3.7: Assessing the Fit of a Line (4 of 4)
Learning Objectives
- Use residuals, standard error, and r 2 to assess the fit of a linear model.
Introduction
Our final investigation into assessing the fit of the regression line focuses on typical error in the predictions.
Previously, we calculated the error in a single prediction by calculating
Residual = Observed value − Predicted value
But we use the regression line to make predictions even when we do not have an observed value, so we need a method for using all of the residuals to compute a typical amount of error.
We ask the question, How do we measure the typical amount of error for predictions from the regression line?
The most common measure of the size of the typical error is the standard error of the regression , which is represented by s e . It is calculated using the following formula:
where SSE stands for the sum of the squared errors.
Finding the standard error of the regression is similar to finding the standard deviation of a distribution of data points from a single quantitative variable. In Summarizing Data Graphically and Numerically , we learned that the standard deviation is roughly a measure of average distance about the mean . Here the standard error is roughly a measure of the average distance of the points about the regression line .
Let’s return to our example where age is used to predict the maximum distance for reading highway signs.
The residual plot for the highway sign data set is shown below. We can visualize the SSE in the formula as simply the sum of the squares of all of the vertical (residual) line segments. After dividing by n − 2, we have the average squared residual. Taking the square root then gives us a measure of the average size of the residuals.
In the case of the highway sign data, the value of s e is 51.35. In the figure below, we added horizontal lines at y = 51.35 and y = −51.35, so the red line represents the typical size of the error.
Comment: When we mark the s e on this residual plot, errors that fall outside of this range are larger than average. We see again that most of the errors that exceed ±51.35 are on the right. This illustrates that predictions of maximum reading distance for older drivers have larger error.
Note: Most statistics software computes r and r 2 and s e . Therefore, our focus is not on calculating but on understanding and interpreting.
Now let’s apply the standard error of the regression as a measurement of typical error.
Example
Highway Sign Visibility
Let’s take another look at the prediction we made earlier using the regression line equation:
Distance = 576 + (−3 * Age)
In a previous example, we predicted the maximum distance that a 60-year-old driver can read a highway sign. We plugged Age = 60 into the equation and found that
Predicted distance = 576 + (−3 * 60) = 396
The question we now ask is, How good is this prediction?
Unfortunately, there is no 60-year-old driver in the original data set of 30 drivers, so we cannot calculate the residual. Instead, we use the s e as a measurement of typical error.
Technology gives s e = 51.35.
So how good is the prediction for the 60-year-old driver? Based on the s e for this data, we estimate that our prediction of 396 feet is off by ±51 feet.
| Intro grade(%) | Upper grade(%) | Predictions | Error (Residual) | Error Squared | |
| Student 1 | 65 | 58 | 59.1 | −1.1 | 1.21 |
| Student 2 | 71 | 63 | 65.4 | −2.4 | 5.76 |
| Student 3 | 72 | 67 | 66.4 | 0.6 | 0.36 |
| Student 4 | 72 | 77 | 66.4 | 10.6 | 112.36 |
| Student 5 | 75 | 63 | 69.6 | −6.6 | 43.56 |
| Student 6 | 83 | 72 | 77.9 | −5.9 | 34.81 |
| Student 7 | 85 | 84 | 80 | 4 | 16 |
| Student 8 | 88 | 83 | 83.2 | −0.2 | 0.04 |
| Student 9 | 94 | 89 | 89.5 | −0.5 | 0.25 |
| Student 10 | 96 | 93 | 91.5 | 1.5 | 2.25 |
Let’s Summarize
- When we use a regression line to make predictions, there is error in the prediction. We calculate this error as Observed data value − Predicted value . A residual is another name for the prediction error.
- We use residual plots to determine whether a linear model is a good summary of the relationship between the explanatory and response variables. In particular, we look for any unexpected patterns in the residuals that may suggest the data is not linear in form.
-
We have two numeric measures to help us judge how well the regression line models the data.
- The square of the correlation coefficient, r 2 , is the proportion of the variation in the response variable that is explained by the least-squares regression line.
- The standard error of the regression, s e , gives a typical prediction error based on all of the data. It roughly measures the average distance of the data from the regression line. In this way, it is similar to the standard deviation, which roughly measures average distance from the mean.
Contributors and Attributions
- Concepts in Statistics. Provided by : Open Learning Initiative. Located at : http://oli.cmu.edu . License : CC BY: Attribution


4: Financial Statements of Business Organizations Last updated Save as PDF Page ID 45729 Lumen Learning Lumen Learning 4.1: Bookkeeping Terms and Phrases 4.2: Bookkeeping Journal Entries 4.3: Relationships Between Financial Statements 4.4: Key Information in Financial Statements 4.5: Introduction to Analyzing Data from Financial Statements 4.6: Reporting Inventories 4.7: Reporting Current Assets 4.8: Reporting Noncurrent Assets 4.9: Reporting Stockholder Equity 4.10: Reporting Unearned Revenue 4.11: Why It Matters- Financial Statements of Business Organizations 4.12: Reporting Expenses 4.13: Introduction to Other Financial Components 4.14: Core Financial Ratios 4.15: Calculating Return on Investment 4.16: Calculating Return on Equity 4.17: Other Key Ratios 4.18: Calculating Profitability Ratios 4.19: Limitations of Financial Statements 4.20: Putting It Together- Financial Statements of Business Organizations 4.21: Assignment- Financial Statements of Business Organizations 4.22: Introduction to Financial Statements 4.23: Transactions 4.24: Accounting Equation 4.25: Account Categories 4.26: Double Entry Bookkeeping System 4.27: Structures of Key Financial Statements 4.28: Introduction to the Bookkeeping Process 4.29: Expanded Accounting Equation


7.3: Energy in Gravitational Systems
Gravitational Potential Energy
We showed in Section 3.2 that our terrestrial model of gravity is a conservative force, and it certainly seems reasonable to assume that universal gravitation is as well, but really we should check to see if this is the case. In Section 3.6 we outlined a procedure for determining whether a force is conservative or not – basically it consists of trying to construct a potential energy function whose gradient equals the force, and if we succeed, then the force is conservative. If we can show that it is impossible to do this, then the force is non-conservative. Let's see what happens when we bring this to bear on gravitation.
Let's start by writing the gravity force in cartesian coordinates:
\[ \overrightarrow F \left(x,y,z\right) = -\dfrac{GMm}{r^2} \widehat r = -\dfrac{GMm}{r^3} \overrightarrow r = -\dfrac{GMm}{\left(x^2+y^2+z^2\right)^\frac{3}{2}} \left(x \widehat i + y \widehat j + z \widehat k \right) \]
Consider next the following partial derivative:
\[ \dfrac{\partial}{\partial x} \left(x^2+y^2+z^2\right)^{-\frac{1}{2}} = -\frac{1}{2}\left(x^2+y^2+z^2\right)^{-\frac{3}{2}}\left(2x\right) = \dfrac{-x}{\left(x^2+y^2+z^2\right)^{\frac{3}{2}}} \]
This is precisely the \(x\)-component of the gravitational force. Obviously partial derivatives with respect to \(y\) and \(z\) yield similar results – the \(y\) and \(z\) components of the gravitational force. This means we can immediately define a potential energy function whose negative gradient is the force:
\[ U\left(x,y,z\right) = \dfrac{-GMm}{\left(x^2+y^2+z^2\right)^\frac{1}{2}} + U_o \;\;\; \Rightarrow \;\;\; U\left(r\right) = \dfrac{-GMm}{r} + U_o\]
We could have saved ourselves a lot of trouble if we happened to know a useful fact from vector calculus: The gradient of a function that is purely a function of \(r\) can be written as:
\[ \overrightarrow \nabla U\left(r\right) = \dfrac{d}{dr} U\left(r\right) \widehat r \]
So:
\[ \overrightarrow F = -\overrightarrow \nabla U\left(r\right) = -\dfrac{d}{dr}\left[\dfrac{-GMm}{r}+U_o\right] \widehat r = -\dfrac{GMm}{r^2} \widehat r \]
As with other potential energy functions for another physical system that we have discussed (the intermolecular forces described by the Lennard-Jones potential), we typically choose our arbitrary constant of integration such that the potential energy falls off to zero at infinity. In the case of our gravitational potential energy, this gives what we will use for the potential energy function henceforth:
\[ U_{grav}\left(r\right) = \dfrac{-GMm}{r} \]
A graph of this function looks like this:
Figure 7.3.1 – Gravitational Potential
This function is significantly different from the "\(mgy\)" that we have been using up to now. To reconcile these two models, we need to make a note of the restriction of our terrestrial model of gravity – it holds for a region very close to the surface of the Earth, \(r=R_{Earth}\). Calling "\(h\)" the height of the object from the ground, the gravitational potential function above is cut-off at ground level. If we restrict ourselves to \(r\) values close to this point, the curve above is very close to a straight line:
Figure 7.3.2 – Gravitational Potential Near Earth's Surface
The negative slope of the potential energy curve is the force, so the slope of the straight line approximating the curve near \(r=R_{Earth}\) is the constant force in that region. Taking the negative derivative gives Newton's gravitational force, and when evaluated at \(r=R_{Earth}\), this force comes out to be \(mg\), as we saw in Equation 7.1.4 .
Bound and Unbound Gravitational Systems
With a graph of the potential energy which goes to zero at infinity, we are naturally drawn again to energy diagrams. There is a problem with jumping straight into this, however. Energy diagrams require 1-dimensional motion, and while \(U\left(r\right)\) is a function of a single variable, the motion is not 1-dimensional. To see where things break down, consider a closed elliptical orbit. As the planet moves toward the perihelion, the value of \(r\) gets smaller. If Figure 7.3.1 were the potential curve for an energy diagram, then when the planet is moving toward smaller values of \(r\), it would keep speeding up indefinitely as it approaches \(r=0\). Okay, so in practice it would hit the surface of the sun before it could "accelerate indefinitely," but still this does not represent orbital motion. In short, there is no part of this potential energy graph that represents the turnaround point that is the perihelion.
