Unit V - BQ #7: Where does the difference quotient come from?

WHERE DOES THE DIFFERENCE QUOTIENT COME FROM?

First of all, a function of any type is needed to figure out where the difference quotient comes from. Take a secant line (a line that touches the function twice) and use the points. The first point is x units on the x-axis and f(x) units in the y-axis, therefore being a coordinate pair of (x, f(x)). The second point is taken, however it has an h units difference from the first and therefore being a coordinate pair of (x+h, (f(x+h)). 

http://cis.stvincent.edu

Plug the two points into the the slope formula [m = (y2-y1)/(x2-x1)] to find the slope.  After the appropriate information has been correctly plugged in, there is some simplification/combining-like-terms that occurs at the denominator. Once the simplification has been complete, the equation left  is [(f(x+h)-f(x)]/h, which is known as the difference quotient. This is used to find the slope of a tangent line, which, unlike a secant line, is a line that only touched a function once.

BQ #6 - Unit U

1. What is a continuity? What is a discontinuity?

A continuity--or better yet a continuous function-- is predictable and does NOT have breaks, holes, nor jumps. Continuous functions can be drawn without lifting the pencil/pen off the paper. On the other hand, a discontinuity is the opposite and are in two families: Removable and Non-Removable Discontinuities. Under Removable discontinuities exists point discontinuity (known as a hole), like in the images below. [Notice, point discontinuities can exist with or without a filled circle]

(http://www.mathwords.com/r/r_assets/r88.gif)

http://images.tutorvista.com/cms/images/113/removable-discontinuity.png

Under Non-Removable Discontinuities exists three more. Jump behavior one of them and is when there is a break or a jump in the function. Here, the left side or the function is different from the right and both sides won't meet at the same f(x) value. See the image first below. Jump discontinuities can exist with either 1. two open holes or 2. one opened and the other closed, but never both closed circles. Another disconinuity under this category is oscillating behavior, which is described as "wiggly" like in the second image. The third and final is infinite discontinuity. It contains--or is composed--of a vertical asymptote that causes unbounded behavior as it will never reach an actual value. See the third image below.

1.

(www.wikimedia.org)

2.

(www.cwladis.com)

3.

(www.milefoot.com)
  

2. What is a limit? When does a limit exist? When does a limit does not exist? What is the difference between a limit and a value?

     A limit is the intended height of a function. The limit will exist at any continuous function as well at Removable Discontinuities, or point discontinuities. The limit DOES NOT EXIST at any of the Non-Removable discontinuities. With jump discontinuities the left and right side of the function, approaching at a specific x value, will not have the same height; therefore there is no limit. With oscillating behavior, the limit won't exists as it doesn't approach any single value. In infinity discontinuity, the vertical asymptote causes the unbounded behavior to occur. Both sides will ATTEMPT to reach a value, going the same direction or opposite, but never will. As f(x) will approach infinity, the limit is unbounded since infinity is not a number.
     As the limit is defined as the intended height of a function, the value is defined as the actual value of a function. "Intended height", in my perspective, can be interpreted as "the calculated/predicted position or point." The value is what determines where the position/point is actually at. The limit and value can 1. both exist and be the same or 2. can both exist and be different (in some point discontinuities). There are functions in which 3. the limit does exist but there is a value of undefined (in some point discontinuities) or 4. where the value exists but there is no limit (in some jump discontinuities)
1.

(www.math-spot.com)

2.

(http://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/4a69dec7-03e0-492f-ac16-4dcd555579c9.gif)

3.

(https://images-blogger-opensocial.googleusercontent.com)

4.

(www.mathwords.com)

3. How do we evaluate limits numerically, graphically, and algebraically?

     To evaluate limits numerically, a number table is used. The value of x is placed in the middle with three other values increasing to the right and decreasing to the left. The x values would increase or decrease by decimal values to the tenths, hundredths, and thousandths places, like in the image below. As f(x) can be predicted, the limit is simply approached, however it can SOMETIMES be reached (in continuous functions only).

