Variation

Section 3.4 Variation

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.

Identify rates as linear, quadratic, exponential, or other (critical thinking)

🔗
Identify data varrying directly or indirectly (critical thinking)

🔗
Solve linear, rational, quadratic, and exponential equations and formulas (skill)

🔗

Section 3.3 presents how to determine the rate at which one variable grows with respect to another. In all of those cases we considered only one variable changing. However, many models have more than one variable or parameter, and we wish to analyze the impact each one has on the result. This section will also present one more type of rate.

🔗

Subsection 3.4.1 Analyzing Models with Multiple Variables

This first example illustrates how a result can be linear with respect to one or more variables and quadratic (or other relationships) with another.

🔗

Example 3.4.1.

Review Model 1.3.3. Consider how lift changes with respect to each of the parameters.

🔗

Consider \(\rho\) the density of air. If the other parameters remain constant (only density changes), then the model looks like

\begin{align*} L & = \frac{1}{2}\rho s C_L v^2\\ & = \left( \frac{1}{2} s C_L v^2 \right)\rho\\ & = k \rho. \end{align*}

All we used is commutativity of multiplication. We can see that lift (\(L\)) changes linearly with respect to air density (\(\rho\)).

🔗

Consider \(s\) the surface area of the airfoil (wing or propeller). If the other parameters remain constant (only surface area changes), then the model looks like

\begin{align*} L & = \frac{1}{2}\rho s C_L v^2\\ & = \left( \frac{1}{2} \rho C_L v^2 \right)s \\ & = k s. \end{align*}

We can see that lift (\(L\)) changes linearly with respect to surface area (\(s\)).

🔗

Consider \(C_L\) the coefficient of lift of the airfoil. If the other parameters remain constant, then the model looks like

\begin{align*} L & = \frac{1}{2}\rho s C_L v^2\\ & = \left( \frac{1}{2} \rho s v^2 \right)C_L \\ & = k C_L. \end{align*}

We can see that lift (\(L\)) changes linearly with respect to the coefficient of lift (\(C_L\)).

🔗

Consider \(v\) the velocity. If the other parameters remain constant, then the model looks like

\begin{align*} L & = \frac{1}{2}\rho s C_L v^2\\ & = \left( \frac{1}{2} \rho s C_L \right)v^2 \\ & = k v^2. \end{align*}

We can see that lift (\(L\)) changes quadratically with respect to velocity (\(v^2\)).

🔗

How can we apply this knowledge? Lift changes linearly with respect to all parameters except for velocity. If greater lift (to handle greater weight) is needed, velocity provides bigger bang for our buck than any other change.

🔗

This second example has us look at the impact on more than just one variable. It also introduces a new relationship.

🔗

Example 3.4.2.

Review Model 1.3.3.

🔗

First consider the impact temperature (\(T\)) has on pressure (\(P\)). If the other parameters remain constant, then the model looks like

\begin{align*} PV & = nRT.\\ P & = \frac{nRT}{V}. & & \text{Divide to isolate } P\\ P & = \frac{nR}{V}T. & & \text{Use commutativity}\\ P & = kT. \end{align*}

Pressure (\(P\)) grows linearly with respect to temperature (\(T\)).

🔗

Notice that we can write that last equation as \(\frac{1}{k}P = T\) which means we can also make the reverse statement: temperature increases linearly with pressure.

🔗

Next consider the impact temperature (\(T\)) has on volume (\(V\)). If the other parameters remain constant, then the model looks like

\begin{align*} PV & = nRT.\\ V & = \frac{nRT}{P}.\\ V & = \frac{nR}{P}T.\\ V & = kT. \end{align*}

Volume (\(V\)) also grows linearly with respect to temperature (\(T\)).

🔗

Finally consider the impact volume (\(V\)) has on pressure (\(P\)). If the other parameters remain constant, then the model looks like

\begin{align*} PV & = nRT.\\ P & = \frac{nRT}{V}.\\ P & = (nRt)\frac{1}{V}.\\ P & = k \cdot \frac{1}{V}. \end{align*}

This pattern is not linear (nor quadratic, nor exponential). If volume is increased the right hand side will decrease (dividing by a larger number). This means that pressure will decrease. Conversely if volume is decreased the right hand side will increase (divide by a smaller number). This means that pressure will increase.

🔗

Definition 3.4.3. Vary Directly.

