BQ#7: Unit V: Derivatives and the Area Problem

1. Explain in detail where the formula for the difference quotient comes from now that you know. Include all appropriate terminology (secant line, tangent line, h/delta x, etc.).

The difference quotient is:

http://coolmath.com/precalculus-review-calculus-intro/precalculus-algebra/images/24-tools-01.gif

 

In order to derive the difference quotient, we must first keep in mind a couple things. A tangent line touches the graph once while a secant graph touches the graph at two points. The first point is (x, f(x)) and the second point is (x+h, f(x+h)). The second point is farther than x and also, it is h units away from x (keep in mind that h is also refered to as delta x). To find where the difference quotient comes from, we would then plug these points into the slope formula, which is (y2-y1/x2-x1). After plugging these points into the equation, you will notice that some stuff cancels; this then leaves the denominator as just being h. When the problem completely simplified, you discover that you get the difference quotient. The difference quotient helps us find the derivative, or the slope of the tangent line.

 

http://cis.stvincent.edu/carlsond/ma109/DifferenceQuotient_images/IMG0470.JPG



BQ#6: Unit U

1. What is a continuity? What is a discontinuity?
A continuity is a continuous function that is predictable. It has no breaks, jumps, or holes. It can be drawn without lifting your pencil off the paper. In a continuous graph, the intended height of the graph is the actual height. The limit and the value are the same.

http://www.math10.com/en/algebra/functions/function-continuity/11.gif

A discontinuity is when a graph is not continuous and it has breaks jumps or holes. The intended height of the graph does not equal the actual height. Discontinuities are broken down into two families. These two families are Removable and Non-Removable Discontinuities. There is one type of discontinuity that is a Removable Discontinuity, and that would be a point discontinuity, or a hole. The three Non-Removable discontinuities are jump, oscillating behavior, and infinite. A jump discontinuity is when there is a jump between two points which was caused by different left/rights. An oscillating discontinuity is a discontinuity that can be described as "wiggly." An infinite discontinuity is caused by a vertical asymptote, and this results in unbounded behavior.


 

 



http://www.ops.org/high/north/Portals/0/ACADEMICS/StaffPages/holleyd/calculushtml01/calc1-4notes8.gif

Point Discontinuity
 





http://upload.wikimedia.org/wikipedia/commons/e/e6/Discontinuity_jump.eps.png


Jump Discontinuity





http://web.cs.du.edu/~rjudd/calculus/calc1/notes/dis6.png


Oscillating Behavior
 





http://dj1hlxw0wr920.cloudfront.net/userfiles/wyzfiles/44bad38c-431e-4382-8fe9-86303561b2a0.gif

Infinite Discontinuity



2. What is a limit? When does a limit exist? When does a limit does not exist? What is the difference between a limit and a value?

A limit is the intended height of a function. A limit exists when the left and right side meet. This is the reason on why a limit still exists at a point.  Both sides intended to go there, so the L/R are the same number. As long as you reach the same height from the left and right, a limit exists. A limit does not exist when there are different L/R. When the limit does not exist, it is because of jumps, breaks, or holes. The value is thr actual height, while the limit is the intended height.


http://www.formyschoolstuff.com/school/math/glossary/images/JumpDiscontinuity.gif
Limit does not exist because of a jump discontinuity; different L/R

http://web.cs.du.edu/~rjudd/calculus/calc1/notes/dis4.png
Limit does not exist because of unbounded behavior

3.  How do we evaluate limits numerically, graphically, and algebraically?
When evaluating a limit, there are 3 possible ways to do this: numerically, graphically, and algebraically. Evaluating a limit numerically means on a table. To do this, we must place the number that x is approaching in the center of the graph, then we must add/subtract .1 from this number. We can then use a calculator to find the f(x) values at these x-values. The table helps us determine the x and f(x), the intended and actual height of the function. An example can be seen below.



http://i1.ytimg.com/vi/uTyYPfwVEGo/maxresdefault.jpg

 

To evaluate a limit graphically, we use a graph. When you have a picture of the graph, it can be done by using your fingers. To do this, we must place our fingers to the left and right of where we want to evaluate the limit.We do this to see if the right and left side meet. If our fingers meet, then there is a limit. The limit exists at the point our fingers meet. If our fingers do not meet, then the limit does not exist. We can also use a calculator to evaluate a limit graphically. To do this, we must simply plug in the function to the calculator.

