Working with Applications

Section 1.3 Working with Applications

This section addresses the following topics.

Interpret data in various formats and analyze mathematical models
Read and use mathematical models in a technical document

This section covers the following mathematical concepts.

Solve linear, rational, quadratic, and exponential equations and formulas (skill)
Read and interpret models (critical thinking)
Use models including linear, quadratic, exponential/logarithmic, and trigonometric (skill)

In life when we figure out processes at work or in science we often express the result in mathematical notation. This includes equations, functions, and other options. These are collectively known as models. They allow us to communicate what we know and calculate results as needed. To succeed in many jobs and to fully enjoy life we need to be proficient at reading and using models.

This section begins by presenting models and illustrates calculating some results from them. It progresses to solving equations (models) as a review of algebra skills. Finally we present tips on how to identify and use models arising in applications. These topics are continued with specific models in later sections.

Subsection 1.3.1 Calculating Results using Models

Model 1.3.1. Ohm’s Law.

Ohm’s Law relates three properties of electricity: voltage, current, and resistance. Voltage, measured in volts (V), is analagous to the amount of pressure to move the electrons. Current, measured in amperes (amps), is how much electricity is moving. Resistance measured in Ohms ($\Omega$), is, as it sounds, the resistance of a material to letting electricity flow.

The relationship is

\begin{equation*} V = IR \end{equation*}

where

$V$ is voltage,
$I$ is current, and
$R$ is the resistance.

Example 1.3.2.

We will consider how voltage must change as the needed current changes. These are measurements in a science model which means we will use significant digits rounding.

Suppose the current is 3.0 amps and the resistance is 8.0 ohms. What is the voltage?

Using the model we can calculate

\begin{equation*} V = 3.0 \cdot 8.0 = 24. \end{equation*}

Thus in this system 24 volts is required.

If the current is increased to 6.0 amps on the 8.0 ohm circuit, then

\begin{equation*} V = 6.0 \cdot 8.0 = 48. \end{equation*}

This result tells us that to double the amps, doubling the voltage is required.

Similarly if we know that the current is 1.7 amps and the resistance is 6.0 ohms, then we can calculate

\begin{equation*} V = 1.7 \cdot 6.0 = 10.2 \approx 10. \end{equation*}

Thus in this system 10 volts is required.

Model 1.3.3. Lift Equation.

Lift is the force that keeps aircraft in the air.The lift equation explains factors that control the strength of lift produced by an airfoil (think wing or propellor). The factors included are air density, surface area of the airfoil, the coefficient of lift, and velocity. Air density is the amount of air per volume; you may see this as highs and lows on a weather map. It is also related to pressurizing aircraft flying at high altitude. The coefficient of lift incorporates multiple factors that are part of the design of the airfoil and how it is in use during flight.

The lift equation is

\begin{equation*} L = \frac{1}{2}\rho s C_L v^2 \end{equation*}

where

$L$ is the lift in units of lbs or Newtons)
$\rho$ is air density in units of slugs per cubic feet or kilograms per cubic meter
$s$ surface area in units of square feet or square meters
$C_L$ is the coefficient of lift which is unitless
$v$ is velocity in units of feet per second or meters per second

Example 1.3.4.

We can consider the impact of changes in air density. Air density generally decreases as altitude increases. This is a science model with measurements, so we will use significant digits rounding.

Suppose the air density is 0.002378 slugs per cubic feet, surface area is 125 ft², $C_L=1.5617\text{,}$ and velocity is 84.4 ^ft⁄_s. Calculate the lift.

\begin{align*} L & = \frac{1}{2} \cdot 0.002378 \cdot 125 \cdot 1.5617 \cdot 84.4^2\\ & \approx 16\underline{5}3.386439\\ & \approx 1650\text{.} \end{align*}

Under these circumstances this airfoil can lift 1650 lbs.

If the air density is reduced to 0.001988 slugs per cubic feet, what is the lift? This represents the same aircraft flying 6000 feet higher.

\begin{align*} L & = \frac{1}{2} \cdot 0.001988 \cdot 125 \cdot 1.5617 \cdot 84.4^2\\ & 13\underline{8}2.225501\\ & \approx 1380\text{.} \end{align*}

If nothing else is changed (like velocity), this airfoil can lift (hold in the air) $1650-1380=270$ lbs less.