Fortunately, we have a nice trick to take care of this shortcoming. When we draw energy diagrams, the kinetic energy comes from the motion along the direction parallel to the one dimension. In this case of measuring energy along the radius, the kinetic energy for the energy diagram can only come from the part of the velocity that is radial . The total kinetic energy is the sum of the radial and tangential parts:
\[ KE_{tot} = KE_{radial} + KE_{tangential} \]
The tangential speed multiplied by the mass and the distance from the center is the angular momentum, which is a constant of the motion, so we have:
\[ \left. \begin{array}{l} KE_{tangential} = \frac{1}{2}mv_{tangential}^2 \\ L = mv_{tangential}r \end{array} \right\} \;\;\; \Rightarrow \;\;\; KE_{tangential} = \dfrac{L^2}{2mr^2} \]
If we now construct the total energy of the system, we have:
\[ E_{tot} = KE_{radial} + KE_{tangential} + U = \frac{1}{2}mv_{radial}^2 + \dfrac{L^2}{2mr^2} - \dfrac{GMm}{r} \]
Note that this can now be treated as 1-dimensional system, by combining the last two terms into a single function of \(r\) that we call an effective potential :
\[ U_{eff}\left(r\right) = \dfrac{L^2}{2mr^2} - \dfrac{GMm}{r} \]
Graphing this allows us to work exactly as before, though we have to keep in mind that whatever we determine the kinetic energy to be is really only a fraction of the kinetic energy. So, for example, at the turnaround points (where we have said the kinetic energy is zero), the kinetic energy is in fact the tangential term. This makes sense for closed orbits, since the orbiting body never actually stops moving entirely – it only stops moving radially . The graph of this effective potential has a rather familiar shape, and including the total system energy makes it an energy diagram.
Figure 7.3.3 – Energy Diagram of a Gravitating System
Figure 7.3.3 labels the turnaround points as the perihelion and aphelion, but there is much more we can extract from this diagram:
- circular motion
If the total energy lies at the bottom of the dip, then according to what we know about about energy diagrams, the kinetic energy is zero. But in this case, it means that the contribution to the kinetic energy by the radial component of velocity is zero. In other words, the orbit is circular. This fits with the fact that there is only one turnaround point, making the perihelion and aphelion the same distance.
- semi-major axis
The semi-major axis is the average of the perihelion and aphelion distances, so it is the value of \(r\) halfway between the two turnaround points on the graph.
- eccentricity
Looking at Figure 7.2.1 , we can write the perihelion and aphelion distances in terms of the eccentricity and the semi-major axis, which we can then invert to get the eccentricity in terms of the perihelion and aphelion distances and the semi-major axis. Doing this gives:
\[ e = \dfrac{r_{max} - r_{min}}{2a} \]
Looking at Figure 7.3.3 , we see that raising the total energy (but keeping it negative, and doing it without increasing the angular momentum, which would change the graph of the effective potential) increases the length of the semi-major axis, but much more of this change comes from the increase of the aphelion distance than from the decrease of the perihelion distance (especially close to the horizontal axis). This results in an increase of eccentricity. How does one increase the total energy without increasing the angular momentum? By giving the orbiting body a "kick" that points radially (inward or outward). This exerts no torque (the force is parallel to the position vector), so the angular momentum doesn't change, but the push increases the total energy, because it adds a radial component of velocity without changing the tangential component.
When a radial kick is given, a polar angle of zero in the same coordinate system will no longer correspond to the aphelion of the orbit – the orbital ellipse will be rotated. If the kick is outward, then at the position of the kick the orbiting body is still getting farther from the gravitating body, which means it is heading for the new aphelion, and the orbit has rotated in the direction of the orbit. That is, if the orbit was clockwise, then the major axis rotates clockwise. For a radially inward kick, the orbiting body is now getting closer to the gravitating body, which means it is coming from the aphelion, so the major has rotated the direction opposite to the orbit direction.
Figure 7.3.4 – Effect of Radial "Kicks" to Orbits
A tangential kick in the direction of motion would also raise the total energy, but it would also increase the angular momentum, increasing the positive term in the effective potential. This has the effect of raising the bottom of the curve, and if this is done properly (at the aphelion), the bottom of the curve comes up faster than the energy line goes up, bringing the eccentricity down. The eccentricity can even be brought all the way down to zero (i.e. a circular orbit) in this way (see Example 7.3.1 ).
- hyperbolic trajectory – unbounded
If the total energy is greater than zero, the orbiting body is not "bound" to the gravitating body. That is, after the orbiting body makes its closest approach, it zooms away, and although it slows down as it departs, it never stops moving away from the gravitating body. The height of the total energy line above the horizontal axis represents the finite kinetic energy of the orbiting body when it gets very far away, since the potential vanishes there. While the tangential part of the kinetic energy also goes to zero very far away, the angular momentum does not vanish – it remains conserved. With a finite speed and a non-zero angular momentum, the motion of the orbiting body must be asymptotically approaching a line that passes by the gravitating body, and we get what is called a hyperbolic trajectory . [ The word "trajectory" is generally preferred over "orbit," because the latter typically implies that the affected body is trapped by the gravitating body. ]
The orbit equation that we found for the elliptical orbit ( Equation 7.2.13 ) still works, but for the hyperbolic trajectory the eccentricity is greater than 1.
Figure 7.3.5 – Hyperbolic Trajectory
- parabolic trajectory – barely (un)bounded
If the total energy of the system equals exactly zero, then the orbiting body just barely runs out of velocity as it reaches infinity. It technically is neither bound nor unbound – it is at the borderline between the two. In this case, the eccentricity equals exactly 1.
An orbiting body is in an orbit where it is four times farther from the gravitating body at its aphelion than at its perihelion. By what percentage must its speed be increased at the aphelion to make its orbit circular?
- Solution
-
Since orbits are closed, if we instantaneously speed up the orbiting body at its aphelion, it will return to that same point with the same speed and moving in the same direction (i.e. its motion will still be perpendicular to the radius). That means that when the orbiting body returns, it will either be once again at its aphelion (if the speed was not increased substantially), or it will be at its perihelion (if the speed was increased a great deal). There is also an "in-between" increase whereby the perihelion and aphelion are the same – a circular orbit. We can compute the required increase in speed by comparing the speed at the aphelion of an eccentric orbit to the speed of a circular orbit whose radius is the same as the semi-major axis of the eccentric orbit. Starting with Equation 7.2.11 , we have:
\[ v_{min} = \left(1-e\right)\dfrac{GMm}{L} = \left(1-e\right)\dfrac{GMm}{mv_{min}r_{max}} = \left(\dfrac{1-e}{1+e}\right)\dfrac{GM}{v_{min}a} \;\;\; \Rightarrow \;\;\; v_{min} = \sqrt{\left(\dfrac{1-e}{1+e}\right)\dfrac{GM}{a}} \nonumber \]
For a circular orbit, the eccentricity is zero, which makes the ratio of the minimum velocity of the eccentric and circular orbits:
\[ \dfrac{v_{circular}}{v_{min}} = \dfrac{\sqrt{\dfrac{GM}{R}}}{\sqrt{\left(\dfrac{1-e}{1+e}\right)\dfrac{GM}{a}}} = \sqrt{\left(\dfrac{1+e}{1-e}\right)\dfrac{a}{R}} \nonumber \]
The new circular orbit must have a radius equal to the aphelion distance of the previous orbit, so:
\[ \dfrac{v_{circular}}{v_{min}} = \sqrt{\left(\dfrac{1+e}{1-e}\right)\dfrac{a}{\left(1+e\right)a}} = \sqrt{\dfrac{1}{1-e}} \nonumber \]
Now we need to know the eccentricity for an orbit where the aphelion distance is 4 times the perihelion distance. Writing these distances in terms of the semi-major axis gives:
\[ \left. \begin{array}{l} r_{min} = \left(1-e\right)a \\ r_{max} = \left(1+e\right)a \end{array} \right\} \;\;\; \Rightarrow \;\;\; 4 = \dfrac{r_{max}}{r_{min}} = \dfrac{1+e}{1-e} \;\;\; \Rightarrow \;\;\; e=\frac{3}{5} \nonumber \]
Plugging this in above gives us that the velocity must be increased by a factor of \(\sqrt{\frac{5}{2}}\) at the aphelion of the eccentric orbit to turn it into a circular orbit. This corresponds to a percentage increase of:
\[ \%\;increase = \left(\sqrt{\frac{5}{2}}-1\right)\times 100\% = \boxed{58\%} \nonumber \]
Suppose an orbiting body is bound by the gravitational attraction of a gravitating body that is a distance \(R\) away. If it is bound, it must be that it possesses insufficient kinetic energy such that when it is added to the (negative) gravitational potential the total energy makes it to zero. We define escape velocity as the minimum speed that an orbiting body must have in order to (barely) go infinitely far away from a gravitational source. Setting the total energy equal to zero, for the case of the object moving at escape velocity, we have:
Notice that whether or not an object can escape a gravitational attraction doesn't depend upon the would-be escaper's mass , but only its velocity . This makes for an interesting discussion when it comes to light. Light has no mass, so we would think that Newton's law of gravitation would indicate that it is unaffected by gravity. But when it comes to escape velocity, the mass does not factor in at all, so is the conclusion that light is unaffected by gravity incorrect?
It turns out that in fact light is affected by gravity (and specifically how this happens requires Einstein's improved theory of gravity), but what is more, we can compute whether light can escape a gravitating body. The speed of light is a well-defined constant, so the question of whether light can escape boils down to how close the origination of the light is to the gravitating body, and of course the mass of that body. The distance from which light will not escape is known as the Schwarzschild radius , and is found by plugging the speed of light (typically designated as \(c\)) into the escape velocity formula:
\[ c = \sqrt{\dfrac{2GM}{R}} \;\;\; \Rightarrow \;\;\; R_S = \dfrac{2GM}{c^2} \]
The Earth's mass is \(5.97 \times 10^{24} kg\), and the speed of light is \(3.0 \times 10^{8} \frac{m}{s}\). Its Schwarzschild radius therefore comes out to equal about 8.8 millimeters. Okay, that is quite small, and since the distance is measured from the center of the Earth, it is clear that we will not witness the phenomenon of light being trapped by the Earth's gravity. For light to be trapped by gravity, the gravitating body must fall inside the Schwarzschild radius. This requires that the gravitating body be incredibly dense, and since the only force available to pull the matter into that kind of density is the same gravity force, generally a very large amount of mass is required. Assuming this occurs, the object thus created is called a black hole . This is a nicely descriptive name, as "black" indicates that light does not escape its gravitational influence. Explaining how appropriate the word "hole" is in this name requires some knowledge of Einstein's theory, which is unfortunately beyond the scope of this work.
Gravitational Slingshots
Let's look at a simple model that demonstrates the process of a gravitational slingshot. This is the phenomenon that was exploited by NASA scientists to get the Voyager probes past to escape velocity for the Sun, despite not being launched with sufficient kinetic energy at the outset. It is also thought to be the likely candidate for so-called "rogue planets" that once orbited the sun, and are now moving through interstellar space, no longer held in orbit.
