BQ#7, UNIT V

Throughout these 10 months, we have learned to memorize and be able to use the difference quotient. It has always been just another song memorized to me, however I know understand how the difference quotient formula is derived. The different quotient is derived because the function is "touched" twice, which is called a secant line, as shown below in orange. A tangent line is when the function is only "touched" once, as shown below in the pink line. 

http://www.mathsisfun.com/definitions/secant-line-.html

We know that the limit of the formula is whenever h=0, and that only happens when there is a tangent line. The slope of the tangent line is usually written as f(x). To derive the difference quotient, we can plug x in for the difference quotient and solve through for whenever there is a tangent line. 

UNIT U- BQ#6

What is a continuity? What is a discontinuity? 
A continuity is a function that does not have any breaks, holes, or jumps on the function. It is predicatable and we can draw it without lifting a pencil from a paper. A discontinuity is the exact opposite, it has a break, a hole, a jump and is drawn with lifting a pencil from the paper. There are 3 different types of discontinuities; point discontinuities, jump discontinuity, oscillating behavior and infinite discontinuities.

http://www.conservapedia.com/Continuous_function

http://www.conservapedia.com/Continuous_function

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 and exists in removable discontinuities and continuous graphs. A limit does not exist at a jump, oscillating or infinite discontinuity because the intended height is never reached. A limit is the intended height, whereas the value is the actual height.

How do we evaluate limits numerically? Graphiclly? Algebraiclly? 
We evaluate limits numerically on a table. We evaluate limits graphically on a graph, were left and right meet in the middle. And algebraically using the substitution method, the dividing/factoring out method and the rationalizing/conjunction method.

BQ# 2 -5 Unit T Concept Intro

• BQ#2
• The period for sine, cosine and tangent and cotangent differ because in sine and cosine, the period is 2pi because 2pi is the distance it takes for the pattern to repeat all over again. Since for sine, the 1st and second quadrant are postive and the 3rd and 4th ones are negative, (+,+,-,-) and for cosine, the 1st and 4th quadrants are postive, the rest negative, (+,-,-,+), it takes the whole unit circle (2pi) to repeat. However, in tangent and cotangent, it only takes half the unit circle (pi) because in tangent and cotangent, the pattern is (+,-,+-), and since the 2nd pair is the same as the 1st, the second pair does not need to be repeated, therefore it only takes up half of the unit circle.
• Both sin and cosine have amplitutes of 1 because that is as far as they can go in the unit circle, since the unit circle has the radius of 1, sine and cosine can only go up to -1 - +1. The ratio for sine is y/r and the ratio for cosine is x/r. r will equal one since we're dealing with the unit circle. Dealing with anything over 1 or less than -1will result in an error. Other ratios will work because they are not over one and are able to be greater than 1 and less than -1.
• BQ#5
• Sine and cosine do not have asymptotes whereas the other ratios do because of several reasons. Because the ratios of sine, cosine, cosecant, secant, tangent, and cotangent differ, so does thier place in the unit circle. Sine and cosine do not have asymptotes because they are both divided by 1 when looking at the ratio. If r=1, then we can not end up with asymptotes because we only get asymptoes when we divide by 0, which is why we get asymptotes in cosecant, secant, tangent and cotanget.
•  BQ#3
• The graphs of sine and cosine differ from the other graphs. Sine starts off going up (positive) and down (negaitive). Cosine is the inverse of sine, which goes down (negative) and up (positive). The cosecant graph is drawn by using the sine graph, however the valleys are used and asymptotes are involved to correctly graph. The secant graph is drawn by using the cosine graph, however the valleys are used and asymptotes are involved to correctly graph. The tangent and cotangent are both alike because thier quadrants stay the same sign (positive/negative), so when it has a negative, the graph goes down and when it has a positive, the graph goes up tp make a curvey, vertical wave.
• BQ#4
• A "normal" tangent graph is uphill and a "normal" cotangent graph is downhill because of a couple reasons. First of all, both tangent and cotangent are both positive in the first quardrant, negative in the second, positive in the third and negative in the fouth. With that being said, back to BQ#5, tangent has asymptotes when the line crosses at 0. The asymptotes hold the line in different quadrants. So when going uphill, the asymptotes are between quadrants 2 and 3 (negative postitive) and so fourth. The same rule would apply to cotangent with asymptotes.

Reflection #1 - Unit Q

24) What does it actually mean to verify a trig identity? To verify a trig identity, it means that it is true and you must prove that it is true by solving through and making x=x. We can't "touch" or move anything on the right side. The left side just equal the right side and if it does, we have verified the trig function.

25)What tips and tricks have you found helpful? A couple tips I found helpful is converting everything to cosine and sin, and working from there. When converting everything to sin and cos, we have more options to figure it out. I also found it helpful to take out a GCF and that helps make the problem simpler.