(from Mrs. Kirch's Unit U SSS Packet)

     To evaluate limits graphically, view the graph and follow the left and right side of the function--use your fingers if you have to--to a specific x-value. If both sides reach the same height/location, then the limit exists. If both sides do not reach the same height/location, then the limit does not exist. The image below is an example of the left and right side approaching the same x-value.

(www.wyzant.com)

     To evaluate limits algebraically, use the direct substitution method and plug in the x-value into the equation. If a numerical answer, 0 over a number (results as 0), or a number over 0 (results as undefined), then the evaluation is complete. However, if a fraction occurs as 0/0, then it is indeterminable and the evaluation continues. Only then one must use the dividing/factoring method or the rationalizing/conjugate method. In the dividing/factoring method, the numerator and denominator are factored in order to have something cancel out from the bottom and to avoid the denominator equaling 0. In the rationalizing/conjugate method, the equation is multiplied by the conjugate of either the numerator or denominator with the radical. 

     If x is ever approaching +/- infinity, then it is also an indeterminable form; however, there is a different way to solve this form. One must divide every term (in the numerator and the denominator) by the highest power of x found in the denominator. Infinity is then substituted in but REMEMBER: 1) a number value divided by infinity is 0; 2) equations without an x in the denominator do not have a limit due to unbounded behaviors; and 3) if the numerator has the same highest degree as the denominator, the limit is based on the ratio of the coefficients. 

BQ #4: Unit T Concept 3: Why is a “normal” tangent graph uphill, but a “normal” COtangent graph downhill?

Why is a “normal” tangent graph uphill, but a “normal” cotangent graph downhill?

The pattern of tangent on the Unit Circle is positive, negative, positive, and negative. From this patter, we know that the period repeats itself within the Unit Circle, or within 2pi. The ratio of tangent is sine/cosine, which means that asymptotes can exist. An asymptote exists when the denominator is 0; and in this case, cosine would need to equal 0. We know that on the Unit Circle, cosine is 0 at 90 degrees (or pi/2) and 270 degrees (or 3pi/2). So within the asymototes, a period would start from the negative and move onto the positive, creating an uphill graph. (Look at the picture below for a visual description)

Cotangent is somewhat similar to tangent. It also has the same patter of positive and negative, however its ratio is cosine/sine. As sine as its denominator, the asymptotes would be located differently. On the Unit Circle, sine is 0 at 0 degrees (0pi) and at 180 degrees (pi). Between the asymptotes, the graph would start with positive and move onto negative, creating a downhill graph.

Overall, the tangent and cotangent graphs depend on where the asymptotes are located at.

BQ #3 : Unit T Concepts 1-3 - How do the graphs of sine and cosine relate to each of the others?

How Do the Graphs of Sine and Cosine Relate to Each of the Others?

TANGENT

(https://www.desmos.com/calculator/hjts26gwst)

     The trig ratio as we all know is sine over cosine. We also know that on a Unit Circle, sine is positive on Quadrants l and ll, cosine is positive on l and lV, and tangent is positive on l and lll. On a graph, sine and cosine have a period of 2pi, while tangent only has 1pi. On the first quadrant (the red shaded section from the snapshot pic), both sine (the red graph) and cosine (the green graph) are positive. When plugged into the ratio to get tangent, algebraically it should be positive as well; and on the graph, too, it shows the tangent graph (the orange graph) positive. On the second quadrant (the green shaded section), sine is only positive while cosine is negative. As a ratio, tangent would be negative, which is proven on the graph as well. For quadrant three (the light orange section), both sine and cosine are negative but together make a positive tangent. As noticed on the graph, tangent is on the positive section in the third quadrant. And finally on the fourth quadrant (blue section), only cosine is positive and creates a negative tangent, which on the graph also shows.
     Tangent is different from sine and cosine as it has asymptotes at pi/2 and 3pi/2. Asymptotes are created when when a fraction, or ratio, has a denominator of 0. In this case, cosine has to be 0; and on a Unit Circle, sine is 0 at pi/2 (90 degrees) and 3pi/2 (270 degrees). On the graph, tangent will move closer to these special restrictions but never actually reach them.

