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3utilize the area .333 in Y=x^2 Re: Solving Arclength in New Math, a sort of form dy+dx rather than area=dy*dx, derivative=dy/dx

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  3utilize the area .333 in Y=x^2 Re: Solving Arclength in New Math, a  sort of form dy+dx rather than area=dy*dx, derivative=dy/dx Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-06-28 18:57 -0700
    Re: 3utilize the area .333 in Y=x^2 Re: Solving Arclength in New Math, a  sort of form dy+dx rather than area=dy*dx, derivative=dy/dx Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-06-28 19:05 -0700
    Re: Archimedes "just not bolted down too tight in the first place" Plutonium flunked the physics lifetime-generation test moroney@world.std.spaamtrap.com (Michael Moroney) - 2018-06-29 03:11 +0000
    method is valid, quick, easy Re: 3utilize the area .333 in Y=x^2 Archimedes Plutonium <plutonium.archimedes@gmail.com> - 2018-06-28 20:17 -0700

#416740 — 3utilize the area .333 in Y=x^2 Re: Solving Arclength in New Math, a sort of form dy+dx rather than area=dy*dx, derivative=dy/dx

Newsgroups: sci.math
Date: Thu, 28 Jun 2018 12:33:53 -0700 (PDT)

Subject: utilize the area .333 in Y=x^2 Re: Solving Arclength in New Math, a
 sort of form dy+dx rather than area=dy*dx, derivative=dy/dx
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Thu, 28 Jun 2018 19:33:53 +0000

utilize the area .333 in Y=x^2 Re: Solving Arclength in New Math, a sort of form dy+dx rather than area=dy*dx, derivative=dy/dx

On Thursday, June 28, 2018 at 2:22:44 PM UTC-5, Archimedes Plutonium wrote: 
> Great progress on solving what real Arclength is. 
> 
> Old Math's arclength is a joke, for if you ever try solving any problem with their crazy formula that only yields indefinite integrals, and about the only example case they can use is Y = x^3/2 (what a joke). Old Math's arclength is a barbed wire nightmare, yet the kooks of Old Math keep on teaching that crap. 
> 
> The Arclength of Y=x^2 in interval 0 to 1 is 1.47 
> 

Now, unless I can enlist the integral 1/3x^3, it will not be feasible to calculate arclength by constant estimates 


> What we do is box in every function you want, box it in its interval, with either a square or rectangle. Most cases will be a rectangle. 
> 
> In our Y=x^2 it is a square 1 by 1 
> 
> We look at the function within that square. 
> 
> If it were Y = x , then our arclength is a snap breeze for it is 1.414, the length of the diagonal. 
> 
> But since our function is Y = x^2, means it is going to be a bit larger of a arclength 
> 
> Keep in mind, no curves exist in mathematics and that arclength is a Summation of tiny straight line segments. In the case of Y=x, we have just one straight line segment of 1.414 
> 
> Now, I am doing this in 10 Grid so I split up the function into ten partitions of .1 length each 
> 
> So, our minimum arclength for functions x, x^2, x^3, x^4, . ., x^10 our minimum arclength is 1.414, and our maximum arclength would be 1+1 = 2, where the function swings from x axis 0 to 1 and from y-axis 0 to 1 
> 
> So the function Y=x^10 approaches the arclength value of 2, never reaching 2, mind you, but perhaps 1.8, or 1.7 just so you get the feel of this. 
> 
> So for Y=x^10, our x =.9 is going to be what .81, then .9x.81= .72, then .9x.72= .64, walking the arclength back to being almost the x-axis from 0 to .9, then shooting straight up to (1,1) for another value of .9 and so the arclength of Y= x^10 is going to be approx 1.8 
> 
> Now, for Y=x^2, our arclength is going to be very close to that of 1.414, not far off. Our .9 value is going to be .81, so not much further to shoot up to (1,1). Our midpoint value is going to be (.5, .25) 
> 
> So now, if we do a gross approximation .5^2 +.25^2 = c^2 and so we have .31 and the square root is .55 
> then, the distance from (.5, .25) to (1,1) would be 
> 
> .5^2 + .75^2 =( .25 + .56)^2 = .81^2 and adding .55+.9 = 1.45 as first approximation which is not far from 1.47 
> 
> But anyone, one can easily see how the overall general pattern to solve all arclength is going to be-- box it in, and compute diagonal length, then the arclength is between diagonal and the dy+dx length, in Y=x^2, between 1.414 and 2 
> 
> In New Math, every function, every every function is turned into a polynomial, even trig functions. So boxing in is going to be easy to solve for what is the minimum and maximum arclength. 
> 
> 

