---- > [!definition] Definition. ([[osculating plane]]) > Let $I$ be an [[open interval]] of $\mathbb{R}$ and $\alpha:I \to \mathbb{R}^{n}$ a [[smooth]] [[parameterized curve|curve]] [[parameterization by arc length|parameterized by arc length]], $\|\alpha'(s)\ \equiv 1$. When $\alpha'(s)$ and $\alpha''(s)$ are both nonzero, the [[velocity vector of a parameterized curve|unit tangent vector]] $t(s):=\alpha'(s)$ and [[unit normal vector to a parameterized curve|unit normal vector]] $n(s)$ are well-defined. Their [[submodule generated by a subset]] determines an [[plane]] in $\mathbb{R}^{n}$, called the **osculating plane at $s$**. > \ > ![[CleanShot 2024-01-15 at [email protected]|400]] > \ > If $\alpha'(s)=0$, we call $s$ a **singular point of order $0$**. $\alpha''(s)=0$, we call $s$ a **singular point of order $1$**. ^9250f4 > [!basicexample] > > Assuming $\alpha: I \to \mathbb{R}^{3}$ is a [[smooth]] curve [[parameterization by arc length|parameterized by arc length]] $s$, with [[curvature form]] $k(s) \neq 0$, for all $s \in I$. > Given $\alpha(s):= \big( a \cos \frac{s}{c}, a \sin \frac{s}{c}, b \frac{s}{c} \big), s \in \mathbb{R},$ where $c^{2}=a^{2} + b^{2}$. > **a.** Show that the parameter $s$ is the [[arc length of a path|arc length]]. > In light of [[parameterization by arc length|the unit-speed parameterizations are those by arc length]], it suffices to verify that $\|\alpha'(s)\|_{2} \equiv 1$. First compute $\alpha'(s)=T(s)=(-\frac{a}{c} \sin \frac{s}{c}, \frac{a}{c} \cos \frac{s}{c}, \frac{b}{c}),$ and then compute $\begin{align} \|\alpha'(s)\|_{2}= & \sqrt{ \left( -\frac{a}{c} \right)^{2} \sin ^{2}\left( \frac{s}{c} \right) + \left( \frac{a}{c} \right)^{2} \cos ^{2}\left( \frac{s}{c} \right)+ \left( \frac{b}{c} \right)^{2} } \\ = & \sqrt{ \frac{a^{2}}{c_{}^{2}} + \frac{b^{2}}{c^{2}} } \\ = & \frac{c_{2}}{c_{2}} \\ = & 1, \end{align}$ where we have used that $a^{2} + b^{2} = c^{2}$. > **b.** The [[curvature of parameterized curve|curvature]] is given by $\begin{align} \|\alpha''(s)\|_{2} = & \|\left( -\frac{a}{c^{2}} \cos \frac{s}{c} , -\frac{a}{c^{2}} \sin \frac{s}{c}, 0\right)\|_{2} \\ = & \sqrt{ \frac{a^{2}}{c^{4}} } \\ = & \frac{a}{c^{2}} . \end{align}$ Notice that this curvature matches that of *some* circle. So, it doesn't characterize the curve completely - for that we need torsion as well. > The [[torsion of a parameterized curve|torsion]] is given implicitly via $\begin{align} b'(t) = \tau(s) n (s) , \end{align}$ i.e., $\frac{d}{dt}\left( \alpha'(s) \times \frac{\alpha''(s)}{k(s)} \right)=\tau(s) \frac{\alpha''(s)}{k(s)}.$ We have computed $\alpha', \alpha''$ and $k$ already. We have $\frac{d}{dt}\big( (-\frac{a}{c} \sin \frac{s}{c}, \frac{a}{c} \cos \frac{s}{c}, \frac{b}{c}) \times c \left( -\frac{a}{c^{2}} \cos \frac{s}{c} , -\frac{a}{c^{2}} \sin \frac{s}{c}, 0\right) \big)=\tau(s) \ c \alpha''(s)$ > > Begin by computing $\begin{align}\alpha''(t) \times \alpha'(t) = & \det \begin{bmatrix}e_{1} & e_{2} & e_{3} \\ -\frac{a}{c^{2}} \cos \left( \frac{s}{c} \right) & -\frac{a}{c^{2}} \sin \left( \frac{s}{c} \right) & 0 \\ -\frac{a}{c} \sin \left( \frac{s}{c} \right) & \frac{a}{c} \cos \left( \frac{s}{c} \right) & \frac{b}{c}\end{bmatrix} \\ = & \begin{bmatrix} -\frac{ab}{c^{3}}\sin\left( \frac{s}{c} \right) \\ \frac{ab}{c^{3}} \cos \left( \frac{s}{c} \right) \\ -\frac{a^{2}}{c^{3}} \cos ^{2}\left( \frac{s}{c} \right) - \frac{a^{2}}{c^{3}} \sin ^{2} \left( \frac{s}{c} \right) \end{bmatrix} \\ = & \big( -\frac{ab}{c^{3}}\sin\left( \frac{s}{c} \right) , \frac{ab}{c^{3}} \cos \left( \frac{s}{c} \right), -\frac{a^{2}}{c^{3}}\big). \end{align}$ Then compute $\begin{align} \|\alpha''(t) \times \alpha'(t)\|^{2} = & \frac{a^{2}b^{2}}{c^{6}} \sin ^{2} \left( \frac{s}{c} \right)+ \frac{a^{2}b^{2}}{c^{3}} \cos ^{2} \left( \frac{s}{c} \right) + \frac{a^{4}}{c^{6}} \\ = & \frac{a^{2}b^{2}}{c^{6}} + \frac{a^{4}}{c^{6}}. \end{align}$ Next compute $\begin{align} \alpha'''(s) = & \big( \frac{a}{c^{3}} \sin (\frac{s}{c}), -\frac{a}{c^{3}}\cos (\frac{s}{c}) , \ 0 \big). \end{align}$ Recalling the [[torsion of a parameterized curve|torsion characterization]], we have $\begin{align} \tau(t)= & \frac{\alpha'''(t) \cdot \big( \alpha''(t) \times \alpha'(t) \big)}{\|\alpha''(t) \times \alpha'(t)\| ^{2}} \\ = & \frac{\big( \frac{a}{c^{3}} \sin (\frac{s}{c}), -\frac{a}{c^{3}}\cos (\frac{s}{c}) , \ 0 \big) \cdot \big( -\frac{ab}{c^{3}}\sin\left( \frac{s}{c} \right) , \frac{ab}{c^{3}} \cos \left( \frac{s}{c} \right), -\frac{a^{2}}{c^{3}}\big) }{\frac{a^{2}b^{2}}{c^{6}} + \frac{a^{4}}{c^{6}}} \\ = & \frac{-\frac{a^{2}b}{c^{6}} \sin ^{2}(\frac{s}{c}) - \frac{a^{2}b}{c^{6}} \cos ^{2}(\frac{s}{c})}{\frac{a^{2}b^{2}+a^{4}}{c^{6}}} \\ = & \frac{-\frac{a^{2}b}{c^{6}}}{\frac{a^{2}b^{2} + a^{4}}{c^{6}}} \\ = & \frac{-a^{2}b}{a^{2}b^{2} + a ^{4}}? \end{align}$ **c. Determine the [[osculating plane]] of $\alpha$.** > The [[osculating plane]] of $\alpha$ at $s$ is the [[submodule generated by a subset]] of $T(s)=:T_{s}$ and $N(s)=:N_{s}$: $\begin{align} \langle \big( T_{s}, N_{s} \big)\rangle = & \{ \beta T_{s} + \gamma N_{s} : \beta,\gamma \in \mathbb{R}\} \\ = & \{ \beta (-\frac{a}{c} \sin \frac{s}{c}, \frac{a}{c} \cos \frac{s}{c}, \frac{b}{c}) + \gamma( -\frac{a}{c^{2}} \cos \frac{s}{c} , -\frac{a}{c^{2}} \sin \frac{s}{c}, 0) : \beta, \gamma \in \mathbb{R} \}. \end{align}$ **d. Show that the lines containing $N(s)$ and passing through $\alpha(s)$ meet the $z$-axis under a constant angle equal to $\frac{\pi}{2}$.** > Such a [[line]] has the form $\alpha(s) + N_{s} t, \ t \in \mathbb{R}$. The line that is the $z$-axis has the form $(0,0,z), z \in \mathbb{R}$. > We need the direction vector of the line, $N_{s}$, to be orthogonal to the $z$-axis, i.e., that $(N_{s}t ) \cdot (0,0 ,z)=0, \text{ for all } t \text{ and } z.$ But this is immediate, because $N_{3}(s)$ equals $0$ as a quantity proportional to $a''(s)$. > **e. Show that the tangent lines to $\alpha$ make a constant angle with the $z$-axis.** > Because $\alpha$ is unit-speed, all of its tangent lines have length $1$. As does the vector $(0,0,1)$, which we choose to represent the $z$-axis. It is therefore sufficient to show that the [[dot product]] $T(s) \cdot (0,0,1)=T_{3}(s)$ is constant. In earlier parts we showed $T_{3}(s)=\alpha_{3}'(s)=\frac{b}{c}$ which indeed is constant. ^46a289 ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```