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1
Technology, Science & Alt Science / Re: Am I correct?
« on: April 05, 2019, 07:03:21 PM »
For any object in space– omitting gravity, IMPOV

An object of any mass (continuous) irrespective of size can be pushed or pulled without any resistance - Right?
Wrong.  You still have to overcome inertia which is the resistance to change in motion of a mass.
if an object of any mass (continuous) irrespective of size can't be pushed or pulled without any resistance then why the total momentum of the system remains sustained after a collision in the presence of resistance - please explain

as you said if i want to overcome inertia, i have to apply a force but aren't all the particles of mass falling towards their common point of the center of gravity which they make so where is inertia?

don't forget i said omitting gravity

Inertia means resistance or holding grounds - Shouldn't B break into parts before sweeping by A if really offer inertial resistance or holding ground against A

2
Technology, Science & Alt Science / Am I correct?
« on: April 04, 2019, 02:23:57 PM »
For any object in space– omitting gravity, IMPOV

An object of any mass (continuous) irrespective of size can be pushed or pulled without any resistance - Right?

It is the gravitating mass, due to which a falling mass shows resistance because of its weight against another force. For Example an object on earth

If this is true then shouldn’t the definition or the concept of inertia, which means resistance, needs revising.

No idea if someone agrees with the following but

An object is said to be in a state of a well-balanced condition if its centroid or center of mass (c.o.m) locates itself at its spontaneous strategic position if left on its own accord. An undisturbed mass at rest is always in well-balanced condition.

An aforementioned object is said to be in a state of an unbalanced condition if its c.o.m off its strategic position due to any means. The course of the shifting of the centroid, which is under duress due to an unbalanced condition, not only moves the object forward but also guides the direction of its motion. A disturbed mass is always at unbalanced condition - Motion. Thus

An object is said to be at rest if its c.o.m remain at its original position.

An object is said to be in state of motion if its c.o.m off its original position.

An object may spin about its undisturbed c.o.m in the direction of applied force when only a part of the mass is disturbed.

A mass may spin and move if its c.o.m and a part of the mass are disturbed.

An object may also be found at fully or partially disturbed and undisturbed conditions

Let A and B are two spherical objects of masses M and m at rest such that A > B and therefore M > m. Ca and Cb are the centre of masses of A and B respectively.

Both A and B maintain their state of rest if Ca and Cb maintain their original position – Undisturbed state

Both A and B lose their state of rest if Ca and Cb loss their original position respectively - Disturbed state

Now, assume A is disturbed and moves with velocity V while B is undisturbed. Following is one the possible conditions when A collides with B.

A pushes B in front of it in its original direction of momentum until both gains V1.

Here B never offers any resistance to A rather it takes the momentum from A – total momentum of the system still remains the same.  Let a and b are the jitters/impulses (shock waves) produced within A and B respectively due to their collision. Cb shifts away from its original position when “b” passes through it, which makes B unstable. Similarly, Ca also shifts towards its original position when “a” passes through it, which makes A less unstable than before.

Although both A and B osculate (juxtaposed) each other but exert no further force on each other after when both masses attains V1. Both Ca and Cb off center, therefore, A is under reduced momentum of while B gained a momentum.

A or B spins if the line action of a and b are truncated – not passes through the centroid.

Push or pull is considered a force. Since A pushes B due to its momentum, therefore, momentum is a Force F.

So force which is push or pull depends upon on both moving mass and its final velocity, not acceleration.

For any object in space – considering gravity

As said, it is the gravitating mass, due to which a falling mass shows resistance because of its weight against another force, therefore the heavier the mass the greater will be its gravity and resistance or the greater the mass the greater the force will be required to displace its c.o.m, therefore, a tiny apple can’t change the strategic position of c.o.m of earth.

The c.o.m of falling mass is below it original position while the c.o.m of flying object is above its original position

The collision effect of the aforementioned A and B depends upon the size, shape, density, and velocity, etc therefore both a and b may or may not reach the center of mass of A and B respectively.  The two outer particles of A and B at tacnode, which collide with other, start exerting a force on the neighboring particles, the said neighboring particles pass a or b to the next connected particles closely and so on and this is how shock wave passes through A or B or M and m respectively.

Both a and b depend upon the mass below, above, right, left, back and in front and how their elasticity or interlocking system, etc is.

It is said that all objects fall at the same rate if this is true then why the damaging/penetrating effect of the same mass is different if fall (at the same rate) from different heights on the ground.

This means force is directly proportional to the mass of the falling object and its final velocity (not acceleration)

Therefore Force F = MV but not F = ma or mg where g=GM/d^2 or 9.8 m/s/s.

Similarly, addition or multiplication of two or more things of different types (e.g. goats and trees) has no useful meaning in mathematics unless totaling them, therefore, I don’t understand why mass and velocity (or mg) are in the multiplication form with each other in the formula of momentum. Why not momentum = M+V if the product of MV is allowed. And the same is applied to all similar mathematical equations.


Is 3 (goats) x 5 (trees) = 15 goat-tree possible or meaningful?

Anyway, shouldn’t force be measured relative to the displacement of c.o.m of moving mass as explained above and or indirectly via relative to standard penetration on the ground or any standard surface?

Addendum: Neither an apple nor earth shows resistance when they feel the same amount of force pulling each other together. Similarly, if an object of any mass (continuous) irrespective of size can be pushed or pulled without any resistance then why the greater the mass the greater force required to move it? Shouldn’t the acceleration of an apple and earth be the same when they feel the same amount of force pulling each other together? 

3
Technology, Science & Alt Science / Re: Question about the calculus
« on: October 19, 2018, 01:26:37 AM »
Anyone who is agreement with the following -  One Vote is enough

Derivatives show us how fast something is changing at any point. For example; the gradient of the graph of y = x2 at any point is twice the value of x thereat. The process of finding the derivation of a gradient / slope of a function y=f(x) or y = x2 is as follow.

