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.
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.
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.
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”
