chapter 2 — graphs · 100 chapter 2 — graphs a) if you have a right triangle, then . b) if you...

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Chapter 2 — Graphs

In this chapter you’ll learn about computing distance on the x–y plane and how that relates to equations of circles. The last section on variation just seems to be thrown into this chapter for lack of a better home.

Section 2.1 — Rectangular Coordinates

Our focus in this section will be a review of the Pythagorean Theorem and its application to finding dis-tances, plus learning how to find the midpoints of line segments, which is also related to distance. The ability to determine distances in the x–y plane is critical if we are to combine algebra with geometric applications.

Topic 1: The Pythagorean Theorem.Topic 2: Finding distance in two dimensions.Topic 3: Midpoint of a line segment.

Practice Problems: 15 – 27, 35 – 43 odd

Topic 1: The Pythagorean Theorem.

Two of the most important theorems in mathematics are related to triangles. One of them, the Triangle Inequality, was discussed in the last chapter.

The other theorem is the famous Pythagorean Theorem. Although named for the mathematician/philoso-pher Pythagoras (c. 585 bc – c. 500 bc) in ancient Greece, this theorem predated Pythagoras by about a thousand years and was known in various parts of the world before his time. Pythagoras coined the word “philosopher” and founded a philosophical society dedicated to learning and the study of rationality, which literally was the application of ratios (fractions and whole numbers) to nature. By modern standards we would regard his phi-losophy as crackpot numerology, but the Pythagoreans didn’t have much to go on in those days and produced clever arguments to support their world view.

The Pythagoreans made one very important contribution which has lasted. The way we define and, with small modifications, tune our musical notes is credited to Pythagoras and involves simple fractions.

Many people don’t realize that the modern form of the Pythagorean Theorem is actually two theorems. Here is a statement of them.

A triangle is a right triangle if and only if it contains sides of length a, b and c such that .

The “if and only if” phrase is a logical statement which means that the theorem works in two directions. Informally:


100 Chapter 2 — Graphs

a) If you have a right triangle, then .

b) If you have then it is a right triangle.

(A right triangle is one in which two sides are perpendicular, or form a 90˚ angle. The long side of a right triangle, always opposite the right angle, is called the hypotenuse.)

The second form, part b), is not as widely known. It is used in construction, for example, in laying out the perpendicular sides of buildings. By staking down a string representing, say, the length of the front of a build-ing, then connecting two strings representing the lengths of the side and the diagonal (hypotenuse), a large but very accurate 90˚ angle is formed. Proving this form of the Pythagorean Theorem is equivalent to deriving the Law of Cosines in trigonometry. We’ll leave it alone here.

The first part, part a), is what most people remember. The funniest adaptation I know of is in the scene in the movie version of The Wizard of Oz when the scarecrow is given his Ph.D. in Thinkology by the wizard. The scarecrow snaps to attention and very quickly rattles off something which sounds smart and resembles the Pythagorean Theorem: “The sum of the square roots of any two sides of an isosceles triangle equals the square root of the remaining side.” The joke is on the scarecrow. Not only is this nonsense, it is impossible for any triangle (due to the Triangle Inequality!). Whether this was in the script or an improvisation by the actor Ray Bolger is unknown to me.

There are over three hundred known ways to prove the Pythagorean Theorem, more than any other math-ematical theorem. The only known contribution to mathematics by a future United States president was a proof of this theorem using trapezoids by James Garfield. Below may be the easiest proof.

Given an arbitrary right triangle with legs of length a and b and hypotenuse of length c, we can form a square imbedded inside a larger square, as follows.

It is easy to show by using simple geometry that the inner object is in fact a true square.The secret to proving is to form an equation using the areas of both squares and the four

triangles. (Note that the area of a right triangle is exactly half of a rectangle.)

101Section 2.1 — Rectangular Coordinates

(Area of large square) – (Area of the 4 triangles) = Area of the small square:

Simplifying the algebra gives the following.

One last simplification gives us the Pythagorean relationship.

Now we can look at ways to apply the Pythagorean Theorem.

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Topic 2: Finding distance in two dimensions.

Find the distance between the points (– 3, 7) and (6, – 4).

The importance of the Pythagorean Theorem lies in its applicability to finding distance in two dimensions. Recall that a method of finding distance is called a metric. The Pythagorean Theorem allows us to find the straight-line (as the crow flies) distance between two points. Similar straight-line methods in higher dimensions adapt the Pythagorean Theorem, and the collection of these particular methods are known as Euclidean metrics, after the ancient Greek geometer, Euclid (c. 325 bc – c. 265 bc).

