BQ#7- Unit V

Where does the formula for the difference quotient comes from?

The difference quotient formula comes from the secant line that touches the graph twice meaning there's two points. The first point is ( x , f(x)) and the second point is (x+h , f(x+h)), from there we find the slope of the secant line which is y2-y1/x2-x1. We substitute the y and x with the given points from the secant line and will turn out to be f(x+h) - f(x)/x + h -x. Then we see that there are variables that cancel each other -x and +x from the denominator then just leaving "h". That's how you get the difference quotient formula f(x+h) - f(x)/h. 

BQ#6- Unit U Concept 1-8

1) What is continuity?
Continuity is predictable and it has no breaks, no holes and no jumps. A continuity function can be drawn with a single, unbroken pencil stroke for example the limit as x approaches a number of f(x) is equal to a number, the intended height. 

What is discontinuity?

Discontinuity is the opposite of continuity meaning it's interrupted and the limit does not exist. There are two families of discontinuity, removable and non removable. Removable has point discontinuity meaning there's a hole in the graph. Non removable has jump-it jumps from one graph the other-,oscillating-the graph is wiggly meaning it never reaches a point- and infinite-when it has vertical asymptotes due to unbounded behavior. 

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 when a value is the actual height of a function. The limit exist when the left and right behavior are the same, when there's no unbounded behavior and when there's no oscillating behavior. A limit doesn't when the left and right bahavior are different, there's an oscillating behavior and there's unbounded behavior. 

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

We evaluate limits numerically, verbally and algebraclly. Numerically means its set on a table to find the intended the height of a function. Verbally is stateting the limit statement as the limit as x approaches a number of f(x) is equal to a number. Algrebraclly has three methods substitution, dividing & factoring out and rationaling & conjugate. 

Direct substitution is when it's substituted by approaching number and has three answers numerical-2-, zero-0/2-, and undefined-1/0. 

When its inderteinate meaning its 0/0 then the dividing & factoring method is use. In this case we factor out both the numerator and denominator to cancel out common terms. Then substitute and get either numerical-2-, zero-0/2-, or undefined-1/0. 

In rationaling & conjugate is when we conjugate the numerator or denominator depending where the square root is at. After common terms are cancel then we substitute to get the answer. 

Work cited 

SSS packet

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

1. 30 Triangle 

The triangle is at quadrant one and label the three sides. Hypotenuse is 2x, adjacent is x radical 3, and opposite is x. To get the values we will need to derive them. For hypotenuse divide by 2x to get one (r value), for adjacent divide by 2x to get radical 3/2 (x value), for opposite divide by 2x to get 1/2 (y value). Finally draw the coordinate planes (0,0) as the orgin, (radical 3/2,0) as the x axis, (radical 3/2,1/2) as the y axis. 

30 triangle

2. 45 triangle

 The triangle is at quadrant one and label the three sides. Hypotenuse is x radical 2, adjacent is x, and opposite is x. Then derive to get the values. For hypotenuse(r) divide by x radical 2 to get one, for adjacent(x) divide by x radical 2 to get 1/radical 2 then rationalize to get radical 2/2, and for opposite(y) divide by x radical 2 to get  1/radical 2 then rationalize to get radical 2/2. Finally draw the coordinates planes, (0,0) as the origin, (radical 2/2) as both the y and x axis. 

45 triangle

3. 60 triangle 

The triangle is at quadrant one and label the three sides. Hypotenuse is 2x, adjacent is x, and opposite is x radical 3. Then derive to get the values. For hypotenuse(r) divide by 2x to get one, for adjacent(x) divide by 2x to get 1/2, and for opposite(y) divide by 2x to get radical 3/2. Finally draw the coordinate planes, (0,0) as the origin, (1/2,0) as the x axis, and (1/2,radical 3/2) as the y axis. 

60 triangle

How does this activity help you to derive the Unit Circle?

This activity helps me derive values that are in the unit circle. Also the values are either positive or negative depending on what quadrant it lies on. All the values and trig functions in quadrant one are positive, in quadrant two the x axis is negative and sine and cosecant are positive, in quadrant three both x and y axis are negative and tangent and cotangent are positive, and for quadrant four the y axis is negative.

THE COOLEST THING I LEARNED FROM THIS ACTIVITY WAS that the triangles and its coordinate planes make up the entire unit circle. Also when dividing with the value of hypotenuse on the adjacent and opposite sides gets us the values from the unit circle. 

THIS ACTIVITY WILL HELP ME IN THIS UNIT BECAUSE now I know what order pair goes in what degree all thanks to the 30, 45, 60 triangles. Also where the values and trig functions are positive or negative. 

