BQ#7

you know there is a tangent line on the graph when the line only touches the graph once as you can see in the picture

on thispicture the line touches the graph twice which meants it is a secant line. this changes which the coordinates are (x+h, f(x+h) )

with this we can derive the quotient formula and by that we plug in the secant coordinates to our slope formula which is y2-y1/x2-x1 when we plug everything in all we need to do is cancel the x's and end up with the formula which is  (f(x+h)-f(x)) /h

BQ6 continuity

Continuity has no breaks no jumps no holes, you can draw it without lifting your pencil , it is predictable and when the limit and value are the same it is continuous.

discontinuity is broken down into two families- removable and non removable they are broken down into two because one has limits and the other doesn't . a discontinuity has limits. the non removable discontinuities re jump infinite and oscillating

A limit is an intended value it exists when it is a continuous function or if it is a hole a limit does not exist when there is a vertical asymptote because it has unbounded behavior or when it is oscillating and when it jumps because it is different from left to right. the difference of a limit and a value is that a limit is an intended height and a value is the actual height

We evaluate limits numerically by using a table, graphically we just go from left to right going toward the same point, algebraically nu using substitution which is plunging in the limit and if we get 0/0 we use different algebraic methods that can work

BQ#4

Tangent Graphs

Tangents graphs g uphill while cotangent graphs go downhill. the reason being is because of their asymptotes. these you can get by your unit circle. the ratio for tangent is y/x and when x=0 we get an asymptote and when that happens is when it is at 90 and 270 degrees which convert to pi/2 and 3pi/3 so knowing this we know that the graph cant touch it but what also has the affect s the quadrants in a unit circle we know that tangent is positive in the 1 and in the 3rd and negative in the 2 and 4th so this shows you were the part of the graph is going to be above or below the x axis

The same goes to cotangent. its ratios are x/y so this time y has to = 0 and when it does it is at the 0 and 180 degree and the asymptotes for that is pi and 2pi even though that have the same four quadrants the asymptote are the ones that make this graph go downhill instead of uphill like tangent

BQ#3

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Tangent the picture graph on the top shows sine cosine and tangent. since tangent is y/x we know that x is cosine so we get asymptotes whenever the denominator = 0 so that means whenever cosine =0 there will be an asymptote. if one of sine or cosine is negative tangent will be negative
Cotangent is on the bottom of the picture. here we know that cotangent is the opposite tangent so the ratio is x/y so now whenever y=0 there will be an asymptote. but if sine and cos are positive the graph is going down hill if its negative it goes up.

secant is on the top picture as you can see wherever cosine =0 there you can find secant it looks like a parabola because there are asymptotes

 for cosecant It is the last picture as you can see it is the same as secant but you can find cosecant whever sine =0 and it too has asymptotes which give it its shape
 

BQ#5

BQ#5
 Sine and Cosine do not have asymptotes the reason being that sine and cosine will never have a denominator equal 0.when the denominator equals zero it means it is undefined and when it is undefined it will give u an asymptote. if we go back to the unit circle we know that sine and cos will always be over one so it can ever be over zero there for not having asymptotes.  but tangent and the others do because they can be over zero.

BQ #2 Unit T Intro

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BQ#2

There is a period for sine and cosine that is 2pi, it is 2pi because it is one revolution of the unit circle, it is how much times it takes to repeat its cycle. When we use A,S,T,C on out graph we know that Sine is positive n quadrant 1 and 2 and will be negative on 3,4 . now for cosine it is positive on quadrant 1 and 4 but will be negative on 2 and 3 . as you can see it takes 4 to repeat the pattern which is why it takes 2pi but with tangent it only takes pi because it is positive negative then positive and negative again so it repeats after the second one.


the fact that it has amplitudes it makes sine and cosine have restrictions. the restriction are 1 and -1 with this restrictions they cant go past a certain paint un like tangent. if we try any number it wont work because of the restriction so its the same like the unit circle.

Reflection #1 unit Q Trig identities

1.  To verify is to prove an identity. you have to use logical steps and identities to show how you can make one side equal the other.

2. the tricks I found help full was to feel my identities next to me and write everything down because it keeps track of what you are doing.
3. I usually just try to convert everything to sine and cos. I try to look at the problem as a whole and see if anything can cancel out or steps I can take to solve it. it takes me a while but after I know something isn't going to work and move on to try something else I can end up solving .

SP7: Unit Q Concept 2 IDENTITIES

This WPP13-14 was made in collaboration with Louis Troung.  Please visit the other awesome posts on their blog by going here.

 

 

WE CAN FIND THE SAME THING USING SOHCAHTOAH

I/D#3: Unit Q Concept 1: Pythagorean Identities

INQUIRY ACTIVITY SUMMARY

     Pythagorean identities come from the Unit Circle and the Pythagorean Theorem. An Identity is an equation that is true no matter what values are chosen. The Pythagorean Theorem kind of looks like an  identity because even if you have 2 sides you can always find the 3rd using the theorem. The Pythagorean theorem is a^2+b^2=c^2 and the Pythagorean identity is cos^2(theta) + sin^2(theta) = 1. but on a unit circle it is x+y=1  we are going to substitute them so a is going to be x,  y is b and 1 = r so 1 is c as you can see. r=1 because r is the radius of the circle and the radius of a unit circle is always going to be 1.