Start with three gravitating particles, two of which initially are in orbit around the third, in opposite directions, roughly in the same (circular) orbital path. One of these orbiting bodies (Jupiter) is much more massive than the other (Voyager), and of course the Sun is much more massive than both of these. At this point Voyager, being in a closed orbit, does not have sufficient energy to escape the Sun.
Figure 7.3.6 – Voyager Far from Jupiter (Sun's Perspective)
At this stage, voyager is quite far from Jupiter (say roughly the same distance as from the Sun), and since the mass of the Sun is so much greater than that of Jupiter (about 1000 times greater), Voyager's motion is, to a very good approximation, governed only by the gravitational force of the Sun. So it merrily continues along its orbital path, and since it is in roughly the same orbit as Jupiter, it is moving at the same speed, which in the diagram we have called "\(v_o\)."
When voyager comes significantly closer to Jupiter, then even though Jupiter has much less mass than the Sun, the much closer proximity to Jupiter makes it such that Voyager's motion is to a very good approximation governed entirely by Jupiter's gravitational force. From Jupiter's perspective, Voyager is coming toward it at a relative speed of \(2v_o\) (before Jupiter's gravity starts to speed it up significantly).
Figure 7.3.7 – Voyager Close to Jupiter (Jupiter's Perspective)
Naturally Voyager is not in exactly the same orbit as Jupiter, or it would crash into it. The slight offset results in Voyager slinging around the planet and coming out the opposite direction. Since Voyager essentially only "knows" about Jupiter's presence, (approximate) conservation of energy in this two-body system results in Voyager emerging with the same speed it had on its way in, but in the opposite direction (this is essentially the same as an elastic collision between two objects, where one has much, much less mass than the other).
So now Voyager is moving at the same speed as before relative to Jupiter, but in the opposite direction. As it gets farther away, Jupiter's gravitational force wanes, and attention again returns to the Sun. But now Voyager has three times the speed it had before relative to the Sun (\(2v_o\) relative to Jupiter, plus Jupiter's \(v_o\), for a total of \(3v_o\)), and it is the same distance away from the Sun as before.
Figure 7.3.8 – Voyager After Jupiter's "Kick" (Sun's Perspective)
Jupiter naturally loses a little of its speed (\(\delta v\) – a completely negligible amount, due to the vast difference in mass it has with Voyager) in this exchange. But Voyager's new speed is now more than enough to escape the Sun's gravitational pull. The overall energy of the system remains conserved, but it is redistributed so that voyager is no longer in a bound state. In the language of Chapter 3, we can say that the total force exerted on Voyager by the solar system (Sun + Jupiter system) is non-conservative, because when Voyager returns to its original position relative to this system, it has more kinetic energy than before, so the work done for that closed path was not zero. The internal energy of the solar system goes down (Jupiter's motion slows), and it goes to the kinetic energy of Voyager.
Exercise
Show that tripling the speed of any satellite in a circular orbit is enough to allow it to escape the gravitating body.
- Solution
-
From Kepler's Third Law for a circular orbit ( Equation 7.2.16 ), we can compute the speed of a satellite in a circular orbit of radius \(R\). It is the circumference divided by the period of the orbit, so:
\[\dfrac{R^3}{T^2}=\dfrac{GM}{4\pi^2}\;\;\;\Rightarrow\;\;\;v=\dfrac{2\pi R}{T}=\sqrt{\dfrac{GM}{R}}\nonumber\]
Comparing this to the formula for escape velocity ( Equation 7.3.12 ), we see that escape velocity is only greater than the circular orbital speed by a factor of \(\sqrt 2\). So increasing the orbital speed by a factor of 3 is more than enough to get the job done.


10.3: Permutations
In this section you will learn to
- Count the number of possible permutations (ordered arrangement) of n items taken r at a time
- Count the number of possible permutations when there are conditions imposed on the arrangements
- Perform calculations using factorials
In Example 10.2.6 of section 10.2, we were asked to find the word sequences formed by using the letters { A, B, C } if no letter is to be repeated. The tree diagram gave us the following six arrangements.
ABC, ACB, BAC, BCA, CAB, and CBA.
Arrangements like these, where order is important and no element is repeated, are called permutations.
A permutation of a set of elements is an ordered arrangement where each element is used once.
How many three-letter word sequences can be formed using the letters { A, B, C, D }?
Solution
There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third.
| 4 | 3 | 2 |
Applying the multiplication axiom, we get \(4 \cdot 3 \cdot 2 = 24\) different arrangements.
How many permutations of the letters of the word ARTICLE have consonants in the first and last positions?
Solution
In the word ARTICLE, there are 4 consonants.
Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. We show as follows:
| 4 | 3 |
Since there are no more restrictions, we can go ahead and make the choices for the rest of the positions.
So far we have used up 2 letters, therefore, five remain. So for the next position there are five choices, for the position after that there are four choices, and so on. We get
| 4 | 5 | 4 | 3 | 2 | 1 | 3 |
So the total permutations are \(4 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 3 = 1440\).
Given five letters { A, B, C, D, E }. Find the following:
- The number of four-letter word sequences.
- The number of three-letter word sequences.
- The number of two-letter word sequences.
Solution
The problem is easily solved by the multiplication axiom, and answers are as follows:
- The number of four-letter word sequences is \(5 \cdot 4 \cdot 3 \cdot 2 = 120\).
- The number of three-letter word sequences is \(5 \cdot 4 \cdot 3 = 60\).
- The number of two-letter word sequences is \(5 \cdot 4 = 20\).
We often encounter situations where we have a set of n objects and we are selecting r objects to form permutations. We refer to this as permutations of n objects taken r at a time , and we write it as nPr .
Therefore, the above example can also be answered as listed below.
- The number of four-letter word sequences is 5 P 4 = 120.
- The number of three-letter word sequences is 5 P 3 = 60.
- The number of two-letter word sequences is 5 P 2 = 20.
Before we give a formula for nPr, we'd like to introduce a symbol that we will use a great deal in this as well as in the next chapter.
\[\mathrm{n} !=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) \cdots 3 \cdot 2 \cdot 1 \nonumber \]
where \(n\) is a natural number.
\[0! = 1 \nonumber \]
Now we define nPr.
The Number of Permutations of n Objects Taken r at a Time
\[\mathrm{ n P r }=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) \cdots(\mathrm{n}-\mathrm{r}+1) \nonumber \]
or
\[\mathrm{ n P r }=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !} \nonumber \]
where \(n\) and \(r\) are natural numbers.
The reader should become familiar with both formulas and should feel comfortable in applying either.
Compute the following using both formulas.
- 6 P 3
- 7 P 2
Solution
We will identify \(n\) and \(r\) in each case and solve using the formulas provided.a. 6 P 3 = \(6 \cdot 5 \cdot 4 = 120\), alternately
\[ 6 \mathrm{P} 3=\frac{6 !}{(6-3) !}=\frac{6 !}{3 !}=\frac{6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1}=120 \nonumber \]
b. 7 P 2 = \(7 \cdot 6 = 42\), or
\[7 \mathrm{P} 2=\frac{7 !}{5 !}=\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}=42 \nonumber \]
Next we consider some more permutation problems to get further insight into these concepts.
In how many different ways can 4 people be seated in a straight line if two of them insist on sitting next to each other?
Solution
Let us suppose we have four people A, B, C, and D. Further suppose that A and B want to sit together. For the sake of argument, we tie A and B together and treat them as one person.The four people are \(\boxed{AB}\) CD. Since \(\boxed{AB}\) is treated as one person, we have the following possible arrangements.
\[ \boxed{AB} CD, \boxed{AB} DC, C \boxed{AB}D, D\boxed{AB}C, CD \boxed{AB}, DC\boxed{AB} \nonumber \]
Note that there are six more such permutations because A and B could also be tied in the order BA. And they are
\[ \boxed{BA}CD, \boxed{BA} DC, C\boxed{BA}D, D\boxed{BA}C, CD\boxed{BA}, DC \boxed{BA} \nonumber \]
So altogether there are 12 different permutations.
Let us now do the problem using the multiplication axiom.
After we tie two of the people together and treat them as one person, we can say we have only three people. The multiplication axiom tells us that three people can be seated in 3! ways. Since two people can be tied together 2! ways, there are 3! 2! = 12 different arrangements
You have 4 math books and 5 history books to put on a shelf that has 5 slots. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books?
Solution
We first do the problem using the multiplication axiom.
Since the math books go in the first three slots, there are 4 choices for the first slot,
3 choices for the second and 2 choices for the third.
The fourth slot requires a history book, and has five choices. Once that choice is made, there are 4 history books left, and therefore, 4 choices for the last slot. The choices are shown below.
| 4 | 3 | 2 | 5 | 4 |
Therefore, the number of permutations are \(4 \cdot 3 \cdot 2 \cdot 5 \cdot 4 = 480\).
Alternately, we can see that \(4 \cdot 3 \cdot 2\) is really same as 4 P3, and \(5 \cdot 4\) is 5 P 2 .
So the answer can be written as ( 4 P 3 ) ( 5 P2) = 480.
Clearly, this makes sense. For every permutation of three math books placed in the first three slots, there are 5 P2 permutations of history books that can be placed in the last two slots. Hence the multiplication axiom applies, and we have the answer ( 4 P 3 ) ( 5 P 2 ).
We summarize the concepts of this section:
1. Permutations
A permutation of a set of elements is an ordered arrangement where each element is used once.
2. Factorial
\[\mathrm{n} !=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) \cdots 3 \cdot 2 \cdot 1 \nonumber \]
Where \(n\) is a natural number.
\[0! = 1 \nonumber \]
3. Permutations of n Objects Taken r at a Time
\[\mathrm{nPr}=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3) \cdots(\mathrm{n}-\mathrm{r}+1) \nonumber \]
or
\[\mathrm{nPr}=\frac{\mathrm{n} !}{(\mathrm{n}-\mathrm{r}) !} \nonumber \]
where \(n\) and \(r\) are natural numbers.


25.2D: Movement of Fluid Among Compartments
How fluid moves through compartments depends on several variables described by Starling’s equation.
- Describe the movement of fluid between extracellular compartments
Key Points
- Interstitial fluid is formed when hydrostatic pressure generated by the heart pushes water out of the capillaries. The water passes from a high concentration outside of the vessels to a low concentration inside of the vessels, but equilibrium is never reached because the constant blood flow.
- Osmotic pressure works opposite to hydrostatic pressure to hold water and substances in the capillaries.
- Hydrostatic pressure is stronger in the arterial ends of the capillaries, while osmotic pressure is stronger at the venous ends of the capillaries.
