Wednesday, May 28, 2014

BQ #7: Unit V

We're all familiar with the difference quotient, but where the heck does it come from and how does that affect derivatives?

Origin of Difference Quotient: f(x+h) -f(x) all divided by h

In this graph, we have the visuals to aid us in our understanding of where the difference quotient is derived from. The first point is (x, f(x)). There is delta x (delta = change) distance between the first and second point meaning the second point is (x + delta x, f(x + delta x)). But for this class we will reference to delta x as 'h' so our second point can also be written as (x+h, f(x+h). The line connecting these two points is called the secant line, much different than the tangent line, which only touches the graph once and is depicted below. Now we want to find the slope of the secant line and for that we are going to use our favorite slope-finding formula: the slope formula! The slope formula is m = y2-y1 all divided by x2-x1. When we plug in our two points from the first graph we have f(x + delta x) - f(x) all over x + delta x - x. Or as we substituted h for delta x, we have f(x+h) - f(x) all over x + h -x. We use the additive process in our denominator which cancels out the 2 x's, leaving only h. This brings us to the difference quotient we all know and love: f(x+h) - f(x) divided by the letter h, that's the difference quotient!

This video by Mathbyfives named Difference (which can be found here) verbally goes through the process in detail and can provide further explanation.

Saturday, May 17, 2014

BQ #6: Unit U

1. Continuities vs. Discontinuities

Continuity in this unit means a continuous functions, no jumps, breaks, strange behavior. A continuous function means a continuous graph -- meaning you are able to draw the graph without lifting your pencil from the paper. Sometimes a graph will show up that seems to demonstrate a discontinuity but it is also a change in function -- a change in function does not make it any less continuous. Discontinuities is much more vast and the main concern of this unit. There are two families of discontinuities. There is removable discontinuities such as point discontinuity, also known as a hole. Then there is non-removable discontinuities such as jump discontinuity, oscillating behavior, and infinite discontinuity, also known as unbounded behavior and occurs at vertical asymptotes.

2. Limits (vs. Values)

A limit is the INTENDED height of a function while a value is the ACTUAL height of a function. When the limit and value of a function are the same then the graph is continuous meaning the absence of of one of discontinuities.
A limit does exist at continuous graphs and removable discontinuities such as point discontinuity (also known as a hole) because although the value is undefined, the limit INTENDED on reaching that height.
A limit does not exist at non-removable discontinuities such as jump discontinuity because of different left/right behavior, oscillating behavior because it does not approach any single value, and infinite discontinuity because of unbounded behavior due to a vertical asymptote.

In this photo we can note the difference between a limit and value. In the first graph, there is a point discontinuity -- the limit exists, but the value is undefined. In the second graph, the limit exists at the point discontinuity as well but the value exists elsewhere (the black dot.) And the last example is one of a jump discontinuity where the limit does not exist but the value exists at one of the one-sided limits.

3. Evaluating Limits through VANG (minus the Verbally -- just reading the limit notation out as "The limit as x approaches 'a number' of f(x) is eqaul to 'L' "


There are three methods to evaluating limits algebraically.
-Direct substitution method
In which you plug in the number that is approaching the limit of f(x) and solve to see what you get.
Your possible answers are numerical, 0/# which is 0, #/0 which is undefined so the limit does not exist because of the presence of a vertical asymptote which means unbounded behavior, or 0/0 which is indeterminate form meaning we must use factoring or rationalizing method.
-Dividing out/Factoring method
We use this method when we get indeterminate form. In this case, we factor both the numerator and denominator and cancel the common term -- removing the 0 from the fraction. Then we use direct substitution with the simplified expression. BUT always be sure to use direct substitution first!!
-Rationalizing/Conjugate method
This method is also helpful with a fraction, especially if it has radicals. You multiply by the conjugate of either numerator or denominator, FOIL the conjugates and DO NOT multiply out the non-conjugate part, leave it factored. Something ought to cancel and then you simplify and once again use direct substitution with the simplified expression.


