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.

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.