReg” from the STAT then CALC menu.
- Use the values returned for a and b to record the model, \(y=a+b\ln(x)\).
- Graph the model in the same window as the scatterplot to verify it is a good fit for the data.
Due to advances in medicine and higher standards of living, life expectancy has been increasing in most developed countries since the beginning of the 20 th century. Table \(\PageIndex{3}\) shows the average life expectancies, in years, of Americans from 1900–2010.
| Year | 1900 | 1910 | 1920 | 1930 | 1940 | 1950 |
|---|---|---|---|---|---|---|
| Life Expectancy(Years) | 47.3 | 50.0 | 54.1 | 59.7 | 62.9 | 68.2 |
| Year | 1960 | 1970 | 1980 | 1990 | 2000 | 2010 |
| Life Expectancy(Years) | 69.7 | 70.8 | 73.7 | 75.4 | 76.8 | 78.7 |
- Let \(x\) represent time in decades starting with \(x=1\) for the year 1900, \(x=2\) for the year 1910, and so on. Let \(y\) represent the corresponding life expectancy. Use logarithmic regression to fit a model to these data.
- Use the model to predict the average American life expectancy for the year 2030.
Solution
- Using the STAT then EDIT menu on a graphing utility, list the years using values \(1–12\) in L1 and the corresponding life expectancy in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logarithmic pattern as shown in Figure \(\PageIndex{3}\):
Figure \(\PageIndex{3}\)
Use the “LnReg” command from the STAT then CALC menu to obtain the logarithmic model,
\(y=42.52722583+13.85752327\ln(x)\)
Next, graph the model in the same window as the scatterplot to verify it is a good fit as shown in Figure \(\PageIndex{4}\):
Figure \(\PageIndex{4}\)
- To predict the life expectancy of an American in the year \(2030\), substitute \(x=14\) for the in the model and solve for \(y\):
\[\begin{align*} y&= 42.52722583+13.85752327\ln(x) \qquad \text{Use the regression model found in part } (a)\\ &= 42.52722583+13.85752327\ln(14) \qquad \text{Substitute 14 for x}\\ &\approx 79.1 \qquad \text{Round to the nearest tenth} \end{align*}\]
If life expectancy continues to increase at this pace, the average life expectancy of an American will be \(79.1\) by the year \(2030\).
Sales of a video game released in the year 2000 took off at first, but then steadily slowed as time moved on. Table \(\PageIndex{4}\) shows the number of games sold, in thousands, from the years 2000–2010.
| Year | 2000 | 2001 | 2002 | 2003 | 2004 | 2005 |
|---|---|---|---|---|---|---|
| Number Sold (thousands) | 142 | 149 | 154 | 155 | 159 | 161 |
| Year | 2006 | 2007 | 2008 | 2009 | 2010 | - |
| Number Sold (thousands) | 163 | 164 | 164 | 166 | 167 | - |
Let \(x\) represent time in years starting with \(x=1\) for the year 2000. Let \(y\) represent the number of games sold in thousands.
- Use logarithmic regression to fit a model to these data.
- If games continue to sell at this rate, how many games will sell in 2015? Round to the nearest thousand.
- Answer a
-
The logarithmic regression model that fits these data is \(y=141.91242949+10.45366573\ln(x)\)
- Answer b
-
If sales continue at this rate, about \(171,000\) games will be sold in the year \(2015\).
Building a Logistic Model from Data
Like exponential and logarithmic growth, logistic growth increases over time. One of the most notable differences with logistic growth models is that, at a certain point, growth steadily slows and the function approaches an upper bound, or limiting value . Because of this, logistic regression is best for modeling phenomena where there are limits in expansion, such as availability of living space or nutrients.
It is worth pointing out that logistic functions actually model resource-limited exponential growth. There are many examples of this type of growth in real-world situations, including population growth and spread of disease, rumors, and even stains in fabric. When performing logistic regression analysis, we use the form most commonly used on graphing utilities:
\(y=\dfrac{c}{1+ae^{−bx}}\)
Recall that:
- \(\dfrac{c}{1+a}\) is the initial value of the model.
- when \(b>0\), the model increases rapidly at first until it reaches its point of maximum growth rate, \((\dfrac{\ln(a)}{b}, \dfrac{c}{2})\). At that point, growth steadily slows and the function becomes asymptotic to the upper bound \(y=c\).
- \(c\) is the limiting value, sometimes called the carrying capacity , of the model.
Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows to an upper limit. We use the command “Logistic” on a graphing utility to fit a logistic function to a set of data points. This returns an equation of the form
\[y=\dfrac{c}{1+ae^{−bx}}\]
Note that
- The initial value of the model is \(\dfrac{c}{1+a}\).
- Output values for the model grow closer and closer to \(y=c\) as time increases.
-
Use the STAT then EDIT menu to enter given data.
- Clear any existing data from the lists.
- List the input values in the L1 column.
- List the output values in the L2 column.
-
Graph and observe a scatter plot of the data using the STATPLOT feature.
- Use ZOOM [9] to adjust axes to fit the data.
- Verify the data follow a logistic pattern.
-
Find the equation that models the data.
- Select “Logistic” from the STAT then CALC menu.
- Use the values returned for \(a\), \(b\), and \(c\) to record the model, \(y=\dfrac{c}{1+ae^{−bx}}\).
- Graph the model in the same window as the scatterplot to verify it is a good fit for the data.
Mobile telephone service has increased rapidly in America since the mid 1990s. Today, almost all residents have cellular service. Table \(\PageIndex{5}\) shows the percentage of Americans with cellular service between the years 1995 and 2012.
| Year | Americans with Cellular Service (%) | Year | Americans with Cellular Service (%) |
|---|---|---|---|
| 1995 | 12.69 | 2004 | 62.852 |
| 1996 | 16.35 | 2005 | 68.63 |
| 1997 | 20.29 | 2006 | 76.64 |
| 1998 | 25.08 | 2007 | 82.47 |
| 1999 | 30.81 | 2008 | 85.68 |
| 2000 | 38.75 | 2009 | 89.14 |
| 2001 | 45.00 | 2010 | 91.86 |
| 2002 | 49.16 | 2011 | 95.28 |
| 2003 | 55.15 | 2012 | 98.17 |
- Let \(x\) represent time in years starting with \(x=0\) for the year 1995. Let \(y\) represent the corresponding percentage of residents with cellular service. Use logistic regression to fit a model to these data.
- Use the model to calculate the percentage of Americans with cell service in the year 2013. Round to the nearest tenth of a percent.
- Discuss the value returned for the upper limit, \(c\). What does this tell you about the model? What would the limiting value be if the model were exact?
Solution
- Using the STAT then EDIT menu on a graphing utility, list the years using values \(0–15\) in L1 and the corresponding percentage in L2. Then use the STATPLOT feature to verify that the scatterplot follows a logistic pattern as shown in Figure \(\PageIndex{5}\):
Figure \(\PageIndex{5}\)
Use the “Logistic” command from the STAT then CALC menu to obtain the logistic model,
\[y=105.73795261+6.88328979e^{−0.2595440013x}\]
Next, graph the model in the same window as shown in Figure \(\PageIndex{6}\) the scatterplot to verify it is a good fit:
Figure \(\PageIndex{6}\)
- To approximate the percentage of Americans with cellular service in the year 2013, substitute \(x=18\) for the in the model and solve for \(y\):
\[\begin{align*} y&= \dfrac{105.7379526}{1+6.88328979e^{-0.2595440013x}} \qquad \text{Use the regression model found in part } (a)\\ &= \dfrac{105.7379526}{1+6.88328979e^{-0.2595440013(18)}} \qquad \text{Substitute 18 for x}\\ &\approx 99.3 \qquad \text{Round to the nearest tenth} \end{align*}\]
According to the model, about 98.8% of Americans had cellular service in 2013.
- The model gives a limiting value of about \(105\). This means that the maximum possible percentage of Americans with cellular service would be \(105%\), which is impossible. (How could over \(100%\) of a population have cellular service?) If the model were exact, the limiting value would be \(c=100\) and the model’s outputs would get very close to, but never actually reach \(100%\). After all, there will always be someone out there without cellular service!
Table \(\PageIndex{6}\) shows the population, in thousands, of harbor seals in the Wadden Sea over the years 1997 to 2012.
| Year | Seal Population (Thousands) | Year | Seal Population (Thousands) |
|---|---|---|---|
| 1997 | 3.493 | 2005 | 19.590 |
| 1998 | 5.282 | 2006 | 21.955 |
| 1999 | 6.357 | 2007 | 22.862 |
| 2000 | 9.201 | 2008 | 23.869 |
| 2001 | 11.224 | 2009 | 24.243 |
| 2002 | 12.964 | 2010 | 24.344 |
| 2003 | 16.226 | 2011 | 24.919 |
| 2004 | 18.137 | 2012 | 25.108 |
Let \(x\) represent time in years starting with \(x=0\) for the year 1997. Let \(y\) represent the number of seals in thousands.
- Use logistic regression to fit a model to these data.
- Use the model to predict the seal population for the year 2020.
- To the nearest whole number, what is the limiting value of this model?
- Answer a
-
The logistic regression model that fits these data is \(y=\dfrac{25.65665979}{1+6.113686306e^{−0.3852149008x}}\).
- Answer b
-
If the population continues to grow at this rate, there will be about \(25,634\) seals in 2020.
- Answer c
-
To the nearest whole number, the carrying capacity is \(25,657\).
Access this online resource for additional instruction and practice with exponential function models.
Visit this website for additional practice questions from Learningpod.
Key Concepts
- Exponential regression is used to model situations where growth begins slowly and then accelerates rapidly without bound, or where decay begins rapidly and then slows down to get closer and closer to zero.
- We use the command “ExpReg” on a graphing utility to fit function of the form \(y=ab^x\) to a set of data points. See Example \(\PageIndex{1}\).
- Logarithmic regression is used to model situations where growth or decay accelerates rapidly at first and then slows over time.