26)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. My thought process when working out a trig function is: 1) I look at the equation and check if there is a GCF to factor out. 2) I look for an identity. 3) If there is an identity, I substitute the identity and hope something will cancel out. If things do cancel out, then I try to look for something simple like costheta, and try to substitute it for 1/sec. When doing that, I hope something else cancels out and if it does, I try to simplify the rest.

SP #7: Unit Q Concept 2: Finding all trig functions when given one trig function and quadrant

This SP7 was made in collaboration with Adrie. Please visit the other awesome posts on her blog by going here

I/D3: Unit Q - Pythagorean Identities

1.) Where does sin^2x+cos^2x=1 come from to begin with? You should be referring to the Unit Circle ratios and the Pythagorean Theorem in your explanations. 

The Pythagorean Theorem is a^2+b^2=c^2. Since a, b and c are variables, we can replace them with any variable of our preference. But, according to the ratios, we use x, y and r, manipulating the theorem into x^2+y^2=r^2. We then have to divide the equation by r^2 to make it equal 1 and we would end off with (x/r)^2+(y/r)^2=1. Using our knowledge from unit O, we remember the trig ratio for cosines is x/r and y/r is the trig ratio for sines. If we simply substitute those variables for cosine and sine, the equation would be cos^2(x)+sin^2(x)=1. Proving this theorem to be an identity, we use one of the "Magic 3" ordered pairs from the Unit Circle and plug it in to our new equation. In example, the ordered pair for a 45 degree angle according to the unit circle is (radical2/2, radical2/2). Plugging it into the equation, we get (radical2)/2)^2 + (radical2)/2)^2, which ends up equaling 1, like every angle on the unit circle should.

2.) Show and explain how to derive the two remaining Pythagorean Identities from cos^2(x)+sin^2(x)=1. Be sure to show step by step. 

To show the left over 2 Pythagorean identities, we just have to manipulate the formula.

If we divide the equation by sin^2, the equation would be: cos^2(x)/sin^2(x)+sin^2(x)/sin(x)=1/sin^2(x). cos^2(x)/sin^2(x) would be substituted by cot^2(x). sin^2(x)/sin(x) would equal 1 since both top and bottom are the same. 1/sin^2(x) would be substituted by csc^2(x). In the end, the equation would end up being cot^2(x)+1=csc^2(x).

Now that we did sine, we do cosine. If we divide the equation by cos^2, the equation would be cos^2(x)/cos^2(x)+sin^2(x)/cos^2(x)=1/cos^2(x). cos^2(x)/cos^2(x) would equal 1 since top and bottom are the same. sin^2(x)/cos^2(x) would be substituted by the ratio tan^2(x). 1/cos^2(x) would be substituted by sec^2(x). In the end, the equation would end up being 1+tan^2(x)=sec^2(x).

INQUIRE ACTIVITY REFLECTION:

"The connections I see so far in Unit N, O, P, and Q so far are..." the use of the unit circle and the trig functions along with the ordered pairs from the unit circle.

"If I had to describe trigonometry in THREE words, they would be..." triangles, ratios and the unit-circle.

WPP #13 & 14: Unit P Concept 6 & 7 - ONE POST

This WPP 13-14 was made in collaboration with Jorge M. Please visit his other awesome posts by clicking here.

Law of Sines-

By the way, that's Adrie (not tyra) going to Farcos ice cream shop.
http://thejasminebrand.com/wp-content/uploads/2013/11/tyra-banks-drake-disneyland-the-jasmine-brand.jpg

Adrie and Drake are on a romantic picnic date together. When thier date is over, the decide to go get ice cream at Farco's icecream shop. But before that, Adrie and Drake play with a frisbee. Drake throws the frisbee 49 feet away to the East of him and Adrie runs after the frisbee. When she gets there, Drake yells out, "FIRST ONE TO FARCOS IS MY NUMBER ONE FAN!" And they both ran. Drake started off at N29*E and Adrie started off at N24*W. Who will get there first? (round to the nearest hundreths)

Law of Cosines-

Drake being attacked by fans.
http://www.popsugar.com/photo-gallery/31330666/Drake-surrounded-fans-during-his-VMAs-performance

The concert just finished and Drake and Adrie had planned hang out at Farco's Ice-Cream Shop. However, they want to keep it on the down low to avoid the crazy crowd of fans. From the stadium where the concert was performed, Drake's and Adrie's course to Farco's shop is at a bearing of 265 degrees and 1850 ft away. The crazy fan crowd heard a rumor that Drake was going to his hotel, which is at a bearing of 115 degrees. After the crowd walked 1780 ft towards the hotel, Farco, the owner of the ice-cream shop, tweeted that Drake and Adrie were in his shop and gave away their location! If the crowd change their route to the shop, how far are they? (Round to the nearest tenth)

BQ# 1: Unit P

1. Law of Sines - Why do we need it?  How is it derived from what we already know?  The derivation must be shown either in a video or in multiple sequential pictures and should include descriptions and information beyond what you can find in the SSS.

