BQ#7: Unit V: Deriving the Difference Quotient

How do we derive the difference quotient?

In the beginning of the year, we learned the difference quotient song. It goes a little something like this: "f of x plus h, minus f of x, divided by the letter h, that's the difference quotient." Here's a visual:

http://images.tutorvista.com/cms/images/39/difference-quotient-formula.png

Even though we've used this formula so often, we never knew where it came from. As out year reaches an end, we finally learn where this formula comes from!

If our graph has a secant line, the two points it hits will be (x, f(x)) and ((x+h), f(x+h)). h can also be represented as delta x, or the change in x. The smaller the delta x, the more the secant and tangent lines resemble one another.

http://upload.wikimedia.org/wikipedia/commons/thumb/6/61/Secant-calculus.svg/250px-Secant-calculus.svg.png

First, we know the slope formula is :

http://0.tqn.com/d/create/1/0/9/p/C/-/slopeformula.jpg

If we were to plug the y and x values into the slope formula, we would get:

http://www.homeschoolmath.net/worksheets/equation_editor.php

Then, since the change of x is so small, almost 0, we can ignore it in the denominator. Therefore, our equation simplifies to:

http://www.homeschoolmath.net/worksheets/equation_editor.php

The smaller the delta x, or h, the more similar the secant and tangent lines become. Therefore, if we take the limit as h approaches 0, we will be able to find the slope of the tangent line. Moreover, evaluating the limit of the difference quotient is also called the derivative.

BQ#6 - Unit U: Introduction to Calculus

http://www.mathwords.com/d/d_assets/d74.gif

1. What is a continuity? What is a discontinuity?
A continuous function will always be predictable. It will have no break, jumps, or holes. You can draw it without lifting your pencil off the paper, and the limit and value will always be the same.
On the other hand, a discontinuity will break all the above rules. There are two families of discontinuities. The first family is a removable discontinuity. The second is a non removable discontinuity. The only removable discontinuity there is, is a point discontinuity. It is called removable because we can cancel the holes when we factor or in other words "remove" the hole.  The non removable discontinuities include: jump discontinuities, oscillating behavior, and infinite discontinuities.
A jump discontinuity will display a jump in the graph that can have either two holes, or one open and one closed point.  A jump discontinuity can never have two closed points because it cannot have two y values for one x value.
Oscillating behavior is when a graph looks "wiggly."
Finally, an infinite discontinuity will occur in the presence of a vertical asymptote, which will lead to unbounded behavior.

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

OSCILLATING
http://www.cwladis.com/math301/lecture%20images/infiniteoscillationdiscontinuityat1.gif

JUMP
http://upload.wikimedia.org/wikipedia/commons/e/e6/Discontinuity_jump.eps.png

INFINITE
http://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/44bad38c-431e-4382-8fe9-86303561b2a0.gif

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 the graph. This differs from a value because a value is the ACTUAL height. A limit exists in continuous functions and point discontinuities. This is because if we approach an x value from the left and right, under these conditions, the left and right values will be the same. A limit does not exist if there is a different left and right value, oscillating behavior, or unbounded behavior.  This applies to non removable discontinuities (jump, oscillating, and infinite). A limit does not exist at a jump discontinuity because there will be a different left and right value. It does not exist at oscillating behavior because it is oscillating. It does not approach any single value. Finally, an infinite discontinuity does not have a limit because of unbounded behavior. It approaches infinity, but infinity is not a number.

3. How do we evaluate limits numerically, graphically, and algebraically?
We evaluate limits numerically on a table. When we evaluate them graphically, we place our fingers on the left and right of the graph, bring them together while tracing the graph, and see if our fingers end up at the same point. If they do, then there is a limit. We can evaluate limits algebraically by using direct substitution, or if that method does not work, the dividing out/factoring method or the rationalizing/conjugate method. When we use direct substitution, we simply plug in the number x approaches into the function. If we get 0/0, and indeterminate answer, then we must use another method. In the factoring method, we factor out the top and bottom and see if something cancels. Then, we use the direct substitution method with the simplified equation. In the rationalizing or conjugate method, we multiply the top and bottom with a conjugate. After FOILing and simplifying, we use the direct substitution method.

BQ#4 - Unit T Concept 3

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

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

 A "normal" tangent graph goes uphill, whereas a "normal" cotangent graph goes downhill. This is due to the ASTC pattern for tangent and cotangent: positive (I), negative (II), positive (III), and negative (IV).  In the last unit, we learned about the ratio identities for tangent and cotangent: tanθ = sinθ/cosθ; cotθ = cosθ/sinθ.
Since asymptotes only exist when there is an undefined, in tangent, cos must equal 0. Thus, cos=0 at 90 degrees and 270 degrees, which is (+)(-)pi/2. Thus, the formula to find the asymptotes for tangent would be b(x-h)= (+)(-)pi/2.
In cotangent, sin must equal 0.  Moreover, sin=0 at 0 degrees and 180 degrees, which is 0 or pi. Therefore, the equation to find the asymptotes for cotangent would be b(x-h)=0 or b(x-h)=pi.
Therefore, he only way for the graph to drawn according to the pattern and without crossing the asymptotes would be in an uphill graph for tangent, and downhill graph for cotangent. 

