5.4: Newton's Second Law
- Distinguish between external and internal forces
- Describe Newton's second law of motion
- Explain the dependence of acceleration on net force and mass
Newton’s second law is closely related to his first law. It mathematically gives the cause-and-effect relationship between force and changes in motion. Newton’s second law is quantitative and is used extensively to calculate what happens in situations involving a force. Before we can write down Newton’s second law as a simple equation that gives the exact relationship of force, mass, and acceleration, we need to sharpen some ideas we mentioned earlier.
Force and Acceleration
First, what do we mean by a change in motion? The answer is that a change in motion is equivalent to a change in velocity. A change in velocity means, by definition, that there is acceleration. Newton’s first law says that a net external force causes a change in motion; thus, we see that a net external force causes nonzero acceleration .
We defined external force in Forces as force acting on an object or system that originates outside of the object or system. Let’s consider this concept further. An intuitive notion of external is correct—it is outside the system of interest. For example, in Figure \(\PageIndex{1a}\), the system of interest is the car plus the person within it. The two forces exerted by the two students are external forces. In contrast, an internal force acts between elements of the system. Thus, the force the person in the car exerts to hang on to the steering wheel is an internal force between elements of the system of interest. Only external forces affect the motion of a system, according to Newton’s first law. (The internal forces cancel each other out, as explained in the next section.) Therefore, we must define the boundaries of the system before we can determine which forces are external. Sometimes, the system is obvious, whereas at other times, identifying the boundaries of a system is more subtle. The concept of a system is fundamental to many areas of physics, as is the correct application of Newton’s laws. This concept is revisited many times in the study of physics.
From this example, you can see that different forces exerted on the same mass produce different accelerations. In Figure \(\PageIndex{1a}\), the two students push a car with a driver in it. Arrows representing all external forces are shown. The system of interest is the car and its driver. The weight \(\vec{w}\) of the system and the support of the ground \(\vec{N}\) are also shown for completeness and are assumed to cancel (because there was no vertical motion and no imbalance of forces in the vertical direction to create a change in motion). The vector \(\vec{f}\) represents the friction acting on the car, and it acts to the left, opposing the motion of the car. (We discuss friction in more detail in the next chapter.) In Figure \(\PageIndex{1b}\), all external forces acting on the system add together to produce the net force \(\vec{F}_{net}\). The free-body diagram shows all of the forces acting on the system of interest. The dot represents the center of mass of the system. Each force vector extends from this dot. Because there are two forces acting to the right, the vectors are shown collinearly. Finally, in Figure \(\PageIndex{1c}\), a larger net external force produces a larger acceleration (\(\vec{a}' > \vec{a}\)) when the tow truck pulls the car.
It seems reasonable that acceleration would be directly proportional to and in the same direction as the net external force acting on a system. This assumption has been verified experimentally and is illustrated in Figure \(\PageIndex{1}\). To obtain an equation for Newton’s second law, we first write the relationship of acceleration \(\vec{a}\) and net external force \(\vec{F}_{net}\) as the proportionality
\[\vec{a}\; \propto\; \vec{F}_{net}\]
where the symbol \(\alpha\) means “proportional to.” (Recall from Forces that the net external force is the vector sum of all external forces and is sometimes indicated as \(\sum \vec{F}\).) This proportionality shows what we have said in words—acceleration is directly proportional to net external force. Once the system of interest is chosen, identify the external forces and ignore the internal ones. It is a tremendous simplification to disregard the numerous internal forces acting between objects within the system, such as muscular forces within the students’ bodies, let alone the myriad forces between the atoms in the objects. Still, this simplification helps us solve some complex problems.
It also seems reasonable that acceleration should be inversely proportional to the mass of the system. In other words, the larger the mass (the inertia), the smaller the acceleration produced by a given force. As illustrated in Figure \(\PageIndex{2}\), the same net external force applied to a basketball produces a much smaller acceleration when it is applied to an SUV. The proportionality is written as
\[a\; \propto\; \frac{1}{m},\]
where m is the mass of the system and a is the magnitude of the acceleration. Experiments have shown that acceleration is exactly inversely proportional to mass, just as it is directly proportional to net external force.
It has been found that the acceleration of an object depends only on the net external force and the mass of the object. Combining the two proportionalities just given yields Newton’s second law .
The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system and is inversely proportion to its mass. In equation form, Newton’s second law is
\[\vec{a} = \frac{\vec{F}_{net}}{m},\]
where \(\vec{a}\) is the acceleration, \(\vec{F}_{net}\) is the net force, and m is the mass. This is often written in the more familiar form
\[\vec{F}_{net} = \sum \vec{F} = m \vec{a}, \label{5.3}\]
but the first equation gives more insight into what Newton’s second law means. When only the magnitude of force and acceleration are considered, this equation can be written in the simpler scalar form:
\[\vec{F}_{net} = ma \ldotp \label{5.4}\]
The law is a cause-and-effect relationship among three quantities that is not simply based on their definitions. The validity of the second law is based on experimental verification. The free-body diagram, which you will learn to draw in Drawing Free-Body Diagrams , is the basis for writing Newton’s second law.
Suppose that the net external force (push minus friction) exerted on a lawn mower is 51 N (about 11 lb.) parallel to the ground (Figure \(\PageIndex{3}\)). The mass of the mower is 24 kg. What is its acceleration?
Strategy
This problem involves only motion in the horizontal direction; we are also given the net force, indicated by the single vector, but we can suppress the vector nature and concentrate on applying Newton’s second law. Since F net and m are given, the acceleration can be calculated directly from Newton’s second law as F net = ma.
Solution
The magnitude of the acceleration a is a = \(\frac{F_{net}}{m}\). Entering known values gives
\[a = \frac{51\; N}{24\; kg} \ldotp\]
Substituting the unit of kilograms times meters per square second for newtons yields
\[a = \frac{51\; kg \cdotp m/s^{2}}{24\; kg} = 2.1\; m/s^{2} \ldotp\]
Significance
The direction of the acceleration is the same direction as that of the net force, which is parallel to the ground. This is a result of the vector relationship expressed in Newton’s second law, that is, the vector representing net force is the scalar multiple of the acceleration vector. There is no information given in this example about the individual external forces acting on the system, but we can say something about their relative magnitudes. For example, the force exerted by the person pushing the mower must be greater than the friction opposing the motion (since we know the mower moved forward), and the vertical forces must cancel because no acceleration occurs in the vertical direction (the mower is moving only horizontally). The acceleration found is small enough to be reasonable for a person pushing a mower. Such an effort would not last too long, because the person’s top speed would soon be reached.
At the time of its launch, the HMS Titanic was the most massive mobile object ever built, with a mass of 6.0 x 10 7 kg. If a force of 6 MN (6 x 10 6 N) was applied to the ship, what acceleration would it experience?
In the preceding example, we dealt with net force only for simplicity. However, several forces act on the lawn mower. The weight \(\vec{w}\) (discussed in detail in Mass and Weight ) pulls down on the mower, toward the center of Earth; this produces a contact force on the ground. The ground must exert an upward force on the lawn mower, known as the normal force \(\vec{N}\), which we define in Common Forces . These forces are balanced and therefore do not produce vertical acceleration. In the next example, we show both of these forces. As you continue to solve problems using Newton’s second law, be sure to show multiple forces.
- The car shown in Figure \(\PageIndex{4}\) is moving at a constant speed. Which force is bigger, \(\vec{F}_{engine}\) or \(\vec{F}_{friction}\)? Explain.
- The same car is now accelerating to the right. Which force is bigger, \(\vec{F}_{engine}\) or \(\vec{F}_{friction}\)? Explain.
Strategy
We must consider Newton’s first and second laws to analyze the situation. We need to decide which law applies; this, in turn, will tell us about the relationship between the forces.
Solution
- The forces are equal. According to Newton’s first law, if the net force is zero, the velocity is constant.
- In this case, \(\vec{F}_{engine}\) must be larger than \(\vec{F}_{friction}\). According to Newton’s second law, a net force is required to cause acceleration.
Significance
These questions may seem trivial, but they are commonly answered incorrectly. For a car or any other object to move, it must be accelerated from rest to the desired speed; this requires that the engine force be greater than the friction force. Once the car is moving at constant velocity, the net force must be zero; otherwise, the car will accelerate (gain speed). To solve problems involving Newton’s laws, we must understand whether to apply Newton’s first law (where \(\sum \vec{F}\) = \(\vec{0}\)) or Newton’s second law (where \(\sum \vec{F}\) is not zero). This will be apparent as you see more examples and attempt to solve problems on your own.
Before manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets.
Calculate the magnitude of force exerted by each rocket, called its thrust T, for the four-rocket propulsion system shown in Figure \(\PageIndex{5}\). The sled’s initial acceleration is 49 m/s 2 , the mass of the system is 2100 kg, and the force of friction opposing the motion is 650 N.
Strategy
Although forces are acting both vertically and horizontally, we assume the vertical forces cancel because there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in Figure \(\PageIndex{5}\).
Solution
Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. We have defined the direction of the force and acceleration as acting “to the right,” so we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with
\[F_{net} = ma\]
where F net is the net force along the horizontal direction. We can see from the figure that the engine thrusts add, whereas friction opposes the thrust. In equation form, the net external force is
\[F_{net} = 4T − f \ldotp\]
Substituting this into Newton’s second law gives us
\[F_{net} = ma = 4T − f \ldotp\]
Using a little algebra, we solve for the total thrust 4T:
\[4T = ma + f \ldotp\]
Substituting known values yields
\[4T = ma + f = (2100\; kg)(49\; m/s^{2}) + 650\; N \ldotp\]
Therefore, the total thrust is
\[4T = 1.0 \times 10^{5}\; N \ldotp\]
Significance
The numbers are quite large, so the result might surprise you. Experiments such as this were performed in the early 1960s to test the limits of human endurance, and the setup was designed to protect human subjects in jet fighter emergency ejections. Speeds of 1000 km/h were obtained, with accelerations of 45 g’s. (Recall that g, acceleration due to gravity, is 9.80 m/s 2 . When we say that acceleration is 45 g’s, it is 45 x 9.8 m/s 2 , which is approximately 440 m/s 2 .) Although living subjects are not used anymore, land speeds of 10,000 km/h have been obtained with a rocket sled.
In this example, as in the preceding one, the system of interest is obvious. We see in later examples that choosing the system of interest is crucial—and the choice is not always obvious.
Newton’s second law is more than a definition; it is a relationship among acceleration, force, and mass. It can help us make predictions. Each of those physical quantities can be defined independently, so the second law tells us something basic and universal about nature.
A 550-kg sports car collides with a 2200-kg truck, and during the collision, the net force on each vehicle is the force exerted by the other. If the magnitude of the truck’s acceleration is 10 m/s 2 , what is the magnitude of the sports car’s acceleration?
Component Form of Newton’s Second Law
We have developed Newton’s second law and presented it as a vector equation in Equation \ref{5.3}. This vector equation can be written as three component equations:
\[\sum \vec{F}_{x} = m \vec{a}_{x}, \sum \vec{F}_{y} = m \vec{a}_{y}, \sum \vec{F}_{z} = m \vec{a}_{z} \ldotp \label{5.5}\]
The second law is a description of how a body responds mechanically to its environment. The influence of the environment is the net force \(\vec{F}_{net}\), the body’s response is the acceleration \(\vec{a}\), and the strength of the response is inversely proportional to the mass m. The larger the mass of an object, the smaller its response (its acceleration) to the influence of the environment (a given net force). Therefore, a body’s mass is a measure of its inertia, as we explained in Newton’s First Law .
