## Thursday, June 5, 2014

### BQ#7: Deriving the Difference Quotient

 http://www.analyzemath.com/calculus/Differentiation/difference_quotient_1.gif

The formula for the  difference quotient comes from the original, algebraic equation for the slope of the line. When trying to find the slope of a tangent line (one intersection), we must also find the slope of a secant line (two intersections). When you plug in the values of the intersections and combine like terms in the denominator, we are left with with difference quotient.

Sources:
• http://www.analyzemath.com/calculus/Differentiation/difference_quotient_1.gif

## Tuesday, May 20, 2014

### BQ6: Unit U Concepts 1-2

1. A continuity is predictable; has no jumps, breaks and holes; and can be drawn without lifting a pencil. A discontinuity is an interruption in a graph.
2. A limit is the intended height of a function. A limit exists when the limit and the value, the actual value of the function, are the same.
3. We evaluate limits numerically with a table, graphically by tracing the ends of a graph, and algebraically  is direct substitution.

## Monday, April 21, 2014

### BQ#4: Unit T Concept 3

A "normal" tangent graph appears to move uphill while a "normal" cotangent graph appears to move downhill because of the location of their asymptotes.

Reviewing what we learned earlier in this unit, we know that asymptotes appear wherever there is an undefined value. Furthermore, we learned that undefined answers are produced whenever we attempt to divide by zero.

In the Unit Circle, the ratio for tangent is "y/x" and the ratio for cotangent is "x/y". Where the denominator of "x" equals zero (pi/2 and 3pi/2) determines where the asymptotes for tangent are located. The same applies to where the value of "y" is equal to zero in a cotangent graph (0pi, pi, 2pi).

## Friday, April 18, 2014

### BQ#3: Unit T Concepts 1-3

The graphs of sine and cosine are related to each of the other trig function graphs in the sense that they contain the measurements found within the Unit Circle. For secant and cosecant, we use the frame of a sine/cosine graph to draw the graphs in the shape of parabolas between the asymptotes. In tangent and cotangent graphs, it is evident that the shape of sine/cosine is depicted but simply cut in half by asymptotes.

### BQ#5: Unit T Concepts 1-3

Sine and cosine do not have asymptotes because we are never going to divide by the value of zero sine "x" and "y". The value of "r" is always set equal to 1 in the Unit Circle. The ratio for sine is "y/r" and the ratio for cosine is "x/r". In the other trig functions, there is a possibility that the denominators "x" and "y" can be valued at zero.

## Thursday, April 17, 2014

### BQ#2: Unit T: Introduction to Trig Graphs

Trig graphs are similar to the Unit Circle in the sense that they both possess the values of the quadrants. The period for sine and cosine is 2π because the sign of the function changes with every two quadrants. The period for tangent and cotangent is π because the sign of the functions changes with every other quadrant.

The fact that sine and cosine have amplitudes of 1 as seen in the Unit Circle means that they start at zero, or the origin of a coordinate plane. Since the other trig functions have amplitudes outside the values of the Unit Circle means that the entirety of their graphs shifts along the y-axis accordingly.

## Friday, April 4, 2014

### Reflection #1

To verify a trig function means to prove that it equals what is on the right side of the equation. In order to verify a trig function, we can use identities and a variety of other algebraic methods. For example, we can factors out a trig function from a problem.

The tips and tricks that I have found helpful are separating and combining fractions. This is helpful because it allows us to reduce a trig function in order to simplify or verify it. Furthermore, altering fractions allows us to substitute trig functions with identities.

When verifying trig functions, you can use identities and substitute one trig function for another. Furthermore. You can combine like terms and simplify. Another way to verify trig functions is to multiply by the conjugate.

## Tuesday, April 1, 2014

### SP#7: Unit Q Concept 2: Findig all Trig Functions When Given One Trig Function & Quadrant

This SP was made in collaboration with my friend Ana G. Check out her other cool blog posts here

 http://www.efunda.com/math/trig_functions/images/trig_relation.gif

## Thursday, March 20, 2014

### I/D#3: Unit Q Concept 1: Using Fundamental Identites to Simplify & Verify Expressions

Inquiry Activity Summary

"sin²x+cos²x=1" comes from the identity (proven factor or formula) of the Pythagorean Theorem. The theorem, when involving the trig ratios of the Unit Circle, is written in the form of "x²+y²=r²". In order to get the set of values to equal "1", we must divide the entire equation by "r²". As a result, we are left with two ratios, "x/r" (cosine), and "y/r" (sine).