SOMETHING I NEVER REALIZED BEFORE ABOUT SPECIAL RIGHT TRIANGLES AND THE UNIT CIRCLE IS that deriving the triangles to get the same values from the unit circle would be easy get. Also that the unit circle was made from the 30, 45, 60 triangles. 

BQ#4 - Unit T Concept 3

Tangent goes uphill because its positive in quadrants one and three and negative in quadrants two and four. Its positive in zero uphill and pi uphill etc. The asymptotes are in zero, pi/2, pi, 3pi/2 and 2pi because x is zero. The tangent graph starts at negative and goes uphill to be positive and to go to the next quadrant.
Cotangent goes downhill because its postive in quadrants one and three and negative in quadrants two and four, just like tangent. Its negative downhill pi/2 and 3pi/2 etc. The asymptotes are in zero, pi, and 2pi because y is zero. The cotangent graph starts positive and goes downhill to be negative its like the reciprocal of tangent.

BQ#3 - Unit T Concepts 1-3

First of all sine and cosine dont have asymptotes when the other graphs do have asymptotes. Tangent related to cosine when cosine is zero because it becomes undefined and the asymptote are in pi/2 and 3pi/2. For cotangent related to sine when sine is zero because it becomes undefined and the asymptotes are in 0 and pi. For secant when cosine is zero then secant is undefined so there's going to be asymptotes at pi/2 and repeat every pi unit. Finally for cosecant when sine is zero then cosecant is undefined like secant, so there's asymptotes at pi and continue every pi unit.

BQ#5 - Unit T Concepts 1-3

The reason for sine and cosine not having asymptotes is because r equals 1 that makes them undefined. Sine is y/r=1 and cosine is x/r=1, there needs to be a zero at the denominator to have asymptotes. Unlike tangent(y/x) and secant(r/x) have asymptotes at 90 and 270 degrees because that's where x=0. Cosecant(r/y) and cotangent(x/y) that have asymptotes at 0 and 180 degrees because that's where y=0.

BQ#2 - Unit T Intro

The quadrants from the unit circle, when placed horizontally in numerical order, create the basis for the trigonometric graphs. Since sine is positive in quadrants A and S and negative in quadrants T and C, when unwrapping it starts positive and positive then negative and negative and start again in another revolution.  For cosecant its positive in  quadrants A and C and negative in quadrants S and T, when unwrapping it starts at positive then negative then negative and end positive. That's how trig functions relate to the unit circle. 

The reason sine and cosine is 2pi because it needs all four quadrants to repeat unlike tangent and cotangent needs only half that why it's just pi. 

Sine and cosine has an amplitude of 1 because it can only be between -1 to 1 unlike the others there's no limit so their amplitude can greater than 1. 

Reflection #1 - Unit Q: Verifying Trig Identities

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

To solve for a trig identity the answer has to be true no matter what beacuse one side of the equation has to match the other side. To do that several steps has to be made. In a regular equation you can work on both sides but in an identity you can't because it won't prove the identity to be true. 

24. What tips and tricks have you found helpful?

Knowing the three types of identities help to verify the equation. The unit circle came in handy too because we had to see were it landed in the unit circle. For concept 4, its easy to plug in equation ones it's verify, to see were it lands.  

24. 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.  
25. When solving an identity i look to see if there's any identities or if can be split apart. From there I see how can manipulate the equation to verify it. There are so many ways to verify its like a puzzle. Ones i see the end of the equation i start to verify to get my answer. 

SP#7:Unit Q Concept 2

This SP7 was made in collaboration with Jorge Barroso.  Please visit the other awesome posts on their blog by going here.

The viewer needs to pay attention on how to use identities and remember how to use SOHCAHTOA from unit O concept 5.

Pert 1:Identities

Part 2:SOHCAHTOA

Part 2

Part 2

Part2

Part 2

I/D#3: unit Q-Pythagorean Identities

• First of all an identity is a proven fact that is always true. The Pythagorean Theorem when using x,y and r, it's x^2+y^2=r^2. Then divide everything by r^2 to get one. It will look like this (x^2/r^2)+(y^2/r^2)=1. Since cosine is x/r and sine is y/r. So it will be replace to be cos^2x+sin^2x=1. That's why it's referred to the Pythagorean Theorem. The magic 5 pairs to make an identity true are (1/2,3/2),(2/2,2/2),(3/2,1/2), (0,1) and (1,0). The connection between units N, O, P, and Q so far are the magic 5 pairs or the unit circle and using sin, cos and tan as well as its reciprocals. If I had to describe trigonometry in THREE words, they would be complicated, long and strategically to solve. 
• 
  
• For further understanding visit this websites, number three is a video explaining the process:
• 1.http://math.tutorvista.com/geometry/pythagorean-identities.html?view=simple
• 
  
• 2.http://www.purplemath.com/modules/idents.htm
• 
  
• 3.http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-27-trig-integrals/

WPP#13&14:unit P concept 6&7

Please see my WPP13-14, made in collaboration with Jorge Barroso, by visiting his blog http://jorgebperiod1.blogspot.com/2014/03/wpp-13-14-unit-p-concepts-6-7.html.  Also be sure to check out the other awesome posts on his blog. 