A 0.400-kg soccer ball is kicked across the field by a player; it undergoes acceleration given by \(\vec{a}\) = 3.00 \(\hat{i}\) + 7.00 \(\hat{j}\) m/s 2 . Find (a) the resultant force acting on the ball and (b) the magnitude and direction of the resultant force.
Strategy
The vectors in \(\hat{i}\) and \(\hat{j}\) format, which indicate force direction along the x-axis and the y-axis, respectively, are involved, so we apply Newton’s second law in vector form.
Solution
- We apply Newton’s second law: $$\vec{F}_{net} = m \vec{a} = (0.400\; kg) \big(3.00\; \hat{i} + 7.00\; \hat{j}\; m/s^{2} \big) = 1.20\; \hat{i} + 2.80\; \hat{j}\; N \ldotp$$
- . Magnitude and direction are found using the components of \(\vec{F}_{net}\): $$F_{net} = \sqrt{(1.20\; N)^{2} + (2.80\; N)^{2}} = 3.05\; N\; and\; \theta = \tan^{-1} \left(\dfrac{2.80}{1.20}\right) = 66.8^{o} \ldotp$$
Significance
We must remember that Newton’s second law is a vector equation. In (a), we are multiplying a vector by a scalar to determine the net force in vector form. While the vector form gives a compact representation of the force vector, it does not tell us how “big” it is, or where it goes, in intuitive terms. In (b), we are determining the actual size (magnitude) of this force and the direction in which it travels.
Find the mass of a car if a net force of −600.0 \(\hat{j}\) N produces an acceleration of −0.2 \(\hat{j}\) m/s 2 .
Strategy
Vector division is not defined, so \(m = \frac{\vec{F}_{net}}{\vec{a}}\) cannot be performed. However, mass m is a scalar, so we can use the scalar form of Newton’s second law, \(m = \frac{F_{net}}{a}\).
Solution
We use m = \(\frac{F_{net}}{a}\) and substitute the magnitudes of the two vectors: F net = 600.0 N and a = 0.2 m/s 2 . Therefore,
\[m = \frac{F_{net}}{a} = \frac{600.0\; N}{0.2\; m/s^{2}} = 3000\; kg \ldotp \nonumber\]
Significance
Force and acceleration were given in the \(\hat{i}\) and \(\hat{j}\) format, but the answer, mass m, is a scalar and thus is not given in \(\hat{i}\) and \(\hat{j}\) form.
Several Forces on a Particle A particle of mass m = 4.0 kg is acted upon by four forces of magnitudes. F 1 = 10.0 N, F 2 = 40.0 N, F 3 = 5.0 N, and F 4 = 2.0 N, with the directions as shown in the free-body diagram in Figure \(\PageIndex{6}\). What is the acceleration of the particle?
Strategy
Because this is a two-dimensional problem, we must use a free-body diagram. First, \(\vec{F}_{1}\) must be resolved into x- and y-components. We can then apply the second law in each direction.
Solution
We draw a free-body diagram as shown in Figure \(\PageIndex{6}\). Now we apply Newton’s second law. We consider all vectors resolved into x- and y-components:
|
\[\sum F_{x} = m a_{x}\] \[F_{1x} - F_{3x} = m a_{x}\] \[F_{1} \cos 30^{o} - F_{3x} = m a_{x}\] \[(10.0\; N)(\cos 30^{o}) - 5.0\; N = (4.0\; kg) a_{x}\] \[a_{x} = 0.92\; m/s^{2} \ldotp\] |
\[\sum F_{y} = m a_{y}\] \[F_{1y} +F_{4y} - F_{2y} = m a_{y}\] \[F_{1} \sin30^{o} + F_{4y} - F_{2y} = m a_{y}\] \[(10.0\; N)(\sin 30^{o}) + 2.0\; N - 40.0\; N = (4.0\; kg) a_{y}\] \[a_{y} =-8.3\; m/s^{2} \ldotp\] |
Thus, the net acceleration is
\[\vec{a} = \big( 0.92\; \hat{i} - 8.3\; \hat{j} \big) m/s^{2},\]
which is a vector of magnitude 8.4 m/s 2 directed at 276° to the positive x-axis.
Significance
Numerous examples in everyday life can be found that involve three or more forces acting on a single object, such as cables running from the Golden Gate Bridge or a football player being tackled by three defenders. We can see that the solution of this example is just an extension of what we have already done.
A car has forces acting on it, as shown below. The mass of the car is 1000.0 kg. The road is slick, so friction can be ignored. (a) What is the net force on the car? (b) What is the acceleration of the car?
Newton’s Second Law and Momentum
Newton actually stated his second law in terms of momentum: “The instantaneous rate at which a body’s momentum changes is equal to the net force acting on the body.” (“Instantaneous rate” implies that the derivative is involved.) This can be given by the vector equation
\[\vec{F}_{net} = \frac{d \vec{p}}{dt} \ldotp \label{5.6}\]
This means that Newton’s second law addresses the central question of motion: What causes a change in motion of an object? Momentum was described by Newton as “quantity of motion,” a way of combining both the velocity of an object and its mass. We devote Linear Momentum and Collisions to the study of momentum.
For now, it is sufficient to define momentum \(\vec{p}\) as the product of the mass of the object m and its velocity \(\vec{v}\):
\[\vec{p} = m \vec{v} \ldotp \label{5.7}\]
Since velocity is a vector, so is momentum.
It is easy to visualize momentum. A train moving at 10 m/s has more momentum than one that moves at 2 m/s. In everyday life, we speak of one sports team as “having momentum” when they score points against the opposing team.
If we substitute Equation \ref{5.7} into Equation \ref{5.6}, we obtain
\[\vec{F}_{net} = \frac{d \vec{p}}{dt} = \frac{d (m \vec{v})}{dt} \ldotp\]
When m is constant, we have
\[\vec{F}_{net} = m \frac{d(\vec{v})}{dt} = m \vec{a} \ldotp\]
Thus, we see that the momentum form of Newton’s second law reduces to the form given earlier in this section.
Explore the forces at work when pulling a cart or pushing a refrigerator, crate, or person. Create an applied force and see how it makes objects move. Put an object on a ramp and see how it affects its motion.


9.18: A Comet Impacts a Planet
Except for seeing meteors enter our atmosphere, no contemporary astronomer had ever observed an impact on another world until 1994. Comet Shoemaker–Levy 9, S-L 9, was discovered in March 1993 by astronomers Eugene and Carolyn Shoemaker and David Levy. Carolyn’s first impression of the discovery photo was it “looked like a squashed comet. ” This description was due to the fact that the comet had been broken apart by the gravitational pull of Jupiter. The largest cometary fragment was 2 kilometers or 1. 2 miles across.
After determining the comet’s orbit, it was found that Shoemaker–Levy 9 fragments would impact Jupiter over several days, 22-26 July 1994. The question became if the impacts would be visible to astronomers and orbiting satellites, like the Hubble Space Telescope.
Over the five-day period, twenty-one (21) distinct impacts were observed by satellites like the Hubble Space Telescope, as well as Earth-based telescopes. Fireballs resembling an atomic bomb mushroom cloud were seen at Jupiter’s horizon. Dark spots were seen in the upper atmosphere of Jupiter after impact; these were visible even in small amateur telescopes. Some of these spots were as big as Earth.  Due to the S-L 9 series of impacts, astronomers were better able to explain rows or chains of craters found on the Moon and other objects, like Jupiter’s moon Ganymede.


11.13: Analyzing Costs
Learning Objectives
- Summarize the concept of different costs analyzed for different purposes
So what matters in one decision, may make no difference at all in another decision. As a manager, you will have the opportunity to make many decision.
Let’s watch some examples:
Now, you can see, that what may be relevant in one decision, will not be in another. We need to use ours skills we have been working on throughout this course to determine whether a particular cost matters when we are looking at different decisions.
You will need different costs for different purposes. But how will you decide which ones are relevant? Every time you come to a new situation, it will be important to evaluate, to get those irrelevant costs, that may cloud your decision making skills, out of the way.
Let’s practice identifying relevant costs. Are you ready? Let’s say you’re taking a fishing trip to Canda and you’re deciding whether it’s better to drive your own car or to fly and ship your fishing equipment. What would we need to look at to determine which made more sense? Might there be other costs besides the actual monetary costs?
- Car: Gas, insurance, maintenance and repairs, and depreciation
- Fly: Cost of the ticket, plus the cost of shipping the fishing stuff.
Which costs do we need to worry about? Well, the flying, both costs are relevant!! If we fly, those happen. But what about the car? Well, what costs would happen whether we drive to Canada or not? The insurance is gonna happen, whether we drive or fly, right? So that cost is irrelevant ! We could figure out the exact amount of depreciation and repairs related to the miles we would be driving to Canada, and factor those in as well while working on that decision.
What might a non-monetary (qualitative) cost be to driving? The hours it takes to get there, as opposed to the flight? If you fly you will get more time to fish! So even if driving is cheaper, you might decide to fly, right? The quality of the trip may be improved by flying!
So what is the take-away here?
Every single decision you need to make as a manager, will require you to analyze the costs, decide which ones are relevant or irrelevant, and then analyze the remaining costs, including the non-monetary (qualitative) ones and come to a conclusion!
We, as managers, need to look at the total cost, then decide what the differential costs are. We then need to reconcile those two sets of costs, to come to the best decision in a given situation! Are you ready to take on that challenge?
Practice Questions


1.3.2: Simplifying, Multiplying and Dividing Rational Expressions
By the end of this section, you will be able to:
- Determine the values for which a rational expression is undefined
- Simplify rational expressions
- Multiply rational expressions
- Divide rational expressions
Before you get started, take this readiness quiz.
- Simplify \(\dfrac{90y}{15y^2}\).
- Multiply \(\dfrac{14}{15}·\dfrac{6}{35}\).
- Divide \(\dfrac{12}{10}÷\dfrac{8}{25}\).
Above are examples of some properties of fractions and their operations. Recall that these fractions where the numerators and denominators are integers are called rational numbers. In this chapter, we will work with fractions whose numerators and denominators are polynomials. We call this kind of expression a rational expression .
A rational expression is an expression of the form \(\dfrac{p}{q}\), where \(p\) and \(q\) are polynomials and \(q\neq 0\).
Here are some examples of rational expressions:
\[−\dfrac{24}{56} \qquad \dfrac{5x}{12y} \qquad \dfrac{4x+1}{x^2−9} \qquad \dfrac{4x^2+3x−1}{2x−8}\nonumber\]
Notice that the first rational expression listed above, \(−\dfrac{24}{56}\), is just a fraction. Since a constant is a polynomial with degree zero, the ratio of two constants is a rational expression, provided the denominator is not zero.
We will do the same operations with rational expressions that we did with fractions. We will simplify, add, subtract, multiply, divide and use them in applications.
Simplify Rational Expressions
Here we recall for example, \(\dfrac12=\dfrac{6}{12}\), which can be understood by imagining dividing a pizza into 2 pieces and eating 1, and imagining dividing the pizza up into 12 pieces and eating 6 of them! In general, \(\dfrac{a}{b}=\dfrac{ca}{cb}\) when \(c\) and \(b\) are not zero. The process of replacing \(\dfrac{ca}{cb}\) by the equivalent expression \(\dfrac{a}{b}\) is called reducing the fraction. It is commonly also called 'canceling' the common factor \(c\).
A fraction is considered simplified if there are no common factors, other than \(1\) and \(-1\), in its numerator and denominator. Similarly, a simplified rational expression has no common factors, other than \(1\) and \(-1\), in its numerator and denominator.