BQ#3 – Unit T Concepts 1-3

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

a. Tangent

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

In the last unit, we learned about ratio identities. The ratio identity of tanθ = sinθ/cosθ. The asymptotes are where the graph is undefined, so cos would have to equal 0 for this to be true. Also, cos=0 at 90 degrees and 270 degrees, which is pos/neg pi/2. This is where the asymptotes are located. Moreover, when sine crosses the x axis, tangent does not because there is an asymptote. When cosine crosses the x-axis, there is another asymptote for tangent. A tangent graph is not continuous because there are asymptotes restricting it.

b. Cotangent

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

The ratio identity of cotθ = cosθ/sinθ. Therefore,in order for there to be asymptotes, sinθ must equal 0. Therefore, there are asymptotes at 0 degrees and 180 degrees, which is 0 and pi. Thus, similarly to tangent, the asymptotes for cotangent correspond to where sine and cosine cross the x axis. It is also not continuous.

c. Secant

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

In a secant graph, they form parabolas above the mountains and below the valleys of a cosine graph. Moreover, it is not continuous because there are asymptotes.

d. Cosecant

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

.A cosecant graph corresponds with a sine graph. A cosecant graph also forms parabolas like a secant graph, but this time they touch the mountains and valleys of a sine graph. This graph is also not continuous because of the asymptotes.

BQ#5 – Unit T Concepts 1-3

Why do sine and cosine not have asymptotes? Yet, the other four types of graphs do?

Asymptotes are where the graphs will equal undefined. In the case of sine and cosine, they both have "r" as a denominator (Please see chart above). And, since it is a unit circle, r=1. This makes it impossible for sine and cosine to be undefined, which explains why they do not have asymptotes.
Moreover, cosecant and cotangent have similar asymptotes because they both have the denominator of "y". The same appies to secant and tangent, which have the same denominator, "x".

BQ#2 – Unit T Concept Intro

How do the trig graphs relate to the Unit Circle?
a. Period?

In sine, the first and second quadrant are positive while the third and fourth quadrant are negative. If we were to unwrap the unit circle and graph it, the 1st and 2nd quadrant would be above the x axis (positive) and the 3rd and 4th quadrant would be below the x axis (negative). This means that for it to have a positive mountain and negative valley, the period must be 2pi because it goes all around the unit circle.Please look at the first picture for a visual.  Cosine is positive in the first and fourth quadrants and negative in the second and third. Cosine also follows the same rules. So, when it is positive, the graph is above the line and when it is negative, it is below the line. Please look at the 2nd picture for a visual. Therefore, it also takes 2pi to have a completed cycle. However, tangent and cotangent are positive in the first and third and negative in the second and fourth quadrant. When we graph this, we can see that it only takes 1pi to complete the cycle. Please see the third picture. 

b. Amplitude

Sine and Cosine graphs have amplitudes of 1 because a unit circle has a radius of 1. If we were to unravel a unit circle, and graph it according to the sin and cosine graph, the amplitudes would be 1. This works for only sine and cosine because they have bounded ranges, which means it has an upper limit and lower limit. The rest of the graphs do not have bounded ranges. They have infinity in the range. 

Reflection #1 - Unit Q: Verifying Trig Identities

1. Verifying a trig function means proving that the equation is true. When we do this, we can never touch the right side because it changes the original problem. It is our goal to make the left side equal the right side.

2. I have found it helpful to have all the identities (reciprocal, ratio, and Pythagorean) memorized so that it saves time. And, this allows me to recognized certain patterns and solve the equations faster. Moreover, I have found it useful to always look for a GCF because it can make an equation tremendously easier to work with.

3. When I verify a trig function, I always keep in mind what the answer is.  This can give me a hint to what I need to do in order to prove it. One of the first things I always do is check if I can pull out a GCF. When applicable, this greatly simplifies the equation.  Next, I check if I can change everything to sin and cos because these two trig functions are the easiest to work with and have ratio identities that may be used. Furthermore, if I see a binomial in the denominator, I try to use conjugates or LCD. If I see a binomial in the numerator, I check to see if separating the equation will help solve it. After determining the first step to take, I use the identities to further simplify the equation.  I have found that after determining the first step, the rest of the equation is fairly simple.  Just utilize the identities to their fullest potential and keep working even though it may look messy. Finally, always keeping the verification answer in mind can help you stay on the right track.  

