BQ #7: Unit V: Derivatives and the Area Problem

1. 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).

 

 To start, let's look at this graph here:




There are two points where the red line (aka the secant line) and the blue curve cross. These two points are known as (x, f(x)) and (x+h, f(x+h)). h is not a point itself, but it represents the change in distance from point x to the second point on the x axis, which is why you sometimes see h notated as delta x (delta meaning 'the change in').

Sometimes, you'll also have points like this:



The purple line (aka the tangent line) is a line that only crosses the blue curve once. A graph like this has an infinite number of tangent lines, which have an infinite number of slopes at different points. But what if we wanted an equation that could give us the slope at any given point of the blue curve?

Let's go back to the secant line and its coordinates. We can easily find the slope of the secant line using the slope formula. Here it is in case you forgot it:



We plug in the two points (x, f(x)) and (x+h, f(x+h)) and get (f(x+h)-f(x))/(x+h)-(x)). The x's in the denominator cancel out, leaving us with (f(x+h)-f(x))/(h). Oh look, it's the difference quotient! It's also known as the derivative after you get far enough in math to start poking at calculus with a stick.



Why is this important? If you want to know the equation of the slope of the tangent line, you have to have h basically be 0. But it can't actually be zero because that would get you an undefined answer. So instead, you take the limit of the difference quotient of the blue curve as h approaches 0, plug in the x value of the coordinate where the tangent line and blue curve cross, and bam, you've got your slope. This can then be used to find all sorts of things, but that's for another time.

Resources:

www.jcu.edu

clas.sa.ucsb.edu

math.about.com

http://cis.stvincent.edu/carlsond/ma109/DifferenceQuotient_images/IMG0470.JPG 

BQ#6: Unit U: Limits, Functions, and the Like

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

A continuous function is one that is predictable, has no breaks/jumps/holes, and is able to be drawn without lifting your pencil from the paper. In other words, you can predict where a continuous function is 'going' on the graph and there aren't any discontinuities (not to be confused with changes in equation) in the graph.

This is an example of a continuous function graph. As you can see, although it does change direction, the graph is still predictable after the change in equation, there aren't any discontinuities, and it is still able to be drawn without lifting pencil from paper.

A discontinuity is what makes a graph not continuous; it is a break, jump, or hole in the graph. There are four different types of discontinuities, separated into two families. The first family is the removable discontinuity family and it contains only point discontinuities, also known as holes.

This is a hole.

The second family is the non-removable discontinuity family, and it contains jump discontinuities, oscillating discontinuities, and infinite discontinuities.



The reason there are two families has to do with the limits (or lack thereof) of the discontinuities, which I will get to in the next question.

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 the function- that is, it's where the limit wants to go. Sometimes it makes it there, and that's called a value. Continuous functions always have their limits and values equal each other. With removable discontinuities, it still reaches where it intends to go, but there isn't a value there; there's a hole. The limit of a point discontinuity doesn't match up with its value (which by the way is undefined), but there is still a limit. For non-removable discontinuities, the two sides of the function don't meet up at all and there is no limit. In this case, the limit Does Not Exist (or DNE for short) because the left side and right side of the function either approach different places (jump), approach so many places it becomes too wiggly to tell (oscillating), or has unbounded behavior caused by the presence of a vertical asymptote (infinite.)

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

 

To evaluate a limit numerically, we use a table like the one below.

 

 

We put what the limit approaches in the top center, go one tenth/hundredth/thousandth before and after it, and plug those numbers into the expression to find f(x) for each x value.

 

To evaluate a limit graphically, we use (big surprise) a graph. We look to see if the left and right sides of the function meet up. If they do, there is a limit (even if there's a hole.) If they don't, there is no limit and the reason depends on the discontinuity.

 

To evaluate a limit algebraically, we have three choices: direct substitution, dividing out/factoring, and rationalizing/conjugate.

