Welcome to Tracey P.'s Math Analysis Blog

Wednesday, June 4, 2014

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

1. Explain in detail where the formula for the difference quotient comes from. Include all appropriate terminology. 

First of all, the equation of the difference quotient is:
It is used to find the slopes of the graph. To derive it, look at the picture below. The first intercepted point is (x, f(x)) and the second is (x+h, f(x+h)). On a side note, the red line is called a secant because it's intercepted at two points and a tangent line is when it intercepts only at one point. Using the two points, plug them into the slope formula: (Y2-Y1)/(X2-X1). Once you have substituted these points, you'll see that the x in the denominator can cancel, leaving an h in the denominator. Once done simplifying, you'll see that it looks exactly like the difference quotient. The difference quotient finds the slope of a tangent line to a graph. This is also known as the derivative.

Work Cited:

Sunday, May 18, 2014

BQ#6: Unit U

1. What is continuity? What is discontinuity?
Continuity means the graph is predictable; it has no breaks, holes, or jumps. It can be drawn without lifting the pencil off the paper.

Discontinuity is when the graph can contain breaks, holes, and jumps. It's separated into two families: Removable Discontinuities and Non-Removable Discontinuities. Point discontinuity is a removable discontinuity, meaning there's a hole in the graph. Non-Removable Discontinuities has Jump Discontinuity, Oscillating Behavior, and Infinite Discontinuity. Jump discontinuity means they have different left and right. You must remember that the jump can have two open circles, one close/open, but they must never be two closed circles. Oscillating behavior means the graph is very wiggly, or unpredictable. The limit DNE because you can't really pinpoint the graph's next destination. Infinite discontinuity is when there's a vertical asymptote leading to unbounded behavior. Unbounded behavior means it increases or decreases without bond. This is because the graph is either going up infinitely or down infinitely.
Jump Discontinuity
Point Discontinuity
Oscillating Behavior
Infinite Discontinuity

 2. What's a limit? When does a limit exist? When does a limit DNE? What's the difference between limit and value?
A limit is the INTENDED height of a function; whereas a value is the ACTUAL height of a function. A limit exists when the intended and actual height are the same. A limit does not exist when the left and right are different, unbounded behavior, and oscillating behavior. 

3. How do we evaluate limits numerically, graphically, and algebraically?
Numerically means you use the table. As the table goes towards the center, it's getting closer to the limit, or y-value. It also helps us find if the limit can be reached or not. Graphically is when we plug the function onto our calculator to graph and then trace to the value you are looking for. To evaluate the limit, you put your finger on a spot to the right and left; where your fingers meet is where the limit is. If your fingers does not meet, then it does not exist. For algebraically, there are three methods: Substitution, Factoring, and Conjugate. Substitution is simply just plugging in the number into the function. You always want to try this method first before doing any other method. If you get 0/0, that means it's an indeterminate form and you have to use another method. The factoring method is when you factor the numerator and denominator and cancel like terms to remove the zero. Then with the simplified expression, use substitution to find your answer. The rationalizing/conjugate method is when you multiply the radical with its conjugate. Remember to not multiply the non-conjugate because something would hopefully cancel. Once something does cancel, you use substitution to get your answer. 

Work Cited:

Monday, April 21, 2014

BQ#4: Unit T Concept 3

Why is a "normal" tangent graph uphill, but a "normal" cotangent graph downhill?
ASYMPTOTES. They are the reasons why tangent goes uphill and cotangent goes downhill. As we previously learned, tangent is sine/cosine and the only way to make an asymptote is for it to be undefined. This means cosine must be zero. We know that cosine equals zero at 90º (0,1) and 270º (0,-1). So, on a graph, the asymptote starts at π/2 and ends at 3π/2. Remember, these aren't all the asymptotes, they are just one period. In the picture below, you can see that the asymptotes are the dotted lines. Quadrants two and three consists of one whole period and quadrants one and four are just the halves because it's a snapshot.
Contrasting with the tangent graph, cotangent's asymptotes are placed in a different spot, resulting in a downhill graph. It's ratio is cosine/sine, which means sine must be undefined for it to have an asymptote. Sine is zero at 0º/360º (1,0) and 180º (-1,0), so that is where the asymptotes are placed to make one period (picture below). The cycle is positive, negative and positive,negative. They repeat twice, so there are two periods only in the picture.
Work Cited:

