3.13: Test Yourself- List in the Correct Order – Answers
The steps are listed in the correct order for each of the following techniques. These are printable flashcards to help you memorize and practice the techniques.
Radial Pulse Technique
- Use the pads of your first three fingers to gently palpate the radial pulse along the radius bone close to the flexor aspect of the wrist
- Press down with your fingers until you can best feel the pulsation
- Note the rate, rhythm, force, and equality when measuring the radial pulse
Carotid Pulse Technique
- Ask the client to sit upright
- Locate the carotid artery medial to the sternomastoid muscle in the middle third of the neck
- Gently palpate the carotid artery one at a time
- Note the rate, rhythm, force, and equality when measuring the carotid pulse
Apical Pulse Technique
- Ask the client to lay flat in a supine position
- Physically palpate the intercostal spaces to locate the landmark of the apical pulse
- Auscultate the apical pulse
- Note the rate and rhythm
Brachial Pulse Technique
- Palpate the bicep tendon in the area of the antecubital fossa
- Move your fingers medial from the tendon and about one inch above the antecubital fossa to locate the pulse
- Note the rate and rhythm
Respiration Technique
- Leave your fingers in place when you are done counting the pulse, and then begin assessing respiration
- Observe the rise and fall of the chest or abdomen
- Count for 30 seconds if the rhythm is regular or for a full minute if it is irregular
- Report respiration as breaths per minute, as well as whether breathing is relaxed, silent, and has a regular rhythm


8.3: Introduction to Simple Linear Regression
- Define independent and dependent variables
- Differentiate between observed and predicted values of the dependent variable
- Compare different linear models using the sum of squared errors
- Motivate the line of best fit and provide means for its computation
- Develop the coefficient of determination
- Conduct linear regression analysis by checking the reasonability of using a linear model, computing \(R^2,\) and finding the line of best fit
- Nuance the predictive and interpretive power of linear regression
Section 8.3 Excel File: (contains all of the data sets for this section)
Review and Preview
In studying bivariate quantitative data, we try to determine whether there is an association between two particular variables or not. If there is an association, a relationship between the variables, we would like to describe the relationship. We are interested in what happens to one variable as the other variable changes. We have discussed several ways to build this understanding: constructing scatter plots, classifying associations, and determining correlation. While more advanced textbooks address nonlinear correlation, we restricted ourselves to linear correlation. Linear correlation assesses the strength of an underlying linear relationship between the two variables of interest. If there is a linear relationship, it seems appropriate to think that there is a linear function that models the relationship. Knowledge of such a function would deepen our understanding of the relationship and allow us to extrapolate regarding values that were not explicitly measured in our collection of the data. In essence, such a function would enable us to make predictions about cases that were not explicitly studied. One of the fundamental motivations of statistical inquiry is to understand the world better so that we may better predict what will happen and act accordingly. This section develops the ideas of constructing a linear function that is the best fit for the data at hand.
When there is a linear relationship between two variables \(x\) and \(y,\) we expect there to be constants \(m\) and \(b\) such that \(y\) \(=mx\) \(+b.\) In this formulation, we refer to \(y\) (the vertical variable) as the dependent variable and \(x\) (the horizontal variable) as the independent variable . Recall that \(m\) is the slope of the line, the amount of change in \(y\) if \(x\) is increased by 1, and \(b\) is the \(y\)-intercept, the \(y\) value of the line when the \(x\) value is \(0.\) Since there is a relationship between the two variables, one variable changes as the other changes; that is, the slope of the line is defined and not \(0;\) we have \(m \neq 0.\)
Remember that our study of correlation did not depend on the ordering of the variables. If there is a linear relationship between \(x\) and \(y,\) there is a linear relationship between \(y\) and \(x\). In which case, we would expect another set of constants say \(M\) and \(B\) such that \(x\) \(=My\) \(+B.\) We would call \(x\) the dependent variable and \(y\) the independent variable. When studying associations, we do not assume causal relationships; do not let the terminology influence your thought in this regard.
When we collect data to understand the relationship, we expect the data to have some natural variation from the equation due to measurement error, natural variation, and the random noise that occurs in reality. As such, when we use collected data to construct a linear function, we are approximating the values of \(m\) and \(b.\) Throughout this section, we will use the hat symbol to indicate approximation values. Thus \(m\) is approximated by \(\hat{m}\) and \(b\) is approximated by \(\hat{b}.\) We will use these values to approximate \(y\) values which we denote \(\hat{y}\) using the following equation.\[\hat{y}=\hat{m}x+\hat{b}\nonumber\]Several fundamental questions arise. How do we pick the values of \(\hat{m}\) and \(\hat{b}?\) How do we know how well we did in picking the values of \(\hat{m}\) and \(\hat{b}?\) Is there an optimal choice of values for \(\hat{m}\) and \(\hat{b}?\) Even if we have the best line, it will not be perfect unless all the points are on the same line. Given that there is some error, how well does the line fit the data? These are questions we will begin to answer. We call the process of finding and evaluating these lines regression analysis .
Modeling Using Linear Functions
When studying bivariate quantitative data, we do not expect, even when there is a linear relationship, that all of the data points fall precisely on the same line. As such, even once we decide upon a linear function to model the data, it is impossible for the function to match the data perfectly. There will necessarily be some error between some of the observed values and the predicted values associated with them. To visualize this consider the following data set along with a scatter plot visualizing it.
Table \(\PageIndex{1}\): Paired values of variables \(x\) and \(y\)
| Observation | \(x\) | \(y\) |
|---|---|---|
| \(1\) | \(7\) | \(5\) |
| \(2\) | \(2\) | \(4\) |
| \(3\) | \(6\) | \(6\) |
| \(4\) | \(3\) | \(3\) |
| \(5\) | \(1\) | \(2\) |
Figure \(\PageIndex{1}\): Scatter plot of \(x\) and \(y\)
Notice that we cannot draw a line that goes through every point of the scatter plot. We could fairly easily plot a line through \(2\) of the points or even \(3\) of the points, but we cannot go through all \(5\) points; we cannot avoid the presence of error in our model! Suppose that we decided upon the linear function \(\hat{y}=0.5x+2\) to model this particular data set and then plotted it below in dark blue. Notice that the line does not go through any of the observed values plotted on the scatter plot. The observed values remain in orange, and the predicted values are colored in a light blue.
Figure \(\PageIndex{2}\): Scatter plot with linear model
The scatter plot labels two particular coordinate pairs on the scatter plot: \((x_3,y_3)\) \(=(6,6)\) and \((x_3,\hat{y}_3)\) \(=(6,5).\) The former is the observed pair; while, the latter is the predicted pair. Notice how the \(x\)-coordinates are the same. The predicted value of \(y_3\) is \(\hat{y}_3.\) We compute \(\hat{y}_3\) using the equation of the line \(\hat{y}_3\) \(=0.5x_3+2\) \(=0.5*6+2\) \(=5.\) We call the difference between \(y_3\) and \(\hat{y}_3\) the error at \(x_3\) which we denote \(e_3.\) In general, we define the error at any observed \(x\) value \(x_i\) as follows.\[e_i=y_i-\hat{y}_i\nonumber\]
- Using the linear function \(\hat{y}=0.5x+2,\) predict the value of \(y\) when \(x=8.\)
- Answer
-
We can predict the value of the variable \(y\) be evaluating the linear function at the indicated \(x\) value. We have a predicted value of \(0.5\cdot 8+2\) \(=6.\)
-
Using the linear function \(\hat{y}=0.5x+2,\) compute the error at each of the \(5\) collected \(x\) values.
Table \(\PageIndex{2}\): Values for variables \(x\) and \(y\)
Observation \(x\) \(y\) \(\hat{y}\) \(e_i=y-\hat{y}\) \(1\) \(7\) \(5\) \(2\) \(2\) \(4\) \(3\) \(6\) \(6\) \(4\) \(3\) \(3\) \(5\) \(1\) \(2\)
- Answer
-
The error is computed by taking the difference between the \(y\) value and the \(\hat{y}\) value at a given \(x\) value. We must determine each \(\hat{y}\) value. We do so in the table that follows.
Table \(\PageIndex{3}\): Computation of predicted values and errors
Observation \(x\) \(y\) \(\hat{y}\) \(e_i=y-\hat{y}\) \(1\) \(7\) \(5\) \(0.5\cdot 7+2=5.5\) \(5-5.5=-0.5\) \(2\) \(2\) \(4\) \(0.5\cdot 2+2=3\) \(4-3=1\) \(3\) \(6\) \(6\) \(0.5\cdot 6+2=5\) \(6-5=1\) \(4\) \(3\) \(3\) \(0.5\cdot 3+2=3.5\) \(3-3.5=-0.5\) \(5\) \(1\) \(2\) \(0.5\cdot 1+2=2.5\) \(2-2.5=-0.5\)
Notice that when the observed value is above the linear function modeling the data that the error is positive and when the observed value is below the line the error is negative. With this basis, we begin to develop the process of determining the best fitting line.
The Line of Best Fit
When we gather information about the world around us, we collect a lot of information. In order to understand best, we try to incorporate as much of the data as we can in our analyses and considerations. We do not collect samples from many people and then only use the results from a handful. Each observation provides important information; we do not want to exclude information without due cause. So, how do we decide upon a line to model our data, when no line perfectly predicts our data in practice? We want to use all of the data in the construction of the line, but how do we achieve such a goal when a line is uniquely determined given two points or a point and a slope?
The answer resides in considering error. We could assess the quality of a line by looking at the error across all observed values, but how are we to assess the totality of the error? If we were sum or average the errors, we would get cancelation between positive and negative errors. Indeed, if we modeled the data from above with the constant function \(\hat{y}=4,\) the sum of the errors and hence the average would be \(0,\) but the function would only go through one point and would indicate that there is no relationship between the variables. This does not match with the reality of the data. It is natural to desire that the measure of the totality of error is \(0\) only when the model perfectly fits the data. We, therefore, must expand our considerations.
Hopefully, we remember a similar discussion surrounding how to measure the dispersion of a data set. We went through several possibilities until we settled on our definition of variance which involved summing the deviations from the mean squared. We will utilize a similar sort of methodology without much motivation; we will consider the sum of the squared errors \((\textbf{SSE})\) as our measure of the totality of the error present in the model. In setting this as our measure, we have that there exists a unique line that minimizes the \(\text{SSE}.\) We can thus find a line and assert that it is one and only best line. We refer to this line as the line of best fit or the least-squares regression line.
-
Compute the \(\text{SSE}\) using the function \(\hat{y}\) \(=0.5x\) \(+2\) on the same data set as before, which we reproduce below.