 

The first method is direct substitution, which is exactly what it sounds like: taking whatever x approaches and plugging it into the expression. You should always use this method first because if you get a number over a number, a number over zero (aka undefined), or zero over a number (aka just plain zero), you're done! But if you get the dreaded zero over zero, that means there's a hole in the graph and you still need to do more work to figure out if the limit exists or not.

 

The second method is the dividing out/factoring method, which is also what it sounds like: looking to see if the numerator or the denominator has something to factor out and cancel. This removes the zero (and by extension, the hole, which is why point discontinuities are in the removable family.) After you've factored and canceled, plug in a la direct substitution and solve.

 

If direct substitution produces an indeterminate answer and factoring doesn't work, your last option is rationalizing/conjugate method. Multiply both the numerator and the denominator by the conjugate of the binomial term. The important thing to remember is NOT to factor the other term with the conjugate as the point of this is to get something to cancel, and it's easier to see if something is able to be canceled if you don't factor the other term with the conjugate. After you've FOILed the conjugates and canceled, use direct substitution and solve like you would normally.

 

Resources:

http://image.tutorvista.com/cms/images/39/graph-of-continuity-of-a-function.JPG

http://www.analyzemath.com/calculus/continuity/continuous_functions.html

http://web.cs.du.edu/~rjudd/calculus/calc1/notes/dis6.png

BQ#4: Unit T Concept 3: Upstair Downstairs...Graphs

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

The parent tangent and cotangent graphs are in different directions because their asymptotes are in different places. According to the Ratio Identities, tan=sin/cos and cot=cos/sin. To get an asymptote, the ratio must be undefined (in other words, the denominator has to equal 0.)

How does all that fit in? Tangent has a denominator of cosine, so the places where cosine equals 0 on the Unit Circle and the tangent graph are 90º (pi/2) and 270º (3pi/2). However, cotangent has a denominator of sin, so the places where there would be an asymptote are 0º (0), 180º (pi), and 360º (2pi). Even though the pattern of positive and negatives are the same (+ - + -), because the asymptotes are in different places, the graphs have to be draw differently in order to not touch the asymptotes and follow the rules.

BQ#5: Unit T Concepts 1-3: What's Different About Sine and Cosine?

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.

Let's start by recapping what the Unit Circle ratios actually are, shall we?



If you will recall, x is the side adjacent to the angle, y is the side opposite to the angle, and r is the hypotenuse of the triangle. In the Unit Circle, r is always equal to 1. 

Now notice how sine and cosine are the only trig functions with a denominator of r. Since r always equals 1, sine and cosine will always be real numbers. For the other trig functions, there are places where the denominator in their ratio will equal 0. This makes the value of the trig function in question undefined, and this is where you get an asymptote. Since r will never equal 0, sine and cosine are never going to be undefined and therefore will also never have asymptotes.

Resources:

http://3.bp.blogspot.com/-_3Jx3hiWnd8/U1HcV67sX9I/AAAAAAAAAQw/U0rSp0q1sSM/s1600/trigratios.png

BQ#3: Unit T Concepts 1-3: More Trig Graphs

How do the graphs of sine and cosine relate to each of the others?  Emphasize asymptotes in your response.
1.) Tangent?
Tangent is equal to sine/cosine, according to the Ratio Identity for tangent anyway. An asymptote is created wherever the denominator of a ratio equals 0 (and therefore is undefined), so whenever cosine's y value equals 0 on the graph, there will be an asymptote.



As you can see in this graph of cosine, the points where the y value is zero are -3pi/2, -pi/2, pi/2, and 3pi/2 (there are more, but let's not worry about that.)



In this graph of tangent, the points where the asymptotes are located are -3pi/2, -pi/2, pi/2, and 3pi/2 (again, there are more, but this is just a snapshot.)

2.) Cotangent?

