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. 

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)