Table \(\PageIndex{4}\): Values for variables \(x\) and \(y\)
Observation \(x\) \(y\) \(\hat{y}\) \(e_i=y-\hat{y}\) \(e_i^2=(y-\hat{y})^2\) \(1\) \(7\) \(5\) \(5.5\) \(5-5.5=-0.5\) \(2\) \(2\) \(4\) \(3\) \(4-3=1\) \(3\) \(6\) \(6\) \(5\) \(6-5=1\) \(4\) \(3\) \(3\) \(3.5\) \(3-3.5=-0.5\) \(5\) \(1\) \(2\) \(2.5\) \(2-2.5=-0.5\)
- Answer
-
We have computed the error for each of the observed \(x\) values in a previous text exercise. All that is left to do is square each of the errors and then add them together.
Table \(\PageIndex{5}\): Computation of predicted values, errors, and squared errors
The sum of the squared errors is thus \(\text{SSE}\) \(=0.25\) \(+1\) \(+1\) \(+0.25\) \(+0.25\) \(=2.75.\)Observation \(x\) \(y\) \(\hat{y}\) \(e_i=y-\hat{y}\) \(e_i^2=(y-\hat{y})^2\) \(1\) \(7\) \(5\) \(5.5\) \(5-5.5=-0.5\) \((-0.5)^2=0.25\) \(2\) \(2\) \(4\) \(3\) \(4-3=1\) \(1^2=1\) \(3\) \(6\) \(6\) \(5\) \(6-5=1\) \(1^2=1\) \(4\) \(3\) \(3\) \(3.5\) \(3-3.5=-0.5\) \((-0.5)^2=0.25\) \(5\) \(1\) \(2\) \(2.5\) \(2-2.5=-0.5\) \((-0.5)^2=0.25\)
-
We can show that the function \(\hat{y}\) \(=0.5x\) \(+2\) is not the best fitting line by finding another line that has a smaller \(\text{SSE}.\) One of the properties of the line of best fit is that it goes through the point \((\bar{x},\bar{y}).\) Let us keep the same slope but adjust the \(y\)-intercept so that our function goes through the point \((\bar{x},\bar{y})\) and then compute the \(\text{SSE}.\)
Table \(\PageIndex{6}\): Values for variables \(x\) and \(y\)
Observation \(x\) \(y\) \(\hat{y}\) \(e_i^2=(y-\hat{y})^2\) \(1\) \(7\) \(5\) \(\) \(2\) \(2\) \(4\) \(\) \(3\) \(6\) \(6\) \(\) \(4\) \(3\) \(3\) \(\) \(5\) \(1\) \(2\) \(\)
- Answer
-
We are going to model our data with the function \(\hat{y}\) \(=0.5x\) \(+b\) so that our function passes through the point \((\bar{x},\bar{y}).\) In order to compute \(b,\) we must compute \(\bar{x}\) and \(\bar{y},\) which we find to be \(\bar{x}=3.8\) and \(\bar{y}=4.\) Knowing the slope and a point through which our line passes is enough to determine the value of \(b.\) We plug the \(x\)- and \(y\)-coordinates into the function and solve for \(b.\) \[\begin{align*}4&=0.5\cdot3.8+b \\[8pt]4&=1.9+b\\[8pt]4-1.9&=b\\[8pt]2.1&=b\end{align*}\]Thus producing the function \(\hat{y}\) \(=0.5x\) \(+2.1\) as our linear model for the data set. From this we can produce a table of the approximated values at each observed \(x\) value and then compute each of the squared errors.
Table \(\PageIndex{7}\): Computation of predicted values and squared errors
The sum of the squared errors is thus \(\text{SSE}\) \(=0.36\) \(+0.81\) \(+0.81\) \(+0.36\) \(+0.36\) \(=2.7.\) Notice how the individual errors increased in magnitude for some values but decreased for others when comparing the error values using the previous model with these error values. Also, note that the sum of the errors is \(0,\) again emphasizing the inadequacy of using the sum or average of the errors as a measure of the totality of error. We finally notice that the \(\text{SSE}\) is smaller with this new model than with the old. We thus say that this model is better than the previous model. The question is, have we found the best model yet?Observation \(x\) \(y\) \(\hat{y}\) \(e_i^2=(y-\hat{y})^2\) \(1\) \(7\) \(5\) \(5.6\) \((-0.6)^2=0.36\) \(2\) \(2\) \(4\) \(3.1\) \(0.9^2=0.81\) \(3\) \(6\) \(6\) \(5.1\) \(0.9^2=0.81\) \(4\) \(3\) \(3\) \(3.6\) \((-0.6)^2=0.36\) \(5\) \(1\) \(2\) \(2.6\) \((-0.6)^2=0.36\)
-
Using technology, we determined that the line of best fit for the data set in question is given by the equation \(\hat{y}\) \(=.5224x\) \(+2.0149\) where the values of \(\hat{m}\) and \(\hat{b}\) are rounded to four decimal places. Compute the \(\text{SSE}.\)
Table \(\PageIndex{8}\): Values for variables \(x\) and \(y\)
Observation \(x\) \(y\) \(\hat{y}\) \(e_i^2=(y-\hat{y})^2\) \(1\) \(7\) \(5\) \(\) \(2\) \(2\) \(4\) \(\) \(3\) \(6\) \(6\) \(\) \(4\) \(3\) \(3\) \(\) \(5\) \(1\) \(2\) \(\)
- Answer
-
We have computed the error for each of the observed \(x\) values in a previous text exercise. All that is left to do is square each of the errors and then add them together.
Table \(\PageIndex{9}\): Computation of predicted values and squared errors
The sum of the squared errors is thus \(\text{SSE}\) \(\approx2.6866.\) We again note that this is the smallest of the sum of squared errors that we have yet to see for this data set. Indeed, this is the smallest attainable value for any linear function! No matter what slope and \(y\)-intercept picked, the sum of the squared errors will be larger than this value. Readers with a background in calculus or linear algebra are encouraged to read or work out the details for why this is!Observation \(x\) \(y\) \(\hat{y}\) \(e_i^2=(y-\hat{y})^2\) \(1\) \(7\) \(5\) \(5.6717\) \((-0.6717)^2\approx0.4512\) \(2\) \(2\) \(4\) \(3.0597\) \(0.9403^2\approx0.8842\) \(3\) \(6\) \(6\) \(5.1493\) \(0.8507^2\approx0.7237\) \(4\) \(3\) \(3\) \(3.5821\) \((-0.5821)^2\approx0.3388\) \(5\) \(1\) \(2\) \(2.5373\) \((-0.5373)^2\approx0.2887\)
Establishing the uniqueness and computational formulas for the line of best fit requires mathematics beyond the scope of this course. The mathematics does, however, produce very elegant results regarding the slope and \(y\)-intercept for the line of best fit, the line that minimizes the sum of the squared errors. We provide the formulas without proof. \[\hat{m}=r\frac{s_y}{s_x}~~~~~~~~~~\hat{b}=\bar{y}-r\frac{s_y}{s_x}\bar{x}=\bar{y}-\hat{m}\bar{x}\nonumber\]
Using the coefficients for the line of best fit provided above, show that every line of best fit, regardless of the data set, goes through the point \((\bar{x},\bar{y}).\) That is, show that when you substitute \(\bar{x}\) in the formula for the line of best fit, the value returned is \(\bar{y}.\)
- Answer
-
The line of best fit is given by the formula \(\hat{y}\) \(=\hat{m}x\) \(+\hat{b}.\) We were given the formulas for \(\hat{m}\) and \(\hat{b}.\) We have \(\hat{y}\) \(=r\frac{s_y}{s_x}x\) \(+\bar{y}-r\frac{s_y}{s_x}\bar{x}.\) When we substitute \(\bar{x}\) into the variable \(x,\) we obtain the following. \[\begin{align*}\hat{y}&=r\frac{s_y}{s_x}\bar{x}+\bar{y}-r\frac{s_y}{s_x}\bar{x}\\[8pt]&=\bar{y}\end{align*}\]
In addition to the fact that the point \((\bar{x}, \bar{y})\) always falls on the line of best fit, we have the sum and the average of all the errors is always \(0\) (when we do not round the numbers). We will not provide a proof of such a fact here, but it is a fact worth noting.
Just as with the computation of the correlation coefficient, we generally rely on technology to compute the slope and \(y\)-intercept for the line of best fit. We provide you with the function in Excel that returns the desired information as an array with the slope in the left cell and the \(y\)-intercept in the right cell. The function name is \(\text{LINEST}\) and takes four arguments: the first is the array of \(y\) values (values of the dependent variable); the second is the array of \(x\) values (values of the independent variable); the third is set of \(\text{TRUE}\) or \(1\); and the fourth is set to \(\text{FALSE}\) or simply \(0.\) Further information is available using the fourth argument but will not be utilized in this course.
Assessing the Line of Best Fit
We have now established that we can find the line of best fit, but another consideration must be made. Just because something is the best does not necessarily mean it is good. Of all the lines that could be used to model the data, we can find the best one, but does this best line actually fit the data well? This is the question we seek to answer and seems closely related to the correlation coefficient. Since the correlation coefficient measures the strength of an apparent linear relationship, we would expect that the closer \(|r|\) is to \(1,\) the better our line of best fit will model the data. This intuition is correct and will be confirmed as we approach the problem from a different direction.
When we are studying bivariate quantitative data (variables \(x\) and \(y\)) we are interested in how one variable changes as the other changes. With this, we may ask how much of the change in one variable can be attributed to the change of the other variable? Inherently, this question requires the development of some method or model which can measure the amount of change in the dependent variable which can be attributed to the model. When making such a measurement, the interest lies in the proportion of the change in one variable that can be attributed to the model, not the raw amount of variation that can be attributed. This allows the measure to be compared across data sets composed of data with vastly different magnitudes and makes the measure value independent of the units of the measurement. A high percentage indicates that the model fits well. Most of the change in \(y\) can be explained as due to the change in the \(x\) variable. If the percentage is low, the model does not fit well. The majority of the change in \(y\) is not understood as due to changes in \(x\) under the model.
