BQ. #7 Unit V

Where does the difference quotient come from?? (f(x+h)-f(x))/h


This graph will help to get a visual on what we are talking about. A secant line is present in this graph. (Secant line- a line that touches the function at two points.) The very first point is know as (x,f(x)) and the second point as (x-h, f(x-h)) since it has an h difference from the first one. (As seen on this graph.)

 You then plug this into the slope formula, m=(y1-y2)/(x2-x). The x's will cancel in the denominator and just leave h on bottom. For the numerator we just leave it as it is because sometimes x is squared so we must use (x+h)^2, The final equation should be (f(x+h)-f(x))/h

The top picture can be found here.

BQ: #6 Unit U

Continuity Vs. Discontinuity

1. Continuity is when the function has no breaks, holes, or  jumps. They can be written/drawn without lifting your pencil off the paper.
 <-- (they look like this)
Discontinuities can't be written without lifting your pencil. Discontinuities are separated into two different families: removable and non- removable discontinuities. The only discontinuity in removable Discontinuities is the point discontinuity.
Point Discontinuity- Known as a hole. The limit still exists but it isn't the same as the actual value since the limit is the intended height of the function and the value is the actual height.
Non Removable Discontinuities:
Jump, Oscillating, Infinite discontinuity (unbounded behavior)

2.Limits and Values

Limits are the intended height of the function and the value is the actual height of the function. When the value and the limit are the same that means that the graph is continuous.  

The limit does exist in Removable Discontinuities because even though the value might be undefined or an actual number, (depending if its just a whole or the actual point is somewhere else on the graph) the graph still has an intended height.

The limit doesn't exist at Non-removable Discontinuities because it doesn't approach a single value at all.

3. V.A.N.G

V- verbally which is more like actually speaking it out so it isn't too hard.

A- algebraically. There are three different ways to evaluate limits.

   -Direct substitution- This concept is easiest in all honesty. Nothing really hard about grabbing your calculator and just plugging in what approaches f(x) into the calculator. You can either get 0/(any number) or (any number)/0 (undefined.) When it comes out being 0/0, this is when you have a problem. This is called indeterminate form and you have to use either dividing out/ factoring out or the rationalizing/ conjugate method.

   -Dividing out/ Factoring out Method- In this method, you have to factor both the numerator and denominator and cancel out some common terms to remove the zero in the denominator. Then use direct substitution with the simplified expression. This is used for removable discontinuities to actually remove the discontinuity.

   -Rationalizing/Conjugate Method- The conjugate is where you change the sign in the middle of two terms (3x+1 changes to 3x-1). Sometimes we use the Conjugate of the denominator while other times we use the conjugate of the numerator. (It depends on where the radical is.) Only foil the conjugate with the original part not the conjugate with the non conjugate. (we hope something will cancel out.)

N-Numerically.

With this concept we just use a graph to find the limit. If the limit can be reached then the limit would be in the middle. If it can't be reached that means it doesn't exist Because it never reaches a certain value. the Limit from the left and the right just get really really really really really really really close.

G-Graphically. You can just plug it into the calculator and hit trace with a certain value. If it is error then the limit can't be reached. But if it does give you a number then the limit can be reached. Another way is to put your left finger on the left side of the graph and your right finger on the right side of the graph. If your fingers meet then the limit exists but if not then the limit does not exist due to different L/R (left / right)

These pictures came from the SSS packets from the wonderful Mrs. Kirch. :) They can be found here.

BQ#4 – Unit T Concept 3 Why is a “normal” tangent graph uphill, but a “normal” cotangent graph downhill?

             

Tangent: Tangent go up because the asymptotes of tangent are found wherever x=0 (tan=y/x) and that is in pi/2 and 3pi/2. Since you can't touch the asymptotes, the only to go through the pattern correctly is to go up in the 1st quadrant since it is positive. (if it were to go down then it would have to go through the asymptote). In quadrants 2-3 they would from positive to negative so the whole period fits in between the asymptotes Pi/2 and 3pi/2. The fourth quadrant is negative so it comes from underneath the 0.

Cotangent: Cotangent goes downward because the asymptotes are found at 0, pi, 2pi. the 1st-2nd quadrants are positive and negative so they can complete a complete period in between the asymptotes 0 and pi.  Same thing applies for in between pi and 2pi with 3rd being positive and 4th being negative.  