- We use the command “LnReg” on a graphing utility to fit a function of the form \(y=a+b\ln(x)\) to a set of data points. See Example \(\PageIndex{2}\).
- Logistic regression is used to model situations where growth accelerates rapidly at first and then steadily slows as the function approaches an upper limit.
- We use the command “Logistic” on a graphing utility to fit a function of the form \(y=\dfrac{c}{1+ae^{−bx}}\) to a set of data points. See Example \(\PageIndex{3}\).
Contributors and Attributions
-
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus .


4: New Vapor Pressure Equation
Distillation Science (a blend of Chemistry and Chemical Engineering)
This is Part IV, New Vapor Pressure Equation of a ten-part series of technical articles on Distillation Science, as is currently practiced on an industrial level. See Part I, Overview for introductory comments, the scope of the article series, and nomenclature. Parts II and III are pre-requisite to Part IV .
Previous Part II, Vapor Pressure deals with the pure component equations normally found in textbooks, but which have limitations in industrial application. Part III, Critical Properties and Acentric Factor develops those parameters used to inter-relate conventional units of temperature, pressure and volume to their reduced property equivalents, as well as the Pfizer acentric factor that is used in modern Equations of State.
Several thermodynamically consistent vapor pressure (VP) equation forms have been developed in the past purposely for organic fluids, and typically have the stipulation that they do not work with either polar or heavily hydrogen-bonded fluids. Being both slightly polar in nature as well as hydrogen-bonded, chlorosilanes and their impurities fall exactly in this exclusion, and therefore make good continuing examples.
In this article, temperature and pressure are used in their reduced form, since this allows better inter-comparison between fluids, is dimensionless, and allows a more global future development. The reduced form is obtained by simply dividing conventional temperature, T, by the fluid’s critical temperature=> T r = T/T c . Similarly, reduced pressure is conventional pressure divided by the fluid’s critical pressure => P r = P/P c . Temperature and pressure must be in absolute units ( e.g., ºK not ºC)
Referring back to Part II, E quation 2-4 , the Clausius-Clapeyron equation is given in reduced format, using \(\psi \nonumber\) as notation for the thermodynamic-based derivative of the natural log of (reduced) vapor pressure with respect to the inverse of (reduced) absolute temperature:
\[\psi = \frac{-d(\ln P_{r})}{d(1/T_{r})} = \frac{\Delta H_{v}}{\Delta ZRT_{c}} \label{4-1} \]
In order to have an improved VP equation form that can be used for the wide range between T b and T c (i.e., from atmospheric pressure to the critical point), that relationship should meet the following criteria:
- The equation form should exactly have a reduced VP value of 1/P c at T b /T c (i.e., yields the known atmospheric boiling point at atmospheric pressure), which is a well-measured lab value and is readily estimated based on structural considerations (see Part III article).
- The equation form should exactly have a reduced VP value of unity at the critical temperature (i.e., match the measured value of the critical point), where the liquid meniscus disappears. This property can also be readily estimated based on structural considerations (see Part III article).
- Give a reasonable value for reduced VP at T r =0.7, corresponding to the definition of the Pfizer acentric factor, which has a basis in molecular complexity (see Part III article).
- Meet the two thermodynamic consistency Reidel tests, as the critical point is approached ( that "α" monotonically drop to its lowest value at critical, and that the temperature derivative of α = zero at critical). See the left-most part of E quation ( \ref{4-4} ) for the definition of "α".
- It must be noted that while this new vapor pressure relationship exactly matches the known atmospheric pressure boiling point, its differential form will not yield the known atmospheric pressure latent heat of vaporization
.Such a VP equation form would allow evaluation of modern Equations of State (see Part V article), Fugacity considerations (see Part VI article), and Binary Interaction Parameters (see Part VII article). Preferably the VP equation form would reasonably fit most fluids (both naturally occurring as well as synthetic), including polar compounds and those with substantial hydrogen bonding. Note that the VP equations given in Part II work over only narrow ranges; however above criteria #1 and #2 require the VP equation form to span a very broad range. In the text "The Properties of Gases and Liquids", by Prausnitz, et al, several alternate VP equation forms are discussed, all of which meet the above four criteria.
In his doctoral thesis at Syracuse University in 1965, under the guidance of Leonard Stiel, Richard Thek proposed exactly such an equation form to fill the technology gap. It was subsequently published by the AIChE in 1966. Unfortunately the computing power at that time was very limited, so equation forms with iterative solutions could not be used; and the internet did not exist to allow ready access to global data taken over the last many decades. Using modern computer processing speed and programming, these limitations have been resolved, and the constants determined for all chlorosilanes and their electronic impurities.
As given by Equation ( \ref{4-2} ) below, the original Thek-Stiel VP equation in its differential form has two terms: the first that governs at lower pressures close to atmospheric, and the second that takes over as the critical point is approached. While Thek set variables “q” and “k” equal to constants in order to avoid iterative solutions, with modern programming these can now be evaluated as true variables. The derivation of the final Modified Thek-Stiel VP Equation is available on request, but is not reproduced in these articles for the sake of brevity.
\[\frac{d(\ln P_{r})}{d(1/T_{r})} = \frac{A}{T_{r}^2}[1-B_{1}T_{r}+B_{2}T_{r}^2-B_{3}T_{r}^3+B_{4}T_{r}^4...]+ \frac{c}{T_{r}^2}(T_{r}^n-k) \label{4-2} \]
The several “B” constants result from the binomial expansion of the general Watson relationship for ΔH v . After dropping the B 4 and higher expansion terms as non-significant mathematically and integrating, the resulting reduced vapor pressure equation is:
\[ \ln (P_{r})= A[B_{0}-T_{r}^{-1}-B_{1}\ln T_{r}+B_{2}T_{r}-(1/2)B_{3}T_{r}^2]+c\left[\frac{T_{r}^{n-1}-1}{n-1} +k(T_{r}^{-1}-1)\right] \label{4-3} \]
- \(A=\dfrac{\Delta H_{vb}}{RT_{c}(1-T_{br})^q} \nonumber\)
- \(B_{0}=1-B_{2}+(\frac{1}{2})B_{3} \nonumber\)
- \(B_{1}=q \nonumber\)
- \(B_{2}= \dfrac{q(q-1)}{2!} = \dfrac{q(q-1)}{2} \nonumber\)
- \(B_{3}=\dfrac{q(q-1)(q-2)}{3!} = \dfrac{q(q-1)(q-2)}{6} \nonumber\)
- \(c=\dfrac{\alpha_{c} -A(1-B_{1}+B_{2}-B_{3})}{1-k} \nonumber\)
- \(n=(1-k)+\dfrac{A}{c}(1-B_{2}+2B_{3}) \nonumber\)
The valuation of equation parameters ΔH vb , q, k, and α c are done iteratively using the techniques in Part X, Convergence Strategy , based on vapor pressure data, the molecule’s structure, and the value of T br (the reduced atmospheric boiling point, T b /T c = T br ). P r must be unity at the critical point (T r = 1), which allows valuing the integration constant of Equation 4-3. The two important thermodynamic derivative functions are:
\[\frac{d(\ln P_{r})}{d(\ln T_{r})} = \alpha = A(T_{r}^{-1}-B_{1}+B_{2}T_{r}-B_{3}T_{r}^2)+c(T_{r}^{n-1}-\frac{k}{T_{r}}) \label{4-4} \]
\[\frac{-d(\ln P_{r})}{d(1/T_{r})} = \frac{\Delta H_{v}}{\Delta ZRT_{c}}= \psi = A(1-B_{1}T_{r}+B_{2}T_{r}^2-B_{3}T_{r}^3)+c(T_{r}^n-k) \label{4-5} \]
The "A" and "B" constants of Equations ( \ref{4-2} ) , ( \ref{4-3} ) , and ( \ref{4-4} ) are the same as Equation ( \ref{4-5} ) . The evaluation of the Reidel derivative function "α" in Equation ( \ref{4-4} ) is used to establish thermodynamic consistency, and allows valuing an equation parameter. Note that the Clapeyron derivative of Equation ( \ref{4-5} ) , \(\psi \nonumber\), is mathematically identical to that in Equation ( \ref{4-1} ) . \(\psi \nonumber\) will be used Part V to establish the latent heat, ΔH v , as well as the saturated vapor and liquid density of the fluids at the various conditions in the distillation column modeling.
Using the results of Part III , the Modified Thek-Stiel VP Equation constants are given below in Table 4-1, for chlorosilanes and electronic impurities found in the manufacture of high purity silicon. For brevity, the values of acentric factor, ω, and the critical Riedel derivative, \(\alpha_{c} \nonumber \), are not given in the table. They can be calculated from the other equation constants. In some instances, an electronic impurity can have either monomeric or dimeric form, but the form is noted in the table. Where a compound is not stable, the reader will note a “hole” in the table (e.g., AlH 3 and Ga 2 H 6 are not stable as liquids or vapors).
Because the solution to the Modified Thek-Stiel VP Equation is necessarily iterative, the application of this VP relationship requires the use of some modern computing tools to perform such iterative calculations (aka “nested loops”). Depending on preference, this could be done as a macro in a spreadsheet program like MS Excel, or a stand-alone program developed in BASIC (or MatLab or Fortran ). As mentioned above, some tips on convergence strategy are given in Part X .
Table 4-1 is rather expansive, to illustrate how general the new VP relationship is. Table organization is primarily done by Periodic Table Group (e.g., Group III, Group IV, and Group V) of the molecule’s core atom, and then by the homologue from hydride to chloride. Methyl chlorosilanes are included as examples of good application to fluids that form the border between classically organic and inorganic. Two of the pentanes are included, which happen to be of industrial concern as electronic impurities, to illustrate that the new VP equation is useful for organic fluids as well.
While volatile Group II, Group VI and Group VII fluids seem to follow the new VP relationship as well, there are no entries given in Table 4-1 since these fluids are generally not of concern in the production of electronic materials. The reader is encouraged to follow the evaluation techniques and stratagems given in these articles to further develop use of the new VP equations.
