this post was submitted on 21 Sep 2024
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Square! (lemmy.zip)
submitted 2 months ago by Maven@lemmy.zip to c/science_memes@mander.xyz
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[–] ltxrtquq@lemmy.ml 4 points 2 months ago (1 children)

A straight line in polar coordinates with the same tangent would be a circle.

I'm not sure that's true. In non-euclidean geometry it might be, but aren't polar coordinates just an alternative way of expressing cartesian?

Looking at a libre textbook, it seems to be showing that a tangent line in polar coordinates is still a straight line, not a circle.

[–] wholookshere@lemmy.blahaj.zone 1 points 2 months ago (1 children)

I’m saying that the tangent of a straight line in Cartesian coordinates, projected into polar, does not have constant tangent. A line with a constant tangent in polar, would look like a circle in Cartesian.

[–] ltxrtquq@lemmy.ml 4 points 2 months ago (1 children)

Polar Functions and dydx

We are interested in the lines tangent a given graph, regardless of whether that graph is produced by rectangular, parametric, or polar equations. In each of these contexts, the slope of the tangent line is dydx. Given r=f(θ), we are generally not concerned with r′=f′(θ); that describes how fast r changes with respect to θ. Instead, we will use x=f(θ)cosθ, y=f(θ)sinθ to compute dydx.

From the link above. I really don't understand why you seem to think a tangent line in polar coordinates would be a circle.

[–] wholookshere@lemmy.blahaj.zone 1 points 2 months ago* (last edited 2 months ago) (1 children)

Sorry that’s not what I’m saying.

I’m saying a line with constant tangent would be a circle not a line.

Let me try another way, a function with constant first derivative in polar coordinates, would draw a circle in Cartesian

[–] ltxrtquq@lemmy.ml 1 points 2 months ago (1 children)

Given r=f(θ), we are generally not concerned with r′=f′(θ); that describes how fast r changes with respect to θ

I think this part from the textbook describes what you're talking about

Instead, we will use x=f(θ)cosθ, y=f(θ)sinθ to compute dydx.

And this would give you the actual tangent line, or at least the slope of that line.

[–] wholookshere@lemmy.blahaj.zone 1 points 2 months ago* (last edited 2 months ago) (1 children)

But then your definition of a straight line produces two different shapes.

Starting with the same definition of straight for both. Y(x) such that y’(x) = C produces a function of cx+b.

This produces a line

However if we have the radius r as a function of a (sorry I’m on my phone and don’t have a Greek keyboard).

R(a) such that r’(a)=C produces ra +d

However that produces a circle, not a line.

So your definition of straight isn’t true in general.

[–] ltxrtquq@lemmy.ml 1 points 2 months ago (1 children)

I think we fundamentally don't agree on what "tangent" means. You can use

x=f(θ)cosθ, y=f(θ)sinθ to compute dydx

as taken from the textbook, giving you a tangent line in the terms used in polar coordinates. I think your line of reasoning would lead to r=1 in polar coordinates being a line, even though it's a circle with radius 1.

[–] wholookshere@lemmy.blahaj.zone 0 points 2 months ago* (last edited 2 months ago) (1 children)

Except here you said here

https://lemmy.ml/comment/13839553

That they all must be equal.

Tangents all be equal to the point would be exponential I thinks. So I assume you mean they must all be equal.

Granted I assumed constant, because that’s what actually produces a “straight” line. If it’s not, then cos/sin also fall out as “straight line”.

So I’ve either stretched your definition of straight line to include a circle, or we’re stretching “straight line”

[–] ltxrtquq@lemmy.ml 1 points 2 months ago (1 children)

Given r=f(θ), we are generally not concerned with r′=f′(θ); that describes how fast r changes with respect to θ

You're using the derivative of a polar equation as the basis for what a tangent line is. But as the textbook explains, that doesn't give you a tangent line or describe the slope at that point. I never bothered defining what "tangent" means, but since this seems so important to you why don't you try coming up with a reasonable definition?

[–] wholookshere@lemmy.blahaj.zone 0 points 2 months ago* (last edited 2 months ago)

My whole point is that a “straight done”, in general, doesn’t exist in the first place. Because in general definitions are actually really hard.

It’s not that it’s important to me. It’s that I’ve spent many parts of my day on the phone with the bank, and never should be taken for more than an asshole on the internet. Sorry if you thought I was more invested than that.