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Section7.1Proportional Models

A proportional relation is the simplest example of a linear relation. So before we progress to general linear relations, we want to be sure that we first understand proportionality.

Definition7.1.1Proportional Relation

Given two covarying variables \(a\) and \(b\), we say \(b\) is proportional to \(a\) and write \(b \propto a\) to mean that the ratio \(b/a\) is a constant. That is, there is some constant \(k\) (called the proportionality constant so that for every state, \begin{equation*}\frac{b}{a} = k.\end{equation*} In other words, \begin{equation*}b \propto a \qquad \Leftrightarrow \qquad b=k \cdot a\end{equation*} for some constant \(k\).

As we think about variables as representing measurable quantities in an experiment, we should recall that each measurement has an associated dimension and units. The proportionality constant \(k\) described in the definition therefore also has a dimension and its units are found by creating the corresponding ratio of units of the measurements.

Example7.1.2 Distance—Rate—Time Models

A classic example of proportional relations is often taught in introductory algebra classes as the rate equation for distance. Under the assumption of constant speed, distance traveled is proportional to the duration of time traveled.

If we let \(d\) measure the distance traveled and let \(t\) measure the duration of time traveled, then \(d \propto t\) with a corresponding relation \begin{equation*}d=rt\end{equation*} with \(r=d/t\) measuring the rate or speed of travel.

The dimension for the distance traveled \(d\) is length. The dimension for the duration of time traveled \(t\) is time. So the dimension measured by the rate \(r\) is a length per time with units given by a unit of length divided by a unit of time. If the units of \(d\) are meters (m) and the units of \(t\) are seconds (s), then the units of \(r\) are meters per second (m/s). If the units of \(d\) are miles (mi) and the units of \(t\) are hour (hr), then the units of \(r\) are miles per hour (mi/hr).

Other quantities behave with a similar rate relation between that of distance and time. For example, hourly workers are paid a monetary value that is proportional to the time worked (with a constant hourly rate). The volume of water added to a reservoir is proportional to the duration of time the water flowed (if the flow rate is constant). Many famous laws of physics and chemistry represent proportional relations. Newton's second law of motion (physics) is often quoted as \begin{equation*}F=ma\end{equation*} which is simply to say that the net force \(F\) applied to a body is proportional to the acceleration \(a\) observed by that body (and mass \(m\) is the proportionality constant).

Calculus was developed to understand relations like that between distance and time when the rate is not constant. That requires dynamic modeling which we will explore later in this course. We will learn about the derivative as the measure of a rate, and we will learn about the integral as the accumulated change resulting from a changing rate.

Subsection7.1.1 Unit Conversions

When a quantity measures an extensive property,  1  a unit of measurement literally refers to a physical unit of that quantity by which you can measure. A metric ruler is an excellent example, where 1 centimeter is a physical unit of length and the total length can be found by subdividing the total length into centimeter units. Different units of measurement give different rules, such as a yard stick where the basic unit is 1 inch.

Unit conversion describes the process of determining the measure in desired units when you are given the measure in different units. Fortunately, for extensive quantities, unit conversion takes advantage of the fact that measurements in different units are proportional to one another, as summarized in the following theorem.

Proof

The proportionality constant for unit conversions is found by measuring the same quantity in both units and computing the appropriate ratio. (This is found in Section 1.3.1 of the Adler textbook, pages 25-27.) In practice, we just create a dimensional equation that allows us to cancel units according to usual rules of algebra.

Example7.1.4

How many centimeters are in a foot? How many feet are in 1 meter?

The official definition of an inch is 2.54 centimeters. One foot has 12 inches. One meter is 100 centimeters. These statements of fact establish dimensional equations which allow us to convert units. \begin{equation*} \begin{gathered} 1 \; \hbox{in} = 2.54 \; \hbox{cm} \\ \frac{1 \; \hbox{in}}{2.54 \; \hbox{cm}} = 1 \\ \frac{2.54 \; \hbox{cm}}{1 \; \hbox{in}} = 1 \end{gathered} \end{equation*} We can multiply by 1 using these unit conversion ratios to cancel old units and insert new units.

