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tech / sci.math / Re: Q on finding the slope of a parametric line

SubjectAuthor
* Q on finding the slope of a parametric lineJulio Di Egidio
+* Re: Q on finding the slope of a parametric lineJulio Di Egidio
|+* Re: Q on finding the slope of a parametric lineFromTheRafters
||`* Re: Q on finding the slope of a parametric lineJulio Di Egidio
|| `- Re: Q on finding the slope of a parametric linePython
|+* Re: Q on finding the slope of a parametric lineMostowski Collapse
||`* Re: Q on finding the slope of a parametric lineJulio Di Egidio
|| +* Re: Q on finding the slope of a parametric lineMostowski Collapse
|| |`* Re: Q on finding the slope of a parametric lineJulio Di Egidio
|| | `* Re: Q on finding the slope of a parametric lineMostowski Collapse
|| |  `* Re: Q on finding the slope of a parametric lineMostowski Collapse
|| |   `* Re: Q on finding the slope of a parametric lineMostowski Collapse
|| |    `- Re: Q on finding the slope of a parametric lineMostowski Collapse
|| `- Re: Q on finding the slope of a parametric lineChris M. Thomasson
|`* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| +* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| |`* Re: Q on finding the slope of a parametric lineFromTheRafters
| | `* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| |  `* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| |   `* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| |    `- Re: Q on finding the slope of a parametric lineJulio Di Egidio
| +* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| |`- Re: Q on finding the slope of a parametric lineJulio Di Egidio
| `- Re: Q on finding the slope of a parametric lineJulio Di Egidio
+- Re: Q on finding the slope of a parametric lineArchimedes Plutonium
+* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
|+* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
||`- Re: Q on finding the slope of a parametric lineArchimedes Plutonium
|`* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| +* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
| |+* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
| ||`* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| || +* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
| || |`* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
| || | `* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
| || |  +* Re: Q on finding the slope of a parametric lineChris M. Thomasson
| || |  |`* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| || |  | `* Re: Q on finding the slope of a parametric lineChris M. Thomasson
| || |  |  `- Re: Q on finding the slope of a parametric lineChris M. Thomasson
| || |  `* Re: Q on finding the slope of a parametric lineMostowski Collapse
| || |   `* Re: Q on finding the slope of a parametric lineHugh Itoh
| || |    `* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
| || |     `* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
| || |      `* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| || |       `- Re: Q on finding the slope of a parametric lineArchimedes Plutonium
| || `- Re: Q on finding the slope of a parametric lineArchimedes Plutonium
| |`* Re: Q on finding the slope of a parametric lineJulio Di Egidio
| | `- Re: Q on finding the slope of a parametric lineArchimedes Plutonium
| `* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
|  `* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
|   `* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
|    `* Re: Q on finding the slope of a parametric lineArchimedes Plutonium
|     `- Re: Q on finding the slope of a parametric lineArchimedes Plutonium
`* Re: Q on finding the slope of a parametric lineChris M. Thomasson
 `* Re: Q on finding the slope of a parametric lineChris M. Thomasson
  `* Re: Q on finding the slope of a parametric lineChris M. Thomasson
   `* Re: Q on finding the slope of a parametric lineJulio Di Egidio
    `- Re: Q on finding the slope of a parametric lineChris M. Thomasson

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Q on finding the slope of a parametric line

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Subject: Q on finding the slope of a parametric line
From: jul...@diegidio.name (Julio Di Egidio)
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 by: Julio Di Egidio - Sun, 24 Apr 2022 14:22 UTC

The slope of a 2D straight line parameterized on t is:
m = (y(t) - y0) / (x(t) - x0) (must be a constant)
but that expression fails (doesn't it?) for t = 0, because x(0) = x0 and y(0) = y0, so we get m = 0/0.

In practice, if we are given expressions y(t) and x(t) and need to find the corresponding slope, (as long as it is in fact a straight line) we are guaranteed that the expression for the slope simplifies to a constant, but, strictly speaking, our derivation is invalid for t = 0.

Do we just say/imply that "the discontinuity is eliminable at t=0" or something of that sort (since m is defined and constant everywhere except at t=0)? Otherwise, how do we correctly find/express the slope of a straight line given parametric equations, or state what we have found?

Thanks in advance for any help,

Julio

Re: Q on finding the slope of a parametric line

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Subject: Re: Q on finding the slope of a parametric line
From: jul...@diegidio.name (Julio Di Egidio)
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 by: Julio Di Egidio - Sun, 24 Apr 2022 15:32 UTC

On Sunday, 24 April 2022 at 16:22:40 UTC+2, Julio Di Egidio wrote:
> The slope of a 2D straight line parameterized on t is:
> m = (y(t) - y0) / (x(t) - x0) (must be a constant) but that
> expression fails (doesn't it?) for t = 0, because
> x(0) = x0 and y(0) = y0, so we get m = 0/0.
>
> In practice, if we are given expressions y(t) and x(t)
> and need to find the corresponding slope, (as long as
> it is in fact a straight line) we are guaranteed that the
> expression for the slope simplifies to a constant, but,
> strictly speaking, our derivation is invalid for t = 0.

In fact, here is a simpler form of the same problem:

Say I have the relation y = m x for m any real number.

I could rewrite it as x = (1/m) y, and now it is not defined for m = 0.

But, it occurs to me, if I need to express x in terms of y, I should rather write:

R := [x = (1/m) y] if m=/=0, else [y = 0]

and I suppose, modulo the clumsy notation, that's it: rather than "eliminable discontinuity" I need "definition by cases".

Correct?