Table 4-1, VP solution constants, by Group and Homologue
|
Fluid |
MW |
A |
B 0 |
B 1 |
B 2 |
B 3 |
c |
n |
k |
|
Group III(A) |
|||||||||
|
B 2 H 6 |
27.670 |
8.9910 |
1.1494 |
0.37848 |
-0.11762 |
0.063573 |
2.8701 |
4.7796 |
0.1197 |
|
B 2 H 5 Cl |
62.119 ǂ |
8.1915 |
1.1494 |
0.37867 |
-0.11764 |
0.063578 |
3.4560 |
3.8431 |
0.1073 |
|
BH 2 Cl as monomer |
42.284 |
8.3872 |
1.1495 |
0.37888 |
-0.11766 |
0.063583 |
3.2767 |
4.0925 |
0.0938 |
|
BHCl 2 as monomer |
82.722 |
8.9910 |
1.1495 |
0.37937 |
-0.11772 |
0.063596 |
2.7577 |
4.9953 |
0.0636 |
|
BCl 3 |
117.169 |
9.5652 |
1.1496 |
0.37988 |
-0.11779 |
0.063609 |
2.2357 |
6.2975 |
0.0291 |
|
AlCl 3 as monomer |
133.341 |
11.9183 |
1.1512 |
0.39300 |
-0.11928 |
0.063892 |
2.2700 |
7.5185 |
0.0291 |
|
Ga 2 H 4 Cl 2 as dimer |
214.383 ǂ |
7.0714 |
1.1506 |
0.38810 |
-0.11874 |
0.063799 |
5.3346 |
2.5583 |
0.0938 |
|
Ga 2 H 2 Cl 4 as dimer |
283.273 ǂ |
10.8499 |
1.1513 |
0.39333 |
-0.11931 |
0.063898 |
3.3656 |
4.9567 |
0.0636 |
|
Ga 2 Cl 6 as dimer |
352.162 ǂ |
11.1760 |
1.1520 |
0.39974 |
0.11997 |
0.063996 |
3.8566 |
4.5874 |
0.0291 |
|
Group IV(A) |
|||||||||
|
SiH 4 |
32.117 |
7.4652 |
1.1492 |
0.37673 |
-0.11740 |
0.063525 |
3.6247 |
3.4716 |
0.0914 |
|
SiH 3 Cl |
66.562 |
8.8830 |
1.1494 |
0.37847 |
-0.11761 |
0.063572 |
2.7614 |
4.9286 |
0.0756 |
|
SiH 2 Cl 2 |
101.007 |
9.9221 |
1.1497 |
0.38068 |
-0.11788 |
0.063629 |
2.1589 |
6.6627 |
0.0598 |
|
SiHCl 3 |
135.452 |
9.9796 |
1.1501 |
0.38395 |
-0.11827 |
0.063708 |
2.5687 |
5.7950 |
0.0446 |
|
SiCl 4 |
169.896 |
10.0240 |
1.1504 |
0.38651 |
-0.11856 |
0.063765 |
2.8465 |
5.3590 |
0.0291 |
|
SiH 3 (CH 3 ) |
46.144 |
9.0535 |
1.1494 |
0.37855 |
-0.11762 |
0.063574 |
2.6380 |
5.1963 |
0.0758 |
|
SiH 2 Cl(CH 3 ) |
80.589 |
10.1402 |
1.1499 |
0.38201 |
-0.11804 |
0.063662 |
2.1796 |
6.7338 |
0.0600 |
|
SiHCl 2 (CH 3 ) |
115.034 |
9.1805 |
1.1503 |
0.38512 |
-0.11840 |
0.063735 |
3.3346 |
4.3854 |
0.0446 |
|
SiCl 3 (CH 3 ) |
149.479 |
10.1187 |
1.1506 |
0.38512 |
-0.11840 |
0.063789 |
2.9357 |
5.2665 |
0.0291 |
|
SiH 2 (CH 3 ) 2 |
60.169 |
8.2747 |
1.1497 |
0.38077 |
-0.11789 |
0.063631 |
3.4554 |
3.9214 |
0.0603 |
|
SiHCl(CH 3 ) 2 |
94.615 |
9.1914 |
1.1503 |
0.38509 |
-0.11840 |
0.063734 |
3.3232 |
4.4012 |
0.0447 |
|
SiCl 2 (CH 3 ) 2 |
129.061 |
10.1007 |
1.1507 |
0.38820 |
-0.11875 |
0.063801 |
3.0279 |
5.1285 |
0.0291 |
|
GeH 4 |
76.642 |
8.1986 |
1.1494 |
0.37858 |
-0.11763 |
0.063575 |
3.3615 |
3.9445 |
0.0914 |
|
GeH 3 Cl |
111.087 |
8.6130 |
1.1496 |
0.37993 |
-0.11779 |
0.063610 |
3.1743 |
4.3025 |
0.0756 |
|
GeH 2 Cl 2 |
145.532 |
8.8966 |
1.1498 |
0.38154 |
-0.11798 |
0.063650 |
3.1184 |
4.4929 |
0.0598 |
|
GeHCl 3 |
179.976 |
9.1230 |
1.1501 |
0.38344 |
-0.11821 |
0.063696 |
3.1490 |
4.5641 |
0.0446 |
|
GeCl 4 |
214.421 |
9.4041 |
1.1503 |
0.38554 |
-0.11845 |
0.063744 |
3.1654 |
4.6724 |
0.0291 |
|
SnH 4 |
122.742 |
9.0298 |
1.1497 |
0.38087 |
-0.11790 |
0.063634 |
3.0671 |
4.5745 |
0.0914 |
|
SnCl 4 |
260.521 |
10.1994 |
1.1506 |
0.38745 |
-0.11867 |
0.063744 |
2.8471 |
5.4354 |
0.0291 |
|
Group V(A) |
|||||||||
|
PH 3 |
33.998 |
7.8041 |
1.1485 |
0.37228 |
-0.11684 |
0.063396 |
2.8050 |
4.3644 |
0.1009 |
|
PH 2 Cl |
68.443 |
8.6412 |
1.1490 |
0.37518 |
-0.11721 |
0.063482 |
2.4920 |
5.2267 |
0.0766 |
|
PHCl 2 |
102.888 |
8.5412 |
1.1495 |
0.37919 |
-0.11770 |
0.063591 |
3.0719 |
4.3813 |
0.0527 |
|
PCl 3 |
137.333 |
7.2366 |
1.1501 |
0.38413 |
-0.11829 |
0.063712 |
4.4289 |
2.9789 |
0.0291 |
|
POCl 3 |
153.331 |
9.7746 |
1.1500 |
0.38331 |
-0.11819 |
0.063693 |
2.4586 |
5.9576 |
0 |
Table 4-2, VP solutions, by Group and Homologue, continued
|
Fluid |
MW |
A |
B 0 |
B 1 |
B 2 |
B 3 |
c |
n |
k |
|
Group V(A),cont’d |
|||||||||
|
AsH 3 |
77.945 |
7.4208 |
1.1484 |
0.37116 |
-0.11670 |
0.063362 |
2.9474 |
4.0298 |
0.1009 |
|
AsH 2 Cl |
112.390 |
8.5242 |
1.1488 |
0.37403 |
-0.11707 |
0.063448 |
2.3972 |
5.3468 |
0.0766 |
|
AsHCl 2 |
146.835 |
9.3548 |
1.1493 |
0.37782 |
-0.11754 |
0.063555 |
2.1821 |
6.2831 |
0.0527 |
|
AsCl 3 |
181.281 |
10.0281 |
1.1499 |
0.38254 |
-0.11810 |
0.063675 |
2.2519 |
6.5171 |
0.0291 |
|
SbH 3 |
124.781 |
8.5409 |
1.1484 |
0.37136 |
-0.11673 |
0.063368 |
2.0491 |
6.0821 |
0.1009 |
|
SbCl 3 |
228.115 |
7.1200 |
1.1511 |
0.39192 |
-0.11916 |
0.063873 |
5.3455 |
2.6317 |
0.0291 |
|
n-pentane |
72.149 |
10.0508 |
1.1504 |
0.38593 |
-0.11849 |
0.063753 |
2.6269 |
5.7673 |
0 |
|
iso-pentane |
72.149 |
9.9797 |
1.1502 |
0.38432 |
-0.11831 |
0.063717 |
2.4514 |
6.0715 |
0 |
Two plots are offered in Figure 4-3 to validate the principles of this Distillation Science article. These plots are for the \(\psi \)(Psi) function of Equation ( \ref{4-5} ) and the \(\alpha \)(alpha) function of Equation ( \ref{4-4} ) , for the chlorosilane homologue. The derivative function \(\phi \) goes through the required minimum, which produces the expected inflection in the vapor pressure curve, typically around T r = 0.80-0.86. Silane’s minimum \(\phi \)occurs at T r =0.67 and DCS’s minimum \(\phi \) occurs at T r = 0.89. The fact that two fluids of this homologue are outliers illustrates why chlorosilane vapor pressures are so poorly fit by VP expressions that are intended for use with hydrocarbons. However, it can be seen how the Reidel \(\alpha \) function does fit the thermodynamic consistency requirements: for all homologue fluids, the \(\alpha \) function asymptotes to a constant value (\(\alpha_{c} \)) as the critical point is approached, with the slope of the \(\alpha \) function going to zero as critical is approached.
Another validation check is shown in Figure 4-4, comparing the vapor pressures of chlorosilanes and chlorophosphines, which are distillation separation applications in all commercial chlorosilane purification.
As expected, the chlorophosphine homologue’s VP curves “nest” around those of the chlorosilanes. There are no cross-overs of the curves, and the location of chlorophosphine vapor pressure curves lie exactly as they are experienced in commercial practice of chlorosilane purification distillation columns.
Since the math in Equation ( \ref{4-3} ) is somewhat involved, an example is provided with calculations.
Example: At 3.50 atmospheres pressure what is the boiling point of trichlorosilane (TCS)?