Pick any two points A and B close to each other on the curve of y =x2. The coordinates of A on the curve are (x, y) or (x, x2). Add Δx at A as usual. When x increases by Δx, then y increases by Δy. The x changes from x to (x +Δx) while y changes from y to (y + Δy) or f(x) to (x+Δx)2. Thus the x and y coordinates of B on the curve are (x + Δx, y + Δy) or ([x+Δx, (x+Δx)2]. Now the instantaneous rate of change is given by


Δy/ Δx = [(x + Δx)2 – x2] / [x + Δx - x]

Δy/ Δx = [x2 + Δx2+2xΔx − x2] / Δx

Δy/ Δx = [2x + Δx] / 1

Reduce Δx close to zero by taking limit (Δx to dx and Δy to dy)

dy /dx = 2x + dx

dy /dx = 2x------Eq1 OR

dy = 2x.dx --Eq2


ABC is an infinitesimal triangle made by dx, dy, and hypotenuse or slope of tangent where point A and C are always on the curve. Length of AB = Base = dx, Length of BC = Perpendicular= dy and Length of Hypotenuse = AC. Angle CAB or BAC is the slope of a tangent

According to the aforementioned Eq1 or Eq2-

•   dy/dx is directly proportional to x or angle CAB is directly proportional to x.

•   dx is indirectly proportional to x OR x is inversely proportional to dx

•   dy is directly proportional to x.dx or dx

The length of dx > dy when Angle CAB < 45 degrees
The length of dx = dy when Angle CAB = 45 degrees
The length of dx < dy when Angle CAB > 45 degrees

The proportionality of both the angle CAB and dy with x are in contradiction with the proportionality of dx and x in the triangle ABC after probing the equation of dy/dx = 2x beyond its derivation on a graph of y = x2. When x increases; dx decreases, dy increases, and angle CAB increases. This means AC also increases and ultimately SECANT when x increases. Our goal is to bring dx, dy and AC to zero (not away from zero either positively or negatively - Point C has to be on the curve) or secant to tangent by reducing them close to zero but here dx heads toward zero but dy and AC increases when x increases on axis.

Although the difference in the length of dx and dy can be noticeable clearly on the graph if we examine the triangle ABC at two different points for a gradient (dy/dx), say when an angle BAC = 0.1 degrees and 89.9 degrees on the curve but UNIT CIRCLE is the best example for observing the change in an angle CAB (say 0.1 and 89.9 degrees) of a triangle ABC for dy and dx and the comparison of their lengths.

RISE = dy = 2x and RUN = dx = 1 (always constant) in a GRADIENT of 1 in 2x which we obtained from the Eq1 of dy/dx=2x /1 at any point on the curve when there is no difference between secant and tangent – No idea how do we get dy/dx = 2x.dx but above said contradiction may be due to the introduction of another curve of y =(x+dx)2 at a point where we seize x or y=x2 deliberately and introduce delta x OR when function y = f(x) changed to y=f(x+Δx)2. The value of x has reached to its maximum value instead of unlimited when a curve y=x2 doesn’t continue anymore at a point where we introduce delta x or dx as y=x2 and y =(x+dx)2 are two different types of curve (two diffrent functions).

Further, integration is the reverse process of differentiation. Although delta x or dx is ignored during the process of derivation of dy/dx becaue of their small values but we can’t ignore them in the process of integration which makes a lot of difference in summation. They can’t be disappeared forever and should resurface during the process of integration or summation.

Similarly, dy is the small vertical change in y, therefore, we take the sum of all the small vertical lengths [dy(s)] not the whole slice or y-coordinate(s) from zero to its value on the curve when we integrate both sides of the equation of dy = 2xdx but it turns into function of x2 or area under the graph – no idea how but summation of vertical lengths on a graph gives vertical length only not curve?

The derivation of the natural relationship of a gradient of 1 in 2x at any point with y=x2 or dy/dx=2x is still unbeknownst to illuminates -

4
Technology, Science & Alt Science / Re: Question about the calculus
« on: October 10, 2018, 05:15:24 PM »
Recap for those who are confused by the previous discussion

Derivation of dy/dx for curve y = x2 is as follow. Pick any two points A and B close to each other on y = f(x)=x2 curve. The coordinates of A on the curve of y = f(x)=x2 are (x, y) or (x, x2). As x changes from x to (x +dx) while y changes from y to y + dy or f(x) to (x+dx)2 threfore the small change in x is dx while the corresponding small change in y is written as dy. Thus the x and y coordinates of B are (x + dx, y + dy) or ([x+dx, (x+dx)2]. ABC is a triangle made by dx, dy, and hypotenuse or slope of tangent where point A is on the curve. Length of AB = Base = dx, Length of BC = Perpendicular= dy and Length of Hypotenuse = AC. Angle CAB or BAC is the slope of a tangent

dy /dx = [(x + dx)2 – x2] / [x+dx-x]

dy /dx = [x2 + dx2+2xdx − x2] / dx

dy /dx = [2x + dx] / 1

dy /dx = 2x + dx

dy /dx = 2x------Eq1

OR dy = 2x.dx --Eq2

According to Eq1- dy/dx is directly proportional to x, dx is indirectly proportional to x and dy is directly proportional to x.

According to Eq2, dy is directly proportional to x.dx or dx - contradiction with the proportionality of dy dx and x - no more dicussion about this unless deemed neccessary

Thus dx decreases, dy increases, and angle CAB increases when x increases. This means AC also increases and ultimately SECANT when x increases. Our goal is to bring dx, dy and AC to zero or secant to tangent  by taking limits but here dx tends to zero but dy and AC increases when x increases – This is mathematically

The difference in the length of dx and dy can also be seen clearly if we analyze the triangle ABC at two different points, say when an angle BAC = 0.1 degrees and 89.9 degrees on the curve – Graphically

UNIT CIRCLE is the best example for observing the change in an angle CAB (say 0.1 and 89.9 degrees) of a triangle ABC for dy and dx and their comparison.