In real life we often measure distance in other ways. When you drive somewhere the odometer is measuring distance along the streets you drive, which invariably do not connect you to your destination in a straight-line. When we measure the distance between cities, for example, we are measuring along the curvature of the Earth, not in a straight line through the three dimensional space we live in. Altering the way we measure distance invariably changes some aspects of the geometry. This became a very important consideration in mid–1800’s mathematics and is very relevant to our modern understanding of physics. Einstein was able to demonstrate that we live in a universe with a geometry warped by gravitational and other forces.

Working the above problem is simply a matter of drawing a right triangle and applying the theorem.

I like to discourage formula approaches when practical because students learn little from formulas and often


make careless errors in applying them. If you have the picture in front of you there is little chance you would get, say, a length of 3 or a height of 3 as some would using a formula without a picture on this problem.

Of course, due to its importance, the Pythagorean Theorem is one of those few formulas you really do need to know. In this case I have labelled the distance between the two points as d, for “distance.”

Now to write the answer, which I’ve also approximated.

Reinforcement Problems

1. Find the distance between (23, – 33) and (– 19, – 48). Give both the exact distance and the approximate distance rounded to two decimal places.

2. Find the distance between (2.3, 4.7) and (– 0.6, 1.9). Give both the exact distance and the approximate distance rounded to two decimal places.

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Topic 3: Midpoint of a line segment.

Find the midpoint of the line segment connecting the points (– 5, 4) and (7, 1).

This is another potentially formula driven problem. The purpose of this course is to make you a better thinker and analyzer, not a formula user, so let’s reason our way to finding midpoints.

What does midpoint mean? It is the point at the exact middle, right? Let’s first try this question on a very easy problem in one dimension before working the two–dimensional problem above. We can often gain insights about more difficult problems by first examining related simpler problems.

What number falls in the middle of 2 and 8? Without knowing anything else, we could just count manually on a number line and arrive at the middle number, 5.

We could also find the distance between 2 and 8, which is 6, take half of the distance, which is 3, then add that to the smaller number or subtract from the larger in order to get 5.

We should try to generalize this process by finding the distance between a and b, where we agree that b is larger than a.

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In the last chapter we saw that absolute value determines distance along a number line. So the distance between a and b is or (it doesn’t matter which), but since b > a, we can avoid the absolute value sign by simply subtracting the smaller number from the larger, in this case, b – a.

Adding half of the distance from a to b to the number a will put us at the midpoint.

Now we’ll simplify this expression by combining fractions.

As it turns out, even if b is less than a, we’d still get the same result. But do you recognize what this says? What do you get when you add two numbers and then divide by 2? Of course, it’s the simple average of two numbers. The average of two numbers is simply the middle of the two numbers.

Now let’s return to the problem at the beginning of this topic, finding the midpoint of the segment connecting the points (– 5, 4) and (7, 1). Looking at the graph below, does it seem reasonable that if we find the midpoints of the vertical and horizontal segments drawn that these would lead us to the midpoint of the diagonal segment? It is an easy geometric argument to show this is true (by use of the two smaller, congruent triangles). And since the vertical and horizontal segments are just like ordinary number lines we can use what we know: the midpoint is the coordinate determined by the average of the x coordinates and the average of the y coordinates.

So we now calculate the midpoint by finding the separate averages of the x and y values.

If you understood this derivation then you learned something of value. If you simply look up formulas to plug numbers in then you have only learned how to plug numbers into a particular formula that you will quickly forget!




3. Find the midpoint of the segment connecting (23, – 33) and (– 19, – 48).

4. Find the midpoint of the segment connecting (2.3, 4.7) and (– 0.6, 1.9)._________________________________________________________________________________________

Solutions to Reinforcement Problems

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105Section 2.4 — Circles

Section 2.4 — Circles

Oddly enough, circles and right triangles are closely related in some respects. The equation of a circle in the plane is an application of the Pythagorean Theorem. In this section we explore various aspects of circles and their corresponding geometry.

Topic 1: Defining circles in the plane.Topic 2: Identifying center and radius from the standard form of the equation of a circle.Topic 3: Identifying center and radius from the general form of the equation of a circle.Topic 4: Finding the equation of a circle from given information.Topic 5: Finding another equation of the circle from given information.

Practice Problems: 7 – 49 odd

Topic 1: Defining circles in the plane.

Everyone knows that circles are perfectly round things, but it is necessary to define circles in a more rigor-ous mathematical sense in order to be able to analyze them. So how would you define a circle? What does a circle have?