In order to continue, we must decide how to measure the change in the \(y\) variable; this is really a question of dispersion. In general, the more the \(y\) variable changes, the greater the spread of the \(y\) variable data. Our most commonly used measure of dispersion has been standard deviation, but as we have seen throughout our bonus discussions, the real statistical power lies not in standard deviation but in variance. Recall that the variance is closely related to the average of square deviations from the mean, but we are not interested in a typical value, rather, we want the total change in the \(y\) variable. As such, we define our measure of change in \(y\) to be the total variation of \(y\) which we can compute with the following for data sets with \(n\) observations, note the similarity to the definition of variance. \[\text{Total Variation}=\sum_{i=1}^n \left(y_i-\bar{y}\right)^2\nonumber\]
We are interested in computing the percentage of the total variation of \(y\) that is explained by using the line of best fit to model the data; we call this percentage the coefficient of determination and denote it using the symbol \(R^2\). To determine the coefficient of determination, we must be able to compute the explained variation in our model. Either the variation is explained or it is not explained. As such, we know that the total variation is equal to the sum of the unexplained variation and the explained variation. The disparity between predicted values and observed values is the source of the unexplained variation. At this point, we recognize that the \(\text{SSE}\) is the unexplained variation. Recall the meaning of the sum of the squared errors and think of the formula that would compute it in general.\[\text{SSE}=\sum_{i=1}^n \left(y_i-\hat{y}_i\right)^2\nonumber\]With these ideas in hand we are ready to derive a computational formula for the coefficient of determination which measures the percent of variation in a data set that is accounted for by using the line of best fit to model the data.\[R^2=\frac{\text{Total Variation}-\text{SSE}}{\text{Total Variation}}=1-\frac{\text{SSE}}{\text{Total Variation}}=1-\frac{\sum\left(y_i-\hat{y}_i\right)^2}{\sum\left(y_i-\bar{y}\right)^2}=\frac{\sum\left(\hat{y}_i-\bar{y}\right)^2}{\sum\left(y_i-\bar{y}\right)^2}\nonumber\]Note that we dropped the indices on the summations in our formula to ease, but remember that for each of the summations, we are summing through all observations, \(1\) through \(n.\) The equivalency of the last computational formula is not obvious but can derived using the definitions of the line of best fit and the correlation coefficient, we encourage the reader to verify the simplification. In simplifying this last formulation, we develop the connection between the correlation coefficient and the coefficient of determination (a connection possibly already surmised given the symbolic representations); namely, that in squaring the correlation coefficient, we obtain the coefficient of determination.
- Optional Derivation Connecting Correlation Coefficient and Coefficient of Determination for the Mathematically Inclined
-
\[\begin{align*}R^2&=\frac{\sum\left(\hat{y}_i-\bar{y}\right)^2}{\sum\left(y_i-\bar{y}\right)^2}\\[8pt]&=\frac{\sum\left(r\frac{s_y}{s_x}x_i+\bar{y}-r\frac{s_y}{s_x}\bar{x}-\bar{y}\right)^2}{\sum\left(y_i-\bar{y}\right)^2}\\[8pt]&=\frac{\sum\left(r\frac{s_y}{s_x}x_i-r\frac{s_y}{s_x}\bar{x}\right)^2}{\sum\left(y_i-\bar{y}\right)^2}\\[8pt]&=\frac{\sum\left(r\frac{s_y}{s_x}\left(x_i-\bar{x}\right)\right)^2}{\sum\left(y_i-\bar{y}\right)^2}\\[8pt]&=\frac{\left(r\frac{s_y}{s_x}\right)^2\sum\left(x_i-\bar{x}\right)^2}{\sum\left(y_i-\bar{y}\right)^2}\\[8pt]&=\frac{r^2\frac{s^2_y}{s^2_x}(n-1)s^2_x}{(n-1)s^2_y}\\[8pt]&=\frac{r^2\frac{\cancel{s^2_y}}{\cancel{s^2_x}}\cancel{(n-1)}\cancel{s^2_x}}{\cancel{(n-1)}\cancel{s^2_y}}=r^2\end{align*}\]
-
Using technology, confirm that the line of best fit matches the formula provided in the earlier text exercise when rounded to \(4\) decimal places. Using the slope and \(y\)-intercept of the line of best fit provided by technology without rounding, determine the \(\text{SSE},\) the \(\text{Total Variation},\) and \(R^2\) by applying the formulas derived thus far in this section. Provide an interpretative sentence explaining the meaning of the coefficient of determination in light of the computed value. Do not round answers in intermediate steps.
Table \(\PageIndex{10}\): Values for variables \(x\) and \(y\)
Observation \(x\) \(y\) \(1\) \(7\) \(5\) \(2\) \(2\) \(4\) \(3\) \(6\) \(6\) \(4\) \(3\) \(3\) \(5\) \(1\) \(2\)
- Answer
-
Using the \(\text{LINEST}\) function in Excel, we confirm the accuracy of the previous text exercise rounded to \(4\) decimal places with a computed slope of \(0.52238806\ldots\) and a \(y\)-intercept of \(2.014925373\ldots.\) We provide a table with values rounded to \(6\) decimals for checking.
Table \(\PageIndex{11}\): Computation of predicted values, squared errors, and variations
Observation \(x\) \(y\) \(\hat{y}\) \(e_i^2=(y-\hat{y})^2\) \((y_i-\bar{y})^2\) \(1\) \(7\) \(5\) \(5.671642\) \((-0.671642)^2\approx0.451103\) \((5-4)^2=1\) \(2\) \(2\) \(4\) \(3.059702\) \(0.940299^2\approx0.884161\) \((4-4)^2=0\) \(3\) \(6\) \(6\) \(5.149254\) \(0.850746^2\approx0.723769\) \((6-4)^2=4\) \(4\) \(3\) \(3\) \(3.582090\) \((-0.582090)^2\approx0.338828\) \((3-4)^2=1\) \(5\) \(1\) \(2\) \(2.537313\) \((-0.537313)^2\approx0.288706\) \((2-4)^2=4\) The sum of the squared errors is thus \(\text{SSE}\) \(\approx2.686567,\) the smallest possible value for this particular data set. The \(\text{Total Variation},\) the summation of the last column is \(10.\) We can compute the coefficient of determination \(R^2\) \(\approx1\) \(-\frac{2.686567}{10}\) \(\approx1-0.268657\) \(\approx0.731343.\) Approximately \(73.1\%\) of the variation present in the \(y\) variable is accounted for using the linear function \(\hat{y}=0.522388x+2.014925.\) This indicates that the linear model fits the data to a certain degree, but there is a decent amount of random variation, error, or noise present.
- Using the results of the previous part of this text exercise and technology, confirm that the square of the correlation coefficient is equal to the coefficient of determination. Compute the correlation coefficient using technology.
- Answer
-
We computed in the previous part that \(R^2\) \(\approx0.731343.\) Using the \(\text{CORREL}\) function in excel, we compute that \(r\) \(\approx 0.855186\) and note that \(0.855186^2\) \(\approx 0.731343\) which is the value that we computed for the coefficient of determination.
The coefficient of determination \(R^2\) can be computed directly using the Excel function \(\text{RSQ}\). The function takes two arrays of numbers, similar to the \(\text{LINEST}\) function, the first array consists of the known \(y\)-values (dependent variable) and the second array consists of the known \(x\)-values (independent variable).
Simple Linear Regression: Predictions and Interpretations
We have yet to conduct simple linear regression outside of a purely mathematical context. Having developed the concepts, we now address the application of these ideas and provide insight to their interpretations. Let us return to a data set that we have started to analyze, the ages of the bride and groom on their wedding day. Using a scatter plot of the data, we have already determined that a linear model would be appropriate. Let us determine the line of best fit and assess how well the model fits the data. In our previous considerations, we had the groom's age on the horizontal axis, the axis traditionally associated with the independent variable. For continuity of presentation, we will continue in this vein of thought. When evaluating the linear model we will be predicting the age of the bride based on the input of a groom's age. If we want to predict the age of a groom based on a particular age of a bride, we will either have to solve for the age of the groom or conduct the linear regression analysis with the variables switched. Either option is fairly straightforward and will produce the same predictions.
- Letting the age of the bride on her wedding day be the dependent variable, determine the line of the best fit and the coefficient of determination for the data set. Explain the results of the linear regression in the context of the problem.
Table \(\PageIndex{12}\): Ages of bride and groom on wedding day
| Married Couple | Groom's Age (years) | Bride's Age (years) |
|---|---|---|
| \(1\) | \(20\) | \(21\) |
| \(2\) | \(26\) | \(20\) |
| \(3\) | \(32\) | \(34\) |
| \(4\) | \(30\) | \(30\) |
| \(5\) | \(21\) | \(22\) |
| \(6\) | \(29\) | \(28\) |
| \(7\) | \(26\) | \(25\) |
| \(8\) | \(34\) | \(34\) |
| \(9\) | \(29\) | \(28\) |
| \(10\) | \(55\) | \(50\) |
| \(11\) | \(30\) | \(26\) |
| \(12\) | \(43\) | \(39\) |
| \(13\) | \(30\) | \(29\) |
| \(14\) | \(24\) | \(22\) |
| \(15\) | \(20\) | \(19\) |
- Answer
-
Using Excel, we have the line of best fit given by \(\hat{y}\) \(=0.878562x\) \(+2.168364\) with a coefficient of determination equal to \(0.938862.\) The computed coefficient of determination indicates that almost \(94\%\) of the variation in the age of a bride on her wedding day can be accounted by modeling the relationship between the ages of the bride and groom with the function \(\hat{y}\) \(=0.878562x\) \(+2.168364,\) where \(x\) is the age of the groom. This is a fairly high percentage which indicates that the model is a good fit. The positive slope indicates a positive association.
- The \(y\)-intercept of a function is the value of the dependent variable when the independent variable is equal to \(0.\) Within the context of our problem \(x=0\) corresponds to the groom's age being \(0.\) The \(y\)-intercept is about \(2.17,\) indicating that the bride would be just a little older than \(2.\) Explain why, contextually, these considerations do not make any sense. What does this say about our model? What does this say about linear regression models in general?
- Answer
-
Infants and toddlers do not get married. Adults get married. It is remiss to try to use the line of best fit to model the relationship where the relationship does not exist. There is a mathematical domain for our function and there is a contextual domain for our relation. If we are trying to understand the reality around us, the contextual domain must be at the forefront of our minds. We do not want to extend our model where the relationship ceases or beyond where our data permits us to engage. As such, we would not want to use our model for any ages less than \(16\) or \(18\) years of age for either the bride or the groom as those are the ages commonly set as the minimum ages for which marriage is legal. This does not say anything negative about our model or models in general; we must be cognizant of when it is appropriate to use the models.  We will develop more nuance as we progress through this section.
- Using the model constructed in part \(1,\) predict the age of the bride when the groom is \(27\) years old.
- Answer
-
We have that the line of best fit is given by \(\hat{y}\) \(=0.878562x\) \(+2.168364.\) We are being asked to predict the age of the bride when the groom is \(27\) years old. This is equivalent to evaluating the function when the \(x\) variable is \(27.\) We predict the bride's age to be \(0.878562\cdot27\) \(+2.168364\) \(=25.88955\) years old; the bride will be just shy of \(26\) years old, when the groom is \(27\) years old.
- Using the model constructed in part \(1,\) predict the age of the groom when the bride is \(32\) years old.
- Answer
-
We have again that the line of best fit is given by \(\hat{y}\) \(=0.878562x\) \(+2.168364.\) We are asked to predict the age of the groom when the bride is \(32\) years old. This is not equivalent to evaluating the function when the \(x\) variable is \(32\) because the \(x\) variable corresponds to the age of the groom, not the bride. We predict the groom's age by solving for \(x\) when our linear equation equals \(32.\) \[\begin{align*}32&=0.878562x+2.168364\\[8pt]32-2.168364&=0.878562x\\[8pt]\frac{32-2.168364}{0.878562}&=x\\[8pt]33.955055&\approx x\end{align*}\]We predict that the groom will be about just shy of \(34\) when the bride is \(32\) years old.