Sunday, April 20, 2014

BQ#3: Unit T Concept 1-3

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

a) Tangent?
First of all, we know that sine (red) starts and ends at zero and cosine starts and ends at 1; both of their periods is 2π. Looking at the picture above, in the first quadrant, sine and cosine are both positive. Since tangent is equal to sine/cosine, you're dividing by two positive numbers so that means tangent must be positive as well. Therefore, tangent (orange) is positive in the picture. In the second quadrant, sine is positive and cosine is negative, so tangent must be negative. In the third quadrant, both sine and cosine are negative, so tangent is positive, In the fourth quadrant, sine is negative and cosine is positive; therefore, tangent is negative. 

Further more, asymptotes are significant in understanding how sine/cosine correlate with tangent. From previous units, we know that tan=sin/cos and in order to have an asymptote, it must be undefined. In this case, cosine must be undefined or equal to zero. From the unit circle, x is zero at 90º (π/2) and 270º (3π/2). Those will be where the asymptotes will be. It's where tangent cannot cross or touch, as you can tell in the picture. Remember there are MANY asymptotes, it goes on forever but the picture above is just a screenshot. 

b) Cotangent?
Cotangent is very similar to tangent because it's equal to cosine/sine. From the picture above, we see that cotangent (blue) is positive in the first and third quadrant because we are either dividing by a positive and positive or a negative and negative. In two and four, cotangent is negative because you're dividing by positive and negative. 

For cotangent to have an asymptote, sine must be equal to zero. Looking at your unit circle, you see that y is equal to zero at 180º (π) and 360º (2π). This will be where the asymptotes will be on the graph. 

c) Secant?
Secant looks very different from tangent and cotangent but it's a similar concept, just a different identity. Secant is equal to 1/cos. Now, for secant, you are just looking at the cosine line (green). Cosine is positive in the first quadrant and is being divided by one, which is positive, secant will be positive. In the second quadrant, cosine is negative; positive divided by negative is negative. Secant is also negative int bird quadrant because you're dividing by a positive and negative. In the fourth quadrant, secant is positive because it's a positive divided by a positive. Looking at the picture above, you can see that a secant graph resembles positive and negative parabolas.

Secant is equal to 1/cos, so cosine must be undefined. As stated in tangent, cosine is equal to zero at 90º (π/2) and 270º (3π/2). That will be where the asymptotes will be placed.

d) Cosecant?
The identity for cosecant is 1/sine. Therefore, we are just looking at the sine line (red) for now. In the first and second quadrant, cosecant is positive because sine is positive. In the third and fourth quadrant, cosecant is negative because sine is negative. It will look like two parabolas but there are an infinite number because the graph goes on forever.

Since cosecant is 1/sine, sine must equal zero for it to have an asymptote. This is similar to cotangent because sine is equal to zero at 180º (π) and 360º (2π). This is where cosecant must not touch or pass.

Work Cited:

Thursday, April 17, 2014

BQ#5: Unit T Concept 1-3 - Graphing All Six Trig Functions

Why do sine and cosine NOT have asymptotes, but the other four trig graphs do? Use unit circle ratios to explain.
Sine and cosine do not have asymptotes because they can't be undefined. In order for a trig function to have an asymptote, it has to be undefined, meaning the denominator has to be zero. Sine is y/r and cosine is x/r. We know from the Unit Circle that the radius (r) is always one; therefore, their denominators can never be undefined.

The other four trig graphs, on the other hand, do have asymptotes. There, however, are restrictions. Cosecant (r/y) and cotangent (x/y) have the same denominator: y. On the unit circle, "y" can be zero at 0º (1,0) and 180º (-1,0). As a result, they can still have asymptotes but they can also have no asymptotes as well. Another example is secant (r/x) and tangent (y/x). They also have the same denominator: x. "X" can be zero at 90º (0,1) and 270º (0,-1). Whenever x is zero, it will have an asymptote. When it's not zero, it won't have asymptotes. These four trig graphs do have asymptotes but there are also times where they don't have asymptotes like sine and cosine.

Wednesday, April 16, 2014

BQ#2: Unit T Intro - Graphing Sine and Cosine

How do trig graphs relate to the Unit Circle?

  • Trig graphs relate to the Unit Circle, because, if unwrapped, it will be a horizontal line with the four quadrants all laid out in a row.
Period? - Why's the period for sine & cosine 2π, whereas the period for tangent/cotangent is π?

