BQ #7: Unit V- Difference Quotient

The difference formula is the slope of a line, which is also known as a secant of a graph. The original slope formula is (y2-y1)/(x2-x1) and we derive the difference formula from this. The two points that we use to plug in are (x, f(x)) and (x+h, f(x+h)). "H" represent the change in x and is also referred to as delta x. Then we would plug in those two coordinates and get f(x+h)-f(x) over (x+h)-x which we need to simplify. The x and -x cancel, so we would get f(x+h)-f(x)/h which is ultimately the difference formula derived from the slope formula.

http://cis.stvincent.edu/carlsond/ma109/diffquot.html

BQ #6: Unit U Concepts 1-7

1. What is a continuity? What is a discontinuity?
A continuous function has certain characteristics. For instance, a continuity does not have any breaks, holes, and/or jumps. In addition, a continuity is predictable so you know where it is headed. Also, you can draw a continuity without lifting up your pencil. Lastly, the continuity exists when the limit and the value are the same. The graph below is a continuity.

http://www.intmath.com/functions-and-graphs/7-continuous-discontinuous-functions.php

A discontinuity is divided into two families; removable discontinuities, and non-removable discontinuity. Discontinuities have breaks, holes, or jumps. A point discontinuity has a hole. A jump discontinuity has different left right and can exist when there is two open circles (A) on the x-axis, one closed circle and one open circle (B). An oscillating behavior discontinuity is very wiggly. Lastly, an infinite discontinuity has a vertical asymptote and it has unbounded behavior.

Point discontinuity (source: http://www.wyzant.com/resources/lessons/math/calculus/limits/continuity)

Jump discontinuity (A) (http://en.wikipedia.org/wiki/Classification_of_discontinuities)

Jump Discontinuity (B) (http://commons.wikimedia.org/wiki/File:Jump_discontinuity_cadlag.svg)

Oscillating Behavior (http://www.milefoot.com/math/calculus/limits/Continuity06.htm)

Infinite discontinuity (source: http://www.wyzant.com/resources/lessons/math/calculus/limits/continuity)

2. What is a limit? When does a limit exist? When does a limit not exist? What is the difference between a limit and a value?
A limit is the intended height of a function which is on the y-axis and it is when you approach a number it gets to a certain height. A limit can infinitely exist on a graph. A limit however, does not exist if there is a jump discontinuity and there is different left and rights are you reach that point from the left and right. In addition, limits do not exist in oscillating discontinuities due to the oscillating behavior because it is unpredictable and we can not determine the intended height. Lastly, limits do not exist when there are infinite discontinuities because of the unbounded behavior due to the vertical asymptote which means that it is going towards infinity. However, the limit is actually reached when the limit and the value are the same. The limit is the intended value and the value is the actual value.

3. How do we evaluate limits numerically, graphically, and algebraically?
When we evaluate limits numerically on a x/y table and we have to use the number that x is approaching in the middle of it. We then add .1 to the number and then we subtract .1 from the number. Then we plug in the formula, and put in the values in the table. When you look at the table, you see that the values on the y-table are getting closer to each other and this gives us the limit of the function.
We evaluate limits on the graph by using our fingers, one to the left and the other to the right of the limit that we are supposed to find. If both our fingers meet, the limit exists at this point and if they don't meet, the limit does not exist. This means that there is a non-removable discontinuity.
We can use direct substitution. dividing out, and rationalizing/conjugate method to evaluate limits algebraically. For direct substitution, you plug in the x and solve and if you get 0/0, you have to use one of the other two methods. The dividing out/factoring method is when you factor the numerator or denominator and get something to cancel. The rationalizing/conjugate method is when you multiply the fraction by the conjugate of the numerator or denominator and something will cancel. Afterwards, you will be able to use direct substitution and plug back in x to solve the problem.

BQ #4: Unit T Concept 3

Why is a "normal" tangent graph uphill, but a "normal" cotangent graph downhill? Use unit circle to explain.

The tangent graph is uphill and the cotangent graph is downhill because of the asymptotes and they are different because of the trig ratios. Tangent has a ration of y/x and cotangent is x/y. Asymptotes exist when the trig function is undefined and the denominator equals zero. Tangent is undefined when it is at pi/2 and 3pi/2 which is where x=0. Cotangent is undefined when y=0 and on the unit circle it would be at 0 and pi.
The signs are the same for both graphs in the quadrants which is positive, positive, negative, negative. The graphs are different because the asymptotes are in different places and the graphs are shifted and that makes tangent go uphill and cotangent go downhill because of the signs.