- Interstitial fluid is removed through the surrounding lymph vessels, and eventually ends up rejoining the blood. Sometimes the removal of tissue fluid does not function correctly and there is a buildup, called edema.
- The Starling equation describes the pressure gradients that drive the movement of water across fluid compartments.
Key Terms
- Starling equation : An equation that illustrates the role of hydrostatic and oncotic forces in the movement of fluid across capillary membranes.
- interstitial fluid : A solution that bathes and surrounds the cells of multicellular animals.
Fluid Movement
Extracellular fluid is separated among the various compartments of the body by membranes. These membranes are hydrophobic and repel water; however, there a few ways that fluids can move between body compartments. There are small gaps in membranes, such as the tight junctions, that allow fluids and some of their contents to pass through membranes by way of pressure gradients.
Formation of Interstitial Fluid
Hydrostatic pressure is generated by the contractions of the heart during systole. It pushes water out of the small tight junctions in the capillaries. The water potential is created due to the ability of the small solutes to pass through the walls of capillaries.
This buildup of solutes induces osmosis. The water passes from a high concentration (of water) outside of the vessels to a low concentration inside of the vessels, in an attempt to reach an equilibrium. The osmotic pressure drives water back into the vessels. Because the blood in the capillaries is constantly flowing, equilibrium is never reached.
The balance between the two forces differs at different points on the capillaries. At the arterial end of a vessel, the hydrostatic pressure is greater than the osmotic pressure, so the net movement favors water and other solutes being passed into the tissue fluid.
At the venous end, the osmotic pressure is greater, so the net movement favors substances being passed back into the capillary. This difference is created by the direction of the flow of blood and the imbalance in solutes created by the net movement of water that favors the tissue fluid.
Removal of Interstitial Fluid
The lymphatic system plays a part in the transport of tissue fluid by preventing the buildup of tissue fluid that surrounds the cells in the tissue. Tissue fluid passes into the surrounding lymph vessels and eventually rejoins the blood.
Sometimes the removal of tissue fluid does not function correctly and there is a buildup, which is called edema. Edema is responsible for the swelling that occurs during inflammation, and in certain diseases where the lymphatic drainage pathways are obstructed.
Starling Equation
The Starling model : Note the concentration of interstitial solutes (orange) increases proportionally to the distance from the arteriole.
Capillary permeability can be increased by the release of certain cytokines, anaphylatoxins, or other mediators (such as leukotrienes, prostaglandins, histamine, bradykinin, etc.) that are released by cells during inflammation. The Starling equation defines the forces across a semipermeable membrane to calculate the net flux.
The solution to the equation is known as the net filtration or net fluid movement. If positive, fluid will tend to leave the capillary (filtration). If negative, fluid will tend to enter the capillary (absorption). This equation has a number of important physiologic implications, especially when disease processes grossly alter one or more of the variables.
Capillary dynamics : Oncotic pressure exerted by the proteins in blood plasma tends to pull water into the circulatory system.
This is a diagram of the Starling model. Note how the concentration of interstitial solutes increases proportionally to the distance from the arteriole.
According to Starling’s equation, the movement of fluid depends on six variables:
- Capillary hydrostatic pressure (Pc)
- Interstitial hydrostatic pressure (Pi)
- Capillary oncotic pressure (πz)
- Interstitial oncotic pressure (πi)
- Filtration coefficient (Kf)
- Reflection coefficient (σ)
The Starling Equation is mathematically described as Flux=Kf[(Pc-Pi)-σ (πz-πi)].
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5: Comparing Associations Between Multiple Variables
Hello everyone, and welcome to the fifth chapter of the University of Southern Queensland’s online, open access textbook on statistics for research students. The aim of this chapter is to discuss associations between three or more variables.
As we have discussed in previous chapters, associations are mathematical relationships between variables. Generally, relationships are framed around a key pair of variables that explain a central effect of interest. For example, we have discussed that reading ability and test performance may be associated. However, there are other variables that could be associated with either reading ability, or test performance, or there may be other variables that are associated with both reading ability and test performance. Such associated variables are often termed covariates. Covariates are often used as additional variables of interest in Multiple Regression and Hierarchical Regression.
There are three main methods of regression analysis that correspond to three different types of models:
- Simple or Basic Regression : a regression model with one independent variable and one dependent variable.
- Multiple Regression : a regression model with two or more independent variables and one dependent variable.
- Hierarchical Regression : a regression model with two or more independent variables entered within two or more blocks of sequential predictors, and one dependent variable.
There are some slides that appear via links within Chapter Five. Please look for these as you review the current chapter.


2.6.3: Cost Leadership
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- Jamie Hammond and Western Technical College
- LibreTexts
Learning Objectives
- Describe the nature of cost leadership.
- Understand how economies of scale help contribute to a cost leadership strategy.
- Know the advantages and disadvantages of a cost leadership strategy.
Table 5.2 Cost Leadership
Firms that compete based on price and target a broad target market are following a cost leadership strategy. Several examples of firms pursuing a cost leadership strategy are illustrated below.
| Despite its name, Dunkin’ Donuts makes more money selling inexpensive coffee than it does from selling donuts. The coffee is often advertised as costing under a dollar, making Dunkin’ Donuts a low-priced alternative to Starbucks. | Payless ShoeSource is a discount retailer tha sells inexpensive shoes for men, women, and children. Their advertising slogans such as “Why pay more when you can Payless?” and “You could pay more, buy why?” consistently preach a low-price message. |
|
Supercuts’ website makes clear their longstanding cost leadership strategy by noting, “A Supercut is a haircut that has kept people looking their best, while keeping money in their pockets, since 1975.” |
Little Debbie snack cakes began when O.D. McKee started selling treats for five cents each in the early 1930s. Little Debbie cakes cost a lot more than five cents today, but they remain cheaper than similar offerings from Entenmann’s, Tastykake, and other snack cake rivals. |
The Nature of the Cost Leadership Strategy
It is tempting to think of cost leaders as companies that sell inferior, poor-quality goods and services for rock-bottom prices. The Yugo, for example, was an extremely unreliable car that was made in Eastern Europe and sold in the United States for about $4,000. Despite its attractive price tag, the Yugo was a dismal failure because drivers simply could not depend on the car for transportation. Yugo exited the United States in the early 1990s and closed down entirely in 2008.
In contrast to firms such as Yugo whose failure is inevitable, cost leaders can be very successful. A firm following a cost leadership strategy offers products or services with acceptable quality and features to a broad set of customers at a low price (Table 5.2). Payless ShoeSource, for example, sells name-brand shoes at inexpensive prices. Its low-price strategy is communicated to customers through advertising slogans such as “Why pay more when you can Payless?” and “You could pay more, but why?” Little Debbie snack cakes offer another example. The brand was started in the 1930s when O. D. McKee began selling sugary treats for five cents. Most consumers today would view the quality of Little Debbie cakes as a step below similar offerings from Entenmann’s, but enough people believe that they offer acceptable quality that the brand is still around eight decades after its creation.
Listeners of the popular radio show Car Talk voted the Yugo as the “worst car of the millennium.”
Wikimedia Commons – public domain.
Perhaps the most famous cost leader is Walmart, which has used a cost leadership strategy to become the largest company in the world. The firm’s advertising slogans such as “Always Low Prices” and “Save Money. Live Better” communicate Walmart’s emphasis on price slashing to potential customers. Meanwhile, Walmart has the broadest customer base of any firm in the United States. Approximately one hundred million Americans visit a Walmart in a typical week (Zimmerman & Hudson, 2006). Incredibly, this means that roughly one-third of Americans are frequent Walmart customers. This huge customer base includes people from all demographic and social groups within society. Although most are simply typical Americans, the popular website http://www.peopleofwalmart.com features photos of some of the more outrageous characters that have been spotted in Walmart stores.
Cost leaders tend to share some important characteristics. The ability to charge low prices and still make a profit is challenging. Cost leaders manage to do so by emphasizing efficiency. At Waffle House restaurants, for example, customers are served cheap eats quickly to keep booths available for later customers. As part of the effort to be efficient, most cost leaders spend little on advertising, market research, or research and development. Waffle House, for example, limits its advertising to billboards along highways. Meanwhile, the simplicity of Waffle House’s menu requires little research and development.
Many cost leaders rely on economies of scale to achieve efficiency. Economies of scale are created when the costs of offering goods and services decreases as a firm is able to sell more items. This occurs because expenses are distributed across a greater number of items. Walmart spent approximately $2 billion on advertising in 2008. This is a huge number, but Walmart is so large that its advertising expenses equal just a tiny fraction of its sales. Also, cost leaders are often large companies, which allows them to demand price concessions from their suppliers. Walmart is notorious for squeezing suppliers such as Procter & Gamble to sell goods to Walmart for lower and lower prices over time. The firm passes some of these savings to customers in the form of reduced prices in its stores.
Advantages and Disadvantages of Cost Leadership
Each generic strategy offers advantages that firms can potentially leverage to enhance their success as well as disadvantages that may undermine their success. In the case of cost leadership, one advantage is that cost leaders’ emphasis on efficiency makes them well positioned to withstand price competition from rivals (Table 5.3). Kmart’s ill-fated attempt to engage Walmart in a price war ended in disaster, in part because Walmart was so efficient in its operations that it could live with smaller profit margins far more easily than Kmart could.
Table 5.3 Executing a Low-Cost Strategy
Using a cost leadership strategy offers firms important advantages and disadvantages. Below we illustrate a few examples in relation to entertainment and leisure.
| Advantages |
High profits can be enjoyed if a cost leader has a high market share. An example is Kampgrounds of America, a chain of nearly 500 low cost camping franchises in the United States.
Low-cost firms such as many municipal golf courses can withstand price wars because high-priced competitors will not want to compete directly with a more efficient rival. |
| Disadvantages |
If perceptions of quality become too low, business will suffer.
Large volumes of sales are a must because margins are slim. The need to keep expenses low might lead cost leaders to be late in detecting key environment trends. Low-cost firms’ emphasis on efficiency makes it difficult for them to change quickly if needed. |
Beyond existing competitors, a cost leadership strategy also creates benefits relative to potential new entrants. Specifically, the presence of a cost leader in an industry tends to discourage new firms from entering the business because a new firm would struggle to attract customers by undercutting the cost leaders’ prices. Thus a cost leadership strategy helps create barriers to entry that protect the firm—and its existing rivals—from new competition.
Challenging a cost leader in a price war may end up destroying a company.
WFIU Public Radio – Closed Sign – CC BY-NC 2.0.