A rational expression is considered simplified if there are no common factors other than \(1\) and \(-1\) in its numerator and denominator. There should be no more "\(-\)"s than necessary.
For example,
\[ \begin{array} {l} \dfrac{x+2}{x+3} \text{ is simplified because there are no common factors of } x+2 \text{ and }x+3. \\ \dfrac{2x}{3x} \text{ is not simplified because \(x\) is a common factor of }2x\text{ and }3x. \\ \end{array} \nonumber\]
We use the Equivalent Fractions Property to simplify numerical fractions. We restate it here as we will also use it to simplify rational expressions.
If \(a\), \(b\), and \(c\) are numbers where \(b\neq 0,c\neq 0,\)
\[\text {then } \dfrac{a}{b}=\dfrac{a\cdot c}{b\cdot c} \text{ and } \dfrac{a\cdot c}{b\cdot c}=\dfrac{a}{b}\nonumber\]
Notice that in the Equivalent Fractions Property, the values that would make the denominators zero are specifically disallowed. We see \(b\neq 0\), \(c\neq 0\) clearly stated.
To simplify rational expressions, we first write the numerator and denominator in factored form. Then we remove the common factors using the Equivalent Fractions Property.
Be very careful as you remove common factors. Factors are multiplied to make a product. You can remove a factor from a product. You cannot remove a term from a sum.
Removing the \(x\)'s from \(\dfrac{x+5}{x}\) would be like cancelling the \(2\)’s in the fraction \(\dfrac{2+5}{2}!\)
Simplifying Rational Expressions: Monomials over Monomials
Now we will look at an exponent property for division. As before, we’ll try to discover a property by looking at some examples.
| Consider | \(\quad\dfrac{x^5}{x^2}\) | and | \(\quad\dfrac{x^2}{x^3}\) |
|---|---|---|---|
| What do they mean? | \(=\dfrac{x\cdot x\cdot x\cdot x\cdot x}{x\cdot x}\) | \(=\dfrac{x\cdot x}{x\cdot x\cdot x}\) | |
| Use the Equivalent Fractions Property. | \(=\dfrac{\cancel{x}\cdot\cancel{x}\cdot x\cdot x\cdot x}{\cancel{x}\cdot \cancel{x}}\) | \(=\dfrac{\cancel{x}\cdot \cancel{x}\cdot 1}{\cancel{x}\cdot \cancel{x}\cdot x}\) | |
| Simplify. | \(=x^3\) |
\(=\dfrac{1}{x}\) (This is not a polynomial!) |
|
| Note. |
\(\quad\dfrac{x^5}{x^2}=\dfrac{x^2x^3}{x^2}=x^3\) Reducing the fraction. |
\(\quad\dfrac{x^2}{x^3}=\dfrac{x^2}{x^2x}=\dfrac{1}{x}\) Reducing the fraction. |
Note that the result when you divide monomials is not necessarily a polynomial!
Simplify \(\dfrac{54a^2b^3}{−6ab^5}\).
Solution
When we divide monomials with more than one variable, we write one fraction for each variable.
| \(\quad \dfrac{54a^2b^3}{−6ab^5}\) | |
|---|---|
| Rewrite as a fraction. | \(=\dfrac{54a^2b^3}{−6ab^5}\) |
| Use fraction multiplication. | \(=\dfrac{54}{−6}\cdot \dfrac{a^2}{a}\cdot\dfrac{b^3}{b^5}\) |
| Simplify and use the Quotient Property. | \(=−9a\cdot\dfrac{1}{b^2}\) |
| Multiply. | \(=−\dfrac{9a}{b^2}\) |
Simplify \(-\dfrac{72a^7b^3}{(8a^{12}b^4}\).
- Answer
-
\(−\dfrac{9}{a^5b}\)
Simplify \(−\dfrac{63c^8d^3}{7c^{12}d^2}\).
- Answer
-
\(\dfrac{−9d}{c^4}\)
Once you become familiar with the process and have practiced it step by step several times, you may be able to simplify a fraction in one step.
Simplify \(\dfrac{14x^7y^{12}}{21x^{11}y^6}\).
Solution
Be very careful to simplify \(\dfrac{14}{21}\) by dividing out a common factor, and to simplify the variables by subtracting their exponents.
| \(\quad\dfrac{14x^7y^{12}}{21x^{11}y^6}\) | |
| Simplify and use the Quotient Property. | \(=\dfrac{2y^6}{3x^4}\) |
Simplify \(\dfrac{28x^5y^{14}}{49x^9y^{12}}\).
- Answer
-
\(\dfrac{4y^2}{7x^4}\)
Simplify \(\dfrac{30m^5n^{11}}{48m^{10}n^{14}}\).
- Answer
-
\(\dfrac{5}{8m^5n^3}\)
Simplifying Rational Expressions: Polynomials over Monomials
Now that we know how to divide a monomial by a monomial, the next procedure is to divide a polynomial of two or more terms by a monomial. The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. It is clear that \(\dfrac{2+3}{7}=\dfrac{2}{7}+\dfrac{3}{7}\) because the left side is \(5\) sevenths and the right side is \(2\) sevenths plus \(3\) sevenths. You may also see this as \(\dfrac17(2+3)=\dfrac17 2+\dfrac17 3=\dfrac27+\dfrac37.\) So in other words, we distribute division over addition (and subtraction) just as we did multiplication. Since variables are place-holders for numbers, it will be the same if instead of numbers we have variables.
For example, the sum \(\dfrac{y}{5}+\dfrac{2}{5}\) simplifies to \(\dfrac{y+2}{5}\). Now we will do this in reverse to split a single fraction into separate fractions. For example, \(\dfrac{y+2}{5}\) can be written \(\dfrac{y}{5}+\dfrac{2}{5}\).
This is the “reverse” of fraction addition and it states that if a , b , and c are numbers where \(c\neq 0\), then \(\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}\). We will use this to divide polynomials by monomials.
To divide a polynomial by a monomial, divide each term of the polynomial by the monomial.
Simplify \(\dfrac{18x^3y−36xy^2}{−3xy}\).
Solution
| \(\quad \dfrac{18x^3y−36xy^2}{−3xy}\) | |
| Rewrite as a fraction. | \(=\dfrac{18x^3y−36xy^2}{−3xy}\) |
| Divide each term by the divisor. Be careful with the signs! | \(=\dfrac{18x^3y}{−3xy}−\dfrac{36xy^2}{−3xy}\) |
| Simplify. | \(=−6x^2+12y\) |
Simplify \(\dfrac{32a^2b−16ab^2}{−8ab}\).
- Answer
-
\(−4a+2b\)
Simplify \(\dfrac{−48a^8b^4−36a^6b^5}{−6a^3b^3}\).
- Answer
-
\(8a^5b+6a^3b^2\)
Simplifying Rational Expressions: Polynomials over Polynomials
Simplify \(\dfrac{x^2+5x+6}{x^2+8x+12}.\)
- Solution
-
\(\quad \dfrac{x^2+5x+6}{x^2+8x+12}\) Factor the numerator and denominator completely. \(= \dfrac{(x+2)(x+3)}{(x+2)(x+6)}\) Simplify by dividing out common factors. \(= \dfrac{\cancel{(x+2)}(x+3)}{\cancel{(x+2)}(x+6)}\)
\(=\dfrac{x+3}{x+6}\)
for the values of \(x\) where \
\(\quad \dfrac{x^2+5x+6}{x^2+8x+12}= \dfrac{(x+2)(x+3)}{(x+2)(x+6)}\) and \(\dfrac{x+3}{x+6}\)
are both defined.
Simplify \(\dfrac{x^2−x−2}{x^2−3x+2}\).
- Answer
-
\(\dfrac{x+1}{x−1}\) for values of \(x\) where \(\dfrac{x^2−x−2}{x^2−3x+2}\) is defined.
Simplify \(\dfrac{x^2−3x−10}{x^2+x−2}\).
- Answer
-
\(\dfrac{x−5}{x−1}\) where \(\dfrac{x^2−3x−10}{x^2+x−2}\) is defined.
Note that in the example it is noted that the equality is valid only where \(\quad \dfrac{x^2+5x+6}{x^2+8x+12}= \dfrac{(x+2)(x+3)}{(x+2)(x+6)}\) and \(\dfrac{x+3}{x+6}\) are defined. The latter is defined everywhere because one can always ad and multiply numbers. The problem is what when \(\quad \dfrac{x^2+5x+6}{x^2+8x+12}= \dfrac{(x+2)(x+3)}{(x+2)(x+6)}\) is not defined. For example, if \(x=-2\) the denominator is 0 (check by evaluating) and so the quotient is undefined whereas \(\dfrac{x+3}{x+6}=\dfrac{-2+3}{-2+6}=\dfrac{1}{4}\) which is defined so that for \(x=-2\), \(\dfrac{x^2+5x+6}{x^2+8x+12}\not=\dfrac{x+3}{x+6}\). Maybe you can spot another value that doesn't give equality. If the two expressions are defined then they are equal, which is most values of \(x\). In the context of an application, it will be necessary to go back and make sure the equality holds for the value of interest. We will see this in action in the next unit.
We now summarize the steps you should follow to simplify rational expressions.
- Factor the numerator and denominator completely.
- Simplify by noting common factors.
Usually, we leave the simplified rational expression in factored form. This way, it is easy to check that we have removed all the common factors.
We’ll use the methods we have learned to factor the polynomials in the numerators and denominators in the following examples.
Every time we write a rational expression, we should make a statement disallowing values that would make a denominator zero. However, to let us focus on the work at hand, we will omit writing it in the examples.
Simplify \(\dfrac{3a^2−12ab+12b^2}{6a^2−24b^2}\).
- Solution
-
\(\quad \dfrac{3a^2−12ab+12b^2}{6a^2−24b^2}\) Factor the numerator and denominator, first factoring out the GCF. \(= \dfrac{3(a^2−4ab+4b^2)}{6(a^2−4b^2)}\)
\(=\dfrac{3(a−2b)(a−2b)}{6(a+2b)(a−2b)}\)
Remove the common factors of \(a−2b\) and \(3\). \(=\dfrac{\cancel{3}(a−2b)\cancel{(a−2b)}}{\cancel{3}\cdot 2(a+2b)\cancel{(a−2b)}}\)
\(=\dfrac{a−2b}{2(a+2b)}\)
Conclude. \(\quad \dfrac{3a^2−12ab+12b^2}{6a^2−24b^2}==\dfrac{a−2b}{2(a+2b)}\)
where both sides are defined.
Simplify \(\dfrac{2x^2−12xy+18y^2}{3x^2−27y^2}\).
- Answer
-
\(\dfrac{2(x−3y)}{3(x+3y)}\)
Simplify \(\dfrac{5x^2−30xy+25y^2}{2x^2−50y^2}\).
- Answer
-
\(\dfrac{5(x−y)}{2(x+5y)}\)
Now we will see how to simplify a rational expression whose numerator and denominator have opposite factors. We previously introduced opposite notation: the opposite of \(a\) is \(−a\) and \(−a=−1\cdot a\).
The numerical fraction, say \(\dfrac{7}{−7}\) simplifies to \(−1\). We also recognize that the numerator and denominator are opposites.
The fraction \(\dfrac{a}{−a}\), whose numerator and denominator are opposites also simplifies to \(−1\).
\[\begin{array} {ll} \text{Let’s look at the expression }b−a. &b−a \\ \text{Rewrite.} &−a+b \\ \text{Factor out }–1. &−1(a−b) \nonumber\end{array} \]
This tells us that \(b−a\) is the opposite of \(a−b\).
