Question

I was playing around with derivatives and equation of a circle and found that the derivative of a circle with no radius is equal to i. Why would it be that the slope at any point around a circle with no radius would be the square root of -1?

Answer 1

like this

Answer 2

just trying to follow your logic

Answer 3

I'm assuming you're integrating the circle of radius 0 centered at the origin. In that case, you want to do implicit differentiation. You get \[2x+2y\frac{dy}{dx}=0\]Thus, \[\frac{dy}{dx}=-\frac{x}{y}\]At the point (0, 0), you're dividing by 0, so the derivative would actually be undefined. That's why you get weird things like \(i\) doing differentiation normally.

Answer 4

A circle with R=0 is a point. So it is not really a function anymore therefore its derivative would be meaningless, at least in real numbers.
KingGeorge si right.

Answer 5

@KingGeorge
what if we use another center and not the orgin?

Answer 6

\[(x-a)^2+(y-b)^2=0\]\[2(x-a)+2(y-b)\frac{dy}{dx}=0\]\[\frac{dy}{dx}=-\frac{x-a}{y-b}\]At the point (a, b) which is our circle, we get \[\frac{0}{0}\]