Model 1.3.5. Ideal Gas Law.

The ideal gas law is a relationship between the volume, pressure, temperature, and number of molecules of an ideal gas. The relationship is

\begin{equation*} PV = nRT \end{equation*}

where

$P$ is the pressure in units of atmospheres (atm) or Pascals (Pa)
$V$ is the volume in units of cubic feet or cubic meters
$n$ is the number of moles (number of molecules, see a chemistry text for details)
$R$ is a constant specific to each gas (e.g., oxygen and nitrogen have different ones) in units that match the other values
$T$ is the temperature in degrees Rankine or Kelvin (these are shifted versions of Fahrenheit and Celsius).

When the number of molecules remains fixed, such as in a closed container, this law can be used to produce the equation

\begin{equation*} \frac{P_1 V_1}{T_1+273} = \frac{P_2 V_2}{T_2+273}. \end{equation*}

where

$P_1\text{,}$ $V_1\text{,}$ and $T_1$ are the initial pressure, volume, and temperature, and
$P_2\text{,}$ $V_2\text{,}$ and $T_2$ are the pressure, volume, and temperature at another time.

Note in both forms of the law the units can be other than those listed (especially different scale like centimeters rather than meters). However, they must always match including the constant $R$ which is looked up in reference books.

For rounding note 273 is three (3) significant digits. 273.15 can be used if more precision is needed.

Example 1.3.6.

Suppose the initial conditions are $P_1=101.3 \text{ Pa}\text{,}$ $V_1=0.125 \text{ m}^3\text{,}$ and $T_1=10.2^\circ \text{ C}\text{.}$ Also $V_2=0.125 \text{ m}^3$ and $T_2=50.7^\circ \text{ C}\text{.}$ Calculate the new pressure. This is a science model with measurements, so we will use significant digits rounding.

\begin{align*} \frac{101.3 \cdot 0.125}{10.2+273} \amp = \frac{P_2 \cdot 0.125}{50.7+273} \text{ Addition maintains precision to tenths}\\ \frac{101.3 \cdot 0.125}{10.2+273} \amp = \frac{P_2 \cdot 0.125}{323.\underline{7}} \text{ Multiplication maintains 3 sigfigs}\\ \frac{12.\underline{6}625}{283.\underline{2}} \amp = \frac{P_2 \cdot 0.125}{323.\underline{7}} \text{ Division maintains 3 sigfigs}\\ 0.044\underline{7}122 \amp \approx 0.00038\underline{6}160 P_2 \text{ Divide to isolate } P_2\\ \frac{0.044\underline{7}122}{0.00038\underline{6}160} \amp \approx \frac{0.00038\underline{6}160 P_2}{0.00038\underline{6}160} \text{ Division maintains 3 sigfigs}\\ 11\underline{5}.786721 \amp \approx P_2\\ 116 \amp \approx P_2\text{.} \end{align*}

If instead $T_2=-10.3^\circ \text{ C}\text{,}$ then we have the following.

\begin{align*} \frac{101.3 \cdot 0.125}{10.2+273} \amp = \frac{P_2 \cdot 0.125}{-10.3+273}\\ \frac{12.\underline{6}625}{283.\underline{2}} \amp = \frac{P_2 \cdot 0.125}{262.\underline{7}}\\ 0.044\underline{7}122 \amp \approx 0.00047\underline{5}828 P_2\\ \frac{0.044\underline{7}122}{0.00047\underline{5}828} \amp \approx \frac{0.00047\underline{5}828 P_2}{0.00047\underline{5}828}\\ 93.\underline{9}6714779 \amp \approx P_2\\ 94.0 \amp \approx P_2 \end{align*}

Checkpoint 1.3.7.

Checkpoint 1.3.8.

Subsection 1.3.2 Calculating Results Requiring Solving

The previous section illustrated calculating model results without solving. This section presents additional example requiring limited solving and finishes with solving before any values have been substituted.