To answer the first question, we start with 1 foot and use unit conversions to find new units. \begin{equation*} \begin{aligned} 1\;\hbox{ft} &= 1\;\cancel{\hbox{ft}} \cdot \frac{12\;\hbox{in}}{1\;\cancel{\hbox{ft}}} = 12\;\hbox{in} \\ &= 12\;\cancel{\hbox{in}} \cdot \frac{2.54\;\hbox{cm}}{1\;\cancel{\hbox{in}}} = 12(2.54)\;\hbox{cm} = 30.48 \;\hbox{cm} \end{aligned} \end{equation*} This could be done in one step by using repeated conversion factors. \begin{equation*} 1\;\hbox{ft} = 1\;\cancel{\hbox{ft}} \cdot \frac{12\;\cancel{\hbox{in}}}{1\;\cancel{\hbox{ft}}} \cdot \frac{2.54\;\hbox{cm}}{1\;\cancel{\hbox{in}}} = 12(2.54)\;\hbox{cm} = 30.48 \;\hbox{cm} \end{equation*}

The second question is answered in a similar way. \begin{equation*} 1\;\hbox{m} = 1\;\cancel{\hbox{m}} \cdot \frac{100\;\cancel{\hbox{cm}}}{1\;\cancel{\hbox{m}}} \cdot \frac{1\;\cancel{\hbox{in}}}{2.54\;\cancel{\hbox{cm}}} \cdot \frac{1\;{\hbox{ft}}}{12\;\cancel{\hbox{in}}} = \frac{100}{2.54(12)}\;\hbox{ft} \approx 3.28 \;\hbox{ft} \end{equation*}

Subsection7.1.2Geometric Similarity

A second application of proportional relations is in regards to geometric similarity. Similarity refers to the idea that two different geometric figures are the same shape but at different scales and perhaps rotated or reflected. The mathematical consequence of this is that any length of a path on one figure is proportional to the length of the corresponding path on the other figure, and the proportionality constant is the same for all such paths representing the scaling factor.

The most common application is applied to the lengths of edges of two similar polygons, usually triangles. If all of the lengths of one polygon are known and you can find the scaling factor, then you can find all of the lengths of the similar polygon.

Example7.1.5

Consider the two similar quadrilaterals \(abcd\) and \(ABCD\) illustrated in the figure below, where the vertices are labeled to show the correspondence between the shapes. We would write \(abcd \sim ABCD\) to indicate that these polygons are similar with the given vertex correspondence.

Suppose that you are given the following lengths: \(ab=3\), \(ad=5\), \(cd=4\), \(AB=4.5\) and \(BC=6\). What are the lengths of the other sides?

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Figure7.1.6Two similar quadrilaterals

Because the shapes are similar, all corresponding lengths are proportional and the ratios \(ab/AB\), \(bc/BC\), \(cd/CD\) and \(ad/AD\) must all have the same value. We look for which of those ratios can be computed numerically. Since \(ab\) and \(AB\) are both given, we can find the common ratio. \begin{equation*}\frac{ab}{AB} = \frac{3}{4.5} = \frac{2}{3} = \frac{bc}{BC} = \frac{cd}{CD} = \frac{ad}{AD}\end{equation*} Knowing the ratio, we can find the missing lengths. \begin{equation*}\begin{aligned} \frac{ad}{AD} &= \frac{5}{AD} = \frac{2}{3} \quad \Rightarrow \quad AD=\frac{15}{2}=7.5 \\ \frac{bc}{BC} &= \frac{bc}{6} = \frac{2}{3} \quad \Rightarrow \quad bc=4 \\ \frac{cd}{CD} &= \frac{4}{CD} = \frac{2}{3} \quad \Rightarrow \quad CD=6 \end{aligned}\end{equation*}

We most commonly encounter similarity relations when dealing with triangles, so the following theorem is stated in the context of triangles, but it easily generalizes to other polygons. This theorem states that when two triangles are similar, then the ratio of edge lengths within a specific triangle equals the corresponding ratio of edge lengths in the other triangle.