In general, we could write the opposite of \(a−b\) as \(b−a\). So the rational expression \(\dfrac{a−b}{b−a}\) simplifies to \(−1\).
The opposite of \(a−b\) is \(b−a\).
\[\dfrac{a−b}{b−a}=−1 \quad a\neq b\nonumber\]
An expression and its opposite divide to \(−1\).
We will use this property to simplify rational expressions that contain opposites in their numerators and denominators. Be careful not to treat \(a+b\) and \(b+a\) as opposites. Recall that in addition, order doesn’t matter so \(a+b=b+a\). So if \(a\neq −b\), then \(\dfrac{a+b}{b+a}=1\).
Simplify \(\dfrac{x^2−4x−32}{64−x^2}.\)
- Solution
-
\(\quad \dfrac{x^2−4x−32}{64−x^2}\) Factor the numerator and the denominator. \(=\dfrac{(x-8)(x+4)}{(8-x)(8+x)}\)
Recognize the factors that are opposites. \(=({\color{red}{-1}})\dfrac{\cancel{(x-8)}(x+4)}{\cancel{(8-x)}(8+x)}\)
Simplify. \(=-\dfrac{x+4}{x+8}\)
Conclude. \(\dfrac{x^2−4x−32}{64−x^2}=-\dfrac{x+4}{x+8}\)
where both sides are defined.
Simplify \(\dfrac{x^2−4x−52}{5−x^2}\)
- Answer
-
\(−\dfrac{x+1}{x+5}\)
Simplify \(\dfrac{x^2+x−2}{1−x^2}\).
- Answer
-
\(−\dfrac{x+2}{x+1}\)
Multiplying Rational Expressions
To multiply rational expressions, we do just what we did with numerical fractions. We multiply the numerators and multiply the denominators. Then, if there are any common factors, we remove them to simplify the result.
If \(p\), \(q\), \(r\), and \(s\) are polynomials where \(q\neq 0\), \(s\neq 0\), then
\[\dfrac{p}{q}\cdot \dfrac{r}{s}=\dfrac{pr}{qs}.\nonumber\]
To multiply rational expressions, multiply the numerators and multiply the denominators.
Remember, throughout this chapter, we will assume that all numerical values that would make the denominator be zero are excluded. We will not write the restrictions for each rational expression, but keep in mind that the denominator can never be zero. So in this next example, \(x\neq 0\), \(x\neq 3\), and \(x\neq 4.\)
Simplify \(\dfrac{2x}{x^2−7x+12}\cdot \dfrac{x^2−9}{6x^2}\).
- Solution
-
Step 1. Factor each numerator and denominator completely. Factor \(x^2-9\) and \(x^2-7x+12\). \(\dfrac{2x}{x^2-7x+12}\cdot\dfrac{x^2-9}{6x^2}\)
\(=\dfrac{2x}{(x-3)(x-4)}\cdot\dfrac{(x-3)(x+3)}{6x^2}\)
Step 2. Multiply the numerators and denominators. Multiply the numerators and denominators. It is helpful to write the monomials first. \(=\dfrac{2x(x-3)(x+3)}{6x^2(x-3)(x-4)}\) Step 3. Reduce the fraction. Recognize factors common to the numerator and denominator and reduce the fraction. "Divide out" the common factors (divide numerator and denominator by those common factors). Leave the result in factored form. \(\dfrac{\cancel{2}\cancel{x}\cancel{(x-3)}(x+3)}{\cancel{2}\cdot 3\cdot \cancel{x}\cdot x\cancel{(x-3)}(x-4)}\)
\(=\dfrac{x+3}{3x(x-4)}\)
Conclude. \(\dfrac{2x}{x^2−7x+12}\cdot \dfrac{x^2−9}{6x^2}=\dfrac{x+3}{3x(x-4)}\)
where both sides are defined.
Note that we will stop writing the qualification that the expressions of interest are defined. It will be tacitly understood for the rest of this chapter.
Simplify \(\dfrac{5x}{x^2+5x+6}\cdot \dfrac{x^2−4}{10x}\).
- Answer
-
\(\dfrac{x−2}{2(x+3)}\)
Simplify \(\dfrac{9x^2}{x^2+11x+30}]\cdot\dfrac{x^2−36}{3x^2}\).
- Answer
-
\(\dfrac{3(x−6)}{x+5}\)
- Factor each numerator and denominator completely.
- Multiply the numerators and denominators.
- Simplify by dividing out the common factors.
Multiply \(\dfrac{3a^2−8a−3}{a^2−25}·\dfrac{a^2+10a+25}{3a^2−14a−5}\).
- Solution
-
\(\begin{array} {ll} &\dfrac{3a^2−8a−3}{a^2−25}·\dfrac{a^2+10a+25}{3a^2−14a−5} \\ & \\ \begin{array} {ll} \text{Factor the numerators and denominators} \\ \text{and then multiply.} \end{array} &\dfrac{(3a+1)(a−3)(a+5)(a+5)}{(a−5)(a+5)(3a+1)(a−5)} \\ & \\ \begin{array} {l} \text{Simplify by dividing out} \\ \text{common factors.} \end{array} &\dfrac{\cancel{(3a+1)}(a−3)\cancel{(a+5)}(a+5)}{(a−5)\cancel{(a+5)}\cancel{(3a+1)}(a−5)} \\ & \\ \text{Simplify.} &\dfrac{(a−3)(a+5)}{(a−5)(a−5)} \\ & \\ \text{Rewrite }(a−5)(a−5)\text{ using an exponent.} &\dfrac{(a−3)(a+5)}{(a−5)^2} \end{array}\),
where \(\dfrac{3a^2−8a−3}{a^2−25}·\dfrac{a^2+10a+25}{3a^2−14a−5}\) is defined (then \(\dfrac{(a−3)(a+5)}{(a−5)^2}\) is also defined).
Simplify \(\dfrac{2x^2+5x−12}{x^2−16}·\dfrac{x^2−8x+16}{2x^2−13x+15}\).
- Answer
-
\(\dfrac{x−4}{x−5}\)
Simplify \(\dfrac{4b^2+7b−2}{1−b^2}·\dfrac{b^2−2b+1}{4b^2+15b−4}\).
- Answer
-
\(−\dfrac{(b+2)(b−1)}{(1+b)(b+4)}\)
Dividing Rational Expressions
Just like we did for numerical fractions, to divide rational expressions, we multiply the first fraction by the reciprocal of the second.
If \(p\), \(q\), \(r\), and \(s\) are polynomials where \(q\neq 0\), \(r\neq 0\), \(s\neq 0\), then
\[\dfrac{p}{q}\div\dfrac{r}{s}=\dfrac{p}{q}\cdot \dfrac{s}{r}.\nonumber\]
To divide rational expressions, multiply the first fraction by the reciprocal of the second.
Once we rewrite the division as multiplication of the first expression by the reciprocal of the second, we then factor everything and look for common factors.
Divide \(\dfrac{p^3+q^3}{2p^2+2pq+2q^2}÷\dfrac{p^2−q^2}{6}\). Use the fact that \(p^3+q^3=(p+q)(p^2-pq+q^2)\).
- Solution
-
So, \(\dfrac{p^3+q^3}{2p^2+2pq+2q^2}÷\dfrac{p^2−q^2}{6}=\dfrac
Simplify \(\dfrac{x^3−8}{3x^2−6x+12}\div\dfrac{x^2-4}{6}\). Use the fact that \(x^3−8=(x-2)(x^2+2x+4)\).
- Answer
-
\(\dfrac{2(x^2+2x+4)}{(x+2)(x^2−2x+4)}\)
Simplify \(\dfrac{2z^2}{z^2−1}\div\dfrac{z^3−z^2+z}{z^3+1}\). Use \(z^3+1=(z+1)(z^2-z+1)\) and don't forget to factor out greatest common factors.
- Answer
-
\(\dfrac{2z}{z−1}\)
- Rewrite the division as the product of the first rational expression and the reciprocal of the second.
- Factor the numerators and denominators completely.
- Multiply the numerators and denominators together.
- Simplify by dividing out common factors.
A complex fraction is a fraction that contains a fraction in the numerator, the denominator or both. Recall that a fraction bar means division. A complex fraction is another way of writing division of two fractions.
Divide \(\dfrac{\dfrac{6x^2−7x+2}{4x−8}}{\dfrac{2x^2−7x+3}{x^2−5x+6}}\).
- Solution
-
\(\begin{array} {ll} &\dfrac{\dfrac{6x^2−7x+2}{4x−8}}{\dfrac{2x^2−7x+3}{x^2−5x+6}} \\ & \\ \text{Rewrite with a division sign.} &\dfrac{6x^2−7x+2}{4x−8}÷\dfrac{2x^2−7x+3}{x^2−5x+6} \\ & \\ \begin{array} {l} \text{Rewrite as product of first times reciprocal} \\ \text{of second.} \end{array} &\dfrac{6x^2−7x+2}{4x−8}·\dfrac{x^2−5x+6}{2x^2−7x+3} \\ & \\ \begin{array} {l} \text{Factor the numerators and the} \\ \text{denominators, and then multiply.} \end{array} &\dfrac{(2x−1)(3x−2)(x−2)(x−3)}{4(x−2)(2x−1)(x−3)} \\ & \\ \text{Simplify by dividing out common factors.} &\dfrac{\cancel{(2x−1)}(3x−2)\cancel{(x−2)}\cancel{(x−3)}}{4\cancel{(x−2)}\cancel{(2x−1)}\cancel{(x−3)}} \\ \text{Simplify.} &\dfrac{3x−2}{4} \end{array}\)
Simplify \(\dfrac{\dfrac{3x^2+7x+2}{4x+24}}{\dfrac{3x^2−14x−5}{x^2+x−30}}\).
- Answer
-
\(\dfrac{x+2}{4}\)
Simplify \(\dfrac{\dfrac{y^2−36}{2y^2+11y−6}}{\dfrac{2y^2−2y−60}{8y−4}}\).
- Answer
-
\(\dfrac{2}{y+5}\)
If we have more than two rational expressions to work with, we still follow the same procedure. The first step will be to rewrite any division as multiplication by the reciprocal. Then, we factor and multiply.
Perform the indicated operations: \(\dfrac{3x−6}{4x−4}·\dfrac{x^2+2x−3}{x^2−3x−10}÷\dfrac{2x+12}{8x+16}\).
- Solution
-
Rewrite the division as multiplication
by the reciprocal.Factor the numerators and the denominators. Multiply the fractions. Bringing the constants to
the front will help when removing common factors.Simplify by dividing out common factors. Simplify.
Perform the indicated operations: \(\dfrac{4m+4}{3m−15}·\dfrac{m^2−3m−10}{m^2−4m−32}÷\dfrac{12m−36}{6m−48}\).
- Answer
-
\(\dfrac{2(m+1)(m+2)}{3(m+4)(m−3)}\)
Perform the indicated operations: \(\dfrac{2n^2+10n}{n−1}÷\dfrac{n^2+10n+24}{n^2+8n−9}·\dfrac{n+4}{8n^2+12n}\).
- Answer
-
\(\dfrac{(n+5)(n+9)}{2(n+6)(2n+3)}\)
- What is a rational expression?
- What is the first goal when aiming at simplifying rational expressions? Why is this the first goal?
- Give an example of dividing a rational expression where the reduced result is a quotient of linear expressions.