Every circle has a center point and a diameter, or radius, measurement.

The radius describes how far it is to the edge of a circle from the center. This gives us a clear mathematical way to define a circle in terms of distance.

A circle is a collection of all points equidistant from a center point.

Let’s apply this definition in a simple way to a circle of radius 5 with the center at the origin, (0, 0).


We seek a way to express this circle in terms of algebra, and since it is a two–dimensional object in the x–y plane, then what we are really looking for is an equation relating x and y together. We can do this by label-ling an arbitrary point on the circle as an (x, y) coordinate and apply what we know about distance. Since the distance from the origin to any point on this circle is 5, then we can measure that distance using the Pythagorean Theorem.

Because this triangle is a right triangle, the Pythagorean Theorem gives us an equation relating both x and y to the radius.

We normally square the radius number when writing the equation.

Since (x, y) represents any point on the circle, we have found its equation describing this circle in terms of x and y.

We can position a circle anywhere in the plane, so we should generalize this idea to include any circle, any center, and any radius. This time we’ll label the center as (h, k) and give the circle a radius of r. Then we measure the distance from the center to any edge point, (x, y), using the distance r.

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Note that I used absolute value to measure the length of the legs of the right triangle. One extends in a direction parallel to the x–axis and the other in a direction parallel to the y–axis. Measuring these lengths is like measuring lengths on a number line, which is why absolute value is a natural application in this instance.

Applying the Pythagorean Theorem once again gives us the following.

Squaring an absolute value expression effectively removes the absolute value sign since the outcome of squaring is, like distance, positive (or zero). Thus we have arrived at what your textbook calls the standard form of the equation of a circle.

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Topic 2: Identifying center and radius from the standard form of the equation of a circle.

Find the center and radius of the circle given by the equation .

This equation is organized in the most user–friendly form for a circle, the so–called standard form.

It will be easy to recognize the center and radius from the equation in the problem if we reorganize it slightly.

Note that the (y – k) form was rewritten as (y – (– 7) ). This shows that – 7 is the y–coordinate of the cen-ter.

Now we can state the answer. It should be written out so that it is clear to the reader what you mean.

Reinforcement Problem

1. Find the center and radius of the circle given by the equation .

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Topic 3: Identifying center and radius from the general form of the equation of a circle.

Find the center and radius of the circle represented by .

This equation is in the form which your textbook calls the general form of the equation of a circle. It is a second degree polynomial equation with two variables in which there is no xy term and in which the coef-



ficients of the x2 and y2 terms are equal. It can be shown that all such equations either represent circles, a single point, or something that can’t be graphed (because there is no real solution). We are only going to worry about circles.

The secret to dealing with these general forms is to find a way to rearrange the terms into the more usable standard form of a circle. This process uses a completing the square technique, very similar to the completing the square process used to solve quadratic equations.

As an overview, we will sort the x and y terms, transfer the constant term to the other side of the equation, then complete the square twice. This gives us the equation of a circle for which we can immediately recognize the center and radius.

First we’ll sort and transfer. Notice that I’ve left some blank space.

Here comes the unusual part of the problem (if you are not used to completing the square). We are going to take half of the linear coefficients (12 for x and – 4 for y), square each, and add those squares to both sides of the equation to keep the equation balanced.

What did that accomplish? This step artificially creates perfect square binomials for both the x and y vari-ables. We’ll simplify and factor.

This can be written as a sum of squared binomials, which is precisely the form of the circle that gives us the information we are looking for.

Now the answer can be stated.




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Topic 4: Finding the equation of a circle from given information.

The center of a circle is at (7, – 4) and the circle is tangent to the x–axis. Find the equation of the circle.

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This problem is an example of curve fitting, which is the art of finding a mathematical equation or function that represents a mathematically oriented situation. It is an important process, one of the ways mathematics is applied in the real world.

In this case it would be good to discuss what tangent means. Tangent actually has two meanings. As a noun, the tangent represents a ratio of two specific sides of a right triangle relative to a specific angle. You would study that in trigonometry. As an adjective, a line is a tangent line which just touches a curve, like a line touching only the edge of a circle. So here is what this example looks like.

It can be shown that tangent lines to circles always run perpendicular to the diameter touching the tangent point. Why is that an important observation here? The line segment I’ve drawn to the tangent point is the short-est distance from the center to the edge, so its length is the radius, which, of course, is 4. (And so the square of the radius is 16.)

We now have both the center and the radius, so we can finish by writing the equation of the circle.


4. The center of a circle is at (– 12, 7) and the circle is tangent to the y–axis. Find the equation of the circle.

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Topic 5: Finding another equation of the circle from given information.