SP#7: Unit Q Concept 2: Finding All Trig Functions

 “This SP7 was made in collaboration with Mellanie T.  Please visit and check out her other cool blog posts by going here”.

First, determine the quadrant the problem will be in by looking at the signs. In his case, it will be in quadrant III. Then, it is best to use the reciprocal identities first because they are the easiest and fastest to find. However, you are not required to use them first. Afterwards, you can use the pythagorean, ratio, or reciprocal identities to find the remaining trig functions. You can also use SOHCAHTOA to solve this problem (as seen in the last image). Draw the triangle in quadrant III and label the sides. Then, use the trig ratios to solve for each trig function.

Please pay special attention to the final answers of the trig functions. There should never be a radical in the denominator. Remember to always rationalize! Furthermore,  it is important to pay attention to the signs in front of each answer. You should write the signs in at the beginning to make sure you do not forget. Good luck!

I/D#3: Unit Q - Pythagorean Identities

INQUIRY ACTIVITY SUMMARY
1. The Pythagorean identity, sin^2x+cos^2x=1, comes from the Pythagorean Theorem. If we arranged the Pythagorean theorem using x, y, and z, we would get x^2 + y^2 = r^2. And, if we made this equation equal one, we would have to divide everything by r^2. So, we would get: (x^2)/(r^2) + (y^2)/(r^2) = 1. We already know the trig ratio for cosine is x/r and the ratio for sine is y/r. So, (x^2)/(r^2)= cos^2 and (y^2)/(r^2) = sin^2. Therefore, for our final answer, we would get: cos^2 + sin^2 = 1, our Pythagorean theorem. Look at the pictures below for a visual.  Also, we can prove this by plugging in values from the unit circle. The example below uses the coordinates of the 30 degree reference angle: (rad 3/2, 1/2).

2. To derive the two remaining Pythagorean Identities from sin^2x+cos^2x=1, divide it by cos^2x and sin^2x. When you divide it by cos^2x, you shuold get 1 + tan^2x = sec^2x. Then, when you divide it by sin^2x, you should get 1 + cot^2x = csc^2x. Look at the pictures below to see, step by step, how I got them.

INQUIRY ACTIVITY REFLECTION
1. "The connections that I see between Units N, O, P, and Q so far are:" how the Pythagorean identities are derived from the Pythagorean Theorem (x^2 + y^2 = r^2) -> ( x^2/r^2 + y^2/r^2 = 1) ->(cos^2x + sin^2 = 1) and how the reciprocal and ratio identities are derived from the trig ratios (cotx= cosx/sinx) -> (x/y) / (y/r) = x/y
2. "If I had to describe trigonometry in THREE words, they would be:" triangles, identities, and ratios.

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

Please see my WPP13-14, made in collaboration with Mellanie T, by visiting her blog here.  And while you're there, be sure to check out her other cool posts as well. ^_^

BQ# 1: Unit P - Oblique Triangles, Law of Sines & Cosines, and Area

3. Law of Cosines

http://www.mathwarehouse.com/trigonometry/images/law-of-cosines/law-of-cosines-formula-and-picture2.gif

We need law of cosines to find certain sides or angles of an oblique triangle when we are given SSS (side-side-side) or SAS (side-angle-side). If we are given an oblique triangle and cut it down the middle of the base, we can create 2 right triangles (see pic below). 

Then, we can label the line h, for height. Now let's say we need to find the length of side a. We can use the distance formula, but to do that, we need the coordinates of angle B and angle C. If cosA=d/c, then d=ccosA. And, if sinA=h/c, then h=csinA. Therefore, angle B will be (ccosA, csinA). For angle C, b is the x value and 0 is the y value (b,0). Now that we have our two coordinates, we can plug these values in the  distance formula (see right): a^2=(ccosA-b^2)^2 + (csinA-0)^2. We need to FOIL (ccosA-b^2)^2 and simolify (csinA-0)^2. Then, we need to factor out c^2 from c^2sin^2A and c^cos^2A. (cos^2 + sin^2A) =1 (This is a trigonometric identity.) Thus, when we take the remaining numbers (highlighted in green), we obtain a^2=b^2+c^2-2bccosA, which is our Law of Cosines formula, which can be rearranged in three different ways (see 1st image).
Moreover, it is similar to the Pythagorean Theorem because the Pythagorean theorem is derived from the law of cosines.  If one of the angles is 90 degrees, then the cos(90)= 0. Therefore, we can leave out the -2bccosC, and it simplifies to a^2+b^2 = c^2.

5. Area Formulas

All three area formulas will work. They just use different values.

References:
http://www.mathwarehouse.com/trigonometry/images/law-of-cosines/law-of-cosines-formula-and-picture2.gif

WPP#12: Unit O Concept 10: Solving Angle of Elevation and Depression Word Problems

I/D2: Unit O - How can we derive the patterns for our special right triangles?