Inquiry Activity Reflection

The connections that I see between Units N, O, P, and Q so far are the use of angels in degrees and the use of geometric properties. Likewise, I have noticed that all of the units make use of the six central trigonometric ratios.

If I had to describe trigonometry in THREE words, they would be: patterns, formulas, and memorization.

## Tuesday, March 18, 2014

### WPP #13 & 14: Unit P Concepts 6 & 7: Applications With Law of Sines and Law of Cosines

This WPP #13 & 14 was made in collaboration with my friend and classmate, Jennifer G. Please visit her blog here to view her other amazing blog posts!

## Monday, March 17, 2014

### BQ #1: Unit P Concepts 2, 4 & 5: Law of Sines (SSA) & Area Formulas

2. Law of Sines
Why is SSA ambiguous? Accurately draw the triangles that would be associated with one of these problems (pick any from the SSS or PQ). Connect your answer to your knowledge of Unit Circle trig function values.
SSA is ambiguous because we are only given one of the three angles as opposed to two in ASA or AAS. Because of this, there is the chance that we may get an answer of one, two, or no possible triangles.

5. Area Formulas

Show that our “traditional” area formula of A=1/2bh, the area of an oblique triangle, and Heron’s area formula will result in the same values. Draw and label a right triangle with angles of 35* and 65*. Have the base be 4 units. Find all the remaining pieces using your knowledge of this chapter and then find the area using all three area formulas.

Our "traditional" area formula, the area of an oblique triangle, and Heron's area formula will all result in the same value for the area of a triangle.

## Tuesday, March 4, 2014

### I/D#2: Unit O: How Can We Derive the Patterns for Special Right Triangles?

Inquiry Activity Summary

The objective of the activity below was to derive or obtain the patterns of special right triangles from equilateral shapes, in this case a larger triangle and a square. The benefits of completing this task is that it provides us with another means of identifying the Unit Circle values.

 Activity page 1 of 2. Click to enlarge.
 Activity page 2 of 2. Click to enlarge.

## 45°, 45°, 90° Triangle

As shown in the Step 1 of the above activity (see page 1), the process of deriving a 45°, °45, 90° congruent triangles of the desired angles. We label the hypotenuse of the smaller triangles "c", and we can label the two legs "a" and "b". Based on what we informed in this activity, we know that the side lengths of the square were valued at 1; thus, both "a" and "b" are equal to 1. To find the value of the hypotenuse, we can use the Pythagorean Theorem (a² + b² = c²). By plugging in the corresponding values, we get c= √2. We can then collectively use the newly-obtained values and a variable (e.g. "n") to triangle from a square first involves the act of dividing, or breaking down, the initial shape. To do so, we slice the square diagonally to from two form a general pattern. This means that by adding a variable to each of the values, the pattern becomes applicable to a square with any given side length!

## 30°, 60°, 90° Triangle

In Step 2 of the activity (see page 2), we are presented with an equilateral triangle with a side length of 1. Referring back to what we have learned in Geometry, we know that a triangle with sides of equal length is also equiangular, meaning that each of the three angles are 60°. To begin this step, me must again breakdown the shape by dividing it into congruent parts--this time by drawing a vertical line through the middle. Again we label out hypotenuse "c" and the two legs of the shape "a" and "b". Using the given information, we know that "a" is equal to 1/2 and "c" is equal to 1. As you may have probably already guessed from Step 1, we use the Pythagorean Theorem to find the missing height. By plugging in the respective values into the equation and solving for "b", we get the answer of √3 /2. In order to translate our values into the normal pattern without fractions, we multiply each of them by 2. By doing so, "a" becomes 1, "b" becomes √3, and "c" becomes 2. As stated before, we can include a variable (e.g. "n") to our answers to form a general pattern. This allows the pattern to be applicable to an equilateral triangle with any given side length!

## Inquiry Activity Reflection

Something I never noticed before about special right triangle is that they can be joined together to form larger, equilateral and equiangular shapes!

Being able to derive these patterns myself aids my learning because it enables me to identify patterns that could prove useful in future practice and exams.