Proof
Example7.1.9

The triangles \(ABC\) and \(PQS\) are similar, with the edge lengths as shown in the figure below. Find the lengths of edges in \(PQS\).

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Figure7.1.10Triangles \(ABC\) and \(PQS\) are similar.

The old way to solve this problem (before the theorem) would be to observe that the edge \(AC\) corresponds to the edge \(PS\) so that the scaling ratio is \(\displaystyle k = \frac{PS}{AC} = \frac{3}{2}\). Then multiplying all of the edges of \(ABC\) by \(k=\frac{3}{2}\) gives the edges of interest: \begin{equation*}\begin{aligned} \frac{PQ}{AB} &= k = \frac{3}{2} \qquad \Rightarrow \qquad PQ = k \, AB = \frac{3}{2}(5) = \frac{15}{2} = 7.5, \\ \frac{QS}{BC} &= k = \frac{3}{2} \qquad \Rightarrow \qquad QS = k \, BC = \frac{3}{2}(3) = \frac{9}{2} = 4.5. \end{aligned} \end{equation*}

The theorem provides a new method where we look at ratios in the triangle \(ABC\) in order to find the same ratios in the triangle \(PQS\). For example, the ratio of edges \(AB : AC\) is the same as the ratio \(PQ : PS\) so that \begin{equation*} \frac{AB}{AC} = \frac{5}{4} \quad \Rightarrow \quad \frac{PQ}{PS} = \frac{PQ}{6} = \frac{5}{4} \quad \Rightarrow \quad PQ = \frac{5}{4}(6) = \frac{15}{2} = 7.5. \end{equation*} Similarly, the ratio of edges \(AC : BC\) must equal the ratio \(PS : QS\) which leads to \begin{equation*} \frac{AC}{BC} = \frac{4}{3} \quad \Rightarrow \quad \frac{PS}{QS} = \frac{6}{QS} = \frac{4}{3} \quad \Rightarrow \quad QS = \frac{3}{4}(6) = \frac{9}{2} = 4.5. \end{equation*} We see that we get the same lengths as the previous method.  2 

Subsection7.1.3Proportional Models

Many mathematical models are constructed using the principle of proportionality. The basic idea is to identify two quantities that are expected to covary in a proportional manner and write an equation capturing that relationship. For some models, the proportionality constant is left unspecified as a parameter. In other cases, the constant is either known or can be computed from the given information. This subsection contains a variety of examples illustrating the power of using proportional models.

Subsubsection7.1.3.1Per Capita Rates

In population biology, we are often interested the relationship between vital rates, such as birth rates and death rates, with the size of the population. That is, we might expect that the birth rate (number of births per unit time) would be larger if the population were larger since there are more individuals potentially giving birth. Similarly, we might expect that the death rate (number of deaths per unit time) in a population would also be larger in a larger population. Models in population biology are often constructed by describing these relations.

The simplest model capturing the idea that the birth rate is larger when the population is larger uses an assumption that the birth rate is proportional to the population size. Using variables, let \(B\) represent the birth rate for a population and let \(P\) represent the size of the population. These are covarying variables. Our assumption is that \(B \propto P\), meaning that we expect that under all conditions, the ratio \(B : P\) is always the same constant.

Using a lowercase \(b\) for this constant, our model is characterized by the relation \begin{equation}B = b \cdot P. \label{men-4}\tag{7.1.1}\end{equation} This proportionality constant has a special name, the per capita birth rate. Here, I am speaking about the role of the constant, not the particular letter \(b\). The phrase per capita has a literal translation of per head. Because \(b\) is defined by the ratio \(\displaystyle b=\frac{B}{P}\), in the case that \(P=1\) (population size is 1), \(b\) would equal \(B\), the number of births per unit time of a single individual.