- Using the model constructed in part \(1,\) when does the model predict that the bride and groom will be exactly the same age? Does this seem like an appropriate use of the model?
- Answer
-
For the last time, we have that the line of best fit is given by \(\hat{y}\) \(=0.878562x\) \(+2.168364.\) We want to find when the model predicts the two ages to be the same, i.e. \(\hat{y}=x.\) To do so, we replace \(\hat{y}\) with \(x\) in the equation and solve. \[\begin{align*}x&=0.878562x+2.168364\\[8pt]x-0.878562x&=2.168364\\[8pt]x(1-0.878562)&=2.168364\\[8pt]x&=\frac{2.168364}{1-0.878562}\approx17.855796\end{align*}\]We predict the bride and groom to be the same age when they are both just shy of \(18\) on their wedding day.
In general, once a person is about \(2\) years of age, the primary focus is on the number of years. As such, our interest might be more of when the model predicts that both the bride and the groom would be in the same year of life. This would seem to be a more appropriate question given the context of the model; although, it is a much harder question to solve for ages of \(17.5\) and \(17.8\) would constitute solutions as well as what we just found.
- When ordering custom clothing or preparing to rent formal wear, many measurements are taken to ensure that the clothes fit well. Two common measurements are height and the length from the center of the back between the scapulae to the tips of the fingers when the arm is fully extended to the side. Let us refer to this latter measurement as an individual's radius. We asked a random selection of \(50\) online elementary statistics students to obtain these measurements for themselves in centimeters and report their findings to analyze as a group. We provide the data in the attached Excel file. Examine the data to check that a linear model is appropriate. If not, explain. If so, find the line of best fit and coefficient of determination using the radius as the independent variable.
- Answer
-
We first create a scatter plot to check if a linear relationship is reasonable. We provide two scatter plots with different scaling.
Figure \(\PageIndex{3}\): Scatter plots of height and radius using different scales and ranges
The scatter plot on the left includes the origin \((0,0)\) while the other scatter plot does not. Both indicate a fairly linear relationship. So we proceed with a linear regression analysis. The coefficient of determination is \(0.9212\) with the linear model defined by \(\hat{y}\) \(=1.381x\) \(+51.3125.\)
In the previous text exercise, we determined the line of best fit and saw that the line fit fairly well. A little more than \(92\%\) of the variation in the height variable was attributed to the difference values of the radius variable through our linear model. We have a nice model to help us understand the relationship between the height and radius of individuals. The possible values of an individual's radius go beyond those collected in our sample. This is one of the reasons that we desired a model; so, that we could estimate values for points where we did not have any data collected. As such, we might be tempted to estimate the height of an individual with a radius of \(40\) centimeters.
Using the line of best fit found in the previous text exercise, estimate the height of an individual with a radius of \(40\) centimeters. Consider the validity of such an estimation.
- Answer
-
We would estimate that individual's height to be \(1.381\cdot40\) \(+51.3125\) \(=106.5525\) centimeters. Both the radius and the height values are within the contextual domain of our variables, but can we use the model in such a way? The predicted height is about three and a half feet tall; a rather short person. In reality, most likely a child. This begs the question: should we use data from a sample of elementary statistics students who are fully formed adults to make predictions about a child? Hopefully, at this stage in our development of statistics, we would be inclined to say no. We would not think that a sample of adults would be representative of children without some significant argumentation explaining why they are fundamentally the same. Our intuition would naturally be that the body structure of children is different than the body structure of adults. We do not want to overgeneralize our results beyond that which we have actually studied. In practice, we must consider both the contextual domain and the extent to which our sample is representative. In general, we do not want to utilize our model too far beyond the values seen in our collected data. Do you want to predict the height of an individual with a radius of \(90\) centimeters? Go right ahead! But, if you want to predict the height of an individual with a radius of \(15,\) best go collect data from individuals around that size.
We conclude this section with one last interpretative guideline. The slope of a linear function describes the rate of change of the function. If the value of \(x\) increases by \(1,\) the value of \(y\) changes in value equal to the slope. In the case of our last text exercise, when we increase the radius by one centimeter, the predicted \(y\) value increases by \(1.381\) centimeters. Are we to interpret this to indicate that if an individual had a radius of \(75\) centimeters and height of \(150\) centimeters and then grew to \(76\) centimeters, the individual's height would be \(151.381\) centimeters? Unfortunately, the answer is no. We built the model by using data from \(50\) individuals. The model predicts the typical relationship between the variables; it does not predict the individual change, nor does it predict the changes in a perfect way. We must temper ourselves from concluding more than we can. We can expect that as individuals increase in radius by \(1\) centimeter, the average gain in height is going to be close to \(1.381\) centimeters, but we cannot make such a claim on the individual level.
This is, in fact, a theme pertaining to the entirety of this textbook. Statistics seeks to understand trends in large groups, and it is almost always inappropriate to use information about a group to infer facts about an individual. If we say one group is shorter than another group on average, that does not necessarily mean that every individual in the first group is shorter than every individual in the second group. If we say that \(80\%\) of some group has a particular disease, that does not necessarily mean that each individual in that group has an \(80\%\) chance of obtaining that disease. If we say that a hypothesis or model predicts a group to have certain parameters, that says nothing about a specific individual in that group. Many issues in modern society arise from people misunderstanding this. People often use facts about a group to inform their thoughts about individuals (such as with stereotyping). People also often ignore facts about groups because of facts they know about an individual; for example, Bob smoked his whole life and lived to be \(100.\) We hope that this text has helped the reader understand how to properly understand facts about groups and why such understanding can be useful.


9.8: Summary of Key Concepts
Summary of Key Concepts
Square Root
The square root of a positive number \(x\) is a number such that when it is squared, the number \(x\) results.
Every positive number has two square roots, one positive and one negative. They opposites of each other.
Principal Square Root \(\sqrt{x}\)
If \(x\) is a positive real number, then
\(\sqrt{x}\) represents the positive square root of \(x\). The positive square root of a number is called the principal square root of the number.
Secondary Square Root \(-\sqrt{x}\)
\(-\sqrt{x}\) represents the negative square root of \(x\). The negative square root of a number is called the secondary square root of the number.
Radical Sign, Radicand; and Radical
In the expression \(\sqrt{x}\),
\(\sqrt{}\) is called the radical sign.
\(x\) is called the radicand.
\(\sqrt{x}\) is called a radical.
The horizonal bar that appears attached to the radical sign, \(\sqrt{}\), is a grouping symbol that specifies the radicand.
Meaningful Expressions
A radical expression will only be meaningful if the radicand (the expression under the radical sign) is not negative:
\(\sqrt{-25}\) is not meaningful and \(\sqrt{-25}\) is not a real number.
Simplifying Square Root Expressions
If \(a\) is a nonnegative number, then
\(\sqrt{a^2} = a\)
Perfect Squares
Real numbers that are squares of rational numbers are called perfect squares.
Irrational Numbers
Any indicated square root whose radicand is not a perfect square is an irrational number.
\(\sqrt{2}, \sqrt{5}\) and \(\sqrt{10}\) are irrational numbers
The Product Property
\(\sqrt{xy} = \sqrt{x}\sqrt{y}\)
The Quotient Property
\(\sqrt{\dfrac{x}{y}} = \dfrac{\sqrt{x}}{\sqrt{y}}\), \(y \not = 0\)
Be Careful
\(\begin{array}{flushleft}
\sqrt{x + y} &\not = \sqrt{x} \sqrt{y} & (\sqrt{16 + 9} &\not = \sqrt{16} + \sqrt{9})\\
\sqrt{x-y} &\not = \sqrt{x} - \sqrt{y} & (\sqrt{25-16} &\not = \sqrt{25} - \sqrt{16})
\end{array}\)
Simplified Form
A square root that does not involve fractions is in simplified form if there are no perfect squares in the radicand.
A square root involving a fraction is in simplified form if there are no
- perfect squares in the radicand,
- fractions in the radicand, or
- square root expressions in the denominator
Radicalizing the Denominator
The process of eliminating radicals from the denominators is called rationalizing the denominator.
Multiplying Square Root Expressions
The product of the square roots is the square root of the product.
\(\sqrt{x}\sqrt{y} = \sqrt{xy}\)
- Simplify each square root, if necessary.
- Perform the multiplication.
- Simplify, if necessary.
Dividing Square Root Expressions
The quotient of the square roots is the square root of the quotient.
\(\dfrac{\sqrt{x}}{\sqrt{y}} = \sqrt{\dfrac{x}{y}}\)
Addition and Subtraction of Square Root Expressions
\(a\sqrt{x} + b\sqrt{x} = (a+b)\sqrt{x}\)
\(a\sqrt{x} - b\sqrt{x} = (a-b)\sqrt{x}\)
Square Root Equation
A square root equation is an equation that contains a variable under a square root radical sign.
Solving Square Root Equation
- Isolate a radical.
- Square both sides of the equation.
- Simplify by combining like terms.
- Repeat step 1 if radical are still present.
- Obtain potential solution by solving the resulting non-square root equation.
- Check potential solutions by substitution.


6.6: Discussion- Recruitment and Selection
A candidate can have every skill you are looking for in a position, and yet still might not be successful at the job. Bringing a new employee into the organization means matching skills to job requirements and matching personalities to workplace culture.
Your HR team has worked hard to build a culture that supports the company’s brand and goals. The organization prides itself on open-door discussions between leaders and subordinates; they provide flexible hours and telecommuting; they’ve established core values and “work lingo” that support their employment brand. Your employee value proposition boasts that working for your company is like having a second family.
You’re an HR manager working with a recruiter to bring a new executive on board in the finance department. As a new leader, this executive will be driving workplace culture as much as he or she will be blending in with it.
Discussion Prompt
What type of interview questions would you ask candidates to ensure that they are a good fit for the company culturally? And what answers to those questions would prompt you to make the candidate an offer of employment? Write one or two questions and their “correct” answers to submit. Then, review the questions and answers of two of your classmate peers and provide comments.
Grading
Share your opinions below and respond to two of your classmates’ thoughts.
| Criteria | Not Evident | Developing | Exemplary | Points |
|---|---|---|---|---|
| Submit your initial response |
0 pts
No post made |
5 pts
Post is either late or off-topic |
10 pts
Post is made on time and is focused on the prompt |
10 pts |
| Respond to at least two peers’ presentations |
0 pts
No response to peers |
2
pts
Responded to only one peer |
5 pts
Responded to two peers |
5 pts |
| Total: | 15 pts |
Contributors and Attributions
- Discussion: Recruitment and Selection. Authored by : Freedom Learning Group. Provided by : Lumen Learning. License : CC BY: Attribution


18.7C: Blood Flow in Skeletal Muscle
Blood flow to an active muscle changes depending on exercise intensity and contraction frequency and rate.