***REMINDER*** These are just snapshots, the whole graphs are infinite.

BQ #3: Unit T: Concepts 1-3. Sin and Cosine Vs other trig functions.

Sine and Cosine:

Sine and cosine both are the only graphs that don't have asymptotes. There are just continuous graphs. The only difference with these two is that the Sine graph starts at 0 and the Cosine graph starts at 1.

 Tangent and Cotangent:

Unlike Sine and Cosine, these graphs are not continuous. They have asymptotes since they don't have 'r' as a denominator (tan=y/x and cot=x/y) The difference between these and Csc and sec is that csc and sec are more of the reciprocal of sin and cos.

                                                                                                                                   

Sec and Csc:

Secant is related to cosine in the way that it is it's reciprocal. Since it is the reciprocal, they both touch at the amplitudes, and that is in fact the only place they touch. When Cos is equal to zero though, THIS IS WHERE THE ASYMPTOTE IS!!
They are both the same values in each quadrant.

The Asymptote is located where the sin is 0. (csc=r/y) They both start at 1. 1st and 2nd quadrant are positive for both and and 3rd and 4th are negative. One period is generally 2pi just as sin and cosine.

BQ #5: Unit T: Concepts 1-3: Why do sine and cosine NOT have asymptotes, but the other four trig graphs do?

Sine and Cosines are the only trig functions that don't have any asymptotes. This is because the value of sin and cosine will never be undefined. Both Sine and Cosine are divided by '"r." r=1 and so the Sin (y/r) and Cosine (x/r) are always divide by 1.

 Now the others aren't as lucky. Since Secant (r/x) and Cosecant (r/y) are divided by x or y. Whenever x or y is equal to 0 then it will be undefined and this is the reason for Asymptotes to be formed. Same goes for Tangent (y/x) and        

Cotangent (x/y)                                                       

(Image here)

BQ#2:Unit T: Intro: Trig graphs relate to the Unit Circle?

The trig graph is simply just the unit circle in a continuos line. For example: the Values for sine are +,+,-,- (from quadrant 1-4 in order.) So that means that the graph for sine will have to start at zero and go all the way up to 1 since that is the highest point that sine can go up to. And then all the way down the -1 since it has to go to the negative in order to finish one of the periods. As seen on the photo below























Period- the period for both Sine and Cosine are 2pi since it needs to complete the cycle of negatives and positives. We already went over he pattern for sine, now Cosine is +,-,-,+. (from quadrant 1-4) so we need all four to just complete 1 period.

***NEVER FORGET THESE GRAPHS ARE INFINITE JUST AS THE UNIT CIRCLE!! THESE ARE JUST ONE PERIOD OF THE WHOLE THING.***

Amplitude: The reason that sine and cosine have an Amplitude of one is because they are over r which is equal to one. Sin=y/r or 1/1 since Sin=1 and so does r. Same applies for cosine and x. 

Reflection Unit Q concept 1-5

1. To verify a trig identity simply means to show how one equation can be manipulated and simplified to equal the other equation. You can't touch the right side of the equation because that is already the simplification of the 1st. equation. When you verify it, it justifies whether the answer would be correct or not.

2. The tips that I really found helpful were the factoring out. GCF is another one that is really helpful to me. Also multiplying by the conjugate and separating of the fraction. GCF more since i tend to have been using it the most. And I cant really explain why or how but it is the easiest one in my head to do.

3. My thought process when verifying a trig identity is 1st. are there any identities here that can be changed or replaced to be simpler. Then, can and should i change into sin and cos? see if there is anything that i can do to cancel and get the minimum amount of trig functions. If they are fractions, then try to get a GCF. I might also separate some fractions. Afterwards i will simplify by using algebra.

Unit Q Concept 2

Please see my SP7, made in collaboration with Vanessa Morales,by visiting her blog here. She also runs a cool blog and really you guys should go check her out.