## Inquiry Activity Summary

The objective of the below in-class activity was to derive the values of the Unit Circle from two types of special right triangles (right). For each of the three examples, we labeled the shapes according to the rules of Special Right Triangles, simplified the values of the sides, and identified the three vertices of each as ordered pairs.
 Activity page 1 of 2. Click to enlarge.
 Activity page 2 of 2. Click to enlarge.

## 30° Triangle

 http://www.montereyinstitute.org /courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3 _text_final_3_files/image099.gif
As seen in the first example of the activity, the shape contains three angles (30°, 60°, 90°). Thus, according to the rules of Special Right Triangles, the hypotenuse (r) is valued at 2x; the horizontal value (x) is x; the vertical value (y) is x√3. Due to the fact that we are instructed to simplify so that the "r" is equal to 1, we divide 2x by 2x. Just as in any given equation, what we do to one side, we must do to the other; we divide the values of the other two sides of the triangle by 2x. We are left with "r"= 1, "x"= √3/2, and "y"= 1/2.

## 45° Triangle

 http://www.montereyinstitute.org /courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3 _text_final_3_files/image036.gif
As seen in the second example of the activity, the shape contains three angles (45°, 45°, 90°). Therefore, according to the rules of Special Right Triangles, the hypotenuse (r) is valued at x√2; the horizontal value (x) is x; the vertical value (y) is also x. As before, we simplify so that the "r" is equal to 1, meaning we divide x√2 by itself. Afterwards, we divide the values of the other two sides of the triangle by x√2. For the "x" and "y" values, we are left with an unacceptable radical in the denominator. To resolve this, we multiply both the top and the bottom by √2 and simplify.In the end we get "r"= 1, "x"= √2/2, and "y"= √2/2. Using this information, we then find the coordinates of the vertices that can be seen on page 2 of the activity.

## 60° Triangle

 http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image108.gif
Similar to the first example, the third shape has three angles (60°, 30°, 90°). Thus, according to the rules of Special Right Triangles, the hypotenuse (r) is valued at 2x; the horizontal value (x) is x; the vertical value (y) is x√3. We are instructed to simplify so that "r" is equal to 1, so we divide 2x by 2x. Like before, we must divide the other two sides of the triangle by 2x. We are left with "r"= 1, "x"= 1/2, and "y"= √3/2. Using this information, we find the vertices of the shape on a coordinate plane (see page 2 of activity).

 http://www.pccmathuyekawa.com/classes-taught/math_7ab/unit%20circle.jpg

By labeling and simplifying the values of the special right triangles, I was able to identify the vertices of the shapes. If I were to plot these points within the first quadrant of a coordinate plane, I would obtain one fourth of the Unit Circle! Using this information, I could simply flip the triangles across the axes of the said coordinate plane as depicted (right) to find the rest of the circle.

The triangles drawn in this activity lie within the first quadrant of a coordinate plane, where both the "x" and the "y" values of the vertex are positive. As shown in the picture to the left, if a triangle was flipped across the x-axis (quadrant IV), the x-value would remain positive, while the y-value would become negative. Conversely, if a triangle was flipped across the y-axis (quadrant II), the x-value would become negative while the y-value would remain positive. If a triangle was flipped both across both axes (quadrant III), then both the "x" and "y" values would become negative.

## Inquiry Activity Reflection

The coolest thing I learned from this activity was that when memorizing the Unit Circle, I only actually have to recall one fourth of it! This activity will help me in this unit because it will help me remember the values and degrees of the unit circle in preparation for the upcoming test. Something I never realized before about special right triangles and the unit circle is that when the shapes are drawn on a coordinate plane, their vertices can be connected to form a circle!

## Works Cited

• http://www.pccmathuyekawa.com/classes-taught/math_7ab/unit%20circle.jpg
• http://kirchmathanalysis.blogspot.com/p/unit-n.html
•  http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image099.gif
• http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image036.gif
• http://www.montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U19_L1_T3_text_final_3_files/image108.gif

## Tuesday, February 11, 2014

### RWA#1: Unit M Concept 6: Graphing Hyperbolas & Identifying All Parts

#### 1. Definition

"Hyperbola - The set of all points such that the difference in the distance from two   points is a constant." (Crystal  Kirch)