- Summarize the factors involved in blood flow to skeletal muscles
Key Points
- The circulatory system is closely associated with skeletal muscle to provide efficient transfer of oxygen and nutrients required for contraction and the removal of inhibitory waste products.
- At rest, skeletal muscle uses approximately 20% of cardiac output, which can rise to 80% during exercise.
- Return of blood to the heart, especially from the legs, is facilitated by the skeletal muscle pump. Deep-lying veins are compressed by skeletal muscle, forcing blood through the circulatory system back to the heart.
- After repeated stimulation, vascular recruitment can lead to an increase in the number of capillaries present in a muscle tissue, facilitating better supply and more efficient removal of waste products.
Key Terms
- skeletal muscle pump : The mechanism whereby skeletal muscles aid the return of blood to the heart by compressing embedded veins.
- vascular recruitment : The increase in the number of capillaries in response to a stimulus; for example, repeated exercise results in an increase in the number of capillaries in a skeletal muscle.
Skeletal muscles are important in maintaining posture and controlling locomotion through contraction. For this reason, they receive approximately 20% of cardiac output at rest, which can increase up to a maximum of approximately 80% with exercise. Due to the requirements for large amounts of oxygen and nutrients, muscle vessels are under very tight autonomous regulation to ensure a constant blood flow, and so can have a large impact on the blood pressure of associated arteries.
Blood vessels are closely intertwined with skeletal muscle tissues lying between the fascicles, or bundles of muscle fibers. Each muscle is supplied by many capillaries. This close association reduces the diffusion distances, allowing for the efficient exchange of oxygen and nutrients required for contraction and the rapid removal of inhibitory waste products.
Skeletal Muscle : Skeletal muscle: 1] Bone, 2] Perimysium, 3] Blood vessel, 4] Muscle fiber, 5] Fascicle, 6] Endomysium, and 7] Epimysium Tendon.
Blood Flow During Exercise
Blood flow within muscles fluctuates as they contract and relax. During contraction, the vasculature within the muscle is compressed, resulting in a lower arterial inflow with inflow increased upon relaxation. The opposite effect would be seen if measuring venous outflow.
This rapid increase and decrease in flow is observed over multiple contractions. If the muscle is used for an extended period, mean arterial inflow will increase as the arterioles vasodilate to provide the oxygen and nutrients required for contraction. Following the end of contractions, this increased mean flow remains to resupply the muscle tissue with required nutrients and clear inhibitory waste products, due to the loss of the inhibitory contractile phase.
Skeletal Muscle Pump
Skeletal muscles also play a key role in the movement of blood around the body. Veins embedded within a muscle are compressed during contraction of that muscle, causing an increase in blood pressure due to the presence of one-way valves within the veins. This increase in pressure drives the blood towards the heart. The skeletal muscles of the legs are particularly important skeletal muscle pumps as they prevent pooling of the blood in the feet and calves due to gravity.
Skeletal Muscle Pump : During contraction of the skeletal muscle the vein is compressed which increases blood pressure. Due to the presence of one way valves the blood can pass only in one direction, back towards the heart.
It is unclear whether the action of skeletal muscle pumps influences arterial flow or if this is maintained purely by the pumping of the heart.
Vascular Recruitment
Following repeated stimulus such as through exercise, the number of capillaries present in a muscle tissue can increase. This vascular recruitment increases the capillary surface area within a muscle, allowing for enhanced oxygen exchange with the muscle fibers, prolonging the period of aerobic respiration and thus muscle output, and facilitating a more rapid removal of inhibitory waster factors such as lactic acid, reducing fatigue.


28.11.3: Figuring for Yourself
Figuring for Yourself
Using the information from Example 28.1, how much fainter an object will you have to be able to measure in order to include the same kinds of galaxies in your second survey? Remember that the brightness of an object varies as the inverse square of the distance.
20.
Using the information from Example 28.1, if galaxies are distributed homogeneously, how many times more of them would you expect to count on your second survey?
21.
Using the information from Example 28.1, how much longer will it take you to do your second survey?
22.
Galaxies are found in the “walls” of huge voids; very few galaxies are found in the voids themselves. The text says that the structure of filaments and voids has been present in the universe since shortly after the expansion began 13.8 billion years ago. In science, we always have to check to see whether some conclusion is contradicted by any other information we have. In this case, we can ask whether the voids would have filled up with galaxies in roughly 14 billion years. Observations show that in addition to the motion associated with the expansion of the universe, the galaxies in the walls of the voids are moving in random directions at typical speeds of 300 km/s. At least some of them will be moving into the voids. How far into the void will a galaxy move in 14 billion years? Is it a reasonable hypothesis that the voids have existed for 14 billion years?
23.
Calculate the velocity, the distance, and the look-back time of the most distant galaxies in Figure 28.21 using the Hubble constant given in this text and the redshift given in the diagram. Remember the Doppler formula for velocity and the Hubble law ( v = H × d , where d is the distance to a galaxy). For these low velocities, you can neglect relativistic effects.
24.
Assume that dark matter is uniformly distributed throughout the Milky Way, not just in the outer halo but also throughout the bulge and in the disk, where the solar system lives. How much dark matter would you expect there to be inside the solar system? Would you expect that to be easily detectable? Hint: For the radius of the Milky Way’s dark matter halo, use R = 300,000 light-years; for the solar system’s radius, use 100 AU; and start by calculating the ratio of the two volumes.
25.
The simulated box of galaxy filaments and superclusters shown in Figure 28.29 stretches across 1 billion light-years. If you were to make a scale model where that box covered the core of a university campus, say 1 km, then how big would the Milky Way Galaxy be? How far away would the Andromeda galaxy be in the scale model?
26.
The first objects to collapse gravitationally after the Big Bang might have been globular cluster-size galaxy pieces, with masses around 10 6 solar masses. Suppose you merge two of those together, then merge two larger pieces together, and so on, Lego-style, until you reach a Milky Way mass, about 10 12 solar masses. How many merger generations would that take, and how many original pieces? (Hint: Think in powers of 2.)


6.7: Managerial Roles
-
- Last updated
- Save as PDF
6. What roles do managers take on in different organizational settings?
In carrying out the responsibilities of planning, organizing, leading, and controlling, managers take on many different roles. A role is a set of behavioral expectations, or a set of activities that a person is expected to perform. Managers’ roles fall into three basic categories: informational roles, interpersonal roles, and decisional roles. These roles are summarized in Table 6.5 . In an informational role , the manager may act as an information gatherer, an information distributor, or a spokesperson for the company. A manager’s interpersonal roles are based on various interactions with other people. Depending on the situation, a manager may need to act as a figurehead, a company leader, or a liaison. When acting in a decisional role , a manager may have to think like an entrepreneur, make decisions about resource allocation, help resolve conflicts, or negotiate compromises.
Managerial Decision Making
In every function performed, role taken on, and set of skills applied, a manager is a decision maker. Decision-making means choosing among alternatives. Decision-making occurs in response to the identification of a problem or an opportunity. The decisions managers make fall into two basic categories: programmed and nonprogrammed. Programmed decisions are made in response to routine situations that occur frequently in a variety of settings throughout an organization. For example, the need to hire new personnel is a common situation for most organizations. Therefore, standard procedures for recruitment and selection are developed and followed in most companies.
| The Many Roles Managers Play in an Organization | ||
|---|---|---|
| Role | Description | Example |
| Information Roles | ||
| Monitor |
|
|
| Disseminator |
|
|
| Spokesperson |
|
|
| Interpersonal Roles | ||
| Figurehead |
|
|
| Leader |
|
|
| Liaison |
|
|
| Decisional Roles | ||
| Entrepreneur |
|
|
| Disturbance handler |
|
|
| Resource allocator |
|
|
| Negotiator |
|
|
Infrequent, unforeseen, or very unusual problems and opportunities require nonprogrammed decisions by managers. Because these situations are unique and complex, the manager rarely has a precedent to follow. The earlier example of the Norfolk Southernemployee, who had to decide the best way to salvage a five-mile-long piece of railroad track from the bottom of Lake Pontchartrain, is an example of a nonprogrammed decision. Likewise, when Hurricane Katrina was forecast to make landfall, Thomas Oreck, then CEO of the vacuum manufacturer that bears his name, had to make a series of nonprogrammed decisions. Oreck’s corporate headquarters were in New Orleans, and its primary manufacturing facility was in Long Beach, Mississippi. Before the storm hit, Oreck transferred its computer systems and call-center operations to backup locations in Colorado and planned to move headquarters to Long Beach. The storm, however, brutally hit both locations. Oreck executives began searching for lost employees, tracking down generators, assembling temporary housing for workers, and making deals with UPS to begin distributing its product (UPS brought food and water to Oreck from Atlanta and took vacuums back to the company’s distribution center there). All of these decisions were made in the middle of a very challenging crisis environment.
Whether a decision is programmed or nonprogrammed, managers typically follow five steps in the decision-making process, as illustrated in Exhibit 6.7 :
- Recognize or define the problem or opportunity. Although it is more common to focus on problems because of their obvious negative effects, managers who do not take advantage of new opportunities may lose competitive advantage to other firms.
- Gather information so as to identify alternative solutions or actions.
- Select one or more alternatives after evaluating the strengths and weaknesses of each possibility.
- Put the chosen alternative into action.
- Gather information to obtain feedback on the effectiveness of the chosen plan.
It can be easy (and dangerous) for managers to get stuck at any stage of the decision-making process. For example, entrepreneurs can become paralyzed evaluating the options. For the Gabby Slome, the cofounder of natural pet food maker Ollie, the idea for starting the company came after her rescue dog began having trouble digesting store-bought pet food after living on scraps. Slome decided that the pet food industry, a $30 billion a year business, was ripe for a natural food alternative. She laments, however, that she let perfect be the enemy of the very good by indulging in “analysis paralysis.” 18
Exhibit 6.7 The Decision-Making Process (Attribution: Copyright Rice University, OpenStax, under CC BY 4.0 license.)
CONCEPT CHECK
- What are the three types of managerial roles?
- Give examples of things managers might do when acting in each of the different types of roles.
- List the five steps in the decision-making process.


5.7: Naming Ionic Compounds
-
- Last updated
- Save as PDF
Learning Objectives
- To use the rules for naming ionic compounds.
After learning a few more details about the names of individual ions, you will be one step away from knowing how to name ionic compounds. This section begins the formal study of nomenclature, the systematic naming of chemical compounds.
Naming Ions
The name of a monatomic cation is simply the name of the element followed by the word ion . Thus, Na + is the sodium ion, Al 3 + is the aluminum ion, Ca 2 + is the calcium ion, and so forth.
We have seen that some elements lose different numbers of electrons, producing ions of different charges (Figure 3.3). Iron, for example, can form two cations, each of which, when combined with the same anion, makes a different compound with unique physical and chemical properties. Thus, we need a different name for each iron ion to distinguish Fe 2 + from Fe 3 + . The same issue arises for other ions with more than one possible charge.