BQ #3: Unit T Concepts 1-3

How do the graphs of sine and cosine relate to each other? Emphasize asymptotes in your response.

A. Tangent? The graph of tangent has asymptotes based on cosine and you need to know that the asymptotes are found. The ratio for tangent is sine/cosine so in a graph where the value is undefined or when cosine is equal to zero. So if cosine equals zero, tangent is undefined and it has asymptotes. Cosine equals pi/2 and 3pi/2, so we know where that the asymptotes lie there.

B. Cotangent? The graph's asymptotes depend on sine and the ratio is cosine/sine. We know that it is undefined when the denominator is equal to zero which is the asymptotes. As a result, when sine(x)= 0, cotangent is undefined. Sine equals zero at 0 and pi, which is where the asymptotes lie.

C. Cosecant? The asymptotes are based on sine since it is the reciprocal and the ratio is 1/sine. We know that the denominator has to be equal to zero which would give us the asymptotes. If sine(x)=0, cosecant is then undefined and we have asymptotes. Sine equals zero at 0 and pi, which is where the asymptotes lie. Also, the positive and negative values are dependent on sine because they share the same ones.

D. Secant? The graph of secant has asymptotes that depend on cosine it is the reciprocal and the ratio is 1/cosine. We know that the denominator has to be equal to zero which would give us the asymptotes. If cosine(x)=0, cosecant is then undefined and we have asymptotes. Sine equals zero at pi/2 and 3pi/2, which is where the asymptotes lie. Also, the positive and negative values are dependent on cosine because they share the same ones.

BQ #5: Unit T- Concepts 1-3

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.
Asymptotes are found when a ratio is divided by zero, or it is undefined. Sine y/r and cosine x/r are never undefined because they are both divided by 'r', where on the Unit Circle 'r' equals 1. However, cosecant r/y, secant r/x, tangent y/x, and cotangent x/y would have asymptotes. The fact that they have y or x as their denominator has a possibility of the denominator being zero. According to the Unit Circle, secant and tangent are undefined at pi/2 or 3pi/2 when x=0. Cotangent and cosecant are undefined at 0 and pi when y=0.

BQ #2: Unit T Concept Intro

BQ #2 

How do the trig graphs relate to the Unit Circle?

Kirch's SSS Packet

a.) Period?- Why is the period for sine and cosine 2pi, whereas the period for tangent and cotangent is pi?

The period for sine and cosine is 2pi because when you go around the Unit Circle, you have to go around the entire circle in order for the patterns to repeat again. Sine's pattern in the quadrants is positive, positive, negative, negative and we see a pattern. Therefore the period would be 2pi because you had to go around the unit circle an entire time to see the repetition. For cosine the same applies but the pattern is positive, negative, negative positive. And you would also have to go around the unit circle entirely to see the pattern. So basically it takes 1 entire time around the Unit Circle to get repeating patterns for both sine and cosine which is 2 pi.

The period for tangent is pi because you only need to go around the Unit Circle half way to get a repeating pattern. So it would be positive, negative, positive, negative to see a distinct pattern. Half of the unit circle is pi.

b.) Amplitude?- How does the fact that sine and cosine have amplitudes of one (and the other trig functions don’t have amplitudes) relate to what we know about the Unit Circle?

The range of the Unit Circle values is from -1 to 1 so it determines the value of the amptitude. Sine's ratio is y/r and r=1 on the unit circle so you would only be limited to using one. Similarily, cosine's ratio is x/r, where r also equals 1, so you could only use 1 as well. Therefore, both sine and cosine have an amplitude of 1.

Reflection# 1: Unit Q: Verifying Trigonometric Identities

1.  In Unit Q, we learned how to verify functions. When we verify functions, it basically means that we use the ratio identities, reciprocal identities, and the pythagorean identities in various order to simplify the equation further. We would only use the identities on the left side and NOT use the identities on the right side. We just want the right side to equal the newly simplified left side.

2. To begin with, when verifying expressions make sure to list out the steps made neatly, so later on you can pinpoint mistakes. I found it useful to convert everything into sine and cosine. When we convert the expressions, then it would lead to canceling out terms and lead to a Pythagorean identity that equals zero. Also, you can factor out a least common denominator because it could lead to finding more properties within the expression.

