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## Inverse Variation | Modeling and Identifying Inverse Variation

Inverse Variation can be modeled after any equation in the form $xy = k$ where $k \neq 0$. The constant of variation is k.

Figure 1

The table to the left models the time it would take to roller blade 36 miles at various speeds. As the rate increases the amount of time to travel 36 miles decreases. If the rate data is examined more closely and focus is placed on the times of 3, 6 and 12 hours, it should be come clear as the rate is doubled, the amount of time is cut in half.

Since the product is 36 for each pair of numbers, this is an inverse variation problem and the product is the constant of variation for this data. Letting x be the rate and y be the time, one equation to model this data is $xy = 36$. A more useful equation is $y = \frac{36}{x}$. Figure 2 is the graph of the inverse variation modeled in this problem.

Figure 2

#### Critical Thinking – Inverse Variation

Why is $y = \frac{36}{x}$ a more useful equation than $xy = 36$?

### Modeling Inverse Variation

This problem involves finding the equation or function that models a given inverse variation. This is a two step process:

1. Multiply the given x- and y-values
2. Substitute the product into $y = \frac{k}{x}$

Before just memorizing this two step process, it would be better to understand what these two steps are ‘doing’ mathematically. In step 1, multiplying the given x- and y-values calculates the constant of variation. The second step is to substitute the constant of variation into one of the acceptable forms of the inverse variation equation.

#### Example 1 – Modeling Inverse Variation

Suppose that x and y vary inversely and x = 7 and y = 5. Write an equation that models the inverse variation.

$xy = k$   Definition of Inverse Variation

$(7)(5) = k$ Substitution Property of Equality – Step 1 from above

$35 = k$ Simplify

$k = 35$   Symmetric Property of Equality

$y = \frac{k}{x}$   Step 2 from above

$y = \frac{35}{x}$   Final Answer

### Identifying Inverse Variation

When given a table of values, it is often important to be able to identify the equation modeling the data. To determine if a table of values is indeed an inverse variation, all of the data pairs must have the same product. This is because of the definition of an inverse variation and how the constant of variation is calculated. So, to identify a table of values as an inverse variation, one must test the products of all the available data. If the products are all equal, then the data models an inverse variation and by find the products, the constant of variation is already known.

#### Example 2 – Identifying Inverse Variation

Is the relationship between the variables in figure an inverse variation? If so, then write the function that models the inverse variation.

Figure 3

As state above, first find the product of all pairs of numbers:

$2(0.7) = 1.4$

$(0.35) = 1.4$

$14(0.1) = 1.4$.

Since all pairs of corresponding numbers form the same product, this table of values models an inverse variation with the constant of variation being 1.4. Therefore the function to models this data:

$y = \frac{1.4}{x}$.

If I don’t write this at the end of every blog article, I should start. Use the comments section for any questions you may have.

## Circle Geometry: Anlge Measures Formed By Intersecting Lines

Figure 1

As I think about circles in a plane and intersecting lines, I can think of more triangle proofs and work with the angle and segment addition postulates. This will lead to both algebraical and geometric types of problems. By algebraic, I mean you will have to use a theorem or property to set up the equation. On the other hand, a geometric type of problem will consist of a proof of a stated given or already known theorem. Before getting into solving any problems, I should discuss the theorem of topic for the geometry article.

### Theorem – Two Lines Intersecting Inside of a Circle

The measure of an angle formed by two lines that intersect inside a circle is half the sum of the measures of the intercepted arcs. Figure 1 illustrates the following equation to represent this theorem:

$m \angle{1} = \frac{1}{2}(x + y)$.

### Example 1 – Find the measure of the missing angle measure.

The first example ends up being a an easy three step equation to solve.

Example 1

To set up the equation I HAVE to use geometry, in this case, Two Lines Intersecting Inside of a Circle Theorem (TLIICT), then I solve the equation using an algebra 1 skill.

$m \angle{1} = \frac{1}{2}(x + y)$TLIICT

(1)$55 = \frac{1}{2}(67 + y)$. Substitution Property of Equality

(2)$110 = 67 + y$ Multiplication Property of Equality

(3)$43 = y$  Subtraction Property of Equality

(4)$y = 43$ Symmetric Property of Equality

### Example 2

The solution to this problem may not present itself right away if one too focused on the current theorem. There is no variable for what you have to solve for first. To find the value of the variable, I must keep in mind that the measure of the angle I am solving for is supplementary to the measure of the angle I can find. Since it is not labeled in the original picture, in figure 2, I label what I can find with what is given, x.