There are two ways to make this distinction. In the simpler, more modern approach, called the Stock system , an ion’s positive charge is indicated by a roman numeral in parentheses after the element name, followed by the word ion . Thus, Fe 2 + is called the iron(II) ion, while Fe 3 + is called the iron(III) ion. This system is used only for elements that form more than one common positive ion. We do not call the Na + ion the sodium(I) ion because (I) is unnecessary. Sodium forms only a 1+ ion, so there is no ambiguity about the name sodium ion .
| Element | Stem | Charge | Modern Name | Common Name |
|---|---|---|---|---|
| iron | ferr- | 2+ | iron(II) ion | ferrous ion |
| 3+ | iron(III) ion | ferric ion | ||
| copper | cupr- | 1+ | copper(I) ion | cuprous ion |
| 2+ | copper(II) ion | cupric ion | ||
| tin | stann- | 2+ | tin(II) ion | stannous ion |
| 4+ | tin(IV) ion | stannic ion | ||
| lead | plumb- | 2+ | lead(II) ion | plumbous ion |
| 4+ | lead(IV) ion | plumbic ion | ||
| chromium | chrom- | 2+ | chromium(II) ion | chromous ion |
| 3+ | chromium(III) ion | chromic ion | ||
| gold | aur- | 1+ | gold(I) ion | aurous ion |
| 3+ | gold(III) ion | auric ion |
The second system, called the common system , is not conventional but is still prevalent and used in the health sciences. This system recognizes that many metals have two common cations. The common system uses two suffixes (- ic and - ous ) that are appended to the stem of the element name. The -ic suffix represents the greater of the two cation charges, and the -ous suffix represents the lower one. In many cases, the stem of the element name comes from the Latin name of the element. Table \(\PageIndex{1}\) lists the elements that use the common system, along with their respective cation names.
| Ion | Name |
|---|---|
| F − | fluoride ion |
| Cl − | chloride ion |
| Br − | bromide ion |
| I − | iodide ion |
| O 2− | oxide ion |
| S 2− | sulfide ion |
| P 3− | phosphide ion |
| N 3− | nitride ion |
The name of a monatomic anion consists of the stem of the element name, the suffix - ide , and then the word ion . Thus, as we have already seen, Cl − is “chlor-” + “-ide ion,” or the chloride ion. Similarly, O 2− is the oxide ion, Se 2 − is the selenide ion, and so forth. Table \(\PageIndex{2}\) lists the names of some common monatomic ions. The polyatomic ions have their own characteristic names, as discussed earlier.
Example \(\PageIndex{1}\)
Name each ion.
- Ca 2 +
- S 2−
- SO 3 2 −
- NH 4 +
- Cu +
Solution
- the calcium ion
- the sulfide ion
- the sulfite ion
- the ammonium ion
- the copper(I) ion or the cuprous ion
Exercise \(\PageIndex{1}\)
Name each ion.
- Fe 2 +
- Fe 3 +
- SO 4 2 −
- Ba 2 +
- HCO 3 −
- Answer a:
- iron(II) ion
- Answer b:
- iron(III) ion
- Answer c:
- sulfate ion
- Answer d:
- barium ion
- Answer e:
- hydrogen carbonate ion or bicarbonate ion
Example \(\PageIndex{2}\)
Write the formula for each ion.
- the bromide ion
- the phosphate ion
- the cupric ion
- the magnesium ion
Solution
- Br −
- PO 4 3 −
- Cu 2 +
- Mg 2 +
Exercise \(\PageIndex{2}\)
Write the formula for each ion.
- the fluoride ion
- the carbonate ion
- the stannous ion
- the potassium ion
- Answer a:
- F -
- Answer b:
- CO 3 2 -
- Answer c:
- Sn 2+
- Answer d:
- K +
Naming Binary Ionic Compounds with a Metal that Forms Only One Type of Cation
A binary ionic compound is a compound composed of a monatomic metal cation and a monatomic nonmetal anion . The metal cation is named first, followed by the nonmetal anion as illustrated in Figure \(\PageIndex{1}\) for the compound BaCl 2 . The word ion is dropped from both parts.
Subscripts in the formula do not affect the name.
Example \(\PageIndex{3}\): Naming Ionic Compounds
Name each ionic compound.
- CaCl 2
- AlF 3
- KCl
Solution
- Using the names of the ions, this ionic compound is named calcium chloride.
- The name of this ionic compound is aluminum fluoride.
- The name of this ionic compound is potassium chloride
Exercise \(\PageIndex{3}\)
Name each ionic compound.
- AgI
- MgO
- Ca 3 P 2
- Answer a:
- silver iodide
- Answer b:
- magnesium oxide
- Answer c:
- calcium phosphide
Naming Binary Ionic Compounds with a Metal That Forms More Than One Type of Cation
If you are given a formula for an ionic compound whose cation can have more than one possible charge, you must first determine the charge on the cation before identifying its correct name. For example, consider FeCl 2 and FeCl 3 . In the first compound, the iron ion has a 2+ charge because there are two Cl − ions in the formula (1− charge on each chloride ion). In the second compound, the iron ion has a 3+ charge, as indicated by the three Cl − ions in the formula. These are two different compounds that need two different names. By the Stock system, the names are iron(II) chloride and iron(III) chloride (Figure \(\PageIndex{2}\)).
| Name of cation (metal) + (Roman Numeral in parenthesis) + Base name of anion (nonmetal) and -ide | |
|---|---|
If we were to use the stems and suffixes of the common system, the names would be ferrous chloride and ferric chloride, respectively (Figure \(\PageIndex{3}\)) .
| "Old" base name of cation (metal) and -ic or -ous + Base name of anion (nonmetal) and -ide | |
|---|---|
|
-ous (for ions with lower charge)
|
-ic (for ions with higher charge)
|
Example \(\PageIndex{4}\):
Name each ionic compound.
- Co 2 O 3
- FeCl 2
Solution
| Explanation | Answer | |
|---|---|---|
| a |
We know that cobalt can have more than one possible charge; we just need to determine what it is.
|
cobalt(III) oxide |
| b |
Iron can also have more than one possible charge.
|
iron(II) chloride |
Exercise \(\PageIndex{4}\)
Name each ionic compound.
- AuCl 3
- PbO 2
- CuO
- Answer a:
- gold(III) chloride
- Answer b:
- lead(IV) oxide
- Answer c:
- copper(II) oxide
Naming Ionic Compounds with Polyatomic Ions
The process of naming ionic compounds with polyatomic ions is the same as naming binary ionic compounds. The cation is named first, followed by the anion. One example is the ammonium sulfate compound in Figure \(\PageIndex{6}\).
Example \(\PageIndex{5}\): Naming Ionic Compounds
Write the proper name for each ionic compound.
- (NH 4 ) 2 S
- AlPO 4 ,
- Fe 3 (PO 4 ) 2
Solution
| Explanation | Answer |
|---|---|
|
a. The ammonium ion has a 1+ charge and the sulfide ion has a 2− charge. Two ammonium ions need to balance the charge on a single sulfide ion. The compound’s name is ammonium sulfide. |
ammonium sulfide |
|
b. The ions have the same magnitude of charge, one of each (ion) is needed to balance the charges. The name of the compound is aluminum phosphate. |
aluminum phosphate |
|
c. Neither charge is an exact multiple of the other, so we have to go to the least common multiple of 6. To get 6+, three iron(II) ions are needed, and to get 6−, two phosphate ions are needed . The compound’s name is iron(II) phosphate. |
iron(II) phosphate |
Exercise \(\PageIndex{5A}\)
Write the proper name for each ionic compound.
- (NH 4 ) 3 PO 4
- Co(NO 2 ) 3
- Answer a:
- ammonium phosphate
- Answer b:
- cobalt(III) nitrite
Figure \(\PageIndex{1}\) is a synopsis of how to name simple ionic compounds.
Exercise \(\PageIndex{5B}\)
Name each ionic compound.
- ZnBr 2
- Al 2 O 3
- (NH 4 ) 3 PO 4
- AuF 3
- AgF
- Answer a:
- zinc bromide
- Answer b:
- aluminum oxide
- Answer c:
- ammonium phosphate
- Answer d:
- gold(III) fluoride or auric fluoride
- Answer e:
- silver fluoride
Summary
- Ionic compounds are named by stating the cation first, followed by the anion.
- Positive and negative charges must balance.
- Some anions have multiple forms and are named accordingly with the use of roman numerals in parentheses.
- Ternary compounds are composed of three or more elements.


24.6: Lorentz transformations and space-time
-
- Last updated
- Save as PDF
Four-dimensional space-time
So far, we have seen that our notions of time intervals (the time between two events) and space intervals (the distance between two locations) depend on our frame of reference. We also saw how space and time are connected, for example by the fact that time-dilation must go hand-in-hand with length contraction. We also concluded that there is no absolute concept of time, and that time is relative (depends on your frame of reference).
In the context of Special Relativity, we introduce the concept of space-time. To describe the location of an object in space-time, we must specify both the location/position coordinates ( \(x\) , \(y\) , \(z\) ) and the time “coordinate”, \(t\) . Since time, \(t\) , has the dimension of time, we usually specify the time coordinate by multiplying it by speed of light, \(ct\) , so that it has dimensions of length. Thus, position in space-time is given by 4 coordinates: \((x,y,z,ct)\) .
Space-time diagrams
It is practically impossible to visualize situations in three dimensions, so four dimensions is hopeless! However, we can gain a lot of insight into Special Relativity models by using “space-time diagrams”. In a space-time diagram, we use only one of the space coordinates (typically \(x\) ) along with the time coordinate, \(ct\) , to define the two axes of a space-time diagram. Space-time diagrams are analogous to “position as a function of time” graphs that one would draw in kinematics, although they are fundamentally different in that, for a space-time diagram, the coordinates should be thought of as independent (one is not plotting a dependent variable as a function of an independent variable).
Figure \(\PageIndex{1}\) shows a space-time diagram for an object that was located at position \(x=x_1\) at time \(t=t_1\) (location \(A\) ), and at position \(x=x_2\) at time \(t=t_2\) (location \(B\) ). The path of an object through space-time, indicated by the line that connects \(A\) and \(B\) , is called the “world line” of the object.
Exercise \(\PageIndex{1}\)
What does the world-line of a stationary particle look like?
- A vertical line.
- A horizontal line.
- A point.
- Answer
-
A pulse of light traveling in the \(x\) direction will always have a world-line that makes a \(45^{\circ}\) angle with the horizontal (space) axis (since \(x=ct\) ). The world line of any object that travels with a speed below the speed of light must always make an angle with the horizontal axis that is greater than \(45^{\circ}\) .
A position in space-time is usually called an “ event ”. We can draw a set of lines, at \(45^{\circ}\) degrees from the horizontal axis, that intersect at an event in space-time. Those lines define two “light cones” corresponding to: (1) locations in space-time in the past that could have had a causal effect on the event (the “past light cone”), and (2), locations in space-time in the future for which the event can have a causal effect (the “future light cone”).