But on the other hand RISE = dy = 2x and RUN = dx = 1 (always constant) in a GRADIENT of 1 in 2x which we obtained from the Eq1 of dy/dx=2x /1 at any point on the curve when there is no difference between secant and tangent – No idea how do we get dy/dx = 2x.dx but contradiction may be due to the introduction of another curve of y =(x+dx)2 at a point where we diminish y=x2  deliberately and introduce delta x.

Curve y=x2 doesn’t continue anymore at a point where we introduce delta x or dx as y=x2 and y =(x+dx)2 are two different types curves no matter how small delta x is which is a portion of the length of known or fixed value of x.

Further, integration is the reverse process of differentiation. Although delta x or dx is ignored during the process of derivation of dy/dx but we can’t ignore them in the process of integration which makes a lot of difference in summation. They can’t be disappeared forever and should resurface during the process of integration or summation.

Similarly, dy is the small vertical change in y, therefore, we take the sum all the small vertical lengths [dy(s)] not the whole y-coordinate(s) from zero to its value on the curve when we integrate both sides of the equation of dy = 2xdx but it turns into function of x2 or area under the graph – no idea how but summation of vertical lengths on a graph gives vertical length only not curve?

The derivation of the natural relationship of a gradient of 1 in 2x at any point with y=x2 or dy/dx=2x is still unbeknownst to illuminates - Anyone who is in agreement with all above.

5
Technology, Science & Alt Science / Re: Question about the calculus
« on: October 07, 2018, 04:44:13 PM »
Let's analyze the equation of “dy/dx  = 2x” in its algebraic form dimensionally.

[ meter / meter ] = [ meter ] ------- is it correct dimensionally?

Although it is impossible [impov], but let's assume dx is a constant - Then dy = c.x OR [meter = meter}

This means when the curve grows or the value of x increases on axis, dy increases, angle CAB or BAC increases but dx doesn’t, as it is constant

This can also be shown in the diagram where dx is constant at 4 different points (1,2,3 and 4 in Fig1 and Fig2) if taken on the curve of y=x2 but then dy increases from dy1 to dy4 and angle CAB also increases from theta 1 to theta 4.

As dx, dy and hypotenuse AC tend to zero after taking limits – But here if dx is constant then dy increases if the curve grows as dy4 > dy3 > dy2 > dy1 can be seen clearly in the Fig2 of diagram.

We can decrease dx further at point 4 or for dy4 in the diagram but then the tip C of dy4 which also decreases upon the compression of dx, goes beyond curve (away from zero negatively) if we have to follow (maintain) the angle BAC or CAB also that increases when x increases as shown in Fig3

Also, dy is directly proportional to dx as dy =2x.dx

This means the height of dy decreases when the length of dx decreases but increases when angle BAC increases – Is it possible simultaneously?

Come on its very easy to notice that dx decreasing when the curve y=x2 grows or x increases. You can judge from the length of dx if taken at point 1, 2, 3 and 4 as shown in Fig1 and Fig2- i think i confused you

Anyhow, since we don’t know the size of a point but for the sake of argument lets consider two adjacent point A and D of the curve of y=x2 as shown in the diagram. When x increases, D starts rolling up over A. Horizontal distance b/t the center of A and the center of D is dx while vertical is dy. AB (dx) decreases, BC (dy) increases when D rolls up over D or when the curve grows or x increases.


imagine what would be the lenght of dy or BC if an angle CAB or BAC = 89.9 degrees, Angle ABC = 90 degrees (already) for a very small value of dx 

6
Technology, Science & Alt Science / Re: Question about the calculus
« on: October 04, 2018, 04:09:04 PM »
Quote
Why is the curve "zero"?
The length of the dx increases when the length of the AB increases and tends to zero or diminish when B approaches the point A which is on the curve. The same is applied to dy and hypotenuse of right-angled triangle ABC. The curve itself, not zero but point A on the curve is a reference point or start point if we measure AB which is equal to dx.
Quote
No, I'm trying to explain that dx is not in any way dependent on x.
The direction of the whole curve (y=x2) changes from the x-axis to a y-axis, therefore, the base of the triangle AB decreases when x increases if we look at the AC on the curve of y=x2 carefully at point 1, 2 3 and 4. The position of the immediate points above A at point 1, 2,3 and point 4 is not the same but slightly changes its position towards y-direction (rolls up on point A) as shown in the diagram when x increases. The center to center distance between two adjacent points (say dx) is not the same in all cases but decreasing. The perpendicular, dy increases when the base, dx decreases even at point level (rolling of a dark point on a lower white one). The diagram is not accurate or to the scale but for the understanding purpose only. 


Although dx starts decreasing when it tends to zero by definition but anyway even if dx is constant when x increases - dx approaches to zero but not increasing or decreasing then all of the following of triangle ABC are increasing as dy = 2xdx due to the trigonometric reasons as shown in the diagram
.
.
• Perpendicular = dy = BC
• Hypotenuse= (dx^2 + dy^2)^0.5 = AC
• Angle of the slop of dy/dx

The position of dy is shown in Fig3 if we shrink dx in Fig2 further to A or zero which we can obviously.

As told our goal is to bring dx, dy and AC to zero not taking them away from zero either positively or negatively.
Quote
Quote
Yes, you are right according to the concept of dimensional analysis but then dy =2xdx is also a curve instead of straight line.
I'm not 100% certain what you mean. Are you saying that the segment dy would be a curve? Why? It's just a straight line with length equal to 2xdx.
- Yes, because you said in the following quote.
Quote
Because y is equal to x squared. Even in the form dy=2xdx you can see that dy is in the form of something with x times something with x. If x represented meters, dy would be measured in square meters. Adding up square meters gives you square meters, no?

EDITED: I had already pointed out (don’t remember when and which forum) that delta y [or delta y=2x. delta x] is a curve because of the multiplication of x with delta x or x but I was told it is not due to the subtraction of coordinates. The reason behind it might be the two different types curves of y=x2 and y =(x+dx)2 as asked earlier.

7
Technology, Science & Alt Science / Re: Question about the calculus
« on: October 02, 2018, 02:47:00 PM »
The tiny right-angled triangle of the subject is ABC where

Base=dx=AB, Perpendicular=dy=BC, Hypotenuse=AC, Angle of the slope AC is BAC or CAB

Our goal is to bring dy, dx and the hypotenuse of right-angled triangle ABC to zero (on the curve) not away from zero either positively or negatively.