In many settings, cost leaders attract a large market share because a large portion of potential customers find paying low prices for goods and services of acceptable quality to be very appealing. This is certainly true for Walmart, for example. The need for efficiency means that cost leaders’ profit margins are often slimmer than the margins enjoyed by other firms. However, cost leaders’ ability to make a little bit of profit from each of a large number of customers means that the total profits of cost leaders can be substantial.
In some settings, the need for high sales volume is a critical disadvantage of a cost leadership strategy. Highly fragmented markets and markets that involve a lot of brand loyalty may not offer much of an opportunity to attract a large segment of customers. In both the soft drink and cigarette industries, for example, customers appear to be willing to pay a little extra to enjoy the brand of their choice. Lower-end brands of soda and cigarettes appeal to a minority of consumers, but famous brands such as Coca-Cola, Pepsi, Marlboro, and Camel still dominate these markets. A related concern is that achieving a high sales volume usually requires significant upfront investments in production and/or distribution capacity. Not every firm is willing and able to make such investments.
Cost leaders tend to keep their costs low by minimizing advertising, market research, and research and development, but this approach can prove to be expensive in the long run. A relative lack of market research can lead cost leaders to be less skilled than other firms at detecting important environmental changes. Meanwhile, downplaying research and development can slow cost leaders’ ability to respond to changes once they are detected. Lagging rivals in terms of detecting and reacting to external shifts can prove to be a deadly combination that leaves cost leaders out of touch with the market and out of answers.
Key Takeaway
- Cost leadership is an effective business-level strategy to the extent that a firm offers low prices, provides satisfactory quality, and attracts enough customers to be profitable.
Exercises
- What are three industries in which a cost leadership strategy would be difficult to implement?
- What is your favorite cost leadership restaurant?
- Name three examples of firms conducting a cost leadership strategy that use no advertising. Should they start advertising? Why or why not?
References
Zimmerman, Ann and Kris Hudson, “Managing Wal-Mart: How US-store chief hopes to fix Wal-Mart,” Wall Street Journal , April 17, 2006.


2.4: Fitting Linear Models to Data
In the real world, rarely do things follow trends perfectly. When we expect the trend to behave linearly, or when inspection suggests the trend is behaving linearly, it is often desirable to find an equation to approximate the data. Finding an equation to approximate the data helps us understand the behavior of the data and allows us to use the linear model to make predictions about the data, inside and outside of the data range.
Example \(\PageIndex{1}\)
The table below shows the number of cricket chirps in 15 seconds, and the air temperature, in degrees Fahrenheit (Selected data from classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010). Plot this data, and determine whether the data appears to be linearly related.
| chirps | 44 | 35 | 20.4 | 33 | 31 | 35 | 18.5 | 37 | 26 |
| Temp | 80.5 | 70.5 | 57 | 66 | 68 | 72 | 52 | 73.5 | 53 |
Solution
Plotting this data, it appears there may be a trend, and that the trend appears roughly linear, though certainly not perfectly so.
The simplest way to find an equation to approximate this data is to try to “eyeball” a line that seems to fit the data pretty well, then find an equation for that line based on the slope and intercept.
You can see from the trend in the data that the number of chirps increases as the temperature increases. As you consider a function for this data you should know that you are looking at an increasing function or a function with a positive slope.
flashback
- What descriptive variables would you choose to represent Temperature & Chirps?
- Which variable is the independent variable and which is the dependent variable?
- Based on this data and the graph, what is a reasonable domain & range?
- Based on the data alone, is this function one-to-one, explain?
- Answer
-
1. a. T = Temperature, C = Chirps (answers may vary)
b. Independent (Chirps) , Dependent (Temperature)
c. Reasonable Domain (18.5, 44) , Reasonable Range (52, 80.5) (answers may vary)
d. NO, it is not one-to-one, there are two different output values for 35 chirps.
Example \(\PageIndex{2}\)
Using the table of values from the previous example, find a linear function that fits the data by “eyeballing” a line that seems to fit.
Solution
On a graph, we could try sketching in a line. Note the scale on the axes have been adjusted to start at zero to include the vertical axis and vertical intercept in the graph.
Using the starting and ending points of our “hand drawn” line, points (0, 30) and (50, 90), this graph has a slope of \(m=\dfrac{60}{50} =1.2\) and a vertical intercept at 30, giving an equation of
\[T(c)=30+1.2c\nonumber \]
where \(c\) is the number of chirps in 15 seconds, and \(T(c)\) is the temperature in degrees Fahrenheit.
This linear equation can then be used to approximate the solution to various questions we might ask about the trend. While the data does not perfectly fall on the linear equation, the equation is our best guess as to how the relationship will behave outside of the values we have data for. There is a difference, though, between making predictions inside the domain and range of values we have data for, and outside that domain and range.
Definition: Interpolation and Extrapolation
Interpolation: When we predict a value inside the domain and range of the data
Extrapolation: When we predict a value outside the domain and range of the data
For the Temperature as a function of chirps in our hand drawn model above,
- Interpolation would occur if we used our model to predict temperature when the values for chirps are between 18.5 and 44.
- Extrapolation would occur if we used our model to predict temperature when the values for chirps are less than 18.5 or greater than 44.
Example \(\PageIndex{3}\)
- Would predicting the temperature when crickets are chirping 30 times in 15 seconds be interpolation or extrapolation? Make the prediction, and discuss if it is reasonable.
- Would predicting the number of chirps crickets will make at 40 degrees be interpolation or extrapolation? Make the prediction, and discuss if it is reasonable.
Solution
With our cricket data, our number of chirps in the data provided varied from 18.5 to 44. A prediction at 30 chirps per 15 seconds is inside the domain of our data, so would be interpolation. Using our model:
\[T(3)=30+1.2(30)=66\text{ degrees}\nonumber \]
Based on the data we have, this value seems reasonable.
The temperature values varied from 52 to 80.5. Predicting the number of chirps at 40 degrees is extrapolation since 40 is outside the range of our data. Using our model:
\[\begin{array} {rcl} {40} &= & {30 + 1.2c} \\ {10} &= & {1.2c} \\ {c} &\approx & {8.33} \end{array}\nonumber \]
Our model predicts the crickets would chirp 8.33 times in 15 seconds. While this might be possible, we have no reason to believe our model is valid outside the domain and range. In fact, generally crickets stop chirping altogether below around 50 degrees.
When our model no longer applies after some point, it is sometimes called model breakdown .
Exercise \(\PageIndex{1}\)
What temperature would you predict if you counted 20 chirps in 15 seconds?
- Answer
-
Add answer text here and it will automatically be hidden if you have a "AutoNum" template active on the page.
Fitting Lines with Technology
While eyeballing a line works reasonably well, there are statistical techniques for fitting a line to data that minimize the differences between the line and data values (Technically, the method minimizes the sum of the squared differences in the vertical direction between the line and the data values.). This technique is called least-square regression , and can be computed by many graphing calculators, spreadsheet software like Excel or Google Docs, statistical software, and many web-based calculators(For example, http://www.shodor.org/unchem/math/lls/leastsq.html ).
Example \(\PageIndex{4}\)
Find the least-squares regression line using the cricket chirp data from above.
Solution
Using the cricket chirp data from earlier, with technology we obtain the equation:
\[T(c)=30.281+1.143c\nonumber \]
Notice that this line is quite similar to the equation we “eyeballed”, but should fit the data better. Notice also that using this equation would change our prediction for the temperature when hearing 30 chirps in 15 seconds from 66 degrees to:
\[T(30) =30.281+1.143(30)=64.571 \approx 64.6\text{ degrees}\nonumber \]
Most calculators and computer software will also provide you with the correlation coefficient , a measure of how closely the line fits the data.
Definition: correlation coefficient
The correlation coefficient is a value, \(r\), between -1 and 1.
\(r > 0\) suggests a positive (increasing) relationship
\(r < 0\) suggests a negative (decreasing) relationship
The closer the value is to 0, the more scattered the data
The closer the value is to 1 or -1, the less scattered the data is
The correlation coefficient provides an easy way to get some idea of how close to a line the data falls.
We should only compute the correlation coefficient for data that follows a linear pattern; if the data exhibits a non-linear pattern, the correlation coefficient is meaningless. To get a sense for the relationship between the value of \(r\) and the graph of the data, here are some large data sets with their correlation coefficients:
Examples of Correlation Coefficient Values
( http://en.Wikipedia.org/wiki/File:Co...n_examples.png )
Example \(\PageIndex{5}\)
Calculate the correlation coefficient for our cricket data.
Solution
Because the data appears to follow a linear pattern, we can use technology to calculate r = 0.9509. Since this value is very close to 1, it suggests a strong increasing linear relationship.
Example \(\PageIndex{6}\)
Gasoline consumption in the US has been increasing steadily. Consumption data from 1994 to 2004 is shown below.(www.bts.gov/publications/nati...ble_04_10.html) Determine if the trend is linear, and if so, find a model for the data. Use the model to predict the consumption in 2008.
| Year | '94 | '95 | '96 | '97 | '98 | '99 | '00 | '01 | '02 | '03 | '04 |
| Consumption (billions of gallons) | 113 | 116 | 118 | 119 | 123 | 125 | 126 | 128 | 131 | 133 | 136 |
Solution
To make things simpler, a new input variable is introduced, \(t\), representing years since 1994.
Using technology, the correlation coefficient was calculated to be 0.9965, suggesting a very strong increasing linear trend.
The least-squares regression equation is:
\[C(t)=113.318+2.209t\nonumber \]
Using this to predict consumption in 2008 (t = 14),
\[C(14)=113.318+2.209(14)=144.244\text{ billions of gallons}\nonumber \]
The model predicts 144.244 billion gallons of gasoline will be consumed in 2008.
Exercise \(\PageIndex{2}\)
Use the model created by technology in example 6 to predict the gas consumption in 2011. Is this an interpolation or an extrapolation?
- Answer
-
1. 54 degrees Fahrenheit
2. 150.871 billion gallons; extrapolation
Important Topics of this Section
- Fitting linear models to data by hand
- Fitting linear models to data using technology
- Interpolation
- Extrapolation
- Correlation coefficient


2.8: Marketing to the Poor
Marketing to the poor: International Development Enterprises (IDE)
Paul Polak didn’t have to do any of this. At age forty-seven, Polak was a successful Colorado psychiatrist with a wife, three daughters and $3 million in real estate. But in his extensive world travels Polak witnessed more and more the debilitating effects of extreme poverty on the world’s rural poor—who often make less than one dollar a day—and became curious about ways to help. Gradually it became clear to Polak that in order to improve the lives of hundreds of millions of subsistence farmers, there was only one place to start: water.