Through this method we use a table that calculates how the limit approaches 'x' from the left and right by first subtracting 1/10 and then adding 1/10.


You may use your calculator by simply plugging in the limit's equation, hit TRACE and then trace to the value you are searching for. OR you can put your finger to the left and to the right of where you want to evaluate the limit --if your fingers meet, that is where the limit exists, if your fingers don't meet then the limit does not exist either due to different left/right behavior (jump discontinuity), it is interrupted by a vertical asymptote which leads to unbounded behavior (infinite discontinuity), or it does not approach any single value (oscillating behavior).

First image may be found here.
All other images may be found here thanks to Mrs. Kirch's Unit U SSS Packet.

Sunday, April 20, 2014

BQ #4: Why is a "normal" tangent graph uphill, but a "normal" cotangent graph downhill?




Let's go back to the unit circle - what appears to be the solid foundation of pre-calculus. 
In the unit circle tangent and cotangent were related by its ratios. Cotangent's ratio was the reciprocal of tangent's ratio. Tangent's ratio is y/x meaning cotangent's ratio would be x/y. Now in this unit we would read that as tangent's ratio being sine/cosine (according to Unit Q's identities) and cotangent's ratio would be cosine/sine
Now what these two graphs have (as well as secant and cosecant) are asymptotes. Remember that this occurs when dividing by 0 leading to undefined. Tangent's asymptotes would occur when cosine equals 0 which would be at pi/2 and 3pi/2. Cotangent's asymptotes would occur when sine equals 0, this would be at 0, pi, and 2pi. (Reflect back on the unit circle, think of these values.) 
Because their asymptotes are placed in different areas, this is going to affect the direction in which they go. Also remember that these graphs WILL NEVER TOUCH the asymptotes, EVER. They will get VERY close but NEVER TOUCH. 
Not only do the asymptotes affect its direction, but look at the different colored areas in the images. Red is the first quadrant of the unit circle, where all trig functions are positive. Green is the second quadrant, where tangent and cotangent will be negative (as well as cosine and secant, but not sine or cosecant.) Orange is the third quadrant, where tangent and cotangent will be positive (the rest of the trig functions will be negative.) Blue is the fourth quadrant, where tangent and cotangent will be negative (as well as sine and cosecant but not cosine or secant.) 

Re-using the images from BQ #3, all thanks again to the wonderful Mrs. Kirch. Her amazing help with these graphs can be found here. 

Saturday, April 19, 2014

BQ #3: How do the graphs of sine and cosine relate to each of the others?

Sine and Cosine

Here we have two trig functions: sine and cosine. Because sine and cosine are always divided by r which is one they will never be divided by 0 hence never being undefined hence never needing an asymptote. There are five possible asymptotes which are depicted as dotted lines at the points of (0,0), (pi/2, 0), (pi, 0), (3pi/2, 0), and (2pi, 0). These points are were asymptotes can be found depending on the trig function selected. The four different colored sections represents one of the four quadrants from ASTC. The red is the first quadrant, green is the second, orange is third, and blue is the fourth. 

Remember that an asymptote results when a ratio is divided by zero, becoming an undefined ratio. 


Tangent's ratio is y/x, meaning that there is a possiblity of an asymptote when x (cosine equals 0). Because of this we know cosine equals 0 at 90* and 270* so pi/2 and 3pi/2, this is where are asymptotes will be located. Now let's look at the first quadrant, remember that in the first quadrant all is positive so it goes in an uphill direction and will NEVER touch the asymptote of pi/2, it simply gets really, really, really close to it. In the second quadrant, both sine and cosine are heading in a downhill direction and in the second quadrant tangent is not positive so it heads downwards but in the third quadrant it is positive. The graph can continue in these two quadrants because there is no asymptote dividing them. And in the fourth quadrant, tangent is not negative hence it's downhill direction. 