It does not matter if the value we desire is by itself, we can solve using arithmetic.

Example 1.3.9.

Recall the model for lift. Suppose we know the weight of the aircraft ($w=239\bar{0}$ lbs), density of air ($\rho=0.001869 \text{ slugs}/\text{ft}^3$), wing surface area ($s=165 \text{ ft}^2$), and velocity ($v=91.1 \text{ ft}/\text{sec}$). Noting that lift must equal weight, what must the coefficient of lift be? This is a science model with measurements, so we will use significant digits rounding.

\begin{align*} L & = \frac{1}{2}\rho s C_L v^2\\ 239\bar{0} & = \frac{1}{2}(0.001869)(165)C_L(91.1)^2. \text{ Multiply.}\\ 239\bar{0} & \approx 12\underline{7}9.675938 C_L. \text{ Multiplication maintains 3 sigfigs}\\ \frac{239\bar{0}}{12\underline{7}9.675938} & \approx \frac{12\underline{7}9.675938 C_L}{12\underline{7}9.675938} \text{ Divide to isolate } C_L.\\ 1.8\underline{6}7660342 & \approx C_L. \text{ Division maintains 3 sigfigs}\\ 1.87 & \approx C_L. \end{align*}

The desired value from the model may be in a denominator. We can solve for this using multiplication and division.

Example 1.3.10.

Recall from Model 1.3.5 that under simplifying assumptions

\begin{equation*} \frac{P_1 V_1}{T_1+273} = \frac{P_2 V_2}{T_2+273}. \end{equation*}

Suppose we know the initial conditions ($P_1=1.00$ atm, $V_1=1.35 \text{ ft}^3\text{,}$ $T_1=51.2^\circ$ F) and also $P_2=1.00$ atm and $V_2=1.39 \text{ ft}^3\text{.}$ What must the new temperature ($T_2$) be? This is a science model with measurements, so we will use significant digits rounding.

\begin{align*} \frac{1.00 \cdot 1.35}{51.2+273} & = \frac{1.00 \cdot 1.39}{T_2+273}. \text{ Perform arithmetic}\\ \frac{1.3\underline{5}}{324.\underline{2}} & = \frac{1.3\underline{9}}{T_2+273}. \text{ More arithmetic}\\ 0.0041\underline{6}40962 & \approx \frac{1.3\underline{9}}{T_2+273}. \text{ Multiply to move } T_2 \text{ out of denominator.}\\ 0.0041\underline{6}40962(T_2+273) & \approx \frac{1.3\underline{9}}{T_2+273}(T_2+273).\\ 0.0041\underline{6}40962(T_2+273) & \approx 1.3\underline{9}. \text{ Divide to undo multiplication.}\\ \frac{0.0041\underline{6}40962(T_2+273)}{0.0041\underline{6}40962} & \approx \frac{1.3\underline{9}}{0.0041\underline{6}40962}. \text{ Division maintains 3 sigfigs}\\ T_2+273 & \approx 33\underline{3}.80593. \text{ Subtract to undo addition.}\\ T_2+273-273 & \approx 33\underline{3}.8059259-273.\\ T_2 & \approx 6\underline{0}.80593. \text{ Subtraction maintains precision to units}\\ T_2 & \approx 61. \end{align*}

Note we use significant digits for rounding because this is a science model.

The previous examples solved for a variable in a model after substituting numbers for the other variables. The next examples illustrate solving first. Note this process is the same as solving after substituting (same algebra) though there may be more steps. We might wish to solve this way, so it is easier to use the model multiple times.

Example 1.3.11.

Solve the equation $V=IR$ for $R\text{.}$ Note, this model is explained in Model 1.3.1.

\begin{align*} V & = IR.\\ \frac{V}{I} & = \frac{IR}{I} & \text{Divide to undo multiplication}.\\ \frac{V}{I} & = R. \end{align*}

Example 1.3.12.

