• wholookshere@lemmy.blahaj.zone
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        1 month 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@lemmy.ml
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          1 month 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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            1 month 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
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              1 month 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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                1 month 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
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                  1 month 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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                    1 month 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.