I/D3: Unit Q - Pythagorean Identities

INQUIRY ACTIVITY SUMMARY

To find where sin^2x+cos^2x=1 we have to use the pythagorean theorem. X^2+Y^2=R^2. First we need to get R^2 to equal one. we do this by dividing R^2 by itself and divide everything else by R^2. we end up getting X^2/R^2+Y^2/R^2. After, we know that X^2/R^2= Cos^2x because we remember our unit circle. Y^2/R^2= Sin^2x. so we know that Cos^2x+Sin^2x=1.

It's good to know these, but these aren't the only trig functions we need to find. We also need the Tan^2x and the Cot^2x. We divide everything by Cos^2x. We come up with Sin^2x/Cos^2x+1=1/Cos^2x. Sin^2x/Cos^2x=Tan^2x since we remember the trig functions. The final answer will be Tan^2x+1=sec^2x.

To find Cot^2x we divide everything by Sin^2x this time.

We get 1+cos^2x+1/Sin^2x. from this we just simplify and we know that Cos/Sin = Cot because of our unit circle.  And 1/Sin is equal to Csc. so in the end we will get 1+Cot^2x=Csc^2x.

INQUIRY ACTIVITY REFLECTION

1. The connections that i see between Units N, O, P, and Q so far are the use of the trig functions throughout all of these. Also we have kept the unit circle through all of this.

2. If i had to describe trigonometry in THREE words, they would be eww (not sure if thats an acceptable word, if not then 'i don't like it'), difficult and helpful.

WPP #13 & 14: Unit P Concept 6 & 7 Law of Sines and Cosines.

This WPP13-14 was made in collaboration with Judith Quintanilla.  Please visit the other awesome posts on their blog by going Here .
Problems.

Law of cosines.

Billy Bob Joe Ray Eugene III decided to go quad racing with Camila. When they leave the lot where they rent the quads, Billy rode 70mph N52E. Camila rode at 80mph. and at an angle of 110 degrees bearing. suppose they rode for 3 hrs. and went in a straight line. How far would they be from each other?

Law of Sines

Camila and Billy Bob Joe Ray Eugene III want to race. They both see a red flag that is 240 ft from Camila and The angle formed at Camila is 54.14 degrees. The angle with Billy's is 58 degrees. How far is Billy from the flag? Is this fair?

Solutions

Cosines

In order to find how far they are we use the law of cosines. we plug x^2=240^2+210^2-2(240)(210)Cos58. and we get x=219.74 miles.

Sines

Sine we Know <A, Side a, and <B, we need to find side b. we get sin54.14/a=sin58/240. we manipulate it to get a by itself so it'll be a=240Sin54.14/Sin58 and we get a=229.36 ft. Billy side was shorter and he won the race!! WOOHOO!!

BQ#1: Unit P: Concept: 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?

The Law of Sines is used when we are trying to derive triangle that is not a right triangle.

We start with a non-right triangle. we drop an imaginary perpendicular line from <B and make two right triangles.

We call the height of the triangles h. Since they are now two right triangles, we can use the basic sig function. SOHCAHTOA. SinA= h/c. SinC=h/c. We use basic algebra to manipulate them.

Since both SinA/a=h and SinC/c=h, this means that SinA/a=SinC/c

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 os 1/2 of the product of 2 sides and the sin of their included angle.

Well we all know that the formula for a triangle is A=1/2bh where b is the base and h is the height.

the height is unknown so we use sinC=h/a.

We need to get "h" by itself so we just multiply the a to the other side so h=aSinc. We now know h so we plug it into the area formula (A=1/2baSinC)

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

Problems

a). Billy Bob Joe Ray Eugene III decided he take a break from his electric banjo to go visit the Empire State Building. When he was walking he saw it. From the ground, it's angle of elevation is 23.5 degrees. He knows that the building is 1250 feet tall. How far is he from the base building? 










http://upload.wikimedia.org/wikipedia/commons/c/c7/Empire_State_Building_from_the_Top_of_the_Rock.jpg





b). When he is on top of the Empire State he notices his favorite pizza place. He is hungry and wants to see how far it is. From the Building to the Pizza Place, the angle of depression is 35degrees. How far is the Pizza Place from the Empire state building?



http://haysvilleonline.com/hol/archives/625-The-Pizza-Place.html

Solutions

a). To find the how far Billy Bob Joe Ray Eugene III is from the building we have to use tan since we know the opposite and the angle. We need to find the adjacent so we use Tan23.5=1250/x because tangent is opposite/ adjacent. We manipulate it so we get x alone on one side so we get x=1250/tan23.5. We plug that in and get 2874.803 feet. So Billy Bob Joe Ray Eugene III still has some walking to do.


b).  To solve this one we have to use tangent because need to find the adjacent. We use Tan35=1250/x. We need to manipulate it to get the x alone. We should get x=1250/sin35.
when we plug it in and we get 1785.185 ft. This means our buddy here has to suck it up and eat a burger from cross the street.