Find the equation of the circle on which the points (27, – 31) and (– 12, – 45) form the endpoints of a diam-eter.

Below is a sketch of this problem. Even if you understand clearly what the problem is demanding, sketch-ing a diagram can help you find errors if you are alert. But if you’re not clear about what is going on, a sketch



is the best place to start.

We will need the center and radius in order to complete the problem, right? Since a specific diameter is given we can use it to find the center of the circle, which is at the midpoint of the diameter.

The center is a little messy because of the fraction. Ordinarily we would calculate the radius using the center and one of the edge points, but in order to avoid fractions, at least for a while, we could first calculate the length of the diameter. Here’s the picture of what’s going on.

The Pythagorean Theorem gives us the diameter.

So the diameter is the square root of this number.

Now we get the radius by taking half of the diameter.

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But we need the square of the radius for the equation of a circle.

Now we can finish the problem.


5. The points (4.6, – 3.7) and (–5.9, – 1.1) form a diameter of a circle. Find the equation of the circle.

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113Section 2.5 — Variation

Section 2.5 — VariationA variation, in the context of algebra, is a proportion. All of us are familiar with simple, direct propor-

tions. For example, if you work twice as long then you expect to get paid twice as much (disregarding potential overtime, of course). Or, if you buy three pounds of steak then you expect to pay three times as much as for one pound of steak.

Variation problems take into account more complicated forms of proportions. As such, variation appears widely in the sciences, especially physics and biological sciences, and also plays a significant role in engineer-ing calculations. In fact, variation explains a lot about the how the world works; you’ll get a glimpse of this as you read on.

Topic 1: Introducing a direct variation.Topic 2: Another direct variation.Topic 3: Inverse variation.Topic 4: Combined variation.Topic 5: Joint variation.Topic 6: Musing on anatomy, physiology, airplanes, and McDonald’s french fries.

Practice Problems: 1 – 43, odd

Topic 1: Introducing a direct variation.

Suppose that a 10 inch round pizza feeds one person. What size pizza would feed four people with a similar appetite?

What do you think? I call the above problem “Davidson’s famous pizza problem.” I have presented it for years in many classes and without fail someone always says, “A 40 inch pizza.” Have you ever seen a 40 inch pizza? Neither have I. But this is a very natural response because we are accustomed to simple direct propor-tions where things grow at the same rate. So a pizza four times as large would need to be four times the size in diameter, right? Of course not!

It’s harder to persuade people that a 20 inch round pizza feeds four people in our problem than it is to convince them that a 20 inch square pizza is four times as large as a 10 inch square pizza. Round pizzas, square pizzas—it doesn’t matter as long as they have the same thickness and toppings per square inch. With the same thickness, we are comparing area.

10 inches

20 inches


So, how can a pizza twice as large feed four times as many people? The question may be deceptive because it is not posed correctly. Only a pizza four times as large could feed four times as many people. Our pizza is not only twice as wide, it is also twice as long, so 2 • 2 = 4 explains this phenomenon.

Since you are to learn about algebraic variation, let’s frame the pizza problem in terms of the language of variation.

The number of people a round pizza feeds varies directly to the square of the diameter. If a 10 inch (diameter) round pizza feeds one person, what size feeds four?

This problem, as stated, has three parts to it. The first sentence describes the algebra equation, or relation-ship, modelling this problem. “Varies directly” tell us what kind of variation it is, which means that it describes how “number of people” and “square of the diameter” relate to each other.

As you can see, I’ve labeled the two quantities, “number of people” and “square of the diameter,” as P and D2, respectively. We could use x and y, but it would be easier to get these mixed up, so I like to use more descrip-tive letters.

You always need to read these very carefully and not overlook a key word such as “square.”Mathematicians symbolize a variational relationship with the tilde symbol, “~”.

What this says is that there is a direct proportional relationship between P and D2. In any direct variation both sides increase or decrease together. In this case it is obvious; make a larger pizza and you can feed more people.

We cannot say that P = D2. Assuming they are equal is the same as assuming that people and square of the diameter are the same, an absurdity. But we can form an algebra equation from this by introducing a third letter, which is a constant. It is traditional to use the letter k.

Constant k is called the constant of variation. (Some people call it the constant of proportionality.) The units of k, when it has units, must convert the right side’s units to the left side’s. In this case k would carry units similar to . In some applications in the sciences or engineering it is useful to know the units of k, but for our purposes, as long as we keep all other units consistent, we can ignore them here. For example, if you used inches in one place and feet in another for the same type of quantity, say, “10 inch round pizza and 2 foot round pizza,” you’d be in trouble without using any appropriate unit adjustment.