So, what I need is to utilize the area 1/3x^3 to be able to go from 1.414 to 1.47 

The area is .333 

Somehow, that area is going to yield that extra .06 I need to go from 1.414 to 1.47 

AP 


zzzzzzzzzzzzzz

END

Newsgroups: sci.math
Date: Thu, 28 Jun 2018 14:57:58 -0700 (PDT)

Subject: Heron's formula is no use Re: 2utilize the area .333 in Y=x^2
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Thu, 28 Jun 2018 21:57:59 +0000

Heron's formula is no use Re: 2utilize the area .333 in Y=x^2

Alright, easy enough to see I need the lengths of A, B when I have C as 1.414 


___C____1 
\  A       / 
   \     /B 

C length is 1.414 for Y=x^2 box with C as diagonal. The function x^2 area is 1/3x^3 is .333 

Area for under C is .500 and length 1.414 

What I need to do is get where A+B = 1.47 a difference of .06 

Now looking at Heron's formula Area = sqrt[S(S-A)(S-B)(S-C)] where S = (A+B+C)/2 is not going to give me anything because although I know the area is .5-.333 = .167 and know C=1.414 is of no help. 

What I need is a height in general, for I know that A+B is a bit larger than 1.414 and hopefully is 1.47 

AP 

Newsgroups: sci.math
Date: Thu, 28 Jun 2018 16:51:38 -0700 (PDT)

Subject: Re: a gnarly patch Re: Heron's formula is no use Re: 2utilize the
 area .333 in Y=x^2
From: Archimedes Plutonium <plutonium....@gmail.com>
Injection-Date: Thu, 28 Jun 2018 23:51:39 +0000

Re: a gnarly patch Re: Heron's formula is no use Re: 2utilize the area .333 in Y=x^2

- hide quoted text -
On Thursday, June 28, 2018 at 6:45:33 PM UTC-5, Archimedes Plutonium wrote: 
> On Thursday, June 28, 2018 at 4:58:05 PM UTC-5, Archimedes Plutonium wrote: 
> > Alright, easy enough to see I need the lengths of A, B when I have C as 1.414 
> > 
> > 
> > ___C____1 
> > \  A       / 
> >    \     /B 
> > 
> > C length is 1.414 for Y=x^2 box with C as diagonal. The function x^2 area is 1/3x^3 is .333 
> > 
> > Area for under C is .500 and length 1.414 
> > 
> > What I need to do is get where A+B = 1.47 a difference of .06 
> > 
> > Now looking at Heron's formula Area = sqrt[S(S-A)(S-B)(S-C)] where S = (A+B+C)/2 is not going to give me anything because although I know the area is .5-.333 = .167 and know C=1.414 is of no help. 
> > 
> > What I need is a height in general, for I know that A+B is a bit larger than 1.414 and hopefully is 1.47 
> > 
> > 
> 
> Let me experimentally try something here, for persistence is the cure of all math. 
> 
> 1.414   is to  A 
> ______           __ 
> .5                   .333 
> 
> 
> lengths for numerator, areas for denominator 
> 
> .333*1.414  = .5A 
> 
> A = .94 
> 
> Now subtract from 1 , then I get .06 and add to 1.414 gives me the 1.47 
> 
> Check to see if it works on Y= x^10 
> 
> 1.414   as to  A 
> _____             ___ 
> .5                   .001 
> 
> 
> .001 = .5A then subtract from 1 is essentially 1 with length 1.4+1 makes no sense. 
> 
> Unless, unless the maximum length is not 1+1, for those are hypoteni 
> 
> So is the maximum length 2.414, where I thought it was 2. 
> 
> 

Let me try it on Y=x^3, integral is 1/4x^4 

1.414   is to A 
_____           ____ 
.5                 .25 


1.414*.25  = .5A 

.3535 = .5A 

A = .707 

Subtract from 1 we have .3 


1.414+ .3 = 1.71 

Is that the arclength of Y= x^3 in interval 0 to 1 

?? 