3. First thing I think about when I look at an identity is to see what identities are present in the problem.  Then I see if there is a greatest common factor to factor out to make it easier to simplify. I also look to see if I need to get a common denominator. If there is a binomial in the denominator then you would multiply by the conjugate. You could also combine or separate denominators.  TIP: Memorize the identities and know where you can use them, it will make it EASIER to solve. 

SP #7: Unit Q Concept 2

This SP7 was made in collaboration with Margie V.  Please visit the other awesome posts on their blog by going here.

Problem

Solutions

1. Using Pythagorean/Ratio/Reciprocal identities (Unit Q Concept 2)

2. Using SOHCAHTOA (Unit O Concept 5)

Although the methods are slightly different, both of the methods led to the same answers. We concluded that we can use both methods to find the values. This proved that the identities deal with SOHCAHTOA.

I/D #3: Unit Q - Pythagorean Identities

Inquiry Activity Summary

1. Where does sin²x+cos²x=1 come from to begin with?


Example with one of the "Magic 3" ordered pairs from the Unit circle.

2.

Inquiry Activity Reflection

1.  The connections that I see between Units N, O, P, and Q so far are all of the concepts deal with triangles or angles of some sort. For example we have worked with right triangles, non-right triangles, obtuse angles, acute angles. For the non-right triangles we had to learn about the Law of Sines and Law of Cosines.  Another connection is that we learned how to utilize sine, cosine, tangent, cosecant, secant, and cotangent along the other units to find missing pieces of information. In addition, all the trigonometry functions related to the unit circle.

2. If I had to describe trigonometry in THREE words, they would be complex, memorization, and concise. 

WPP #13 & 14: Unit P Concept 6 & 7

This WPP13-14 was made in collaboration with Margie V.  Please visit the other awesome posts on their blog by going here.

http://inspirationalstorytellers.com/the-daisy-and-the-oak-tree/

The Problem

a) Danielle is due east from an oak tree. Danielle is looking at Andrew at a bearing of S32W. Andrew is looking at the same oak tree with a bearing of N15E. Andrew is also 52 feet away from the oak tree. How many feet apart are Andrew and Danielle?

b) Andrew and Danielle are now together and they decide to go on a date. They go on a date and now it is time to part. They leave from the same point. Their paths diverge by a bearing of 088 degrees. If Danielle walks for 2.7 miles and Andrew walks for 3 miles, how far apart are they at this time?

The Solutions

BQ# 1: Unit P Concepts 1-5

1. Law of Sines: In the Law of Sines, a proportion is set up and you can find an angle or a side of a non-right triangle. The Law of Sines works when you have the angle-side- angle triangle, angle-angle-side triangle, or side-side-angle. The pictures below will show how the Law of Sines is derived.

5. Area Formulas: The area formulas for non-right triangles are shown below with the angles of 35*-65*-80* and a base of 4. You will get the area to be 4.2 units squared for all three area formulas.

a)Using the "traditional" area formula:

b) Using the area of an oblique triangle:

c) Using the Heron's formula:

WPP #12: Unit O Concept 10: Angles of Elevation and Depression

http://gomighty.com/wp-content/themes/gomighty/lib/goal_images/files/Hot-Air-Balloon.jpg

The Problem

a) Andrew is at ground level and measures the angle of elevation to the hot air balloon to be 37*. If, at this point he is 57 feet away from the hot air balloon, what is the height of the hot air balloon from the ground? (round to the nearest foot)

b) Andrew is now in the hot air balloon and measures the angle of depression which is 25*. Andrew knows that he is 240 feet high up in the air than the base of the forest. How far away horizontally is Andrew from the forest? (round to the nearest foot)

The Solutions

I/D #2: Unit O: Derive the Special Right Triangles

Inquiry Activity Summary

Derive the 45-45-90 triangle from an square with a side length of 1

Derive the 30-60-90 triangle from an equilateral triangle with a side length of 1

Inquiry Activity Reflection

Something I never noticed before about the special right triangles is where the triangle comes from. I did not know that the 45-45-90 degree triangle came from a square and that the 30-60-90 degree triangle came from half an equilateral triangle. It was interesting to find out that so much can be found with the Pythagorean Theorem.

