• wholookshere@lemmy.blahaj.zone
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    1 day 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 day 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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        18 hours 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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          17 hours 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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            2 hours 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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              2 hours 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.