Quadratics

Section 5.1 Quadratics

This section addresses the following topics.

Interpret data in various formats and analyze mathematical models

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Read and use mathematical models in a technical document

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This section covers the following mathematical concepts.

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

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Read and interpret models (critical thinking)

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In Section 3.4 we learned to identify data that has a quadratic relation. This section presents algebraic notation for quadratics with emphasis on the forms we will use in this book. You should be able to identify a model as quadratic by looking at the equation.

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One of comparing different rates is to enable us to enable us to provide more specific statements than “slow” or “fast.” Quadratic is faster than linear. In the next chapter we will learn that exponential is faster than quadratic (or any polynomial). Frequently descriptions of rates in casual conversation and also in the media are lacking in detail or are even inaccurate. This is part of the concept “Identify rates as linear, quadratic, exponential, or other.”

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Subsection 5.1.1 Algebraic Forms of Quadratics

Quadratic refers to any expression or equation that has a non-zero squared term. The three most common forms are below. All three rows have the same quadratic.

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Table 5.1.1. Algebraic Notation for Quadratics

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Example	Form	Description
\(y=5x^2-17x-12\)	\(y=ax^2+bx+c\text{,}\) \(a \ne 0\)	Standard
\(y=(5x+3)(x-4)\)	\(y=(a_1 x -b_1)(a_2 x -b_2)\text{,}\) \(a_i \ne 0\)	Factored
\(y=5\left( x -\frac{17}{10} \right)^2-\frac{1489}{100}\)	\(y=a(x-h)^2+k\text{,}\) \(a \ne 0\)	Convenient

For this class the form we will see the most is

\begin{equation*} y=k x^2 \end{equation*}

where \(k\) has some meaning in each model.

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When reading models quadratics may look a little different. Table 5.1.2 shows examples of equations that are quadratic and some that are not (but might look like it).

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Table 5.1.2. Quadratic and Non-quadratic

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Quadratic	Non-quadratic
\(11x^2+32x-3\)	\(5x+3\)
\(2(x-3)^2+7\)	\(y=x^3+7x^2-5x+3\)
\(y=23-3x^2\)	\(y=\frac{17}{x^2}\)
\(x(6x-5)=21\)	\(x^2(x-5)=7\)
\(y=(x+3)(x-5)\)	\(y=0x^2+3x+2\)

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Subsection 5.1.2 Quadratic Models

Now that we know how to recognize models (equations) as quadratic, we will present a few.

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Model 5.1.3. Load Factor.

The maximum load factor on an aircraft is

\begin{equation*} n_{max}=\left(\frac{v}{v_s}\right)^2 \end{equation*}

where

\(n\) is the load factor,

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\(v\) is the velocity, and

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\(v_s\) is the stall speed.

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Load factor is a measure of the force exerted on the aircraft by a maneuver. Maximum load factor is the greatest load factor an aircraft would experience if at the given speed, the aircraft executed a maximum performance maneuver.

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Load factor is measured as a multiple of the force of gravity. Thus a load factor of 2 means the object is subject to a force twice as strong as earth’s gravity. The expression “pulling g’s” refers to experiencing a load factor greater than one.

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Example 5.1.4.

What is the maximum load factor if the stall speed is 54 kias, and the current speed is 95 kias? What if the current speed is 105 kias? How many g’s does the maximum load factor increase between those two speeds? Recall kias is nm/hour. Round to one decimal place.

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We use the formula

\begin{align*} n_{max} & = \left(\frac{95}{54}\right)^2\\ & \approx (1.759259259)^2\\ & \approx 3.094993140\\ & \approx 3.1. \end{align*}

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\begin{align*} n_{max} & = \left(\frac{105}{54}\right)^2\\ & \approx (1.9444444449)^2\\ & \approx 3.780864196\\ & \approx 3.8. \end{align*}

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The increase of 10 kias increased the g’s by \(3.8-3.1=0.7\text{.}\)

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Model 5.1.5. Kinetic Energy.

The kinetic energy of an object in motion is given by

\begin{equation*} E = \frac{1}{2}m v^2 \end{equation*}

where

\(E\) is the energy (in foot pounds or Joules)

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\(m\) is the mass (think weight) of the object (in pound mass or grams)

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\(v\) is the velocity of the object (in feet or meters per second).

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Example 5.1.6.

What is the kinetic energy of a glider weighing 1323 lbs and flying at 34 kias, at 54 kias? How much did it increase? Use significant digits.

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First, we need to convert the speeds to feet per second. Table 1.1.2 suggests we can multiply.

\begin{gather*} \frac{34 \text{ nm}}{\text{hr}} \cdot \frac{6076 \text{ ft}}{\text{nm}} \cdot \frac{\text{hr}}{3600 \text{ sec}} \approx 5\underline{7}.38444444.\\ \frac{54 \text{ nm}}{\text{hr}} \cdot \frac{6076 \text{ ft}}{\text{nm}} \cdot \frac{\text{hr}}{3600 \text{ sec}} \approx 9\underline{1}.14. \end{gather*}

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Now, we can calculate the kinetic energy.

\begin{align*} E_{34} & = \frac{1}{2}(1323 \text{ lbs})\left(\frac{5\underline{7}.38444444 \text{ ft}}{\text{sec}} \right)^2\\ & \approx 2.\underline{1}78302608 \times 10^6\\ & \approx 2.2 \times 10^6.\\ E_{54} & = \frac{1}{2}(1323 \text{ lbs})\left(\frac{9\underline{1}.14 \text{ ft}}{\text{sec}} \right)^2\\ & \approx 5.\underline{4}94749485 \times 10^6\\ & \approx 5.5 \times 10^6. \end{align*}

The increase was \(5.5 \times 10^6 - 2.2 \times 10^3=3.3 \times 10^3\) foot pounds.

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Model 5.1.7. Potential Energy.

Potential energy of an object is given by

\begin{equation*} E=mgh \end{equation*}

where

\(m\) is the mass of the object (in pound mass or grams)

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\(g\) is the gravitational constant (in feet per seconds squared or meters per second squared)

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\(h\) is the height of the object (in feet or meters)

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Potential energy can be thought of as the energy that was expended to move the object higher that can be regained by letting gravity pull it back down. The height may be calculated as above ground level, or above sea level, or any other reference convenient for a question.

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Subsection 5.1.3 Inversely Quadratic Relations

Table 5.1.2 has \(y=17/x^2\) as an example that is not quadratic. Instead \(y\) varies inversely with the square of \(x\text{.}\) While not quadratic, as illustrated below by the tables of differences, they can be solved using the same techniques. Solving quadratics will be demonstrated in the next section.

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Table 5.1.8. Quadratic Relation

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\(x\)	\(3x^2\)	1st Difference	2nd Difference
1	3
2	12	9
3	27	15	6
4	48	21	6
5	75	27	6
6	108	33	6

Table 5.1.9. Inversely Quadratic Relation

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\(x\)	\(\frac{3}{x^2}\)	1st Difference	2nd Difference
1	3
2	3/4	-9/4
3	1/3	-5/12	11/6
4	3/16	-7/48	13/48
5	3/25	-27/400	47/600
6	1/12	-11/300	37/1200

In Table 5.1.8 the second differences are all 6, so that is quadratic. In contrast in Table 5.1.9 the second differences are all different.

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Model 5.1.10. Gravitational Attraction.

Two objects exert a gravitational pull on each other related by their masses and the distance between them. The relationship is

\begin{equation*} F = G \cdot \frac{m_1 m_2}{r^2} \end{equation*}

where

\(F\) is the resulting gravitational force in Newtons

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\(G\) is the gravitational constant

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\(m_1, m_2\) are the masses of the two objects in kilograms

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\(r\) is the distance between the two objects in meters

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The gravitational constant is \(G=6.6743 \times 10^{-11} \frac{\text{m}^3}{\text{kg} \cdot \text{s}^2}\text{.}\) The constant has not been determined to any greater precision.

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Example 5.1.11.

What force does the earth exert on the moon? The mass of the earth is approximately \(5.97219 \times 10^{24}\) kg, and the mass of the moon is approximately \(7.34767 \times 10^{22}\) kg. The average distance between the earth and moon is 382,500 km.

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We need to convert the kilometers to meters so units match (gravitational constant). Because it is kilo the conversion is \(382,500 \text{ km} \cdot \frac{1000 \text{ m}}{\text{km}} = 382,500,000 \text{ m}\text{.}\) Using the model

\begin{align*} F & = (6.6743 \times 10^{-11}) \frac{\text{m}^3}{\text{kg} \cdot \text{s}^2} \left(\frac{(5.97219 \times 10^{24} \text{ kg})(7.34767 \times 10^{22} \text{ kg})}{(382500000 \text{ m})^2} \right)\\ & \approx 2.001824977 \times 10^{20} \frac{\text{m}\text{ kg}}{\text{s}^2}\\ & \approx 2.002 \times 10^{20} \text{N} \end{align*}

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Exercises 5.1.4 Exercises