Being able to derive these patterns myself aids in my learning because if I forget the rules for the special right triangles I will be able to derive them from a square and the equilateral triangle with the side lengths of 1. I also will know how to use the Pythagorean theorem to determine the side and substitute the values for the variable n, so any other multiples of those values can be used. 

I/D #1: Unit N Concept 7- Unit Circle Derivation

Inquiry Activity Summary

1. 30* Right Triangle

2. 45* Right Triangle

3. 60* Right Triangle



4. This activity showed me where the values from the unit circle came from which was the special right triangles. Getting the three triangles let me know that they are the same in each quadrant so it leads to knowing the entire unit circle

Different angles in different quadrants

5. The triangles in the videos are all in the first quadrant and the ordered pairs are positive. Redrawing the 30* angle in the second quadrant made the x values become negative. The 45* angle in the third quadrant makes both x and y values become negative.The 60* angle in the fourth quadrant makes the y values become negative.

Inquiry Activity Reflection

1. The coolest thing I learned from this activity was learning where the numbers for the unit circle come from. I now know if I'm given some sort of right angle that I can solve for the points.

2. This activity will help me in this unit because this is the first time that I have been introduced to the unit circle and it is a great way to remember the values of the points of the graph.

3. Something I never realized before about special right triangles and the unit circle is the points are directly correlated and that a triangle will let me know the points on the graph.

RWA1: Unit M Concepts 4-6: Conic Sections in Real Life

1. Defintion:
 A parabola is the set of all points the same distance from a point, known as the focus, and a line, known as the directrix.

(http://www.mathsisfun.com/geometry/parabola.html)
2. Description:
The algebraic equation or formula for a parabola is (x-h)²=4p(y-k) or (y-k)²=4p(x-h). A parabola is a curve and it is arch-like or in other words "U-shaped" which is symmetrical to the other side of the graph. The parabola has a vertex which the bottom point of a parabola and it is shown as the (h,k) in the equation and it is the point where the axis of symmetry divides the parabola. The axis of symmetry splits the parabola where both sides are a mirror image of each other and this axis of symmetry can be vertical or horizontal, the axis of symmetry depends on the direction of the parabola going up/down (x=#) or parabola going left/right (y=#). In order to decide if it goes up/down or left/right you look at the squared term and if the x value is squared then the parabola goes either up/down and if the y value is squared then the parabola goes right/left. To get the direction of the parabola you have to find the p value and if it is negative it goes down/left and if the p value is positive it goes up/right.

Also, the focus is p value units away from the vertex and also lies on the axis of symmetry and the focus will always be "inside" the parabola. The closer the focus is to the the vertex the skinnier the graph and the further away the focus is from the vertex, the fatter the graph. The directrix is the opposite direction form the focus and is p units away from the vertex in the opposite direction and the value would be perpendicular to the axis of symmetry. So, if the axis of symmetry is y=# then the directrix would be x=#.


3. Real World Application:
Flashlights is where parabolas go into action and it helps to focus light into a certain beam. The shape of the flashlights reflectors is a parabola and the light bulb is at the focus of the parabola. The rays given off from the focus point and the light is reflected parallel in straight lines to the axis of symmetry. "Since light is a wave, if a light source is  placed at the focus of a paraboloid the result will be a focused beam of light emerging outward along the direction of the axis."(http://jwilson.coe.uga.edu/EMAT6680Fa08/Wisdom/EMAT6690/Parabolanjw/reflectiveproperty.htm)

(http://jwilson.coe.uga.edu/EMAT6680Fa08/Wisdom/EMAT6690/Parabolanjw/reflectiveproperty.htm)

Therefore the shape of the flashlight reflectors allow light to go in the direction of the axis and the more the light is aimed in different places, the more different the lighting is. Also the lighting would be different due to where the filament is located and it can bend the light waves and change where the light goes. So if we "offset the filament from the focus and change the beam entirely" would aim the light in different directions. (http://www.pleacher.com/mp/mlessons/calculus/appparab.html)

(http://www.pleacher.com/mp/mlessons/calculus/appparab.html)


4. Works Cited: 

• http://www.mathsisfun.com/geometry/parabola.html
• http://jwilson.coe.uga.edu/EMAT6680Fa08/Wisdom/EMAT6690/Parabolanjw/reflectiveproperty.htm
• http://www.youtube.com
• http://www.pleacher.com/mp/mlessons/calculus/appparab.html