is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.



Simple Definition of Curvature

Before we start formally defining curvature, we must at least have a basic idea of what it should be. So curvature is how curvy something is. What I mean is, if something doesn't curve at all, like y = x, the curvature is 0. If something is extremely curvy, then it's curvature is really high. Now, I know that's very hand-wavy, so let's get to some real math.


So, we need to define a few things. 

MathJax TeX Test Page The unit tangent vector is defined as: $$T(t) = \dfrac{r'(t)}{|r'(t)|}$$ Because $|T(t)|$ is constant (it's always 1), by a theorem we proved before (which isn't on this site, sorry), $T(t)$ is perpendicular to $T'(t)$. So, now we have our unit normal vector. The unit normal vector is defined as: $$N(t) = \dfrac{T'(t)}{|T'(t)|}$$ This is always perpendicular to the normal vector.

Another way

MathJax TeX Test Page Sometimes, it's not good to differentiate with t, after all we don't want to define curvature based on time. If something goes across a curve at 2t speed, that would change the curvature. So, let's define it based on arclength, denoted s. $$s = \int_0^{s}\mathrm{\sqrt{f'(t)^2 + g'(t)^2}}\, \mathrm{d}t$$ If we take the derivative with respect to s on both sides, we get $$1 = \sqrt{(f'(s))^2 + (g'(s))^2}$$ Note that arclength also equals the magnitude of r'(s), so $$|r'(s)| = 1$$ That makes it equivalent to the unit tangent vector, so we call it $T(s)$.


MathJax TeX Test Page Let $\theta$ equal the angle between the unit tangent vector and $\hat{i}$. So we can define curvature, denoted k as $$K = |\dfrac{d\theta}{ds}|$$ Now, we can use this to derive another formula for curvature. If C is the graph y = f(x), then the curvature is $$K = \dfrac{|y''|}{(1 + (y')^2)^\frac{3}{2}}$$ Let's say we have x = f(t), and y = g(t). Curvature then equals $$K = \dfrac{|f'(t)g''(t) - g'(t)f''(t)|}{(f'(t)^2 + g'(t)^2)^{3/2}}$$ You can prove that by dividing $\frac{d\theta}{dt}$ over $\frac{ds}{dt}$

3D Curvature Formula

MathJax TeX Test Page If x = f(t), y = g(t), z = h(t), $$K = \dfrac{|r'(t) \times r''(t)|}{|r'(t)|^3}$$ Note: this is similar to the 2D formula. The denominator is equal to the $\sqrt{f'(t)^2 + g'(t)^2}^3$, which is equal to the denominator of the 2D curvature formula. The numerator is just like the 2D version, just 3D.
David Witten
Tangent and Normal Components of Acceleration

Tangent and Normal Components of Acceleration

Tangent Spaces