#### 2. Descriptions

 http://www.mathwarehouse.com/hyperbola/images/compare-hyperbola-graphs.gif

### Key Parts

The center is written as an ordered pair (x,y), and it marks the midpoint of the entire  graph. It can be identified algebraically by identifying the "h" and "k" values from the equation. Note that "x" is always partnered with "h", and "y" is always partnered with "k". It can be identified graphically as the point directly in between the drawings of the two asymptotes. The transverse axis is written as an equation. It connects the asymptotes and contains the two vertices; it is perpendicular to the conjugate axis. If “x” comes before “y”, then the axis is written as “y= ”. Vice versa, if “y” comes before “x”, then the axis is written as “x=”. The axis usually appears as a solid line connecting the two graphs of the asymptotes. The conjugate axis is written as an equation; it runs in between the asymptotes, contains the two co-vertices, and is perpendicular to the transverse axis. If “x” comes after “y”, then the axis is written as “x= ”. Vice versa, if “y” comes after “x”, then the axis is written as “y=”. The conjugate axis is usually drawn as a dashed line that runs in between but never touches the asymptotes.

The vertices are written as ordered pairs. They determine the starting point of each asymptote and lie along the transverse axis. The “x” value of the vertices can be found by moving “a” units from the center along the transverse axis. The center and the vertices have a common “y” value”. The vertices can be found at the point when the asymptotes touch the transverse axis. The co-vertices are written as ordered pairs and they lie along the conjugate axis. The co-vertices and the center share the same “x” value. The “y” value of the vertices can be found by moving “b” units from the center along the conjugate axis. The co-vertices can be found at the ends of the conjugate axis that runs between the two asymptotes. The foci are written as ordered pairs and are located in between each of the asymptotes and determine their width. The further the foci are from the asymptote, the thinner the asymptote becomes and the higher the eccentricity (vice versa). The “x” values of the foci can be found by moving “c” units away from the center along the transverse axis. The foci and the center share the same “y” value. The foci are identified as the points within the opening of the vertices that are “c” units away from the center.

The value of "a" is written as a numerical value that represents the distance between one vertex and the center.  In a standard form equation, the squared value of “a” can be found underneath the first fraction. Graphically, the value can be identified by counting the number of units between the center and a vertex. "b" is also written as a numerical value that represents the distance between one co-vertex and the center. In a standard form equation, the squared value of “b” can be found underneath the second fraction. It can be identified graphically by counting the number of units between one co-vertex and the center. The value of "c" is written as a numerical value and it measures the distance between one foci and the center. After obtaining the values of “a” and “b” from the equation, you can identify “c” plugging in the appropriate numbers into: a^2 + b^2 = c^2. Graphically, it can be found by counting the number of units between the center and one foci.

Eccentricity is written as a numerical value. It measures how much a conic section isn't circular. It can be found algebraically by dividing the value of "c" by "a". Because the graph depicts a hyperbola, we know that the eccentricity is greater than one. The asymptotes are written as equations; they start at the vertices and extend rightward and leftward away from the center. If “x” comes before “y” in the equation, the asymptote open left and right; vice versa, the asymptotes open up and down. The asymptotes can be identified by finding two points on the line, or by finding the center and the slope using b/a for horizontal graphs, and a/b for vertical graphs. The standard form is the typical way of writing the equation of a hyperbola. You can find it by completing the square and using the information provided to fill in all the needed parts. You can identify it graphically by following the said steps to complete the necessary parts.  Set everything equal to "1" to complete the equation. To learn more about how to graph hyperbolas from standard form, watch the above video here!

#### 3.Real World Application

 http://www.a-levelphysicstutor.com/images/optics/lnss-ccv-ray02.jpg

Hyperbolas, which are found within several variations of math and science, can be applied to everyday life! One of the most common, and possibly the most overlooked example is the use of hyperbolas in glasses and telescopes. Of the two types of lenses that are manufactured, the one that applies the properties of hyperbolas is known as a concave lens. This lens helps people to see objects from greater distances.

Similar to our graphs of hyperbolas, the lens has two curves that cave inwards in opposite directions. Likewise, there is a focus point within each of the curves. Concave lenses diffracts, or changes the direction of light waves. As a result, a virtual image of what is actually seen appears on the opposite side of the focal point, nearer the lens. To learn more about concave lenses and their application of hyperbolas, click here.

#### 4.  Works Cited

• http://www.lessonpaths.com/learn/i/unit-m-conic-section-applets/hyperbola-from-the-definition-geogebra-dynamic-worksheet