In a similar way, if we assume that the death rate of a population is proportional to the size of the population, then we should consider the per capita death rate \(d\) which leads to a model relation between the population death rate \(D\) and the size of the population \(P\) given by \begin{equation}D = d \cdot P. \label{men-5}\tag{7.1.2}\end{equation} These models are often characterized by saying the population has a constant per capita rate.

Example7.1.11

The United States Census Bureau's summary of births reports a birth rate of 14.0 births per thousand individuals in the population for the year 2008. The population that year was 304.1 million individuals. How many births were recorded in 2008?

This vital statistic information allows us to compute the per capita birth rate based on a population of 1000. \begin{equation*} b = \frac{B}{P} = \frac{14.0}{1000} = 0.0140 \end{equation*} For the full population \(304.1\times 10^{6}\), we can compute the birth rate for the year. \begin{equation*} B = b \cdot P = 0.0140 (304.1 \times 10^{6}) = 4.26 \times 10^{6} \end{equation*} There were about 4.26 million births during the year 2008.

Checking the bureau's direct report, we can find there were 4.248 million births during the year. The discrepancy comes from reporting the per capita birth rate with only three significant digits. A more accurate value would have been \(b=13.97\) which leads to a matching result.

Subsubsection7.1.3.2Cost and Unit Cost

Many goods and services are purchased with a fixed unit cost. This is another way of saying that the total cost is proportional to the measure of the goods or services purchased. Someone working at a fixed hourly rate receives a payment that is proportional to the number of hours worked. Grocery stores give prices for produce in dollars per pound, meaning that the cost of purchase is proportional to the weight.

When multiple items are purchased, the total cost is the sum of the individual costs. In this way, if there are different unit costs, we can still compute the total cost.

Example7.1.12Material Costs

You are finishing a room that is 10 feet by 18 feet and a ceiling that is 8 feet high. There are two windows (3 feet by 5 feet) and one door (3 feet by 7 feet). If wall paper costs $0.60 per square foot and carpet costs $1.25 per square foot, what will be the total cost of materials (ignoring unused scraps)?

Notice that cost information is given per square foot. These prices are the proportionality constants between the cost and the area. So we first need to find the areas and then compute the cost.

The carpet is a rectangle that is 10 feet by 18 feet. The area is computed as length times width, \begin{equation*}A_{\hbox{carpet}} = L \cdot W = 10(18) \; \hbox{ft}^2 = 180 \; \hbox{ft}^2,\end{equation*} and the carpet cost is proportional, \begin{equation*}C_{\hbox{carpet}} = 1.25 \; \hbox{dollars/ft}^2 \cdot A_{\hbox{carpet}} = 225 \; \hbox{dollars}.\end{equation*} Notice how the unit price divides by a dimension of area which when multiplied by the area (in the same units) cancel. If the price was given in different units, we would first need to perform unit conversion.

The walls consist of four rectangles which are then missing three smaller rectangles (for the door and windows). Two walls have width 10 feet and height 8 feet for area \(80 \; \hbox{ft}^2\) each; the other walls have length 18 ft and height 8 ft for area \(144 \; \hbox{ft}^2\) each. The door removes \(21 \; \hbox{ft}^2\) and each window removes \(15 \; \hbox{ft}^2\). The total area needing wallpaper is \begin{equation*}A_{\hbox{wallpaper}} = [2(80) + 2(144) - 21 - 2(15)] \; \hbox{ft}^2 = 397 \; \hbox{ft}^2,\end{equation*} with a resulting proportional cost \begin{equation*}C_{\hbox{wallpaper}} = 0.60 \; \hbox{dollars/ft}^2 \cdot A_{\hbox{wallpaper}} = 238.20 \; \hbox{dollars}.\end{equation*} The total cost is the sum \begin{equation*}C = C_{\hbox{carpet}} + C_{\hbox{wallpaper}} = 463.20 \; \hbox{dollars}.\end{equation*}