-
a. Multiply \(\dfrac{7}{4}·\dfrac{9}{10}\) and explain all your steps.
b. Multiply \(\dfrac{n}{n−3}·\dfrac{9}{n+3}\) and explain all your steps.
c. Evaluate your answer to part b. when \(n=7\). Did you get the same answer you got in part a.? Why or why not -
a. Divide \(\dfrac{24}{5}÷6\) and explain all your steps.
b. Divide \(\dfrac{x^2−1}{x}÷(x+1)\) and explain all your steps.
c. Evaluate your answer to part b. when \(x=5\). Did you get the same answer you got in part a.? Why or why not?
Simplify:
a. \(\dfrac{5p^2}{p^2−5p−36}\cdot\dfrac{p^2−16}{10p}\)
b. \(\dfrac{2y^2−10y}{y^2}+\dfrac{10y+2}{5}\cdot\dfrac{y+5}{6y}\)
Key Concepts
-
Determine the values for which a rational expression is undefined.
- Set the denominator equal to zero.
- Solve the equation.
-
Equivalent Fractions Property
If \(a\), \(b\), and \(c\) are numbers where \(b\neq 0\), \(c\neq 0\), then
\(\quad\dfrac{a}{b}=\dfrac{a·c}{b·c}\) and \(\dfrac{a·c}{b·c}=\dfrac{a}{b}.\) -
How to simplify a rational expression.
- Factor the numerator and denominator completely.
- Simplify by dividing out common factors. (The resulting equality holds for values of the variable where the initial expression is defined.)
-
Opposites in a Rational Expression
The opposite of \(a−b\) is \(b−a\).
\(\quad\dfrac{a−b}{b−a}=−1 \qquad a\neq b\)
An expression and its opposite divide to \(−1\). -
Multiplication of Rational Expressions
If \(p\), \(q\), \(r\), and \(s\) are polynomials where \(q\neq 0\), \(s\neq 0\), then
\(\quad\dfrac{p}{q}·\dfrac{r}{s}=\dfrac{pr}{qs}\) -
How to multiply rational expressions.
- Factor each numerator and denominator completely.
- Multiply the numerators and denominators.
- Simplify by dividing out common factors.
-
Division of Rational Expressions
If \(p\), \(q\), \(r\), and \(s\) are polynomials where \(q\neq 0\), \(r\neq 0\), \(s\neq 0\), then
\(\quad\dfrac{p}{q}÷\dfrac{r}{s}=\dfrac{p}{q}·\dfrac{s}{r}\) -
How to divide rational expressions.
- Rewrite the division as the product of the first rational expression and the reciprocal of the second.
- Factor the numerators and denominators completely.
- Multiply the numerators and denominators together.
- Simplify by dividing out common factors.


12: Retail Pricing and Sales Strategies Last updated Save as PDF Page ID 44970 12.1: Price Elasticity Calculations 12.2: Introduction to Product Pricing Methods 12.3: Cost-Oriented Pricing Equations 12.4: Break-Even Point 12.5: Competition-Oriented Pricing 12.6: Demand-Oriented Pricing 12.7: Introduction to Price Adjustment Techniques 12.8: Markdown Optimization Software 12.9: Reducing Product Markdowns 12.10: Liquidating Markdown Merchandise 12.11: Why It Matters- Retail Pricing and Sales Strategies 12.12: Introduction to Pricing Techniques 12.13: Price Discrimination 12.14: Loss Leader 12.15: Price Lining 12.16: History of Odd Pricing 12.17: Putting It Together- Retail Pricing and Sales Strategies 12.18: Assignment- Retail Pricing 12.19: Introduction to Pricing in Retail 12.20: Defining Value 12.21: Retail Pricing Strategies 12.22: Pricing and the Retail Mix 12.23: Introduction to Pricing Objectives 12.24: External Factors and Retail Pricing 12.25: Pricing Strategies 12.26: Price Elasticity


6.3: The Uniform Distribution
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The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive.
Example 5.3.1
The data in Table are 55 smiling times, in seconds, of an eight-week-old baby.
| 10.4 | 19.6 | 18.8 | 13.9 | 17.8 | 16.8 | 21.6 | 17.9 | 12.5 | 11.1 | 4.9 |
| 12.8 | 14.8 | 22.8 | 20.0 | 15.9 | 16.3 | 13.4 | 17.1 | 14.5 | 19.0 | 22.8 |
| 1.3 | 0.7 | 8.9 | 11.9 | 10.9 | 7.3 | 5.9 | 3.7 | 17.9 | 19.2 | 9.8 |
| 5.8 | 6.9 | 2.6 | 5.8 | 21.7 | 11.8 | 3.4 | 2.1 | 4.5 | 6.3 | 10.7 |
| 8.9 | 9.4 | 9.4 | 7.6 | 10.0 | 3.3 | 6.7 | 7.8 | 11.6 | 13.8 | 18.6 |
The sample mean = 11.49 and the sample standard deviation = 6.23.
We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution.
Let \(X =\) length, in seconds, of an eight-week-old baby's smile.
The notation for the uniform distribution is
\(X \sim U(a, b)\) where \(a =\) the lowest value of \(x\) and \(b =\) the highest value of \(x\).
The probability density function is \(f(x) = \frac{1}{b-a}\) for \(a \leq x \leq b\).
For this example, \(X \sim U(0, 23)\) and \(f(x) = \frac{1}{23-0}\) for \(0 \leq X \leq 23\).
Formulas for the theoretical mean and standard deviation are
\(\mu = \frac{a+b}{2}\) and \(\sigma = \sqrt{\frac{(b-a)^{2}}{12}}\)
For this problem, the theoretical mean and standard deviation are
\(\mu = \frac{0+23}{2} = 11.50\) seconds and \(\sigma = \sqrt{\frac{(23-0)^{2}}{12}} = 6.64\) seconds.
Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example.
Exercise 5.3.1
The data that follow are the number of passengers on 35 different charter fishing boats. The sample mean = 7.9 and the sample standard deviation = 4.33. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. State the values of a and \(b\). Write the distribution in proper notation, and calculate the theoretical mean and standard deviation.
| 1 | 12 | 4 | 10 | 4 | 14 | 11 |
| 7 | 11 | 4 | 13 | 2 | 4 | 6 |
| 3 | 10 | 0 | 12 | 6 | 9 | 10 |
| 5 | 13 | 4 | 10 | 14 | 12 | 11 |
| 6 | 10 | 11 | 0 | 11 | 13 | 2 |
Answer
\(a\) is zero; \(b\) is \(14\); \(X \sim U (0, 14)\); \(\mu = 7\) passengers; \(\sigma = 4.04\) passengers
Example 5.3.2
Exercise <IP_ADDRESS>
a. Refer to Example . What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds?
Answer
a. Find \(P(2 < x < 18)\).
\(P(2 < x < 18) = (\text{base})(\text{height}) = (18 – 2)\left(\frac{1}{23}\right) = \left(\frac{16}{23}\right)\).
Figure 5.2.1.
Exercise <IP_ADDRESS>
b. Find the 90 th percentile for an eight-week-old baby's smiling time.
Answer
b. Ninety percent of the smiling times fall below the 90 th percentile, \(k\), so \(P(x < k) = 0.90\)
\[P(x < k)= 0.90\]
\[(\text{base})(\text{height}) = 0.90\]
\[(k−0)\left(\frac{1}{23}\right) = 0.90\]
\[k = (23)(0.90) = 20.7\]
Figure 5.3.2.
Exercise 5.3.3
c. Find the probability that a random eight-week-old baby smiles more than 12 seconds KNOWING that the baby smiles MORE THAN EIGHT SECONDS .
Answer
c. This probability question is a conditional . You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby has smiled for more than eight seconds.
Find \(P(x > 12 | x > 8)\) There are two ways to do the problem. For the first way , use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds.
Write a new \(f(x): f(x) = \frac{1}{23-8} = \frac{1}{15}\)
for \(8 < x < 23\)
\(P(x > 12 | x > 8) = (23 − 12)\left(\frac{1}{15}\right) = \left(\frac{11}{15}\right)\)
Figure 5.3.3.For the second way , use the conditional formula from Probability Topics with the original distribution \(X \sim U(0, 23)\):
\(P(\text{A|B}) = \frac{P(\text{A AND B})}{P(\text{B})}\)
For this problem, \(\text{A}\) is (\(x > 12\)) and \(\text{B}\) is (\(x > 8\)).
So, \(P(x > 12|x > 8) = \frac{(x > 12 \text{ AND } x > 8)}{P(x > 8)} = \frac{P(x > 12)}{P(x > 8)} = \frac{\frac{11}{23}}{\frac{15}{23}} = \frac{11}{15}\)
Figure 5.3.4.Exercise 5.3.2
A distribution is given as \(X \sim U(0, 20)\). What is \(P(2 < x < 18)\)? Find the 90 th percentile.
Answer
\(P(2 < x < 18) = 0.8\); 90 th percentile \(= 18\)
Example 5.3.3
The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive.
Exercise <IP_ADDRESS>
a. What is the probability that a person waits fewer than 12.5 minutes?
Answer
a. Let \(X =\) the number of minutes a person must wait for a bus. \(a = 0\) and \(b = 15\). \(X \sim U(0, 15)\). Write the probability density function. \(f(x) = \frac{1}{15-0} = \frac{1}{15}\) for \(0 \leq x \leq 15\).
Find \(P(x < 12.5)\). Draw a graph.
\(P(x < k) = (\text{base})(\text{height}) = (12.5−0)\left(\frac{1}{15}\right) = 0.8333\)
The probability a person waits less than 12.5 minutes is 0.8333.
Figure 5.3.5.Exercise <IP_ADDRESS>
b. On the average, how long must a person wait? Find the mean, \(\mu\), and the standard deviation, \(\sigma\).
Answer
b. \(\mu = \frac{a+b}{2} = \frac{15+0}{2} = 7.5\). On the average, a person must wait 7.5 minutes.
\(\sigma = \sqrt{\frac{(b-a)^{2}}{12}} = \sqrt{\frac{(12-0)^{2}}{12}} = 4.3\). The Standard deviation is 4.3 minutes.
Exercise <IP_ADDRESS>
c. Ninety percent of the time, the time a person must wait falls below what value?
Note <IP_ADDRESS>.1
This asks for the 90 th percentile.
Answer
c. Find the 90 th percentile. Draw a graph. Let \(k =\) the 90 th percentile.
\(P(x < k) = (\text{base})(\text{height}) = (k−0)\left(\frac{1}{15}\right)\)
\(0.90 = (k)\left(\frac{1}{15}\right)\)
\(k = (0.90)(15) = 13.5\)
\(k\) is sometimes called a critical value.
The 90
th
percentile is 13.5 minutes. Ninety percent of the time, a person must wait at most 13.5 minutes.
Exercise 5.3.4
The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive.
- Find \(a\) and \(b\) and describe what they represent.
- Write the distribution.
- Find the mean and the standard deviation.
- What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours?
- What is the 65 th percentile for the duration of games for a team for the 2011 season?
Answer
- \(a\) is \(447\), and \(b\) is \(521\). a is the minimum duration of games for a team for the 2011 season, and \(b\) is the maximum duration of games for a team for the 2011 season.
- \(X \sim U(447, 521)\).
-
\(\mu = 484\), and \(\sigma = 21.36\)
Figure 5.3.1.
- \(P(480 < x < 500) = 0.2703\)
- 65 th percentile is 495.1 hours.
Example 5.3.4
Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. Let \(X =\) the time, in minutes, it takes a nine-year old child to eat a donut. Then \(X \sim U(0.5, 4)\).
a. The probability that a randomly selected nine-year old child eats a donut in at least two minutes is _______.