> Do we just say/imply that "the discontinuity is eliminable at t=0" or something of that sort (since m is defined and constant everywhere except at t=0)? Otherwise, how do we correctly find/express the slope of a straight line given parametric equations, or state what we have found?
>
> Thanks in advance for any help,

Julio

Re: Q on finding the slope of a parametric line

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From: nom...@afraid.org (FromTheRafters)
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Subject: Re: Q on finding the slope of a parametric line
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 by: FromTheRafters - Sun, 24 Apr 2022 16:31 UTC

Julio Di Egidio wrote :
> On Sunday, 24 April 2022 at 16:22:40 UTC+2, Julio Di Egidio wrote:
>> The slope of a 2D straight line parameterized on t is:
>> m = (y(t) - y0) / (x(t) - x0) (must be a constant) but that
>> expression fails (doesn't it?) for t = 0, because
>> x(0) = x0 and y(0) = y0, so we get m = 0/0.
>>
>> In practice, if we are given expressions y(t) and x(t)
>> and need to find the corresponding slope, (as long as
>> it is in fact a straight line) we are guaranteed that the
>> expression for the slope simplifies to a constant, but,
>> strictly speaking, our derivation is invalid for t = 0.
>
> In fact, here is a simpler form of the same problem:
>
> Say I have the relation y = m x for m any real number.
>
> I could rewrite it as x = (1/m) y, and now it is not defined for m = 0.
>
> But, it occurs to me, if I need to express x in terms of y, I should rather
> write:
>
> R := [x = (1/m) y] if m=/=0, else [y = 0]
>
> and I suppose, modulo the clumsy notation, that's it: rather than "eliminable
> discontinuity" I need "definition by cases".
>
> Correct?
>
>> Do we just say/imply that "the discontinuity is eliminable at t=0" or
>> something of that sort (since m is defined and constant everywhere except at
>> t=0)? Otherwise, how do we correctly find/express the slope of a straight
>> line given parametric equations, or state what we have found?
>>
>> Thanks in advance for any help,
>
> Julio

As a point/slope formula neither x nor y can be zero. Both x in terms
of y and y in terms of x must be true. IOW I think you would have to
capture both 'bad' cases.

Re: Q on finding the slope of a parametric line

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Subject: Re: Q on finding the slope of a parametric line
From: jul...@diegidio.name (Julio Di Egidio)
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 by: Julio Di Egidio - Sun, 24 Apr 2022 16:46 UTC

On Sunday, 24 April 2022 at 18:31:24 UTC+2, FromTheRafters wrote:
> Julio Di Egidio wrote :
> > On Sunday, 24 April 2022 at 16:22:40 UTC+2, Julio Di Egidio wrote:
> >> The slope of a 2D straight line parameterized on t is:
> >> m = (y(t) - y0) / (x(t) - x0) (must be a constant) but that
> >> expression fails (doesn't it?) for t = 0, because
> >> x(0) = x0 and y(0) = y0, so we get m = 0/0.
> >>
> >> In practice, if we are given expressions y(t) and x(t)
> >> and need to find the corresponding slope, (as long as
> >> it is in fact a straight line) we are guaranteed that the
> >> expression for the slope simplifies to a constant, but,
> >> strictly speaking, our derivation is invalid for t = 0.
> >
> > In fact, here is a simpler form of the same problem:
> >
> > Say I have the relation y = m x for m any real number.
> >
> > I could rewrite it as x = (1/m) y, and now it is not defined for m = 0.
> >
> > But, it occurs to me, if I need to express x in terms of y, I should rather
> > write:
> >
> > R := [x = (1/m) y] if m=/=0, else [y = 0]
> >
> > and I suppose, modulo the clumsy notation, that's it: rather than "eliminable
> > discontinuity" I need "definition by cases".
> >
> > Correct?
> >
> >> Do we just say/imply that "the discontinuity is eliminable at t=0" or
> >> something of that sort (since m is defined and constant everywhere except at
> >> t=0)? Otherwise, how do we correctly find/express the slope of a straight
> >> line given parametric equations, or state what we have found?
> >>
> >> Thanks in advance for any help,
>
> As a point/slope formula neither x nor y can be zero. Both x in terms
> of y and y in terms of x must be true. IOW I think you would have to
> capture both 'bad' cases.

In the "parametric" example indeed I said t=0, not x nor y. But parametric was a red herring: look at the "simpler" example, that's already and directly the problem, that whenever I take the "factor" to the other side, I incur into an issue that seems should be avoidable by some algebraic property...

Julio

Re: Q on finding the slope of a parametric line

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Subject: Re: Q on finding the slope of a parametric line
From: burse...@gmail.com (Mostowski Collapse)
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 by: Mostowski Collapse - Sun, 24 Apr 2022 17:17 UTC

The usual way to define it as a set of pairs
is to use some algebraic form:

0 = a*x + b*y.

You can cover all straight lines going through
zero, also horizontal and vertical lines.

You can also derive this parametrization:

x = t*cos(phi)

y = t*sin(phi)

where phi = arg(b,-a).

ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 17:32:19 UTC+2:
> On Sunday, 24 April 2022 at 16:22:40 UTC+2, Julio Di Egidio wrote:
> > The slope of a 2D straight line parameterized on t is:
> > m = (y(t) - y0) / (x(t) - x0) (must be a constant) but that
> > expression fails (doesn't it?) for t = 0, because
> > x(0) = x0 and y(0) = y0, so we get m = 0/0.
> >
> > In practice, if we are given expressions y(t) and x(t)
> > and need to find the corresponding slope, (as long as
> > it is in fact a straight line) we are guaranteed that the
> > expression for the slope simplifies to a constant, but,
> > strictly speaking, our derivation is invalid for t = 0.
> In fact, here is a simpler form of the same problem:
>
> Say I have the relation y = m x for m any real number.
>
> I could rewrite it as x = (1/m) y, and now it is not defined for m = 0.
>
> But, it occurs to me, if I need to express x in terms of y, I should rather write:
>
> R := [x = (1/m) y] if m=/=0, else [y = 0]
>
> and I suppose, modulo the clumsy notation, that's it: rather than "eliminable discontinuity" I need "definition by cases".
>
> Correct?
> > Do we just say/imply that "the discontinuity is eliminable at t=0" or something of that sort (since m is defined and constant everywhere except at t=0)? Otherwise, how do we correctly find/express the slope of a straight line given parametric equations, or state what we have found?
> >
> > Thanks in advance for any help,
>
> Julio