Quote
It's not "constant"
So when x increases - dx approaches to zero but decrease - right, not constant or increases. If this happens then both BC and AC decrease due to the reduction in dx further in the equation of dy = 2xdx.

But the angle BAC or CAB increases when x increases on the axis

Now if the angle BAC or CAB  increases (say to 89.9 degrees), dy also increases (going away from zero) in order to maintain the three sides of right-angled triangle ABC.

Thus the height of dy decreases when dx decreases but increases when angle BAC increases – Is it possible simultaneously?

Hence, there is an issue with the adjustment of dy in the right-angled triangle ABC when x increases.
 
Quote
Because y is equal to x squared. Even in the form dy=2xdx you can see that dy is in the form of something with x times something with x. If x represented meters, dy would be measured in square meters. Adding up square meters gives you square meters, no?
Yes, you are right according to the concept of dimensional analysis but then dy =2xdx is also a curve instead of straight line.

8
Technology, Science & Alt Science / Re: Question about the calculus
« on: October 02, 2018, 02:00:56 AM »
You may be frustrated with my constant disagreement but I appreciate your patience

If the length of dx is constant and tends to zero then the size of the tiny triangle ABC increases when the value of x increase on axis as BC=dy = 2xdx or slope=dy/dx=2x.

Two possibilities
 
1-   The tip (point C) of the perpendicular dy=BC go beyond zero or curve (as shown in Fig3) as length of the BC=dy increases when x increase on axis in the general equation of dy = 2xdx and the angle of the slope of AC as well

2-   Taking limits doesn’t make sense to me if we keep the tip (point C) of the perpendicular dy=BC on a curve due to the increase in the size of the triangle ABC

As dy = 2xdx therefore when x=4 on axis, then dy = 8dx. Since dx tends to zero therefore the infinitesimal value of dy at x=4 will be 8 multiplied by dx. This is the LINEAR length of dy which is the perpendicular=BC of tiny triangle ABC not the y coordinate at x=4 on axis. How come the summation of the LINEAR length of all these tiny perpendiculars turns into x2 (second-degree curve) in the process of integration? – This is my question or confusion.

9
Technology, Science & Alt Science / Re: Question about the calculus
« on: October 01, 2018, 08:58:37 PM »
Here is the equation of subject

dy/dx= 2x or dy = 2xdx

Where dx=AB=Base, tends to zero. dy=BC=Perpendicular and AC= Hypotenuse of the earlier mentiond tiny triangle ABC are > AB

Since the length of AC and BC are greater than AB which tends to zero, therefore, we can't ignore AC and BC

Possibilities:

1-   When x increases - dx approaches to zero but constant 

All of the following in the equation of dy = 2xdx increase due to the trigonometric reasons
.
•   Perpendicular = dy = BC
•   Hypotenuse= (dx^2 + dy^2)^0.5 = AC
•   Angle of the slop of dy/dx

No need to explained aforementioned further
 
2-   When x increases - dx approaches to zero but decrease

Both BC and AC may decrease but the angle of the slope of dy/dx increases. This means dy increases if the angle of slope (say 89.9 degrees) of dy/dx increases. Thus the height of dy needs adjustment with the wider angle

Integration: As dy = 2xdx therefore we get y=x2 if we integrate both sides of the equation of dy = 2xdx. Again I don’t see any relationship of y = x2 with the summing of the length of all dy(s). The integral of 2 is 2x +c --- no idea how?

IMPOV - There may be natural relationship existed between the curve (any degree) with the equation of the tangent to that curve (e.g. y=x2 with 2x) but limiting dx to zero has nothing to do with the slope of the tangent dy/dx

10
Technology, Science & Alt Science / Re: Question about the calculus
« on: September 30, 2018, 12:05:31 PM »
Quote
Yes, it does.
-

Great - when x increases then this means BC and AC also increase and ultimately the size of the whole tiny triangle ABC for each point as we go along the curve. let's determine the slope at point 1 and point 4. Triangle ABC taken at point 1 will be smaller than triangle at the point 4.

AC and BC of a triangle taken at point 4 are greater than AC and BC of a triangle taken at point 1. This means we have more points on AC of the triangle taken at point 4 as compared to AC of the triangle taken at point 1 if point C of both the triangle (taken at point 1 and point 4) on (or approaches) the curve. This means AC of the tiny triangle or secant increases when x increase instead of turning into a tangent at a point where the slope of the curve is to be determined.

remember our goal is to bring PQ (two points you know) or secant to a tangent at a point on the curve

We get Fig 3 as shown in the diagram if the number of points on AC of both the triangles is equal which is again impossible


11
Technology, Science & Alt Science / Re: Question about the calculus
« on: September 30, 2018, 09:57:07 AM »
Quote
How do you adjust the length of dy=BC = 2x.dx of the tiny triangle ABC so that point C remains (or approaches) on curve of y = x2 when x increases if dx is not the same (or say same) for all dy(s)?

What do you mean "if dx is not the same (or say same) for all dy(s)"? What does that mean?

The length of dy is 2xdx+dx^2. Dy/dx is 2x+dx. The limit of that as dx tends to 0 is the derivative, which is 2x. Usually when we use dy and dx it is implied that we've already taken the limit, and we're talking about infinitesimal quantities. So we might as well say that for these infinitesimal quantities, dy=2xdx. What that tells you is that as dx gets smaller and smaller, the length of dy tends to become equal to 2xdx. Where the your issue?

I asked because there is an isssue of the adjustment of dy with dx in the tiny triangle of ABC if x increases. As you said the length of dy tends to become equal to 2xdx. Doesn't the length of dy increase when the value of x increases in the equation 2xdx as explained earlier if dx is constant. Hence the size of the tiny triangle ABC also increases when x and dy increase simultaneously. You can also see the different lengths of dy1, dy2, dy3, and dy4 and their respective tiny triangles for points 1, 2,3 and 4 in the diagram.