We need the law of sines to solve for triangles that are not 90* triangles. With the triangle having sides a, b and c, and angles A, B and C, C can be cut down to make 2 90* triangles. Using opposite over hypotenuse, the trig functions are rearranged. When everything is then equal to height, and everything is simplified, the law of sines is done.

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 is derived from the original a=(b)(h) but the change is that instead of all 3 sides given, only 2 sides are given, with 1 angle (SAS). There are 3 given formulas we can use, depending on what information is given to us, but it is just a matter of manipulating the formula to get what we need.

Either: Area = 1 ab sin C 2

Or: Area = 1 bc sin A 2

Or: Area = 1 ca sin B 2

REFRENCES:

http://math.ucsd.edu/~wgarner/math4c/derivations/trigidentities/lawofsines.htm

http://www.mathsisfun.com/algebra/trig-area-triangle-without-right-angle.html

Unit O Concept 10- Solving Elevation and Depression Word Problems

I am having the time of my life at Drake's concert. The stage is exactly 5 feet away from me. The angle of elevation from me, to him is 30*. How many feet do I have to jump to make it on stage?

Photo taken by Adrie

In the picture above, I used cos35=5. I then multiplied by x, then divided by cos35. I plugged it into my calculator and got 5.019 feet.

After the concert, I stood outside Drake's tour bus to take a picture. When he came out, a million things ran across my head, but most importantly, I wondered how far he would have to lean in to kiss me. With Drake being 11 inches taller than me, and the angle of elevation being 21*, how far does he have to lean in to give me a kiss?

Photo photoshopped by Adrie

In the picture above, I used sin21. I multiplied x out and then divided by sin21. After plugging 11/sin21, I got 30.695 inches.

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

INQUIRY ACTIVITY SUMMARY

In order to get a 30-60-90 triangle, you ut an equatlateral triangle straight through the middle. Assumimg the legs in an equalateral triangle equal 1, when it is cut in half b equals 1/2. Then we must use the pythagorean theorem to solve for c. As seen above, we end up with radical 3/2 for b. Since b and c are both under 2, we multiply all sides by 2. Since not all triangles will have the side length of 1, we also mutliply everything by n.

In order to get a 45-45-90 triangle, you cut a  square diagonaly in half. In a 45-45-90 triangle, 2 sides are congruent. All sides are equal to 1. Using the pythagorean therom, seen above, c is equal to radical 2. Since not all triangles will have the side length of 1, we also mutliply everything by n.

INQUIRY ACTIVITY REFLECTION

“Something I never noticed before about special right triangles is…” I didn't know from just a square, we could get a 45-45-90 degree triangle which then we use the pythagorean theorem to figure out c.

“Being able to derive these patterns myself aids in my learning because…” if I become a cheesebucket and forget "c" then I could simply solve using the pythagorean theorem.

I/D #1: Unit N: Concept 7: Unit Circle & Its Relation to Special Rights Triangle

INQUIRY ACTIVITY SUMMARY

In class, Mrs. Kirch gave us a worksheet which had pictures of 30*, 45*, and 60* triangles.

First we labeled the triangle following the given rules of the Special Right Triangle. The hypotenuse (r) is 2x, (y) is x, and (x) is x radical 3. After that, we had to make the hypotnuse equal to 1. Since the hypotnuse is 2x, we divided all 3 sides by 2x. R simplified to 1. In Y, the x's canceled and simplified to 1/2. In x, radical three/2x, the x's canceled out, simplifying to radical 3/2. We then drew the coordinate plane with the triangle lying in quadrant 1, making the origin (0,0). Since a point consists of (x,y) we substitued x for radical 3/2, and substituted 1/2 for y (highlighted in blue). 

   Following the same steps, we then solved for the 45* triangle. The hypotenuse (r) is x radical 2, and both (y) and (x) are x. We continued to make the hypotnuse equal 1. Since the hypotnuse is x radical 2, we had to divide all 3 sides by x radical 2. We divided x by x radical 2 but since we can not have a radical as the denominator, we rationalized by multiplying radical 2 on both top and bottom which simplified to radical 2/2. After, we drew a coordinate place with the origin being (0,0).  And since we know an ordered pair is (x,y) we get the ordered pair of (radical 2/2, radical 2/2) (highlighted in green).

 The last angle we solved was the 60* triangle. We followed the same steps as before. The hypotenuse (r) is 2x, (y) is x radical 3 and (x) is x. Then, we have to make the hypotnuse equal 1. Since the hypotnuse is 2x, we have to divide all 3 sides y 2x. r simplified to 1, (y)'s x radical 3/2x simplified to radical 3/2, (x)'s x will simplify to be 1/2. We drew the coordinate plane setting the origin to (0,0). Knowing that an ordered pair is (x,y), the 60* triangles ordered pai is (1/2, radical 3/2) (highlighted in green).