Re: Q on finding the slope of a parametric line

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Subject: Re: Q on finding the slope of a parametric line
From: jul...@diegidio.name (Julio Di Egidio)
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 by: Julio Di Egidio - Sun, 24 Apr 2022 18:11 UTC

On Sunday, 24 April 2022 at 19:18:01 UTC+2, Mostowski Collapse wrote:
> ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 17:32:19 UTC+2:
<snip>
> The usual way to define it as a set of pairs
> is to use some algebraic form:
>
> 0 = a*x + b*y.
>
> You can cover all straight lines going through
> zero, also horizontal and vertical lines.

Right, I see, but it's the exact opposite that you are saying: instead of the relations here we bring it back to a functional form in two variables. Indeed, cool idea...

> You can also derive this parametrization:
>
> x = t*cos(phi)
>
> y = t*sin(phi)
>
> where phi = arg(b,-a).

I do not understand what you mean by arg(b,-a).

Julio

Re: Q on finding the slope of a parametric line

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Subject: Re: Q on finding the slope of a parametric line
From: burse...@gmail.com (Mostowski Collapse)
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 by: Mostowski Collapse - Sun, 24 Apr 2022 18:38 UTC

The relation L ⊆ R x R would be:

{ (x,y) e R x R | 0 = a*x + b*y }

There is a corner case a=0 & b=0 which is not a straight
line, but the full plane R x R. The form with 0 = ... sees
a curve as a finding roots problem. Which is not what one

usually expects. But the "roots" of a multivariate polynomial
are the algebraic curves. See also:

classically studying zeros of multivariate polynomials
https://en.wikipedia.org/wiki/Algebraic_geometry

The arg(_,_) function is from here:
https://en.wikipedia.org/wiki/Argument_%28complex_analysis%29

ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 20:11:15 UTC+2:
> On Sunday, 24 April 2022 at 19:18:01 UTC+2, Mostowski Collapse wrote:
> > ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 17:32:19 UTC+2:
> <snip>
> > The usual way to define it as a set of pairs
> > is to use some algebraic form:
> >
> > 0 = a*x + b*y.
> >
> > You can cover all straight lines going through
> > zero, also horizontal and vertical lines.
> Right, I see, but it's the exact opposite that you are saying: instead of the relations here we bring it back to a functional form in two variables. Indeed, cool idea...
> > You can also derive this parametrization:
> >
> > x = t*cos(phi)
> >
> > y = t*sin(phi)
> >
> > where phi = arg(b,-a).
> I do not understand what you mean by arg(b,-a).
>
> Julio

Re: Q on finding the slope of a parametric line

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Subject: Re: Q on finding the slope of a parametric line
From: jul...@diegidio.name (Julio Di Egidio)
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 by: Julio Di Egidio - Sun, 24 Apr 2022 19:15 UTC

On Sunday, 24 April 2022 at 20:38:11 UTC+2, Mostowski Collapse wrote:
> ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 20:11:15 UTC+2:
> > On Sunday, 24 April 2022 at 19:18:01 UTC+2, Mostowski Collapse wrote:

> > > x = t*cos(phi)
> > > y = t*sin(phi)
> > > where phi = arg(b,-a).
> >
> > I do not understand what you mean by arg(b,-a).
>
> The arg(_,_) function is from here:
> https://en.wikipedia.org/wiki/Argument_%28complex_analysis%29

The arg function takes one complex argument. I guess you meant arg(b-ia).

Julio

Re: Q on finding the slope of a parametric line

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 by: Python - Sun, 24 Apr 2022 20:32 UTC

Stronzo Julio Di Egidio wrote:
> On Sunday, 24 April 2022 at 18:31:24 UTC+2, FromTheRafters wrote:
>> Julio Di Egidio wrote :
>>> On Sunday, 24 April 2022 at 16:22:40 UTC+2, Julio Di Egidio wrote:
>>>> The slope of a 2D straight line parameterized on t is:
>>>> m = (y(t) - y0) / (x(t) - x0) (must be a constant) but that
>>>> expression fails (doesn't it?) for t = 0, because
>>>> x(0) = x0 and y(0) = y0, so we get m = 0/0.
>>>>
>>>> In practice, if we are given expressions y(t) and x(t)
>>>> and need to find the corresponding slope, (as long as
>>>> it is in fact a straight line) we are guaranteed that the
>>>> expression for the slope simplifies to a constant, but,
>>>> strictly speaking, our derivation is invalid for t = 0.
>>>
>>> In fact, here is a simpler form of the same problem:
>>>
>>> Say I have the relation y = m x for m any real number.
>>>
>>> I could rewrite it as x = (1/m) y, and now it is not defined for m = 0.
>>>
>>> But, it occurs to me, if I need to express x in terms of y, I should rather
>>> write:
>>>
>>> R := [x = (1/m) y] if m=/=0, else [y = 0]
>>>
>>> and I suppose, modulo the clumsy notation, that's it: rather than "eliminable
>>> discontinuity" I need "definition by cases".
>>>
>>> Correct?
>>>
>>>> Do we just say/imply that "the discontinuity is eliminable at t=0" or
>>>> something of that sort (since m is defined and constant everywhere except at
>>>> t=0)? Otherwise, how do we correctly find/express the slope of a straight
>>>> line given parametric equations, or state what we have found?
>>>>
>>>> Thanks in advance for any help,
>>
>> As a point/slope formula neither x nor y can be zero. Both x in terms
>> of y and y in terms of x must be true. IOW I think you would have to
>> capture both 'bad' cases.
>
> In the "parametric" example indeed I said t=0, not x nor y. But parametric was a red herring: look at the "simpler" example, that's already and directly the problem, that whenever I take the "factor" to the other side, I incur into an issue that seems should be avoidable by some algebraic property...

