Curvature

Intuitively, curvature measures how "curved" a curve is. Suppose that \(\gamma(t)\) is a pamameterized curve. As \(t\) changes to \(t+\Delta t\), the curve moves away from its tangent \(\gamma'(t)\) by

\[(\gamma(t+\Delta t) - \gamma(t))\cdot \hat{\mathbf n}, \|\hat{\mathbf n}\| = 1, \hat{\mathbf n}\cdot \gamma' = 0\]

Then, Taylor's theorem gives

\[\gamma(t+\Delta t) = \gamma(t) + \gamma'(t)\Delta t + \frac{1}{2}\gamma''(t)\Delta t^2 + rem\]

By Taylor's remainder theorem, \(rem\) vanishes faster than \(\Delta t^2\).

Therefore, we have

\[(\gamma(t) + \gamma'(t)\Delta t + \frac{\Delta t^2}{2}\gamma''(t)-\gamma(t))\cdot\hat{\mathbf n} = \frac{\Delta t^2}{2}\gamma''(t)\cdot\hat{\mathbf n} \]

Then, note that \(\gamma'\cdot \gamma'' = 0\) and known that \(\gamma\) is unit-speed, hence \(\gamma'' \parallel \hat{\mathbf n}\). So that

\[ \frac{\Delta t^2}{2}\gamma''(t)\cdot\hat{\mathbf n} = \frac{\Delta t^2}{2}\|\gamma''(t)\| \]

Therefore, we can derive the definition:

If \(\gamma\) is a unit-speed curve, its curvature \(\kappa(t)\) at the point \(\gamma(t)\) is defined as \(\|\gamma''(t)\|\).

Cross Product

For 3D vectors \(\mathbf a, \mathbf b \in \mathbb R^3\), cross product is a 3D vector s.t. \(\mathbf a\times \mathbf b \in \mathbb R^3\) s.t. \(\forall w\in\mathbb R^3. (\mathbf a\times \mathbf b)\cdot \mathbf w = \det(\begin{bmatrix}\mathbf a&\mathbf b&\mathbf w\end{bmatrix})\).

Note that \(\det(A) = \det(A^T)\), and for \(3\times3\) matrices, switching two adjacent rows will flip the sign, then switch twice won't change the determinant.

\[\det(\begin{bmatrix}\mathbf a&\mathbf b&\mathbf w\end{bmatrix}) = \det(\begin{bmatrix}a_1&b_1&w_1\\a_2&b_2&w_2\\a_3&b_3&w_3\end{bmatrix}) = \det(\begin{bmatrix}w_1&w_2&w_1\\a_1&a_2&a_3\\b_1&b_2&b_3\end{bmatrix})\]

Therefore, we can rewrite the cross product as

\[\begin{align*} \mathbf a\times \mathbf b &= \begin{bmatrix} \det(\begin{bmatrix}a_2&a_3\\b_2&b_3\end{bmatrix})\\ -\det(\begin{bmatrix}a_1&a_3\\b_1&b_3\end{bmatrix})\\ \det(\begin{bmatrix}a_1&a_2\\b_1&b_2\end{bmatrix}) \end{bmatrix} = \begin{bmatrix} a_2b_3-a_3b_2\\ a_3b_1-a_1b_3\\ a_1a_2-a_2b_1 \end{bmatrix} \\ &= (a_2b_3-a_3b_2)\mathbf e_1 - (a_1b_3-a_3b_1)\mathbf e_2 + (a_1a_2-a_2b_1)\mathbf e^3 \end{align*}\]

Properties of Cross Product

Linearity

\(\mathbf a\times (\mathbf b + \mathbf c) = \mathbf a \times \mathbf b + \mathbf a \times \mathbf c\)
\(\lambda \mathbf a \times \mathbf b = \lambda (\mathbf a \times \mathbf b) = \mathbf a \times \lambda\mathbf b\)
\(\mathbf a \times \mathbf b + \mathbf c \times \mathbf d = (\mathbf a - \mathbf c) \times (\mathbf b - \mathbf d) + \mathbf a \times \mathbf d + \mathbf c \times \mathbf b\)

Plane Normal

Claim Cross product is perpendicular to both of its vectors.

\[(\mathbf a \times \mathbf b) \cdot \mathbf a = (\mathbf a \times \mathbf b) \cdot \mathbf b = 0\]

proof. Note that

\[(\mathbf a \times \mathbf b) \cdot \mathbf a = \det(\begin{bmatrix}\mathbf a&\mathbf b&\mathbf a\end{bmatrix}), (\mathbf a \times \mathbf b) \cdot \mathbf b = \det(\begin{bmatrix}\mathbf a&\mathbf b&\mathbf b\end{bmatrix})\]

\(\begin{bmatrix}\mathbf a&\mathbf b&\mathbf a\end{bmatrix}, \begin{bmatrix}\mathbf a&\mathbf b&\mathbf b\end{bmatrix}\) are both linearly dependent, hence their determinant are both 0.