For two quantities \(a\) and \(b\text{,}\) if increasing \(b\) increases \(a\text{,}\) then \(a\) is said to vary directly with \(b\text{.}\)

🔗

Definition 3.4.4. Vary Indirectly.

For two quantities \(a\) and \(b\text{,}\) if increasing \(b\) decreases \(a\text{,}\) then \(a\) is said to vary indirectly with \(b\text{.}\)

🔗

Example 3.2.31 illustrated a model where the increase in one variable caused a decrease in the other. We used an experiment (plugging in numbers) to discover the inverse relationship.

🔗

If two quantities vary directly, the model may be linear, quadratic, or exponential. If two quantities vary inversely, the model may be neither linear nor quadratic. In contrast the next example illustrates two quantities can vary inversely, and the model is exponential.

🔗

Example 3.4.5. Decreasing Exponential.

Review the data in the table below. First, notice that \(a\) decreases as \(n\) increases which means \(a\) varies inversely with \(n\text{.}\) Second, note that the data is exponential with a ratio of \(1/2\text{.}\) All exponential models that are decreasing have this property.

🔗

\(n\)	\(a=\frac{1}{2^n}\)	Ratio
1	1/2
2	1/4	1/2
3	1/8	1/2
4	1/16	1/2
5	1/32	1/2
6	1/64	1/2

🔗

Subsection 3.4.2 More Model Usage

In the section above and in Section 3.3 we determined growth rates (e.g., linear, quadratic exponential, and direct and inverse variation) by generating data and checking the consecutive differences and/or the consecutive ratios. In this section we learn a method to directly calculate the relationship.

🔗

Model 3.4.6.

Pressure is a measure of the amount of force spread over an area.

\begin{equation*} P = \frac{F}{A} \end{equation*}

where

\(P\) is pressure,

🔗
\(F\) is the force in units of pounds (lbs) or Newtons (N),

🔗
\(A\) is the area in units of square inches, or meters, or similar.

🔗

🔗

Example 3.4.7.

Walking on snow (or mud) is difficult because our feet tend to puncture the snow (high pressure applied to low strength snow). We can use snowshoes to avoid this problem: they reduce pressure with minimal increase in force (weight). This is a science model with measurements, so we will round using significant digits.

🔗

Determine the pressure Guido exerts on the snow without and with snow shoes. Suppose Guido weighs 172 lbs.

🔗

Without snowshoes each foot has an area of 22 in². What is the pressure he exerts on the snow?

\begin{align*} P & = \frac{172 \text{ lbs}}{2 \cdot 22 \text{ in}^2}\\ & = \frac{172 \text{ lbs}}{44 \text{ in}^2}\\ & \approx 3.909090 \text{ lbs}/\text{in}^2\\ & \approx 3.9 \text{ psi}. \end{align*}

If he wears snowshoes that have a surface area of 144 in², what is the pressure?

\begin{align*} P & = \frac{172 \text{ lbs}}{2 \cdot 144 \text{ in}^2}\\ & = \frac{172 \text{ lbs}}{288 \text{ in}^2}\\ & \approx 0.597222 \text{ lbs}/\text{in}^2\\ & \approx 0.60 \text{ psi}. \end{align*}

🔗

The next example illustrates how we can construct another model when we are interested in only some of the variables.

🔗

Hydraulic system with two pistons of different sizes. — Figure 3.4.8. Hydraulic Press
🔗

Model 3.4.9. Hydraulics.

Consider the situation in Figure 3.4.8. The pressure exerted on a fluid by a piston is the ratio of the force exerted and the area of the piston. On the left that is \(P=\frac{F_1}{A_1}\text{.}\) The same is true on the right \(P=\frac{F_2}{A_2}\text{.}\) Because the hydraulic fluid is contiguous the pressure is the same on both sides. Thus

\begin{equation*} \frac{F_1}{A_1} = \frac{F_2}{A_2}\text{.} \end{equation*}

🔗

Example 3.4.10.

If the left piston has area \(16 \text{ cm}^2\text{,}\) and 5.0 N of force is exerted, what is the force exerted on the second piston if it has area \(25 \text{ cm}^2\text{?}\) This is a science model with measurements, so we will round using significant digits.