From Part III , Table 3-3, the pertinent property values for TCS are: T c =479.15°K and P c =41.15 atm. MW, T b , V c and Z c are not needed for this example. From Part IV , the VP solution constants for TCS are: A=9.9796; B 0 =1.1501; B 1 =0.38395; B 2 =-0.11827; B 3 =0.063708; c=2.5687; n=5.7950; and k=0.0446.
As all advanced VP equations are, determining the TCS boiling point at 3.50 atmospheres is iterative; so it is nice to have a VP plot to get a good first guess. Using above Figure 4‑4, for
\ln (VP) =\ln (3.50) = 1.2528, it looks like 1/T= 2.89E-3 might be a good first guess => T=346°K; so first guess T r =346/479.15 =0.7221. Plugging that first guess value of T r into the T-S VP equation with the constants above:
\( \ln (P_{r})= A[B_{0}-T_{r}^{-1}-B_{1}\ln T_{r}+B_{2}T_{r}-(1/2)B_{3}T_{r}^2]+c[\frac{T_{r}^{n-1}-1}{n-1} +k(T_{r}^{-1}-1)] \nonumber\) yields a reduced VP value of P r = 0.08273; so VP=0.08273*41.15 = 3.404 atm, which is just a bit less than the desired 3.50 atm. So T needs to increase a little from the 346°K initial first guess.
Re-doing the TCS VP equation with T= 347°K yields a VP of 3.50 atm (note: 347.05°K is the answer, if the solution is worked out to two decimal places of temperature). So the boiling temperature at 3.50 atmospheres is 347°K or 73.9°C.


14.3: Soil Classification
Soil Classifications
OSHA 1926 Subpart “P” Appendix A
Stable Rock
Stable rock is defined as a natural mineral matter that can be excavated with vertical sides and remain intact while exposed.
Type "A"
Type A soil is cohesive soil with the following characteristics:
- An unconfined compressive strength of 1.5 tons per sq. ft. or greater.
- Soils like clay, silty clay, sandy clay, clay loam and in some cases silty clay loam and sandy clay loam are classified as Type A.
- Cemented soils such as caliche and hardpan are classified as Type A soils.
- Soils cannot be classified as Type A if any of the following conditions exist:
- The soil is fissured.
- The soil is subject to vibration from heavy traffic, pile driving or other similar effects.
- The soil has been previously disturbed.
- The soil is part of a sloped, layered system where the layers dip into the excavation on a slope of four horizontal to one vertical or greater.
- The material is subject to other factors that would require it to be classified as a less stable material.
Type "B"
Type B soil is cohesive soil with the following characteristics:
- Unconfined compressive strength greater than 0.5 tsf but less than 1.5 tsf.
- Granular cohesion less soils like angular gravel, silt, silt loam, and in some cases silty clay loam and sandy clay loam are classified as Type B soils.
- Soils would be classified as Type B if any of the following conditions exist:
- Previously disturbed soils except those which would otherwise be classified as Type C soil.
- Soil that meets the unconfined compressive strength or cementation requirements for Type A soil, but is fissured or subject to vibration.
- Dry rock that is not stable.
- The soil is part of a sloped, layered system where the layers dip into the excavation on a slope of four horizontal to one vertical, but only if the material would otherwise be classified as Type B soil.
Type "C"
Type C soil is cohesive soil with the following characteristics:
- An unconfined compressive strength of 0.5 tsf or less.
- Type C soils are granular soils including gravel, sand and loamy soil.
- Submerged soil or soil from which water is freely seeping.
- Submerged rock that is not stable.
- The soil is part of a sloped, layered system where the layers dip into the excavation on a slope of four horizontal to one vertical or steeper.
Soil Site Analysis
Each soil and rock deposit shall be analyzed and classified by a competent person as one of the four types identified above, Stable Rock, or Type A, B, or C.
The competent person shall use at least one visual test and one manual test to perform the soil deposit analysis.
Visual tests are done by observing samples of the soil that are excavated and samples taken
from the sides of the excavation. Appendix A, of Subpart P, lists the appropriate procedures for performing visual tests.
Manual tests are performed to determine the quality and type of the soil deposit. Some of the most common manual tests are: Plasticity, Dry Strength, Thumb Penetration and the Drying test. Appendix A, of Subpart P, lists the appropriate procedures for performing manual tests.


2.1: Vectors and Linear Combinations
It is a remarkable fact that algebra, which is about equations and their solutions, and geometry are intimately connected. For instance, the solution set of a linear equation in two unknowns, such as \(2x + y = 1\text{,}\) can be represented graphically as a straight line. The aim of this section is to further this connection by introducing vectors, which will help us to apply geometric intuition to our thinking about linear systems.
Vectors
A vector is most simply thought of as a matrix with a single column. For instance,
are both vectors. Since the vector \(\mathbf v\) has two entries, we say that it is a two-dimensional vector; in the same way, the vector \(\mathbf w\) is a four-dimensional vector. We denote the set of all \(m\)-dimensional vectors by \(\mathbb R^m\text{.}\) Consequently, if \(\mathbf u\) is a 3-dimensional vector, we say that \(\mathbf u\) is in \(\mathbb R^3\text{.}\)
While it can be difficult to visualize a four-dimensional vector, we can draw a simple picture describing the two-dimensional vector \(\mathbf v\text{.}\)
In this chapter, we will uncover geometric information in a matrix like this, which will lead to an intuitive understanding of the insights we previously gained into the solutions of linear systems.
We think of \(\mathbf v\) as describing a walk we take in the plane where we move two units horizontally and one unit vertically. Though we allow ourselves to begin walking from any point in the plane, we will most frequently begin at the origin, in which case we arrive at the the point \((2,1)\text{,}\) as shown in the figure.
There are two simple algebraic operations we can perform on vectors.
- Scalar Multiplication
-
We multiply a vector \(\mathbf v\) by a real number \(a\) by multiplying each of the components of \(\mathbf v\) by \(a\text{.}\) For instance,
\begin{equation*} -3\left[\begin{array}{r} 2 \\ -4 \\ 1 \\ \end{array}\right] = \left[\begin{array}{r} -6 \\ 12 \\ -3 \\ \end{array}\right]. \end{equation*}The real number \(a\) is called a scalar .
- Vector Addition
-
We add two vectors of the same dimension by adding their components. For instance,
\begin{equation*} \left[\begin{array}{r} 2 \\ -4 \\ 3 \\ \end{array}\right] + \left[\begin{array}{r} -5 \\ 6 \\ -3 \\ \end{array}\right] = \left[\begin{array}{r} -3 \\ 2 \\ 0 \\ \end{array}\right]. \end{equation*}
Preview Activity 2.1.1. Scalar Multiplication and Vector Addition.
Suppose that
-
Find expressions for the vectors
\begin{equation*} \begin{array}{cccc} \mathbf v, & 2\mathbf v, & -\mathbf v, & -2\mathbf v, \\ \mathbf w, & 2\mathbf w, & -\mathbf w, & -2\mathbf w\text{.} \\ \end{array} \end{equation*}
and sketch them below.
- What geometric effect does scalar multiplication have on a vector? Also, describe the effect multiplying by a negative scalar has.
- Sketch the vectors \(\mathbf v, \mathbf w, \mathbf v + \mathbf w\) below.
- Consider vectors that have the form \(\mathbf v + a\mathbf w\) where \(a\) is any scalar. Sketch a few of these vectors when, say, \(a = -2, -1, 0, 1, \) and \(2\text{.}\) Give a geometric description of this set of vectors.
-
If \(a\) and \(b\) are two scalars, then the vector
\begin{equation*} a \mathbf v + b \mathbf w \end{equation*}
is called a linear combination of the vectors \(\mathbf v\) and \(\mathbf w\text{.}\) Find the vector that is the linear combination when \(a = -2\) and \(b = 1\text{.}\)
- Can the vector \(\left[\begin{array}{r} -31 \\ 37 \end{array}\right]\) be represented as a linear combination of \(\mathbf v\) and \(\mathbf w\text{?}\)
The preview activity demonstrates how we may interpret scalar multiplication and vector addition geometrically.
First, we see that scalar multiplication has the effect of stretching or compressing a vector. Multiplying by a negative scalar changes the direction of the vector. In either case, we see that scalar multiplying the vector \(\mathbf v\) produces a new vector on the line defined by \(\mathbf v\text{,}\) as shown in Figure \(\PageIndex{5}\).
To understand the sum \(\mathbf v + \mathbf w\text{,}\) we imagine walking from the origin with the appropriate horizontal and vertical changes given by \(\mathbf v\text{.}\) From there, we continue our walk using the horizontal and vertical changes prescribed by \(\mathbf w\text{,}\) after which we arrive at the sum \(\mathbf v + \mathbf w\text{.}\) This is illustrated on the left of Figure \(\PageIndex{6}\) where the tail of \(\mathbf w\) is placed on the tip of \(\mathbf v\text{.}\)
Alternatively, we may construct the parallelogram with \(\mathbf v\) and \(\mathbf w\) as two sides. The sum is then the diagonal of the parallelogram, as illustrated on the right of Figure \(\PageIndex{6}\).
We have now seen that the set of vectors having the form \(a\mathbf v\) is a line. To form the set of vectors \(a\mathbf v+\mathbf w\text{,}\) we can begin with the vector \(\mathbf w\) and add multiples of \(\mathbf v\text{.}\) Geometrically, this means that we begin from the tip of \(\mathbf w\) and move in a direction parallel to \(\mathbf v\text{.}\) The effect is to translate the line \(a\mathbf v\) by the vector \(\mathbf w\text{,}\) as shown in Figure \(\PageIndex{7}\).
At times, it will be useful for us to think of vectors and points interchangeably. That is, we may wish to think of the vector \(\left[\begin{array}{r} 2 \\ 1 \end{array}\right]\) as describing the point \((2,1)\) and vice-versa. When we say that the vectors having the form \(a\mathbf v + \mathbf w\) form a line, we really mean that the tips of the vectors all lie on the line passing through \(\mathbf w\) and parallel to \(\mathbf v\text{.}\)
Observation 2.1.4.
Even though these vector operations are new, it is straightforward to check that some familiar properties hold.