COTANGENT

(https://www.desmos.com/calculator/hjts26gwst)

     Cotangent is the inverse of tangent, meaning the ratio will be cosine over sine. With the previous information from before, both sine and cosine are positive in the first quadrant and will result with a positive cotangent, which is shown on the graph from the screenshot pic. On the second quadrant, cosine is negative while sine is positive, which will result with a negative cotangent on the graph (along with cosine). Moving to the third quadrant, both sine and cosine are negative which created a positive cotangent and will graphically be above the other two. Finally on the fourth quadrant, cosine is positive but with a negative sine, cotangent is negative and underneath with sine. 
     Cotangent as well contains aymptotes, however, at pi and 2pi. For cotangent and because of its inverse ratio, sine has to equal to 0 in order for the asymptotes to exists. On the Unit Circle, sine is 0, pi (180 degrees), and back to 0 which is also 2pi (360 degrees). On the graphs, the cotangent will never reach the asymptotes but it will get closer and closer to it. 

SECANT

(https://www.desmos.com/calculator/hjts26gwst)

     Secant is the inverse of cosine, which also has a period of 2pi and start at the amplitudes. If we remember, cosine and secant are positive at l and lV and negative at ll and lll. When graphing them, it shows and profs the same results. On quadrant one and four, the secant graph (the purple graph) is above with the positive cosine but going in an inverse direction. For quadrants two and three, they are under the negative cosines and as well going on the inverse direction. 

     Secants is also a trig ratio with asymptotes and is seen through its ratio 1/cosine. Cosine (the denominator in this case) needs to equal 0 again which is found in pi/2 and 3pi/2, which on the unit circle it would have been 90 and 270 degrees. 

COSECANT

(https://www.desmos.com/calculator/hjts26gwst)

     Cosecant is the inverse of sine, which has a period of 2pi. Using the Unit Circle, sine and cosecant are positive at l and ll while negative at lll and lV. Graphing cosecants would be somewhat similar to graphing sine. It would be positive on the first and fourth and have a parabola at the peak of the amplitudes with an inverse direction. It would be negative at lll and lV and underneath of the sine graph with same parabolas: negative and heading at an inverse direction of the amplitude. 

     Cosecants like secants, tangent, and cotangents, have asymptotes. The trig ratio is 1/sine at which sine has to equal 0. Sine is only negative at 0, pi (180 degrees), and 2pi (360 degrees and same as 0). At these asymptotes, cosecant graphs would not be able to overlap them but only get nearer. 

BQ #5: Unit T Concept 1-3 - Why do sine and cosine NOT have asymptotes, but the other four trig graphs do?

(http://literacy.purduecal.edu/STUDENT/mrrieste/trigratios.png)

     First of all, in order for a trig function to have an asymptote, the DENOMINATOR MUST BE 0, or undefined. Looking at the trig ratios from the Unit Circle (as seen on the picture above), sine and cosine have denominators of r, or the hypotenuse. On the unit circle. r will always be 1; r is not 0 and therefore these two trig functions won't have asymptotes. Even outside the Unit Circle, r will never be 0 since it represents the hypotenuse of a right triangle. 

     Cosecant and cotangent relate since both ratios have denominators of y. In order for cosecant and cotangent to have asymptotes, y would need to equal to 0. Same goes for secant and tangent;  both share a denominator of x and x needs to be 0 in order to have an asymptote. 

BQ #2: Unit T - Concept Intro

How do the trig graphs relate to the Unit Circle?

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

     On the Unit Circle, sine has a patter throughout the quadrants in order as positive, positive, negative, negative, and repeat. Cosine has a similar patter with positive, negative, negative, positive, and repeat. For sine and cosine, it take the WHOLE Unit Circle to complete until the pattern repeats, meaning it takes 2pi.
And when actually graphed, it would take 2pi units for the period to complete itself before starting again.

