Unit U BIG QUESTION - BQ#6

1. What is continuity? What is discontinuity?
          Continuity is when a function is continuous, meaning it can be drawn without lifting your pencil off the page.
          Discontinuity is when a function is discontinuous, meaning it has either a hole, a break in the graph, a jump, or if it oscillates.


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 that exists anywhere on a graph as long as you reach the same height from both the left and the right.
           A limit does not exist if the left hand limit and the right hand limit are not equal.
           A limit is the intended height of a function while a value is the actual height of the function.

3. How do we evaluate limits numerically, graphically, and algebraically?
           We evaluate limits numerically by making a table that gets really close to the limit but the x-values never actually touch it; graphically by looking at the graph either drawn out or on a calculator and writing out the limits; and algebraically by directly substituting into the function.

SP 11: Unit R Concept 3 Type 2- Trig of an inverse trig function (use sum and difference to prove)

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.
• 
  

     A "normal" tangent graph is uphill while a normal cotangent graph is downhill, not because it is the inverse of the other, but because of the positive/negative sign that corresponds to tangent in each quadrant. For example, tangent (and cotangent) are positive in the first and third quadrants, and negative in the second and fourth quadrants. They are arranged this way because of the unit circle ratios for these trig functions tan=y/x and cot=x/y. So for tangent, the graph will go--in one period--from negative (below the x-axis) to positive (above the x-axis). And for cotangent, the inverse of that rule applies.


Tangent

 Cotangent

BQ #3: Unit T Concepts 1-3 graphing tangent, cotangent, secant, and cosecant

How do the graphs of sine and cosine relate to each of the others? Emphasize asymptotes in your response.

These pictures show the relationship between sin and cos with all the other trig graphs. Sin and cos are alike in how they continuously swing. Unlike the other graphs with an asymptote that don't continue the way sine and cosine do. The other graphs run along the asymptotes without ever actually touching them. 

BQ#5-Unit T Concepts 1-3

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.
         Sine and cosine will never have asymptotes because asymptotes only happen when the trig ratio is undefined, in neither sine nor cosine will the ratio be divided by 0 because r will always equal 1.

BQ #2: Unit T intro

How do 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?

The period for sine and cosine is 2pi because in relation to the unit circle, it takes a full rotation (2pi) for the positive/negative pattern to repeat, whereas it only takes half of a full rotation (pi) for the sign pattern to repeat.

              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?

Sine and cosine are the two trig functions whose inverse trig function can only be between -1 and 1, if it were more than the absolute value of 1, the function would bust and it wouldn't be possible.

Reflection #1 Unit Q: Verifying Trig Identities

"What does it means to verify a trig function?"
When we are given a problem that is set equal to something, we must prove that it equals the other side.

"What tips and tricks have you found helpful?"
Making sure that it is all a single trig function before trying to do anything else.

"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."
1. Check for anything that is squared, if a function is squared, try to change it using a pythagorean identity. 2. Try to change it to all being one trig function. 3. Verify to make one side equal to the other.

SP #7: Unit Q Concept 2

Please see my SP7, made in collaboration with Joe Castillo, by visiting their blog here.  Also be sure to check out the other awesome posts on their blog!

ID#3: Unit Q Concept 1

Where does sin2x+cos2x=1 come from?
Thinking back to the unit circle, this rule comes from the pythagorean theorem which states "a2+b2=c2".The pythagorean theorem can be written to be x^2+y^2=r^2, this is possible because when we look at a triangle from the unit circle two of its legs are x and y and the hypotenuse is r. But again, when we think back to the unit circle, we want the equation to equal 1, so we divide everything by r to get (x/r)^2 + (y/r)^2 = 1. because we have our identities memorized, we realized that (x/r) and (y/r) have corresponding trig functions. With that being said, we know that Cos= x/r and Sin= y/r. when we plug that back into the equation, we are left with sin2x+cos2x=1. that is why this formula is referred to as the pythagorean identity.

