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.
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
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.