- Commutativity
-
\(\mathbf v + \mathbf w = \mathbf w + \mathbf v\text{.}\)
- Distributivity
-
\(a(\mathbf v + \mathbf w) = a\mathbf v + a\mathbf w\text{.}\)
Sage can perform scalar multiplication and vector addition. We define a vector using the
vector
command; then
*
and
+
denote scalar multiplication and vector addition.
v = vector([3,1]) w = vector([-1,2]) print (2*v) print (v + w)
2.1.2 Linear combinations
Linear combinations, which we encountered in the preview activity, provide the link between vectors and linear systems. In particular, they will help us apply geometric intuition to problems involving linear systems.
The linear combination of the vectors \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) with scalars \(c_1,c_2,\ldots,c_n\) is the vector
The scalars \(c_1,c_2,\ldots,c_n\) are called the weights of the linear combination.
Activity 2.1.2.
In this activity, we will look at linear combinations of a pair of vectors,
with weights \(a\) and \(b\text{.}\)
The diagram below can be used to construct linear combinations whose weights \(a\) and \(b\) may be varied using the sliders at the top. The vectors \({\mathbf v}\) and \({\mathbf w}\) are drawn in gray while the linear combination $$ a{\mathbf v} + b{\mathbf w} $$ is in red.
- The weight \(b\) is initially set to 0. Explain what happens as you vary \(a\) with \(b=0\text{?}\) How is t his related to scalar multiplication?
- What is the linear combination of \(\mathbf v\) and \(\mathbf w\) when \(a = 1\) and \(b=-2\text{?}\) You may find this result using the diagram, but you should also verify it by computing the linear combination.
- Describe the vectors that arise when the weight \(b\) is set to 1 and \(a\) is varied. How is this related to our investigations in the preview activity?
- Can the vector \(\left[\begin{array}{r} 0 \\ 0 \end{array} \right]\) be expressed as a linear combination of \(\mathbf v\) and \(\mathbf w\text{?}\) If so, what are weights \(a\) and \(b\text{?}\)
- Can the vector \(\left[\begin{array}{r} 3 \\ 0 \end{array} \right]\) be expressed as a linear combination of \(\mathbf v\) and \(\mathbf w\text{?}\) If so, what are weights \(a\) and \(b\text{?}\)
- Verify the result from the previous part by algebraically finding the weights \(a\) and \(b\) that form the linear combination \(\left[\begin{array}{r} 3 \\ 0 \end{array} \right]\text{.}\)
- Can the vector \(\left[\begin{array}{r} 1.3 \\ -1.7 \end{array} \right]\) be expressed as a linear combination of \(\mathbf v\) and \(\mathbf w\text{?}\) What about the vector \(\left[\begin{array}{r} 15.2 \\ 7.1 \end{array} \right]\text{?}\)
- Are there any two-dimensional vectors that cannot be expressed as linear combinations of \(\mathbf v\) and \(\mathbf w\text{?}\)
This activity illustrates how linear combinations are constructed geometrically: the linear combination \(a\mathbf v + b\mathbf w\) is found by walking along \(\mathbf v\) a total of \(a\) times followed by walking along \(\mathbf w\) a total of \(b\) times. When one of the weights is held constant while the other varies, the vector moves along a line.
The previous activity also shows that questions about linear combinations lead naturally to linear systems. Let's ask how we can describe the vector \(\mathbf b=\left[\begin{array}{r} -1 \\ 4 \end{array} \right]\) as a linear combination of \(\mathbf v\) and \(\mathbf w\text{.}\) We need to find weights \(a\) and \(b\) such that
Equating the components of the vectors on each side of the equation, we arrive at the linear system
This means that \(\mathbf b\) is a linear combination of \(\mathbf v\) and \(\mathbf w\) if this linear system is consistent.
To solve this linear system, we construct its corresponding augmented matrix and find its reduced row echelon form.
which tells us the weights \(a=-2\) and \(b=3\text{;}\) that is,
In fact, we know even more because the reduced row echelon matrix tells us that these are the only possible weights. Therefore, \(\mathbf b\) may be expressed as a linear combination of \(\mathbf v\) and \(\mathbf w\) in exactly one way.
This example demonstrates the connection between linear combinations and linear systems. Asking if a vector \(\mathbf b\) is a linear combination of vectors \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) is the same as asking whether an associated linear system is consistent.
In fact, we may easily describe the linear system we obtain in terms of the vectors \(\mathbf v\text{,}\) \(\mathbf w\text{,}\) and \(\mathbf b\text{.}\) Notice that the augmented matrix we found was \(\left[ \begin{array}{rr|r} 2 & 1 & -1 \\ 1 & 2 & 4 \end{array} \right].\) The first two columns of this matrix are \(\mathbf v\) and \(\mathbf w\) and the rightmost column is \(\mathbf b\text{.}\) As shorthand, we will write this augmented matrix replacing the columns with their vector representation:
This fact is generally true so we record it in the following proposition.
The vector \(\mathbf b\) is a linear combination of the vectors \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) if and only if the linear system corresponding to the augmented matrix
is consistent. A solution to this linear system gives weights \(c_1,c_2,\ldots,c_n\) such that
The next activity puts this proposition to use.
Activity 2.1.3. Linear combinations and linear systems.
-
Given the vectors
\begin{equation*} \mathbf v_1 = \left[\begin{array}{r} 4 \\ 0 \\ 2 \\ 1 \end{array} \right], \mathbf v_2 = \left[\begin{array}{r} 1 \\ -3 \\ 3 \\ 1 \end{array} \right], \mathbf v_3 = \left[\begin{array}{r} -2 \\ 1 \\ 1 \\ 0 \end{array} \right], \mathbf b = \left[\begin{array}{r} 0 \\ 1 \\ 2 \\ -2 \end{array} \right]\text{,} \end{equation*}
we ask if \(\mathbf b\) can be expressed as a linear combination of \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{.}\) Rephrase this question by writing a linear system for the weights \(c_1\text{,}\) \(c_2\text{,}\) and \(c_3\) and use the Sage cell below to answer this question.
-
Consider the following linear system.
\begin{equation*} \begin{alignedat}{4} 3x_1 & {}+{} & 2x_2 & {}-{} x_3 & {}={} & 4 \\ x_1 & & & {}+{} 2x_3 & {}={} & 0 \\ -x_1 & {}-{} & x_2 & {}+{} 3x_3 & {}={} & 1 \\ \end{alignedat} \end{equation*}
Identify vectors \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) \(\mathbf v_3\text{,}\) and \(\mathbf b\) and rephrase the question "Is this linear system consistent?" by asking "Can \(\mathbf b\) be expressed as a linear combination of \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{?}\)"
-
Consider the vectors
\begin{equation*} \mathbf v_1 = \left[\begin{array}{r} 0 \\ -2 \\ 1 \\ \end{array} \right], \mathbf v_2 = \left[\begin{array}{r} 1 \\ 1 \\ -1 \\ \end{array} \right], \mathbf v_3 = \left[\begin{array}{r} 2 \\ 0 \\ -1 \\ \end{array} \right], \mathbf b = \left[\begin{array}{r} -1 \\ 3 \\ -1 \\ \end{array} \right]\text{.} \end{equation*}
Can \(\mathbf b\) be expressed as a linear combination of \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{?}\) If so, can \(\mathbf b\) be written as a linear combination of these vectors in more than one way?
- Considering the vectors \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\) from the previous part, can we write every three-dimensional vector \(\mathbf b\) as a linear combination of these vectors? Explain how the pivot positions of the matrix \(\left[\begin{array}{rrr} \mathbf v_1 & \mathbf v_2 & \mathbf v_3 \end{array} \right]\) help answer this question.
-
Now consider the vectors
\begin{equation*} \mathbf v_1 = \left[\begin{array}{r} 0 \\ -2 \\ 1 \\ \end{array} \right], \mathbf v_2 = \left[\begin{array}{r} 1 \\ 1 \\ -1 \\ \end{array} \right], \mathbf v_3 = \left[\begin{array}{r} 1 \\ -1 \\ -2 \\ \end{array} \right], \mathbf b = \left[\begin{array}{r} 0 \\ 8 \\ -4 \\ \end{array} \right]\text{.} \end{equation*}
Can \(\mathbf b\) be expressed as a linear combination of \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{?}\) If so, can \(\mathbf b\) be written as a linear combination of these vectors in more than one way?
- Considering the vectors \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\) from the previous part, can we write every three-dimensional vector \(\mathbf b\) as a linear combination of these vectors? Explain how the pivot positions of the matrix \(\left[\begin{array}{rrr} \mathbf v_1 & \mathbf v_2 & \mathbf v_3 \end{array} \right]\) help answer this question.
Example 2.1.13.
Consider the vectors \(\mathbf{v}=\left[\begin{array}{r}-1 \\ 1\end{array}\right]\) and \(\mathbf{w}=\left[\begin{array}{r}2 \\ -2\end{array}\right]\), as shown in Figure \(\PageIndex{9}\).
These vectors appear to lie on the same line, a fact that becomes apparent once we notice that \(\mathbf{w}=-2 \mathbf{v}\). Intuitively, we think of the linear combination
\[
c \mathbf{v}+d \mathbf{w}
\]
as the result of walking \(c\) times in the \(\mathbf{v}\) direction and \(d\) times in the \(\mathbf{w}\) direction. With these vectors, we are always walking along the same line so it would seem that any linear combination of these vectors should lie on the same line. In addition, a vector that is not on the line, say \(\mathbf{b}=\left[\begin{array}{l}3 \\ 0\end{array}\right]\), should be not be expressible as a linear combination of \(\mathbf{v}\) and \(\mathbf{w}\).
We can verify this by checking
\[
\left[\begin{array}{rr|r}
-1 & 2 & 3 \\
1 & -2 & 0
\end{array}\right] \sim\left[\begin{array}{rr|r}
1 & -2 & 0 \\
0 & 0 & 1
\end{array}\right]
\]
This shows that the associated linear system is inconsistent, which means that the vector \(\mathbf{b}=\left[\begin{array}{l}3 \\ 0\end{array}\right]\) cannot be written as a linear combination of \(\mathbf{v}\) and \(\mathbf{w}\).