WPP #13 & 14: Unit P Concepts 6 & 7

This WPP was made in collaboration with Joe Castillo. Please visit the other awesome posts on their blog by going here


Anthony Dufegt has survived an extremely hot day so far today and decides he's had enough of this constant struggle against the heat alone, so he decides to go to the pool. He walks out of his house and turns N 30° W then turns 100° due north, after walking for 60 meters more, he finally arrives at the pool. But alas, before he can get in, he ponders how many meters he would've walked if he could've just walked in a straight line instead. After he has finished his "underwater aerobics" class, he decides to take a different route and walks east 30 meters then turns at a bearing of 150° and continues to walk for another 80 meters until he gets home. Again, as he arrives home, he wonders how much he would've walked if had, instead, walked in a straight line. 

Part 1-Law of Sines (going to the pool)

Part 2-Law of Cosines (coming back from the pool)

BQ #1: Unit P Concepts 1 & 4: Law of Sines and Area Formulas

1. Law of Sines: Why do we need it?  How is it derived from what we already know?
We use the law of sines to solve for non-right triangles.This formula helps us find missing sides or angles when we only have AAS (angle, angle, side), ASA (angle side angle), or SSA (side, side, angle). We use the trigonometric function of sine to derive the formula and form two right triangles then we are able to derive for the formula. But before we do, we will have to add other variables to make it possible, like 'h' for the height (perpendicular to the base). If we draw a perpendicular line, we will get our 2 right triangles, which we need to derive for the formula. The following pictures go over the derivation of the formula.

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 area of a triangle formula: A=1/2 bh. This formula allows us to find the area of a triangle without having to know what the value of "h" is. The formula for an oblique triangle is similar to the formula of a regular triangle because they both have the same purpose and the same root or origin. The only difference is that in the original formula we must know the value of "h", but in the oblique triangle formula we don't need to know the value of "h" because instead we will be using one of the angle measures. The picture below will go over the derivation of the formula.

WPP #12: Unit O Concept 10: solving angle of elevation and depression word problems

Introduction: The PETCO across from Jose's bakery. 

Problem: The PETCO across from Jose's bakery

Jose is standing atop his bakery looking out towards the PETCO across the street that he often visits to buy neat things for his dog Simba. The street between the two buildings is 12 meters long. He looks down at the entrance to see a customer walking in with a dog that reminds him of Simba. He looks down at an angle of depression of 38°, this makes him wonder how tall PETCO is from his level of elevation. He then looks up at an angle of elevation of 34°, staring directly at the top of PETCO. How tall is PETCO from the height of his building and what is the total height of PETCO?

Solution: The PETCO from across Jose's bakery

PART 1 ANSWER: First we solve for how tall PETCO is from his level of elevation: tan38=x/12. in order to solve for x, we must multiply the entire problem by 12 then solve, which equals 8.09.

PART 2 ANSWER: then we solve for the rest of the building's height: tan34=x/12. to solve for x, we must also multiply the entire problem by 12 then solve, equaling 9.38. 

So to get the total height of PETCO, we must add the two distances of 8.09 meters and 9.38 meters, equaling 17.47 meters.

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

Inquiry Activity Summary

30-60-90 Triangles

Before we can solve a 30-60-90 triangle, we need to know how to get 30-60-90 triangles. On this worksheet we are given a 60-60-60 triangle (equiangular), and in order to get a 30-60-90, we must cut down the middle of it, giving us 2 triangles, both 30-60-90. To derive the patterns for 30-60-90 triangles, we must first know what to use to solve for each side. One way to get the side measures is using a special pattern. This pattern says that the shortest side is 'n', the hypotenuse is '2n', and the height is 'n√3'. But because this triangle is a right triangle, we can also use the Pythagorean theorem. But we need two sides in order to solve for one missing side; unfortunately we are only given one side measure: the hypotenuse (equals 1), that isn't an issue in this case because in these specific triangles, the shortest side is always half the size of the hypotenuse. Now we have the size of 2 sides; now we can use the Pythagorean theorem: a²+b²=c² . For this example, I'll label the short leg "b".  a²+(⅟₂)² =1² so simplified, a= ^√3/_2. 

