BQ# 7: Unit V

     The origins of the the difference quotient comes from a graph and using an old equation from early this year.

Here is the graph f(x) and the line that's barely touching it at x is the tangent line. So the coordinates for the graph is ( x, f(x) ). 

This is another graph that has a secant line going through it. It still has the original point as the first one but it has another point in it. Since we moved a little bit from the original to the new point, then it's a change in the graph. That change can be written as delta x or as i put it h. So the new coordinates for this graph is ( x, f(x) ), ( x+h, f(x+h) ).

Highlighted in blue is the two coordinates from the graph. Highlighted in green is the slope formula that will help find the difference quotient. The one in pink has everything plugged in and in the denominator we see that the xs' cancel so there's only an  h left in the denominator. The last one in purple is the difference quotient and that is how we get the equation.

BQ #6 : Unit U Concepts 1-8

1) A continuity graph is a graph that's predictable, it has not breaks, holes, and jumps. Also the limit and the value are the same, the limit is the intended height while the value is the actual height. There are two groups, the removable and non-removable, a continuous graph is in the removable group.

A continuous graph is in the removable group because it has a limit. This can be considered a continuous graph because it has no breaks, hole, or jumps.

 Discontinuity graphs are in the non-removable group because these graphs have no limits. These graphs have jumps, breaks, and not predictable at one point. There are three types of discontinuity graphs, jump discontinuity, oscillating behavior, and infinite discontinuity.

Highlighted in green would be our jump discontinuity, orange is oscillating behavior, and in blue is infinite discontinuity. None of these graphs have a limit but only jump discontinuity can have a value. The other two have no limit or value. 

2) In this unit a limit is the intended height of a function. It only exits in a point discontinuity graph.

This graph is a point discontinuity and it has a limit. Now the first one is continuous but is still has a limit, the second one has a hole but it still has a limit because it's the intended height.

All of these don't have a limit so the limit does not exist for them.

Just like in the first question the difference between a limit and a value is that the limit is the intended height while the value is the actual height.

3)

To evaluate a limit numerically we do it by using a table that we can plug in to our graphing calculator. We can find the limit using the graphing calculator by tracing the graph. Then we can tell if the limit was reached.

 

When it's done graphically we can either use our fingers to see where the limit is if we come to the middle from the left and right. Then we just look at what type of discontinuity it is and if the limit exists or not. 

A

B

C

When it's done algebraically we can use three methods but the one we will always try first is the substitution method, know as picture A. Ifa problem can't be done using that method then we use the dividing/ factoring method. If those two don't work then we use the rationalizing/ conjugate method.

BQ#4: Unit T Concepts 1-3

Why is a "normal" tangent graph uphill, but a "normal" tangent graph downhill? Use unit circle ratios to explain.

    The main reason why their graphs are different is because of their asymptotes and their location. The graphs are based on their Unit Circle ratios, and the graphs can't touch the asymptotes. The asymptotes themselves are based on the Unit Circle.

Both tangent and cotangent have their asymptotes based on their Unit Circle ratios. The ratio for tangent is y/x and in order to how and asymptote x has to 0 so it can be undefined. If we look at the graph, we see that 90 degrees and 270 degrees has the ratio to be undefined. 90 and 270 degrees in radians are pi/2 and 3pi/2, and those are the asymptotes. The same thing for cotangent except that the ratios are switched, it's x/y. So in order for cotangent to have asymptotes, y must be 0 and 180 and 360 degrees have those points. The asymptotes for cotangent, in radians, is pi and 2pi.

 Tangent:

If we look at it already graphed we can see that 2pi and 3pi/2 are the asymptotes and the graphs can't touch them. Also, based on the Unit Circle, the four quadrants and if they are positive and negative. Both quadrants 1 and 3 for tangent is positive so it's above the x-axis. Quadrants 2 and 4 are negative so the graph is below the x-axis.

Cotangent:

The asymptotes for cotangent is pi and 2pi and just like tangent, the graphs can't touch them. The Unit Circle and it's four quadrants are just like tangent, 1 and 3 is positive and 2 and 4 are negative. Expect that the graphs of cotangent will not be the same as tangent because the asymptotes make the graph go downhill to follow the positive- negative of the quadrants.

BQ# 3: Unit T Concepts 1-3

How do the graphs of sine and cosine relate to each trig graph?

Sine will be in Green.
Cosine will be in Orange.
Tangent will be in Blue.
Cotangent will be Yellow.
Cosecant will be in Pink.
Secant will be in Purple.

Tangent:

To see where the tangent graph will be we must first know about it's ratio. Tangent's ratio is y/x or sin/cos, cosine will determine where tangent will be. Where sine and cosine are both positive or negative then tangent will just be positive. If one of them is negative then tangent is negative. Tangent's asymptotes is determined by cosine, if cosine is 0 on the x-axis then that is one of tangent's asymptote. 