\what you can notice about how cosine =x and sine= y that it really does relate because the equation is cos^2(theta) + sin^2(theta) = 1 and you can see cos = x in the equation and sin=y and 1 =1. just like the Pythagorean theorem they are squares and like the unit circle they are x and y. and this is one of the identities.  

 2nd identity

 

3rd identity
1. the connection I have seen between Units N,O,P, and Q are that we can find angles on the unit circle and the sides and angles inside triangles that are missing using law of sines and cosines
2) If I had to describe trigonometry in THREE words, they will be hard, understandable, and progressive.

Wpp13-14:concepts6-7 applications law of sines and cosine

WPP 13 & 14
 This WPP13-14 was made in collaboration with Louis Troung.  Please visit the other awesome posts on their blog by going here.
 Stephanie decides to go on a vacation with Louis after all her soccer tournaments were over.they end up going to a beachside house. there are two boats but they are both different colors one is blue and the other is red. Stephanies favorite color is blue just like louises so they are going to race for it. Stephanie is 10 feet away from Louis . Stephanie knows she ahs to turn
S30E and Louis need to turn N32E to get to the boat. which one of them is closer to the boat?

this is the visual we get from the word problem

 once we start using the hints we got for the angles we can now see that this triangle is an angle side angle and we can figure this out by using law of sines

as you can see in this picture we are going to figure out the missing angle that is left there bcause it will be the one pair we need to use law of sines we are going to subtract 180 from 32 and 30 because all triangle angles add up to 180 so that's how we are going to find out the missing angle after that you will just stat using the law of sines as you see in the picture leaving us with Louis being closer thus him getting to the boat faster.


On a nice day Stephanie and louis decide to go boating on our boats ! Stephanie and louis are 10 miles apart . Stephanie and louis know that louis has a bearing of 043 to Stephanie. The light house is 40 miles away from louis to the east I him .  Stephanie wants to go to the light house. How far is she from the light house ?

BQ#1 Unit P: Concepts 1 and 4 law of sinces and oblique trangles

Concept 1: Law Of Sine

We need to use law of sines when we need to find sides of a triangle when it isn't a special right triangle. But we can only use it if the trangle is AAS or ASA


We can derive the law of sines with this video on top .

Concept 4 Area Of Oblique Triangles.

we can derive this from the area of a triangle. we have information about the triangle but not about the height so we are missing that part for the equation so what we can do is use the area of oblique triangles. the equation is A=1/2 *(2 sides)*sine of the angle that is given. we can use this because as you can see on the picture above we used sine of c anf that will be sinc= h/a . after that we will multiply a to both sides and get H=asinec  and since we needed h we plug that in into the area of the triangle and that's how we get our equation. we can work out a problem on the pictures on bottom

WPP#12 Unit O Concept 10 finding elevation and depression

Stephanie is playing a soccer game and when she kicks the ball she accidentally hits a girl in the face. this girl is 5.5 feet tall that's from her foot to her head the angle of elevation the ball went was 24.5. knowing this how far did the ball travel .

Stephanie has won a toy helicopter at the soccer game and it was 15 feet up in the air. she wanted too ply with it by making it land like an airplane, the toy helicopter is 20 feet away from where she wants to land it what is the angle of depression

ID#2 Unit O Concept 7-8

INQUARY AVTIVITY SUMMARY

 

 

So basically in this paper we are going to derive an equilateral square and triangle. With these we are going to get a 45-45-90 triangle and the 30-60-90 triangle. when we cut the in half we will get these special right triangles and for us to solve the missing side we are logically going to use the Pythagorean  theorem.

45-45-90 Triangle

In this picture you can see what we are starting off with, it is an equilateral square.

Here we can see that we already cut it in have so since all sides are 90 when you cut it in half your other angles are going to be 45 degrees.

Here I am doing the work for the use of the Pythagorean theorem. we will use a^2+ b^2 = c^2 since we already have both sides it is obviously one and since we have to take the radical to get c alone it will end up being radical 2 for out hypotenuse answer.

Here we can see that the both sides will always be the same number so we can go ahead and substitute it got "n" and then for the hypotenuse we will always have to multiply with radical 2 so for example we have one side of the 45 degree angle 5 the other side must me 5 and the hypotenuse 5 radical 2.

  30-60-90- Triangle

Here we can see how we have our equilateral triangle

now here we have already but the triangle in half and we have all the degrees there too. all sides on an equilateral triangle are 60 degrees and when you cut it in half we will have 30 and 30 degrees on the sides.

So when we have to do the Pythagorean theorem we can see that the bottom will have to split into 2 so one divided by 2 is one half. when we have that we can plug everything in. we edn up getting radical 3 over 2 for side b.

