BQ#7 - Unit V

Explain in detail where the formula for the difference quotient comes from now that you know! Include all appropriate terminology (secant line, tangent line,h/delta x, etc).

In this unit, we have been utilizing the difference quotient to solve derivatives. This raises the question: "what is a derivative?" Well, the derivative is the slope of the tangent line to a graph (the graph is most commonly referred to as "f(x)" while the derivative is noted as "f'(x)." To better understand what a tangent line is, we should look into secant lines. A secant line is a line that passes through a graph at two points; however, if you were to move those two points closer and closer together, until they meet, you would receive a tangent line - a line that touches the graph only once. In order for a secant line to turn into a tangent line, we must alter the slope. 

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

In the diagram above, we see two lines. Ultimately, our goal is to alter the red secant line in order to create a tangent line. Remember, this involves using slopes! This is where the slope formula () comes into play. However, for this to work, we need points to plug into the formula. Looking at the picture above, we notice that our two points will be (x, f(x)) and (x+h, f(x+h)) {"h" in this equation is solely the difference in space between the two points}. WOO! We now have our two points to plug into the slope formula. The work for this is shown below!

In the picture above, we can see the two points plugged into the formula for a slope. Nothing in the numerator can cancel out. However, we CAN cancel out the "x" on the bottom! This leaves us with our difference quotient!!!! 

Sources: http://cis.stvincent.edu/carlsond/ma109/diffquot.html
http://math.about.com/od/allaboutslope/ss/Find-Slope-With-Formula-JW.htm

BQ#6 - Unit U

1. What is continuity? What is discontinuity?

A continuity is a continuous function. This is defined as a function that is predictable, not having any breaks, jumps, or holes, which can be drawn without lifting your pencil from the paper. This is also seen when the limit and value are the same. A discontinuity is a function in which the intended height and the actual height are not the same. There are two different types of discontinuities: removable and non-removable discontinuities. Removable discontinuities are defined as point discontinuities (basically holes). These are removable due to the fact that the value is removed and placed elsewhere on the graph. On the other hand, non-removable discontinuities include: jump discontinuities, oscillating behavior, and the infinite discontinuity. Jump discontinuities include the use of closed and open circles to represent the value along the functions. A closed circle represents a value while the open circle represents the fact that there is no value at the given point. Oscillating behavior is defined as a non-removable discontinuity because it does not approach an specific point on the graph. Lastly, infinite discontinuities are defined as non-removable because of the vertical asymptotes. The vertical asymptotes create unbounded behavior which then leads to increases or decreases without bound (a discontinuity).

Kirch Math Analysis

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. This exists if you travel along a function from the left and right side toward some specific value of x, as long as the function meets or the heights from the right and the left are the same, then the limit exists. A limit do not exist at the three non-removable discontinuities (jump, infinite, and oscillating behavior). Jump discontinuities do not have limits because the left ad right behaviors are different. In other words, if you were to trace the function coming from both the left and right sides, you would not end up at the same value, leading to no limit. Limits also cease to exist when we encounter vertical asymptotes (part of infinite discontinuities). This is due to the fact that vertical asymptotes lead to unbounded behavior that cannot be determined as far as limits go. Furthermore, limits do not exist when oscillating behavior takes places. This is the result of not approaches any specific value. Without a specific value it is impossible to determine a limit for a given function.

http://img1.wikia.nocookie.net/__cb20130409141027/socio/images/0/05/The_limit_does_not_exist.gif

The different between a limit and a value is that a limit is the intended height of the function while the value is the actual height of the function.

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

To evaluate a limit numerically, we practice the use of limit tables. Before we get started, if the problem involves a fraction, we can use our prior knowledge of holes. In other words, we can factor out the top and see if a value cancels out with the bottom. If something cancels then we know that we will have a hole. The next step in this process is to then add or subtract 1/10 to our limit; this creates your boundaries for the table. From here, we fill in x values that move close to and away from the limit. After setting this up, we simply plug the equation into our calculators and hit trace until we receive all of our values. 