     Tangent and cotangent, however, have a patter of positive, negative, positive, and negative. If one would notice, the pattern repeated itself within the Unit Circle and it took half of the circle to finish one pattern. If one remembers, half of the unit circle is 1pi. When graphing tangent and cotangent, it would only take 1pi units for the period to go through. 

     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 trig ratio for sine is (y/r) and cosine (x/r) and on the Unit Circle, r will always be 1. On the Unit Circle, the center is (0,0) and radius is 1, therefore that is how much possible sine and cosine can extend to, from -1 to 1. This is where their restrictions are also implied as anything outside of -1 and 1 will result with ERROR or Not Possible. The other trig functions, however, don't have an amplitude of 1 since their ratio is not over r (or 1). They are not restricted like sine and cosine, meaning they are free to move part -1 and 1.

Reflection #1: Unit Q - Verifying Trig Identities

1.) To verify a trig identity, it means both sides of an equation has to equal to each other. This requires simplification on the left side only, as the right side is not to be touched. In most or in all cases, the right side is already simplified to one or two trig identities, so rewriting it would create complications and more work. Throughout the verification and simplification, identities (definition from Mrs. Kirch: proven facts and formulas that are always true) are used for substitution, canceling, etc. The relations of the 11 identities all have some sort of pattern that go/relate with one another.

2.) There are a few tips and trick to verifying a trig identity. If the problem looks simple, start by checking if an identity would work. Usually from there, one identity will lead to another and trig functions will start canceling out perfectly. But if the problem is more complex, like those in Concept 5, the process is a bit different. For fractions, separation might work best but in some cases multiplying the conjugate works as well. In fractions, mainly, the problem looks most complicated, however one should not be afraid them as the fraction will simplify and decrease its size along the way. In other situations and problems, factoring a common multiple might help, while in the other hand powering up/down will also do the trick. Trig identities are also usually found where there is any addition/subtraction of number values. A main thing to remember is that one CANNOT divide by a trig function.   

3) When a trig identity is given to be verified, I feel less tense as the answer is technically already given but needs to be simplified/proven. I always look at the right side at the "answer" and start thinking how I would want to rearrange the problem to get there [to the answer]. I usually start simple by checking if an identity can be substituted in order to cancel out. I also search for some sort of relation that holds the problem together and that can be used to break it down. I prefer factoring since in many cases it would simplify to a trig identity. I try avoiding multiplying binomials/conjugates, squaring binomials, etc as it involves more work and can have several sections were mistakes can be made. 

SP #7: Unit Q Concept 2 - Finding Trig Functions Using Identities

Please see my SP #7, made in collaboration with Victoria Ventura, by visiting their blog here.  Also be sure to check out the other awesome posts on their blog.

I/D #3: Unit Q - Pythagorean Identities

INQUIRE ACTIVITY SUMMARY

1.) Where does sin^2x+cos^2x=1 come from to begin with? You should be referring to the Unit Circle ratios and the Pythagorean Theorem in your explanations. 

The Pythagorean Theorem is considered an "identity" (def: a proven fact or formula that is always true). The Theorem a^2+b^2=c^2 can also be written with different variable like x,y, and r to get: x^2+y^2=r^2. In order to get the equation equal 1, r^2 would need to be divided on both sides to get: (x/r)^2+(y/r)^2=1. From this new equation we can get information using our knowledge from the Unit Circle. If we remember, x/r is the trig ratio for cosines while y/r is the ratio for sines. The equation can be written as: cos^2(x)+sin^2(x)=1. This is referred to as a Pythagorean Identity since it was first derived from the Pythagorean Theorem, which is an identity that was simply rewritten but never lost its value. To prove that this equation is an identity and always true, we can take one of the "Magic 3" ordered pairs from the Unit Circle. For example, the ordered pair for a 45 degree angle is (radical2/2, radical2/2). When the ordered pair is plugged into the equation: (radical2)/2)^2 + (radical2)/2)^2, the answer WILL result to get 1. 