  • The period for sine and cosine is 2π because when looking at sine, it can be positive in quadrants one and two and negative in quadrant three and four. So, if we lay that circle in a straight line it will look like the picture below. The part shaded red is the first quadrant, green is second, orange is third, and blue is fourth. Quadrants one and two are positive so it's shaped as a "hill" and quadrants three and four are negative so it's a "valley". A period is how long the cycle takes to repeat itself, which in this case is 2π. The same goes for cosine because it's positive in quadrants one and four and negative in two and three, you can clearly see it drawn out in the picture below. The two quadrants in the middle are negative and the sides are positive. As you can see, it only shows half of the curve in quadrant 1 and 2, so together it will be a "hill".

    Sine Graph
    Cosine Graph
    ●  The period for tangent and cotangent is π. Let's look at tangent; it's positive in quadrants 1 and 3 and negative in 2 and 4. On a line it will be positive, negative, positive, negative (picture below). The cycle repeats faster because it's positive then negative. It's period is at 180º which is π.

Tangent Graph

Amplitude? - How does the fact that sine and cosine have amplitudes of one relate to what we know about the Unit Circle?
  • Amplitudes are half the distance between the highest and lowest points on the graph. Sine and cosine have amplitudes of one relate to the Unit Circle because they have to be between 1 and -1. This means that the highest and lowest points for either sine or cosine are between 1 and -1; it can't be anything greater. The other trig functions, such as tangent, don't have amplitudes of one because it's ratio is sin/cos. Meaning they don't have a low and high point (you can refer back to the picture above).

Work Cited:

Thursday, April 3, 2014

Reflection#1: Unit Q - Verifying Trig Identities

1. What does it mean to verify a trig identity?
Verifying a trig identity means to prove that one side of the equation is equal to the other side, by using identities, of course. Also, you must remember that you must never touch the right side but just rearrange the left to make it equal to that side. The right side must always remain the same; untouched.

2. What tips and tricks have you found helpful?
A HUGE tip for you would be to memorize the identities! Memorizing the identities would make it so much easier to solve the problems because you won't have to constantly look back on your packet and waste your time searching for the correct identity. Working on practice problems will not only help you memorize the identities but it would also help you on your problem-solving, since there is usually more than one way to solve these kinds of problems. When working on a problem, see if you can change all the trig functions to sine or cosine. If there were fractions in the problem, you can look for the LCD or just multiply by the conjugate, whichever would make it simpler.

3. Explain your thought process and steps you take in verifying a trig identity?
My thought process, first, would be to see if there are any identities I could use that would cancel out with. A light bulb usually turns on when I see a squared trig function because it could be indicating that it's one of the pythagorean identities. If there are fractions, I will see if there's anything that could cancel if I use a certain identity. If not, then I try separating it. Like I said before, I would also try to  look for a LCD or multiply by a conjugate.

Wednesday, March 26, 2014

SP#7: Unit Q Concept 2 - Finding all Trig Function Values Using Identities

Please see my SP7, made in collaboration with Vanessa C. by visiting her blog HERE. Also, be sure to check out the other fabulous posts on her blog.

Wednesday, March 19, 2014

I/D#3: Unit Q - Pythagorean Identities

Inquiry Activity Summary:
1. The Pythagorean Theorem is an identity that can always be proven true no matter how much it is manipulated. It is evident in the Unit Circle when we take the sine and cosine of the 45º, for instance. The ordered pair would be (√2/2, √2/2) and the sin45 is √2/2 and the cos45 is √2/2. If we plug it into the Pythagorean theorem, it would be one equals one, which is true. Sin2x+cos2x=1 comes from the pythagorean theorem, which is x2+y2=r2. Next, you have to manipulate the pythagorean theorem so that there is only one "r". This is because in the unit circle, r is equal to one. To do this, you divide everything by r2. Once you have done this, you will get (x2/r2)+(y2/r2)=r. However, you can arrange that to make it look simpler: (x/r)2+(y/r)2=1. Now, you know that x/r is equal to cosine and y/r is equal to sine. As a result, you just substitute or replace it with cos and sin. Then, it would look like cos2x+sin2x=1.