AP

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#416741

In something like Y=x^10, I could see the arc length in interval 0 to 1 being a tiny tiny bit bigger than 2. say 2.01, but not no 2.4



On Thursday, June 28, 2018 at 8:57:53 PM UTC-5, Archimedes Plutonium wrote:
> Newsgroups: sci.math
> Date: Thu, 28 Jun 2018 12:33:53 -0700 (PDT)
> 
> Subject: utilize the area .333 in Y=x^2 Re: Solving Arclength in New Math, a
>  sort of form dy+dx rather than area=dy*dx, derivative=dy/dx
> From: Archimedes Plutonium <plutonium....@gmail.com>
> Injection-Date: Thu, 28 Jun 2018 19:33:53 +0000
> 
> utilize the area .333 in Y=x^2 Re: Solving Arclength in New Math, a sort of form dy+dx rather than area=dy*dx, derivative=dy/dx
> 
> On Thursday, June 28, 2018 at 2:22:44 PM UTC-5, Archimedes Plutonium wrote: 
> > Great progress on solving what real Arclength is. 
> > 
> > Old Math's arclength is a joke, for if you ever try solving any problem with their crazy formula that only yields indefinite integrals, and about the only example case they can use is Y = x^3/2 (what a joke). Old Math's arclength is a barbed wire nightmare, yet the kooks of Old Math keep on teaching that crap. 
> > 
> > The Arclength of Y=x^2 in interval 0 to 1 is 1.47 
> > 
> 
> Now, unless I can enlist the integral 1/3x^3, it will not be feasible to calculate arclength by constant estimates 
> 
> 
> > What we do is box in every function you want, box it in its interval, with either a square or rectangle. Most cases will be a rectangle. 
> > 
> > In our Y=x^2 it is a square 1 by 1 
> > 
> > We look at the function within that square. 
> > 
> > If it were Y = x , then our arclength is a snap breeze for it is 1.414, the length of the diagonal. 
> > 
> > But since our function is Y = x^2, means it is going to be a bit larger of a arclength 
> > 
> > Keep in mind, no curves exist in mathematics and that arclength is a Summation of tiny straight line segments. In the case of Y=x, we have just one straight line segment of 1.414 
> > 
> > Now, I am doing this in 10 Grid so I split up the function into ten partitions of .1 length each 
> > 
> > So, our minimum arclength for functions x, x^2, x^3, x^4, . ., x^10 our minimum arclength is 1.414, and our maximum arclength would be 1+1 = 2, where the function swings from x axis 0 to 1 and from y-axis 0 to 1 
> > 
> > So the function Y=x^10 approaches the arclength value of 2, never reaching 2, mind you, but perhaps 1.8, or 1.7 just so you get the feel of this. 
> > 
> > So for Y=x^10, our x =.9 is going to be what .81, then .9x.81= .72, then .9x.72= .64, walking the arclength back to being almost the x-axis from 0 to .9, then shooting straight up to (1,1) for another value of .9 and so the arclength of Y= x^10 is going to be approx 1.8 
> > 
> > Now, for Y=x^2, our arclength is going to be very close to that of 1.414, not far off. Our .9 value is going to be .81, so not much further to shoot up to (1,1). Our midpoint value is going to be (.5, .25) 
> > 
> > So now, if we do a gross approximation .5^2 +.25^2 = c^2 and so we have .31 and the square root is .55 
> > then, the distance from (.5, .25) to (1,1) would be 
> > 
> > .5^2 + .75^2 =( .25 + .56)^2 = .81^2 and adding .55+.9 = 1.45 as first approximation which is not far from 1.47 
> > 
> > But anyone, one can easily see how the overall general pattern to solve all arclength is going to be-- box it in, and compute diagonal length, then the arclength is between diagonal and the dy+dx length, in Y=x^2, between 1.414 and 2 
> > 
> > In New Math, every function, every every function is turned into a polynomial, even trig functions. So boxing in is going to be easy to solve for what is the minimum and maximum arclength. 