🔗

We notice this model is a proportion. We can use the relationship

\begin{align*} \frac{5.0 \text{ N}}{16 \text{ cm}^2} \amp = \frac{F_2}{25 \text{ cm}^2}.\\ \frac{5.0 \text{ N}}{16 \text{ cm}^2} \cdot (25 \text{ cm}^2) \amp = F_2.\\ 7.8125 \amp = F_2.\\ 7.8 \text{ N} \amp \approx F_2. \end{align*}

🔗

The next example illustrates how we can determine how one variable varies with respect to another by direct calculation. Specifically, consider how changing the size of the second piston affects the resulting force.

🔗

Example 3.4.11.

If the second piston’s area is changed from \(25 \text{ cm}^2\) to \(36 \text{ cm}^2\text{,}\) what is the increase or decrease in force?

🔗

Solution.

To determine if it increases or decreases we should calculate the result for the second piston size. We set up the same problem as above.

\begin{align*} \frac{5.0 \text{ N}}{16 \text{ cm}^2} \amp = \frac{F_2}{36 \text{ cm}^2}.\\ \frac{5.0 \text{ N}}{16 \text{ cm}^2} \cdot (36 \text{ cm}^2) \amp = F_2.\\ 11.25 \amp = F_2.\\ 11 \text{ N} \amp \approx F_2. \end{align*}

🔗

The piston is larger and the force is also larger. Thus we know that force and area vary directly. The equation also shows us that they increase linearly, because all terms are linear. In particular \(F_2 = A_2 \left(\frac{F_1}{A_1}\right)\) is in the form of a line with ratio \(F_1/A_1\) and no shift.

🔗

Model 1.3.5 gives two forms of the ideal gas law. The second is produced from the first in the same way as the hydraulics model is produced from the pressure definition. To illustrate consider the case of the air/water mixture in a pressure cooker. First recall the ideal gas law \(PV=nRT\text{.}\) The pressure cooker changes the temperature of the gas, but the volume, number of molecules and gas constant do not change. Thus we have \(P_1 V=nRT_1\) and \(P_2 V=nRT_2\text{.}\) In both cases we can solve for the shared variables.

\begin{align*} P_1 V & = nRT_1\\ P_1 V \cdot \frac{1}{VT_1} & = nRT_1 \cdot \frac{1}{VT_1}\\ \frac{P_1}{T_1} & = \frac{nR}{V} \end{align*}

Similarly \(\frac{P_2}{T_2} = \frac{nR}{V}\text{.}\) Because both expressions equal \(nR/V\) we can set them equal giving \(\frac{P_1}{T_1} = \frac{P_2}{T_2}\) which is the other model.

🔗

The example above produces a second form of the ideal gas law. The next example produces a second form of Ohm’s Law. In many applications the voltage is fixed. For example it might be 12V from a car battery. Thus if we are considering current and resistance in two places we have \(V=I_1 R_1\) and \(V=I_2 R_2\text{.}\) Because voltage is the same (same source of electricity) we have \(I_1 R_1 = I_2 R_2\text{.}\)

🔗

Example 3.4.12.

If the current is 3 amps when the resistance is 8 Ohms, what will the current be when the resistance is 6 Ohms? 4 Ohms?

🔗

Solution.

We can write

\begin{align*} I_1 R_1 \amp = I_2 R_2\\ (3 \text{ amps})(8 \Omega) \amp = I_2(6 \Omega)\\ \frac{(3 \text{ amps})(8 \Omega)}{6 \Omega} \amp = I_2\\ 4 \text{ amps} \amp = I_2. \end{align*}

The second case can be written

\begin{align*} (3 \text{ amps})(8 \Omega) \amp = I_2(4 \Omega)\\ \frac{(3 \text{ amps})(8 \Omega)}{4 \Omega} \amp = I_2\\ 6 \text{ amps} \amp = I_2. \end{align*}

🔗

Notice as the resistance decreased, the current increased. Current varies inversely with resistance.

🔗

Subsection 3.4.3 Limitations of Models

In Example 3.4.2 we solved the model equation to see how one value changes with respect to another. We considered how temperature impacts pressure if the other properties do not change. We also considered how temperature impacts volume if the other properties do not change. We might ask how how volume impacts temperature if the other properties do not change.

🔗

We will not however, because this is impossible. If volume is reduced (e.g., we press down on the handle of a tire pump and do not let the air out), then the pressure will increase. There is no way around that. The increase of pressure will then result in an increase of temperature. This still matches the model we have.

🔗

When using a model, mathematics is used to understand what is meant, but we must understand some of the background so we do not draw false conclusions.

🔗