Wednesday, June 4, 2014

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.).

First of all, we must remember that the difference quotient is just [f(x +h) - f(x]/h. But how is it derived?

Let's say we have a graph like the one below.
 (http://www.sophia.org/tutorials/unit-v-concept-1--2?cid=embedplaylist)
Our goal would be to find the slope of tangent line at a certain point. A tangent line that is a line that touches the graph only once.  A secant line would be a line that touches the graph at two points. Logically, we would then try to find the slope of the secant line using the slope formula: m = (y2 - y1)/(x2 - x1). Since we don't have any concrete numbers, let's use our first point as (x, f(x)). Then. we can pick another point that is farther than x and let's call it (x + h, f(x + h)) because we moved h units away from x. Sometimes, h might also be referred as delta x because we had two x values and the change in between the two x values would be the same as moving h units away.
 (http://www.sophia.org/tutorials/unit-v-concept-1--2?cid=embedplaylist)

We simply plug in the variables into the slope formula to get m = (f(x+h) - f(x))/((x + h) - x). This simplifies to the the much familiar difference quotient: (f(x + h) - f(x))/h.

To find the slope of the tangent line to a graph at a certain point, we must find the limit as h approaches 0. That is because as h decreases the slope of the secant line becomes increasingly similar to that of the tangent line. To do this, we simply substitute 0 into h, and solve.
 (http://www.sophia.org/tutorials/unit-v-concept-1--2?cid=embedplaylist)
 (http://www.sophia.org/tutorials/unit-v-concept-1--2?cid=embedplaylist)

 (http://www.sophia.org/tutorials/unit-v-concept-1--2?cid=embedplaylist)

Therefore, the derivative is simply just the slope of the tangent line while the difference quotient is the mathematical definition of the derivative.
 (http://en.wikipedia.org/wiki/Numerical_differentiation)
References:
http://www.sophia.org/tutorials/unit-v-concept-1--2?cid=embedplaylist
http://en.wikipedia.org/wiki/Numerical_differentiation

Saturday, May 17, 2014

BQ #6: Unit U

1. What is a continuity? What is a discontinuity?

A continuous function is a function is predictable, has no discontinuities (breaks, jumps, or holes), and it can be drawn without lifting you pencil. That just means that the function will go wherever you expect it to and its graph is a smooth line with every value of x having a value. In addition, if the function is continuous, then the intended height of the graph (limit) will equal the the actual height of the graph (value).

 (http://www.mathsisfun.com/calculus/continuity.html)

A discontinuity would therefore be where the function is not continuous, where all the breaks, jumps, or holes are. That would make the limit of the graph not equal to the value of the function. There are two families of discontinuities: removable and non-removable. The only removable discontinuity is point discontinuity, which we have previously known as a hole. The three non-removable discontinuities are jump discontinuity, oscillating behavior, and infinite discontinuity. They are grouped as non-removable discontinuities because it is only at those 3 discontinuities that the limit DOES NOT EXIST (DNE). The limit still exists at a hole because the function still intended to go to that point, but it actually didn't. The limit doesn't exist at a jump discontinuity because the left and right limits do not match and for infinite discontinuity, there exists a vertical asymptote which results in unbounded behavior.

Not Continuous             Not Continuous           Not Continuous
(Hole)                           (Jump)             (Infinite/Vertical Asymptote/Unbounded Behavior)

Oscillating behavior is just when the graph gets all "wiggly". The graph doesn't have a limit because it doesn't approach any single value.