> > 
> > 
> 
> So, what I need is to utilize the area 1/3x^3 to be able to go from 1.414 to 1.47 
> 
> The area is .333 
> 
> Somehow, that area is going to yield that extra .06 I need to go from 1.414 to 1.47 
> 
> AP 
> 
> 
> zzzzzzzzzzzzzz
> 
> END
> 
> Newsgroups: sci.math
> Date: Thu, 28 Jun 2018 14:57:58 -0700 (PDT)
> 
> Subject: Heron's formula is no use Re: 2utilize the area .333 in Y=x^2
> From: Archimedes Plutonium <plutonium....@gmail.com>
> Injection-Date: Thu, 28 Jun 2018 21:57:59 +0000
> 
> Heron's formula is no use Re: 2utilize the area .333 in Y=x^2
> 
> Alright, easy enough to see I need the lengths of A, B when I have C as 1.414 
> 
> 
> ___C____1 
> \  A       / 
>    \     /B 
> 
> C length is 1.414 for Y=x^2 box with C as diagonal. The function x^2 area is 1/3x^3 is .333 
> 
> Area for under C is .500 and length 1.414 
> 
> What I need to do is get where A+B = 1.47 a difference of .06 
> 
> Now looking at Heron's formula Area = sqrt[S(S-A)(S-B)(S-C)] where S = (A+B+C)/2 is not going to give me anything because although I know the area is .5-.333 = .167 and know C=1.414 is of no help. 
> 
> What I need is a height in general, for I know that A+B is a bit larger than 1.414 and hopefully is 1.47 
> 
> AP 
> 
> Newsgroups: sci.math
> Date: Thu, 28 Jun 2018 16:51:38 -0700 (PDT)
> 
> Subject: Re: a gnarly patch Re: Heron's formula is no use Re: 2utilize the
>  area .333 in Y=x^2
> From: Archimedes Plutonium <plutonium....@gmail.com>
> Injection-Date: Thu, 28 Jun 2018 23:51:39 +0000
> 
> Re: a gnarly patch Re: Heron's formula is no use Re: 2utilize the area .333 in Y=x^2
> 
> - hide quoted text -
> On Thursday, June 28, 2018 at 6:45:33 PM UTC-5, Archimedes Plutonium wrote: 
> > On Thursday, June 28, 2018 at 4:58:05 PM UTC-5, Archimedes Plutonium wrote: 
> > > Alright, easy enough to see I need the lengths of A, B when I have C as 1.414 
> > > 
> > > 
> > > ___C____1 
> > > \  A       / 
> > >    \     /B 
> > > 
> > > C length is 1.414 for Y=x^2 box with C as diagonal. The function x^2 area is 1/3x^3 is .333 
> > > 
> > > Area for under C is .500 and length 1.414 
> > > 
> > > What I need to do is get where A+B = 1.47 a difference of .06 
> > > 
> > > Now looking at Heron's formula Area = sqrt[S(S-A)(S-B)(S-C)] where S = (A+B+C)/2 is not going to give me anything because although I know the area is .5-.333 = .167 and know C=1.414 is of no help. 
> > > 
> > > What I need is a height in general, for I know that A+B is a bit larger than 1.414 and hopefully is 1.47 
> > > 
> > > 
> > 
> > Let me experimentally try something here, for persistence is the cure of all math. 
> > 
> > 1.414   is to  A 
> > ______           __ 
> > .5                   .333 
> > 
> > 
> > lengths for numerator, areas for denominator 
> > 
> > .333*1.414  = .5A 
> > 
> > A = .94 
> > 
> > Now subtract from 1 , then I get .06 and add to 1.414 gives me the 1.47 
> > 
> > Check to see if it works on Y= x^10 
> > 
> > 1.414   as to  A 
> > _____             ___ 
> > .5                   .001 
> > 
> > 
> > .001 = .5A then subtract from 1 is essentially 1 with length 1.4+1 makes no sense. 
> > 
> > Unless, unless the maximum length is not 1+1, for those are hypoteni 
> > 
> > So is the maximum length 2.414, where I thought it was 2. 
> > 
> > 
> 
> Let me try it on Y=x^3, integral is 1/4x^4 
> 
> 1.414   is to A 
> _____           ____ 
> .5                 .25 
> 
> 
> 1.414*.25  = .5A 
> 
> .3535 = .5A 
> 
> A = .707 
> 
> Subtract from 1 we have .3 
> 
> 
> 1.414+ .3 = 1.71 
> 
> Is that the arclength of Y= x^3 in interval 0 to 1 
> 
> ?? 
> 
> AP