Notice that the reduced row echelon form of the coefficient matrix
\[
\left[\begin{array}{ll}
\mathbf{v} & \mathbf{w}
\end{array}\right]=\left[\begin{array}{cc}
-1 & 2 \\
1 & -2
\end{array}\right] \sim\left[\begin{array}{cc}
1 & -2 \\
0 & 0
\end{array}\right]
\]
tells us to expect this. Since there is not a pivot position in the second row of the coefficient matrix \(\left[\begin{array}{ll}\mathbf{v} & \mathbf{w}\end{array}\right]\), it is possible for a pivot position to appear in the rightmost column of the augmented matrix
\[
\left[\begin{array}{ll|l}
\mathbf{v} & \mathbf{w} & \mathbf{b}
\end{array}\right]
\]
for some choice of \(\mathbf{b}\).
Summary
This section has introduced vectors, linear combinations, and their connection to linear systems.
- There are two operations we can perform with vectors: scalar multiplication and vector addition. Both of these operations have geometric meaning.
- Given a set of vectors and a set of scalars we call weights, we can create a linear combination using scalar multiplication and vector addition.
-
A solution to the linear system whose augmented matrix is
\begin{equation*} \left[\begin{array}{rrrr|r} \mathbf v_1 & \mathbf v_2 & \ldots & \mathbf v_n & \mathbf b \end{array}\right] \end{equation*}
is a set of weights that expressex \(\mathbf b\) as a linear combination of \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\text{.}\)
Exercises 2.1.4Exercises
Consider the vectors
-
Sketch these vectors below.
- Compute the vectors \(-3\mathbf v\text{,}\) \(2\mathbf w\text{,}\) \(\mathbf v + \mathbf w\text{,}\) and \(\mathbf v - \mathbf w\) and add them into the sketch above.
-
Sketch below the set of vectors having the form \(2\mathbf v + t\mathbf w\) where \(t\) is any scalar.
-
Sketch below the line \(y=3x - 2\text{.}\) Then identify two vectors \(\mathbf v\) and \(\mathbf w\) so that this line is described by \(\mathbf v + t\mathbf w\text{.}\) Are there other choices for the vectors \(\mathbf v\) and \(\mathbf w\text{?}\)
Shown below are two vectors \(\mathbf v\) and \(\mathbf w\)
- Express the labeled points as linear combinations of \(\mathbf v\) and \(\mathbf w\text{.}\)
- Sketch the line described parametrically as \(-2\mathbf v + t\mathbf w\text{.}\)
Consider the vectors
- Find the linear combination with weights \(c_1 = 2\text{,}\) \(c_2=-3\text{,}\) and \(c_3=1\text{.}\)
-
Can you write the vector \({\mathbf 0} = \left[\begin{array}{r} 0 \\ 0 \end{array}\right]\) as a linear combination of \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{?}\) If so, describe all the ways in which you can do so.
-
Can you write the vector \({\mathbf 0} = \left[\begin{array}{r} 0 \\ 0 \end{array}\right]\) as a linear combination using just the first two vectors \(\mathbf v_1\) \(\mathbf v_2\text{?}\) If so, describe all the ways in which you can do so.
-
Can you write \(\mathbf v_3\) as a linear combination of \(\mathbf v_1\) and \(\mathbf v_2\text{?}\) If so, in how many ways?
Nutritional information about a breakfast cereal is printed on the box. For instance, one serving of Frosted Flakes has 111 calories, 140 milligrams of sodium, and 1.2 grams of protein. We may represent this as a vector
One serving of Cocoa Puffs has 120 calories, 105 milligrams of sodium, and 1.0 grams of protein.
- Write the vector describing the nutritional content of Cocoa Puffs.
- Suppose you eat \(a\) servings of Frosted Flakes and \(b\) servings of Cocoa Puffs. Use the language of vectors and linear combinations to express the total amount of calories, sodium, and protein you have consumed.
- How many servings of each cereal have you eaten if you have consumed 342 calories, 385 milligrams of sodium, and 3.4 grams of protein.
- Suppose your sister consumed 250 calories, 200 milligrams of sodium, and 4 grams of protein. What can you conclude about her breakfast?
Consider the vectors
- Can you express the vector \(\mathbf b=\left[\begin{array}{r} 10 \\ 1 \\ -8 \end{array}\right]\) as a linear combination of \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{?}\) If so, describe all the ways in which you can do so.
- Can you express the vector \(\mathbf b=\left[\begin{array}{r} 3 \\ 7 \\ 1 \end{array}\right]\) as a linear combination of \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{?}\) If so, describe all the ways in which you can do so.
- Show that \(\mathbf v_3\) can be written as a linear combination of \(\mathbf v_1\) and \(\mathbf v_2\text{.}\)
-
Explain why any linear combination of \(\mathbf v_1\text{,}\) \(\mathbf v_2\text{,}\) and \(\mathbf v_3\text{,}\)
\begin{equation*} a\mathbf v_1 + b\mathbf v_2 + c\mathbf v_3, \end{equation*}
can be rewritten as a linear combination of just \(\mathbf v_1\) and \(\mathbf v_2\text{.}\)
Consider the vectors
For what value(s) of \(k\text{,}\) if any, can the vector \(\left[\begin{array}{r} k \\ -2 \\ 5 \end{array}\right]\) be written as a linear combination of \(\mathbf v_1\) and \(\mathbf v_2\text{?}\)
Provide a justification for your response to the following statements or questions.
- True of false: Given two vectors \(\mathbf v\) and \(\mathbf w\text{,}\) the vector \(2\mathbf v\) is a linear combination of \(\mathbf v\) and \(\mathbf w\text{.}\)
- True or false: Suppose \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) is a collection of \(m\)-dimensional vectors and that the matrix \(\left[\begin{array}{rrrr} \mathbf v_1 & \mathbf v_2 & \ldots & \mathbf v_n \end{array}\right]\) has a pivot position in every row. If \(\mathbf b\) is any \(m\)-dimensional vector, then \(\mathbf b\) can be written as a linear combination of \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\text{.}\)
- True or false: Suppose \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) is a collection of \(m\)-dimensional vectors and that the matrix \(\left[\begin{array}{rrrr} \mathbf v_1 & \mathbf v_2 & \ldots & \mathbf v_n \end{array}\right]\) has a pivot position in every row and every column. If \(\mathbf b\) is any \(m\)-dimensional vector, then \(\mathbf b\) can be written as a linear combination of \(\mathbf v_1,\mathbf v_2,\ldots,\mathbf v_n\) in exactly one way.
- True or false: It is possible to find two 3-dimensional vectors \(\mathbf v_1\) and \(\mathbf v_2\) such that every 3-dimensional vector can be written as a linear combination of \(\mathbf v_1\) and \(\mathbf v_2\text{.}\)
A theme that will later unfold concerns the use of coordinate systems. We can identify the point \((x,y)\) with the tip of the vector \(\left[\begin{array}{r}x\\y\end{array}\right]\text{,}\) drawn emanating from the origin. We can then think of the usual Cartesian coordinate system in terms of linear combinations of the vectors
The point \((2,-3)\) is identified with the vector
If we have vectors
we may define a new coordinate system, such that a point \(\{x,y\}\) will correspond to the vector
For instance, the point \(\{2,-3\}\) is shown on the right side of Figure 2.1.8
-
Write the point \(\{2,-3\}\) in standard coordinates; that is, find \(x\) and \(y\) such that
\begin{equation*} (x,y) = \{2,-3\}\text{.} \end{equation*}
-
Write the point \((2,-3)\) in the new coordinate system; that is, find \(a\) and \(b\) such that
\begin{equation*} \{a,b\} = (2,-3)\text{.} \end{equation*}
- Convert a general point \(\{a,b\}\text{,}\) expressed in the new coordinate system, into standard Cartesian coordinates \((x,y)\text{.}\)
- What is the general strategy for converting a point from standard Cartesian coordinates \((x,y)\) to the new coordinates \(\{a,b\}\text{?}\) Actually implementing this strategy in general may take a bit of work so just describe the strategy. We will study this in more detail later.


8.3: BEFORE YOU MOVE ON
Key Concepts
National and personal identities do not magically happen; they are built on and influenced by immediate and past events, environments, traditions, and cultural legacies. Artists capture and document not only the physical conditions of a society but also the emotional and mental conditions. They construct a sense of who we were and are as a person and as a nation. Society’s identity is always fluid. When we see identity as static, we record people with stereotypes and do not see them for who they are. Art is one way to challenge static notions of identity by engaging the viewer in visual narratives that are unfamiliar to them, and that educate and challenge their previously held notions.
Since the 1970s, postmodern theories have challenged historical and traditional notions of ethnic and cultural identity by developing a model that views identity as being multifaceted, fluid, and socially constructed. Some scholars contend that we are in a period of post identity and post-ethnicity, repudiating the old essentialist view of identity. Globalization of people, the Internet, and travel have all brought about fluid cultures which may have contributed to people’s more fluid sense of identity, and also to their interest in researching their heritage, culture, and ethnic identity. Heritage is the treasure and symbols of pride for an individual, country, and nation. Many works of art are seen as part of national heritage because they help citizens appreciate their past. Art provides life to the past, something that can be visualized, touched, walk through, and identified as being part of a legacy and culture.
Test Yourself
- On the surface Kim Sooja’s art seems simple, but underneath it is an enigma of traditions that make a metaphoric identity statement; for example, her use of fabric as an art form evokes intimacy and honor of her culture and history. Discuss and identify at-least two artists whose work makes a personal and historical statement. Be specific as you reference each image associated with your essay. (minimum of 500 words).
- A number of circumstances throughout history have compelled artists to confront the context of social issues, select at-least two works of art that best describe an event or issue. Discuss the problems associated with the issue, and how the event and art shaped the legacy or identity of the country or nation. Describe the power the work communicates, discuss the significance of the work and how it convey a message, and identity of the people in that period of time. At the end of your essay make commentary on why you selected the art works what you think about the art. (Attach selected work with captions.) Answer to the question is located throughout the chapter).