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 the function.A limit only exists when the left and right side limits match each other (or when they meet). That is why a limit still exist at a point discontinuity; the left and and right hand limits are the same number because that is where both sides intended to go. A limit doesn't exist at the 3 non-removable discontinuities because either their left and right limits don't match, of unbounded behavior, or of oscillating behavior.A limit, as stated before, is the intended height of a graph, while the value is the actual height the graph reaches.
(Limit DNE b/c left and right limits don't match)                       (Limit DNE b/c unbounded behavior)
(Limit DNE b/c oscillating behavior)

3. How do we evaluate limits numerically, graphically, and algebraically?
One way to evaluate limits is numerically, meaning using a table. We simply put the number that x is approaching in the middle of the table, add and subtract 0.1 from that number for the two outermost values, and slowly work our way towards that number. Then, we can just use a calculator to figure out the f(x) values on the graph at those x-values. The middle f(x) value should be blank because unknown, but if we compare the the values approaching the number from the left and right, we see that the values are approaching a certain number, which will be where the limit is. Depending on the function though, the limit might or might not be reached.
To evaluate limits graphically, we simply use...a graph. We could use a graphing calculator or if we are given the graph, we can use our left and right hands to see if the right and left hand side meets. With a graphing calculator, all you need to do is to set up your table, type in the values close to the limit yuo want to evaluate, and plug in your function. If you are given a graph (like the one below), you just put a finger on a spot to the left and to the right of where you want to evaluate the limit.

If your fingers meet, then there is a limit and that is where it is.
-So the limit as x approaches -4 of f(x) is equal  to 2 (point discontinuity).

If your fingers don't meet, then the limit DNE.
-So the limit as x approaches 1 of f(x) of f(x) DNE because the left and right limits don't match (jump discontinuity).

Finally, to evaluate limits algebraically,we just use direct substitution, the dividing out/factoring method, or the rationalizing/conjugate method. For direct substitution, we just simply substitute the number is approaching into x and simplify to get the answer. However, if it ends up to be a number over 0, then it is undefined and the limit DNE because of unbounded behavior, If you end up with 0/0, then it is indeterminate and you must use either the factoring method or the rationalizing method next.
 (http://www.onlinemathlearning.com/limits-calculus.html)
The dividing out/factoring method is just basically trying to factor the numerator and denominator to cancel some common terms to remove the zero, which is what is causing substitution to fail. Once you have the simplified expression, you just use direct substitution to solve.
 (http://nfva.info/LarsonCalc/Ch1_files/ch1_1_3w.htm)
And if direct substitution and the dividing out/factoring method both fail, then our final resort would be the rationalizing/conjugate method. We would just multiply the function by the conjugate of the denominator or the conjugate of numerator, depending on where the radical is. We would leave the non-conjugate part factored, since we want something to cancel.And we would just foil the multiplied conjugates, simplify, and eliminate any common factors in the numerator and denominator. Then, we just use direct substitution to solve for the answer.
 (http://faculty.eicc.edu/bwood/math150supnotes/supplemental2.htm)

Works Cited:
http://www.mathsisfun.com/calculus/continuity.html
http://www.analyzemath.com/calculus/continuity/continuous_functions.html
http://www.onlinemathlearning.com/limits-calculus.html
http://tutorial.math.lamar.edu/Classes/CalcI/OneSidedLimits.aspx
http://nfva.info/LarsonCalc/Ch1_files/ch1_1_3w.htm
http://faculty.eicc.edu/bwood/math150supnotes/supplemental2.htm

Monday, April 21, 2014

BQ #4: Unit T Concept 3

4. Why is a "normal" tangent graph uphill, but a "normal" cotangent graph downhill?

From our knowledge of the Unit Circle, we know that tangent and cotangent is positive in the 1st quadrant (0 to π/2) and in the 3rd quadrant (π to 3π/2) and negative in the 2nd and 4th quadrants (π/2 to π and 3π/2 to 2π). Cosine is positive in the 1st and 4th quadrants, but negative in the 2nd and 3rd quadrants.So both of their graphs will be positive and negative in the same places of the graph.

However, we must remember that tangent and cotangent are different ratios.Tangent is equal to sine/cosine while cotangent is equal to cosine/sine. And from the previous concept, we know that tangent will have asymptotes whenever they are undefined, which is whenever their denominator is equal to 0. Therefore, tangent would have asymptotes whenever cosine=0, which is at π/2 and 3π/2. Of course, those are the only asymptotes in the 4 quadrants that we are looking at, but there are more asymptotes at the same intervals as well.