             For further explanation: 

To evaluate limits graphically, we use graphs and limit statements. This is relatively simple. We place our fingers to the left and right of the point we want to evaluate and move them towards one another. If they touch, then we have a limit. However, if the fingers do not make contact, there is no limit. While doing so, we will be expected to be able to write limits and values. To write our limits, we will phrase it as "f(x) as x approaches  # = __" and to compose our values we write "f(#)=__." This will help us to further analyze what is taking place on the graph. 

             For further explanation: 

To evaluate limits algebraically, we can use direct substitution, dividing out/factoring method, or rationalizing/conjugate method. The first option, direct substitution is the easiest. For this form of evaluation, we simply substitute the given value into the equation and solve. This will result in one of four possible methods: a numerical answer, 0/# resulting in 0, #/0 resulting in an undefined answer, or 0/0 which is indeterminate form. If we receive indeterminate form, we will then have to use the dividing out/factoring method, or rationalizing/conjugate method. When using the dividing out/factoring method, we must factor out the numerator and denominator, then cancel out like terms. After this step, we use direct substitution to find our answer. Why does this work? This works because we are getting rid of the hole in the problem, thereby resulting in a viable answer. Our other option is the rationalizing/conjugate method. Through this method, we multiply the numerator and the denominator by the conjugate of the side that has a radical. When doing so we foil the side from which we derived the conjugate from while the other side remains in parenthesis. This allows us to late cancel out like terms. Once this is done we practice direct substitution to receive our answer. 

             For further explanation:

And that is it for this BQ! 

http://media.tumblr.com/tumblr_mb6qjolp0h1qdvgu1.gif

Sources: 

kirchmathanalysis.blogspot.com

https://www.youtube.com/watch?v=8z4aofW85K4

https://www.youtube.com/watch?v=55Udw8r3Txw

https://www.youtube.com/watch?v=l7Tcay720vw

http://img1.wikia.nocookie.net/__cb20130409141027/socio/images/0/05/The_limit_does_not_exist.gif

http://media.tumblr.com/tumblr_mb6qjolp0h1qdvgu1.gif

BQ#4 – Unit T Concept 3

Why is a “normal” tangent graph uphill, but a “normal” Cotangent graph downhill? Use unit circle ratios to explain.

Tangent and cotangent are both positive in quadrant 1, negative in quadrant 2, positive in quadrant 3, and negative in quadrant 4. To start off with, we know that asymptotes exist whenever we divide by zero. For the graph of tangent, we need to look back at our ratios from the unit circle. Because tangent is equal to "y/x", tangent is going to have asymptotes wherever x equals zero (if "x" equals "0", we will receive an undefined answer). If we look at the graph, we know that the x value equals zero at pi/2 and 3pi/2. These two points will be the points of the asymptotes. With these asymptotes, we know that the graph has to be positive in quadrant 1 and quadrant 3 (ASTC) while it is negative in quadrant two and four. This outcome can only take place if the graph is positive in quadrant one, negative in quadrant two, so on and so forth. This makes the tangent graph uphill!

Cotangent is going to have asymptotes wherever sine equals zero. Yet again we need to look back at our unit circle ratios. Cotangent is positive in quadrant 1, negative in quadrant 2, positive in quadrant 3, and negative in quadrant 4. We also know that cotangent is equals to "y/x" (could be replaced with cos/sin). By looking at the graph, we see that this leads us to zero, pi, and 2pi. These values will become our asymptotes (that is where cotangent will become undefined). Because of the asymptotes and the sin graph, we will have to create a graph that goes from positive to negative to positive to negative. This can only be achieved one way with these asymptotes. this makes the normal cotangent graph downhill!

BQ#3 – Unit T Concepts 1-3

How do the graphs of sine and cosine relate to each of the others?

