Origin of Difference Quotient: f(x+h) -f(x) all divided by h

This video by Mathbyfives named Difference Quotient.mov (which can be found here) verbally goes through the process in detail and can provide further explanation.

We're all familiar with the difference quotient, but where the heck does it come from and how does that affect derivatives?

Origin of Difference Quotient: f(x+h) -f(x) all divided by h

In this graph, we have the visuals to aid us in our understanding of where the difference quotient is derived from. The first point is (x, f(x)). There is delta x (delta = change) distance between the first and second point meaning the second point is (x + delta x, f(x + delta x)). But for this class we will reference to delta x as 'h' so our second point can also be written as (x+h, f(x+h). The line connecting these two points is called the secant line, much different than the tangent line, which only touches the graph once and is depicted below. Now we want to find the slope of the secant line and for that we are going to use our favorite slope-finding formula: the slope formula! The slope formula is m = y2-y1 all divided by x2-x1. When we plug in our two points from the first graph we have f(x + delta x) - f(x) all over x + delta x - x. Or as we substituted h for delta x, we have f(x+h) - f(x) all over x + h -x. We use the additive process in our denominator which cancels out the 2 x's, leaving only h. This brings us to the difference quotient we all know and love: f(x+h) - f(x) divided by the letter h, that's the difference quotient!

This video by Mathbyfives named Difference Quotient.mov (which can be found here) verbally goes through the process in detail and can provide further explanation.

Origin of Difference Quotient: f(x+h) -f(x) all divided by h

This video by Mathbyfives named Difference Quotient.mov (which can be found here) verbally goes through the process in detail and can provide further explanation.

A limit is the INTENDED height of a function while a value is the ACTUAL height of a function. When the limit and value of a function are the same then the graph is continuous meaning the absence of of one of discontinuities.

A limit

A limit

In this photo we can note the difference between a limit and value. In the first graph, there is a point discontinuity -- the limit exists, but the value is undefined. In the second graph, the limit exists at the point discontinuity as well but the value exists elsewhere (the black dot.) And the last example is one of a jump discontinuity where the limit does not exist but the value exists at one of the one-sided limits.

In which you plug in the number that is approaching the limit of f(x) and solve to see what you get.

Your possible answers are numerical, 0/# which is 0, #/0 which is undefined so the limit does not exist because of the presence of a vertical asymptote which means unbounded behavior, or 0/0 which is indeterminate form meaning we must use factoring or rationalizing method.

We use this method when we get indeterminate form. In this case, we factor both the numerator and denominator and cancel the common term -- removing the 0 from the fraction. Then we use direct substitution with the simplified expression. BUT always be sure to use direct substitution first!!

This method is also helpful with a fraction, especially if it has radicals. You multiply by the conjugate of either numerator or denominator, FOIL the conjugates and DO NOT multiply out the non-conjugate part, leave it factored. Something ought to cancel and then you simplify and once again use direct substitution with the simplified expression.

Numerically

Through this method we use a table that calculates how the limit approaches 'x' from the left and right by first subtracting 1/10 and then adding 1/10.

Graphically

You may use your calculator by simply plugging in the limit's equation, hit TRACE and then trace to the value you are searching for. OR you can put your finger to the left and to the right of where you want to evaluate the limit --if your fingers meet, that is where the limit exists, if your fingers don't meet then the limit does not exist either due to different left/right behavior (jump discontinuity), it is interrupted by a vertical asymptote which leads to unbounded behavior (infinite discontinuity), or it does not approach any single value (oscillating behavior).

First image may be found here.

All other images may be found here thanks to Mrs. Kirch's Unit U SSS Packet.

Let's go back to the unit circle - what appears to be the solid foundation of pre-calculus.

In the unit circle tangent and cotangent were related by its ratios. Cotangent's ratio was the reciprocal of tangent's ratio. Tangent's ratio is y/x meaning cotangent's ratio would be x/y. Now in this unit we would read that as tangent's ratio being sine/cosine (according to Unit Q's identities) and cotangent's ratio would be cosine/sine.

Now what these two graphs have (as well as secant and cosecant) are asymptotes. Remember that this occurs when dividing by 0 leading to undefined. Tangent's asymptotes would occur when cosine equals 0 which would be at pi/2 and 3pi/2. Cotangent's asymptotes would occur when sine equals 0, this would be at 0, pi, and 2pi. (Reflect back on the unit circle, think of these values.)