“You could see how essential water was to alleviating poverty,” says Polak. “If you wanted to do anything, you had to start with these small farmers and irrigation. The power to control water is absolutely crucial to them. That fact should shape all development policy.”
Inspired to help however he could, in 1981 Polak formed International Development Enterprises (IDE), a nonprofit that develops and facilitates the sale of affordable, simple, income-enhancing products to the poor, with a special expertise in water technologies for small-scale irrigation and safe drinking water. Over the years IDE has helped millions of rural farmers in the world’s poorest countries increase their agricultural productivity, providing them with a basis for food security, income generation, integration with markets, and the beginnings of an upward spiral out of poverty.
Needless to say, IDE would be nowhere if they couldn’t sell their products. But how do you sell to the world’s poorest people? How do you design products for them, and then get them to buy them? As with any business, marketing is essential to IDE’s success and at the same time comes fraught with challenges.
The first challenge, according to Polak, is molding IDE’s products around customer need—that is, making them so cheap that the world’s poorest people can afford them. Says Polak, “We focus on people who make less than one dollar a day. If things aren’t extremely cheap, they can’t afford them. Affordability is the key issue; it’s the absolute bottom line. In just about every product we sell to the poor there seems to be a threshold point of about one-fifth the conventional price; when we reach that threshold, sales take off.”
To reach that threshold, IDE develops products that can be made inexpensively by using locally available materials. An example is the IDE treadle pump, two million of which have been installed worldwide since its introduction in 1985. The pump consists of two metal cylinders with pistons operated by a natural walking motion on two treadles, like a stairmaster; the individual walks up and down on the treadles and in the process brings water up to the surface. The pump is manufactured locally in simple metalworking shops, and the treadles and support structure are made of bamboo or other inexpensive, locally available material. The unit sells for $25-50, and enables farmers to generate more than $100 a year in extra income.
A second challenge has to do with the attitudes and inclinations of the poor themselves. Many poor people aren’t willing to cough up what is to them a large sum of money for devices they’re not sure will work. Says Polak, “We feel that the best way for poor farmers to prosper is by growing labor-intensive, high-cash crops. But if the farmers grow rice like they usually do, they’ll have enough to eat; they’re averse to risk because they want to avoid going hungry. Poor farmers are very risk averse. So we have to lower the risk—find ways to show them that moving to a high-income crop is a good thing.”
IDE goes about this in a variety of ways. “First, we don’t ask them to give up rice and move straight to high-value crops right away. We start off by helping them improve their yield on staple crops through procedures like the use of urea granules planted in the ground between rice plants, which give them bigger yields and cost less than fertilizer because they don’t wash away. Once they’ve seen that we can help them grow enough rice to feed their family, they’re much more open to growing higher-value crops.”
Another IDE risk-mitigation strategy involves the marketplace itself. Because a poor farmer can never predict the market price of any crop she grows, IDE doesn’t promote just one high-value crop, instead promoting packages of four or five. “That way,” says Polak, “even though you can’t predict the final value at market, farmers can play the odds. Maybe on one crop they’ll make out like bandits, three will do OK, and one they’ll feed to the pigs.”
And what about getting the word out? How do you make poor rural farmers aware of your product? When it comes to promotional techniques and advertising, IDE takes the only route available to them: they get out in the village streets and push the product. “Let’s say that we want to sell treadle pumps,” says Polak. “We’ll put on a demonstration in a village fair that draws a lot of people. We’ll have a three-rickshaw procession: I’ll be on the first rickshaw, and I’ll have a microphone and will shout, ‘Come see this demonstration!’ The second rickshaw will have someone demonstrating how to work a treadle pump, and a third rickshaw has somebody handing out leaflets.”
“In Bangladesh we sometimes use a troubadour group. It’s a little three-person orchestra that plays a song about a treadle pump. And also in Bangladesh, because there’s a very big market there, we made a ninety-minute movie with the treadle pump as the main part of the story. We played that movie to an audience of over one million in a year.”
Indeed, numbers in the millions frequently come up when talking about Polak and IDE. IDE’s efforts around the world are estimated to create more than $200 million of additional income each year for the rural poor, and the number of lives touched reaches the tens of millions.


19.6: Resonance in an AC Circuit
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Learning Objectives
By the end of the section, you will be able to:
- Determine the peak ac resonant angular frequency for a RLC circuit
- Explain the width of the average power versus angular frequency curve and its significance using terms like bandwidth and quality factor
In the RLC series circuit of Figure 15.4.1 , the current amplitude is, from Equation 15.4.7 ,
\[I_0 = \dfrac{V_0}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}}. \label{15.15} \]
If we can vary the frequency of the ac generator while keeping the amplitude of its output voltage constant, then the current changes accordingly. A plot of \(I_0\) versus \(\omega\) is shown in Figure \(\PageIndex{1}\).
In Oscillations , we encountered a similar graph where the amplitude of a damped harmonic oscillator was plotted against the angular frequency of a sinusoidal driving force (see Forced Oscillations ). This similarity is more than just a coincidence, as shown earlier by the application of Kirchhoff’s loop rule to the circuit of Figure 15.4.1 . This yields
\[L\dfrac{di}{dt} + iR + \dfrac{q}{C} = V_0 \, \sin \, \omega t, \label{15.16} \]
or
\[L\dfrac{d^2q}{dt^2} + R\dfrac{dq}{dt} + \dfrac{1}{C}q = V_0 \, \sin \, \omega t, \nonumber \]
where we substituted dq (t)/ dt for i (t). A comparison of Equation \ref{15.16} and, from Oscillations , Damped Oscillations for damped harmonic motion clearly demonstrates that the driven RLC series circuit is the electrical analog of the driven damped harmonic oscillator.
The resonant frequency \(f_0\) of the RLC circuit is the frequency at which the amplitude of the current is a maximum and the circuit would oscillate if not driven by a voltage source. By inspection, this corresponds to the angular frequency \(\omega_0 = 2\pi f_0\) at which the impedance Z in Equation \ref{15.15} is a minimum, or when
\[\omega_0 L = \dfrac{1}{\omega_0 C} \label{resonantfrequency1} \]
and
\[\omega_0 = \sqrt{\dfrac{1}{LC}}.\label{resonantfrequency2} \]
This is the resonant angular frequency of the circuit. Substituting \(\omega_0\) into Equation 15.4.5 , Equation 15.4.7 , and Equation 15.4.8 , we find that at resonance,
\[\phi = tan^{-1}(0) = 0, \, I_0 = V_0/R, \, and \, Z = R. \nonumber \]
Therefore, at resonance, an RLC circuit is purely resistive, with the applied emf and current in phase.
What happens to the power at resonance? Equation 15.5.18 tells us how the average power transferred from an ac generator to the RLC combination varies with frequency. In addition, \(P_{ave}\) reaches a maximum when \(Z\), which depends on the frequency, is a minimum, that is, when \(X_L = X_C\) and \(Z = R\). Thus, at resonance, the average power output of the source in an RLC series circuit is a maximum. From Equation 15.5.18 , this maximum is \(V_{rms}^2 /R\).
Figure \(\PageIndex{2}\) is a typical plot of \(P_{ave}\) versus \(\omega\) in the region of maximum power output. The bandwidth \(\Delta \omega\) of the resonance peak is defined as the range of angular frequencies \(\omega\) over which the average power \(P_{ave}\) is greater than one-half the maximum value of \(P_{ave}\). The sharpness of the peak is described by a dimensionless quantity known as the quality factor Q of the circuit. By definition,
\[Q = \dfrac{\omega_0}{\Delta \omega}, \label{15.18} \]
where \(\omega_0\) is the resonant angular frequency. A high Q indicates a sharp resonance peak. We can give Q in terms of the circuit parameters as
\[Q = \dfrac{\omega_0L}{R}. \label{15.19} \]
Resonant circuits are commonly used to pass or reject selected frequency ranges. This is done by adjusting the value of one of the elements and hence “tuning” the circuit to a particular resonant frequency. For example, in radios, the receiver is tuned to the desired station by adjusting the resonant frequency of its circuitry to match the frequency of the station. If the tuning circuit has a high Q , it will have a small bandwidth, so signals from other stations at frequencies even slightly different from the resonant frequency encounter a high impedance and are not passed by the circuit. Cell phones work in a similar fashion, communicating with signals of around 1 GHz that are tuned by an inductor-capacitor circuit. One of the most common applications of capacitors is their use in ac-timing circuits, based on attaining a resonant frequency. A metal detector also uses a shift in resonance frequency in detecting metals (Figure \(\PageIndex{3}\)).
Example \(\PageIndex{1}\): Resonance in an RLC Series Circuit
- What is the resonant frequency of the circuit of Example 15.3.1 ?
- If the ac generator is set to this frequency without changing the amplitude of the output voltage, what is the amplitude of the current?
Strategy
The resonant frequency for a RLC circuit is calculated from Equation \ref{resonantfrequency2}, which comes from a balance between the reactances of the capacitor and the inductor. Since the circuit is at resonance, the impedance is equal to the resistor. Then, the peak current is calculated by the voltage divided by the resistance.
Solution
- The resonant frequency is found from Equation \ref{resonantfrequency2}: \[\begin{align*} f_0 &= \dfrac{1}{2\pi} \sqrt{\dfrac{1}{LC}} \\[4pt] &= \dfrac{1}{2\pi}\sqrt{\dfrac{1}{(3.00 \times 10^{-3} H)(8.00 \times 10^{-4}F)}} \\[4pt] &= 1.03 \times 10^2 \, Hz. \end{align*} \]
- At resonance, the impedance of the circuit is purely resistive, and the current amplitude is \[I_0 = \dfrac{0.100 \, V}{4.00 \, \Omega} = 2.50 \times 10^{-2}A. \nonumber \]
Significance
If the circuit were not set to the resonant frequency, we would need the impedance of the entire circuit to calculate the current.
Example \(\PageIndex{2}\): Power Transfer in an RLC Series Circuit at Resonance
- What is the resonant angular frequency of an RLC circuit with \( R = 0.200 \, \Omega, \, L = 4.00 \times 10^{-3} H\), and \(C = 2.00 \times 10^{-6}F\)?
- If an ac source of constant amplitude 4.00 V is set to this frequency, what is the average power transferred to the circuit?
- Determine Q and the bandwidth of this circuit.
Strategy
The resonant angular frequency is calculated from Equation \ref{resonantfrequency2}. The average power is calculated from the rms voltage and the resistance in the circuit. The quality factor is calculated from Equation \ref{15.19} and by knowing the resonant frequency. The bandwidth is calculated from Equation \ref{15.18} and by knowing the quality factor.
