• wholookshereEnglish
          3·
          2 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@lemmy.mlEnglish
            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.

            • wholookshereEnglish
              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.mlEnglish
                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.

                • wholookshereEnglish
                  1·
                  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.mlEnglish
                    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.