What an idiot you are Julio...

Re: Q on finding the slope of a parametric line

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 by: Chris M. Thomasson - Sun, 24 Apr 2022 20:44 UTC

On 4/24/2022 11:11 AM, Julio Di Egidio wrote:
> On Sunday, 24 April 2022 at 19:18:01 UTC+2, Mostowski Collapse wrote:
>> ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 17:32:19 UTC+2:
> <snip>
>> The usual way to define it as a set of pairs
>> is to use some algebraic form:
>>
>> 0 = a*x + b*y.
>>
>> You can cover all straight lines going through
>> zero, also horizontal and vertical lines.
>
> Right, I see, but it's the exact opposite that you are saying: instead of the relations here we bring it back to a functional form in two variables. Indeed, cool idea...
>
>> You can also derive this parametrization:
>>
>> x = t*cos(phi)
>>
>> y = t*sin(phi)
>>
>> where phi = arg(b,-a).
>
> I do not understand what you mean by arg(b,-a).

I assume it is the argument of a complex number.

Re: Q on finding the slope of a parametric line

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Subject: Re: Q on finding the slope of a parametric line
From: plutoniu...@gmail.com (Archimedes Plutonium)
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 by: Archimedes Plutonium - Sun, 24 Apr 2022 21:04 UTC

On Sunday, April 24, 2022 at 9:22:40 AM UTC-5, ju...@diegidio.name wrote:
> The slope of a 2D straight line parameterized on t is:
> m = (y(t) - y0) / (x(t) - x0) (must be a constant)
> but that expression fails (doesn't it?) for t = 0, because x(0) = x0 and y(0) = y0, so we get m = 0/0.
>
> In practice, if we are given expressions y(t) and x(t) and need to find the corresponding slope, (as long as it is in fact a straight line) we are guaranteed that the expression for the slope simplifies to a constant, but, strictly speaking, our derivation is invalid for t = 0.
>
> Do we just say/imply that "the discontinuity is eliminable at t=0" or something of that sort (since m is defined and constant everywhere except at t=0)? Otherwise, how do we correctly find/express the slope of a straight line given parametric equations, or state what we have found?
>
> Thanks in advance for any help,
>
> Julio

Hi, Julio, I am so happy to see you back in sci.math posting. I love reading your posts especially when you put the Cretins in their place.

I suppose this may help in your question.

In my mathematics, the only real true functions of math are Polynomials, polynomial functions.

And there are plenty plenty of Y = mx + b functions, all straight line functions.

So when I come to mathematics with the definition that to be a function, you have to be a Polynomial. That eliminates the horrible philosophy of slope or gradient of y-y_0 / x - x_0

So when you make all functions be Polynomials, you never run into the sad, stupid silly Old Math of philosophy worrying about 0/0.

So in Old Math, their set-up is a screw-up.

In New Math-- all functions are Polynomials, and only Polynomials and so we never run into a idiocy of 0/0

And thanks Julio, for now I have to use this post to add to my Mathopedia, that in Old Math, those silly people have to stumble with 0/0, while New Math totally bypasses the foolishness with the statement-- Functions all are polynomials and that forever banishes 0/0.

Thanks Julio

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 by: Archimedes Plutonium - Sun, 24 Apr 2022 23:00 UTC

Question please, for Julio. For I have always shyed away from parametric equations in my entire life.

If all valid and true functions are only Polynomials, then, does that not eliminate parametric equations completely off the face of all of mathematics? Do parametric equations evaporate into thin air, once a mathematician accepts the idea-- the only valid functions in mathematics are polynomials.

I see parametric equations in Old Math only because they foolishly accepted anything as a function, for a moron off the street could dream up a new function. While in New Math-- polynomials are the end all and be all of functions.

So my question is-- if you accept that polynomials are the only functions in all of mathematics, then parametric equations disappear altogether as being fakery.

I do not know the answer, for I never dealt much with parametric hoopla

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 by: Archimedes Plutonium - Sun, 24 Apr 2022 23:06 UTC

On Sunday, April 24, 2022 at 6:00:21 PM UTC-5, Archimedes Plutonium wrote:
> Question please, for Julio. For I have always shyed away from parametric equations in my entire life.
>
> If all valid and true functions are only Polynomials, then, does that not eliminate parametric equations completely off the face of all of mathematics? Do parametric equations evaporate into thin air, once a mathematician accepts the idea-- the only valid functions in mathematics are polynomials.
>
> I see parametric equations in Old Math only because they foolishly accepted anything as a function, for a moron off the street could dream up a new function. While in New Math-- polynomials are the end all and be all of functions.
>
> So my question is-- if you accept that polynomials are the only functions in all of mathematics, then parametric equations disappear altogether as being fakery.
>
> I do not know the answer, for I never dealt much with parametric hoopla

I suspect the answer is going to be from someone who has worked long and hard in parametric equations, that the answer is that parametric equations was a "polynomial seeking equation". That parametric equations were foolish and silly mathematics trying to Become a polynomial equation. Sort of like Plato's being and becoming.

