• ltxrtquq@lemmy.ml
    link
    fedilink
    English
    arrow-up
    4
    ·
    2 months ago

    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
      link
      fedilink
      English
      arrow-up
      1
      ·
      2 months ago

      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
        link
        fedilink
        English
        arrow-up
        4
        ·
        2 months ago

        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
          link
          fedilink
          English
          arrow-up
          1
          ·
          edit-2
          2 months ago

          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
            link
            fedilink
            English
            arrow-up
            1
            ·
            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

            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
              link
              fedilink
              English
              arrow-up
              1
              ·
              edit-2
              2 months ago

              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
                link
                fedilink
                English
                arrow-up
                1
                ·
                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
                  link
                  fedilink
                  English
                  arrow-up
                  1
                  arrow-down
                  1
                  ·
                  edit-2
                  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
                    link
                    fedilink
                    English
                    arrow-up
                    1
                    ·
                    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?