Solve the lift equation $L=\frac{1}{2}\rho S C_L v^2$ for $S\text{.}$

\begin{align*} L \amp = \frac{1}{2}\rho S C_L v^2.\\ 2L \amp = 2\frac{1}{2}\rho S C_L v^2. & \text{ Multiply to undo division.}\\ 2L \amp = \rho S C_L v^2.\\ \frac{2L}{\rho} \amp = \frac{\rho S C_L v^2}{\rho}. & \text{Divide to undo multiplication.}\\ \frac{2L}{\rho} \amp = S C_L v^2.\\ \frac{2L}{\rho C_L} \amp = \frac{S C_L v^2}{C_L}.\\ \frac{2L}{\rho C_L} \amp = S v^2.\\ \frac{2L}{\rho C_L v^2} \amp = \frac{S v^2}{v^2}.\\ \frac{2L}{\rho C_L v^2} \amp = S. \end{align*}

Notice that the steps are the same algebra as if there were numbers. Also we could divide by $v^2$ and we are not concerned with the square as part of solving for $S\text{.}$

Checkpoint 1.3.13.

Solve the lift equation $L=\frac{1}{2}\rho S C_L v^2$ for $\rho\text{.}$

Subsection 1.3.3 Process Overview

Above we started with a model and were asked to do something with it. Normally we will start with a problem which does not identify a model to use. Very few problems you encounter in lifte come pre-labled with models. This section presents how to start with a problem and work to a solution by identifying the model first.

Our first task is to read the problem to understand it.

Read the problem description a few times.
- If you can paraphrase it, you understand it enough.
- Drawing a picture and labeling parts may help.
Identify what we are asked to do.
Identify the information we are given. Note distinctions like measurements and rates.
Identify any units. These often help us set up a model.
Write everything! We do not model in our heads.

Next we write the mathematical model (equation or function).

Use the description to determine which application type (e.g., percent, proportion, linear model, etc.). Note units can suggest this (e.g., meters and meters squared indicate something was squared).
Do not insert any numbers yet.
Do not do any calculations yet.

Now we will have a model that matches our situation and possibly some numbers to insert.

Insert numbers into the model. You may have to calculate some of these (e.g., you are given two points but not the slope you need).
Solve for the desired value. Note it may help to do some calculations with the numbers first.
State your answer and use units appropriately.

Finally we should check that our answer makes sense. We should not have negative prices (usually) or distances larger than the earth (when working with terrestrial problems).

Example 1.3.14.

You moved across town and rented a 20 foot moving truck for the day. You want to make sure the bill you received is correct. If you paid $81.03, for how many miles were you charged? Assume there were no extra fees.

Ad containing an image of the truck, truck size information, and pricing information. $39.95 plus $0.79 per mile.

Solution.

We want to compare the bill we received to the price listed in the add. The question is about how many miles (not how much money).

We are given the price per mile ($0.79 per mile). There is also a fixed cost for the rental ($39.95). Adding the fixed cost and the milage cost will give us the total.

Our model is $C = \$39.95+\$0.79m$ where $C$ is the total cost and $m$ is the number of miles.

We know the total cost, which will leaves $m$ in the equation, the number of miles, which is what we want to calculate. We can use the solving technique in Section 3.1

\begin{align*} \$81.03 & = \$39.95+\$0.79m\\ -\$39.95+\$81.03 & = -\$39.95+\$39.95+\$0.79m\\ \$41.08 & = \$0.79m\\ \frac{\$41.08}{\$0.79} & = \frac{\$0.79}{\$0.79}m\\ 52 & = m\text{.} \end{align*}

Mathematics in Trades and Life

Section 1.3 Working with Applications

Subsection 1.3.1 Calculating Results using Models

Model 1.3.1. Ohm’s Law.

Example 1.3.2.

Model 1.3.3. Lift Equation.

Example 1.3.4.

Model 1.3.5. Ideal Gas Law.

Example 1.3.6.

Checkpoint 1.3.7.

Checkpoint 1.3.8.

Subsection 1.3.2 Calculating Results Requiring Solving

Example 1.3.9.

Example 1.3.10.

Example 1.3.11.

Example 1.3.12.

Checkpoint 1.3.13.

Subsection 1.3.3 Process Overview

Example 1.3.14.

Exercises 1.3.4 Exercises

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