Proving Curvature Circle Theorem: c = \kappa=\frac{1}{r}, E(s)=C(s)+rN(s)

[americanforest]

Here is the problem:

Show that if $$c$$ is a curve with $$\kappa=\frac{1}{r}$$ (r is a positive constant) that $$c$$ is moving on a circle of radius r.

He gives a hunt to use the formula $$E(s)=C(s)+rN(s)$$. I don't know where he got this equations and I have no idea what the function E is supposed to represent. I'm sure C and S are position and arclength respectively. So first I showed that $$\frac{dE}{ds}=0$$ with the definitions of T and N vectors as related to curvature K.

Then he gives a hint to show $$absolute value(C-E)=r$$ which I have no idea how to show, and then from that to explain why that makes C a circle or radius r?

I know that the equation for a circle is nx^2+ny^2=r^2 but I don't see where that will get me here.

Any help?

I know this isn't in the correct format but this is more of a rigorous proof than a problem with given information...

 

[lippyka]

Solution:To show that c is a circle of radius r, we must first prove that the absolute value of (C - E) = r. We can do this using the equation E(s) = C(s) + rN(s), which states that the position vector E is equal to the position vector C plus the normal vector N multiplied by the radius r. Taking the absolute value of both sides yields |E(s)| = |C(s)| + |rN(s)|. Since the normal vector has a magnitude of 1, this simplifies to |E(s)| = |C(s)| + r. Then, subtracting |E(s)| from both sides gives |C(s)| = |E(s)| - r, which can be rewritten as |C(s) - E(s)| = r. Now that we have shown that |C(s) - E(s)| = r, we can use this result to show that c is a circle of radius r. Recall that the curvature of c is given by \kappa = \frac{1}{r}. This implies that the radius of curvature of c is r. By definition, the radius of curvature of a curve is the radius of the circle that best fits the curve at a given point. Therefore, since the radius of curvature of c is r, it follows that c is a circle of radius r.