• ltxrtquq@lemmy.ml
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    2 months ago

    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
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      2 months ago

      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
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        2 months ago

        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
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          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.