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CLP-4 Vector Calculus

Section 1.5 A Compendium of Curve Formula

In the following \(\vr(t)=\big(x(t)\,,\,y(t)\,,\,z(t)\big)\) is a parametrization of some curve. The vectors \(\hat\vT(t),\ \hat\vN(t),\ \) and \(\ \hat\vB(t)\ \) are the unit tangent, normal and binormal vectors, respectively, at \(\vr(t)\text{.}\) The tangent vector points in the direction of travel (i.e. direction of increasing \(t\)) and the normal vector points toward the centre of curvature. The arc length from time \(0\) to time \(t\) is denoted \(s(t)\text{.}\) The binormal \(\ \hat\vB(t)=\hat\vT(t)\times \hat\vN(t)\ \) is perpendicular to the plane that fits the curve best at \(\vr(t)\text{.}\) Some formulae use an arc length parametrization, which is denoted \(\vr(s)\text{.}\)
the velocity
\(\displaystyle \vv(t)=\diff{\vr}{t}(t)=\diff{s}{t}(t)\,\hat\vT(t)\)
the unit tangent vector
\(\hat\vT(t)=\frac{\vv(t)}{|\vv(t)|}\) (general parametrization)
\(\hat\vT(s)=\diff{\vr}{s}(s)\) (arc length parametrization)
the acceleration
\(\displaystyle \va(t)=\difftwo{\vr}{t}(t)=\difftwo{s}{t}(t)\,\hat\vT(t) +\ka(t)\big(\diff{s}{t}(t)\big)^2\hat\vN(t)\)
the speed
\(\displaystyle \diff{s}{t}(t) = |\vv(t)| = \big|\diff{\vr}{t}(t)\big|\)
the arc length
\(\displaystyle s(T) = \int_0^T\! \diff{s}{t}(t)\,\dee{t} = \int_0^T\! \sqrt{x'(t)^2\!+\!y'(t)^2\!+\!z'(t)^2}\,\dee{t}\)
the curvature
\(\ka(t) = \big|\diff{\hat\vT}{t}(t)\big|/\diff{s}{t}(t) =\displaystyle{ \frac{|\vv(t)\times\va(t)|}{(\diff{s}{t}(t))^3} }\)
\(\ka(s) = \big|\diff{\phi}{s}(s)\big| = \big|\diff{\hat\vT}{s}(s)\big|\)
the unit normal vector
\(\displaystyle \hat\vN(t) = \diff{\hat\vT}{t}(t)/\big|\diff{\hat\vT}{t}(t)\big| \qquad \hat\vN(s) = \diff{\hat\vT}{s}(s)/\ka(s)\)
the radius of curvature
\(\displaystyle \rho(t)=\frac{1}{\ka(t)}\)
the centre of curvature
\(\displaystyle \vr(t)+\rho(t)\hat\vN(t)\)
the torsion
\(\displaystyle \displaystyle \tau(t)=\frac{\big(\vv(t)\times\va(t)\big) \cdot \diff{\va}{t}(t)} {|\vv(t)\times\va(t)|^2}\)
the binormal
\(\displaystyle \displaystyle \hat\vB(t)=\hat\vT(t)\times \hat\vN(t)=\frac{\vv(t)\times\va(t)}{|\vv(t)\times\va(t)|}\)
Under arclength parametrization (i.e. if \(t=s\)) we have \(\hat\vT(s)=\frac{d\vr}{ds}(s)\) and the Frenet-Serret formulae
\begin{align*} \diff{\hat\vT}{s}(s)&=\phantom{-}\ka(s)\ \hat\vN(s)\cr \diff{\hat\vN}{s}(s)&=\phantom{-}\tau(s)\ \hat\vB(s)-\ka(s)\ \hat\vT(s)\cr \diff{\hat\vB}{s}(s)&=-\tau(s)\ \hat\vN(s)\cr \end{align*}
which in matrix form is
\begin{align*} \diff{}{s} \left[ \begin{matrix}\hat\vT(s) \\ \hat\vN(s)\\ \hat\vB(s)\end{matrix} \right] =\left[\begin{matrix} 0 & \ka(s) & 0 \\ -\ka(s) & 0 &\tau(s) \\ 0 &-\tau(s) & 0 \end{matrix}\right] \left[\begin{matrix}\hat\vT(s) \\ \hat\vN(s)\\ \hat\vB(s)\end{matrix}\right] \end{align*}
When the curve lies entirely in the \(xy\)-plane the curvature is given by
\begin{gather*} \ka(t) =\frac{\big| \diff{x}{t}(t)\ \difftwo{y}{t}(t)-\diff{y}{t}(t)\ \difftwo{x}{t}(t) \big|}{\Big[\big(\diff{x}{t}(t)\big)^2 +\big(\diff{y}{t}(t)\big)^2\Big]^{3/2}} \end{gather*}
When the curve lies entirely in the \(xy\)-plane and the \(y\)-coordinate is given as a function, \(y(x)\text{,}\) of the \(x\)-coordinate, the curvature is
\begin{gather*} \ka(x) =\frac{\big|\difftwo{y}{x}(x)\big|} {\Big[1+\big(\diff{y}{x}(x)\big)^2\Big]^{3/2}} \end{gather*}
Notice that this follows from the previous formula since \(\diff{x}{x}=1\) and \(\difftwo{x}{x}=0\text{.}\)