Unit P concept 3-5

3. Law of Cosines
Why do we need it? How is it derived from what we already know?

We need law of cosines to find the third side when knowing two sides and an angle (SSA) or knowing all three sides (SSS) to find the angles.

We got a triangle label ABC that's not a right angle. Angle A's point is (0,0), angle B's point is (ccosA,csinA), ccosA is the x value because cosA equals d/c and csinA is the y value because sinA equals h/c. Angle C's point is (b,0), b is the distance across from angle B between angle A and C. Draw a line down angle B to make h, h is the drawn height of the triangle. With that it makes a 90 degrees angle. Then label side c and d also ccosA. Side c is the distance across from angle C between angle A and B. Side d is the distance across from angle B in just the right triangle. Side a is the distance between angle B and C (see picture 1). Use the Pythagorean Theorem in triangle CBD but substitute h to csinA and r to ccosA. Foil to get b^2-2bccosA+c^2cos^2A (see picture 2). Put cos^2A + sin^2A in parenthesis, since since cos^2A + sin^2A equals one. Finally get the formula (see picture 3), it can also be used to produce other letters statements (see picture 4). That's how we derive law of cosines.

Picture 1

Picture 2

Picture 3

Picture 4

Refrence:
http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

5. Area formulas
 Draw out a right triangle with angle A equal to 35 and angle B equal to 65. Side b equal to 4. Use the law of sines to get side b and c. Use the area of an oblique triangle, and Heron’s area formula to get both areas (see picture 5).

Picture 5

 Work cited

http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm

WPP#12 Unit O concept 10: Solving angle of elevation and depression word problems

A. Juanito at ground level measures the angle of elevation to the top of the cliff is 44 degrees. If he is 7 feet away from the cliff, what is the height of the cliff?

B. Junito climbs to the top of cliff and wants to glide down the cliff. The angle of depression is 47 degrees. The landing site is 35 feet apart, how tall is the from that side?

Remember opposite, adjacent, and hypotenuse and SOHCAHTOA.

I/D#2 unit O-How can we derive the patterns for our special right triangles?

For the 30-60-30 triangle label all the sides by 1. Divide the triangle in two. That will divide the 60 degrees in to 30 degrees, the adjacent in half and make a 90 degree. Use the Pythagorean theorem to get the opposite side. Multiply by 2 to leave the radical alone and get 1 in the adjacent side and get 2 in the hypotenuse side. 

 For a 45-45-90 triangle label all the sides by 1. Divide the square to get two triangles. That will divide the 90 degrees into 45 degrees. Use the the Pythagorean theorem to get the hypotenuse side.  

 We use "n" to reference the missing value.

Something I never noticed before about special right triangles is, why we use "n" for?

Being able to derive these patterns myself aids in my learning because it gives a picture of both triangles and its formula. 

This are pictures of both triangles and showing how to get their formulas.

30-60-30 triangle

45-45-90 triangle

RWA#1:unit M concept 5

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

 

"An ellipse is the set of points for which the sum of the distances from the foci is a fixed constant."

The equation for an ellipse is (x-h)^2/a^2+(y-k)^2/b^2=1. To identify an ellipse it has both sides squared, its adding and different coefficients. To get the equation is by completing the square. First group the Xs and Ys and move the constant to the other side. Next find the GCF of X and Y and put on the other as well. Then factor both perfect square trinomials and simplify the other side. Finally divide everything by what it equal to get 1 and to reduce the the fractions.

An ellipse looks like a stretch circle because its eccentricity is between 0 and 1. To get the key points of an ellipse it must be in standard form. An ellipse must have its shape skinny or fat, a center, a= with its two vertices and major axis, b= with its two co vertices and minor axis and c with its two foci and eccentricity. If the bigger number lies under x its fat but if it's under y its skinny. The center is (h,k). The major axis is horizontal if the bigger number is under the x term and if its under the y term its 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 focus determines how stretch out is the ellipse, if the focus is closer to zero it makes a circle shape but if its close to one it makes a parabola shape. To find a missing value use the equation c^2=a^2-b^2.

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

Work cited

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

http://kirchmathanalysis.blogspot.com/

WWP#10: Unit L concept 9-14: probability

Create your own Playlist on LessonPaths!