The constant of variation is a constant that depends on the specific problem at hand. If, for example, the above pizza is a thin crust, then a thick crust pizza would probably result in a different k value since a person could not normally eat the same size pizza with a thicker crust as compared to a thin crust.

The second part of the problem, “If a 10 inch (diameter) round pizza feeds one person,” describes a calibra-tion, or measurement that will allow us to calculate k. In algebraic terms it states that, “When D = 10 inches, then P = 1 person.” We can substitute these numbers in to P = k D2 and find the value for k.

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Now we can write the equation of variation for this problem.

This equation will now allow us to make computations about this brand of pizza and people with this appetite for any size of pizza or any number of people. In this sense, we have created a mathematical model.

The last part of the problem, “what size feeds four?” asks us to apply the equation we derived. This is equivalent to saying, “Find D when P = 4.”

This is a quadratic equation, but a very simple one. We can solve it by applying the square root to both sides.

Algebraically, there are two answers, 20 and – 20, but a negative length is impossible, so we throw it out.Be sure to write out the answer in plain English.

Reinforcement Problem1. The weight of trees of the same species and similar shape varies directly to the cube of the height. Sup-

pose that a 9 foot tall Christmas tree weighs 40 pounds. How much would a similar 20 foot tall tree weigh? (Round to the nearest pound.)

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Topic 2: Another direct variation.

The period (length of time for one back and forth swing) of a pendulum varies directly to the square root of the length of the pendulum. If a 42 inch pendulum has a period of 1.6 seconds, what is the period for a similar 30 inch pendulum? (Round to one decimal place.)

This relationship between the period and length of a pendulum is approximate but works quite accurately provided that the arc of the swing is fairly small. We derive this relationship in both our physics courses and our differential equations course, but here we’ll just used the derived relationship.

I will use T for the time of the period, and L for the length of the pendulum. With a direct variation, we

Section 2.5 — Variation


are comparing these letters this way.

It would be just fine to skip on to the next step, creating an equation by introducing the constant of variation, k.

Every time I’ve given a pendulum problem on a test several people carelessly have read “square root” as “square,” thus screwing up the mathematics. It would be sensible to believe that you, too, are vulnerable to this kind of reading mistake if you rush through these problems.

The second statement, the calibration of our constant k, states that T = 1.6 seconds when L = 42 inches. Inserting these into our initial equation allows us to find k for this problem.

Why mess with decimal numbers at this point? I must make a serious point here about precision.

For the greatest precision, use all the decimal places available and only round off your final answer to the desired number of decimal places.

In this case, the problem asks that the answer be rounded to one decimal place. If we were to start rounding off to one decimal place now before the problem is completed then we will have introduced some fairly major er-rors into the calculation and will lose a lot of precision. Rather than mess with a lot of decimals, I like to work symbolically and save the calculator work to the very end.

The decimal value of k is not terribly interesting to us in this instance anyway. Let’s just feed what we have into the first equation to produce the equation of variation for this situation.

Now we are ready to apply this equation to answer a new question, namely, what is the period of a 30 inch pendulum?

Don’t forget to write the answer in plain English after rounding to one decimal place.

Reinforcement Problem2. The stopping distance for a car varies directly to the square of its speed. If a car going 40 miles per hour

can stop in 100 feet, what is the stopping distance for a car going 100 miles per hour?_________________________________________________________________________________________

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Topic 3: Inverse variation.

The frequency of a vibrating string varies inversely to the length of the string. If the low string (note E) on a double bass has a frequency of 41.2 Hz with a string length of 42 inches, what string length produces note B above the E, a frequency of 61.7 Hz?

For those of you interested in a little more background to this problem, the double bass is also known as the string bass, or contrabass, the largest stringed instrument in orchestras. “Hz” stands for “Hertz,” named for a 19th century German physicist, Heinrich Hertz (1857 – 94), who pioneered important work concerning light and electromagnetic waves. The units of Hz are 1/sec., but would be thought of as vibrations per second in this context. So the lowest note on a conventional double bass, E, vibrates about 41.2 times per second, about an octave higher than we humans can hear. (Some basses are fitted with an attachment in order to sound an even lower note, C.)

Pythagoras realized this inverse relationship between string length and pitch and either he or his followers devised a tuning system based on string length which defined the notes we use today, although tuning systems have undergone refinements through the centuries.