2.) Show and explain how to derive the two remaining Pythagorean Identities from cos^2(x)+sin^2(x)=1. Be sure to show step by step. 

     a) If the equation were to be divided by cos^2, the equation would look like:  cos^2(x)/cos^2(x)+sin^2(x)/cos^2(x)=1/cos^2(x). The first part (highlighted in blue) will cancel out to equal 1 since both numerator and denominator are the same. The second part (highlighted in green) can be substituted by a ratio identity of tan^2(x). The third part (highlighted in light blue) can be substituted by a reciprocal identity of sec^2(x). In the end, the derivation will result as 1+tan^2(x)=sec^2(x).

     b) If the equation were to be divided by sin^2, the equation would look like:  cos^2(x)/sin^2(x)+sin^2(x)/sin^2(x)=1/sin^2(x). The first part (highlighted in purple) can be substituted by a ratio identity of cot^2(x). The second part (highlighted in orange) cancels out to equal 1 since both numerator and denominator are the same. The third part (highlighted in red) can be substituted by a reciprocal identity of csc^2(x). In the end, the derivation will result as cot^2(x)+1=csc^2(x).

INQUIRE ACTIVITY REFLECTION

"The connections I see so far in Unit N, O, P, and Q so far are..." the repeating use of the trig functions of sine, cosine, tangent, cosine, cosecant, and cotangent from the Unit Circle AND the use of triangles.

"If I had to describe trigonometry in THREE words, they would be..." triangles, triangles, and triangles!...Just kidding; it would be complex, overwhelming, and triangles. 

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

Please see my WPP13-14, made in collaboration with Adrie Garcia, by visiting their blog here.  Also be sure to check out the other awesome posts on their blog.

BQ #1: Unit P Concept 1 & 4 - Law of Sines and Area Formula

1. Law of Sines - Why do we need it?  How is it derived from what we already know?

The law of sines is needed to figure out the missing parts (angles and sides) of a non-right triangle. Non-right triangles are more complex to solve for since they are not like special right triangles and have a special side relationships (ex: 45-45-90 triangle has a relationship of (x)-(x)-(x)radical 2).
The deviation to this law is shown through the process and pictures below:
First, there is the non-right triangle ABC. An imaginary perpendicular line is drawn to cut angle B. The imaginary line is labeled as h.

The perpendicular line created (2) right triangles. From there, the basic trig functions of sin (opposite/hypotenuse) can be used for angle A and C. The trig functions are then rearranged to equal to h.  

Since both functions are now equal to h, they can be equal to one another. Both are then over the corresponding sides used (sides a and c). Simplification can be done and the law is finished. (*Note: this implies the same for B if the perpendicular line was cut either from angle A or C)

4. Area formulas - How is the “area of an oblique” triangle derived?  How does it relate to the area formula that you are familiar with?

The area of an oblique triangle is derived from the triangle area equation and a trig function. The area of any triangle is (1a) A=1/2bh, where b is the base and h is the height. However, in an oblique triangle, the height is not given and the sin trig function must be used.
With the oblique triangle provided as an example, triangle ABC is cut perpendicularly in half from angle B to create 2 right triangles with the sharing side of h. Side h must be used with sin, in this example sin C is used with h/a (sin A could have been used as well). (1b)The sin trig function is rearranged to equal h.

Since the sin trig function now equals to h, it can be substituted into the (1a) area of triangles. (*Note: the perpendicular line from the beginning can cut any angle and create in total 3 ways to find the area with the same concept).

     It relates to the formula we are familiar with since it is mostly substitution, the new equation has parts from the old one. The triangle is still multiplied by 1/2 and consists of 2 sides. However in the new one, it also consists of an angle which is in between the two sides used. In the original one, meanwhile, uses the base and the perpendicular height. 