2. The second pythagorean identity is 1+tan2x=sec2x. To get this, you divide the first identity by cos2x so that it will cancel and is equal to one. It will look like this: 1+(sinx/cosx)2=(1/cosx)2. We know that sine is equal to y/r and cosine is equal to x/r, so you can cancel out the r's and end up with y/x. This would mean that it's tangent, so you substitute it in. Next, 1/cosx is secant because cosine is x/r but it's the denominator, so you take the reciprocal of it. The third pythagorean identity: 1+cot2x=csc2x. To get this, you divide the first identity by sin2x so that the sin2x can cancel and equal one. So far it will look like this: 1+(cosx/sinx)2=(1/sinx)2. We know that x/r is cosine and y/r is sine, so we can cancel out the r's and get x/y, which is cotangent. 1/sinx is (1/ y/r), to get rid of the fraction you do the reciprocal which is r/y, and that would be cosecant.

Inquiry Activity Reflection:
1. "The connections that I see between Units N, O, P, and Q so far are..." they all relate to the Unit Circle and they incorporate the Pythagorean Theorem one way or another

2. "If I had to describe trigonometry in three words, they would be..." Ratios, "Unit Circle", FUN.

Monday, March 17, 2014

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

This WPP 13-14 was made in collaboration with Diana Pham. Please visit the other awesome posts on her blog by going here.

Create your own Playlist on LessonPaths!

Saturday, March 15, 2014

BQ#1: Unit P Concept 2 & 5 - Law of Sines & Area of an Oblique Triangle

2. Law of Sines
SSA (Side Side Angle) is considered ambiguous because there is no one sure solution. The problem only gives us one angle of the triangle so there is some ambiguity of what the triangle looks like. There could be one possible solution, two possible solutions, or no solutions.
This connects to the Unit Circle trig function values because the Law of Sines can give us the measures for each line segment on the Unit Circle. In Unit N, we learned that the unit circle has a radius that is measured to be one, however, it is not always the case. As a result, the Law of Sines help us correlate the angles with the length of their sides.

4. Area Formulas
The "area of an oblique" triangle is derived from the regular area equation, A=1/2bh. In the triangle below, you can see that there is perpendicular line through the triangle, creating two triangles. If we are looking at angle C, we know from previous units that SinC=h/a , therefore, h=aSinC. Now, just substitute it into the regular area equation: A=1/2b(aSinC). The equation we just found, however, is only if the problem gives you angle C. The equation for angle A: 1/2bcSinA. Equation for Angle B: 1/2acSinB. We use the same process to get the area for angles A and B. This relates to the area we are familiar with because we have to use that equation in order to get the area for an oblique triangle. Like I said before, you plug what h (height) is into the normal area equation.
SSS Packet

Tuesday, March 4, 2014

I/D #2: Unit O Concept 7-8 - Deriving Patterns for the SRTs

Inquiry Activity Summary:
1. For a 30-60-90 triangle, we must cut it straight down because it was currently an equilateral triangle, meaning all the angles are the same. Since we cut it in half, the length will be cut in half as well, making it 1/2. Since it was given that the hypotenuse side is one and we found out that the length is 1/2, we just need to find the height. To find the height, you use the pythagorean theorem which is a^2+b^2=c^2 (look at picture for mathematical explanation). To get rid of the fraction, you multiply everything by two. We use "n" as if it were equal to one. As a result, the hypotenuse is 2n, the side across the 60 degrees angle is n radical 3 and the last side is n. We use n because it's a variable that represents 1 since it's in all the terms.

2.  We can derive a 45-45-90 triangle from a square by cutting it diagonally. Cutting it diagonally will split the square in half as well as the 90 degree angles, making it 45 degrees. We know that the sides opposite from the 45 degree angles are still 1 because we didn't cut any sides in half, just the angles. To get the hypotenuse, you use the pythagorean theorem. (1)^2 + (1)^2 = c^2, C should equal radical 2 according to the work shown below. Also, we use n as a constant or a variable which represents 1. Using "n" represents the relationship between the sides and angles.

Inquiry Activity Reflection:
1. "Something I never noticed before about special right triangles is..." That there is a reason why the sides are the way they are, it's not just memorization.
2. "Being able to derive these patterns myself aids my learning because..." There is logical reasoning in deriving the patterns and I won't have to memorize it like I did before.

Friday, February 21, 2014

I/D #1 - Unit N Concept 7: How do SRTs & the UC relate?