Solution
a. 0.5714
Exercise <IP_ADDRESS>
b. Find the probability that a different nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes.
The second question has a conditional probability. You are asked to find the probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solve the problem two different ways (see Example ). You must reduce the sample space. First way : Since you know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is 1.5 minutes.
Write a new \(f(x)\):
\(f(x) = \frac{1}{4-1.5} = \frac{2}{5}\) for \(1.5 \leq x \leq 4\).
Find \(P(x > 2|x > 1.5)\). Draw a graph.
Figure 5.3.2.\(P(x > 2|x > 1.5) = (\text{base})(\text{new height}) = (4 − 2)(25)\left(\frac{2}{5}\right) =\) ?
Answer
b. \(\frac{4}{5}\)
The probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes is \(\frac{4}{5}\).
Second way: Draw the original graph for \(X \sim U(0.5, 4)\). Use the conditional formula
\(P(x > 2 | x > 1.5) = \frac{P(x > 2 \text{AND} x > 1.5)}{P(x > 1.5)} = \frac{P(x>2)}{P(x>1.5)} = \frac{\frac{2}{3.5}}{\frac{2.5}{3.5}} = 0.8 = \frac{4}{5}\)
Exercise 5.3.5
Suppose the time it takes a student to finish a quiz is uniformly distributed between six and 15 minutes, inclusive. Let \(X =\) the time, in minutes, it takes a student to finish a quiz. Then \(X \sim U(6, 15)\).
Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes.
Answer
\(P(x > 8) = 0.7778\)
\(P(x > 8 | x > 7) = 0.875\)
Example 5.3.5
Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix a furnace is uniformly distributed between 1.5 and four hours. Let \(x =\) the time needed to fix a furnace. Then \(x \sim U(1.5, 4)\).
- Find the probability that a randomly selected furnace repair requires more than two hours.
- Find the probability that a randomly selected furnace repair requires less than three hours.
- Find the 30 th percentile of furnace repair times.
- The longest 25% of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25% of repair times.) What percentile does this represent?
- Find the mean and standard deviation
Solution
a. To find \(f(x): f(x) = \frac{1}{4-1.5} = \frac{1}{2.5}\) so \(f(x) = 0.4\)
\(P(x > 2) = (\text{base})(\text{height}) = (4 – 2)(0.4) = 0.8\)
Figure 5.3.3. Uniform Distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time \(x\) is greater than twob. \(P(x < 3) = (\text{base})(\text{height}) = (3 – 1.5)(0.4) = 0.6\)
The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between \(x = 1.5\) and \(x = 3\). Note that the shaded area starts at \(x = 1.5\) rather than at \(x = 0\); since \(X \sim U(1.5, 4)\), \(x\) can not be less than 1.5.
Figure 5.3.4. Uniform Distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time \(x\) is less than threec.
Figure 5.3.5. Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times.
\(P(x < k) = 0.30\)
\(P(x < k) = (\text{base})(\text{height}) = (k – 1.5)(0.4)\)
\(0.3 = (k – 1.5) (0.4)\)
; Solve to find \(k\):
\(0.75 = k – 1.5\), obtained by dividing both sides by 0.4
\(k = 2.25\)
, obtained by adding 1.5 to both sides
The 30 th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less.
d.
Figure 5.3.6. Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times.
\(P(x > k) = 0.25\)
\(P(x > k) = (\text{base})(\text{height}) = (4 – k)(0.4)\)
\(0.25 = (4 – k)(0.4)\)
; Solve for \(k\):
\(0.625 = 4 − k\),
obtained by dividing both sides by 0.4
\(−3.375 = −k\),
obtained by subtracting four from both sides:
\(k = 3.375\)
The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer).
Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 3.375 hours is the 75 th percentile of furnace repair times.
e. \(\mu = \frac{a+b}{2}\) and \(\sigma = \sqrt{\frac{(b-a)^{2}}{12}}\)
\(\mu = \frac{1.5+4}{2} = 2.75\) hours and \(\sigma = \sqrt{\frac{(4-1.5)^{2}}{12}} = 0.7217\) hours
Exercise 5.3.6
The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. Let \(X =\) the time needed to change the oil on a car.
- Write the random variable \(X\) in words. \(X =\) __________________.
- Write the distribution.
- Graph the distribution.
- Find \(P(x > 19)\).
- Find the 50 th percentile.
Answer
- Let \(X =\) the time needed to change the oil in a car.
- \(X \sim U(11, 21)\).
-
Figure 5.3.7.
- \(P(x > 19) = 0.2\)
- the 50 th percentile is 16 minutes.
Chapter Review
If \(X\) has a uniform distribution where \(a < x < b\) or \(a \leq x \leq b\), then \(X\) takes on values between \(a\) and \(b\) (may include \(a\) and \(b\)). All values \(x\) are equally likely. We write \(X \sim U(a, b)\). The mean of \(X\) is \(\mu = \frac{a+b}{2}\). The standard deviation of \(X\) is \(\sigma = \sqrt{\frac{(b-a)^{2}}{12}}\). The probability density function of \(X\) is \(f(x) = \frac{1}{b-a}\) for \(a \leq x \leq b\). The cumulative distribution function of \(X\) is \(P(X \leq x) = \frac{x-a}{b-a}\). \(X\) is continuous.
Figure 5.3.8.The probability \(P(c < X < d)\) may be found by computing the area under \(f(x)\), between \(c\) and \(d\). Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height.
Formula Review
\(X =\) a real number between \(a\) and \(b\) (in some instances, \(X\) can take on the values \(a\) and \(b\)). \(a =\) smallest \(X\); \(b =\) largest \(X\)
\(X \sim U(a, b)\)
The mean is \(\mu = \frac{a+b}{2}\)
The standard deviation is \(\sigma = \sqrt{\frac{(b-a)^{2}}{12}}\)
Probability density function: \(f(x) = \frac{1}{b-a} \text{for} a \leq X \leq b\)
Area to the Left of \(x\): \(P(X < x) = (x – a)\left(\frac{1}{b-a}\right)\)
Area to the Right of \(x\): P(\(X\) > \(x\)) = (b – x)\(\left(\frac{1}{b-a}\right)\)
Area Between \(c\) and \(d\): \(P(c < x < d) = (\text{base})(\text{height}) = (d – c)\left(\frac{1}{b-a}\right)\)
Uniform: \(X \sim U(a, b)\) where \(a < x < b\)
- pdf: \(f(x) = \frac{1}{b-a}\) for \(a \leq x \leq b\)
- cdf: \(P(X \leq x) = \frac{x-a}{b-a}\)
- mean \(\mu = \frac{a+b}{2}\)
- standard deviation \(\sigma = \sqrt{\frac{(b-a)^{2}}{12}}\)
- \(P(c < X < d) = (d – c)\left(\frac{1}{b-a}\right)\)
References
McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995.
Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes.
| 1.5 | 2.4 | 3.6 | 2.6 | 1.6 | 2.4 | 2.0 |
| 3.5 | 2.5 | 1.8 | 2.4 | 2.5 | 3.5 | 4.0 |
| 2.6 | 1.6 | 2.2 | 1.8 | 3.8 | 2.5 | 1.5 |
| 2.8 | 1.8 | 4.5 | 1.9 | 1.9 | 3.1 | 1.6 |
The sample mean = 2.50 and the sample standard deviation = 0.8302.
The distribution can be written as \(X \sim U(1.5, 4.5)\).
Exercise 5.3.7
What type of distribution is this?
Exercise 5.3.8
In this distribution, outcomes are equally likely. What does this mean?
Answer
It means that the value of x is just as likely to be any number between 1.5 and 4.5.
Exercise 5.3.9
What is the height of \(f(x)\) for the continuous probability distribution?
Exercise 5.3.10
What are the constraints for the values of \(x\)?
Answer
\(1.5 \leq x \leq 4.5\)
Exercise 5.3.11
Graph \(P(2 < x < 3)\).
Exercise 5.3.12
What is \(P(2 < x < 3)\)?
Answer
0.3333
Exercise 5.3.13
What is \(P(x < 3.5 | x < 4)\)?
Exercise 5.3.14
What is \(P(x = 1.5)\)?
Answer
zero
Exercise 5.3.15
What is the 90 th percentile of square footage for homes?
Exercise 5.3.16
Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet.
Answer
0.6
Exercise 5.3.17
What is \(a\)? What does it represent?
Exercise 5.3.18
What is \(b\)? What does it represent?
Answer
\(b\) is \(12\), and it represents the highest value of \(x\).
Exercise 5.3.19
What is the probability density function?
Exercise 5.3.20
What is the theoretical mean?
Answer
six
Exercise 5.3.21
What is the theoretical standard deviation?
Exercise 5.3.22
Draw the graph of the distribution for \(P(x > 9)\).
Answer
Figure 5.3.9.Exercise 5.3.23
Find \(P(x > 9)\).
Exercise 5.3.24
Find the 40 th percentile.
Answer
4.8
Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years.
Exercise 5.3.25
What is being measured here?
Exercise 5.3.26
In words, define the random variable \(X\).
Answer
\(X\) = The age (in years) of cars in the staff parking lot
Exercise 5.3.27
Are the data discrete or continuous?
Exercise 5.3.28
The interval of values for \(x\) is ______.
Answer
0.5 to 9.5
Exercise 5.3.29
The distribution for \(X\) is ______.
Exercise 5.3.30
Write the probability density function.
Answer
\(f(x) = \frac{1}{9}\) where \(x\) is between 0.5 and 9.5, inclusive.
Exercise 5.3.31
Graph the probability distribution.
-
Sketch the graph of the probability distribution.
Figure 5.3.10.
-
Identify the following values:
- Lowest value for \(\bar{x}\): _______
- Highest value for \(\bar{x}\): _______
- Height of the rectangle: _______
- Label for x -axis (words): _______
- Label for y -axis (words): _______
Exercise 5.3.32
Find the average age of the cars in the lot.
Answer
\(\mu\) = 5
Exercise 5.3.33
Find the probability that a randomly chosen car in the lot was less than four years old.
-
Sketch the graph, and shade the area of interest.
Figure 5.3.11.
- Find the probability. \(P(x < 4) =\) _______
Exercise 5.3.34
Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old.
-
Sketch the graph, shade the area of interest.
Figure 5.3.12.
- Find the probability. \(P(x < 4 | x < 7.5) =\) _______
Answer
- Check student’s solution.
- \(\frac{3.5}{7}\)
Exercise 5.3.35
What has changed in the previous two problems that made the solutions different
Exercise 5.3.36
Find the third quartile of ages of cars in the lot. This means you will have to find the value such that \(\frac{3}{4}\), or 75%, of the cars are at most (less than or equal to) that age.
-
Sketch the graph, and shade the area of interest.
Figure 5.3.13.
- Find the value \(k\) such that \(P(x < k) = 0.75\).
- The third quartile is _______
Answer
- Check student's solution.
- \(k = 7.25\)
- \(7.25\)
Glossary
- Conditional Probability
- the likelihood that an event will occur given that another event has already occurred
Contributors
-
Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at<EMAIL_ADDRESS>.


5.7: The Covariant Derivative
In the preceding section we were able to estimate a nontrivial general relativistic effect, the geodetic precession of the gyroscopes aboard Gravity Probe B, up to a unitless constant 3\(\pi\). Let’s think about what additional machinery would be needed in order to carry out the calculation in detail, including the 3\(\pi\).
First we would need to know the Einstein field equation , but in a vacuum this is fairly straightforward:
\[R_{ab} = 0.\]
Einstein posited this equation based essentially on the considerations laid out in Section 5.1 .