I think the deal with parametric equations is that it was a attempt in a sea-sewer full of all types of equations allowed into mathematics when polynomials are the only Valid equation, and that parametrics was an attempt to make the most obnoxious stench filled equations not polynomial, to have some features be polynomial-like. Like in religion, where one religion tries to convert others in a different religion. When the only true equation is polynomials.

AP

Re: Q on finding the slope of a parametric line

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Subject: Re: Q on finding the slope of a parametric line
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 by: Chris M. Thomasson - Mon, 25 Apr 2022 00:12 UTC

On 4/24/2022 7:22 AM, Julio Di Egidio wrote:
> The slope of a 2D straight line parameterized on t is:
> m = (y(t) - y0) / (x(t) - x0) (must be a constant)
> but that expression fails (doesn't it?) for t = 0, because x(0) = x0 and y(0) = y0, so we get m = 0/0.
>
> In practice, if we are given expressions y(t) and x(t) and need to find the corresponding slope, (as long as it is in fact a straight line) we are guaranteed that the expression for the slope simplifies to a constant, but, strictly speaking, our derivation is invalid for t = 0.
>
> Do we just say/imply that "the discontinuity is eliminable at t=0" or something of that sort (since m is defined and constant everywhere except at t=0)? Otherwise, how do we correctly find/express the slope of a straight line given parametric equations, or state what we have found?
>
> Thanks in advance for any help,

Probably misunderstanding you here... Think in terms of two 2d vectors,
and the angle between them. atan2(0, 0) is undefined when x _and_ y are
both zero. So, taking the angle between points:

vec2 p0 = {0, 0}
vec2 p1 = {0, 0}
vec2 dif = p1 - p0
angle = atan2(dif.y, dif.x)

angle is undefined. So you can get zero radians, or undefined. Depends
on what the implementation decides to do.

Re: Q on finding the slope of a parametric line

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Subject: Re: Q on finding the slope of a parametric line
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 by: Mostowski Collapse - Mon, 25 Apr 2022 00:28 UTC

I guess I have switched the arguments, its should be:

phi = atan2(-a,b)
https://en.wikipedia.org/wiki/Atan2

You can now solve the equation:

0 = a*x + b*y

= a*t*cos(phi)+b*t*sin(phi)

= a*t*b/c+b*t*(-a)/c
where c = sqrt(a^2 + b^2) since phi = atan2(-a,b)

= 0

ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 21:15:53 UTC+2:
> On Sunday, 24 April 2022 at 20:38:11 UTC+2, Mostowski Collapse wrote:
> > ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 20:11:15 UTC+2:
> > > On Sunday, 24 April 2022 at 19:18:01 UTC+2, Mostowski Collapse wrote:
>
> > > > x = t*cos(phi)
> > > > y = t*sin(phi)
> > > > where phi = arg(b,-a).
> > >
> > > I do not understand what you mean by arg(b,-a).
> >
> > The arg(_,_) function is from here:
> > https://en.wikipedia.org/wiki/Argument_%28complex_analysis%29
> The arg function takes one complex argument. I guess you meant arg(b-ia).
>
> Julio

Re: Q on finding the slope of a parametric line

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 by: Mostowski Collapse - Mon, 25 Apr 2022 00:33 UTC

Example: A 45° Line (Respectively a -135° Line):

0 = x - y (or alternatively 0 = y - x)
https://www.wolframalpha.com/input?i=0+%3D+x+-+y

LoL

Mostowski Collapse schrieb am Montag, 25. April 2022 um 02:28:39 UTC+2:
> I guess I have switched the arguments, its should be:
>
> phi = atan2(-a,b)
> https://en.wikipedia.org/wiki/Atan2
>
> You can now solve the equation:
> 0 = a*x + b*y
> = a*t*cos(phi)+b*t*sin(phi)
>
> = a*t*b/c+b*t*(-a)/c
> where c = sqrt(a^2 + b^2) since phi = atan2(-a,b)
>
> = 0
> ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 21:15:53 UTC+2:
> > On Sunday, 24 April 2022 at 20:38:11 UTC+2, Mostowski Collapse wrote:
> > > ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 20:11:15 UTC+2:
> > > > On Sunday, 24 April 2022 at 19:18:01 UTC+2, Mostowski Collapse wrote:
> >
> > > > > x = t*cos(phi)
> > > > > y = t*sin(phi)
> > > > > where phi = arg(b,-a).
> > > >
> > > > I do not understand what you mean by arg(b,-a).
> > >
> > > The arg(_,_) function is from here:
> > > https://en.wikipedia.org/wiki/Argument_%28complex_analysis%29
> > The arg function takes one complex argument. I guess you meant arg(b-ia).
> >
> > Julio

Re: Q on finding the slope of a parametric line

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Subject: Re: Q on finding the slope of a parametric line
From: plutoniu...@gmail.com (Archimedes Plutonium)
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 by: Archimedes Plutonium - Mon, 25 Apr 2022 00:40 UTC

240th book of science by AP

Parametric Equations of Old Math were phony props, to bolster up the fact that only Polynomials are true functions of mathematics.

And so when practitioners of math call a function which is fakery and need to graph it, they break that phony and fake function into parts trying to converge on a polynomial.

For example in Stewart 2003 "Calculus": on page 654 looks at the fake function x = y^4 - 3 y^2 and Stewart breaks that down into a parametric of y= t and x = t^4 -3t^2.

So, at the start x = y^4 - 3y^2 is not a function in New Math. However Y = x^4 -3x^2 is a function for it is a polynomial, but it is not the same as X = y^4 - 3y^2.

And the exercise that Stewart goes through in obtaining what he calls a Parametric equation, is simply the exercise that Stewart is a failure to recognize that only polynomials as functions exist. And so when Stewart runs into a equation that is not polynomial, he wiggles and squeezes out parametric equations that are polynomial.

