# What is curvature? (multivariable calculus)

if space curve C=<f(t),g(t),h(t)>, and

v=$\frac{dC}{dt}$=<$\frac{df(t)}{dt}$,$\frac{dg(t)}{dt}$,$\frac{dh(t)}{dt}$>

Why is curvature defined this way? κ$\equiv$$\frac{d\widehat{T}}{dS}$

$\hat{T}$=unit tangent vector
S=arc length

to elaborate, for a space curve, i understand what $\frac{dT}{dt}$ is, but what is $\frac{d\widehat{T}}{dS}$? please explain this to me in an intuitive way, as in what it graphically represents.

wiki says that the "cauchy defined the curvature C as the intersection point of two infinitely close normals to the curve"

okay so how is this any different from $\frac{dT}{dt}$?

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lurflurf
Homework Helper
Is this what you mean?
$$\kappa = \left\|\frac{d\mathbf{T}}{ds}\right\|$$

So at a point we fit the curve (locally) to a circle. The curvature is the reciprocal of the radius of that circle. We think of a small circle as very curved so it has high curvature. A large circle is slightly curved so it has low curvature.

Is this what you mean?
$$\kappa = \left\|\frac{d\mathbf{T}}{ds}\right\|$$

So at a point we fit the curve (locally) to a circle. The curvature is the reciprocal of the radius of that circle. We think of a small circle as very curved so it has high curvature. A large circle is slightly curved so it has low curvature.

alright, i have no problem with the curvature being defined that way, but could you show me how $$\kappa = \left\|\frac{d\mathbf{T}}{ds}\right\|$$ means that? ie.. could you show me how $$\kappa = \left\|\frac{d\mathbf{T}}{ds}\right\|$$ directly leads 1/r?

WannabeNewton