Desmos.com - via Kirch Math Analysis

In the graph above, we see three different three different lines: sine(red), cos(blue), and tan (orange). In Quadrant I (the blue section), we see that both the sine and cosine values are positive. Looking back at our identities we know that, tangent equals sine/cosine. if sine and tangent are positive, they will continue to yield positive results. Therefore, tangent is positive in the first quadrant. In the second quadrant, sine is positive and cosine is negative. Knowing that tan = sin/cos, then tangent is going to have to be negative. In quadrant three both sine and cosine are negative. If we divide a negative by a negative we will come to find a positive answer. In quadrant IV, we have sin being negative and cos being positive. If we divide a negative by a positive, tangent is going to be negative. To find our asymptotes, we have to figure out where cosine will equal "0." By looking at the graph we see that cosine equals zero at (pi/2, 0) and (3pi/2, 0). We can see this asymptote as the orange line can not touch the points given. 

desmos.com - via Kirch Math Analysis

In quadrant I, both sine and cosine are positive; this will lead to positive cotangent! Having a positive cotangent creates a graph that starts from the axis and goes up. From here, we should think of our identities. We know that "cot = cos/sin." This means that we will have asymptotes where sin = 0 (0 on bottom = undefined = asymptote). In Quad II, sin is positive while cos is negative. Therefore, if we divide cos by sin, we will receive a negative cotangent. This leads to a negative shift in the graph. In Quad III, cotangent is going to be positive because sin and cos are both negative. As we all know, two negatives make a positive. The final quadrant involves a negative sin value and a positive cos value. Dividing the two together would result in a negative turn in the graph. 

desmos.com - via Kirch Math Analysis

In Quadrant I cosine and secant are both positive. Thanks to our identities, we know that secant is the reciprocal of cosine. So where cos is (0,1), secant is going to be the reciprocal of the value (1). Since we know that secant is the  reciprocal of cos, we know that the asymptotes will be located where cos = 0. In Quadrant II, cosine is negative so secant is negative. We must not forget that the ACTUAL graph will not make contact with the asymptote boundaries. We can see that cosine is positive in Quadrant IV. Therefore, if we follow our ongoing trend, secant is going to have to be positive. We draw the graph from the high and low points of the valleys. These types will always be parabolas that have their set boundaries.

desmos.com - via Kirch Math Analysis

For our cosecant graph, asymptotes and sine values are VERY important. Wherever sin(x) = 0 there will be asymptotes for the cosecant graph. In quadrant 1, our sine value is positive; the same goes for quadrant II. So if sin is positive in both quadrants that means that cosecant will be above the x-axis (we draw csc from the vertex). In quadrant III and IV, we see that our sine value is negative. This means that our cosecant graph will also be negative. As you can see, the entire cosecant graph depends on the vertex and asymptotes of the sine graph. Why does this happen? We know that cosecant = 1/sin, wherever sin(x) = 0. This occurs because when sin=0 we receive the asymptotes that set our cosecant graph's boundaries.



Resources - http://kirchmathanalysis.blogspot.com/p/unit-t.html (Big Question Help)

BQ#5 – Unit T Concepts 1-3

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do?

To answer this question, we first need to look at the ratios of the trig functions we are dealing with! These ratio include: sine, cosine, cosecant, secant, tangent, and cotangent.

Kirch Math Analysis

When looking at the ratios, we see that sine and cosine are the only two trig functions that have "r" as the denominator. From our previous work with the unit circle, we know that "r" is equal to "1" as that is it is the variable for the radius of the circle. Knowing that this denominator will always be "1," we can conclude that there will never be an undefined value. {to receive and undefined value, we must have a "0" in the denominator; however, as there is a "1," the chance of this happening is eliminated.}The remaining four trigonometric functions contains "x" and "y" values in the denominator. These values could possible have a "0" in the denominator which would lead to the creation of asymptotes.