For cotangent our ratio is x/y, meaning we will have our asymptotes where y =0 (sine) and those locations would be at 0*, 180*, and 360*. On our graph these would be the values of  0, pi/2, 2pi. Because both sine and cosine are positive in the first quadrant as everything is, cotangent is positive as well. Yet in the second quadrant, sine is positive and cosine is negative leading contangent to continue in the negative direction crossing the x-axis when cosine does as well. The first and second quadrants already contain one period of cotangent. A similar process continues off from the asymptote of pi in the third and fourth quadrants. Because sine and cosine are negative in third quadrant, cotangent will be positive and because sine is negative and cosine is positive in the fourth quadrant, cotangent will be negative. 


Remember that secant is the reciprocal of cosine's ratio which will be r/x meaning that there will be asymptotes where cosine is equal to 0, similar to tangent's asymptotes. A similar pattern also follows here.
In the first quadrant, both sine and cosine are positive and so will secant. But however in the second quadrant, sine and cosine are both negative, as will secant and will continue to be negative because although cosine is positive in the third quadrant, sine is negative and a negative and positive will result in a negative. In the fourth quadrant, both sine and cosine are moving in an uphill direction and secant will also be positive. Once again notice how none of secant's graph is touching the asymptotes and how they develop at the mountains and valleys of cosine's graph.


Cosecant is the inverse of sine meaning its ratio will be r/y. This being said, cosecant will have asymptotes wherever sine equals 0, also similar to cotangent (oooh connections!) In the first quadrant secant will remain positive because all functions are positive in the first quadrant. Yet although cosine is negative and sine is positive in the second quadrant, cosecant will still be positive because it is positive in the sine quadrant of the unit circle. The graph continues into the third and fourth quadrant even after having gone through its period and it's direction is enforced by the unit circle's positive or negative values of the cosecant function. However more importantly because cosecant is the inverse of sine, it relies on the sine graph to be drawn because once again notice, like secant, it is drawn on the mountains and valleys of its corresponding reciprocal of the sine function. 

All images made available thanks to the amazing and wonderful Mrs. Kirch on Desmos, you can view and animate as well here.

Tuesday, April 15, 2014

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

Image found here.

Image found here. (thank you Google!)

Sine and cosine do not have asymptotes because they are always divided by 1 (by "r")

Yet once we move away from being divided by "r" we start reaching asymptote area.
If you reference to the photos you will notice undefined under tangent. Do not forget about their reciprocal ratios of cosecant (r/y), secant (r/x) and tangent (x/y).
Remember that we get asymptotes when our circle ratios equal undefined which results when you divide by 0. 
For sine and cosine you will never be dividing by 0 as you may for all other four circle ratios. But because you'll never be dividing by 0, you will never reach undefined, so you'll never have asymptotes! 

BQ #2: Unit T: Concept Intro: How do trig graphs relate to the Unit Circle?

Sine and Cosine In these images you see how the Unit Circle translates and unravels itself into a sine curve graph. 

Period: The period for sine and cosine is 2pi because it takes four quadrants (ASTC) to repeat the pattern. Sine's pattern according to ASTC from the unit circle is + + - -

 While cosine's pattern is + - - +
It merely takes all of the unit circle (which is 2pi at 360* to complete sine and cosine's patterns.
Graphs are merely snapshot of the graph, these graphs are infinite as circles are - no ends, no beginning but for the sake of this class we will only be graphing one single period.

 Amplitude: Remember that sine equals y/r and cosine equals x/r and that r=1. In the unit circle, the values cannot be bigger or smaller than 1, thus x and y always equal 1. 1 divided by 1 equals *drumroll please* ONE!!!
Meanwhile all the other trig functions have asymptotes because they do not have "r" such as tangent and cotangent (y/x and x/y respectively.)

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Meanwhile with tangent and cotangent.....
Period: Tangent and cotangent have a period of just pi. According to ASTC from the Unit Circle, the pattern this trig function has is + - + -. Meaning the pattern is completed in the first two quadrants which is 180* aka pi aka half of the circle.
Just remember again to keep in mind that a period is one time through their cycle, their pattern. 

Images found here.