12
Technology, Science & Alt Science / Re: Question about the calculus
« on: September 30, 2018, 03:31:57 AM »
How do you adjust the length of dy=BC = 2x.dx of the tiny triablge ABC so that point C remains (or approaches) on curve of y = x2 when x increases if dx is not the same (or say same) for all dy(s)?

Doesn't point C of dy=BC = 2x.dx out of a curve of y=x2 as shown Fig 3 if x increases whether dx is the same or not for all dy(s) ?

13
Technology, Science & Alt Science / Re: Question about the calculus
« on: September 30, 2018, 01:53:18 AM »
Quote
No, dx cannot be the same for dy1, dy2, dy3. I don't have the time to draw a diagram with these on it right now.

so it means you are ok with 1, 2 and 3 as explained earlier when x increases. 

INTEGRATION PROBLEM - Following are three sides of a previously mentioned tiny triangle

Base = dx, Perpendicular = dy, Hypotenuse= (dx^2 + dy^2)^0.5

Now come to the original derivation of dy /dx = [(x + dx)2 – x2] / [x+dx-x]
dy /dx = [x2 + dx2+2xdx − x2] / dx
dy / dx = [2xdx + dx^2] / dx

dy /dx = [2x + dx] / 1, After simplification and taking limits, it reduces to

dy/dx = 2x + dx where dx tends to zero but not zero

OR dy = 2xdx + dx2 -------- Eq1

OR dy = 2xdx ------- Eq2

This means the length of the tiny perpendicular of dy of a very infinitesimally small right-angled triangle = 2x multiplied by the tiny base of dx + dx2.

Although dx and dx2 are very small but their summation makes a lot of difference in integration, therefore, should we ignore dx or dx2 in integration which is the reverse process of differentiation?

Similarly, how come we get y = 2x2/2 = x2after integrating both sides of Eq2 when we are taking the sum of all the tiny perpendicular dy(s) at various points on the curve but it turns into an area of a graph? - we are not integrating the Y-Coordinates but the tiny perpendicular dy(s).



14
Technology, Science & Alt Science / Re: Question about the calculus
« on: September 29, 2018, 11:26:29 PM »
Thank you for your reply and suggestion but

Can dx be the same for dy1, dy2, dy3….........dyn if we reduce it to zero as shown in the diagram, not the scale?


IMPOV – NO

I’m trying to get you above and beyond the conclusion of dy/dx =2x where dx is not equal to zero but tends to zero, therefore, now imagine, if there is a dx then are dy (perpendicular) and hypotenuse also which form a tiny triangle ABC at a required point on the curve. This is because the lengths of dy and hypotenuse are always greater than dx. The three sides of the aforementioned tiny triangle are

Base = dx, Perpendicular = dy, Hypotenuse= (dx^2 + dy^2)^0.5

Now when x increases in the equation of dy = 2xdx [dy/dx= 2x], then all of the following also increase due to the trigonometric reasons if we keep the value of dx constant I mean not zero but almost zero, close to zero or limiting it to zero.
.
1-   Perpendicular = dy
2-   Hypotenuse= (dx^2 + dy^2)^0.5
3-   Angle of the slop of dy/dx


So when x increases

1-   The length of the hypotenuse also increases. This means we have more points on hypotenuse when both x and dy increase simultaneously. Now if we zoom point A further then dx can be reduced to zero further for smaller slopes but not for larger slopes due to the increase in the length of dy(s) if we start decreasing the value of x. This is why dx also increases when x increases due to the minimum limit from zero towards infinity set by trigonometry.

2-   Point C of the tiny triangle doesn’t coincide with curve due to its increased length as dy = 2x.dx but crosses the curve as shown in Fig 3 if dx is constant for all the dy(s) or even reduce to zero further.

3-   This means secant also increases and can’t be turned into a tangent

15
Technology, Science & Alt Science / Re: Question about the calculus
« on: September 28, 2018, 06:35:59 PM »
EDITED: it is said there is no difference b/t secant and tangent in limiting secant to the tangent but actually, it does depend upon the location of picking two very small positive numbers on the aforementioned curve. The length of the tiny perpendicular dy increases when x increases as dy = 2xdx and hence the tiny hypotenuse and the angle of slope of secant to the tangent are also increased due to the trigonometric ratio for the same value of dx, however, more points on hypotenuse or secant means dx taken at P1 must greater than the dx taken at P as dx taken at P approaches zero quickly due to fewer points on the secant as compared to more points on the secant at P1 - I previously said perpendicular and base increases due to some inexplicable reasons - sorry about that

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Technology, Science & Alt Science / Re: Question about the calculus
« on: September 27, 2018, 04:53:50 PM »
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This of course does not make any sense. You can't divide 0 by 0. But remember, we're trying to compute a limit, not substitute dx for 0. We want to know what number this function tends to as dx approaches 0, but without actually taking it to 0.

Their difference is 2xdx+dx^2. That's the change of y for a change dx, and we call it dy for that reason (the delta of something is usually the change in something). If I walk dx steps to the right, I will have to take 2xdx+dx^2 steps up to get to follow the function. The slope I walked up to is then clearly dy/dx=2x+dx. The derivative is simply the limit as dx tends to 0, and it corresponds to determining the slope for an infinitesimally small step, to determine the slope at the particular point you are on.

Agreed but the angle of the hypotenuse/slope of the two similar right-angled triangles is the same no matter if the ratio of the rise to run is infinitesimally small (dy/dx) where dx not equal to zero but tends to zero say 1/infinity or bigger (x/y).

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We multiply the height by the width for each slice, and then we add them all together.
Here is the confusion - the original equation is dy/dx = [2xdx+dx^2]/dx. After simplification and taking limit, it reduces to dy/dx = 2x where dx tends to zero. Now dy = 2xdx after rearranging. This means the length of the tiny perpendicular dy of a very infinitesimally small right-angled triangle = 2x multiplied by the tiny base dx. We are integrating all the tiny perpendicular dy(s) if we integrate both sides, not the full heights (y-coordinates) with tiny width dx of the slices- I don't see any rectangle of height y with tiny width dx?