The entire exercise of fetching Parametric equations is to make hideous ugly nonfunctions turned into polynomials.

Now in some cases like that of many trigonometry equations, they should have been turned into a polynomial over an interval using the Lagrange transform, but then many such are turned into more fake functions like Stewart's example on page 652 of x^2 + y^2 = 1 where Stewart calls that a parametric equation of x= cos(t) and y = sin(t).

But in general, the exercise of looking for and finding parametric equations in Old Math is because of the blindness of the person in never realizing that the only functions in true mathematics are Polynomials. And this seeking parametric equations is a substitute for not finding the Lagrange transformation of a given nonfunction over a prescribed interval as a polynomial.

AP

Re: Q on finding the slope of a parametric line

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 by: Mostowski Collapse - Mon, 25 Apr 2022 00:42 UTC

But atan2(_,_) drops out of the equation, also sin(_) and cos(_),
you only need sqrt(_) if you want this parametrization:,
with t=1 you land on the unit circle, centered at zero:

x = t*cos(phi)

y = t*sin(phi)

Give it a go with by this variant, with t=1 you land again
on the unit circle, centered at zero:

x = t*b/c

y = t*(-a)/c

Where c = sqrt(a^2 + c^2), or forget about c, with t=1
you land on a circle with radius c, centered at zero:

x = t*b

y = t*(-a)

Mostowski Collapse schrieb am Montag, 25. April 2022 um 02:33:54 UTC+2:
> Example: A 45° Line (Respectively a -135° Line):
>
> 0 = x - y (or alternatively 0 = y - x)
> https://www.wolframalpha.com/input?i=0+%3D+x+-+y
>
> LoL
> Mostowski Collapse schrieb am Montag, 25. April 2022 um 02:28:39 UTC+2:
> > I guess I have switched the arguments, its should be:
> >
> > phi = atan2(-a,b)
> > https://en.wikipedia.org/wiki/Atan2
> >
> > You can now solve the equation:
> > 0 = a*x + b*y
> > = a*t*cos(phi)+b*t*sin(phi)
> >
> > = a*t*b/c+b*t*(-a)/c
> > where c = sqrt(a^2 + b^2) since phi = atan2(-a,b)
> >
> > = 0
> > ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 21:15:53 UTC+2:
> > > On Sunday, 24 April 2022 at 20:38:11 UTC+2, Mostowski Collapse wrote:
> > > > ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 20:11:15 UTC+2:
> > > > > On Sunday, 24 April 2022 at 19:18:01 UTC+2, Mostowski Collapse wrote:
> > >
> > > > > > x = t*cos(phi)
> > > > > > y = t*sin(phi)
> > > > > > where phi = arg(b,-a).
> > > > >
> > > > > I do not understand what you mean by arg(b,-a).
> > > >
> > > > The arg(_,_) function is from here:
> > > > https://en.wikipedia.org/wiki/Argument_%28complex_analysis%29
> > > The arg function takes one complex argument. I guess you meant arg(b-ia).
> > >
> > > Julio

Re: Q on finding the slope of a parametric line

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 by: Mostowski Collapse - Mon, 25 Apr 2022 00:50 UTC

Maybe we should switch attention to ellipse:

0 = a*x^2 + b*y^2 -a*b

AP brain farto thinks, its an oval.
I guess the following parametrization works:

x = sqrt(b) cos(t)

y = sqrt(a) sin(t).

Mostowski Collapse schrieb am Montag, 25. April 2022 um 02:42:21 UTC+2:
> But atan2(_,_) drops out of the equation, also sin(_) and cos(_),
> you only need sqrt(_) if you want this parametrization:,
> with t=1 you land on the unit circle, centered at zero:
> x = t*cos(phi)
>
> y = t*sin(phi)
> Give it a go with by this variant, with t=1 you land again
> on the unit circle, centered at zero:
>
> x = t*b/c
>
> y = t*(-a)/c
>
> Where c = sqrt(a^2 + c^2), or forget about c, with t=1
> you land on a circle with radius c, centered at zero:
>
> x = t*b
>
> y = t*(-a)
> Mostowski Collapse schrieb am Montag, 25. April 2022 um 02:33:54 UTC+2:
> > Example: A 45° Line (Respectively a -135° Line):
> >
> > 0 = x - y (or alternatively 0 = y - x)
> > https://www.wolframalpha.com/input?i=0+%3D+x+-+y
> >
> > LoL
> > Mostowski Collapse schrieb am Montag, 25. April 2022 um 02:28:39 UTC+2:
> > > I guess I have switched the arguments, its should be:
> > >
> > > phi = atan2(-a,b)
> > > https://en.wikipedia.org/wiki/Atan2
> > >
> > > You can now solve the equation:
> > > 0 = a*x + b*y
> > > = a*t*cos(phi)+b*t*sin(phi)
> > >
> > > = a*t*b/c+b*t*(-a)/c
> > > where c = sqrt(a^2 + b^2) since phi = atan2(-a,b)
> > >
> > > = 0
> > > ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 21:15:53 UTC+2:
> > > > On Sunday, 24 April 2022 at 20:38:11 UTC+2, Mostowski Collapse wrote:
> > > > > ju...@diegidio.name schrieb am Sonntag, 24. April 2022 um 20:11:15 UTC+2:
> > > > > > On Sunday, 24 April 2022 at 19:18:01 UTC+2, Mostowski Collapse wrote:
> > > >
> > > > > > > x = t*cos(phi)
> > > > > > > y = t*sin(phi)
> > > > > > > where phi = arg(b,-a).
> > > > > >
> > > > > > I do not understand what you mean by arg(b,-a).
> > > > >
> > > > > The arg(_,_) function is from here:
> > > > > https://en.wikipedia.org/wiki/Argument_%28complex_analysis%29
> > > > The arg function takes one complex argument. I guess you meant arg(b-ia).
> > > >
> > > > Julio

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Subject: Re: Q on finding the slope of a parametric line
From: jul...@diegidio.name (Julio Di Egidio)
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 by: Julio Di Egidio - Mon, 25 Apr 2022 13:00 UTC

On Monday, 25 April 2022 at 01:00:21 UTC+2, Archimedes Plutonium wrote:
> Question please, for Julio.