http://s3.amazonaws.com/KA-youtube-converted/XIsPC-f2e2c.mp4/XIsPC-f2e2c.png

To evaluate a limit algebraically, we must first use something called direct substitution. In this method, we must substitute the number it is approaching into te equation. If the answer ends up being 0/0(indeterminate form), only then can we use the dividing out/factoring method and the rationalizing/conjugate method. The dividing out/factoring method involves factoring out both the numerator and denominator in order to cancel out some terms. When the expression is simplified, you can then plug in the number into what you have left. The rationalizing/conjugate method is used when all of the other methods fail. To do this, we must multiply the function by the numerator or denominator, depending which one has the radical. Then, we must then FOIL, but we must leave the non-conjugate denominator factored. We must then use direct substitution on the simplified expression.



http://www.drcruzan.com/Images/Mathematics/Limits/Example_SumAndPowerLimit.png
Direct Substitution
 

http://www.drcruzan.com/Images/Mathematics/Limits/Example_RationalLimit_02.png
Dividing out/Factoring

 

http://dq1ouwfo9m6uw.cloudfront.net/datastreams/f-d%3A703a064c524d17a02ea802a0e5079348%2BEQUATION%2BEQUATION.1
Rationalizing/conjugate



BQ#4: Unit T Concept 1-3

From the Unit Circle, we know that tangent and cotangent are both positive in the first and third quadrants and negative in the second and fourth quadrants. Tangent=sine/cosine while cotangent =cosine/sine. Tangent would have asymptotes wherever cosine equals zero (whenever it is undefined) In contrast, cotangent would have asymptotes wherever sine equals zero. The graphs of tangent and cotangent are different because of their different asymptotes. Even though the asymptotes are different, they both still have to be positive in the first and third quadrant and negative in the second and fourth quadrant. Tangent is going uphill and cotangent is going downhill in order to not touch the asymptotes and also to follow the positive/negative rules of the quadrants in the Unit Circle.

(Red=Quadrant 1, Green=Quadrant 2, Orange= Quadrant 3, Blue= Quadrant 4)

 

Tangent

 

Cotangent

 

BQ#3: Unit T Concepts 1-3

Tangent

Sine and cosine relate to tangent because when we look back at the Unit Circle, sine and cosine are positive in quadrant one. Based on our knowlwdge of identities, we know that tangent=sine/cosine and since sine and cosine are both positive in the first quadrant, then a positive divided by a positive will result in a positive, which is what tangent is in this quadrant as well. This will cause the tangent graph to be above the x-axis in the first quadrant. In the second quadrant, sine is positive and cosine is negative. If tangent=sin/cos, then this means that tangent will be negative in the second quadrant, since a positive divided by a negative will result in a negative. In the picture, we see that the tangent graph is below the x-axis. In the third quadrant, both sine and cosine are negative, so when you divide two negatives, you get a positive, and tangent is positive as a result of that. In the picture, we see that tangent is once again above the x-axis. In quadrant four, sine is negative and cosine is positive. When divided, this gives you a negative, so tangent will be negative and this will cause it to be below the x-axis. Also, because tangent is sine/cosine, tangent will have asymptotes wherever cosine equals zero on the graph.
(Red=Quadrant 1, Green=Quadrant 2, Orange= Quadrant 3, Blue= Quadrant 4)

Cotangent

In quadrant one, sine and cosine are both positive, so because cotangent=cosine/sine, when this is divided, cotangent will be positive, or above the x-axis. In quadrant two, sine is positive and cosine is negative, so a negative divided by a positive gives you a negative, so cotangent is negative in this quadrant, or below the x-axis. In quadrant three, cotangent will be positive because sine and cosine are bothe negative, so when they are divided, you get a positive. Cotangent will be above the x-axis here. In quadrant four, cosine is positive and sine is negative, so when they are divided, cotangent will result in being negative, or below the x-axis. Also, Cotangent will have asymptotes wherever sine equals zero. Asymptotes only occur when we have undefined, or whenever we divide by zero, so whenever sine=0, there will be an asymptote.

 

 

Secant





In quadrant one,cosine is positive, so secant is positive as well. Secant is the reciprocal of cosine. The cosine graph has very small plotted points, so the reciprocal of those numbers is very large. This is why secant goes up so high. The points plotted in the secant graph will be the reciprocals of the cosine graph. Secant will have asymptotes wherever cosine equals zero. In quadrant two, cosine is negative, and so is secant, so secant will be below the x-axis.  Therefore, secant also has a point here as well. In quadrant three, cosine is negative, so secant is negative too, so secant will be below the x-axis. In quadrant four, cosine is positive, as well as secant, so secant will be above the x-axis.