Solution
- The resonant angular frequency is \[ \begin{align*} \omega_0 &= \sqrt{\dfrac{1}{LC}} \\[4pt] &= \sqrt{\dfrac{1}{(4.00 \times 10^{-3}H)(2.00 \times 10^{-6}F)}} \\[4pt] &= 1.12 \times 10^4 \, rad/s. \end{align*} \]
- At this frequency, the average power transferred to the circuit is a maximum. It is \[P_{ave} = \dfrac{V_{rms}^2}{R} = \dfrac{[(1/\sqrt{2})(4.00 \, V)]^2}{0.200 \, \Omega} = 40.0 \, W. \nonumber \]
- The quality factor of the circuit is \[Q = \dfrac{\omega_0L}{R} = \dfrac{(1.12 \times 10^4 \, rad/s)(4.00 \times 10^{-3}H)}{0.200 \, \Omega} = 224. \nonumber \] We then find for the bandwidth \[\Delta \omega = \dfrac{\omega_0}{Q} = \dfrac{1.12 \times 10^4 \, rad/s}{224} = 50.0 \, rad/s. \nonumber \]
Significance
If a narrower bandwidth is desired, a lower resistance or higher inductance would help. However, a lower resistance increases the power transferred to the circuit, which may not be desirable, depending on the maximum power that could possibly be transferred.
Exercise \(\PageIndex{1}\)
In the circuit of Figure 15.4.1 , \(L = 2.0 \times 10^{-3}H, \, C = 5.0 \times 10^{-4} F\), and \(R = 40 \, \Omega\).
- What is the resonant frequency?
- What is the impedance of the circuit at resonance?
- If the voltage amplitude is 10 V, what is i ( t ) at resonance?
- The frequency of the AC generator is now changed to 200 Hz. Calculate the phase difference between the current and the emf of the generator.
- Answer
-
a. 160 Hz; b. \(40 \Omega\); c. \((0.25 A) \, \sin \, 10^3t\); d. 0.023 rad
Exercise \(\PageIndex{2}\)
What happens to the resonant frequency of an RLC series circuit when the following quantities are increased by a factor of 4: (a) the capacitance, (b) the self-inductance, and (c) the resistance?
- Answer
-
a. halved; b. halved; c. same
Exercise \(\PageIndex{3}\)
The resonant angular frequency of an RLC series circuit is \(4.0 \times 10^2 \, rad/s\). An ac source operating at this frequency transfers an average power of \(2.0 \times 10^{-2} W\) to the circuit. The resistance of the circuit is \(0.50 \, \Omega\). Write an expression for the emf of the source.
- Answer
-
\(v(t) = (0.14 \, V) \, sin \, (4.0 \times 10^2 t)\)


11.5: Bond Energies and Chemical Reactions
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Learning Objectives
The Learning Objectives of this Module are to:
- Define bond energy .
- Calculate enthalpy change or heat of reaction.
- Determine if a chemical process is exothermic or endothermic.
What happens when you take a basketball, place it halfway up a playground slide, and then let it go? The basketball rolls down the slide. What happens if you do it again? Does the basketball roll down the slide? It should.
If you were to perform this experiment over and over again, do you think the basketball would ever roll up the slide? Probably not. Why not? Well, for starters, in all of our experience, the basketball has always moved to a lower position when given the opportunity. The gravitational attraction of Earth exerts a force on the basketball, and given the chance, the basketball will move down. We say that the basketball is going to a lower gravitational potential energy . The basketball can move up the slide, but only if someone exerts some effort (that is, work) on the basketball. A general statement, based on countless observations over centuries of study, is that all objects tend to move spontaneously to a position of minimum energy unless acted on by some other force or object.
Bond Energy
A similar statement can be made about atoms in compounds. Atoms bond together to form compounds because in doing so they attain lower energies than they possess as individual atoms. A quantity of energy, equal to the difference between the energies of the bonded atoms and the energies of the separated atoms, is released, usually as heat. That is, the bonded atoms have a lower energy than the individual atoms do. When atoms combine to make a compound, energy is always given off, and the compound has a lower overall energy. In making compounds, atoms act like a basketball on a playground slide; they move in the direction of decreasing energy.
We can reverse the process, just as with the basketball. If we put energy into a molecule, we can cause its bonds to break, separating a molecule into individual atoms. Bonds between certain specific elements usually have a characteristic energy, called the bond energy, that is needed to break the bond. The same amount of energy was liberated when the atoms made the chemical bond in the first place. The term bond energy is usually used to describe the strength of interactions between atoms that make covalent bonds. A C–C bond has an approximate bond energy of 80 kcal/mol, while a C=C has a bond energy of about 145 kcal/mol. The C=C bond is stronger than C-C (as discussed in relation to bond length in Section 4.4). For atoms in ionic compounds attracted by opposite charges, the term lattice energy is used. For now, we will deal with covalent bonds in molecules.
Although each molecule has its own characteristic bond energy, some generalizations are possible. For example, although the exact value of a C–H bond energy depends on the particular molecule, all C–H bonds have a bond energy of roughly the same value because they are all C–H bonds. It takes roughly 100 kcal of energy to break 1 mol of C–H bonds, so we speak of the bond energy of a C–H bond as being about 100 kcal/mol. Table \(\PageIndex{1}\) lists the approximate bond energies of various covalent bonds.
| Bond | Bond Energy (kcal/mol) |
|---|---|
| C–H | 100 |
| C–O | 86 |
| C=O | 190 |
| C–N | 70 |
| C–C | 85 |
| C=C | 145 |
| C≡C | 200 |
| N–H | 93 |
| H–H | 105 |
| Br-Br | 46 |
| Cl–Cl | 58 |
| O–H | 110 |
| O=O | 119 |
| H–Br | 87 |
| H–Cl | 103 |
When a chemical reaction occurs, the atoms in the reactants rearrange their chemical bonds to make products. The new arrangement of bonds does not have the same total energy as the bonds in the reactants. Therefore, when chemical reactions occur, there will always be an accompanying energy change . The enthalpy change, for a given reaction can be calculated using the bond energy values from Table \(\PageIndex{1}\).
Enthalpy Change or Heat of Reaction, Δ H
Energy changes in chemical reactions are usually measured as changes in enthalpy. In this process, one adds energy to the reaction to break bonds, and extracts energy for the bonds that are formed.
\[\text{enthalpy change} ≈ \sum (\text{bonds broken}) - \sum (\text{bonds formed}) \nonumber \]
The ≈ sign is used because we are adding together average bond energies (i.e., over many different molecules). Hence, this approach does not give exact values for the enthalpy change, Δ H .
Let’s consider the reaction of 2 mols of hydrogen gas (H 2 ) with 1 mol of oxygen gas (O 2 ) to give 2 mol water:
\[\ce{2H_2(g) + O_2(g) \rightarrow 2H_2O(g)} \nonumber \]
or shown graphically:
with these average bond energies involved:
- \(\ce{H–H}\) = 105 kcal/mol
- \(\ce{O=O}\) = 119 kcal/mol
- \(\ce{O–H}\) = 110 kcal/mol
In this reaction, 2 H–H bonds and 1 O=O bonds are broken, while 4 O–H bonds (two for each H 2 O) are formed. The energy changes can be tabulated and calculated as follows:
| Bonds Broken (kcal/mol) | Bonds Formed (kcal/mol) | ||
|---|---|---|---|
| 2 H–H | 2 mol x 105 kcal/mol = 210 kcal | 4 O–H | 4 mols x 110 kcal/mol = 440 kcal |
| 1 O=O | 1 mol x 119 kcal/mol = 119 kcal | ||
| Total = 329 kcal | Total = 440 kcal |
\[\begin{align*} \Delta H &≈ \sum (\text{bonds broken}) - \sum (\text{bonds formed}) \\[4pt] &≈ 329 \: \text{kcal} - 440 \: \text{kcal} \\[4pt] &≈ −111 \: \text{kcal} \end{align*} \nonumber \]
The enthalpy change (ΔH) of the reaction is approximately −111 kcal/mol. This means that bonds in the products (440 kcal) are stronger than the bonds in the reactants (329 kcal) by about 111 kcal/mol. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes. This excess energy is released as heat, so the reaction is exothermic . Hence, we can re-write the reaction with the heat released (111 kcal) on the product side of the equation, as follows:
We can also re-write the reaction equation with the ΔH information (see below). Note that an exothermic reaction has a negative ΔH value.
\[\ce{2H_2(g)+O_2(g) \rightarrow 2H_2O(g)} \ \: \: \: \: \: \Delta H = -111 \: \text{kcal} \nonumber \]
Example \(\PageIndex{1}\)
What is the enthalpy change for this reaction? Is the reaction exothermic or endothermic?
\[\ce{H_2(g) + Br_2(g) \rightarrow 2HBr(g)} \nonumber \]
Solution
Step 1- First look at the equation and identify which bonds exist on in the reactants (bonds broken).
- one H-H bond and
- one Br-Br bond
Step 2- Do the same for the products (bonds formed)
- two H-Br bonds
Step 3- Identify the bond energies of these bonds from Table \(\PageIndex{1}\):
- H-H bonds: 105 kcal/mol
- Br-Br bonds: 46 kcal/mol
Step 4- Set up the table (see below) and apply the formula for enthalpy change.
| Bonds Broken (kcal/mol) | Bonds Formed (kcal/mol) | ||
|---|---|---|---|
| 1 H–H | 1 mol x 105 kcal/mol = 105 kcal | 2 H– Br | 2 mols x 87 kcal/mol = 174 kcal |
| 1 Br–Br | 1 mol x 46 kcal/mol = 46 kcal | ||
| Total = 151 kcal | Total = 174 kcal |
\[\begin{align*} \Delta H &≈ 151 \: \text{kcal} - 174 \: \text{kcal} \\[4pt] &≈ −23 \: \text{kcal} \end{align*} \nonumber \]
Step 5- Since \(ΔH\) is negative (−23 kcal), the reaction is exothermic .
Exercise \(\PageIndex{1}\)
Using the bond energies given in the chart above, find the enthalpy change for the thermal decomposition of water:
\[\ce{H_2(g) + Cl_2(g) \rightarrow 2HCl(g)} \nonumber \]
Is the reaction written above exothermic or endothermic? Explain.
- Answer
-
ΔH = −43 kcal
Since ΔH is negative (−43 kcal), the reaction is exothermic .