- Throughout history building were constructed in a manner to symbolize power; spirituality; and godlessness. Structures house institutions that guide, influence and shape a society’s morals, values, politics, religious and social conditioning. Select 4 structures that best symbolize the identity or culture of a society. Describe its impact on influencing a nation, significance to the nation and how the structure contributes to national or individual identity. At the end of your essay discuss why you selected the structures and the aesthetics of the building. (Attach selected structures with captions.)
- Compare and contrast four works of art that best describe a personal or national identity. Discuss with specifics how the artist is able to capture the character of the person or nation. At the end of your essay add a commentary why you selected the works and their significance. (Attach selected works with captions.)
KEY TERMS
Baroque : a style of architecture and art that originating in Italy in the early seventeenth century
Bottari : Cloth wrapped and tied around clothes , fabric, or/and items into a bundle for carry
Grave stele : is a stone or wooden slab, generally taller than it is wide, erected usually in Greek cemeteries as a monument, for funerary or commemorative purposes.
Hypogeum : an underground prehistoric burial site
Impressionism : is a nineteenth-century art movement that developed in France during the late nineteenth century by a group of artists called the Anonymous Society of Painters, Sculptors
Impressionist : A painter whose painting have characteristics of the impressionism movement, emphasizing accurate depiction of light in its changing qualities, uses small, thin, yet visible brush strokes, open composition,
Individualism : emphasizes potential of man and self development own beliefs. The Individualism during the Renaissance period became a prominent theme in Italy
Industrial Revolution : period during the late eighteenth and nineteenth centuries in western Europe and the United States when industry quickly developed due to the invention of steam powered engines and the growth of factories. Fundamental changes occurred in agriculture, textile and metal manufacture, transportation, economic and policies, and had a major impact on how people lived
Obas : The title of “oba,” or king, is passed on to the firstborn son of each successive king of Benin, Africa at the time of his death
Renaissance Period : a period of time from the fourteenth to the seventeenth century in Europe. The era bridged the time between the Middle Ages and modern
Tableau : is an incidental scene, as of a group of people
Tower tomb : are mausoleums, built in 1067 and 1093


17.2: Policies for Reducing Poverty
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Social Insurance
Social insurance are government-sponsored programs, such as Medicare, that provide benefits to people based on individual contributions to that program.
learning objectives
- Describe the characteristics of social insurance programs
Social insurance has been defined as a program where risks are transferred to and pooled by an organization (often governmental) that is legally required to provide certain benefits. It is any government-sponsored program with the following four characteristics:
- The benefits, eligibility requirements, and other aspects of the program are defined by statute;
- Explicit provision is made to account for income and expenses (often through a trust fund);
- It is funded by taxes or premiums paid by (or on behalf of) participants (although additional sources of funding may be provided as well); and
- The program serves a defined population, and participation is either compulsory, or the program is subsidized heavily enough that most eligible individuals choose to participate.
Social insurance differs from welfare in that the beneficiary’s contributions to the program are taken into account. A welfare program pays recipients based on need, not contributions. Medicare is an example of a social insurance program, while Medicaid is an example of a welfare one.
In the United States, Social Security, Medicare, and unemployment insurance are among the most well-known forms of social insurance.
Social Security
Social Security in the U.S. is primarily the Old-Age, Survivors, and Disability Insurance (OASDI) federal insurance program. Social Security is funded through payroll taxes called Federal Insurance Contributions Act tax (FICA) and/or Self Employed Contributions Act Tax (SECA). Tax deposits are collected by the Internal Revenue Service (IRS) and are formally entrusted to the Social Security Trust Funds. Social Security provides monetary benefits to retirees, their spouses and surviving dependent children, and disabled workers.
Social Security Poster : Social Security is one of the best-known social insurance programs in the United States. It provides benefits to retirees, surviving family members, and disabled workers who have contributed to the Social Security Trust Fund through payroll taxes.
Medicare
Medicare is a national program that guarantees access to health insurance for Americans aged 65 and older, younger people with disabilities, and people with certain chronic diseases. Medicare is funded through revenue from FICA and SECA payroll taxes, as well as through premiums paid by Medicare enrollees and general fund revenue from the federal government.
Unemployment Insurance
Unemployment insurance provides a monetary benefit to workers who have become unemployed through no fault of their own. Benefits are generally paid by state governments, and are funded in large part by state and federal payroll taxes levied against employers. These payroll taxes were established by the Federal Unemployment Tax Act (FUTA), and allow the IRS to collect federal employer taxes used to fund state workforce agencies. FUTA covers the costs of administering the Unemployment Insurance and Job Service programs in all states. In addition, FUTA pays one-half of the cost of extended unemployment benefits (during periods of high unemployment) and provides for a fund from which states may borrow, if necessary, to pay benefits.
Public Assistance
Public assistance is the provision of a minimal level of social support for all citizens.
learning objectives
- Define and describe different types of public assistance
Public Assistance
Public assistance, also referred to colloquially as welfare, is the provision of a minimal level of social support for all citizens. In most developed countries, public assistance is provided by the government, charities, social groups, and religious groups. It is funded by government agencies and private organizations.
Public assistance systems vary by country, but welfare is usually provided to individuals who are unemployed, those with an illness or disability, the elderly, those with dependent children, and veterans. Individuals must meet specific criteria to be eligible to receive public assistance.
In the United States, the funds for public assistance are given at a flat rate to each state based on population. Each state has to meet certain criteria to ensure that individuals receiving public assistance are being encouraged to work themselves out of welfare. The goal of public assistance is to support individuals who are in need of help while encouraging them to seek employment and better their lives.
Forms of Public Assistance
Public assistance is offered in a variety of forms including:
- Monetary payments: individuals are paid bi-weekly or monthly based on their income level. Individuals must apply for monetary public assistance and meet specific criteria. Monetary payments will be lessened or stopped once the individual’s income reaches a certain level. An example of monetary payments is Temporary Assistance for Needy Families (TANF), which provides a cash benefit to families in need.
- Subsidy: government funded programs that provide assistance to citizens on federal, state, local, and private levels. Subsidies help to provide food, housing, education, healthcare, and financial support to individuals in need. Examples include Medicaid.
- Vouchers: are bonds given out by the government or other welfare organizations. A voucher is worth a certain monetary value and can only be spent on specific goods.
- Housing assistance: provided by the government to ensure that individuals have shelter. In some cases individuals will receive free housing while other will receive housing at a discounted rate. Housing assistance is based on an individual’s level of income.
- Universal healthcare: health care coverage that provides health care and financial protection to all citizens. It provides a specific package of benefits to all members of society with the goal of providing financial risk protection, improved access to health services, and improved health outcomes.
- Public assistance is provided by the government, charities, social groups, and religious groups. It is funded by government agencies and private organizations.
- Public assistance systems vary by country, but welfare is usually provided to individuals who are unemployed, those with an illness or disability, the elderly, those with dependent children, and veterans.
- Forms of public assistance include monetary payments, subsidies, vouchers, housing assistance, and universal healthcare.
Key Points
- Social insurance programs share four characteristics: they have well-defined eligibility requirements and benefits, have provisions for program income and expenses, are funded by taxes or premiums paid by participants, and have mandatory or heavily subsidized participation.
- Social insurance programs differs from welfare programs in that they take participant contributions into account. Welfare benefits are based on need, not contributions.
- Social Security, Medicare, and unemployment insurance are three well-known social insurance programs in the United States.
Key Terms
- social insurance : Any government-sponsored program where risks are transferred to and pooled by an organization that is legally required to provide certain benefits.
- public assistance : Payment made to disadvantaged persons by government in order to alleviate the burdens of poverty, unemployment, disability, old age, etc.
- subsidy : Financial support or assistance, such as a grant.
- voucher : A piece of paper that entitles the holder to a discount, or that can be exchanged for goods and services.
LICENSES AND ATTRIBUTIONS
CC LICENSED CONTENT, SPECIFIC ATTRIBUTION
- Medicare (United States). Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Medicare_(United_States) . License : CC BY-SA: Attribution-ShareAlike
- Unemployment benefit. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Unemployment_benefit%23United_States . License : CC BY-SA: Attribution-ShareAlike
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- public assistance. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/public_assistance . License : CC BY-SA: Attribution-ShareAlike
- Universal healthcare. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Universal_healthcare . License : CC BY-SA: Attribution-ShareAlike
- Unemployment insurance. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Unemployment_insurance . License : CC BY-SA: Attribution-ShareAlike
- Welfare. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Welfare . License : CC BY-SA: Attribution-ShareAlike
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7.6: Anabolism of Glucose- Photosynthesis
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Learning Objectives
- Describe the process of photosynthesis
The Importance of Photosynthesis
The processes of all organisms—from bacteria to humans—require energy. To get this energy, many organisms access stored energy by eating food. Carnivores eat other animals and herbivores eat plants. But where does the stored energy in food originate? All of this energy can be traced back to the process of photosynthesis and light energy from the sun.
Photosynthesis is essential to all life on earth. It is the only biological process that captures energy from outer space (sunlight) and converts it into chemical energy in the form of G3P (
Glyceraldehyde 3-phosphate) which in turn can be made into sugars and other molecular compounds. Plants use these compounds in all of their metabolic processes; plants do not need to consume other organisms for food because they build all the molecules they need. Unlike plants, animals need to consume other organisms to consume the molecules they need for their metabolic processes.
The Process of Photosynthesis
During photosynthesis, molecules in leaves capture sunlight and energize electrons, which are then stored in the covalent bonds of carbohydrate molecules. That energy within those covalent bonds will be released when they are broken during cell respiration. How long lasting and stable are those covalent bonds? The energy extracted today by the burning of coal and petroleum products represents sunlight energy captured and stored by photosynthesis almost 200 million years ago.