Cotangent is equal to cosine/sine, also according to the Ratio Identity for Tangent. As stated before, an asymptote is found where the denominator equal 0. In this case, the denominator is sine, so the asymptotes are wherever sine's y value equals 0 on the graph.



As you can see in this graph of cosine, the points where the y value is zero are -2pi, -pi, 0, pi, and 2pi.



In this graph of cotangent, the points where the asymptotes are located are 0, 3.14, and 6.28- all of which are (approximately) the same points where sine equals 0.

3.) Secant?
Secant is equal to 1/cos, according to the Reciprocal Identity for secant. (We're done with Ratio Identities for now.) Secant has asymptotes too, which are located at the beginning and the end of each period.



Additionally, the minima and/or maxima of the secant graph also corresponds with the peaks and valleys of the cosine graph.

4.) Cosecant?

Cosecant is equal to 1/sin, according to the Reciprocal Identity for cosecant. Cosecant also has asymptotes (what a surprise), which are located at the beginning and the end of each period.



Additionally, the minima and/or maxima of the cosecant graph also corresponds with the peaks and valleys of the sine graph.

Resources:
http://www.mathsisfun.com/algebra/trig-sin-cos-tan-graphs.html
http://www.purplemath.com/modules/triggrph3.htm

BQ# 2: Unit T Concept Intro: Trig Graphs

How do the trig graphs relate to the Unit Circle?

A trig graph can be seen as an 'unwrapped' Unit Circle. The quadrants of a Unit Circle correspond with the sections on a trig graph. Whether or not the section of the graph is positive or negative -in other words, above or below the x axis- depends on the trig function graphed. For example, since sine is positive in Quadrants I and II and negative in Quadrants III and IV, the pattern of the sections of the trig graph for sine is + + - -, or above the x axis for two sections and below for the next two sections.

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 it takes a length of 2pi along the x axis of the graph to complete the pattern of positives and negatives for sine and cosine. This length is also the length of the circumference of the Unit Circle (how coincidentally convenient!) The period for tangent and cotangent, on the other hand, is pi because it only takes a lengths of pi along the x axis of the graph to complete the pattern of positives and negatives for tangent and cotangent. Since tangent (and cotangent) is positive in Quadrant I, negative in Quadrant II, positive in Quadrant III, and negative in Quadrant IV, it only takes half of the circle to repeat the pattern.

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?
Sine and cosine have amplitudes because their waves are restricted. On the Unit Circle, sine and cosine cannot be bigger than 1, but neither can they be smaller than -1. The other trig functions don't have these restrictions in the Unit Circle, so they don't have restrictions in their graph either in the form of amplitudes.

Reflection #1: Unit Q: Verifying Trig Identities

1.) What does it actually mean to verify a trig identity?

To verify a trig identity means to simplify an expression with trig functions until it equals something. When someone asks you to 'verify a trig identity', they mean 'look, here's two expressions that are set equal to each other. Make one side look like the other side.' Usually this consists of simplifying the more complex side to that it has the same terms as the simpler side. It also usually involves substituting trig identities to get things to cancel (and therefore simplify.)

2.) What tips and tricks have you found helpful?

I personally like to convert everything to sine and cosine. It makes it easier for me to see what cancels and what can be substituted (particularly with the Pythagorean Identities.) I also find it helpful to use the Reciprocal Identities to convert everything to the same trig function. Again, it makes it easier for me to see what cancels. Another tip is to try to get things to cancel to 1 (when there's multiplication or division) or 0 (when there's addition or subtraction.)

3.) Explain your thought process and steps you take in verifying a trig identity.Do not use a specific example, but speak in general terms of what you would do no matter what they give you.

Like I explained in the second answer, I like to try to use the Reciprocal Identities to cancel things out. I don't like to convert everything to sine and cosine until later on in the solving process. This is because converting to sine and cosine too early in the solving process can make things needlessly complicated. My last resorts are squaring the expression or multiplying by a conjugate (because those are almost guaranteed to make things very complicated.)