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

(www.acclaimimages.com)

Problems

1. Farco decided to put a flag vertically on top of his ice-cream building. After having it set up, he wonders how high up is the the tip of the flag from the ground. He stands 5 feet away from the building and at ground level the measure of elevation is 76 degrees. a) How high up is the tip of the flag in the nearest foot?
2. Farco decides to stand on top of the building, which is 15 ft high, and leans against the flag post. From that view, he sees an army of starving children on the other side of the street in the Pie Store. If the distance of Farco's position to the bottom of the Pie store is radical 62725 away, b) what is the angle of depression from the top of the building to the corner to the nearest hundreth?

Solutions

1.

2.

I/D #2: Unit O - How can we derive the patterns for our special right triangles?

INQUIRY ACTIVITY SUMMARY
1. 30-60-90 Triangle

     The 30-60-90 triangle comes from an equilateral triangle, which has equilateral sides with equiangles of 60 degree. The equilateral triangle is cut in half, making: (2) triangles, (2) 30 degree angles, and (2) 90 degree angle. To explain the relationship between the sides and angles, an equilateral triangle with the sides of 1 can be used. If the process would be applied to this example, and focusing on ONE of the two triangle, the angles would remain the same as mentioned. The side cut in half will equal to 1/2, while the other untouched side (the hypotenuse in this case) would remain as 1. However, one side remains missing with its  value, and in order to find it the Pythagorean Theorem (a^2 + b^2 = c^2) is used.

     Using the Pythagorean Theorem, c would be the hypotenuse while a and b would be the other two sides that are across the 30 and 60 degree angles. Side a would equal to 1/2 while c would equal to 1. Plug those into the equation to solve for b will result with b equaling to radical 3 over 2. Across the 30 degree angle, a= 1/2; across the 60 degree angle, b= radical 3/2; and across the 90 degree angle c= 1. The relationship for a 30-60-90 triangle is 1/2-radical 3/2-1. HOWEVER, not all triangles are the same and not all derive from an equilateral triangle with the side of 1. The variable n can be used on the relation to represent any value for a triangle, creating  (n/2)-n(radical 3)/2-(n). (And to make life simpler, the fraction can be removed my multiplying the relation by 2 to make (n)-n(radical 3)-2n..

2. 45-45-90 Triangle

     The 45-45-90 triangle comes from cutting a square directly in half diagonally. Cutting the square diagonally in half will create: (2) triangles, (4) 45 degree angles, and (2) 90 degree angles. To explain the relationship between the sides and angles, a square with the sides of 1 can be used. Cutting the square and focusing on ONE triangle, the right triangle created will have two sides with the length of 1 with a missing hypotenuse. The sides of 1 will be across the 45 degree angles, and the hypotenuse can be found using the Pythagorean Theorem.

     With the Pythagorean Theorem, side a will be 1 while side b will also be 1. After squaring the sides, adding them, and finding the square root, the hypotenuse will equal to radical 2. So a 45-45-90 triangle has a relationship of 1-1-radical 2. HOWEVER, not all triangles will derive from a square of 1. As in the 30-60-90 triangle, n can be used on the relation to represent any value of the triangle, creating (n)-(n)-n(radical 2).

INQUIRY ACTIVITY REFLECTION
“Something I never noticed before about special right triangles is…” that these special right triangles derived from equilateral triangles with equiangles. The 45-45-90 right triangle came from a square, a quadrilateral with the same length in sides and had (4) 90 right angles. The 30-60-90 right triangle derives from an equilateral triangle that had (3) 60 degree angles.
“Being able to derive these patterns myself aids in my learning because…” it gives me a better understanding where the numbers and relations actually come from and now know that a mathematician didn't randomly chose numbers. Also, in case I forget the relation during an important test, I could remember where and how these patterns are derived from.

I/D #1: Unit N Concept 7 - How do SRT and UC relate?