Figure \(\PageIndex{2}\) shows the light cones associated with an event, \(A\) , in space-time. The past light cone is the only region of space-time in which a different event could have had an impact on the event \(A\) . For example, the event \(A\) might be that “the object is at position \(x=x_1\) at time \(t=t_1\) ”, so that the past light cone corresponds to the only locations in space-time that the object could have been in the past. Similarly, the future light-cone defines the locations in space-time upon which the event \(A\) could have an effect. For example, this could define the possible locations of the object in the future. The regions outside the light cones can never have an effect on the event \(A\) ; they are not causally connected. A signal or object would need to travel faster than the speed of light in order to have an effect on something outside of its light cone. There are locations in space-time, in the future of our Universe, that we cannot influence, no matter what we do.
When two events in space-time are within each other’s light cones, we say that the space-time interval between them (the line that you draw from one event to the other) is “time-like”. Time-like events are such that all observers, in any frame of reference, will agree that one event happened before the other. Thus, events that are causally related must have a time-like interval between them (they are connected by a line that makes an angle greater than \(45^{\circ}\) with the horizontal axis).
Two events that are outside of each other’s light cones are said to be “space-like”. Events that are connected by space-like intervals cannot be causally related (one cannot cause the other). Observers in different frames of reference will disagree on the time ordering of space-like events. For example, when Alice observed the two clocks on the platform to emit pulses of light at the same time, Brice disagreed; those two events are connected by a space-like interval.
Finally, the space-time interval between events that are on each other’s light-cone (connected by a line that makes a \(45^{\circ}\) angle with the x-axis), is said to be “light-like” or “null”.
Lorentz transformations
In this section, we consider how to transform the space-time coordinates, \((x,y,z,ct)\) , as measured in a frame of reference, \(S\) , to coordinate \((x',y',z',c't)\) , as measured in a frame of reference, \(S'\) , that is moving with a constant speed, \(v\) , relative to the frame, \(S\) . For simplicity, we assume that frame \(S'\) is moving with speed \(v\) in the positive \(x\) direction, as measured in frame, \(S\) , and that the origin of the two coordinate systems coincided at time \(t=0\) . Figure \(\PageIndex{3}\) shows an illustration of how the two frames of reference are related (note that these are actual coordinate systems, not space-time diagrams).
If we ignore any of Special Relativity, then the coordinates in \(S'\) are easily related to those in the \(S\) frame of reference using the “Galilean transformations”:
\[\begin{aligned} x' &= x - vt\\[4pt] y' &= y\\[4pt] z' &= z\\[4pt] t' &= t\end{aligned}\]
and this corresponds to transformations that we have implicitly used before considering Special Relativity. These equations also allow us to relate the speeds measured in different frames of reference. Suppose that an object has a velocity, \(\vec u=(u_x,u_y,u_z)\) , as measured in the frame of reference, \(S\) . We can obtain the components of the velocity vector, \(\vec u'\) , as measured in the frame of reference, \(S'\) , by taking the time derivatives of the above equations:
\[\begin{aligned} u'_x &= \frac{dx'}{dt'}=\frac{dx'}{dt}=\frac{d}{dt}(x-vt)=\frac{dx}{dt}-v=u_x - v\\[4pt] u'_y &=\frac{dy'}{dt'}=u_y\\[4pt] u'_z &=\frac{dz'}{dt'}=u_z\end{aligned}\]
which is trivial, since \(t=t'\) . The transformation above are equivalent (identical) to the rules for transforming velocity that we derived in Section 3.4 for kinematics. In Galilean relativity, time is an absolute quantity that does not depend on the frame of reference. In Special Relativity, the time coordinate is different between different frames of reference, so we cannot simply convert a time derivative in \(t'\) to a derivative in \(t\) . Instead, we must use the Lorentz transformations.
We can use the formulas for length contraction and time dilation to derive the Lorentz Transformations. Referring to Figure \(\PageIndex{3}\), \(x\) refers to the distance between a point in space-time and the origin of the \(x\) axis in the frame, \(S\) , as measured in frame, \(S\) . Similarly, \(x'\) , is the distance to the point in space-time as measured in frame \(S'\) , from the origin of \(S'\) . In frame, \(S\) , the distance \(x'\) will be contracted to the length \(x'/\gamma\) , so that the Galilean transformation for the \(x\) coordinate is modified as follows:
\[\begin{aligned} x' &= x - vt\quad\text{(Galilean)}\\[4pt] \frac{x'}{\gamma} &= x - vt\\[4pt] \therefore x'&=\gamma (x-vt)\quad\text{(Lorentz)}\end{aligned}\]
The \(y\) and \(z\) coordinates are the same between frames of references, since all of the length contraction will take place in the direction of the relative motion between frames of reference, which we chose to be in the \(x\) direction.
We can obtain the equation for the time coordinate by considering that, in the \(S'\) frame of reference, it is the \(x\) coordinate that is contracted to \(x/\gamma\) . In the \(S'\) frame of reference, the distance between the origins of the two systems is \(vt'\) (note the prime on \(t\) ). We can thus write the contracted distance \(x\) , in the \(S'\) frame of reference:
\[\begin{aligned} \frac{x}{\gamma}&=vt'+x'\\[4pt] t'&= \frac{1}{v}\left(\frac{x}{\gamma}-x'\right)\end{aligned}\]
We can eliminate \(x'\) from the last equation using the Lorentz transformation for \(x'\) that we just found:
\[\begin{aligned} t'&= \frac{1}{v}\left(\frac{x}{\gamma}-x'\right)\\[4pt] t'&= \frac{1}{v}\left(\frac{x}{\gamma}-\gamma x+\gamma vt\right)\\[4pt] \frac{t'}{\gamma}&=\frac{1}{v}\left(\frac{x}{\gamma^2}-x+vt\right)\\[4pt] &=\frac{1}{v}\left(x\left(1-\frac{v^2}{c^2}\right)-x+vt\right)\\[4pt] &=\frac{1}{v}\left(-\frac{v^2}{c^2}x+vt\right)\\[4pt] &=t-\frac{vx}{c^2}\\[4pt] \therefore t'&=\gamma\left(t-\frac{vx}{c^2} \right)\end{aligned}\]
where we wrote out the \(\gamma\) factor out explicitly in the fourth line. We can summarize the Lorentz transformations as follows:
\[\begin{aligned} x' &=\gamma (x-vt)\\[4pt] y' &= y\\[4pt] z' &= z\\[4pt] t' &=\gamma\left(t-\frac{vx}{c^2} \right)\end{aligned}\]
and the inverse relations are easily found:
\[\begin{aligned} x &=\gamma (x'+vt')\\[4pt] y &= y'\\[4pt] z &= z'\\[4pt] t &=\gamma\left(t'+\frac{vx'}{c^2} \right)\end{aligned}\]
Note that the Lorentz transformations reduce to the Galilean transformations when the speed, \(v\) , between frames of reference is small (so that \(\gamma \sim 1\) ).
Example \(\PageIndex{1}\)
In a frame, \(S\) , a pulse of light is emitted (at the speed of light) in the positive \(x\) direction, at \(t=0\) , from the origin. The pulse is then absorbed at time \(t\) , at position \(x=d\) . Use the Lorentz transformation to show that, in a frame, \(S'\) , moving in the positive \(x\) direction with speed \(v\) , relative to \(S\) , the pulse also traveled at the speed of light.
Solution
In order to use the Lorentz transformations, we need to define “events”, with coordinates in space-time, that we can then convert from one frame of reference to another. Let \(A\) be the event that corresponds to the emission of the pulse of light, and \(B\) the event that corresponds to the absorption of the pulse. In frame, \(S\) , the coordinates of these events are:
\[\begin{aligned} x_A&=0\\[4pt] t_A&=0\\[4pt] x_B&=d\\[4pt] t_B&=\frac{d}{c}\end{aligned}\]
where in the last line, we used the fact that, in frame, \(S\) , the pulse travels at the speed of light. Applying the Lorentz transformations, we can find the coordinates of the same events in frame, \(S'\) :
\[\begin{aligned} x'_A&=\gamma (x_A-vt_A)=0\\[4pt] t'_A&=\gamma\left(t_A-\frac{vx_A}{c^2} \right)=0\\[4pt] x'_B&=\gamma (x_B-vt_B)=\gamma \left(d-v\frac{d}{c}\right)\\[4pt] t'_B&=\gamma\left(t_B-\frac{vx_B}{c^2} \right)=\gamma\left(\frac{d}{c}-\frac{vd}{c^2} \right)\end{aligned}\]
The speed, \(v'_p\) , of the pulse of light in frame, \(S'\) , is given by:
\[\begin{aligned} v'_p&=\frac{(x'_B-x'_A)}{(t'_B-t'_A)}=\frac{\gamma \left(d-v\frac{d}{c}\right)}{\gamma\left(\frac{d}{c}-\frac{vd}{c^2} \right)}\\[4pt] &=\frac{\left(d-v\frac{d}{c}\right)}{\left(\frac{d}{c}-\frac{vd}{c^2} \right)}=c\frac{\left(\frac{d}{c}-v\frac{d}{c^2}\right)}{\left(\frac{d}{c}-\frac{vd}{c^2} \right)}=c\end{aligned}\]
which is the speed of light, as expected.
Discussion
In this example, we showed how to use the Lorentz transformations, by clearly defining “events” and their coordinates in space-time. We saw that the Lorentz transformation are consistent with Einstein’s second postulate and that the speed of light is the same all frames of reference. This of course makes sense, as we derived the Lorentz transformations from time dilation and length contraction, which are consequences of the postulate.
Einstein’s second postulate states that the speed of light is independent of the frame of reference. Consider two points in space-time corresponding to the emission ( \(A\) ) and the absorption ( \(B\) ) of a pulse of light. In the reference frame, \(S\) , the distance squared in space between these two events must be equal to the distance (squared) that light traveled between the time of emission and absorption:
\[\begin{aligned} (x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2&= c^2(t_B-t_A)^2\\[4pt] \therefore \Delta x^2 + \Delta y^2 +\Delta z^2&=c^2\Delta t^2\end{aligned}\]
where \((x_A,y_A,z_A,ct_A)\) and \((x_B,y_B,z_B,ct_B)\) are the space-time coordinates of events \(A\) and \(B\) . The above equation must hold in all frame of references (e.g. adding a prime to each coordinate), since it is a statement that the speed of light is \(c\) .
