• wholookshere@lemmy.blahaj.zone
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    3 months ago

    Only true in Cartesian coordinates.

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

    EDIT: it is still a “straight” line. But then the result of a square on a surface is not the same shape any more.

    • ltxrtquq
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      3 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
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        3 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
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          3 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
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            3 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
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              3 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
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                3 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
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                  3 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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                    3 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”