Is that even you? I don't think so, for a couple of years now.

> For I have always shyed away from parametric equations in my entire life.

Yet, for example, physics is full of it: take the 2D coordinates of a point, (x,y), and now have those change in time, whence (x(t),y(t)), which is a trajectory, i.e. a line, parametrized on time. And what's more ubiquitous than that?

> If all valid and true functions are only Polynomials

Would you discard stuff as simple as f(x) = sqrt(x)?

I think where we differ is that I don't shy away from irrational numbers and functions: in fact, I don't shy away from (actual, proper) *infinity* to begin with, and would rather fix that in standard mathematics... Indeed, at least in applications, I seem to be able to, by rather keeping around the constants that blow up as scale factors: a little algebraic "trick" that allows me to compute and draw over *closed* intervals and without taking any limits even when those intervals go infinite (*). But the issue remains that, while I am pretty sure of the correctness of the results I get in my applications, I am not quite sure about the formal justification and treatment.

In fact, the problem is how to properly generalize that approach to a number system: e.g. I'd like, with 'a' a constant (a parameter), y=ax and its inverse x=y/a to be defined for all a (and "inverses" *are* ubiquitous, up to and including category theory), which, as said, I am obtaining by letting k=1/a and keeping k around as a scale factor (a "unit"). Indeed, thinking in the direction of "numbers with units" seems interesting, but I am still not sure what that actually brings and how general it is: which is my line of research at this stage...

(*) E.g. here I scale by gamma(beta) and keep the whole thing finite even when beta = 1, the speed of light:
<https://jp-diegidio.github.io/STUDY.Physics.SpecialRelativity/InertialFrames/App/>
(Pretty reminiscent of an actual "infinity boundary" in action, isn't it, Archimedes?)

Your thoughts?

Julio

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 by: Archimedes Plutonium - Mon, 25 Apr 2022 21:40 UTC

On Monday, April 25, 2022 at 8:00:54 AM UTC-5, ju...@diegidio.name wrote:
> On Monday, 25 April 2022 at 01:00:21 UTC+2, Archimedes Plutonium wrote:
> > Question please, for Julio.
> Is that even you? I don't think so, for a couple of years now.

The same person who writes science books.

> > For I have always shyed away from parametric equations in my entire life.
> Yet, for example, physics is full of it: take the 2D coordinates of a point, (x,y), and now have those change in time, whence (x(t),y(t)), which is a trajectory, i.e. a line, parametrized on time. And what's more ubiquitous than that?
> > If all valid and true functions are only Polynomials
> Would you discard stuff as simple as f(x) = sqrt(x)?
>

Yes, that maybe the most simple equation discarded Y = x^-1/2. Now can we take the Power Rules to coax out a derivative or integral? Y' = -1/2 (x^-3/2)

Old Math allowed admittance and entry of anything, and called it mathematics. New Math has a prime essential concern-- it has to be Logical and simple..

So in New Math, we want to have all of Calculus be extraordinarily simple. So simple that 1 day classroom lesson we learn that the only functions that exist (not equations but functions) are polynomials. And learn that the Power Rules unlock every derivative and every integral of every function that exists.

No longer do we have a 1,000 classroom lectures teaching what the derivative and integral of nonpolynomials. No longer do we have the first 300 pages of a Freshman Calculus text teaching rules and techniques-- Simpson rule, trapezoid rule. No longer do we have 10,000 rules to memorize in order to do 10,000 different functions.

Whenever a person wants a derivative or integral of a nonfunction such as Y = sqrt(x). They simply must convert with Lagrange transform, convert over a specified interval into a polynomial then apply the power rule.

AP can teach Calculus in junior year High School. Old Math never teaches true and proper calculus, no matter what age the student.

I need to convert sqrt(x) from interval 1 to 2 and see what the derivative is. Is it anywhere near that of -1/2(x^-3/2)?

AP math, the student need only learn the Power Rules and be finished with all of Calculus. Old Math, the student is never finished with calculus, but having to learn new rules for every new function.

> I think where we differ is that I don't shy away from irrational numbers and functions: in fact, I don't shy away from (actual, proper) *infinity* to begin with, and would rather fix that in standard mathematics... Indeed, at least in applications, I seem to be able to, by rather keeping around the constants that blow up as scale factors: a little algebraic "trick" that allows me to compute and draw over *closed* intervals and without taking any limits even when those intervals go infinite (*). But the issue remains that, while I am pretty sure of the correctness of the results I get in my applications, I am not quite sure about the formal justification and treatment.
>

There was a recent article in New Scientist, 16Apr2022 "How Big is Infinity" by Timothy Revell. Sadly there is no new news. For Revell just repeats the sad and silly history of "infinity". The only time there is new news on infinity, is when the article asks where the borderline is from finite then infinite numbers. Apparently Revell never studied Huygen's tractrix. Nor did Revell ever read sci.math to learn that a borderline exists between finite and infinite. No, Revell tracks through the same mountain of silly ideas Cantor, Hilbert, Cohen, Godel. Revell has no new ideas on infinity, for Revell does not even understand what it means to be "finite". I think the problem is for Revell, he breathes too much dirty air in London and unable to think about infinity clearly and straightforward.