45-45-90 Triangles

Deriving the pattern for a 45-45-90 triangle is simple and somewhat similar to the previous triangle. In order to get a 45-45-90 triangle, we must split the square in half diagonally, we do so because each angle in a square is 90˚ and splitting the opposites in half gives us 45˚ on both sides; leaving us with a 45-45-90 triangle. Because this triangle is also a right triangle, we can also use the Pythagorean theorem to solve for missing sides. In this activity we have the side lengths. For any of these triangles, "a" and "b" are the same lengths, and because of this fact about 45-45-90 triangles, filling in the Pythagorean formula is easier. 1²+1²=c² is the result of substituting in the formula, fully simplified: c=√2.

INQUIRY ACTIVITY REFLECTION

Something I never noticed about special right triangles is... that the easiest way to solve for sides is to use the Pythagorean theorem.

Being able to derive these patterns myself aids in my learning because... it gave me a visual representation which helped me better my understanding of special right triangles.

I/D# 1: Unit N Concept 7-How do SRT and UC relate?

A) 30 degree triangle

This activity is meant to show us where and how we get the plotted points of the Unit Circle's special three degrees (30, 60, 90) by using the rules of Special Right Triangles. Starting the activity requires us to label all three sides.This includes labeling the Hypotenuse "r" because the hypotenuse is the longest side and because the longest measurement from the center to any side of the circle equals 1 in a unit circle; labeling the horizontal value "x" because it is on the x axis; and then the vertical value as "y". The Hypotenuse being equal to 1 is a given and in my picture it is labeled so. Next we must find the other sides, we do so using certain formulas like: y=1/2r and x=1/2(r)√3 , using these formulas (substituting r for 1) we should get y=1/2 and x=√3/2 .Then we draw the graph on the triangle starting from (0,0) where the 30º opens is the origin (0,0), where the 90º opens will be (√3/2,0), and where the 60º opens we graph as (√3/2,1/2). Its important to know what this means because the whole point of this activity is to show why the points are being labelled what they are. This is also supposed to help you see that the unit circle is just the 3 special triangles repeating themselves in different quadrants. So instead of viewing it as a circle, we see focus specifically on the triangles.

B) 45 degree triangle

For the 45º triangle, we also start with the same basics. You label the sides accordingly to the special rules of a 45º triangle. Since it is a 45 degree triangle it varies very little. This means that the hypotenuse is still equal to 1 (because it is the hypotenuse). Since this is a 45º triangle, the two legs are the same length which means both the horizontal and vertical value use the same formula to get the value. The formula they use is x=1/2(r)√2. Now although they use the same formula, we cannot label the horizontal and vertical sides x, only the horizontal leg (the one on the x axis) can be labelled "x", the horizontal side (the leg parallel to the y axis) must still be labelled "y". After solving the formula, x=√2/2 . Then we sketch the graph plotting the opening of the 45º angle at the origin (0,0), the 90º at (√2/2,0), and the other 45º angle at (√2/2,√2/2). This again is relative to the unit circle because as you go into different quadrants, you will notice that they're the same points just that the x and y are negative or positive in respect to the quadrants.

C) 60 degree triangle

The 60º triangle is the same as the 30º one. It uses the same rules just switched. The horizontal value is still x and the vertical value is still y. Knowing this, we use the same procedures as in the 30º triangle. First we label the hypotenuse r and set it equal to 1. We then use the same formulas: y=1/2r and x=1/2(r)√3 except we switch the x and the y giving us x=1/2r and y=1/2(r)√3. If you notice, its the same as with the 30 degree triangle, so we've already done the work. Either way, solved out, x=1/2 and y=√3/2. Once you do the graph you realize it is the same points as a 30 degree triangle but with the x and y switched. 60º opens at the origin which is (0,0), where 90º opens is at (1/2,0), and where 30º opens is graphed at (1/2,√3/2).

D) How does this help derive the unit circle?

This activity helps you because it gives you a simple way to solve for the lengths of sides instead of having to memorize them.