 

Cosecant

In quadrant one and two, sine is positive, so cosecant is positive as well. The cosecant graph will touch the mountain of the sine graph and as it goes higher, it gets closer to the asymptotes. In quadrant three and four, sine is negative, so cosecant will be negative as well, or below the x-axis. The cosecant graph is shaped where the asymptotes are and the asymptotes are where the sine graph is. Cosecant is one over sine, so whenever sine equals zero, cosecant will be divided by zero. Dividing by zero means getting undefined, and undefined means that there will be an asymptote.





BQ#5: Unit T Concepts 1-3

Sine and cosine graphs do not have asymptotes because their denominator is 1, going back to Unit Circle ratios.
Sin=y/r
Cos=x/r
Csc=r/y
Sec=r/x
Tan=y/x
Cot=x/y
As we see above, the only ratios that have r as the denominator are sine and cosine. R means that the numerator will always have 1 as the denominator, unlike all of the other trig functions. The other trig functions can be either greater or less than 1, so they have asyptotes. Because sine and cosine have the denominator as being 1, they will always be a real number. The other trig functions, however, can end up with 0 as their denominator, causing these functions to be undefined. The result of these undefined functions are what cause these trig functions to have asymptotes

BQ#2: Unit T Concept Intro

1. Trig graphs relate to the unit circle because if we were to unwrap the circle, it would still have quadrant 1,2,3, and 4. This "unwraping" would be as if it were in a graph. For example, sine is positive in the first and second quadrant of the unit circle, so in the graph that section would be above the x-axis. These values would be positive from 0 to pi and the values would be negative until it would get to 2pi because quadrant 3 and 4 in the unit circle are negative. This cycle goes on forever, this is just a portion of it. For cosine, it is pretty much the same idea. Cosine is positive in the 1st quadrant, negative in the 2nd and 3rd quadrant, and positive in the fourth quadrant. If the unit circle were to be unwrapped once again, the line would be above the x-axis until it reached 90 degrees, or pi/2, and then it would be below the x- axis until 270 degrees, or 3pi/2, was reached and finally, it would be positive from 3pi/2 to 0. This would be one complete period. The period of sine and cosine is 2pi because it will take the entire distance of the unit circle to reach one complete period whereas tangent and cotangent only take half of that, which is just pi to complete a period. To continue further explanation, tangent is positive in the 1st quadrant, negative in the 2nd quadrant, positive in the 3rd quadrant, and negative in the 4th quadrant. This means that from 0 to pi/2, the graph is positive, from 0 pi/2 to pi, the graph is negative, from pi to 3pi/2, the graph is positive, and from 3pi to 0, the graph is negative.
(pictures below for further explanation)

Sine

 

Cosine

 

 

Tangent

 

 

2. Amplitudes are half the distance between the highest and lowest points on the graph. Sine and Cosine have amplitudes because the trig ratio for sine is y/r and for cosine it is x/r, and we know that r will always be 1. The center of the unit circle is (0,0) and its radius is one, therefore, this is how much sine and cosine can extend to. This is where the 1 and -1 come from; these are the only numbers that work, anything that is outside these numbers will not funcion. The rest of the four trig function's ratio is not over 1, so they are not as restricted. Because these values are not restricted and go on forever, they cannot have an amplitude. They will always be from negative infinity to positive infinity, so this does not allow them to have a highest and lowest point.

Unit Q: Verifying Trigonometry Reflection

1. Verifying means to solve a problem. To verify something, you must check to so see if it is correct. It meant to solve and go to several steps in order to achieve the correct answer. Verifying means checking different identities and solving them to see if you end up with the correct answer in both sides.
2. Tips would be to turn everything into sin/cos to make everything easier. Finding the common denominator also. Multiplying the conjugate would also make it easier, and that is the denominator binomial. Factoring and separating fractions are also tips to make the problem easier.
3. I look at the problem to see if I can turn anything into sin/cos. sometimes when a problem looks hard, it may already be an identity that equals 1, so that makes it a lot simpler. Memorizing the identities really helps with this because this way you know what it equals. You keep substituting what you know in order to get your final answer. The answer will either have to be as simple as it can be, or equal to the problem you are verifying.