I.D#2. :Unit O :Concept.7-8. How can we derive the patterns for our special right triangles?

INQUIRY ACTIVITY SUMMARY:

In order to derive the pattern for a 45-45-90 triangle from from a square with a side length of 1, we first need to cut the square from corner to corner, as shown in the picture, to get the triangle. Since the square has a side length of 1, we know that the x and y both also equal one so we can label the sides 1.

Now you have the sides, so all you need no is the hypotenuse. To find it we will use the pythagorean theorem. a^2+b^2=c^2. plug in the numbers we have. Since we know a and b both equal 1 then we use 1+1=c^2 since 1^2 still equals 1. We end up having C^2=2. so we have to cancel he square by finding the square root of c^2 which will equal C... But what you do to one side, you do to the other so it'll end up being the square root of 2.

Now that we have the  numbers, we will now add the variable because the sides are not always gonna equal 1. You can use any letter for the variable, i used X. Since both a and b both equal 1 we can just use X for both of those sides. To find the hypotenuse we have to use the pythagorean theorem. we plug in X^2+X^2= 2X^2

Now that we're done with the 45-45-90, we can move on to the 30-60-90 from an equilateral triangle with a side length of 1 and angles of 60-60-60. We start by cutting it right down the middle.

 Since we cut it right down the middle. the top corner is also cut in half so instead of a 60, its now 2 30's. And also the bottom is cut in half so instead of being 1 is it 2 .5's or 1/2's.

We, again, have to use Pythagorean theorem to find the the Y or b. First we have to manipulate a (1/2)^2+b^2=1. Once we do, it will end up being b^2=1-(1/4). That means b will equal to radical3/2.

since these are just for the side, the letter variables you use are the exact same thing. Hypotenuse=n

x=n/2 y=nradical3/2. To get rid of the fractions all you do is multiply each side by 2. Hyp.=2n  x=n  y=n radical3. N is just what you use when The hypotenuse is unknown. so the completed triangle should look like this.

INQUIRY ACTIVITY REFLECTION

Something I never noticed before about special right triangles they are the exact same things as the Unit circle.

Being able to derive these patterns myself aids in my learning because it makes me independent and i don't have to rely on anything because i have all this in my head.

I/D #1: Unit:N Special Right Triangles and the Unit Circle. How do they relate?

Inquiry Activity Summary

To derive the unit circle values, in class we would use visuals to help us remember. We used the Special Right Triangles to find the dimensions of the UC.

We First have to label the triangle according to the rules of SRT. We will use the 30 degrees triangle for this example. Since we are using a unit circle we know that the radius is one. so we have to divide it by 2X to make it one. And what you do to one, you do to the others. then you just use algebra to solve it. Once you have the X and Y you plot it at the top right corner.

 The 30 degrees triangle has angles 30,60,90 and the sides are r=1, x=radical3/2, y=1/2.

 The 45 degree triangle has the angles of 45,45,90. The sides are r=1, x=radical2/2, y=radical2/2

The 60 degree triangle has the angles 60,30,90. the sides are r=1, x=1/2, y=radical3/2. The 60 is the same as the thirty but just switched t the Y and X.

The points the top right corner that you write are the points on the circle that fits the degrees. Each of these triangles lie in the 1st. quadrant.

You can use the SRT concept for every quadrant but you have to change the sign in either the X or Y or both depending on which quadrant you want it in.

“The coolest thing I learned from this activity was if you can't memorize the unit circle, you can just use the SRT to get the answers.”

“This activity will help me in this unit because it will make it easier to complete the unit circle.”

“Something I never realized before about special right triangles and the unit circle is how they really do relate to one another and if you cant remember one, use the other for help.”