Figure 2

Let x = to the value of the vertical angles formed not labeled y

$m \angle{1} = \frac{1}{2}(x + y)$

$x = \frac{1}{2}(60 + 70)$

$x = \frac{1}{2}(130)$

$x = 65$

$y= 180 - x$

$y= 180 - 65$

$y= 115$

In this post, I have discussed the Lines Intersecting Inside of a Circle Theorem and worked out two problems. I worked out the problems differently in each example. Which method do you prefer to read and why? I look forward to your answers.

## How to Solve Logarithmic Equations

In this video there are two problems modeled. The first example works with common logs and the second example models an equation with a natural log. The keys to solving any type of logarithmic equation: being able to write the given equation in exponential form or being able to take the log of each side.

To be able to write any logarithmic function in exponential form knowing the definition of logarithm and natural logarithm is key. So here they are…

Definition: Logarithm – The log to the base b of a positive number y i s defined:

If $y = b^x$, then $log_b \hspace{0.1 cm}y = x$.

Definition: Natural Logarithm – If $y = e^x$ , then $log_e \hspace{0.1 cm}y = x$ which is also written as $ln \hspace{0.1 cm}y = x$. The natural logarithmic function is the inverse, written as $y = ln \hspace{0.1 cm}x$.

In other words, if $y = e^x$, then $y = ln \hspace{0.1 cm}x$.

Now that you have reviewed the definition of log and natural log, read through these two examples, then watch the embedded algebra 2 video.

### Example 1 Solving a Logarithmic Equation

This a problem that I assigned to my Algebra 2 class for a review before a quiz on solving logarithmic equations:

$7^{x-3} = 25$ Given

$log 7^{x-3} = log 25$ Take the log of each side

$(x-3)log 7 = log 25$ Product Property

$x-3 = \frac{log25}{log7}$ Division Property

$x = \frac{log25}{log7}+3$ Addition Property

$x \approx 4.6542$ Use a calculator

As you can see in this example, you can easily solve an equation with multiple log in it. All you need to do is use the properties of logarithms and that taking the log of each side is a legal mathematical move. The properties for common logs are used with natural logs.

### Example 2 Solving a Natural Logarithmic Equation

Solve the following equation.

$2 \cdot ln \hspace{0.1 cm}x + 3 \cdot ln \hspace{0.1 cm} 2 = 5$ Given

$ln \hspace{0.1 cm}x^2 + ln \hspace{0.1 cm}2^3 = 5$ Power Property

$ln \hspace{0.1 cm}x^2 + ln \hspace{0.1 cm}8 = 5$ Simplify

$ln \hspace{0.1 cm}8x^2 = 5$ Product Property

$8x^2 = e^5$ Write in Exponential Form

$x^2 = \frac{e^5}{8}$ Division Property

$x = \sqrt{ \frac{e^5}{8}}$ Square Root Property

$x \approx 4.3072$ Use a calculator

Solving logarithmic equations with natural log is easy if you can use the properties of natural logs and you know how to write a natural logarithmic equation in exponential form.

## Art Meets Geometry – Two Point Perspective Drawing

"My Perspective of a Box"

Here is a drawing I did to show to my High School Geometry classes. At the time of this post, the class is starting a unit on 3-Dimensional shapes. Prior to this chapter, most of our work has been in a single plane. Learning how to draw simple 3-D shapes in both one point and two point perspective makes the transition to working with intersecting planes smoother for some students. Other students would do just as well without actually making the drawings. The image to the right is my little sketch after I drew my box. Be sure to click the image to see it in full size. Below is a picture I drew back in 2008 sometime.

How to Draw a Box in Two-Point Perspective:

1. On a sheet of paper, draw horizontal line the entire length of a sheet of paper, place one point to the left of center and one point the right of center. Preferably closer to the side of the paper.
2. Draw a vertical line of about 1 inch long above or below the box. The bigger the vertical line, the bigger the box.
3. Draw light line segments connecting each end of the vertical line to both points. You should draw 4 line segments in this step.
4. Draw two vertical segments, one on each side of the first vertical segment.
5. Connect the new vertical segments two the vanishing points. You should draw four line segments for this step.
6. Clean up the unnecessary lines and used dashed segments for the hidden edges.

Mr. Pi's Double Two Point Perspective