Now for cotangent, it is equal to cosine/sine, So it would have asymptotes whenever sine=0, which is at 0, π, and 2π.
Notice how the asymptotes for the two graphs are different. Yet, they both still have to be positive in the 1st and 3rd quadrants while negative in the 2nd and 4th quadrants. So in order for the graphs to be drawn according to those rules and to avoid touching the given asymptotes, tangent is drawn going "uphill" while cotangent is drawn being "downhill". The only reason the graphs are different is because of their different asymptotes, which are dictated when their ratios are undefined (denominator=0).

Friday, April 18, 2014

BQ #5: Unit T Concepts 1-3

5. Why do sine and cosine NOT have asymptotes, but the other four trig graphs do?

To understand why sine and cosine do not have asymptotes but the others do, we need to remember our Unit Circle ratios.
 http://literacy.purduecal.edu/STUDENT/mrrieste/coordinates.html
Looking at the ratio of sine and cosine, we notice that both of their denominators is r while all the other trig functions have a denominator of x or y. From our knowledge about the Unit Circle, we know that r stands for the hypotenuse of the triangle, which in our case will always be equal to 1. We also know that the other trig functions have either the x-value or the y-value of the triangle as their denominator.

Since, r is a constant that will always be 1, sine and cosine will always be real numbers. However, that is not the case with the other trig functions. There are times when sine (y-value) or cosine (x-value) will equal 0. That would make their ratios have a denominator of 0, causing them to become undefined because it's not impossible to have something over nothing. As a result, for tangent, cotangent, cosecant, and secant, will have asymptotes, lines that the graph can never cross. Where these asymptotes are will be dictated by their denominator in the ratio. For example, tangent and secant both have a denominator of x (cosine) so wherever cosine is equal 0, there will be an asymptote. For cosecant and cotangent, they have a denominator of y, so their asymptotes will be wherever y (sine) is equal to 0. That is why the 4 graphs will always look like they're approaching the asymptotes, but they can never touch or cross it because that is where they are undefined. Sine and cosine are under no such restrictions, since their denominator will always be 1, making it impossible for their ratios to become undefined at any value.

Thursday, April 17, 2014

BQ #3: Unit T Concepts 1-3

3. How do the graphs of sine and cosine relate to each of the others?

a) Tangent?
First, let's sketch the sine and cosine graphs.We know that sine and cosine both have a period of 2π with sine starting and ending at 0 while cosine starts and ends at its amplitude. Looking back to the Unit Circle, we know that sine is positive in the 1st quadrant (0 to π/2) and in the 2nd quadrant (π/2 to π) and negative in the 3rd and 4th quadrants (π to 2π). Cosine is positive in the 1st and 4th quadrants, but negative in the 2nd and 3rd quadrants.

Now according to the ratio identities, tangent is equal to sine/cosine. So we would be correct to assume that if sine or cosine was negative, then tangent would also be negative. Following this train of thought, that would mean that if both sine and cosine were positive or negative, then tangent would be positive. Therefore, since sine and cosine are both positive in the 1st quadrant (0 to π/2), then tangent would also be positive. Or since cosine is negative while sine is positive in the 2nd quadrant (π/2 to π), then cosine would cause tangent to be negative as well. In the 3rd quadrant, both sine and cosine is negative, so that cancels out to make tangent positive. In the 4th quadrant, cosine is positive, but sine is negative, causing tangent to be negative.

In addition, since tan= sin/cos, then the tangent graphs would have asymptotes whenever cosine=0, making the ratio undefined. According to the Unit Circle, we know that cos=0 when the x value is 0, so at π/2 (90*) and 3π/2 (270*). The asymptotes then keep repeating itself at the same intervals since the domain goes on and on from left to right. Tangent will never be able to touch and cross the asymptotes, no matter how close it gets to it.

b) Cotangent?
Going back to the sine and cosine graphs, cotangent is going to be very similar to the tangent graph. Cotangent, since it's the reciprocal of tangent, is equal to cosine/sine.So we would be correct to assume again that if sine or cosine was negative, then cotangent would also be negative. As a result, that would mean that if both sine and cosine were positive or negative, then cotangent would also be positive. Therefore, since sine and cosine are both positive in the 1st quadrant (0 to π/2), then cotangent would also be positive. Or since cosine is negative while sine is positive in the 2nd quadrant (π/2 to π), then cosine would cause cotangent to be negative as well. In the 3rd quadrant, both sine and cosine is negative, so that cancels out to make cotangent positive. In the 4th quadrant, cosine is positive, but sine is negative, causing cotangent to be negative.