BQ#2 – Unit T Concept Intro

How do trig graphs relate to the unit circle?
Trig graphs are largely related to the unit circle! This can be seen in the formation of the graphs and the curves that accompany specific trigonometric functions. Each trig function correlates to a quadrant on the unit circle. This creates zones of positive and negative values. These values are reflected throughout trig graphs as the curvature of the lines follows the specific guides. The "all students take calculus" technique still applies! If we were to unravel the unit circle, the positive and negative quadrants would still apply to the trig functions.

Period?- Why is the period for sine and cosine 2π, whereas the period for tangent and cotangent is π ?
The period for sine and cosine is 2π because that is how long it take to repeat the pattern! The pattern that applies to sine is "++--" while cosine refers to "+--+" In order to complete these patterns (or make a full rotation around the circle) you would need to use all four quadrants. As we know from our unit circle knowledge, a full revolution around the circle is 360 degrees OR 2π! That being said, it would take 2π to complete the necessary pattern. The period for tangent and cotangent is π because you only need half of the unit circle to create a period. This is because the tan and cot pattern can be created with only half of the unit circle; half of the unit circle is equal to π.

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?

In order to answer this question, we should think back to what we know about the unit circle. From previous lessons, we have learned that the unit circle has a radius of 1. We have learned that the trig function for sine is equal to "y/r" and cosine is equal to "x/r." If "r" - the radius - is 1, that means that the numerator would have to be a value that is less than 1. We also know that sine and cosine have limitations within the circle. The two must be smaller than both -1 and 1. If sine and cosine cannot pass the value of "1" in either direction on the unit circle, they won't be able to pass them on a regular graph (remember that a trig graph is just the unit circle unraveled). That being said, the two trig functions are restricted to the y values of 1 and -1. This is the equivalent to having amplitudes of 1 and -1.

Reflection #1 - Unit Q: Verifying Trig Identities

I - What does it actually mean to verify a trig identity? 

Verifying a trig identity is one of the easier things to do throughout this concept. What it means, is that you have to take the given problem, and simplify it so that the left side is equal to the right side. In other words, you verify that one side equals the other even if they are formatted differently. 

II - What tips and tricks have you found helpful?

Starting off this unit, you will most likely feel extremely confused and probably won't want to touch an identity. It gets easier! Sort of... Something that really helps get problems done quickly is knowing all of your identities (ratio, reciprocal, and Pythagorean)! Knowing identities by heart will speed up the process of simplifying as you know what you want to get to in order to solve the problem. This will also prevent you from having to search through the SSS to find ways to work out the problems. One of the most important tips any Math Analysis student could give incoming identity solvers is to not give up! It is obvious that this is one, if not the most, of the more challenging units of the year; that just means that you need to push through it to succeed. Study, do your PQ's, and practice in order to do well on the test. As far as tricks go, don't be afraid of pulling out a GCF, substituting an identity, or playing with fractions! These steps will help you conquer the beast that is mathematical identities. 

III - Explain your thought process and steps you take in verifying a trig identity.

As soon as I look at my problem, the first thing I need to do is see what I'm working with. "Do I need to verify or solve?" The next step would be to then see if I should be pulling out a factor from the problem. If i need to, then I proceed to pull out the factor and attempt to set the problem equal to zero so that I can use the Zero Product Property. The next step would be to go on to see if anything can be simplified into a Pythagorean Identity. If something can be, then I go ahead and substitute one for the other. This may help us cancel out some parts of our problem. Depending on the problem, this may be enough to find your answer. However, dealing with fractions can cause some difficulty. When I come across fractions, I do the same steps as earlier, but then also take into the account the denominator. If there is a binomial on the bottom, we can multiply by the conjugate in an attempt to make our fractions a little easier to work with. If there is a monomial in the denominator, we can go ahead and split the fraction into two separate fractions. From these two steps, we can go on to factor our problem! These should all be accompanied by and explanation on the right hand side of the paper! With that, good luck to the future math analysis students!