Because their asymptotes are placed in different areas, this is going to affect the direction in which they go. Also remember that these graphs WILL NEVER TOUCH the asymptotes, EVER. They will get VERY close but NEVER TOUCH.

Not only do the asymptotes affect its direction, but look at the different colored areas in the images. Red is the first quadrant of the unit circle, where all trig functions are positive. Green is the second quadrant, where tangent and cotangent will be negative (as well as cosine and secant, but not sine or cosecant.) Orange is the third quadrant, where tangent and cotangent will be positive (the rest of the trig functions will be negative.) Blue is the fourth quadrant, where tangent and cotangent will be negative (as well as sine and cosecant but not cosine or secant.)

Re-using the images from BQ #3, all thanks again to the wonderful Mrs. Kirch. Her amazing help with these graphs can be found here.

In the unit circle tangent and cotangent were related by its ratios. Cotangent's ratio was the reciprocal of tangent's ratio. Tangent's ratio is y/x meaning cotangent's ratio would be x/y. Now in this unit we would read that as tangent's ratio being sine/cosine (according to Unit Q's identities) and cotangent's ratio would be cosine/sine.

Now what these two graphs have (as well as secant and cosecant) are asymptotes. Remember that this occurs when dividing by 0 leading to undefined. Tangent's asymptotes would occur when cosine equals 0 which would be at pi/2 and 3pi/2. Cotangent's asymptotes would occur when sine equals 0, this would be at 0, pi, and 2pi. (Reflect back on the unit circle, think of these values.)

Because their asymptotes are placed in different areas, this is going to affect the direction in which they go. Also remember that these graphs WILL NEVER TOUCH the asymptotes, EVER. They will get VERY close but NEVER TOUCH.

Not only do the asymptotes affect its direction, but look at the different colored areas in the images. Red is the first quadrant of the unit circle, where all trig functions are positive. Green is the second quadrant, where tangent and cotangent will be negative (as well as cosine and secant, but not sine or cosecant.) Orange is the third quadrant, where tangent and cotangent will be positive (the rest of the trig functions will be negative.) Blue is the fourth quadrant, where tangent and cotangent will be negative (as well as sine and cosecant but not cosine or secant.)

Re-using the images from BQ #3, all thanks again to the wonderful Mrs. Kirch. Her amazing help with these graphs can be found here.

Remember that an asymptote results when a ratio is divided by zero, becoming an undefined ratio.

Tangent's ratio is y/x, meaning that there is a possiblity of an asymptote when x (cosine equals 0). Because of this we know cosine equals 0 at 90* and 270* so pi/2 and 3pi/2, this is where are asymptotes will be located. Now let's look at the first quadrant, remember that in the first quadrant all is positive so it goes in an uphill direction and will NEVER touch the asymptote of pi/2, it simply gets really, really, really close to it. In the second quadrant, both sine and cosine are heading in a downhill direction and in the second quadrant tangent is not positive so it heads downwards but in the third quadrant it is positive. The graph can continue in these two quadrants because there is no asymptote dividing them. And in the fourth quadrant, tangent is not negative hence it's downhill direction.

Cotangent

For cotangent our ratio is x/y, meaning we will have our asymptotes where y =0 (sine) and those locations would be at 0*, 180*, and 360*. On our graph these would be the values of 0, pi/2, 2pi. Because both sine and cosine are positive in the first quadrant as everything is, cotangent is positive as well. Yet in the second quadrant, sine is positive and cosine is negative leading contangent to continue in the negative direction crossing the x-axis when cosine does as well. The first and second quadrants already contain one period of cotangent. A similar process continues off from the asymptote of pi in the third and fourth quadrants. Because sine and cosine are negative in third quadrant, cotangent will be positive and because sine is negative and cosine is positive in the fourth quadrant, cotangent will be negative.

Secant

Remember that secant is the reciprocal of cosine's ratio which will be r/x meaning that there will be asymptotes where cosine is equal to 0, similar to tangent's asymptotes. A similar pattern also follows here.

In the first quadrant, both sine and cosine are positive and so will secant. But however in the second quadrant, sine and cosine are both negative, as will secant and will continue to be negative because although cosine is positive in the third quadrant, sine is negative and a negative and positive will result in a negative. In the fourth quadrant, both sine and cosine are moving in an uphill direction and secant will also be positive. Once again notice how none of secant's graph is touching the asymptotes and how they develop at the mountains and valleys of cosine's graph.

In the first quadrant, both sine and cosine are positive and so will secant. But however in the second quadrant, sine and cosine are both negative, as will secant and will continue to be negative because although cosine is positive in the third quadrant, sine is negative and a negative and positive will result in a negative. In the fourth quadrant, both sine and cosine are moving in an uphill direction and secant will also be positive. Once again notice how none of secant's graph is touching the asymptotes and how they develop at the mountains and valleys of cosine's graph.

Cosecant is the inverse of sine meaning its ratio will be r/y. This being said, cosecant will have asymptotes wherever sine equals 0, also similar to cotangent (oooh connections!) In the first quadrant secant will remain positive because all functions are positive in the first quadrant. Yet although cosine is negative and sine is positive in the second quadrant, cosecant will still be positive because it is positive in the sine quadrant of the unit circle. The graph continues into the third and fourth quadrant even after having gone through its period and it's direction is enforced by the unit circle's positive or negative values of the cosecant function. However more importantly because cosecant is the inverse of sine, it relies on the sine graph to be drawn because once again notice, like secant, it is drawn on the mountains and valleys of its corresponding reciprocal of the sine function.

All images made available thanks to the amazing and wonderful Mrs. Kirch on Desmos, you can view and animate as well here.

Image found here.

Image found here. (thank you Google!)

Sine and cosine do not have asymptotes because they are always divided by 1 (by "r")

Yet once we move away from being divided by "r" we start reaching asymptote area.

If you reference to the photos you will notice undefined under tangent. Do not forget about their reciprocal ratios of cosecant (r/y), secant (r/x) and tangent (x/y).

Remember that we get asymptotes when our circle ratios equal **undefined which results when you divide by 0. **

For *sine and cosine* you will never be dividing by 0 as you may for all other four circle ratios. But because you'll never be dividing by 0, you will never reach undefined, so you'll never have asymptotes!

Period: The period for sine and cosine is 2pi because it takes four quadrants (ASTC) to repeat the pattern. Sine's pattern according to ASTC from the unit circle is + + - -

It merely takes all of the unit circle (which is 2pi at 360* to complete sine and cosine's patterns.

Graphs are merely snapshot of the graph, these graphs are infinite as circles are - no ends, no beginning but for the sake of this class we will only be graphing one single period.

Meanwhile all the other trig functions have asymptotes because they do not have "r" such as tangent and cotangent (y/x and x/y respectively.)

**continue scrolling**

**continue scrolling**

**here we go, but continue scrolling just in case**

Meanwhile with tangent and cotangent.....

**Period: **Tangent and cotangent have a period of just pi. According to ASTC from the Unit Circle, the pattern this trig function has is + - + -. Meaning the pattern is completed in the first two quadrants which is 180* aka pi aka half of the circle.

Just remember again to keep in mind that a period is one time through their cycle, their pattern.

Just remember again to keep in mind that a period is one time through their cycle, their pattern.

This SP #7 was made in collaboration with Anthony Lopez. Please visit the other awesome and non-cheesebuckety posts on their blog by going here.

**Also, both Anthony (cheesebucket) and I worked together on this post but someone *cough*Anthony*cough* has poor penmanship so in order to spare you the trouble of trying to decipher it, here it is in my writing.

##
Identity Work

*What is this problem about?*

This quadrant discusses how to find ALL trig. functions when you are given one trig. functions and quadrant, using the identities from concept 1. It allows us to apply these fundamental identities to more difficult concepts such as this one.

*What must the viewer pay special attention to in order to understand? *

You first have to decide the quadrant in which the values lie in by using ASTC quadrants, this will only determine the signs (very important!!) But also please remember you must using reciprocal, ratio, or Pythagorean identities to find the remaining values (don't rely on just SOHCAHTOA, we will use this to find our values.) AND you can only have ONE unknown trig function within solving the identity no matter what identities you use. Also make sure that you properly rationalize all your answers!!

##
SOHCAHTOA Work (Verification)

####
We use SOHCAHTOA to verify our answers from the identities we had previously used. We found the hypotenuse value by using the Pythagorean theorem and then simply used the values from the triangle in our ratios to verify that what we had originally gotten with the identities was correct.

**Also, both Anthony (cheesebucket) and I worked together on this post but someone *cough*Anthony*cough* has poor penmanship so in order to spare you the trouble of trying to decipher it, here it is in my writing.

This quadrant discusses how to find ALL trig. functions when you are given one trig. functions and quadrant, using the identities from concept 1. It allows us to apply these fundamental identities to more difficult concepts such as this one.

You first have to decide the quadrant in which the values lie in by using ASTC quadrants, this will only determine the signs (very important!!) But also please remember you must using reciprocal, ratio, or Pythagorean identities to find the remaining values (don't rely on just SOHCAHTOA, we will use this to find our values.) AND you can only have ONE unknown trig function within solving the identity no matter what identities you use. Also make sure that you properly rationalize all your answers!!

Let's go back to oh like maybe three weeks?? (I'm awful with time) where we first learned about the Unit Circle. In the Unit Circle we learned that the ratio for cosine is (x/r) and the ratio of sine is (y/r). Now let's reflect back on Pythagorean Theorem which is a^2 + b^2 = c^2, but in context of the Unit Circle it would be x^2 + y^2 = r^2, but in the Unit Circle r always equaled 1 and in order to make this true and correct, we would divide by r^2 on both sides, leaving us with (x/r)^2 + (y/r)^2 = 1. But wait...(x/r) and (y/r) look similar...wait....isn't that cosine and sine?

2.

In order to derive the remaining two Pythogorean Identities we must divide the whole thing by cos^2x to find the tangent derivation which will lead us to tan^2x + 1 = sec^2x

- The most obvious one would be the role of sine and cosine in this Unit Q from the Unit Circle and its properties which can be seen in the reciprocal identities which are similar to the inverse of these trig. functions from Unit O and P.
- In Unit P Concept 3 when using the distance formula for Law of Cosines with SSS or SAS, one of the formulas is a^2 = c^2(sin^2A + cos^2A) - 2bcCosA + b^2 and we know that cos^2A + sin^2A = 1.

mentally-draining (this counts as one word), stressful, and insightful.

Problem: Scott and Phil agreed to meet at the record store so Phil can show Scott what he's been up to in life. They realize they're 15 feet apart. Scott heads to the store at an angle of N25E while Phil is heading off at an angle of N55W. Considering that Phil is a slowpoke, how far is he from the record store?

Solution:

Problem: Phil and Scott decide to go on a boat ride to relax and just ride the waves out after such a long day. Phil gets to the harbor before Scott however and told him to amount the SS Kirch at 5pm. At 5pm, Scott gets aboard the SS Cheesbucket. Phil's boat, the SS Kirch, is traveling at 80 mph and is at a bearing of 220* while Scott's boat is traveling at 115 mph at a bearing of 335*. It takes four hours for both "men" to realize they aren't aboard the same boat (they're huge ships and cell reception isn't exactly great out in the waters.) How far apart are the boats when Phil and Scott finally realize they're aboard different ships?

Solution:

Now Scott's boat, the SS Cheesebucket is traveling at 115mph for 4 hours then it has traveled a distance of 460 miles. Phil's boat, SS Kirch travels at 80 mph for 4 hours so really has traveled 320 miles.

Our triangle is SAS, making sure our included angle is between two sides. Now that we have this information we can plug it into our equation of Law of Cosines, **a^2 = b^2 + c^2 - 2bcCosA**, which we plug straight into our calculator and then square root to find that the distance between Scott and Phil is a tragic 662.13 miles apart...someone better start swimming now.

In an SSA triangle, our three angles are not carved in stone as AAS or ASA triangles where even though we had two angles, in reality we had all three because all one had to do was add the two and subtract it from 180 to get the third angle's values, meaning there was no ambiguity there.

With SSA we only know one angle's value! (and two sides) There is a possibility however that with only this information that there may be no triangle at all! We always start the problem assuming there is two possible triangles, until you "hit a wall" which will lead to only one possible triangle or no possible triangle. A "wall" can either be if SinA=1.1 or 2.8, something that is not possible. Remember from our unit circle trig functions, we know that sin and cos are not compatible and do not work with numbers greater than one,

In this triangle, we are solving for angle B and C, and side b. We have information for both angle A and side a, our magic pair, and we have information for side C but not side c, which is our bridge. We set our magic pair and bridge in proportion to one another to solve for the angle of C and use inverse sin to find it. But because we are assuming we have two triangles, and we are using sin

In a normal triangle where the area could be found with the formula of A = 1/2bh, we know that the perpendicular height of the triangle is h and b is the base. While the oblique triangles (where all sides are different lengths) area is** one half of the product of two sides** and the **sine of their included angle <- very important to realize, it must be in between the two sides. **

We know that sinC=h/a from concept 1, so once you mulitply 'a' on both sides h (height) = asinC, and plugging 'h' into that A=1/2bh formula we really have A=1/2b(asinC)

Our dear friend Phil has been struck with the wish to travel around the country, he's been bitten by the wander bug.

##
The Problems

a) Phil packs his bags and takes a roadtrip up to Seattle. There he finds Seattle's Space Needle of**605 feet** casting a **410 foot** shadow at which Phil is standing at the end of. If Phil looks **at the top** of the building (and avoids getting his eyes burnt by the sun) what is the **angle** of Phil's eyes to the top of the building? (to the nearest hundredth of a degree AND assume Phil's eyes are **5 feet above ground level** since he's a short little fellow.(http://upload.wikimedia.org/wikipedia/commons/3/38/BMX_aloft_and_Space_Needle_03.jpg)

b) Once Phil gets tired of all the Space Needle this and Space Needle that, and the city overall in general, he heads over to Yosemite where Moro Rock lies standing tall at **300 feet**** **where he can enjoy the panoramic view. But the mighty rock is slanted at its highest point, Phil estimates the angle of depression from where he is standing (at 300 feet) to the bone-crushing bottom to be **53*****.** Should Phil fall, like the klutz he is, how long is his path to certain doom? **(Round to the nearest foot.) **

http://triggerpit.com/wp-content/uploads/2012/01/sequoia-national-park-moro-rock.jpg

##
The Solutions

b) We know that the height is 300ft and the angle of depression to be 53*. We have the opposite side of 300 and we are looking for x which is the hypotenuse. Opposite and hypotenuse, SOH - Sin. So this time it's sine of 53. So sin 53 = 300/x, we multiply by x on both sides which leads to us dividing 300 by sin 53 which equals approximately 758 feet.

##

a) Phil packs his bags and takes a roadtrip up to Seattle. There he finds Seattle's Space Needle of

http://triggerpit.com/wp-content/uploads/2012/01/sequoia-national-park-moro-rock.jpg

a) Because it is from Phil's eye level of 5 feet we subtract 5 from 605 giving us 600. Its height is now 600 and the length is 410 which gives us the opposite (600) and adj (410) side. So we are looking for the angle (x). We use tan because of TOA and we use the inverse to undo tan to find our angle so it is tan -1 x (600/410) which is 55.65 degrees.

First of all, make sure you recognize that this is

We cut straight down the triangle because if 180* is the total sum of the triangle's angle we are left with 30* at the top, 60* at the side, and 90* at the bottom of the triangle from 30*.

And since it is an equilateral triangle, all of the sides are of equal length of 1. But since we cut the triangle in half, the bottom side's length is cut to 1/2. With this information, we're able to find the height since the hypotenuse (c) equals 1 is already given to us and we know that one of the values is 1/2, let's call 1/2 (a) for now. If we plug these values into the Pythagorean Theorem of a^2 + b^2 = c^2,

If we give 1 a variable instead, like n, we are left with 30* - 1/2n, 60* - radical 3/ 2, and 90* - 1n. But we use the variable

This time we are given an equilateral square whose sides' lengths are 1. Now a squares' total angle sum is 360* meaning each corner has an angle of 90*. We cut

To find our pattern, we use a variable - which represents any number. Let's use 'n' for now. By using n to represent our two equal sides of 1, our hypotenuse now equals n radical 2; the variable does allow for the relationship to stay consistent and allows us to expand it should the case ever be that the side length equal a number greater than one

Subscribe to:
Posts (Atom)