However, since cot= cos/sin then the cotangent graphs would have asymptotes whenever sine=0, making the ratio undefined. According to the Unit Circle, we know that sin=0 when the y value is 0, so at 0 or 2π (360*), and π (180*). The asymptotes then keep repeating itself at the same intervals since the domain goes on and on from left to right. Cotangent will never be able to touch and cross the asymptotes, no matter how close it gets to it. Furthermore, because of the different asymptotes that act as boundary, cotangent graphs go "downhill" instead of "uphill" like tangent graphs do.

c) Secant?
How do secant and cosine relate? Well, looking at the cosine graph alone, we know that cosine is positive in the 1st and 4th quadrants, but negative in the 2nd and 3rd quadrants. Since secant is the reciprocal of cosine and correlates to it, secant is also positive in the 1st and 4th quadrants, and negative in the 2nd and 3rd quadrants.

Also, since secant is the reciprocal of cosine, then the secant graph is basically the cosine graph flipped. For example, since the the slope of the cosine graph is negative in the 1st quadrant, making the values very small fractions. So if we were to take the reciprocal of that for secant, we would end up with these huge numbers.

However, since sec=1/cos, then there would be an asymptote whenever cos=0, making the ratio undefined. We know that cos=0 when the x value is 0, so at π/2 (90*) and 3π/2 (270*) there would be an asymptote for secant, which repeats themselves again forever at the same intervals. The secant graph then is at the peak of each cosine mountains and at the bottom of each cosine valley.

d) Cosecant?
How does cosecant relate to sine? Looking at the sine graph alone, we know that sine is positive in the 1st and 2nd quadrant, but negative in the 3rd and 4th quadrant. Since cosecant is the reciprocal of sine (csc= 1/sin), then cosecant would be positive wherever sine is positive and negative wherever sine is negative. Cosecant would also have to touch the peaks of the sine mountains and the bottom of the sine valleys. Then, it spreads out from both sides.

Then, we know that that csc= 1/sin. So cosecant would basically the sine graph flipped. However, cosecant will have asymptotes wherever sin=0, since that would make the ratio undefined. Sin=0 at 0 or 2π (360*), and π (180*). The cosecant graphs would spread out from the peak/valley and approach as close it can possible can to the asymptotes (because they keep on repeating over and over again at the same intervals) on either side but will never touch or cross them, instead the y-value would keep increasing/decreasing (going to +/- infinity).

Wednesday, April 16, 2014

BQ #2: Unit T Concept Intro

2. How do the trig graphs relate to the Unit Circle?

If we "unwrap" the unit circle and put it onto the horizontal axis of a standard graph, we see the values of the trig functions would correspond. For example, sine is only positive in the first and second quadrants of the unit circle, so when we unwrap the values of sin would be positive from 0 to 180 degrees or 0 to π. Then, after π, all the sine values will become negative until the graph reaches 2π and it becomes zero. Also, we know that sine of 0 degrees (1,0), 180 degrees (-1,0), and 360 degrees (1,0) is all equal to zero since it's just y/r. That is why sine graphs start and end at zero while being zero again in the middle.

In addition, cosine is only positive in the first and fourth quadrants of the unit circle. So from 0 to π/2, the values of cosine would be positive, become negative from π/2 to 3π/2, and become positive from 3π/2 to 2π. However, the graph starts and ends at 1 because the cosine of 0 and 360 degrees (1, 0) is equal to 1.