We can define, \(s\) , the ”space-time interval”, between events, \(A\) and \(B\) :
\[\begin{aligned} s^2 = \Delta x^2 + \Delta y^2 +\Delta z^2&-c^2\Delta t^2\end{aligned}\]
which turns out to be “Lorentz invariant” (meaning that this value is the same in all reference frames). The space-time interval can be thought of as a “distance” in space-time that is the same in all reference frames. If the events \(A\) and \(B\) corresponds to the emission and absorption of light, then \(s=0\) , and we say that the interval between \(A\) and \(B\) is light-like or null. If \(s<0\) , the events are on a time-like interval, and if \(s>0\) , the events are separated by a space-like interval. Since \(s\) does not depend on the frame of reference, all observers will agree on whether events are separated by time or space-like intervals.
We can visualize the effect of Lorentz transformations on space-time diagrams, as in Figure \(\PageIndex{4}\), which shows the space-time diagrams for a reference frame, \(S\) , and a second reference frame, \(S'\) , moving with speed \(v\) in the \(x\) direction relative to \(S\) .
The effect of the Lorentz transformation on a space-time diagram is to tilt both the space and time axes “inwards” 1 , by an angle, \(\alpha\) , given by:
\[\begin{aligned} \tan\alpha = \frac{v}{c}\end{aligned}\]
Figure \(\PageIndex{4}\) shows a light-like interval between two points, \(A\) and \(B\) , and how to determine the space-time coordinates in the two reference frames. You can think of space-time as the sheet of paper on which events happen. You can then draw different coordinate systems on that piece of paper to describe the position (in space and time) of different events.
Example \(\PageIndex{2}\)
Use a space-time diagram to show how two events at different locations that are simultaneous in one frame of reference are not simultaneous in a reference frame that is moving relative to the one where the events are simultaneous. This is an illustration of the relativity of simultaneity that we uncovered at the beginning of the chapter when examining Alice on a train platform and the two pulses of light.
Solution
Let \(S\) be the frame of reference where events, \(A\) and \(B\) , are simultaneous. These events are connected by a space-like interval, since they are separated in space, but not in time. There is no way for one event to have caused the other. In the frame, \(S\) , these events are on a horizontal line in a space-time diagram.
Let us define a second reference frame, \(S'\) , that is moving with speed \(v\) , relative to \(S\) . We have illustrated space-time diagrams for the the two reference frames, and the events \(A\) and \(B\) , in Figure \(\PageIndex{5}\).
From the space-time diagram, it is clear that, in the frame of reference, \(S'\) , the event \(B\) happened first. If the frame of reference \(S'\) was moving in the opposite (negative \(x\) ) direction, event \(A\) would occur first, and the axes of \(S'\) would be tilted in the other direction (so that the opening angle between the axes is greater than \(90^{\circ}\) ).
Discussion
This example illustrates how space-time diagrams can be used to qualitatively model events in space-time between two different reference frames. In particular, we showed how two events that are simultaneous in one frame ( \(S\) ) are not simultaneous in a different frame. For events connected by space-like intervals, there exist frames of reference where the events are simultaneous, or where either one happened first. If two events are separated by a time-like interval, there is no frame of reference in which one will disagree in the ordering of the events (although observers in different frames of reference will still measure different lengths of time between events due to time-dilation). For time-like events, the moving frame of reference would have to go faster than the speed of light for the time ordering to be different. This would violate causality, and is a good argument as to why nothing can go faster than the speed of light!
josh's thoughts
This chapter is where the bubble of intuitive reality is popped, and students (like you and me) are given the oppourtunity to challenge our understanding of how the universe operates. As amazing and exciting as this is, it can also be incredibly frustrating. Many students rely on intuition to guide them as they problem solve, but hit a wall in special relativity. To avoid this issue, I suggest drawing spacetime diagrams and using Lorentz transformations. Practicing with these tools will help make the process of understanding the strange consequences of Einstein’s postulates less awkward.
In addition to the practical advice I have given, I reccommend embracing the strangeness of reality. Throughout history, scientists have ventured into the unknown in attemps to discover and decode the universe. In many cases, the answer to a question has posed more questions than answers, and humanity is given the oppourunity to further understand the world we live in. As a student, you are participating in a process of understanding which allows us to continue the adventure that is scientific inquiry. Confusion can be frustrating, but don’t let it discourage you, being confused means that you’re only a few steps away from understanding!
Lorentz addition of velocities
In the previous section, we reviewed the Galilean velocity transformations, that allow us to convert a velocity, \(\vec u\) , as measured in one frame of reference, to a velocity, \(\vec u'\) , measured in a different frame of reference. We now derive the equivalent relations based on the Lorentz transformation. Again, we assume that frame, \(S'\) , moves in the positive \(x\) direction with speed, \(v\) , relative to frame, \(S\) .
The \(x\) component of the velocity vector, \(\vec u'\) , for some object in the \(S'\) frame of reference is given by:
\[\begin{aligned} u'_x = \frac{d}{dt'} x'\end{aligned}\]
In Galilean relativity, we could simply replace the derivative over \(t'\) by a derivative over \(t\) , since the two are equivalent. This is no longer the case. However, we can use the Chain Rule and the Lorentz transformations to convert a derivative over \(t'\) to a derivative over \(t\) :
\[\begin{aligned} \frac{d}{dt'} &= \frac{dt}{dt'}\frac{d}{dt}\\[4pt] &= \frac{d}{dt'} \gamma\left(t'+\frac{vx'}{c^2} \right)\frac{d}{dt}\\[4pt] &=\gamma \left(1+\frac{v}{c^2}\frac{dx'}{dt'}\right)\frac{d}{dt}\\[4pt] &=\gamma \left(1+\frac{vu'_x}{c^2}\right)\frac{d}{dt}\end{aligned}\]
where we recognized that \(\frac{dx'}{dt'}=u'_x\) . The \(x\) component of the velocity, as measured in the \(S'\) frame of reference, is then given by:
\[\begin{aligned} u'_x &= \frac{d}{dt'} x' = \gamma \left(1+\frac{vu'_x}{c^2}\right)\frac{d}{dt}x'\\[4pt] &=\gamma \left(1+\frac{vu'_x}{c^2}\right)\frac{d}{dt}\gamma (x-vt)\\[4pt] &=\gamma^2 \left(1+\frac{vu'_x}{c^2}\right)(u_x -v)\\[4pt] \frac{u'_x}{\gamma^2}&=u_x -v +\frac{vu'_xu_x}{c^2}-\frac{v^2u'_x}{c^2}\\[4pt] u'_x\left (1-\frac{v^2}{c^2}\right)&=u_x -v +\frac{vu'_xu_x}{c^2}-\frac{v^2u'_x}{c^2}\\[4pt] u'_x \left( 1- \frac{vu_x}{c^2}\right) &=u_x -v \end{aligned}\]
\[\therefore\mu_{x}^{'}=\frac{\mu_{x}-v}{1-\frac{v\mu_{x}}{c^{2}}}\]
where we made use of the Lorentz transformation: \(x'=\gamma (x-vt)\) . We can proceed in a similar way to determine the \(y\) and \(z\) components. Note that, unlike the Galilean case, all of the velocity components must transform, since the time derivative is involved for each component. Intuitively, we expect all components of velocity to be affected, since one needs to guarantee that the total speed is always below \(c\) . The velocity transformations for all components are given by the following:
\[\begin{aligned} u'_x&=\frac{u_x -v}{1- \frac{vu_x}{c^2}}\\[4pt] u'_y &=\frac{u_y}{\gamma\left(1- \frac{vu_x}{c^2}\right)}\\[4pt] u'_z &=\frac{u_z}{\gamma\left(1- \frac{vu_x}{c^2}\right)}\end{aligned}\]
and the reverse transformations are given by:
\[\begin{aligned} u_x &=\frac{u'_x +v}{1+ \frac{vu_x}{c^2}}\\[4pt] u_y &=\frac{u'_y}{\gamma\left(1+ \frac{vu_x}{c^2}\right)}\\[4pt] u_z &=\frac{u'_z}{\gamma\left(1+ \frac{vu_x}{c^2}\right)}\end{aligned}\]
Example \(\PageIndex{3}\)
An archer can shoot a very fast arrow with a speed of \(0.5c\) . The archer is on a train moving with speed, \(v=0.7c\) , and fires an arrow in the direction of motion. What is the speed of the arrow, as measured in the frame of reference of the ground?
Solution
Let the train be the frame of reference, \(S'\) , moving in the positive \(x\) direction with speed \(v=0.7c\) relative to the frame, \(S\) , which corresponds to the ground. The speed of the arrow, as seen from the train ( \(S'\) ), is given by:
\[\begin{aligned} u'_x = 0.5c\end{aligned}\]
The speed of the arrow, as measured from the ground, is thus given by:
\[\begin{aligned} u_x &=\frac{u'_x +v}{1+ \frac{vu_x}{c^2}}\\[4pt] &=\frac{(0.5c) +(0.7c)}{1+ \frac{(0.7c)(0.5c)}{c^2}}\\[4pt]` &=\frac{(1.2c)}{1+(0.7)(0.5)}=\frac{1.2}{1.35}c=0.89c\end{aligned}\]
Discussion
By using the Lorentz transformations for velocity, we see that the arrow does not exceed the speed of light. Had we used Galilean relativity, we would have concluded that the arrow has a speed of \(1.2c\) when measured from the ground.


3: Two-Dimensional Kinematics
The arc of a basketball, the orbit of a satellite, a bicycle rounding a curve, a swimmer diving into a pool, blood gushing out of a wound, and a puppy chasing its tail are but a few examples of motions along curved paths. In fact, most motions in nature follow curved paths rather than straight lines. Motion along a curved path on a flat surface or a plane (such as that of a ball on a pool table or a skater on an ice rink) is two-dimensional, and thus described by two-dimensional kinematics.
-
- 3.0: Prelude to Two-Dimensional Kinematics
- Motion not confined to a plane, such as a car following a winding mountain road, is described by three-dimensional kinematics. Both two- and three-dimensional kinematics are simple extensions of the one-dimensional kinematics developed for straight-line motion in the previous chapter. This simple extension will allow us to apply physics to many more situations, and it will also yield unexpected insights about nature.
-
- 3.2: Vector Addition and Subtraction- Graphical Methods
- A vector is a quantity that has magnitude and direction. Displacement, velocity, acceleration, and force, for example, are all vectors. In one-dimensional, or straight-line, motion, the direction of a vector can be given simply by a plus or minus sign. In two dimensions (2-d), however, we specify the direction of a vector relative to some reference frame (i.e., coordinate system), using an arrow having length proportional to the vector’s magnitude and pointing in the direction of the vector.
-
- 3.3: Vector Addition and Subtraction- Analytical Methods
- Analytical methods of vector addition and subtraction employ geometry and simple trigonometry rather than the ruler and protractor of graphical methods. Part of the graphical technique is retained, because vectors are still represented by arrows for easy visualization. However, analytical methods are more concise, accurate, and precise than graphical methods, which are limited by the accuracy with which a drawing can be made.
-
- 3.4: Projectile Motion
- Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of falling objects is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible.
-
- 3.5: Addition of Velocities
- Velocities in two dimensions are added using the same analytical vector techniques. Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame. Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move relative to one another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s).