# Thread: Find K and R = 1/K

1. ## Find K and R = 1/K

Find the curvature and radius of curvature.

y = 2x^2 + 3, x = -1

r (t) = < t, 2t^2 + 3, 0 >

r'(t) = < 1, 4t, 0 >

The magnitude of r'(t) = root {16t^2 + 1}.

Is this correct so far?

2. ## Re: Find K and R = 1/K

Wky did you title your post "Find K and R = 1/K" ?

3. ## Re: Find K and R = 1/K

Originally Posted by DenisB
Wky did you title your post "Find K and R = 1/K" ?
I wanted the title to stand out. Am I right so far?

4. ## Re: Find K and R = 1/K

Am I correct here?

yes

6. ## Re: Find K and R = 1/K

I suspect that the "K" in the title was supposed to be "kappa", $\kappa$, the curvature, and R the radius of curvature. The first, and easiest, step of one method of finding those would be to find the length of the tangent vector.

7. ## Re: Find K and R = 1/K

I will continue with this problem today. If I get stuck or cannot finish, I will post that as well.

8. ## Re: Find K and R = 1/K

Originally Posted by romsek
yes
Let's see how far I can go with this problem.

9. ## Re: Find K and R = 1/K

Originally Posted by HallsofIvy
I suspect that the "K" in the title was supposed to be "kappa", $\kappa$, the curvature, and R the radius of curvature. The first, and easiest, step of one method of finding those would be to find the length of the tangent vector.
Yes, K = curvature and R = radius of curvature.

10. ## Re: Find K and R = 1/K

If a curve is given by a vector function, $\vec{r}(t)= x(t)\vec{i}+ y(t)\vec{j}+ z(t)\vec{k}$, then the tangent vector at any point is the derivative: $\vec{T}(t)= x'(t)\vec{i}+ y'(t)\vec{j}+ z'(t)\vec{k}$. If, in particular, the parameter, t, is the arc-length, s, then that tangent vector will have length 1 and is the "unit tangent vector". If t is not the arc-length then we still have relation $s= \int |T|dt$.

With arc-length as parameter, so that the tangent vector has unit length, the derivative of the tangent vector is perpendicular to the tangent vector (Since $|\vec{T|}= \sqrt{\vec{T}\cdot \vec{T}}= 1$, $\vec{T}\cdot \vec{T}= 1$, a constant, so $\frac{d\vec{T}\cdot \vec{T}}{ds}= 2\vec{T}\cdot\frac{d\vec{T}}{ds}= 0$) and is called the "normal vector" to the curve at that point. The "curvature" is the length of that normal vector. The "radius of curvature" is 1 over the curvature.

Where t is not the arc-length, we have, by the chain rule, $\frac{d\vec{T}}{ds}= \frac{d\vec{T}}{dt}\frac{dt}{ds}$. And, since $s= \int \left| \frac{d\vec{T}}{dt}\right| dt$, $\frac{ds}{dt}= |\vec{T}|$, $\frac{d\vec{T}}{ds}= \frac{d\vec{T}}{dt}\frac{ds}{dt}$.

Cool.