But just knowing that a certain tensor vanishes identically in the space surrounding the earth clearly doesn’t tell us anything explicit about the structure of the spacetime in that region. We want to know the metric. As suggested at the beginning of the chapter, we expect that the first derivatives of the metric will give a quantity analogous to the gravitational field of Newtonian mechanics, but this quantity will not be directly observable, and will not be a tensor. The second derivatives of the metric are the ones that we expect to relate to the Ricci tensor \(R_{ab}\).
The Covariant Derivative in Electromagnetism
We’re talking blithely about derivatives, but it’s not obvious how to define a derivative in the context of general relativity in such a way that taking a derivative results in well-behaved tensor.
To see how this issue arises, let’s retreat to the more familiar terrain of electromagnetism. In quantum mechanics, the phase of a charged particle’s wavefunction is unobservable, so that for example the transformation \(\Psi \rightarrow − \Psi\) does not change the results of experiments. As a less trivial example, we can redefine the ground of our electrical potential, \(\Phi \rightarrow \Phi + \delta \Phi\), and this will add a constant onto the energy of every electron in the universe, causing their phases to oscillate at a greater rate due to the quantum-mechanical relation
\[E = hf.\]
There are no observable consequences, however, because what is observable is the phase of one electron relative to another, as in a double-slit interference experiment. Since every electron has been made to oscillate faster, the effect is simply like letting the conductor of an orchestra wave her baton more quickly; every musician is still in step with every other musician. The rate of change of the wavefunction, i.e., its derivative, has some built-in ambiguity.
For simplicity, let’s now restrict ourselves to spin-zero particles, since details of electrons’ polarization clearly won’t tell us anything useful when we make the analogy with relativity. For a spin-zero particle, the wavefunction is simply a complex number, and there are no observable consequences arising from the transformation
\[\Psi \rightarrow \Psi' = e^{i \alpha} \Psi\]
where \(\alpha\) is a constant. The transformation \(\Phi \rightarrow \Phi + \delta \Phi\) is also allowed, and it gives \(\alpha (t) = (\frac{q \delta \Phi}{\hbar})t\), so that the phase factor e i\(\alpha\)(t) is a function of time \(t\). Now from the point of view of electromagnetism in the age of Maxwell, with the electric and magnetic fields imagined as playing their roles against a background of Euclidean space and absolute time, the form of this time-dependent phase factor is very special and symmetrical; it depends only on the absolute time variable. But to a relativist, there is nothing very nice about this function at all, because there is nothing special about a time coordinate. If we’re going to allow a function of this form, then based on the coordinate-invariance of relativity, it seems that we should probably allow α to be any function at all of the spacetime coordinates. The proper generalization of \(\Phi \rightarrow \Phi - \delta \Phi\) is now A b → A b − \(\partial_{b} \alpha\), where A b is the electromagnetic potential four-vector ( section 4.2 ).
Exercise \(\PageIndex{1}\)
Self-check: Suppose we said we would allow \(\alpha\) to be a function of t, but forbid it to depend on the spatial coordinates. Prove that this would violate Lorentz invariance.
The transformation has no effect on the electromagnetic fields, which are the direct observables. We can also verify that the change of gauge will have no effect on observable behavior of charged particles. This is because the phase of a wavefunction can only be determined relative to the phase of another particle’s wavefunction, when they occupy the same point in space and, for example, interfere. Since the phase shift depends only on the location in spacetime, there is no change in the relative phase.
But bad things will happen if we don’t make a corresponding adjustment to the derivatives appearing in the Schrödinger equation. These derivatives are essentially the momentum operators, and they give different results when applied to \(\Psi'\) than when applied to \(\Psi\):
\[\begin{split} \partial_{b} \Psi &\rightarrow \partial_{b} (e^{i \alpha} \Psi) \\ &= e^{i \alpha} \partial_{b} \Psi + i \partial_{b} \alpha (e^{i \alpha} \Psi) \\ &= (\partial_{b} + A'_{b} - A_{b}) \Psi' \end{split}\]
To avoid getting incorrect results, we have to do the substitution \(\partial_{b} \rightarrow \partial_{b} + ieA_{b}\), where the correction term compensates for the change of gauge. We call the operator \(\nabla\) defined as
\[\nabla_{b} = \partial_{b} + ieA_{b}\]
the covariant derivative . It gives the right answer regardless of a change of gauge.
The Covariant Derivative in General Relativity
Now consider how all of this plays out in the context of general relativity. The gauge transformations of general relativity are arbitrary smooth changes of coordinates. One of the most basic properties we could require of a derivative operator is that it must give zero on a constant function. A constant scalar function remains constant when expressed in a new coordinate system, but the same is not true for a constant vector function, or for any tensor of higher rank. This is because the change of coordinates changes the units in which the vector is measured, and if the change of coordinates is nonlinear, the units vary from point to point.
Consider the one-dimensional case, in which a vector v a has only one component, and the metric is also a single number, so that we can omit the indices and simply write v and g. (We just have to remember that v is really a covariant vector, even though we’re leaving out the upper index.) If v is constant, its derivative \(\frac{dv}{dx}\), computed in the ordinary way without any correction term, is zero. If we further assume that the coordinate x is a normal coordinate, so that the metric is simply the constant g = 1, then zero is not just the answer but the right answer. (The existence of a preferred, global set of normal coordinates is a special feature of a one-dimensional space, because there is no curvature in one dimension. In more than one dimension, there will typically be no possible set of coordinates in which the metric is constant, and normal coordinates only give a metric that is approximately constant in the neighborhood around a certain point. See Figure 5.3.7 for an example of normal coordinates on a sphere, which do not have a constant metric.)
Now suppose we transform into a new coordinate system X, which is not normal. The metric G, expressed in this coordinate system, is not constant. Applying the tensor transformation law, we have \(V = v \frac{dX}{dx}\), and differentiation with respect to X will not give zero, because the factor \(\frac{dX}{dx}\) isn’t constant. This is the wrong answer: V isn’t really varying, it just appears to vary because G does.
We want to add a correction term onto the derivative operator \(\frac{d}{dX}\), forming a covariant derivative operator \(\nabla_{X}\) that gives the right answer. This correction term is easy to find if we consider what the result ought to be when differentiating the metric itself. In general, if a tensor appears to vary, it could vary either because it really does vary or because the metric varies. If the metric itself varies, it could be either because the metric really does vary or . . . because the metric varies. In other words, there is no sensible way to assign a nonzero covariant derivative to the metric itself, so we must have \(\nabla_{X}\)G = 0. The required correction therefore consists of replacing \(\frac{d}{dX}\) with
\[\nabla_{X} = \frac{d}{dX} - G^{-1} \frac{dG}{dX} \ldotp\]
Applying this to G gives zero. G is a second-rank contravariant tensor. If we apply the same correction to the derivatives of other second-rank contravariant tensors, we will get nonzero results, and they will be the right nonzero results. For example, the covariant derivative of the stress-energy tensor T (assuming such a thing could have some physical significance in one dimension!) will be \(\nabla_{X} T = \frac{dT}{dX} − G^{-1} (\frac{dG}{dX})T\).
Physically, the correction term is a derivative of the metric, and we’ve already seen that the derivatives of the metric (1) are the closest thing we get in general relativity to the gravitational field, and (2) are not tensors. In 1+1 dimensions, suppose we observe that a free-falling rock has \(\frac{dV}{dT}\) = 9.8 m/s 2 . This acceleration cannot be a tensor, because we could make it vanish by changing from Earthfixed coordinates X to free-falling (normal, locally Lorentzian) coordinates x, and a tensor cannot be made to vanish by a change of coordinates. According to a free-falling observer, the vector v isn’t changing at all; it is only the variation in the Earth-fixed observer’s metric G that makes it appear to change.
Mathematically, the form of the derivative is \((\frac{1}{y}) \frac{dy}{dx}\), which is known as a logarithmic derivative, since it equals \(\frac{d(\ln y)}{dx}\). It measures the multiplicative rate of change of y. For example, if y scales up by a factor of k when x increases by 1 unit, then the logarithmic derivative of y is ln k. The logarithmic derivative of e cx is c. The logarithmic nature of the correction term to \(\nabla_{X}\) is a good thing, because it lets us take changes of scale, which are multiplicative changes, and convert them to additive corrections to the derivative operator. The additivity of the corrections is necessary if the result of a covariant derivative is to be a tensor, since tensors are additive creatures.
What about quantities that are not second-rank covariant tensors? Under a rescaling of contravariant coordinates by a factor of k, covariant vectors scale by k −1 , and second-rank covariant tensors by k −2 . The correction term should therefore be half as much for covariant vectors,
\[\nabla_{X} = \frac{d}{dX} - \frac{1}{2} G^{-1} \frac{dG}{dX} \ldotp\]
and should have an opposite sign for contravariant vectors.
Generalizing the correction term to derivatives of vectors in more than one dimension, we should have something of this form:
\[\begin{split} \nabla_{a} v^{b} &= \partial_{a} v^{b} + \Gamma^{b}_{ac} v^{c} \\ \nabla_{a} v_{b} &= \partial_{a} v_{b} - \Gamma^{c}_{ba} v_{c}, \end{split}\]
where \(\Gamma^{b}_{ac}\), called the Christoffel symbol, does not transform like a tensor, and involves derivatives of the metric. (“Christoffel” is pronounced “Krist-AWful,” with the accent on the middle syllable.) The explicit computation of the Christoffel symbols from the metric is deferred until section 5.9 , but the intervening sections 5.7 and 5.8 can be omitted on a first reading without loss of continuity.
An important gotcha is that when we evaluate a particular component of a covariant derivative such as \(\nabla_{2} v^{3}\), it is possible for the result to be nonzero even if the component v 3 vanishes identically. This can be seen in example 5 and example 21 .
Example 9: Christoffel symbols on the globe
As a qualitative example, consider the geodesic airplane trajectory shown in Figure 5.6.4, from London to Mexico City. In physics it is customary to work with the colatitude , \(\theta\), measured down from the north pole, rather then the latitude, measured from the equator. At P, over the North Atlantic, the plane’s colatitude has a minimum. (We can see, without having to take it on faith from the figure, that such a minimum must occur. The easiest way to convince oneself of this is to consider a path that goes directly over the pole, at \(\theta\) = 0.)
At P, the plane’s velocity vector points directly west. At Q, over New England, its velocity has a large component to the south. Since the path is a geodesic and the plane has constant speed, the velocity vector is simply being parallel-transported; the vector’s covariant derivative is zero. Since we have v \(\theta\) = 0 at P, the only way to explain the nonzero and positive value of \(\partial_{\phi} v^{\theta}\) is that we have a nonzero and negative value of \(\Gamma^{\theta}_{\phi \phi}\).
By symmetry, we can infer that \(\Gamma^{\theta}_{\phi \phi}\) must have a positive value in the southern hemisphere, and must vanish at the equator.
\(\Gamma^{\theta}_{\phi \phi}\) is computed in example 10 .
Symmetry also requires that this Christoffel symbol be independent of \(\phi\), and it must also be independent of the radius of the sphere.
Example 9 is in two spatial dimensions. In spacetime, \(\Gamma\) is essentially the gravitational field (see problem 7 ), and early papers in relativity essentially refer to it that way. 9 This may feel like a joyous reunion with our old friend from freshman mechanics, g = 9.8 m/s. But our old friend has changed. In Newtonian mechanics, accelerations like g are frame-invariant (considering only inertial frames, which are the only legitimate ones in that theory). In general relativity they are frame-dependent, and as we saw earlier, the acceleration of gravity can be made to equal anything we like, based on our choice of a frame of reference.