INQUIRY ACTIVITY SUMMARY

1.) 30 degrees

     There were several part that were needed to be labeled and solved for. The hypotenuse is equaled to 1, and due to the rules of the 30,60,90 right triangle: the side adjacent to 30 (which is x on the pic) has a value of x radical 3; the side opposite (which is y) has a value of x; and the hypotenuse (r) as 2x. To find an actual number value, trig functions (sin and cos) must be used. The sin of 30 degrees solves for the opposite side (y) and is found with y/r; substituting the variables will result with x/2x and simplified will equal to 1/2. The cos of 30 degrees solves for the adjacent side (x) and is found with x/r; substituting the variables will result with x radical 3/2x and simplified will equal to radical 3/2.  If the triangle were to be placed in a Unit Circle, with the origin being located at the labeled angle measure, three coordinate pairs will be needed and two are found with the values solved. The origin is (0,0), while only going across the x axis is (radical 3/2,0), and going directly above from the previous coordinate is (radical 3/2,1/2).

2.) 45 degrees

     The instructions are the same but differ in the number values. Due to the 45,45,90 right triangle rules, x is the side adjacent to the given degree with value of x; the y is opposite to the angle but similar to the adjacent side as it, too, has the value of x; and the hypotenuse, r, is equaled to 1 and x radical 2. The trig functions of sin and cos are also used with sin 45 degrees = y/r and cos 45 = x/r. Solving for the sin and cos is similar with this SRT since the same values are used to substitute, which is x/x radical 2 and simplified equaling to radical 2/2. If the triangle were to be placed in the Unit Circle, with the origin being located at the labeled angle measure, there would be three coordinate pairs. The origin would be (0,0), shifting to the right would be at (radical 2/2,0), and moving directly upward will get to (radical 2/2,radical 2/2)

3.) 60 degrees

     The instructions and requirements are as well the same from the previous triangles, and it will be closely identical to the 30 degree triangle. Same as before: x is adjacent but with the value of x; y would be opposite to the angle with the value of x radical 3; and the hypotenuse, r, is 1 and 2x. Sin and cos will be used to find the length of the sides. The sin of 60 degrees (y/r) would be x radical 3/2x and simplified will equal to radical 3/2. The cos of 60 degrees (x/r) would be x/2x and simplified will equal to 1/2. If the triangle were to be placed in a Unit Circle, with the origin being located at the labeled angle measure, three coordinate pairs will be (0,0) at the origin, (1/2,) going across the x axis, and (1/2, radical 3/2,) shifting directly up.

4. This activity helps derive the unit circle by figuring out the certain point of the circle with the special right triangles. Figuring out the lengths of these triangles, if placed on the circle, will every time give the same coordinate pairs that are points on the Unit Circle if done correctly. 

5.) 

     The quadrant in which the original triangles from the beginning are draw in the first one. Depending on which quadrant the SRT is found, the value will change by its sign (positive or not). If the triangle is found on Quadrant II, then x from the coordinate pair will be negative. In Quadrant III, both x and y will be negative. Finally in Quadrant IV, only y will be negative. In the three pictures provided above, the triangles in a different location from Quadrant I will change in its sign due new location. In Quadrant I, all trig functions are positive, while in Quadrant II only sin and csc are positive and the rest negative. Meanwhile in Quadrant III, tan and cot are the only positive. Finally in the Quadrant IV, only cos and sec are positive.

INQUIRY ACTIVITY REFLECTION
1.) “THE COOLEST THING I LEARNED FROM THIS ACTIVITY WAS..." that special right triangles are actually beneficial and do a bit more than just being a perfect shape.
2.) “THIS ACTIVITY WILL HELP ME IN THIS UNIT BECAUSE..." I can refer back to it if I ever forget how to fill out the Unit Circle for the coordinate pairs section.
3.) “SOMETHING I NEVER REALIZED BEFORE ABOUT SPECIAL RIGHT TRIANGLES AND THE UNIT CIRCLE IS…” how they relate to one another. I would have not have guessed that the coordinates on the Unit Circle could also come from different shape.

RWA #1: Unit M Concepts 4-6 - Conic Sections in real life (parabola)

1. Definition of Parabola
     According to Mrs. Kirch, the definition of a parabola is a set of all points that are equidistant from a point (focus) and a line (directrix). The definition can best be explained/viewed from an image used in this page linked here.