INQUIRY ACTIVITY SUMMARY
1. We can derive the 30-60-90 triangle from an equilateral triangle with a side length of 1 by using the information we already know about equilateral triangles. An equilateral triangle has the same length on all sides (1) and has equal degrees as well (60 degrees). If we cut this triangle down the middle, the side length that is cut will become 1/2 for the 30 degree angle. And, the degree formed will become a 90 degree angle, which will have a opposite side length of 1 still. Since we know the hypotenuse and one leg, we can use the Pythagorean Theorem to solve for the second leg.  The work for this is shown below.  In the end, we get the leg to be rad 3/2. If we were to multiply all these side lengths by 2, to make the numbers easier to work with, we would get: 1 (30 degree), 2 (90 degree), and rad 3 (60 degree). If we were to replace these values with n, we would get: n, 2n, and n rad 3. The "n" is a variable that depends on the triangle and shows the relation between the sides, the ratio.

2. We can derive the 45-45-90 triangle from a square with a side length of 1 by using the information we already know about squares. All the sides are equal and all the angles are 90 degrees. Therefore, if we cut the square along its diagonal, the two legs of the triangle formed will remain 1. Also, the right angle will be cut in half, so the angles would be 45 degrees. In order to find the length of the hypotenuse, we can use the Pythagorean Theorem. The work shown below gives the answer of c= rad 2. Therefore, if we get our sides as: 1 (45 degree), 1 (45 degree), rad 2 (90 degree). If we replace the values with n, we get: n, n, and n rad 2. The "n" is a variable that depends on the triangle and displays the ratio of the sides, or, in other words, the relationship between them.

 INQUIRY ACTIVITY REFLECTION
1. "Something I never noticed before about special right triangles is:" how they are derived from basic shapes, follow the same rules, and make logical sense as well as mathematical.
2. "Being able to derive these patterns myself aids in my learning because:" it allows me to fully comprehend where these values come from, not just accept them blindly.

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

INQUIRY ACTIVITY SUMMARY
1. For a 30 degree special right triangle, the rule is

30°	60°	90°
n	n√3	2n

Therefore, when we simply the hypotenuse to "1," by dividing each value by 2n, we get: r = 1 (hypotenuse), x = √3/2 (horizontal), and y = 1/2 (vertical). The work for dividing is shown below.

After finding the values, we can plot the special triangle onto a coordinate plane. If this triangle is put in the first quadrant, its coordinates would be (0,0); (√3/2,1/2); (√3/2,0).

 
2. For a 45 degree special right triangle, the rule is

45°	45°	90°
n	n	n√2

When we simply the hypotenuse to "1," by dividing by n√2, we get: r = 1 (hypotenuse), x = √2/2 (horizontal), and y = √2/2 (vertical). The work for dividing is shown below.

After finding the values, we can plot the special triangle onto a coordinate plane. If this triangle is put in the first quadrant, its coordinates would be (0,0); (√2/2,√2/2); (√2/2,0).

3. For a 60 degree special right triangle, the rule is the same as a 30 degree special triangle:

30°	60°	90°
n	n√3	2n

So, when we simply the hypotenuse to "1," by dividing each value by 2n, we get: r = 1 (hypotenuse), x = 1/2 (horizontal), and y = √3/2 (vertical). This is very similar to the 30 degree triangle. The only difference is that the x and y values are switched. The work for dividing is shown below.

After finding the values, we can plot the special triangle onto a coordinate plane. If this triangle is put in the first quadrant, its coordinates would be (0,0); (1/2, √3/2); (1/2, 0).

4. This activity helped me derive the Unit Circle by showing me that the Unit Circle is just made up of special right triangles. The 30, 45, and 60 degree triangles have the same coordinates as the first quadrant of the Unit Circle. And, if we use the reference angles of these triangles, we can get the coordinates of the remaining points on the Unit Circle.

5. The triangle in this activity lies in the first quadrant.  However, in the second quadrant, the x values are negative. In the third quadrant, both the x and y values are negative. Finally, in the fourth quadrant, the y values are negative. The first image shows a 30 degree special triangle in Quadrant II, notice how the x value is negative. The second image portrays a 45 degree special triangle in Quadrant III, notice how both the x and y value of the coordinates are negative. Finally, the third picture displays a 60 degree special triangle in Quadrant IV, notice how the y value is negative.

INQUIRY ACTIVITY REFLECTION
1. "The coolest thing I learned from this activity was:" that the triangles are in all the quadrants. They are just flipped around, so the coordinate signs change.
2. "This activity will help me in this unit because:" it helped me memorize the unit circle
3. "Something I never realized before about special right triangles and the unit circle is:" that the unit circle is completely made up of special triangles.