Back to the problem at hand, an inverse variation is a comparison between one variable and the recipro-cal of another (or another raised to a power). The hallmark of an inverse variation is that when one quantity is increased the other is decreased. Thus if you hold your finger down on the string of a double bass somewhere along the fingerboard, the string length that vibrates is shortened but the pitch, or frequency, is raised.

For this problem I’ll let F = frequency and L = string length. Here is our comparison.

To create the initial equation we need to put k somewhere. It doesn’t matter where as long as k is multiplying or dividing something. Different placement of the constant will result in different values for k, but you can still get the right answer for the problem. Most people would place k as the numerator of the fraction.

Now let’s find k by substituting our initial values for frequency and string length.

In this case the arithmetic is rather simple and gives us an exact short decimal for k, so this time I’ll multiply it out.

Here’s our equation of variation for the lowest string in a standard 42 inch double bass.

Now we are ready to solve the problem that was posed: What string length produces a frequency of 61.7 Hz?



I suppose that problems like this are why the authors of your textbook placed this section in with the rational equations.

Even if we round this off to one decimal place, 28.0, it comes out to 28 inches, two-thirds of the original 42 inches.

Pythagoras would have predicted this.

Reinforcement Problem3. The amount of light from a light source that we perceive varies inversely to the square of our distance

from the light source and the object. Suppose that a flashlight produces an illumination of 10 footcandles at a distance of 12 feet. What is the illumination at a distance of 50 feet?

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Topic 4: Combined variation.

The volume of a gas varies directly to its temperature (measured in Kelvin) and inversely to its pressure. Suppose a helium balloon with a volume of 4 gallons is released at sea level at a temperature of 300 K. At an altitude of 18,000 feet the air pressure is half that at sea level. (Air pressure is 14.7 pounds per square inch at sea level.) If the temperature is 272 K at 18,000 feet, what is the volume?

This problem is based on a gas law and gives us a mix of direct and inverse variations. That makes this problem a combined variation.

For the gas law to work we must use a temperature scale that is proportional in itself, which is why we need the Kelvin scale. A temperature of 300 K, for example, means there is three times the heat energy present as at 100 K. Absolute zero is 0 K. And if you are wondering, 300 K is 80˚ F, and 272 K is 30˚ F.

Our 4 gallon balloon would correspond in size to a spherical balloon about a foot in diameter.With those measurements out of the way, let’s set up the starting equation. I will use the constants V for

volume, T for temperature and P for pressure. Since T varies directly to V, it is a numerator; since P varies inversely to V, it is a denominator.

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As before, our next step involves plugging in some given measurements in order to find k.

It turns out that k calculates to an exact three decimal representation. (Otherwise, I would be inclined to leave all the calculator work to the end.)

I always think it a good idea, for organizational purposes, to write the equation of variation before apply-ing it.

So, let’s find out how large the balloon will be at 18,000 feet.

An answer of “V = 7.25” won’t mean much to your readers. We should clearly communicate this re-sult.

Since volume varies directly to the cube of the diameter, the balloon has swelled to a diameter of nearly 15 inches. If it hasn’t popped yet, it will soon as it continues to rise. All helium balloons expand when they rise until they either break or lose enough helium in order to sink.

Reinforcement Problem4. The electrical resistance of a wire varies directly to the length of the wire and inversely to the square of

the diameter of the wire. Suppose that a 100 foot wire with a diameter of 2 mm has a resistance of 1.5 ohms. What would the resistance be for a similar wire 200 feet long and a diameter of 3 mm?

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Topic 5: Joint variation.

The force exerted by the wind on a surface varies jointly with the area of the surface and the square of the speed of the wind. A window with an area of 12 square feet will sustain a force of 26 pounds in a 20 mile per hour wind. What is the force on a 6 square foot window in a hurricane wind of 74 miles per hour?

A variation like this is why people need to board up windows in a hurricane, or why umbrellas can get away from us in a gust of wind. For example, since the force varies directly to the square of the speed, the force on an object quadruples when the wind speed is doubled. A wind speed that is tripled would increase the force by a factor of nine, and so on.

This problem compares one quantity, force, to the product of two quantities, area and square of the speed.



A joint variation, then, means that we will be multiplying two or more variables together, but as a direct varia-tion.

I will use the following variables: F for force, A for area, and s for speed. Along with our constant of variation, k, here is the initial equation of variation.

Now we can insert our initial, given measurement, “A window with an area of 12 square feet will sustain a force of 26 pounds in a 20 mile per hour wind,” and find k.