To compute the covariant derivative of a higher-rank tensor, we just add more correction terms, e.g.,
\[\nabla_{a} U_{bc} = \partial_{a} U_{bc} - \Gamma^{d}_{ba} U_{dc} - \Gamma^{d}_{ca} U_{bd}\]
or
\[\nabla_{a} U_{b}^{c} = \partial_{a} U_{b}^{c} - \Gamma^{d}_{ba} U_{d}^{c} + \Gamma^{c}_{ad} U_{b}^{d} \ldotp\]
With the partial derivative \(\partial_{\mu}\), it does not make sense to use the metric to raise the index and form \(\partial^{\mu}\). It does make sense to do so with covariant derivatives, so \(\nabla^{a} = g^{ab} \nabla_{b}\) is a correct identity.
Comma, Semicolon, and Birdtracks Notation
Some authors use superscripts with commas and semicolons to indicate partial and covariant derivatives. The following equations give equivalent notations for the same derivatives:
\[\partial_{\mu} X_{\nu} = X_{\nu,\; \mu}\]
\[\nabla_{a} X_{b} = X_{b;a}\]
\[\nabla^{a} X_{b} = X_{b}^{;a}\]
Figure 5.6.5 shows two examples of the corresponding birdtracks notation. Because birdtracks are meant to be manifestly coordinate-independent, they do not have a way of expressing non-covariant derivatives. We no longer want to use the circle as a notation for a non-covariant gradient as we did when we first introduced it in section 2.1 .
References
9 “On the gravitational field of a point mass according to Einstein’s theory,” Sitzungsberichte der Königlich Preussischen Akademie der Wissenschaften 1 (1916) 189, translated in arxiv.org/abs/physics/9905030v1.


41.2: Statistical Questions
Lesson
Let's look more closely at data and the questions they can help to answer.
Exercise \(\PageIndex{1}\): Pencils on A Plot
- Measure your pencil to the nearest \(\frac{1}{4}\)-inch. Then, plot your measurement on the class dot plot.
- What is the difference between the longest and shortest pencil lengths in the class?
- What is the most common pencil length?
- Find the difference in lengths between the most common length and the shortest pencil.
Exercise \(\PageIndex{2}\): What's in the Data?
Ten sixth-grade students at a school were each asked five survey questions. Their answers to each question are shown here.
\(\begin{array}{lllllllllll}{\text{data set A}}&{0}&{1}&{1}&{3}&{0}&{0}&{0}&{2}&{1}&{1}\\{\text{data set B}}&{12}&{12}&{12}&{12}&{12}&{12}&{12}&{12}&{12}&{12}\\{\text{data set C}}&{6}&{5}&{7}&{6}&{4}&{5}&{3}&{4}&{6}&{8}\\{\text{data set D}}&{6}&{6}&{6}&{6}&{6}&{6}&{6}&{6}&{6}&{6}\\{\text{data set E}}&{3}&{7}&{9}&{11}&{6}&{4}&{2}&{16}&{6}&{10}\end{array}\)
-
Here are the five survey questions. Match each question to a data set that could represent the students’ answers. Explain your reasoning.
- Question 1: Flip a coin 10 times. How many heads did you get?
- Question 2: How many books did you read in the last year?
- Question 3: What grade are you in?
- Question 4: How many dogs and cats do you have?
- Question 5: How many inches are in 1 foot?
- How are survey questions 3 and 5 different from the other questions?
Exercise \(\PageIndex{3}\): What Makes a Statistical Question?
These three questions are examples of statistical questions :
- What is the most common color of the cars in the school parking lot?
- What percentage of students in the school have a cell phone?
- Which kind of literature—fiction or nonfiction—is more popular among students in the school?
These three questions are not examples of statistical questions :
- What color is the principal’s car?
- Does Elena have a cell phone?
- What kind of literature—fiction or nonfiction—does Diego prefer?
-
Study the examples and non-examples. Discuss with your partner:
- How are the three statistical questions alike? What do they have in common?
- How are the three non-statistical questions alike? What do they have in common?
- How can you find answers to the statistical questions? How about answers to non-statistical questions?
-
What makes a question a statistical question?
Pause here for a class discussion.
-
Read each question. Think about the data you might collect to answer it and whether you expect to see
variability
in the data. Complete each blank with “Yes” or “No.”
-
How many cups of water do my classmates drink each day?
- Is variability expected in the data? ______
- Is the question statistical? _____
-
Where in town does our math teacher live?
- Is variability expected in the data? ______
- Is the question statistical? _____
-
How many minutes does it take students in my class to get ready for school in the morning?
- Is variability expected in the data? ______
- Is the question statistical? _____
-
How many minutes of recess do sixth-grade students have each day?
- Is variability expected in the data? ______
- Is the question statistical? _____
-
Do all students in my class know what month it is?
- Is variability expected in the data? ______
- Is the question statistical? _____
-
How many cups of water do my classmates drink each day?
Exercise \(\PageIndex{4}\): Sifting for Statistical Questions
- Your teacher will give you and your partner a set of cards with questions. Sort them into three piles: Statistical Questions, Not Statistical Questions, and Unsure.
- Compare your sorting decisions with another group of students. Start by discussing the two piles that your group sorted into the Statistical Questions and Not Statistical Questions piles. Then, review the cards in the Unsure pile. Discuss the questions until both groups reach an agreement and have no cards left in the Unsure pile. If you get stuck, think about whether the question could be answered by collecting data and if there would be variability in that data.
-
Record the letter names of the questions in each pile.
- Statistical questions:
- Non-statistical questions:
Are you ready for more?
Tyler and Han are discussing the question, “Which sixth-grade student lives the farthest from school?”
- Tyler says, “I don’t think the question is a statistical question. There is only one person who lives the farthest from school, so there would not be variability in the data we collect.”
- Han says: “I think it is a statistical question. To answer the question, we wouldn’t actually be asking everyone, 'Which student lives the farthest from school?' We would have to ask each student how far away from school they live, and we can expect their responses to have variability.”
Do you agree with either one of them? Explain your reasoning.
Summary
We often collect data to answer questions about something. The data we collect may show variability , which means the data values are not all the same.
Some data sets have more variability than others. Here are two sets of figures.
Set A has more figures with the same shape, color, or size. Set B shows more figures with different shapes, colors, or sizes, so set B has greater variability than set A.
Both numerical and categorical data can show variability. Numerical sets can contain different numbers, and categorical sets can contain different categories or types.
When a question can only be answered by using data and we expect that data to have variability, we call it a statistical question . Here are some examples.
- Who is the most popular musical artist at your school?
- When do students in your class typically eat dinner?
- Which classroom in your school has the most books?
To answer the question about books, we may need to count all of the books in each classroom of a school. The data we collect would likely show variability because we would expect each classroom to have a different number of books.
In contrast, the question “How many books are in your classroom?” is not a statistical question. If we collect data to answer the question (for example, by asking everyone in the class to count books), the data can be expected to show the same value. Likewise, if we ask all of the students at a school where they go to school, that question is not a statistical question because the responses will all be the same.
Glossary Entries
Definition: Categorical Data
A set of categorical data has values that are words instead of numbers.
For example, Han asks 5 friends to name their favorite color. Their answers are: blue, blue, green, blue, orange.
Definition: Dot Plot
A dot plot is a way to represent data on a number line. Each time a value appears in the data set, we put another dot above that number on the number line.
For example, in this dot plot there are three dots above the 9. This means that three different plants had a height of 9 cm.
Definition: Numerical Data
A set of numerical data has values that are numbers.
For example, Han lists the ages of people in his family: 7, 10, 12, 36, 40, 67.
Definition: Statistical Question
A statistical question can be answered by collecting data that has variability. Here are some examples of statistical questions:
- Who is the most popular musical artist at your school?
- When do students in your class typically eat dinner?
- Which classroom in your school has the most books?
Definition: Variability
Variability means having different values.
For example, data set B has more variability than data set A. Data set B has many different values, while data set A has more of the same values.
Practice
Exercise \(\PageIndex{5}\)
Sixth-grade students were asked, “What grade are you in?” Explain why this is not a statistical question.
Exercise \(\PageIndex{6}\)
Lin and her friends went out for ice cream after school. The following questions came up during their trip. Select all the questions that are statistical questions.
- How far are we from the ice cream shop?
- What is the most popular ice cream flavor this week?
- What does a group of 4 people typically spend on ice cream at this shop?
- Do kids usually prefer to get a cup or a cone?
- How many toppings are there to choose from?
Exercise \(\PageIndex{7}\)
Here is a list of questions about the students and teachers at a school. Select all the questions that are statistical questions.
- What is the most popular lunch choice?
- What school do these students attend?
- How many math teachers are in the school?
- What is a common age for the teachers at the school?
- About how many hours of sleep do students generally get on a school night?
- How do students usually travel from home to school?
Exercise \(\PageIndex{8}\)
Here is a list of statistical questions. What data would you collect and analyze to answer each question? For numerical data, include the unit of measurement that you would use.
- What is a typical height of female athletes on a team in the most recent international sporting event?
- Are most adults in the school football fans?
- How long do drivers generally need to wait at a red light in Washington, DC?
Exercise \(\PageIndex{9}\)
Describe the scale you would use on the coordinate plane to plot each set of points. What value would you assign to each unit of the grid?
- \((1,-6), (-7,-8), (-3,7), (0,9)\)
- \((-20,-30), (-40,10), (20,-10), (5,-20)\)
- \((\frac{-1}{3},-1), (\frac{2}{3},-1\frac{1}{3}),(\frac{-4}{3},\frac{2}{3}),(\frac{1}{6},0)\)
(From Unit 7.3.3)
Exercise \(\PageIndex{10}\)
Noah’s water bottle contains more than 1 quart of water but less than \(1\frac{1}{2}\) quarts. Let \(w\) be the amount of water in Noah’s bottle, in quarts. Select all the true statements.
- \(w\) could be \(\frac{3}{4}\).
- \(w\) could be \(1\).
- \(w>1\)
- \(w\) could be \(\frac{4}{3}\).
- \(w\) could be \(\frac{5}{4}\).
- \(w\) could be \(\frac{5}{3}\).
- \(w>1.5\)
(From Unit 7.2.2)
Exercise \(\PageIndex{11}\)
Order these numbers from least to greatest:
\(|-17|\qquad |-18|\qquad -18\qquad |19|\qquad 20\)
(From Unit 7.1.7)


Chapter 6: Kinetics of Point Masses
In the previous chapter we have discussed and developed a toolbox to analyse the kinematics of point masses, determining their motion, velocity and acceleration in various coordinate systems. In this chapter we extend our toolbox by determining the effect of forces on acceleration, thus analysing the kinetics of point masses.
The outline of this chapter is as follows. We first introduce Newton’s second law (Sec. 6.1) and use it to define mass (Sec. 6.2). We define force vectors and their properties governed by Newton’s third law (Sec. 6.3). Then we describe in detail the step-by-step procedure (Sec. 6.4) to determine and solve the equations of motion. The procedure starts from a sketch (Sec. 6.5) and choosing the coordinate system (Sec. 6.6). Then constraint equations are determined, free body diagrams are used to determine the resultant forces, and Newton’s second law is used to determine the equations of motion along each coordinate axis (Sec. 6.7-6.10), which are then solved (Sec. 6.11.1) to obtain the motion or forces (Sec. 6.11.2). Inertial reference frames are introduced (Sec. 6.12) and finally we discuss the properties of various forces (Sec. 6.13).