SP#7: Unit Q Concept 2: Finding All Trig Function Values Using Identities

 

 

This SP was made in collaboration with Leslie E. Please visit the other awesome posts on their blog by going here

 

Identities

 

 

SOH CAH TOA

The viewer must need to understand what identities are and how to use them for this problem. The viewer needs to be aware of the substituting that goes on while trying to find all trig functions. The viewer must be familiar with SOH CAH TOA in order to do the problem the other way. They need to pay attention to the ratios when solving it with SOH CAH TOA.

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

INQUIRY ACTIVITY SUMMARY

1. The Pythagorean Theorem is an identity because it is a proven fact that is always true. The Pythagoreom Theorem is a^2+b^2=c^2, but in the Unit Circle, these letters are mentioned as x, y and r, making a slight shift to the original theorem. With these letters, the Pythagorean Theorem is now x^2+y^2=r^2. In order to make the Pythagorean Theorem equal to one, we must divide both sides of this equation by r^2, as seen in the picture below.

 

Then, going back to the Unit Circle,  the ratio for cosine was x/r and the ratio for sine was y/r. These ratios can then be plugged into (x/r)^2+(y/r)^2=1 in order to get sin^2x+cos^2x=1. This is referred to as a Pythagorean Identity because it is just simply the Pythagorean Theorem rearranged in a different way. To show that this is identity is true, we can use one of the "Magic 3" ordered pairs from the Unit Circle. If we have a 45 degree angle, we know that the ordered pair is (radical2/2, radical2/2). When this is plugged into the equation, it will be radical2/2^2+radical2/2^=1. This is true because when radical2/2 is squared, it results to being 1/2 and then when it is added to the other 1/2, it is one. Therefore, the identity is true.

2.To derive the identity with Secant and Tangent, cos^2x must be divided by both sides in the equation of sin^2x+cos^2x=1. You will get:

 

In the unit circle, sine has a ratio of y/r and cosine has a ratio of x/r. When the ratios are divided,with sine being the numerator and cosine being the denominator, you get y/x, which is the ratio for tangent. The two cosines will then cancel and result in being 1. 1/(x/r)^2 (the ratio of cosine) is equal to sec^2. This is how you get tan^2x+1=sec^2x.

To derive the identity with Cosecant and Cotangent, we must now divide sin^2x+cos^2x=1 by sin^2x.

 

The sines will cancel, leaving us with 1. The ratio of x/y equals cotangent (refer to picture for further explanation). 1/sin^2x equals csc^2x because of reciprocal identities. This ends up being 1+cot^2x=csc^2x.   

INQUIRY ACTIVITY REFLECTION

 

1. "The connections that I see between units N, O, P, and Q so far are..." that they all somehow connect to the Unit Circle and also that they somehow involve triangles as well.

 

2. "If I had to describe trigonometry in three words, they would be..." complicated, intricate, and time-consuming. 

WPP #13 & 14: Unit P Concept 6 & 7

Please see my WPP 13-14, made in collaboration with Leslie E,  by visiting their blog here. Also be sure to check out the other awesome posts on their blog.

BQ#1: Unit P Concepts 1 and 4: Law of Sines and Area of an Oblique Triangle

 

1. Law of Sines
We need the Law of Sines to help us find sides or angles that are not necessarily right triangles. Since the Pythagorean Theorem can only be used for right triangles, the Law of Sines helps us find the sides for triangles that are not right triangles.

 

http://etc.usf.edu/clipart/36700/36740/tri19_36740_lg.gif

To derive the Law of Sines, we first need to draw an imaginary line down angle B.




http://etc.usf.edu/clipart/36700/36738/tri17_36738_lg.gif



This imaginary line can be labeled h, as the picture above shows. This allows for two right triangles to be formed. We can now use SOH CAH TOA to help us figure out the rest (just SOH in this case)

Sin A=h/c

Sin C=h/a

Since both of these have h as a common variable, we can simplify to get c sin A=a sinC
Then, divide by ac and then you will get sinA/a=sin C=c.
(this applies to angle B too, if a perpendicular line was drawn from either angle A or C.)