I am inclined to leave k as it is for the moment. It does calculate to 0.005416666667. I don’t want to write all those numbers down and have to punch them in again later, and if we round it off to 0.005 or 0.0054, we’ll create rounding errors before we are done that may cause us to lose a lot of precision, depending on how many significant digits are used. You might say that I am motivated by both precision and laziness in the way I do this!

So, here is the completed equation of variation.

Now we can complete the problem by inserting 6 for A and 74 for s. (Hurricanes have sustained winds of at least 74 miles per hour, incidentally.)

Actually, this comes out to 177.97. It would be reasonable to give three significant digits and round this off to 178.

Reinforcement Problem5. The amount of weight that a beam can support varies jointly to the width and to the square of the thick-

ness of the beam, and inversely to its length. Suppose that an 8 foot long “two by four” has a width of 3 3/4 inches and a thickness of 1 3/4 inches, and can support 300 pounds. If you turn this board on its side, i.e., switch width and thickness measurements, how much weight will it support?

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Topic 6: Musing on anatomy, physiology, airplanes, and McDonald’s french fries.

Variation is one of my favorite topics in algebra. It explains a lot about our world, especially its geometry. In this section I want to share with you some of the insights which variation provides.

My first daughter weighed 7 1/2 pounds and was 20 inches long when she was born. By the time she reached 5 feet tall (60 inches), what would be the prediction of her weight? If weight varied directly to length, or height, then when she was three times as long we would expect her to weigh three times as much, about 22 or 23 pounds. That can’t be right!

Weight ought to vary according to the cube of the height. Why? Weight and volume are closely related because weight occupies volume, which carries the units of length • length • length. So since my daughter was three times as long, should she have been expected to weigh 3 • 3 • 3 = 27 times as much? That would be about 200 pounds, roughly twice as much as reality. So what is wrong with the math here?

Weight does vary directly to the cube of the length or height, but only if the proportions and weight density remain the same. All humans change proportions as they grow. Imagine what adults would be like if they had the same proportions as newborn infants. Just our heads, alone, would be enormous!

Let’s assume a comparison of two animals of the same species, but these animals share the same propor-tions. I will use easy numbers for sake of comparison.

Length Skin Area Weight small animal 1 ft. 1 ft.2 1 pound large animal 2 ft.

The skin area measures the amount of skin covering for the animal, and since we would measure in terms of area, we would measure in terms of feet • feet = square feet. How much would the larger animal have? It is twice as long, so it has 2 • 2 = 4 times as much skin area.

What about weight? We observed above that weight, since it depends upon volume, a cubic measurement, must vary according to the cube of the length. So our larger animal weighs 2 • 2 • 2 = 8 times as much.

Length Skin Area Weight small animal 1 ft. 1 ft.2 1 pound large animal 2 ft. 4 ft.2 8 pounds

Now here’s where it gets interesting and informative. If this animal is a mammal, then it produces its own internal heat, much as we maintain a constant body temperature. The larger animal must heat eight times as much body weight, so it produces eight times the heat. What happens to the heat? Heat escapes through the skin. Humans sweat, dogs pant, and so forth, so different mammals can differ on how they lose heat, but in all cases heat loss varies according to area, not volume, since heat is lost across an area such as skin.

Which animal will run hotter? We can produce an index of sorts for comparison purposes by dividing weight by skin area. This compares heat production to ability to lose heat, as you’ll see.

The small animal has an index of .



The large animal has an index of . The large animal runs hotter since it produces twice as much heat as its ability to lose heat. It will feel

warmer, on the whole. Because of this, the larger animal will probably be less active in order to produce less heat than the smaller one, just as small children are a lot more active than adults. (Heat production may not be the sole explanation for the higher energy of children!)

In a sense, this variation explains why most large mammals have very little fur. Elephants may have the least, with only a hair sticking up here and there. Elephants like to splash about in ponds, and their huge ears are their principle way of cooling off; these act similar to car radiators, only they have miles of blood capillaries to circulate blood across the surface of the ears. Polar bears do have a lot of fur, but they are also happy and comfortable in frigid water that would kill humans rather quickly.

On the other end of the spectrum, small mammals such as mice and rats are very furry. (One exception is the naked mole rat, which lives underground where the temperature is rather constant.) These small rodents constantly forage for high energy food such as insects and nuts in order to sustain their body temperature since heat is lost so easily. The smallest mammal is a species of shrew which must eat three times its body weight every day just to stay alive.

Most large mammals are vegetarians. Heat production is not a priority, so they can afford to eat low energy foods such as leaves and grasses. Lions, among the largest carnivores, do eat higher energy food, but spend close to 23 hours per day sleeping.