Inquiry Activity Summary:
The unit circle activity is to help us develop our understanding of the special right triangles and how it correlates with the unit circle. There are three special right triangles: 30º, 45º, and 90º. The three sides of the triangle has to be simplified in such a way that the hypotenuse is equal to one because when inserting the triangle into the unit circle, the hypotenuse would be the radius, which always have to be one.

1. Rules for 30º triangle:
  • Hypotenuse (r) is 2x *Remember to change it to 1 by dividing 2x*  
  • Horizontal value (x) is x *When divided by 2x, it will make 1/2* 
  • Vertical value (y) is x radical 3 *When divided by 2x, it will be radical 3/2*
  • The 30º angle must be at the origin (0,0)

2. Rules for 45º triangle:
  • Hypotenuse is radical 2 but remember to divide by radical 2 to make it 1
  • Horizontal value is 1 over radical 2, when divided it'll be radical 2 /2
  • Vertical value will be the same as horizontal value
  • 45º angle must be at origin

3. Rules for 60º triangle:
  • Hypotenuse is 2x, divide by 2x and it'll become 1
  • Horizontal value is x radical 3, after dividing it'll be radical 3 / 2
  • Vertical value is x, after dividing it'll be 1/2
  • 60º must be at origin

4. The special right triangles activity helps us derive the unit circle because the points from the triangles will form a circle if all the points are connected. But of course, the 30º, 45º, & 60º are only in the first quadrant. So, you would have to find the other coordinate pairs, however, once you know quadrant 1, you basically know all the other quadrants.

5. The triangles drawn in this activity lies in quadrant 1. The values will change with positive or negative depending on which quadrant you're in. If you're in the 2nd quadrant, then your x value will be negative and your y value is positive. If it's in the 3rd, both the x and y value will be negative. If it's in the 4th, x value will be positive and y value negative.
In this 2nd quadrant, I showed the 30º reference angle,
the actual angle is 150º. As you can see, the coordinate pair
farthest to the right, is similar to the 30º angle in quadrant I.
The only difference is the x value is negative.
In this 3rd quadrant, I showed the 45º reference angle, the angle
is 225º. The coordinate pair closest to the bottom has both a negative
y and x value. It's also similar to 45º in quadrant I.
In quadrant IV, I showed the 60º reference angle and the angle
would be 300º. The coordinate pair closest to the bottom has a negative
y value and a positive x value. This angle is similar to the 60º angle
in quadrant I.

Inquiry Activity Reflection:
"The coolest thing I learned from this activity was..." The unit circle consists of special right triangles and it's not just a bunch of numbers you have to memorize; there's a pattern.

"This activity will help me in this unit because..." I'll be able to comprehend the unit circle without having to memorize random points.

"Something I never realized before about special right triangles and the unit circle is..." That just by knowing the first quadrant, you pretty much memorized the entire unit circle.

Monday, February 10, 2014

RWA#1: Unit M Concept 5 - Conic Sections in Real Life (Parabola)

1. Definition: The set of all points that are equal distance from a point (focus) and a line (directrix)

2. Algebraically, the standard equation is (x-h)^2=4p(y-k) or (y-k)^2=4p(x-h), depending if the graph goes right, left, up, or down. The parabola will go up if x appears first and p is positive. It will go down if x appears first and p is negative. It will go right if y is first and p is positive, and left is when y appears first and p is negative. Your vertex is (h,k) and the directrix is a dotted line that is p units away from the vertex, in the opposite direction of the focus. The axis of symmetry will be perpendicular to the directrix and it will always go through the vertex and the focus. In addition, you don't have to find the eccentricity because it is always be equal to one. 
    Graphically, the parabola goes in four directions depending on the x and y, as mentioned above. The vertex is in between the focus and directrix (FVD or DVF). The focus will always be on the inside of the parabola. As seen on the picture above, the point from the focus to a point on the parabola to the directrix will be the same distance.

Here's a cool video to learn more on parabolas: This video will show you that there are parabolas in video games such as Mario. It shows how the parabola would become skinnier/fatter if he ran while jumping or jumped normally.

3. Real World Application: Not only does parabolas appear in video games (shown in the video above), but they also appear on bridges, roller coasters, and satellite dishes. The picture displayed below is a Roman aqueduct, it would help channel water and was a major component in the Roman water supply. A more sophisticated example would be the satellite dishes. The structure of the dish enables it to transmit and detect signals from a specific source. When the signal bounces off of the dish it creates a parabola.

4. Work Cited: