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Post by sawdy on Aug 9, 2017 23:07:14 GMT -6
Okay, so this has likely gotten a bit tedious for everyone, so it would probably be easier to have anyone interested just send a message... I've been learning a lot. At times I need to take a break from reading it as it takes a lot of brain power for me, but I get back into it. I am not a math whiz, so I can't generally comment, but I appreciate your explanation. Please keep posting. 😀 I still own a scientific calculator- although I only use it to add up the scores when my family plays Ticket to Ride. 🤣. From the example I gave, you can see that although I did take math in high school, it has definitely been years since I review it. Your explanations are really helping me. 👍 |
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Post by yardstick on Aug 10, 2017 13:09:43 GMT -6
Well, thank you for saying so.
I'll try to move it along a little faster, but I am finding it difficult to not go too slow for those skilled at math, while not going too fast for those who are weaker in the topic.
I'll try to cover a bit more trig, then show how calculus is used with the basic functions, and then start applying the two to demonstrate the 5th-10th dimensions.
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Post by whatif on Aug 10, 2017 13:13:53 GMT -6
You're doing a great job, yardstick! Keep up the good work!
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Post by yardstick on Aug 10, 2017 20:57:50 GMT -6
(continued) Alright, lets pick up where I left off. I was talking about using the basic trig functions sine, cosine and tangent and also pythagorean theorem. Some of the stuff I typed before might have been a little confusing, because the alpha looks like an a. Most math instruction uses theta, θ, to represent the angle of interest, so that is what we will use. I am also going to walk you through some of the other interesting trig relationships and functions before applying calculus to them, but first i want to make sure everyone understands that at the most basic level, all Trigonometry is, is the relationship between circles and triangles. I will draw you what I mean in a moment. I need to divert for one moment and discuss two terms: supplementary and complementary, with respect to angles. Supplementary means that two angles add up to 180 degrees. Complementary means that two angles add up to 90 degrees. Also, does everyone remember that the sum of all the angles in a triangle is 180 degrees? You will see why i brought up supplementary and complementary in a moment. I will also be using the shorthand for sine cosine and tangent: sin, cos and tan respectively. So here is a review pic:  The angle of interest is marked with θ. Its opposite side is b, and its adjacent side is a. The hypotenuse is c. So next, lets overlay a circle:  So when we do this, we see that the side a goes in the X-axis direction, and the side b goes in the Y-axis direction and the hypotenuse c is essentially the radius of the circle. So lets take our equations and have a closer look at them:  Because of what we notice from the last pic, we can substitute the X for the a, the Y for the b and the R for the c. b = c sin θ ----> y = r sin θ a = c cos θ -----> x = r cos θ y tan θ = --- <<---------- slope m! x (continued...) |
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Post by yardstick on Aug 10, 2017 21:26:33 GMT -6
So now we have 3 formulas derived from the circle overlaying the triangle:
x = r cos θ
y = r sin θ
y
tan θ = ---
x
But what about pythagorean theorem?
a2 + b2 = c2 What if we substitute the values of a, b and c into this?
a2 = y2 = (r sin θ)2
b2 = x2 = (r cos θ)2
c2 = r2 and if we plug in these values to the pythagorean theorem:
x2 + y2 = r2 this is the equation of a circle!
(r sin θ)2 + (r cos θ)2 = r2
we can now expand the left side:
r2 sin2θ +r2 cos2θ = r2
then all we have to do is divide both sides of the equation by r2 to get:
sin2θ +cos2θ = 1 which is the first trig identity. (Strangely enough, I have yet to see a math teacher go from pythagorean theorem to equation of a circle to first trig identity. Go figure.)
I just need to cover a couple more things and then we can move back to a little calculus.
recall that
x = r cos θ
y = r sin θ
y
tan θ = ---
x
and my brief mention of supplementary and complementary angles?
well, co-sine is short for complementary to sine
and here is where I introduce to you the next 3 basic trig functions:
secant, cosecant and cotangent: sec, csc and cot, which are all inverses...to the first three basic trig functions. That is:
1
cosecant (csc) = -------- = sin-1 (remember negatives in the exponent?)
sine
1
secant (sec) = ----------- = cos-1
cosine
1
cotangent (cot) = ------------- = tan-1
tangent
and you probably noticed this: co-secant + secant = 90 degrees co-tangent + tangent = 90 degrees
Okay, I'll cover how to do calculus with trig functions next time, because the method is not the same. I will make it as easy as drawing a picture though.
(continued...)
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Post by yardstick on Aug 14, 2017 19:29:54 GMT -6
(continued)
okay, I need to pick back up where I left off, do a couple more things, and then move on to 'mechanical' part, where I hope to adequately demonstrate the applications of the math I talked about - for showing 5th-10th dimensional stuff.
Recall that I did a demonstration of how to get from pythagorean theorem to the first trig identity. Now I need to demonstrate how to get to the second and third trig identities, and then review what I have given everyone, and then go to the mechanical stuff.
so starting with the first trig identity:
sin2θ +cos2θ = 1
and dividing everything by cos2θ gives:
sin2θ + cos2θ = 1
------ ------ ----- ------->> tan2θ + 1 = sec2θ
cos2θ cos2θ cos2θ
How did I get this? Well, we know from previous posts that sine/cosine = tangent, and we know that anything (cosine) divided by itself = 1; and we know that the inverse of cosine is exactly opposite of the inverse of secant, therefore the inverse of cosine is secant. It naturally follows that the we can use the squares of these functions in lieu of the functions themselves, and get the squared results.
Similarly, dividing everything by sin2θ will give us:
sin2θ + cos2θ = 1
------ ------ ----- ------->> 1 + cot2θ = csc2θ
sin2θ sin2θ sin2θ
okay, still need to cover how to tell whether you are working with sine or cosine, as well as how to take derivatives using calculus, because it is not the same as a regular polynomial equation; but both of these are as simple as drawing a circle.
(continued...)
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Post by yardstick on Aug 17, 2017 9:33:08 GMT -6
(continued) Okay, I am going to try to add a little more here. I promise, I am almost done with the pure math stuff. Remember how to get the circumference of a circle? C = 2πr, right? where π = pi = 3.14... So a full circle of 360 degrees = 2π, when the radius is 1 (as it is in the unit circle): C = 2π(1) = 2π What happens if we only travel halfway (180 degrees) around the circle's circumference? we get half of 2π ----> π How about 1/4th of the way? π/2 (90 degrees) We can use two ways to denote the value of the angle θ: in degrees, and in radians (the π nomenclature). The conversion is simple: pi (π) 360 actual degrees x ---------- = radians or actual radians x ----------- if we want to go from radians to degrees. 360 pi (π) Now, recall that y = sine θ and x = cosine θ If we want to demonstrate how cosine θ = 1, on the unit circle (a circle with a radius of 1 unit), we would need to figure out what angle θ gives us that value. Same for sine θ = 1. We use the value 1 because it is directly applicable to the unit circle, and is the easiest to work with. You will probably need your scientific calculator for this next part. for cosine θ = 1, we can solve for θ, by taking the inverse of cosine on both sides of the equation: x = cosine θ, for x = 1: 1 = cos θ cos θ = 1 cos -1(cosine θ) = cos -1(1) the inverse of cosine 'undoes' the cosine, but what we do to one side of the equation, we have to do to the other side also. θ = cos -1(1) θ = 0 degrees If you plug 1 into your inverse cosine function on your calculator, you should get 0; that is, the cosine angle to get a value of 1 is 0 degrees. Similarly, if we solve the sine function y = sine θ, (and letting y = 1) 1 = sine θ sin θ = 1 sin-1(sin θ) = sin-1(1) θ = sin-1(1) θ = 90 degrees, which is π/2 What happens on the other side of the circle though, at -1? x = cosine θ, for x =-1: -1 = cos θ cos θ = -1 cos -1(cosine θ) = cos -1(-1) θ = cos -1(-1) θ = 180 degrees, which is π y = sine θ, (y = -1) -1 = sine θ sin θ = -1 sin -1(sin θ) = sin -1(-1) θ = sin -1(-1) θ = 270 degrees, which is 3π/2 And we can show these graphically like this:  Some study of this sketch will show that sine tends to oscillate on the y axis, and cosine tends to oscillate on the x axis from 1 to -1; and that these oscillating values are due to the projection of the point on the circle's circumference onto the axes. So when the point on the circle is at (1,0), x = 1 and y = 0 , the value of θ = 0 degrees or 0 radians, whether you use sine (y value = 0) or cosine (x value = 1) to calculate it. (0,1), x = 0 and y = 1, θ = 90 degrees, or π/2 radians (-1,0), x = -1, y = 0, θ = 180 degrees, or π radians (0,-1), x = 0, y = -1, θ = 270 degrees, or 3π/2 radians It turns out that there are two common triangles, that when overlayed on the unit circle, give us some patterns for their projected sine and cosine values; or alternately, some patterns for their projected angles: 45-45-90 triangle, which when overlayed gives angle values of π/4 radian increments and projected values in fractions using 1 and √2 30-60-90 triangle, which when overlayed gives angle values of π/6 radian increments and projected values in fractions using 1, 2, and √3 (continued...) |
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Post by yardstick on Aug 17, 2017 9:56:01 GMT -6
(continued) Now, I gave that sketch in the last post, because I really wanted everyone to see the next sketch, since it is really helpful when trying to figure out how to take a derivative (or it's opposite, the integral) of an oscillating function, like sine or cosine (or the other 4):  I used the Greek letter delta (Δ) to represent derivative, even though it represents average change in slope, rather than instantaneous change in slope. The elongated s ( ∫) is the integral sign. We wont get into integrals in this, but they are essentially the opposite math function of a derivative. Like +, - or multiplication and division. While derivatives represent slope, or the rate of change of something, the integral represents size. That is, derivative is location or direction, and integral is magnitude. Unlike polynomial equations, trig functions are oscillating functions. They always have values between 1 and -1 when using the unit circle. This is due to the projection of the (x,y) coordinate point located on the circle radius onto the X and Y axes, as previously described. The arrows in the sketch point to what trig function to convert to when solving a derivative or integral. That is, which way do you proceed around the circle to get to the correct derivative or intregral. Let's use an example. d --- sin x (the derivative of sin x with respect to x) = cosine x dx d --- -cosine δ (the derivative of -cosine δ with respect to δ) = sin δ dδ Derivatives move counter-clockwise around the unit circle when solving. Integration goes the opposite way. Also note that the location of sine, cosine, -sine, -cosine is different than in the first sketch. Sine and Cosine and their respective negatives are reflected (mirrored) across a 45 degree angle in this sketch, compared to the first sketch. We can also use a little shorthand when writing the above derivatives: f(x) = sin x (function) f'(x) = cosine x (derivative of function) f(δ) = -cosine δ (function) f'(δ) = sine δ (derivative of function) and also even more shorthand: y = sin x (function - recalling dependent and independent variables) y' = cosine x (derivative) λ = -cosine δ λ' = sine δ You can also use the original independent variable, and put a dot over it to indicate the first derivative you take.  A second derivative is done just like a first derivative, except you take the derivative of the first derivative: y = sin x (function - recalling dependent and independent variables) y' = cosine x (first derivative) y'' = -sine x (second derivative) λ = -cosine δ λ' = sine δ λ'' = cosine δ One way of proving that the trig functions oscillate is to take several derivatives. You will see that eventually, you get back to the original function: λ = -cosine δλ' = sine δ λ'' = cosine δ λ''' = -sine δ λ'''' = -cosine δ Second and subsequent derivatives simply add more dots to the original symbol:  this symbol is for the second derivative with respect to x. edit - I must give a caveat to taking derivatives or integrals of the trig functions other than sine and cosine, as the other 4 require special techniques such as Product Rule and Quotient Rule; which are basically shortcuts to ease the problem solving. (continued...) |
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Post by watchmanjim on Aug 24, 2017 22:35:57 GMT -6
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Post by uscgvet on Aug 25, 2017 10:14:30 GMT -6
Yea, this is a really good find. Using Base 60 is such an interesting way to handle division by 3. |
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Post by yardstick on Aug 26, 2017 20:02:55 GMT -6
Saw it! Commented on it on another thread! They have an easier trig table because they are using base 60, not base 10 for their numbering. |
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Post by yardstick on Aug 26, 2017 20:04:31 GMT -6
Yea, this is a really good find. Using Base 60 is such an interesting way to handle division by 3. Base 12 and base 36 are also good that way. I believe the Maya used base 12. |
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Post by uscgvet on Aug 27, 2017 12:44:43 GMT -6
Yea, this is a really good find. Using Base 60 is such an interesting way to handle division by 3. Base 12 and base 36 are also good that way. I believe the Maya used base 12. Yes Base 12 would also make since due to giants likely having 6 fingers. |
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Post by yardstick on Aug 28, 2017 0:06:15 GMT -6
Base 12 and base 36 are also good that way. I believe the Maya used base 12. Yes Base 12 would also make since due to giants likely having 6 fingers. interesting point! |
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Post by yardstick on Aug 29, 2017 21:46:22 GMT -6
(continued) Okay! at long last! The next lesson! Some of you are no doubt groaning; and others have the eye-roll going, but I will press on! Now I am going to apply the subject matter I talked about before to real life. I am going to keep it as simple as possible, but mechanically, I hope you will all find it obvious. Let's start with a mechanism called a crank-slider, which as a motorized crank, which drives an arm, connected to a second arm (linkage) which is in turn connected on the other end to a block. The block has its movement restricted to only oscillate (trig!) left to right. Here is a sketch:  As the crank revolves around its center point, its connecting arm (linkage) causes the block to slide back and forth. So in order to demonstrate multi-dimensional math, we have to analyze the movement of the crank, the connecting arm, and the block. Let us begin by labeling each, so we know what is what:  So we have labeled: r 1 - the 'arm' between the crank and the slider θ 1 - the angle between the crank and the slider r 2 - the crank arm θ 2 - the angle that the crank arm is at at any one moment r 3 - the 'linkage' arm θ 1 - the angle between the linkage arm and the horizontal So, what are some direct observations we can make with the sketch and the knowledge we have right now? 1. the r 2 and r 3 arm lengths are fixed. They do not vary in length. 2. the θ 1 angle is fixed, it does not vary in its angle. 3. The θ 2 angle varies as the crank rotates. 4. The θ 3 angle varies as the crank rotates. 5. the r 1 arm length varies as the crank rotates. (continued...) |
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