This activity helped me derive the ordered pairs for the unit circle, using the 30*, 45* and 60* triangles. 

The quadrants which I showed lie on the first quadrant. The number of the values do not change if we put it in Quadrant II, Quadrant III or in Quadrant IV, only the signs change.

As you can see highlighted in this picture, the signs change varied of what quadrant the triangle is in.

The coolest thing I learned from this activity was... learning that the unit circle is not a bunch of random numbers I had to memorize, it actually makes sense knowing why/how the ordered pairs are there.

This activity will help me in this unit because… now that I know the magic 5, which are located in the 1st quadrant, I know the rest of the ordered pairs.

Something I never realized before about special right triangles and the unit circle is… that the special triangles overlapped each other. Wait, I did not even know there were triangles in the unit circle.

RWA# 1: Unit M Concept 4-6 : Conic sections in real life

Ellipses

1) The mathematical definition of an ellipse is, "The set of all points, such that the sum of the distance from 2 points known as the oci, is a constant." -Mrs. Kirch

2) An ellipse includes the center, major and minor axis, 2 verticies, 2 co-verticies, 2 foci and and eccentricity along with a, b and c.

Here is what the equation of an ellipse looks like: 

(http://www.descarta2d.com/BookHTML/Chapters/ell.html)

Here is how an ellipse looks like graphically: 

(http://www.mathwarehouse.com/ellipse/equation-of-ellipse.php) 

We can automatically tell how our graph is going to look like graphically when the equation is given. We can tell if the ellipse is "fat" when the biggest number comes first(under x) and the ellipse will be skinny if the bigger number is under y. We can also figure out the center by using (x,y) and plugging it into (h,k) when x goes with h, and y goes with k. Both verticies will be the same number away from the center, along with the co-verticies and the 2 foci. 

To figure out how the equation will graph, we must find some "hints" that can lead to other answers. First, once the equation is given, we can find out the center, which is (h,k). After finding out the center, we can find out "a" and "b". The square root of "a" will always be the bigger number when dealing with ellipses and the smaller number will be "b". To figure out c, we can use "a^2-b^2=c^2. Once c is found, we can find the eccentricity which would be "c/a". The eccentricity of an ellipse is between 0 to 1. The verticies, co-verticies, foci, major and minor axis depend on whether the bigger number is under x or y.  

For further explanation, refer to the video below. 

3) Below, there is a picture of a soccer/track field in the shape of an ellipse. The midpoint would be the point in the middle which separates the middle of the field. The major axis would go horizontally, and the minor axis would go vertically. In this case, algebraically, the bigger number would come first, under x.

(https://www.google.com/search?q=racetrack&safe=active&client=firefox-a&hs=hai&rls=org.mozilla:en-US:official&source=lnms&tbm=isch&sa=X&ei=46r6UsiUH6O92wW4poC4Cw&ved=0CAoQ_AUoAg&biw=1280&bih=647#facrc=_&imgdii=_&imgrc=bWTXj9M8I81LGM%253A%3BV553ddJQAu7xVM%3Bhttp%253A%252F%252Fus.123rf.com%252F400wm%252F400%252F400%252Fexperimental%252Fexperimental1001%252Fexperimental100100014%252F6252442-football-soccer-field-pitch-vector-along-with-racetrack.jpg%3Bhttp%253A%252F%252Fwww.123rf.com%252Fphoto_6252442_football-soccer-field-pitch-vector-along-with-racetrack.html%3B1200%3B936)

4) Work Cited: 

(http://www.descarta2d.com/BookHTML/Chapters/ell.html)

 (http://www.mathwarehouse.com/ellipse/equation-of-ellipse.php)

((https://www.google.com/search?q=racetrack&safe=active&client=firefox-a&hs=hai&rls=org.mozilla:en-US:official&source=lnms&tbm=isch&sa=X&ei=46r6UsiUH6O92wW4poC4Cw&ved=0CAoQ_AUoAg&biw=1280&bih=647#facrc=_&imgdii=_&imgrc=bWTXj9M8I81LGM%253A%3BV553ddJQAu7xVM%3Bhttp%253A%252F%252Fus.123rf.com%252F400wm%252F400%252F400%252Fexperimental%252Fexperimental1001%252Fexperimental100100014%252F6252442-football-soccer-field-pitch-vector-along-with-racetrack.jpg%3Bhttp%253A%252F%252Fwww.123rf.com%252Fphoto_6252442_football-soccer-field-pitch-vector-along-with-racetrack.html%3B1200%3B936)
https://www.youtube.com/watch?v=5nxT6LQhXLM