Back in the 50’s a low budget sci-fi film, Attack of the 50 Foot Woman, presumed that a woman had an accident with an encounter with an alien that caused her to grow to 50 feet tall overnight. (Naturally her clothes grew as well!) By coincidence, she was distraught about her cheating husband and inclined towards revenge, which lead to her stomping about town, creating mayhem while local scientists on the ground were forced to come up with Nobel Prize winning science on the spot in order to shrink her back down to size.

It is interesting to ponder the problems a human would have if she suddenly found herself in this predica-ment. To use easy numbers, suppose that she was originally 5 feet tall and weighed 100 pounds. Now she is ten times as tall, so she weighs 10 • 10 • 10 = 1,000 times as much. Tipping the scale at 100,000 pounds would make any lady pretty mad!

We’ve already discussed the heat problem. Our tall person would produce 1,000 times as much heat, but with only 10 • 10 = 100 times as much skin, would be producing heat at ten times the amount that her body could normally get rid of. She would overheat very quickly.

Bone strength becomes a problem. Large mammals must have much thicker bones than small mammals in order to support the increased weight. The strength of a bone, for starters, depends upon the cross sectional area since bones break across a cross section. Like skin area, the cross sectional areas of her bones have only increased by a factor of 100 compared to her weight increase by a factor of 1,000. Right there the bones are ten times as weak. But longer bones will also break easier. Their strength also varies inversely to the length, so actually our 50 foot tall woman ends up with a skeletal structure 10 times as strong as her 5 foot tall earlier self, but supporting 1,000 times the weight. The poor lady would crumble in a heap.

On top of her other problems, our 50 foot tall person would also quickly suffocate. Our lung capacity depends on the area of the membranes inside it, so she could only pull in one tenth of the oxygen she would need.

Come to think of it, insects can only grow so large. They get their oxygen through tiny holes in the abdo-mens, so air circulates in across a cross section area, so the air requirement varies according to the cube of the

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length while the ability to draw in air varies according to the square of the length. (There are other complications. Larger air holes would make it easier for an insect to drown. More air holes would weaken the exoskeleton.)

I have been able to verify the heat exchange problem due to size with several of my past human acquain-tances. If you undergo a large weight loss then expect to feel colder than normal for a while. If you undergo a large weight gain then expect to feel warmer than usual. Let’s leave it at that. I think, however, that our bodies eventually adjust to such radical changes, but for the people I’ve known these periods of temperature differences lasted for several months.

I’ll mention one other variational consideration when it comes to animals. Smaller animals are stronger by weight, in general. Have you ever known of an elephant climbing trees? But if you watch a squirrel you’ll see that the little guy doesn’t break stride between running along the ground and climbing up a tree. If we could jump like an insect we wouldn’t need stairs. The energy for strength is partly related to oxygen intake, muscle contractions, ability for cells to convert food to energy, and so forth, none of which varies according to volume or weight.

There is a weight to area comparison with birds and airplanes. The lift of an airplane or a bird comes from air passing across its wings. A heavier bird needs proportionally larger wings or the ability to flap them faster, which requires greater strength. Several birds, notably the ostrich, are simply too heavy to fly. An ostrich would need huge wings in order to become airborne, not to mention the strength to flap them. Hummingbirds, on the other hand, have small wings in comparison to body size and are the only birds with the strength capability to flap their wings fast enough for us to hear an audible hum.

A similar constraint affects airplane design. A heavier plane must either go faster or have sufficiently larger wings in order to sustain lift. Increasing wing size to match an increase in weight becomes impractical at a point.

Finally, let’s take a look at McDonald’s french fries. Why are they so popular with so many people? If you listen to company hype, it’s because of they use the freshest, highest quality Idaho potatoes, the best frying oil, etc. Those may be factors, but for someone like me who really enjoys fries because of the salt and grease taste, size has a lot to do with it. Someone else who prefers the taste of potatoes over grease and salt is more likely to prefer a larger cut of french fries. As has often been the case with the examples I’ve brought up in this topic, the comparison is one of volume versus area.

The amount of potato flavor in a fry is unquestionably related to the amount of potato there—its volume. But when you dump the fry in the grease and salt it down afterwards the amount of salt and grease is absorbed across the area of the fry. Since volume grows at a faster rate than surface area when you increase the size of anything, larger fries have a higher potato to salt/grease ratio than smaller fries. Conversely, smaller fries have a higher salt/grease ratio than larger fries. If more of the public prefers the salt/grease taste over potato, then McDonald’s long ago decision to cut smaller fries than most of its competitors was an act of genius.

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