 Cotangent:

It's almost the same thing for cotangent except that it's sine that needs to be 0. If both sine and cosine are positive or negative then cotangent will be positive. If one of them is negative then cotangent is negative. The reason why cotangent is downhill is because of the asymptotes and where they are found. Since it has to be positive in quadrant 1, the cotangent graph has to be above the x-axis. Same thing for quadrant 2 but it's negative so it's below the x-axis.

Cosecant:

Where sine is 0, the asymptotes for secant will be those points. If sine is positive in two quadrants then cosecant will be positive, if it's negative, like in quadrants 3 and 4, then it's negative. Cosecant relates to the graphs by it's shape, the shape is between two asymptotes. So the asymptotes basically determine the shape and the asymptotes is where sine is 0.

 Secant:

Where cosine is 0 the asymptotes of secant can be found. The way the secant graph looks like, positive or negative, is the same as the cosine graph. The shape  of is is also determined by the location of the asymptotes and the cosine graph.

BQ# 5: Unit T Concepts 1-3

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 asymptotes because asymptotes occur when the denominator of the ratios is 0 (undefined). The other trig graphs can be divided by 0 and it can be undefined while sine and cosine can't be divided by 0. They can only be divided by 1 because 1 is their restriction on the Unit Circle and on the Unit Circle itself it goes from 1 to -1 on both axis. They both can't be divided by 0 because it's not undefined but no solution. The other four especially tangent and cotangent have no restrictions and if their ratios equal undefined then it's their asymptote.

BQ#2: Unit T Concept Intro

A)   When is comes down to sine and cosine, their periods are are 2pi and it's like that due to their similarity to the Unit Circle.

Sine in the Unit circle is positive in the first and second quadrant and negative in the third and fourth quadrant. If we start from zero and do one complete rotation to get it to be positive again, then then it was a 360 degree or 2pi rotation. So when we stretch Unit Circle into a line, the cyclical will be stretched out too because the first two quadrants will be above the x-axis and the last two will be below the x-axis. 

The same thing goes for cosine since quadrant I and IV are positive and quadrant II and III are negative. We need to start at 0 degrees and as we move around in the Unit Circle well reach the positive when we get to 360 degrees. When it reaches 360 degrees the it made one complete rotation. If the stretch out the Unit Circle to a straight line for cosine then we will start above the x-axis since quad. I is positive and for quads II and III it will be below the x-axis. As we approach quad IV to make a cyclical then it go back up above the x-axis.

For tangent it's different because it's only half the trip to get a cyclical. Quadrants I is positive and quadrant II is a negative so we already have our cyclical. On the Unit Circle, it will start at 0 degrees and when we reach a positive again it will be at 180 degrees and the radian value is pi.

B)    Sine and cosine only have an amplitude of 1 because they have a restriction of 1 and -1. If we use any number and use it in their ratio, it will not work except those numbers less than 1. We only use the 1 in the Unit Circle then sine and cosine will work because it is in their restriction,  if it wasn't then it will be undefined and will not work.

BQ #1: Unit P Concepts 1 and 4: Law of Sines and Area of Obliques

Concept 1: LAW OF SINES
     We need the Law of Sines when are working on a triangle that is not a right triangle. When we have that we use the Law of Sines to solve it but only if the the triangle is an AAS or an ASA. How to derive the Law of Sines will be shown in these following pictures:

If we are given a triangle the A,B, and C as their angles and a, b, and c as their sides we can just split the triangle in half to form two triangles.

 

Now if we just focus on angle A, angle C, side a, side c and the height, h, I can show you how they can equal each other. If we take the sine of angle A, it will be h/c but since this is a part of a triangle we don't know what h is. So c will be multiplied to both side and it will give us h = c(sin of angle A). The same thing will happen if we do the sine of C, except it's h/a and we multiply a to both side and we'll get h= a(sine of angle C). We can get the height in two way but in the end the height will be the same for the both of them. Since c(sine of angle A) = a(sine of angle C), and we cross multiply, then ( sine of angle A)/a =  (sine of angle C/c.

If we stay with the same angles and side but with a different height then I can show how sine of B can be the same as the other two.

In this version of the triangle we'll use the sine of angle B and it is k/c but we don't know what k is so we multiply c to both side and we'll get k= c(sine of angle B ). On the other triangle it's the sine of angle C and that's k/ b. Just like the other one we multiply b to both side and get k= b(sine of angle C). But if  c(sine of angle B ) = k and b(sine of angle C)= k then they are both the same. Which mean that if we cross multiply them then (sine of angle B )/b = (sine of angle C)/c and if that is true then (sine of angle A)/a will be the same and all three will work and we'll get the same answer.

Concept 4: AREA OF OBLIQUES.
     The "area of obliques" is derived from the area of a triangle which is A = 1/2 bh. The base is b and the height is h in a right triangle. But if we don't have a right triangle then it will almost be the same but that depends on the side and angle given to us. If we have angle A, side b, and side c given to us then the area formula will be A = 1/2 b(c(sine of angle A). The formula can be rewritten to work with the two other angle. The following picture will show which formula will work for that problem.