For tangent, it is only positive in the first and third quadrant of the unit circle. So from 0 to π/2 the values are positive, negative between π/2 to π, becomes positive again from π to 3π/2, and negative from 3π/2 to 2π. Or you could look at the trig ratios since tan= sin/cos. So whenever sine or cosine is negative, tangent will also be negative. Or if both functions are positive/negative, then tangent will be positive.We also have to make sure that we add the asymptotes, since tangent would become undefined whenever cosine is equal zero. Sine and cosine will never have asymptotes since their denominator of r will always be 1 and not undefined.

a) Period? Why is the period for the sine and cosine 2π, whereas the period for tangent and cotangent is π?

b) Amplitude? How does the fact that sine and cosine have amplitudes of 1 (and the other trig functions don't have amplitudes) relate to what we know about the Unit Circle?

Amplitudes are half the distance between the highest and lowest points on the graph. Sine and cosine both have an amplitude of 1 because the Unit Circle is restricting their values. On the Unit Circle the highest and lowest values for sine and cosine would be –1 and +1. They could be anything from –1 and +1, but not anything higher or any lower because there were no other points beyond them. That is why we had no solution every time sine or cosine was equal to number greater than 1 or less than -1. Tangent and cotangent, on the other hand, never had any such restrictions since they were equal to the ratio of sine and cosine.Since tangent and cotangent go on forever (to positive and negative infinity), then they won't be able to have or find an amplitude. Its range will always be (negative infinity, positive infinity), making no definite highest and lowest point.Secant and cosecant will also have no amplitudes because they will also be under no restrictions from negative infinity to a certain point, and another certain point to positive infinity. Therefore, they too do not have definite highest and lowest points.

Works Cited:
http://www.purplemath.com/modules/triggrph.htm

Thursday, April 3, 2014

Reflection #1: Unit Q-Verifying Trig Identities

1. What does it mean to verify a trig identity?
To verify a trig identity just means that you are just verifying, or proving, that the equation is true. That means you have to use logical steps to show that one side of the equation equals another. However, you can only work on one side of the identity at a time.

2. What tips and tricks have you found helpful?
You should probably be knowledgeable in all the identities. If you knew all your ratio, reciprocal, and Pythagorean identities, then it would be quicker and easier for you to verify. Pythagorean identities could especially be helpful when you have a squared trigonometric function. There are several ways to verify a trig identity, so a different combinations of techniques would still get you the same answer. You should start off with the more complicated side first. You look for a GCF (greatest common factor), LCD(least common denominator), multiply by the conjugate, factor, or substitute an identity. If you have a monomial denominator, then you could separate the fractions or if you have a binomial denominator, you can try combining the fraction. As a last resort, you can convert everything to sine and cosine.

3. Explain your thought process and steps you take in verifying a trig identity.
I always first look if the identity could have a GCF that I could factor out or FOIL. If not, I look to see if substituting an identity would help me. Like if a function is squared, then I'll check if I can use a Pythagorean Identity. If the identity is a fraction, I'll check to see if I'll either multiply the conjugate, separate it into fractions if it has a monomial denominator, combine other fractions with a binomial denominator with a LCD. If nothing else works, I would just convert everything to sine and cosine and try to cancel everything out so that it is equal to the other side.

Wednesday, March 26, 2014

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

This SP7 was made in collaboration with Katie.  Please visit the other awesome posts on her blog by going here.

Problem:
tanθ = -7/5
cosθ > 1

1. Solving other trig values using SOHCAHTOA (Unit O Concept 5)
 1. Draw triangle. See that it is clearly in Quadrant 4. According to ASTC rule, only cos and sec are positive in Quadrant 4 2. Use Pythagorean Theorem to solve for missing side.    3. Solve for trig values. Sine= opposite/hypotenuse; Cosine= adjacent/hypotenuse; Tangent= opposite/adjacent; Cosecant= hypotenuse/opposite; Secant= hypotenuse/adjacent; Cotangent= adjacent/hypotenuse
2. Solving other trig values by using identities (Unit Q Concept 2)