(https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiFbHi5R27wuWOgJunN81YIE-3iOL2ArH6sCgU6U8ZTVaoXOcvwWMuoirPfF5kO5rHN7nrGUknniz5ua4tJWUDTbGgMePRxQLJueD1caTwZVVSbSQ4LoCFQO5UBhOfyhiqVQNzsnUCVnf0/s1600/CIRCLE+RADIANS.jpg)

The above image shows that looking at the light blue triangles, the points are all the same and only change in their sign in respect of whatever quadrant they are in, they all correspond to 30º triangles. The green triangles are also the same in their points corresponding to 45º triangles except with changes in the sign (also because of the quadrant they are in). Lastly, the red triangulars' points are the same as well, corresponding to 60º triangles.

E) How do quadrants affect the values? 

Each quadrant is a simple point. If the the triangle lies in quadrant range I, than the the values will be all positive, if it lies in quadrant II, the values will be (negative, positive) (-x, +y), if in quadrant III the signs will all be negative, if they lie in quadrant IV the values change to being (positive, negative) (x,-y).

                                              Inquiry Activity Reflection

A) The coolest thing I've learned from this activity was the formulas used to solve for the length of each side.

B) This activity helped me in this unit because if i get stuck on problems like these involving the unit circle, I will know what to do in order to solve them.

C) Something I never realized before about SRT and UC was how to actually get the lengths of the sides of these triangles.

RWA #1: Unit m Concept 5

"An ellipse is a set of all points, such that the sum of the distance from two points is constant."

(https://www.uwgb.edu/dutchs/MATHALGO/Ellipses.HTM)

(http://en.m.wikipedia.org/wiki/Ellipse)

The formula for an ellipse is (x-h)² /b² + (y-k)² /a²=1, if it is "skinny"; or (x-h)² /a² + (y-k)² /b²=1, if it is fat. You identify an ellipse by noticing that: both x and y are squared, the equation adds both x and y terms on one side, and they have different coefficients.To get the equation of an ellipse, first group the Xs and Ys and move the constant to the other side of the "equals" sign; next, take out the GCF of X and put it out side the parenthesis (same procedure for Y) and put the GCF of both on the other side of the "equals" sign well; then you must complete the square for x and y; then, factor both perfect square trinomials and simplify the opposite side; finally, divide the "x and y" part of the equation by what it is equal to to get the equation to equal 1, and reduce the fractions. 

(http://www.clausentech.com/lchs/dclausen/algebra2/lecture_notes/conics/ellips4.gif)

An ellipse looks like a stretched circle because a circle's eccentricity is 0, but an ellipses' eccentricity is between 0 and 1. It is much easier to get the key points of an ellipse when it is in standard form. All ellipses consist of: being shaped either like a "skinny" or "fat' circle, a center, two vertices, two co-vertices, it's two foci, and it's eccentricity. If the bigger number in the equation lies under the x it will be fat, but if it is under y it will be skinny. The center of an ellipse is (h,k). The major axis is horizontal if the bigger number is under the x term and if its under the y term it is vertical. The major axis connects the two vertices together (length of 2a). The minor axis connects the two co-vertices together (length of 2b). The foci determines how stretched out the ellipse is, if the focus is close to zero it makes the ellipse more circularly shaped, but if it is closer to 1, it stretches out more. To find a missing value such as a, b, or c; use the equation c² =a² -b²--CAUTION: this formula will only work if you have two of the mentioned values.

(http://csep10.phys.utk.edu/astr161/lect/history/eccentricity.gif)

Ellipses are used in the real world, like our solar system. Johannes Kepler discovered that each planet travels around the sun in an elliptical orbit with the sun being the focus. This also applies to an atom, with the electrons orbiting elliptically and the nucleus being the focus. In Lithotripsy, a medical procedure for treating kidney stones, the patient is placed in an elliptical tank of water, the kidney stone at one focus and high-energy shock waves generated at the other focus are concentrated on the stone, to destroy it.

(http://astro.wsu.edu/worthey/astro/html/lec-ellipse.html)

Work cited

http://britton.disted.camosun.bc.ca/jbconics.htm

http://kirchmathanalysis.blogspot.com/

SP#6 Unit K Concept 10: Writing a repeating decimal as a rational number using geometric series.

The viewer needs to know the the geometric series formula. To get the ratio, divide the second term with the first term. To check if its right just plug the fraction into the calculator.