Similarly, if dx tends to zero in differentiation then it tends to infinity in integration but we ignore dx or dx^2 in integration which makes a lot of difference.

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it is said there is no difference b/t secant and tangent in limiting secant to the tangent but actually it does depend upon the location of picking two very small positive numbers on the aforementioned curve. The angle of the slope of a tangent increases when x increases and vice versa as dy =2xdx. This means secant never turns into tangent when x increases and hence we can’t approach to zero fully as we thought due to restriction of minimum limit (length of dy=2xdx) set by the trigonometry (base and perpendicular of right angle triangle increases or decrease with certain ratio).

I'm not sure what you're referring to and why it's relevant, can you post a link to where you saw that so I can understand what you're saying a bit better?

Sorry, I should have written IMPOV - there is a huge difference in taking dy/dx at P and P1 as shown in the diagram.



The length of the tiny perpendicular dy increases when x increases as dy = 2xdx, therefore, dy at P and P1 are not the same and hence their dx are also not the same. This means the length of the dx taken at P1 must be greater than the dx taken at P and this why dx at P1 doesn't approach to zero fully due to the restriction set by trigonometry.

Similarly, as said y=x2 and y =(x+dx)2 two different types curves no matter how small delta x is, say 1/infinity which is a portion of the distance of known value of x, therefore, I don't understand why the Y-Coordinate of second point Q (close P) is taken as (x+dx)2 on the curve of y=x2. Since value of x doesn't stop but continue at a point where dx started in the general derivation of dy/dx, therefore, we can't introduce another curve of y =(x+dx)2 in the middle of a continuous curve of y=x2. if we introduce dx at a point where x stops then it means we are started graphing another curve of y =(x+dx)2 at a point where x stopped and therefore it's not a curve of y=x2 anymore beyond the point where the valuve of x is stopped or where dx started.

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Technology, Science & Alt Science / Re: Question about the calculus
« on: September 27, 2018, 05:30:47 AM »
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When we "shrink" dx to 0 we take its limit, we don't substitute dx for 0. The limit of 2x+dx as dx approaches 0 is 2x. What does that mean? Saying that a number is the limit of a function as x approaches to x0 means that you can get as close to that number as you want by approaching x0.

As asked, we introduce delta x or dx right at the start in order to derive the equation of slope dy/dx at any point on the curve but we ignore delta x or dx just before the derivation of dy/dx – it does not make sense to me?

Further, delta x or dx is ignored during the process of derivation of dy/dx but can it be ignored in the process of integration which makes a lot of difference in summation?

Similarly, we are integrating all the dy(s) only when we integrate both sides of the equation dy = 2xdx but it turns into the area under the graph – no idea how? -

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The formal definition involves picking two very small positive numbers, ε and δ. The limit of f(x) as x approaches x0 is a number for which you can pick ε and δ to be as small as you like, and you will still be able to find an x whose distance to x0 is smaller than δ so that the distance of f(x) to that limit will be less than ε.

It is said there is no difference b/t secant and tangent in limiting secant to the tangent but actually it does depend upon the location of picking two very small positive numbers on the aforementioned curve. The angle of the slope of a tangent increases when x increases and vice versa as dy =2xdx. This means secant never turns into tangent when x increases and hence we can’t approach to zero fully as we thought due to restriction of minimum limit (length of dy=2xdx) set by the trigonometry (base and perpendicular of right angle triangle increases or decrease with certain ratio).

This can easily be observed if we take two positive numbers at a larger value of x where the angle of the slope of tangent is greater.

Also, aren’t both y=x2 and y =(x+dx)2 two different types curves no matter how small delta x is which is a portion of the distance of known value of x if yes then why we are taking the y-coordinate as (x+dx)2 of the second point chosen close to the first point?.

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Technology, Science & Alt Science / Question about the calculus
« on: September 24, 2018, 09:50:08 AM »
Can someone clarify the following

The vertical change between two points is called the RISE while the horizontal change is called the RUN. Hence slope is the ratio of the vertical and horizontal changes between two points on a surface or a line. So no slope when there is no rise and run at any given point.

We all know how to draw a graph of y = f(x)=x2 curve. Take two points P and Q close to each other on y = f(x)=x2 curve.

The coordinates of P on the curve of y = f(x)=x2 are (x, y) or (x, x2).

Since it is said x changes from x to (x +dx) where dx is the symbol we use for a small change, or small increment in x while y changes from f(x) to (x+dx)2. The corresponding small change in y is written as dy.

Thus the x and y coordinates of Q are (x + dx, y + dy).

Now dy /dx = [(x + dx)2 – x2] / [x+dx-x]
dy /dx = [x2 + dx2+2xdx − x2] / dx
dy /dx = [2x + dx] / 1
dy /dx = 2x + dx
dy /dx = 2x
When dx is shrunk to zero then it is disappeared on RHS but it's not on LHS – no idea why?

Similarly, when dx heads towards zero then the coordinates of Q are also becoming equal to the coordinates of P but they are not.

Moreover, we introduce delta x or dx right at the start in order to derive the equation of slope dy/dx at any point on the curve but we ignore delta x or dx just before the derivation of dy/dx – Does it make sense?

Anyhow, this means, RISE = dy =2x while RUN = dx = 1 but is it possible for x to be present in such small change in y or dy and 1 in small change in x or dx?

Further, integration is the reverse process of differentiation. Although delta x or dx is ignored during the process of derivation of dy/dx but can it be ignored in the process of integration which makes a lot of difference in summation?

Similarly, we are integrating all the dy(s) not the y-coordinates when we integrate both sides of the equation of dy = 2xdx but it turns into the area under the graph – no idea how? -

As y-coordinate of the chosen point Q is (x+dx)2 therefore

Aren’t both y=x2 and y =(x+dx)2 two different types curves no matter how small delta x is which is a portion of the distance of known value of x. Since we don’t know the limit of the distance of the x, therefore, can we fit delta x within the distance of x?