You finally get:

 

 http://www.clarku.edu/~djoyce/trig/lawofsines.jpg



4. Area Formulas
The area of an oblique triangle is derived from the area formula which is A=1/2bh

http://www.compuhigh.com/demo/lesson07_files/oblique.gif

 

To find the height, a verical line is drawn and labeled h. The trig functions, which are sinA=h/c, sinB=h/b and sinC=h/a can be simplified by their denominators in order to make them equal to h. Then, this is plugged into the area of a triangle. H is then replaced with either asinC or csinA. Finally, you end up with the area of an oblique triangle, which is:

 

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgZrTRmOJrogy3QUkgKj4h6OGuoWqoGscJ0J0Vx9qiKdr2IOoRiD67ngtI6YPhH06z0jxjEKdDKrUYCPBh8lVllMYlDLUvRHJ3G3KjA0Mm_OFJp-z-5Au4VdDeRht56fsINaXJ00Sijqu4/s400/hi.bmp

 

The area of an oblique triangle related to the area I am familiar with because it is what was used to derive the area of an oblique triangle. The area of a triangle is essential in finding the area of an oblique triangle because you need to plug the height into the familiar area equation in order to derive the area of an oblique triangle. 

 

 

 

WPP#12: Unit O Concept 10: Solving Angles of Elevation and Depression Word Problems

http://image.yaymicro.com/rz_1210x1210/0/9d9/baby-boy-climbing-on-bookcase-9d9831.jpg

 

The Problem

A.) Sasha is walking to a shelf to get the flour she needs to bake a recipe. She is 124 ft. away from the shelf and the angle of elevation is 24 degrees. She wants to know the height of the shelf so that she will know how high she has to reach for the flour. What is the height of the shelf?

B.) Sasha needs to get on top of the shelf in order to get her flour. When she is on top, she realizes that she had to climb 55.2 ft. to get there. She spots a flour sack on the floor and notices that she did not have to go through the trouble of climbing to get a flour sack. She then decides that she wants to jump on the flour sack that is on the ground to avoid getting hurt by the floor. The angle of depression to the flour sack is 42 degrees. How long will her fall be?

The Solution

 

A.)

 

 

B.)

 

 



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

INQUIRY ACTIVITY SUMMARY

 

In this activity, we were given a square that told us to derive the pattern for 45-45-90 triangles. We were  only given the side length, which was 1. We were also given an equilateral triangle and we were asked to derive the pattern for 30-60-90 triangles. The information we were given was that the side of the equilateral triangle was 1. This activity is meant for us to understand why special right triangles have these patterns.

 

30-60-90 Triangle

To derive a 30-60-90 Triangle, we first need to label the given information, which is that the side lengths of the equilateral triangle are 1. Then, we need to split the equilateral triangle in half. This will create two right triangles. Since the side lengths of the equilateral triangle are one, the two right triangles now have a side that is 1/2. We then need to solve the missing side using the Pythagorean Theorem. We know that one side is 1/2 and the hypotenuse is 1. We then need to square both sides. 1/2 squared is 1/4 and 1 squared is 1. Then, we need to subtract 1/4 from both sides. We will get 3/4, and finally we need to take the square root of it. The top stays as radical 3, but the bottom is 2. The final side ends up being radical 3/2. Lastly, we need to multiply the sides by two to get rid of the fractions. We will end up with radical 3, 1, and 2. A 30-60-90 Triangle has"n" in its pattern simply because "n" is a variable that represents any number. If any number were placed in to replace "n", the pattern would still work. (refer to pictures below for steps)

 

 

 

 



45-45-90 Triangle

To derive a 45-45-90 Triangle, we first need to label the given information, which is that the side lengths of the square are 1. Then, we need to split the square into two right triangles by cutting the square diagonally. We know that the lengths of both legs of the triangle are 1, so we need to use the Pythagorean Theorem to solve for the hypotenuse. We need to square 1 twice, and then add it. Since 1 squared is simply 1, 1+1 equals 2. Then, we need to take the square root of two, so the hypotenuse ends up being radical 2. The side lengths are now 1,1,radical 2. If 'n' is multiplied to this, it ends up being n, n, n radical 2. The relationship between the sides is the same, so n applies to any number because it is just a constant. " n" basically just represents a number that can be substituted in (refer to pictures below for steps).

 

 

 

 

 

 

 

 

 

 



 

 

INQUIRY ACTIVITY REFLECTION

 

1. "Something I never noticed before about special right triangles is" that the Pythagorean Theorem helps derive the patterns for special right triangles.
 

2. "Being able to derive these problems myself aids in my learning because" now I know why these patterns are there, so I do not have to memorize things anymore and can use this logical reasoning instead to help me solve problems.