[b][size=150]Curvature of a curve[/size][/b][br][br]Suppose a [b]smooth oriented curve[/b] in [math]\mathbb{R}^3[/math] is parametrized by a smooth vector-valued function [math]\vec{r}(t)=\langle f(t),g(t),h(t)\rangle[/math] with [math]a\leq t \leq b[/math] and a positive orientation. Then the [b]unit tangent vector[/b] [math]\vec{T}(t)[/math] is defined as follows:[br][br][math]\vec{T}(t)=\frac{\vec{r'}(t)}{|\vec{r'}(t)|}=\frac{\vec{v}(t)}{|\vec{v}(t)|}[/math][br][br]where [math]\vec{v}(t)=\vec{r'}(t)[/math] is the tangent vector of the curve at the point with position vector [math]\vec{r}(t)[/math].[br][br]We want to define a quantity that describes how "sharp" the bending of the curve is, or more precisely, how much the curve direction changes over a small distance along the curve. Moreover, we would like to define this quantity such that it is independent of the parametrization. Hence, we consider the reparametrization by arc length and define this quantity, called [b]curvature[/b], as follows:[br][br][math]\kappa(s)=\left|\frac{d\vec{T}}{ds}\right|[/math][br][br]where [math]s=s(t)[/math] is the arc length reparametrization i.e. [math]s=\int_a^t |\vec{r'}(u)| \ du[/math]. In other words, it is the magnitude of the rate of change of [math]\vec{T}[/math] with respect to arc length. Moreover, we can rewrite the definition as follows:[br][br][math]\kappa=\left|\frac{d\vec{T}}{ds}\right|=\left|\frac{\frac{d\vec{T}}{dt}}{\frac{ds}{dt}}\right|=\frac 1{|\vec{v}|}\left|\frac{d\vec{T}}{dt}\right|[/math] (because by FTC, [math]\frac{ds}{dt}=|\vec{r'}|=|\vec{v}|[/math])[br][br]Therefore, [math]\kappa(t)=\frac{|\vec{T'}(t)|}{|\vec{r'}(t)|}[/math] i.e. the formula for curvature in terms of the given parametrization.[br][br]

[u]Example[/u]:[br][br]Consider a line parametrized by [math]\vec{r}(t)=\vec{r}_0+t\vec{v}=\langle x_0+at,y_0+bt,z_0+ct\rangle[/math]. Then [math]\vec{v}(t)=\vec{r'}(t)=\langle a,b,c \rangle[/math] i.e. a constant vector. Hence[br][br][math]\vec{T}(t)=\frac{\langle a,b,c \rangle}{\sqrt{a^2+b^2+c^2}}[/math] and [math]\vec{T'}(t)=\vec{0}[/math],[br][br]which implies that [math]\kappa=0[/math] i.e. zero curvature anywhere on the line.[br][br]

[u]Example[/u]:[br][br]Consider a circle on the plane [math]z=1[/math], centered at [math](0,0,1)[/math] with radius [math]R[/math]. It can be parametrized by [math]\vec{r}(t)=\langle R\cos t, R\sin t, 1\rangle[/math] with [math]0\leq t \leq 2\pi[/math]. Then [math]\vec{v}(t)=\vec{r'}(t)=\langle -R\sin t, R\cos t, 0\rangle[/math]. We have[br][br][math]\vec{T}(t)=\frac 1{R}\langle -R\sin t, R\cos t,0\rangle=\langle -\sin t, \cos t,0\rangle[/math],[br][br]which implies that [math]\vec{T'}(t)=\langle -\cos t, -\sin t, 0\rangle[/math].[br][br]Therefore, [math]\kappa=\frac{|\vec{T'}(t)|}{|\vec{v}(t)|}=\frac 1R[/math] i.e. constant positive curvature on the circle.[br][br][br][u]An alternative curvature formula[/u][br][br]Suppose [math]\vec{a}(t)=\vec{v'}(t)[/math]. We have[br][br][math]\vec{a}=\frac{d}{dt}\vec{v}=\frac{d}{dt}(|\vec{v}|\vec{T})=\left(\frac{d}{dt}|\vec{v}|\right)\vec{T}+|\vec{v}|\frac{d\vec{T}}{dt}[/math][br][br][math]\vec{v}\times\vec{a}=(|\vec{v}|\vec{T})\times\left[\left(\frac{d}{dt}|\vec{v}|\right)\vec{T}+|\vec{v}|\frac{d\vec{T}}{dt}\right]=(|\vec{v}|\vec{T})\times\left(\frac{d}{dt}|\vec{v}|\right)\vec{T}+(|\vec{v}|\vec{T})\times|\vec{v}|\frac{d\vec{T}}{dt}[/math][br][math]=|\vec{v}|^2\vec{T}\times \frac{d\vec{T}}{dt}[/math][br][br]Therefore, we have[br][br][math]|\vec{v}\times\vec{a}|=|\vec{v}|^2\left|\vec{T}\times\frac{d\vec{T}}{dt}\right|=|\vec{v}|^2\left|\vec{T}\right| \left|\frac{d\vec{T}}{dt}\right|\sin \theta[/math][br][br]where [math]\theta[/math] is the angle between [math]\vec{T}[/math] and [math]\frac{d\vec{T}}{dt}[/math]. Since [math]|\vec{T}|=1[/math], by previous exercise, [math]\vec{T}[/math] and [math]\frac{d\vec{T}}{dt}[/math] are orthogonal i.e. [math]\theta=\frac{\pi}2[/math]. Hence [math]|\vec{v}\times\vec{a}|=|\vec{v}|^2\left|\frac{d\vec{T}}{dt}\right|[/math] and we have[br][br][math]\kappa=\frac 1{|\vec{v}|}\left|\frac{d\vec{T}}{dt}\right|=\frac{|\vec{v}\times\vec{a}|}{|\vec{v}|^3}[/math][br][br]

[b][size=150]Principal unit normal vector[/size][/b][br][br]The unit vector in the direction of [math]\frac{d\vec{T}}{ds}[/math] is the [b]principal unit normal vector[/b], denoted by [math]\vec{N}[/math]. Therefore, we have [math]\frac{d\vec{T}}{ds}=\kappa \vec{N}[/math]. Moreover, [math]|\vec{T}|=1[/math] implies that [math]\vec{T}[/math] and [math]\frac{d\vec{T}}{ds}[/math] are orthogonal. Hence [math]\vec{T}[/math] and [math]\vec{N}[/math] are orthogonal. [br][br][u]Note[/u]: In practice, we can find the principal unit normal vector by computing the unit vector in the direction of [math]\frac{d\vec{T}}{dt}[/math] because both [math]\frac{d\vec{T}}{ds}[/math] and [math]\frac{d\vec{T}}{dt}[/math] are pointing towards the same direction.[br][br]The plane containing [math]\vec{T}[/math] and [math]\vec{N}[/math] at a point [math]P[/math] on the curve is called the [b]osculating plane[/b] at [math]P[/math]. And the circle contained in the osculating plane with center [math]C[/math] such that [math]\overrightarrow{PC}[/math] is in the same direction as [math]\vec{N}[/math] and radius [math]\frac 1{\kappa}[/math] is called the [b]osculating circle[/b] at [math]P[/math]. It is the circle that best approximates the curve at [math]P[/math].[br][br]In the applet below, you can see how [math]\vec{T}, \vec{N}[/math], osculating plane, osculating circle, and curvature change when you drag the slider to move the point [math]P[/math] along the curve.[br][br][br]