Endothermic and Exothermic Reactions
Endothermic and exothermic reactions can be thought of as having energy as either a reactant of the reaction or a product. Endothermic reactions require energy, so energy is a reactant. Heat flows from the surroundings to the system (reaction mixture) and the enthalpy of the system increases (Δ H is positive). As discussed in the previous section, in an exothermic reaction, heat is released (considered a product) and the enthalpy of the system decreases (Δ H is negative).
In the course of an endothermic process, the system gains heat from the surroundings and so the temperature of the surroundings decreases (gets cold). A chemical reaction is exothermic if heat is released by the system into the surroundings. Because the surroundings is gaining heat from the system, the temperature of the surroundings increases. See Figure \(\PageIndex{1}\).
Endothermic Reaction : When \(1 \: \text{mol}\) of calcium carbonate decomposes into \(1 \: \text{mol}\) of calcium oxide and \(1 \: \text{mol}\) of carbon dioxide, \(177.8 \: \text{kJ}\) of heat is absorbed. Because the heat is absorbed by the system, the \(177.8 \: \text{kJ}\) is written as a reactant . The Δ H is positive for an endothermic reaction.
\[\ce{CaCO_3} \left( s \right) \rightarrow \ce{CaO} \left( s \right) + \ce{CO_2} \left( g \right) \: \: \: \: \: \Delta H = +177.8 \: \text{kJ} \nonumber \]
Exothermic Reaction : When methane gas is combusted, heat is released, making the reaction exothermic. Specifically, the combustion of \(1 \: \text{mol}\) of methane releases 890.4 kilojoules of heat energy. This information can be shown as part of the balanced equation in two ways. First, the amount of heat released can be written in the product side of the reaction. Another way is to write the heat of reaction (Δ H ) information with a negative sign, \(-890.4 \: \text{kJ}\).
\[\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( l \right) \: \: \: \: \: \Delta H = -890.4 \: \text{kJ} \nonumber \]
Example \(\PageIndex{2}\)
Is each chemical reaction exothermic or endothermic?
- CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(ℓ) + 213 kcal
- N 2 (g) + O 2 (g) + 45 kcal → 2NO(g)
Solution
- Because energy (213 kcal) is a product, energy is given off by the reaction. Therefore, this reaction is exothermic.
- Because energy (45 kcal) is a reactant, energy is absorbed by the reaction. Therefore, this reaction is endothermic.
Exercise \(\PageIndex{2}\)
Is each chemical reaction exothermic or endothermic?
- H 2 (g) + F 2 (g) → 2HF (g) + 130 kcal
- 2C(s) + H 2 (g) + 5.3 kcal → C 2 H 2 (g)
- Answer
-
a. The energy (130 kcal) is produced, hence the reaction is exothermic
b. The energy (5.3 kcal) is supplied or absorbed to react, hence, the reaction is endothermic
Energy Diagrams
Endothermic and exothermic reactions can be visually represented by energy-level diagrams like the ones in Figure \(\PageIndex{2}\) . In endothermic reactions, the reactants have higher bond energy (stronger bonds) than the products. Strong bonds have lower potential energy than weak bonds . Hence, the energy of the reactants is lower than that of the products . This type of reaction is represented by an "uphill" energy-level diagram shown in Figure \(\PageIndex{2A}\). For an endothermic chemical reaction to proceed, the reactants must absorb energy from their environment to be converted to products.
In an exothermic reaction, the bonds in the product have higher bond energy (stronger bonds) than the reactants. In other words, the energy of the products is lower than the energy of the reactants, hence is energetically downhill, shown in Figure \(\PageIndex{2B}\). Energy is given off as reactants are converted to products. The energy given off is usually in the form of heat (although a few reactions give off energy as light). In the course of an exothermic reaction, heat flows from the system to its surroundings, and thus, gets warm.
| Endothermic Reactions | Exothermic Reactions |
|---|---|
| Heat is absorbed by reactants to form products. | Heat is released. |
| Heat is absorbed from the surroundings; as a result, the surroundings get cold. | Heat is released by the reaction to surroundings; surroundings feel hot. |
| Δ H rxn is positive | Δ H rxn is negative |
| The bonds broken in the reactants are stronger than the bonds formed in the products | The bonds formed in the products are stronger than the bonds broken in the reactants |
| The reactants are lower in energy than the products | The products are lower in energy than the reactants |
| Represented by an "uphill" energy diagram | Represented by an "downhill" energy diagram |
Key Takeaways
- Atoms are held together by a certain amount of energy called bond energy.
- Energy is required to break bonds. Energy is released when chemical bonds are formed because atoms become more stable.
- Chemical processes are labeled as exothermic or endothermic based on whether they give off or absorb energy, respectively.


9.2: Trapezium Rule
The trapezium rule is a simple extension of the mid-point integration rule. Instead of calculating the area of a sequence of rectangles, we calculate the area of a sequence of trapeziums, as shown in Fig. \(\PageIndex{1}\). As before, we divide the integration region into \(N\) equal segments; however, we now consider the end-points of these segments, rather than their mid-points. These is a set of \(N+1\) positions \(\{x_0, \dots x_{N}\}\), such that
\[x_n = a + n\, \Delta x ,\quad \Delta x \equiv \frac{b-a}{N}.\]
Note that \(a = x_0\) and \(b=x_{N}\). By using the familiar formula for the area of each trapezium, we can approximate the integral as
\[\begin{align}\mathcal{I}^{\mathrm{trapz}} &= \sum_{n=0}^{N-1} \Delta x \left(\frac{f(x_n) + f(x_{n+1})}{2}\right) \\ &= \Delta x \left[\left(\frac{f(x_0) + f(x_1)}{2}\right)+ \left(\frac{f(x_1) + f(x_2)}{2}\right) + \cdots + \left(\frac{f(x_{N-1}) + f(x_{N})}{2}\right)\right] \\ &= \Delta x \left[\frac{f(x_0)}{2} + \left(\sum_{n=1}^{N-1} f(x_n)\right) + \frac{f(x_{N})}{2}\right].\end{align}\]
9.2.1 Python Implementation of the Trapezium Rule
In Scipy, the trapezium rule is implemented by the
trapz
function. It usually takes two array arguments,
y
and
x
; then the function call
trapz(y,x)
returns the trapezium rule estimate for \(\int y \;dx\), using the elements of
x
as the discretization points, and the elements of
y
as the values of the integrand at those points.
Note
Ma
tlab
compatibility note
: Be careful of the order of inputs! For the Scipy function, it's
trapz(y,x)
. For the Matlab function of the same name, the inputs are supplied in the opposite order, as
trapz(x,y)
.
Note that the discretization points in
x
need not be equally-spaced. Alternatively, you can call the function as
trapz(y,dx=s)
; this performs the numerical integration assuming equally-spaced discretization points, with spacing
s
.
Here is an example of using
trapz
to evaluate the integral \(\int_0^\infty \exp(-x^2) dx = 1\):
>>> from scipy import * >>> x = linspace(0,10,25) >>> y = exp(-x) >>> t = trapz(y,x) >>> print(t) 1.01437984777
9.2.2 Numerical Error for the Trapezium Rule
From a visual comparison of Fig. 9.1.1 (for the mid-point rule) and Fig. \(\PageIndex{1}\) (for the trapezium rule), we might be tempted to conclude that the trapezium rule should be more accurate. Before jumping to that conclusion, however, let's do the actual calculation of the numerical error. This is similar to our above calculation of the mid-point rule's numerical error. For the trapezium rule, however, it's convenient to consider a pair of adjacent segments, which lie between the three discretization points \(\left\{x_{n-1}, x_n, x_{n+1}\right\}\).
As before, we perform a Taylor expansion of \(f(x)\) in the vicinity of \(x_{n}\):
\[f(x) = f(x_n) + f'(x_n) (x-x_n) + \frac{f''(x_n)}{2} (x-x_n)^2 + \frac{f'''(x_n)}{6} (x-x_n)^3 + \cdots \]
If we integrate \(f(x)\) from \(x_{n-1}\) to \(x_{n+1}\), the result is
\[\begin{align} \mathcal{I}_n & \equiv \int_{x_n - \Delta x}^{x_n + \Delta x} f(x) dx \\ &= \; 2 f(x_n) \Delta x \;\;+\; f'(x_n) \int_{x_n - \Delta x}^{x_n + \Delta x} (x-x_n) dx \; +\; \frac{f''(x_n)}{2} \int_{x_n - \Delta x}^{x_n + \Delta x} (x-x_n)^2 dx \; +\; \cdots \end{align}\]
As before, every other term on the right-hand side integrates to zero. We are left with
\[\mathcal{I}_n \; = \; 2 f(x_n) \Delta x \;+\; \frac{f''(x_n)}{3} \Delta x^3 + O(\Delta x^5) \;+\; \cdots \]
This has to be compared to the estimate produced by the trapezium rule, which is
\[\mathcal{I}_n^{\mathrm{(trapz)}} = \Delta x \left[ \frac{f(x_{n-1})}{2} + f(x_n) + \frac{f(x_{n+1})}{2}\right].\]
If we Taylor expand the first and last terms of the above expression around \(x_{n}\), the result is:
\[\begin{align} \mathcal{I}_n^{\mathrm{(trapz)}} & = \frac{\Delta x}{2} \left[f(x_n) - f'(x_n) \Delta x + \frac{f''(x_n)}{2} \Delta x^2 - \frac{f'''(x_n)}{6} \Delta x^3 + \frac{f''''(x_n)}{24} \Delta x^4 + O(\Delta x^4)\right] \\ &\quad + \;\;\Delta x \; f(x_n) \\ &\quad + \frac{\Delta x}{2} \left[f(x_n) + f'(x_n) \Delta x + \frac{f''(x_n)}{2} \Delta x^2 + \frac{f'''(x_n)}{6} \Delta x^3 + \frac{f''''(x_n)}{24} \Delta x^4 + O(\Delta x^4)\right] \\ & = 2 f(x_n) \Delta x + \frac{f''(x_n)}{2} \Delta x^3 + O(\Delta x^5). \end{align}\]
Hence, the numerical error for integrating over this pair of segments is
\[\mathcal{E}_n \;\equiv\; \left|\mathcal{I}_n - \mathcal{I}_n^{\mathrm{trapz}}\right| \;=\; \frac{|f''(x_n)|}{6} \Delta x^3 \;\sim\; O\left(\frac{1}{N^3}\right).\]
And the total numerical error goes as
\[\mathcal{E}_{\mathrm{total}} \sim O\left(\frac{1}{N^2}\right),\]
which is the same scaling as the mid-point rule! Despite our expectations, the trapezium rule isn't actually an improvement over the mid-point rule, as far as numerical accuracy is concerned.