Plants, algae, and a group of bacteria called cyanobacteria are the only organisms capable of performing photosynthesis. Because they use light to manufacture their own food, they are called photoautotrophs (“self-feeders using light”). Other organisms, such as animals, fungi, and most other bacteria, are termed heterotrophs (“other feeders”) because they must rely on the sugars produced by photosynthetic organisms for their energy needs. A third very interesting group of bacteria synthesize sugars, not by using sunlight’s energy, but by extracting energy from inorganic chemical compounds; hence, they are referred to as chemoautotrophs.
The importance of photosynthesis is not just that it can capture sunlight’s energy. A lizard sunning itself on a cold day can use the sun’s energy to warm up. Photosynthesis is vital because it evolved as a way to store the energy in solar radiation (the “photo-” part) as high-energy electrons in the carbon-carbon bonds of carbohydrate molecules (the “-synthesis” part). Those carbohydrates are the energy source that heterotrophs use to power the synthesis of ATP via respiration. Therefore, photosynthesis powers 99 percent of Earth’s ecosystems. When a top predator, such as a wolf, preys on a deer, the wolf is at the end of an energy path that went from nuclear reactions on the surface of the sun, to light, to photosynthesis, to vegetation, to deer, and finally to wolf.
Key Points
- Photosynthesis evolved as a way to store the energy in solar radiation as high-energy electrons in carbohydrate molecules.
- Plants, algae, and cyanobacteria, known as photoautotrophs, are the only organisms capable of performing photosynthesis.
- Heterotrophs, unable to produce their own food, rely on the carbohydrates produced by photosynthetic organisms for their energy needs.
Key Terms
- photosynthesis : the process by which plants and other photoautotrophs generate carbohydrates and oxygen from carbon dioxide, water, and light energy in chloroplasts
- photoautotroph : an organism that can synthesize its own food by using light as a source of energy
- chemoautotroph : a simple organism, such as a protozoan, that derives its energy from chemical processes rather than photosynthesis


3.3: What are the educational milestones of the 20th century?
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Veronica Montalvo
INTRODUCTION AND THE GROWTH OF EDUCATION
In the United States, the start of the twentieth century marked a movement towards the inclusion of more people than ever into the educational system. In 1900, close to six percent of teenagers graduated from high school (Thattai, 2001). States attempted to increase that number by making that goal more accessible through the construction of more high schools in both urban and rural localities (Wolfe, 2001). Laws were passed that made school mandatory for children until elementary school; later it became obligatory until the child became sixteen years old (Thattai, 2001). However, a good and equal education was not yet widely available to all Americans. Marginalized groups hovered at the fringes of the educational system. African-Americans received unequal and inferior educations as compared to that of whites, as did other minorities and students with limited-English-proficiency (LEP). Women were discriminated against, as well as the handicapped, in being fully included in the educational system. Major developments in the twentieth-century education system include various rulings and acts that promoted a fair and equal education for Americans that had been neglected and marginalized.
G.I. BILL of 1944
Established in 1944, the G.I. Bill was designed to provide assistance to veterans returning from World War II. Although the educational provisions within the Bill were originally intended as “another form of unemployment relief,” the educational system experienced a drastic change in its perception by Americans (Clark, 1998, p. 173). Prior to the influx of veterans in colleges and universities, a college education was perceived as being a privilege of the wealthy and a mark of high status. The veterans returning from WWII were viewed as everyday men, and to see and hear about them going to college broke down that perception and made the dream of a college education more of a reality for all Americans (Clark, 1998). According to Hess and McGuinn (2002), “Education gained a new prominence after World War II” (p. 76). More students were graduating from high school and going to college after the G.I. Bill passed (Hess and McGuinn, 2002).
Aside from changing the perception of college education and increasing enrollments, the G.I. Bill led to the growth of community colleges and vocational schools (Wolfe, 2001). The 2.2 million veterans utilizing the educational benefits of the G.I. Bill (Bound and Turner, 2002) demanded a curriculum with classes more similar to what they had been exposed to in the military. Practicality was a vital aspect in their desired education, and schools responded by creating vocational programs to suit their needs (Clark, 1998). The G.I. Bill impacted future generations by making college more accessible to the average citizen and by causing the expansion of the college curriculum.
BROWN V. BOARD OF EDUCATION, 1954
In 1954, the Supreme Court ruled that segregation in public schools was unconstitutional in Brown v. Board of Education . The decision marked the beginning of more educational opportunities for African-Americans, but these opportunities were still severely limited. The Supreme Court declared that schools must desegregate, but did not make a deadline or any guidelines as to when it must be complete (Carson, 2004). A decade after the decision, less than 10% of black students were attending an integrated school (Ravitch, 2000). Throughout the ‘60s and ‘70s, the government used funds to force schools to desegregate or risk losing their funding (Carper, 2001), and by 1980, the practice of legal segregation for the most part had ended (Thattai, 2001).
Despite the long delay in achieving the goal of desegregation, the message that Brown delivered impacted the educational system and future policy in civil rights (Carson, 2004). Brown revealed the potential for other minority groups to begin the struggle for equal educational access (Gándara, Moran, and García, 2004). Equal and fair education began to be viewed as “the birthright of a free citizenry” (Hess and McGuinn, 2002, pp. 76–77), and more groups began demanding that right.
ELEMENTARY AND SECONDARY EDUCATION ACT OF 1965
The Elementary and Secondary Education Act of 1965 sought to equalize the education opportunities for all children. As the start of federal government grants to states for education (Hess and McGuinn, 2000), the law attempted “to compensate in some sense for disadvantages due to inequities in the social system” (Wolfe, 2001). President Johnson hoped that increasing and equalizing education for all children would one day rid the country of poverty (Easley, 2005). Funds were provided to the states for services for poor and minority students (Baker, 2001) as well as for “school library resources, textbooks, and other instructional materials for school children” ( Digest of Education Statistics , 2006). In 1967, Title VII authorized grants to be given for programs for LEP students (Baker, 2001). More than giving funds for different programs for the poor and minorities, ESEA along with the Civil Rights Act of 1964 aided in providing funds for the desegregation of southern schools who had refused to do so (Ravitch, 2000), more than a decade after the Brown v. Board of Education ruling. ESEA served to make an equal education more obtainable for all classes and races.
TITLE IX OF 1972
The place of women within the educational system was limited in terms of opportunities and the material being taught. The Civil Rights Movement during the ‘60s and ‘70s led women to protest more and demand equality in education. The Civil Rights Act of 1964 included the requirement that grants be given to the states in order for them to work towards that gender equality in education, and in 1972, Title IX of the Education Amendments was passed (Marshall, 2002). Title IX required that there be gender equality in schools that received money from the government and that “curricula should not stereotype girls’ and boys’ interests and careers” (Marshall, 2002, p. 712). In spite of the worthwhile objectives of Title IX, it was a feeble mandate. No penalties existed for schools that refused to comply, and a way to even supervise them was not set up until a quarter of the century after the passage of Title IX (Marshall, 2002).
In 1974, the Women’s Educational Equity Act (WEEA) was passed. This act provided a financial backbone to Title IX with funds granted to the schools and agencies in order to fulfill the goals of Title IX ( Women’s Educational Equity ). Title IX and WEEA made great strides in opening up more doors to education for women in the twentieth century.
INDIVIDUALS WITH DISABILITIES ACT OF 1975
The place of handicapped children in schools improved in the twentieth century. Almost five percent of students have a mental disability (Gunning, 2008), and with the 1975 Education for All Handicapped Children’s Act, they are able to receive a better education (Keyes, Hanley-Maxwell, and Capper, 1999). The name later changed to the Individuals with Disabilities Act (IDEA), but the goal of guaranteeing an education to disabled students remained the same (Keyes, Hanley-Maxwell, and Capper, 1999). A “free appropriate public education” or FAPE requires that “special education and related services are provided at public expense” (McLaughlin and Thurlow, 2003, p. 436). Disabled children must be placed in the least restrictive environment possible, meaning that their educational environment is as close to that of their nondisabled classmates as possible (Keyes, Hanley-Maxwell, and Capper, 1999). Individualized Education Programs (IEPs) are used to monitor the learning disabled student’s progress, and in 1997, assessments became mandatory for disabled students in order to make the schools more accountable for the students’ progress (McLaughlin and Thurlow, 2003). If necessary, accommodations in the assessments may be used (Gunning, 2008). Thanks to IDEA, more disabled students are graduating than if IDEA had not been passed ( Special Education and the Individuals with Disabilities Education Act ).
CONCLUSION
The twentieth century achieved a great deal in terms of including more groups that had been previously excluded or marginalized. A great deal still remains to be done in making education truly equal and accessible to every American, but the foundation has been laid for more work in the twenty-first century. Each American deserves to have equal educational opportunities, and as society progresses and more policies are created, that goal may be achievable.
References
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Carper, J. C. (2001). The changing landscape of U.S. education. Kappa Delta Pi Record , 37 no3, 106-110.
Carson, C. (2004). Two cheers for Brown v. Board of Education. The Journal of American History , 9:1, n.p.
Clark, D. A. (1998). “The two Joes meet—Joe College, Joe Veteran”: the G.I. Bill, college education, and postwar American culture. History of Education Quarterly ,38 no2, 165-189.
Digest of Educational Statistics: Chapter 4 Federal Programs for Education and Related Activities . 2006. Retrieved February 1, 2008 from Department of Education web site: http://nces.ed.gov/programs/digest/d06/ch_4.asp .
Easley II, J. (2005). The political tension of education as a public good: the voice of a Martin Luther King, Jr., scholar. Education and Urban Society , 37, 490-505.
Gándara, P., R. Moran, E. Garcia. (2004). Legacy of Brown: Lau and language policy in the United States. Review of Research in Education , 28, 27-46.
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Ravitch, D. (2000). Left back: a century of battles over school reform . New York City, NY: Touchstone.
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Thattai, D. (2001, November). A history of public education in the United States. Journal of Literacy and Education in Developing Societies ,1:2. Retrieved February 1, 2008 from www.servintfree.net/~aidmn-ejournal/.
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