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#416742 — Re: Archimedes "just not bolted down too tight in the first place" Plutonium flunked the physics lifetime-generation test

Math Failure Archimedes "just not bolted down too tight in the first place" Plutonium fails:

>AP writes: Google has abuse, but what good is that when it cannot recognize
>a 26 year long stalker Moroney with his ad hominem bully stalking-- never 

So you are back to causing trouble again, Failure Plutonium?

[toc] | [prev] | [next] | [standalone]

#416744 — method is valid, quick, easy Re: 3utilize the area .333 in Y=x^2

A test run on the quarter circle


> On Thursday, June 28, 2018 at 6:45:33 PM UTC-5, Archimedes Plutonium wrote: 
> > On Thursday, June 28, 2018 at 4:58:05 PM UTC-5, Archimedes Plutonium wrote: 
> > > Alright, easy enough to see I need the lengths of A, B when I have C as 1.414 
> > > 
> > > 
> > > ___C____1 
> > > \  A       / 
> > >    \     /B 
> > > 
> > > C length is 1.414 for Y=x^2 box with C as diagonal. The function x^2 area is 1/3x^3 is .333 
> > > 
> > > Area for under C is .500 and length 1.414 
> > > 
> > > What I need to do is get where A+B = 1.47 a difference of .06 
> > > 
> > > Now looking at Heron's formula Area = sqrt[S(S-A)(S-B)(S-C)] where S = (A+B+C)/2 is not going to give me anything because although I know the area is .5-.333 = .167 and know C=1.414 is of no help. 
> > > 
> > > What I need is a height in general, for I know that A+B is a bit larger than 1.414 and hopefully is 1.47 
> > > 
> > > 
> > 
> > Let me experimentally try something here, for persistence is the cure of all math. 
> > 
> > 1.414   is to  A 
> > ______           __ 
> > .5                   .333 
> > 
> > 
> > lengths for numerator, areas for denominator 
> > 
> > .333*1.414  = .5A 
> > 
> > A = .94 
> > 
> > Now subtract from 1 , then I get .06 and add to 1.414 gives me the 1.47 
> > 
> > Check to see if it works on Y= x^10 
> > 
> > 1.414   as to  A 
> > _____             ___ 
> > .5                   .001 
> > 
> > 
> > .001 = .5A then subtract from 1 is essentially 1 with length 1.4+1 makes no sense. 
> > 
> > Unless, unless the maximum length is not 1+1, for those are hypoteni 
> > 
> > So is the maximum length 2.414, where I thought it was 2. 
> > 
> > 
> 
> Let me try it on Y=x^3, integral is 1/4x^4 
> 
> 1.414   is to A 
> _____           ____ 
> .5                 .25 
> 
> 
> 1.414*.25  = .5A 
> 
> .3535 = .5A 
> 
> A = .707 
> 
> Subtract from 1 we have .3 
> 
> 
> 1.414+ .3 = 1.71 
> 
> Is that the arclength of Y= x^3 in interval 0 to 1 


Alright the Quarter circle area is .785

So we have

1.414  is to  A
______         ____
.5                 .285

The .285 is .785-.5


So we have

.402/.5 = A

.805 = A

Subtract from 1 we have .2

So add .2 + 1.414 = 1.61

Not bad, not bad at all for the quarter circle is arclength 1.57

So it appears the method is valid, quick, easy. And what I like most of all, the mind is in full comprehension of each step and knows what is going on. No-one in Old Math can say that with their thousands of obnoxious tricks and trig you have to shovel. Does anyone know why math professors love shovelling tricks and trig?

AP

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