> In fact, the problem is how to properly generalize that approach to a number system: e.g. I'd like, with 'a' a constant (a parameter), y=ax and its inverse x=y/a to be defined for all a (and "inverses" *are* ubiquitous, up to and including category theory), which, as said, I am obtaining by letting k=1/a and keeping k around as a scale factor (a "unit"). Indeed, thinking in the direction of "numbers with units" seems interesting, but I am still not sure what that actually brings and how general it is: which is my line of research at this stage...
>
> (*) E.g. here I scale by gamma(beta) and keep the whole thing finite even when beta = 1, the speed of light:
> <https://jp-diegidio.github.io/STUDY.Physics.SpecialRelativity/InertialFrames/App/>
> (Pretty reminiscent of an actual "infinity boundary" in action, isn't it, Archimedes?)
>
> Your thoughts?
>
Well, Julio, I thank you for launching my 240th book of science for I knew that Calculus forced a Upper Bound on mathematics so that the only valid functions are Polynomials. The only functions that solves all derivatives and integrals with one Rule-- the power rule. So that is an upper.

But when you posted your question on how to justify y_2-y_1/ x_2-x_1 as the slope in Y = mx + b. Your question of how to avoid the 0/0 or even just k/0. How to avoid that Logical snafu. I saw that your question is the lower bound for why all functions to be valid are polynomial functions. Because the instant you admit the only Valid functions are polynomials, there never is a problem a snafu with slope being 0/0 or k/0. Once you admit all functions are polynomials, you never run into the question of slope as 0.

Thanks Julio for my book 240 is launched all because of your question in sci.math.

AP

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 by: Archimedes Plutonium - Mon, 25 Apr 2022 22:14 UTC

Math should be simple and easy. Only when math professors get in there, they seek fame and fortune and thus they look for "hard things" incomprehensible things to garner fame, yet leaves students in agony.

Math is easy, but given the centuries after Newton, math was made more and more crazy with every math professor wanting fame and leaving math a tangled mess of illogical methods.

We see that today in that in 1900, with Planck's Quantum of Quantum Mechanics means math had to be discrete, not continuous. Yet there are the bozo the clowns Cohen pursuing ever more continuity when physics found discreteness.

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 by: Archimedes Plutonium - Mon, 25 Apr 2022 22:23 UTC

On Monday, April 25, 2022 at 8:00:54 AM UTC-5, ju...@diegidio.name wrote:

>
> (*) E.g. here I scale by gamma(beta) and keep the whole thing finite even when beta = 1, the speed of light:
> <https://jp-diegidio.github.io/STUDY.Physics.SpecialRelativity/InertialFrames/App/>
> (Pretty reminiscent of an actual "infinity boundary" in action, isn't it, Archimedes?)
>
> Your thoughts?
>

Are you saying with your Gamma, you have a borderline between finite numbers and infinite numbers? It is not clear to me what your program is with Gamma and speed of light?

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 by: Archimedes Plutonium - Mon, 25 Apr 2022 22:37 UTC

On Monday, April 25, 2022 at 5:23:50 PM UTC-5, Archimedes Plutonium wrote:
> On Monday, April 25, 2022 at 8:00:54 AM UTC-5, ju...@diegidio.name wrote:
>
> >
> > (*) E.g. here I scale by gamma(beta) and keep the whole thing finite even when beta = 1, the speed of light:
> > <https://jp-diegidio.github.io/STUDY.Physics.SpecialRelativity/InertialFrames/App/>
> > (Pretty reminiscent of an actual "infinity boundary" in action, isn't it, Archimedes?)
> >
> > Your thoughts?
> >
> Are you saying with your Gamma, you have a borderline between finite numbers and infinite numbers? It is not clear to me what your program is with Gamma and speed of light?

Alright, so, if we take all of Special Relativity to mean-- it makes no difference if the coil in Faraday law is moving and magnet is stationary or vice versa. Plus the idea that speed of light is a maximum.

So the idea of speed of light a maximum must link up with the idea that there is a maximum finite number-- 1*10^604 where Huygen's tractrix area converges upon the associated circle area. Where pi has 3 zero digits in a row.

So, I have never tried to assemble the speed of light with the number 1*10^604. I may as well get started.

AP

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 by: Chris M. Thomasson - Mon, 25 Apr 2022 23:29 UTC

On 4/24/2022 5:12 PM, Chris M. Thomasson wrote:
> On 4/24/2022 7:22 AM, Julio Di Egidio wrote:
>> The slope of a 2D straight line parameterized on t is:
>>      m = (y(t) - y0) / (x(t) - x0)  (must be a constant)
>> but that expression fails (doesn't it?) for t = 0, because x(0) = x0
>> and y(0) = y0, so we get m = 0/0.
>>
>> In practice, if we are given expressions y(t) and x(t) and need to
>> find the corresponding slope, (as long as it is in fact a straight
>> line) we are guaranteed that the expression for the slope simplifies
>> to a constant, but, strictly speaking, our derivation is invalid for t
>> = 0.
>>
>> Do we just say/imply that "the discontinuity is eliminable at t=0" or
>> something of that sort (since m is defined and constant everywhere
>> except at t=0)?  Otherwise, how do we correctly find/express the slope
>> of a straight line given parametric equations, or state what we have
>> found?
>>
>> Thanks in advance for any help,
>
> Probably misunderstanding you here... Think in terms of two 2d vectors,
> and the angle between them. atan2(0, 0) is undefined when x _and_ y are
> both zero. So, taking the angle between points:
>
> vec2 p0 = {0, 0}
> vec2 p1 = {0, 0}
> vec2 dif = p1 - p0
> angle = atan2(dif.y, dif.x)
>
> angle is undefined. So you can get zero radians, or undefined. Depends
> on what the implementation decides to do.

One p0 is no longer equal to p1, everything works.

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