2. Properties:

 (http://youtu.be/r-KmkpxVtGg)

     Algebraically:

(encrypted-tbn2.gstatic.com)

(encrypted-tbn1.gstatic.com)

     In standard form, parabolas are written as (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h) (however in the picture to the left, four ways are shown and the only difference is simply because of the negative sign) . The variables h and k is the vertex (origin) of the parabola as h represents x and k to y. (Notice how on the image to the top left, h is paired with x and k with y). Also, p has to be on the side that is NOT being squared. (The image on the left represents p with an a, which is the same thing as explained in the video from above). The number of units of p represents how far is the focus point and the directrix line of the graph. The sign of p, too, (if it is positive or negative) determines if the graph will go up or down or left or right. Meanwhile, the variable that is squared can determine the direction of the parabola: horizontal or vertical. 

     The image on the right is an example of a problem solving to rewrite the equation in standard form. The side being squared was found by solving-the-square and the binomial is by itself. In the example, 4 represents p (or a), and notice how it is on the other side of the squared binomial and sitting alone, undistributed to the other binomial. The video from above can help explain a bit more on rewriting to standard form. 

     Graphically: 

(encrypted-tbn0.gstatic.com)

(img.docstoccdn.com)

     The variable that is squared, as mentioned before, determines the direction of the parabola. As represented on the image to the right, if x is squared then the graph will go vertically; and if y is squared, then the graph will go vertically. The sign of p determines another part of the direction: if positive, the graph will move positive units either up or right; if negative, then the graph will move negative units either down or left. The axis of symmetry is always a line that spits the parabola in half from the vertex. The directrix is a line that is perpendicular to the axis of symmetry. As shown on the image to the left, the directrix is not within the parabola, but found outside and is found by p units away from the vertex. The focus point is within the graph and is found with p units away from the vertex. The distance away from the focus to the vertex can also help determine how wide or narrow the parabola would be. If the focus is far, then the parabola would be wide; and if the focus is close, then the parabola would be narrow. 

   

3. RWA:

(visual.merriam-webster.com/astronomical) 

(parkerlab.bio.uci.edu)

     In the real world, parabolas can be found within the science, astronomical department.  The images above show a radio telescope which, according to Merriam Webster Dictionary, is an "instrument used to capture, concentrate and analyze radio waves emanating from a celestial body or a region of the celestial sphere." Radio waves hit the parabolic reflector (the curved, bowl-like part of the telescope) which has a parabola shape. The waves automatically hits to the center of the reflector to a secondary reflector (considered the focus) that sends the waves down to the receiver. Anywhere where a wave would hit the parabolic reflector (a position on the parabola), it will always then aim to the secondary reflector (focus) since it was structured to have the same distance. 

4. Work Cited:

• Conic Parabola Video - http://youtu.be/r-KmkpxVtGg
• Standard From Equations (pic) - https://encrypted-tbn2.gstatic.com/images?q=tbn:ANd9GcRf8C72WS3K4tZSQoWR_RF9hTYjCHXWeyeweyXqYD0PtWD3Gk2uWw
• Parabola Graphs (pic) - http://img.docstoccdn.com/thumb/orig/41803311.png
• Standard Form Example (pic) - https://encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcQKIXbJTTqeqSGF8weMMOnzNdoSH7DA-jc1CG6zvl7_tQWSdcov
• Parabola Parts w/ graph (pic)- https://encrypted-tbn0.gstatic.com/images?q=tbn:ANd9GcQYmQdKDw1dEQKMsjpDpEbOr84_BF6-hOHW9Y_bP83bhlhapZIm
• Merriam Webster parabolic reflector (pic) - http://visual.merriam-webster.com/astronomy/astronomical-observation/radio-telescope.php
• Radio telescope (pic) - http://parkerlab.bio.uci.edu/nonscientific_adventures/otherplacesinCA.htm

WPP #10: Unit L Concepts 9-14 - Probabilities with one event, independent AND events (with/without replacements), OR events (non/mutually exclusive), and with combinations

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