Both y=x2 and y =(x+dx)2 works fine for all numerical values in which we set the limit of x but since we don’t know the maximum limit of x, therefore, isn’t it wrong to say y=f(x)=x2=(x+dx)2? Aren’t we actually started graphing another equation of y=(x+dx)2 (unbeknownst) instead of y=x2 after point P by choosing another point Q at a distance of delta x from P on the curve of y=f(x)=x2 in which the maximum limit of x is unknown?

We can see the difference if we start graphing both y=(x+dx)2 and y=x2 right from the origin at the same time.

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So a couple questions:

1. Is the outer midnight circle as represented or traced out by (x0,x1,x2,...) larger in circumference than the inner circle as represented or traced out by (y0, y1, y2,...)?

2. If yes to question 1.  Would there be more midnights than noons because the Circumference of outer circle > Circumference of inner circle and hence the # of Midnights > # of Middays?
The diagram is the top view

Only Midnights (x,x1,x2,x3………) are touching the outer circle
Only Middays (y,y1,y2,y3………..) are touching the inner circle
 
Any point on OUTER circle represents Midnight
Any point on INNER circle represents Midday

The circumference of outer circle MUST EQUAL TO the circumference of inner circle if the occurrence of Midnights is EQUAL TO the occurrence of Middays BUT -

Circumference of outer circle is greater than the Circumference of inner circle – RIGHT. So

So there are two possibilities as said earlier

1-   The # of Midnights are more than the # of Middays OR the occurrence of Midnights is more than the occurrence of Middays
2-   There is a orbital DRAG of Midnight in the space

1 of 2 is not possible – we all know

So if there is an orbital DRAG of Midnight then there is an orbital DRAG of Midday as well

Since orbital velocity of Midnight (x,x1,x2...) > orbital velocity of Midday (y,y1,y2....)
.

Therefore orbital GRAG of Midnight is more than the orbital DRAG of Midday in order to make the occurrence of Midnight = occurrence of Middays.

Technically, shadow growth on earth is either stopped or very very slow during this drag w/o spin.

Question:
Would this DRAG w/o spin affect SEASONS or Calender due to the change in the position of the earth in its orbit as asked earlier?

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A tangent line to a circle is a line that touches the circle at exactly one point – Wikipedia

Fig #1: when the earth rotates on its axis but has zero orbital motion.

Tangent A touches C exactly at x or Anti-Noon x
Tangent B touches C exactly at y or Noon y

A common point of C and tangent A is x
A common point of C and tangent B is y

Noon and Midnight are happening in this case INSTANTLY.

Fig #2: When the earth rotates on its axis and has also orbital motion

As said
Tangent A touches C exactly at x or Anti-Noon x
Tangent B touches C exactly at y or Noon y - Similarly,

Tangent A1 touches C exactly at x1 or Anti-Noon x1
Tangent B1 touches C exactly at y1 or Noon y1
Tangent A2 touches C exactly at x2 or Anti-Noon x2
Tangent B2 touches C exactly at y2 or Noon y2
Tangent A3 touches C exactly at x3 or Anti-Noon x3
Tangent B3 touches C exactly at y3 or Noon y3 .... and son on

When there is a midnight x , there is a noon y and hence
When there is midnight x1,, there is a is noon y1
When there is midnight x2,, there is a is noon y2
When there is midnight x3,, there is a is noon y3 and so on

Original Diagram of the question

The outer circle is traced by all the Midnights (x, x1, x2, x3……) on earth in space due to the spin of earth on axis and its orbital motion

The inner circle is traced by all the Middays (y, y1, y2, y3……) on earth in space due to the spin of earth on axis and its orbital motion 

If both Noons and Midnights were instant and equal in numbers then, the

Circumference of outer circle must be equal to the Circumference of inner circle but
Circumference of outer circle > Circumference of inner circle and hence
# of Midnights > # of Middays or
Occurrence of midnights > Occurrence of middays
Since the instantaneous

Orbital velocity of Point x > Orbital velocity of Point y
Orbital velocity of Point x1 > Orbital velocity of Point y1
Orbital velocity of Point x2 > Orbital velocity of Point y2
Orbital velocity of Point x3 > Orbital velocity of Point y3 and so on

Therefore, the

Drag of Point x > Drag of Point y in space
Drag of Point x1 > Drag of Point y1 in space
Drag of Point x2 > Drag of Point y2 in space
Drag of Point x3 > Drag of Point y3 in space and so on

The earth spins on its axis at the speed of 30 m/sec.
The travelling speed of earth is 30,000 m/sec in its orbit.
As  30,000 m/sec >>>> 30 m/sec , therefore either # of Midnights > # of Middays (which is impossible)
or there must be the happening of drag of midday and midnight in the orbit of earth if the

Circumference of outer circle > Circumference of an inner circle


I called this Drag or gentle ride of midnight (or midday) in space DURATION between the happening of x and x1 (or y and y1)

Although the orbital speed of midnight > middays but the DURATION between the two successive midnights (say x1 and x2) is same and = DURATION between the two successive middays (say y1 and y2)
 
Shadow growth stops (or very very slow) due to the gentle ride or transferring of noon or midnight to next point on earth during this DURATION

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The speed the surface of the earth is travelling and the orbital circumferences have nothing to do with the number of occurrences of anything

I think you didn't follow the post properly - As mentioned earlier the speed of the earth's spin is 30 m/sec and earth's traveling speed is 30,000 m/sec in its orbit. Midnights are touching the outer circle. Middays are touching the inner circle - Right?

So you agree Midnight is traveling in the orbital circumference for a while (as you quoted) not just touching the outer circle instantly but not Midday - Right?

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By equating Arc I and Arc II you are referring to their measure or extent (Length).  Each point has no measure so no quantity of points will give you measure.  Each point is an instant in both position and time and has no duration.  Each point is undergoing a continuous change in time and position.

To me it’s like asking how many apples equal an orange?


"The occurrence of Midnights (Anti-Noons) on outer circles are more than the Middays (Noons) on inner circle"


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Because Arc I is at a greater radius than Arc II and circumference 2 x pi x radius.