 

Here we can see that we really don't want to use the radial 3 over 2 because it will make it difficult so we are going to multiply everything by 2 to make it easier when we do we are going to add an n to represent any number so when we have that we and up getting one for side a radial 3 for side b and 2 for the hypotenuse but got this triangle there are different degrees so the 60 side as you can see has the radical 3 so it will always have to do with that. when we have n equal to 6 side b will be 6 radial 3 and the hypotenuse will be 12. when we do the problem backward for example and b equals 8 we will have to go back to the a side by dividing by radial 3 which will give us 8 rad 3 over 3 and the hypotenuse 16 rad 3 over 3 .

INQUIRY ACTIVITY REFLECTION

1. Something I never noticed before about special right triangles was that they all had the same ration between each other and how we found that out.
2. Being able to derive these patters myself aids in my learning because i know how i got there and how those triangle were made

 

I/D 1 Unit N Concepts

1. A triangle with one right angle is called a right triangle. The side opposite the right angle is called the hypotenuse of the triangle. The other two sides are called legs. The other two angles have no special name, but they are always complementary. Do you see why? The total angle sum of a triangle is 180 degrees, and the right angle is 90 degrees, so the other two must sum to 90 degrees. One is the right triangle formed when an altitude is drawn from a vertex of an equilateral triangle, forming two congruent right triangles. The angles of the triangle will be 30, 60, and 90 degrees, giving the triangle its name: 30-60-90 triangle. The ratio of side lengths in such triangles is always the same: if the leg opposite the 30 degree angle is of length x , the leg opposite the 60 degree angle will be of x , and the hypotenuse across from the right angle will be 2x . Here is a 30-60-90 triangle pictured below. Which side is which? The side opposite the 30 degree angle will have the shortest length. The side opposite the 60 degree angle will be sqrt(3) times as long, and the side opposite the 90 degree angle will be twice as long. The triangle below diagrams this relationship.

30-60-90 triangle side ratios proof: Proving the ratios between the sides of a 30-60-90 triangle

(http://www.khanacademy.org/math/geometry/right_triangles_topic/special_right_triangles/v/30-60-90-triangle-side-ratios-proof)

2. 45-45-90 triangle side ratios: Showing the ratios of the sides of a 45-45-90 triangle are 1:1:sqrt(2)

(http://www.khanacademy.org/math/geometry/right_triangles_topic/special_right_triangles/v/45-45-90-triangle-side-ratios)

The other common right triangle results from the pair of triangles created when a diagonal divides a square into two triangles. Each of these triangles is congruent, and has angles of measures 45, 45, and 90 degrees. If the legs opposite the 45 degree angles are of length x , the hypotenuse has a length of x . This ratio holds true for all 45-45-90 triangles. 45-45-90 triangles are also often called isosceles right triangles.

3.

4.This Activity help us derive the unit circle by knowing the lengths . of we forget something about the unit circle

RWA #1 Unit M Concept 5 elipses

1. Ellipse - The set of all points such that the sum of the distance from two points is a constant.

The ellipse


(http://www.physics.unlv.edu/~jeffery/astro/ellipse/ellipse_001.png)

2.  To find the ellipse it has to be in standard conic form which is (x-h)^2/a^2+ (y-k)^2/b^2 = 1. It doesn't matter in what order a^2 and b^2 are they can be under whichever equation. h and k have to be in a specific one. H always has to be with x and k always has to be with y. another thing is that to determine the size of the ellipse if it's skinny or fat has to be based on x or y. If there ia a bigger number under x then it's fat, but if it's under the y then it's skinny. As you can see on the picture whatever d and d 2 are when they are added they equal 2a.
    When graphed h,k will be center points. What ever numbers are on the bottom like we said then it  place will be a and the second one b.vertices come from a, co-vertices from b, and foci from c and we find c from a^2-b^2=c^2. If a has bigger numbers then it's our major axis and b will be minor axis. The foci will be on the major axis but a little before the vertices points. To determine the eccentricity of an ellipse,is e = c/a. So it's whatever number is c comes out to be divided by whatever number a is. If it is still not clear the video on youtube below this will end up explaining how to solve the problem.

(http://www.youtube.com/watch?v=o9EXI24plxo)

3.You can find ellipses in a training machine. "An elliptical training machine simulates the motion of running or climbing to provide the user with a healthful exercise without any impact on the joints." this helps the people to have better results little by little.
  "The foot of a user describes the shape of an ellipse as the machine is used. An elliptical machine can be motor-driven or user-driven, as well as dual action, where handlebars and leg mounts interdependently provide motion for each other. The elliptical trainer provides a stationary exercise for those who wish to avoid joint injury as a result of running."
(http://www.ehow.com/info_8522010_real-life-uses-ellipses.html) this ellipse figure really helps the people that are trying to avid injuries so ellipses have a big impact on that.

4.

•  http://www.physics.unlv.edu/~jeffery/astro/ellipse/ellipse_001.png
• http://www.youtube.com/watch?v=o9EXI24plxo
• http://www.ehow.com/info_8522010_real-life-uses-ellipses.html

WPP10 Unit L Concepts 9-14 probability

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