My question is in reference to Midnight and Noon -

Each point on the outer circle represents the MIDNIGHT on an outer surface of the earth at the equator - Right
Each point on the inner circle represents the MIDDAY on an outer surface of the earth at the equator - Right

There is a Midnight for every Noon if both Noon and Midnight are instant - Right.

Therefore, shouldn't Arc I = Arc II if both Midnight and Midday are instant and equal in numbers but Arc I > Arc II - Any special reasons???

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Midnight and midday don't have durations
- if they don't have DURATIONS, then my question is why Arc I is greater than Arc II

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Epitrochoid/ Generator is just humanly manipulated web application -  It can miss many things including what I am talking about.

Back to the original question:

Do you agree midnight (Arc I in the diagram) is a "DURATION" due to its longer path in space and greater velocity in the earth's orbit for any instant noon in/on Arc II if not then why there is a drag of midnight OR Arc I > Arc II?

I think here is the confusion (may be wrong): You consider the earth's drag w/o spin on outer circle due to its greater velocity but ignore earth is also dragged on an inner circle due to its velocity (not zero) though less as compared to on outer circle.

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I pose a challenge for you to demonstrate what it is you are claiming with the noon point having duration. Again, don’t worry about scale.  Just show a point having a noon position

I'm novice in astronomy. It's not about the challenge but it's all about positive discussion and IMPOV. I may be wrong unless someone agrees.

Figure on the Top Right:) - NOON is an INSTANT when the stationary earth (zero orbital speed) rotates around its axis. "When there is an instant noon, there is an instant midnight as well"

Figure on the Left:)

Disc #1: It rotates about its own center which is the center of the sun. The rotation speed of this disc at R=R1=R2... is equivalent to the orbital speed of the outer surface of the earth at the equator (noons on inner circle). Disc#2 rests on Disc#1 as shown.

Disc #2 represents earth with radius r. It also rotates about its axis. The surface of the earth at the equator moves at a speed of 460 meters per second - approx

Figure on bottom Right:)

There are two types of motion of the earth
1. Earth's rotation w.r.t its axis (nearly 460 m/s) - 
2. Earth's rotation w.r.t sun (nearly 30,000 m/s) -

Both r and R1 are in a straight line at any noon. Since 30,000 m/s is way greater 460 m/s, therefore, an earth orbits around the sun faster than its spin around its axis. Thus line r of earth transfers to another R2 but =R1 on the inner circle but without axial spin (Earth doesn't have a chance to spin completely at this particular moment due to the greater orbital speed) before it spun to an afternoon when the the earth moves forward in its orbit. Now r is in a straight line with R2 not R1. Hence,

Noon is a "DURATION" between B and A when an earth doesn't have a chance to spin around its axis while simply moves/ drags fast forwards in its orbit.

---:)) Even if a noon is an instant, midnight should also be an instant not multiples midnights for single/instant noon - Isn't midnight a "DURATION" due to its longer path for instant noon according to you in the previous discussion.

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The most simplified form of "Instant Noon" is when the earth spins around its axis and is stationary (it doesn't revolve around the sun (i.e orbital speed is zero) - instant Noon occurs when the orbital speed < speed of axial rotation of the earth or equal to zero.

Draw a radius r of earth at any Noon. Let R1 is the radius of the inner circle in the diagram. Both r and R1 are in a straight line at any noon. Since the orbital speed of earth > its axial rotation speed, therefore, the line r of earth transfers to another R2 but =R1 on the inner circle without axial spin (Earth doesn't have a chance to spin at that point due to the greater orbital speed). Now r is in a straight line with R2 not R1. So the earth simply moves forward in its orbit on the inner circle b/w R1 and R2 but w/o spin. This means Noon is a DURATION not Instant if it drags forward in its orbit w/o any spin. it happens when the orbital velocity of earth is greater than the speed of its axial rotation.




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IMPOV
Instant Noon occurs when orbital velocity = or < axial speed. Shadows growth doesn’t stop in instant noon.

Duration Noon occurs when orbital speed > axial speed.
Noon point is dragged forward in the orbit but on the constant orbital radius (on the inner circle) for some DURATION before it spun again to afternoon due to the greater orbital velocity. Shadow growth is stopped when the sun is overhead during this "constant noon".

Would the position of earth change in its orbit for seasons due to aforementioned "Duration Noon" (when the earth (or each noon) is dragged and forward in orbit)?

Imagine the said diagram (3 circles) is a circular disc. The center of this disc is the center of the sun. The disc is rotating about its center such that the rotating speed of middle circle on a disc is equivalent to an orbital speed of earth.

The earth as shown in the diagram is another small disc. This small disc rotating about its own axis – spinning of earth

Now imagine both the bigger disc (for orbital speed) and smaller disc (for axial speed) at the same time – would you be able to find “duration noon” now?

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OK, here is the confusion -  Since the orbital velocity of the earth is greater than its axial rotational speed, therefore, Noon/ Midnight is the DURATION (time span). Noon or Midnight doesn't occur instantly at a particular point on the surface of the earth due to the combined effect of orbital and axial motion of the earth. Sorry, a couple of more question

Can you calculate aforesaid DURATION on the inner and outer circles?
Would the said DURATION affect sidereal and solar time/seasons etc?
Since shadow of any vertical pole would not grow during noon (DURATION) therefore would we be able to notice or measure such DURATION on earth?

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The average orbital radius on the midnight side = (149,597,871 + 6,378 km) and The average orbital radius on the midday side = (149,597,871 - 6,378 km)

Here is the caption of Question

Does time dilate during nighttime when the earth orbits the sun?  - never indicated by our mechanical clocks

Any point on the outer circle, which represents midnight if connected to the center of the sun via a straight line passes through the noon (inner circle). When there is midnight, there is a noon, therefore, total duration/occurrence of midnights must be equivalent to the total duration/occurrence of noons and hence their lengths. We can't have multiples noons for one midnight at particular instant due to the combined effect of axial and orbital motion of the earth.

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