Geometric Meaning

Lemma \((\mathbf a \times \mathbf b)\cdot (\mathbf c \times \mathbf d) = \det(\begin{bmatrix}\mathbf a\cdot\mathbf c&\mathbf b\cdot\mathbf c\\\mathbf a\cdot\mathbf d&\mathbf b\cdot\mathbf d\end{bmatrix})\).

proof. By linearity, it is sufficient to show only when \(\mathbf a, \mathbf b, \mathbf c, \mathbf d \in \{\mathbf e_1, \mathbf e_2, \mathbf e_3\}\). Then, we can list all possible combinations and prove this claim.

Corollary (Lagrange's Identity in 3D). \(\|\mathbf a\times \mathbf b\|^2 = \det(\begin{bmatrix}\mathbf a\cdot\mathbf a&\mathbf a\cdot\mathbf b\\\mathbf b\cdot\mathbf a&\mathbf b\cdot\mathbf b\end{bmatrix})\).

proof. By the lemma above.

proof. Using Lagrange's Identity in 3D, we can then have

\[\begin{align*} \|\mathbf a\times \mathbf b\|&= (\mathbf a\times \mathbf b)\cdot(\mathbf a\times \mathbf b)\\ &= \det(\begin{bmatrix}\mathbf a\cdot\mathbf a&\mathbf a\cdot\mathbf b\\\mathbf b\cdot\mathbf a&\mathbf b\cdot\mathbf b\end{bmatrix})\\ &= (\mathbf a\cdot\mathbf a) (\mathbf b\cdot\mathbf b) - (\mathbf a\cdot\mathbf b)(\mathbf b\cdot\mathbf a)\\ &= \|\mathbf a\|^2 \|\mathbf b\|^2 - (\mathbf a\cdot\mathbf b)^2\\ &= \|\mathbf a\|^2 \|\mathbf b\|^2 - \|\mathbf a\|^2 \|\mathbf b\|^2 \cos^2\theta\\ &= \|\mathbf a\|^2 \|\mathbf b\|^2 (1 - \cos^2\theta)\\ &= \|\mathbf a\|^2 \|\mathbf b\|^2 \sin^2\theta\\ \|\mathbf a\times \mathbf b\| &= \|\mathbf a\| \|\mathbf b\|\sin \theta \end{align*}\]

Therefore, \(\|\mathbf a\times \mathbf b\|\) can be understood are the area of the parallelogram the area of the parallelogram that \(\mathbf a, \mathbf b\) span. Also, \(\mathbf a\times \mathbf b, \mathbf a, \mathbf b\) forms a signed basis.

Curvature for 3D

Note that our definition for curvature depends on \(\gamma\) being unit-speed. However, find a unit-speed parameterization is not always easy even though we know it exist.

Claim Let \(\gamma(t)\) be a regular curve in \(\mathbb R^3\), then its curvature is \(\kappa = \frac{\|\gamma''\times \gamma'\|}{\|\gamma'\|^3}\).

proof. Since \(\gamma\) is regular, take \(s\) be a unit-speed map for \(\gamma\).

\[\begin{align*} \gamma' &= \frac{d\gamma}{dt} = \frac{d\gamma}{ds}\frac{ds}{dt}&\text{chain rule}\\ \implies \frac{d\gamma}{ds} &= \frac{d\gamma}{dt}/\frac{ds}{dt}\\ \kappa &= \|\frac{d^2\gamma}{ds^2}\|\\ &= \|\frac{d}{ds}(\frac{d\gamma}{dt}/\frac{ds}{dt})\|\\ &= \|\frac{d}{dt}(\frac{d\gamma}{dt}/\frac{ds}{dt}) / \frac{ds}{dt}\|\\ &= \| (\frac{ds}{dt}\frac{d^2\gamma}{dt^2} - \frac{d^2s}{dt^2}\frac{d\gamma}{dt})(\frac{ds}{dt})^{-3}\| \end{align*}\]

Note that \(s\) is the unit-length map, so that \((\frac{ds}{dt})^2 = (s')^2 = \|\gamma'\|^2 = \gamma'\cdot \gamma'\), differentiating both side gives

\[\frac{d}{dt}((\frac{ds}{dt})^2) = \frac{ds}{dt} \frac{d^2s}{dt^2} = \gamma'\cdot\gamma''\]

Then, we can insert these back to get

\[\begin{align*} \kappa &= \| (\frac{ds}{dt}\frac{d^2\gamma}{dt^2} - \frac{d^2s}{dt^2}\frac{d\gamma}{dt})(\frac{ds}{dt})^{-3}\|\\ &= \| ((\frac{ds}{dt})^2\frac{d^2\gamma}{dt^2} - \frac{d^2s}{dt^2}\frac{ds}{dt}\frac{d\gamma}{dt})(\frac{ds}{dt})^{-4}\|\\ &= \|(\gamma'\cdot \gamma')\gamma'' - (\gamma'\cdot \gamma'')\gamma'\|\|\gamma'\|^{-4}\\ &= \frac{\|(\gamma'\cdot \gamma')\gamma'' - (\gamma'\cdot \gamma'')\gamma'\|}{\|\gamma'\|^{4} } \end{align*}\]

Using vector triple product identity

\[\mathbf a\times (\mathbf b\times \mathbf c) = (\mathbf a\cdot \mathbf c)\mathbf b - (\mathbf a \cdot \mathbf b)\mathbf c\]

\[\|(\gamma'\cdot \gamma')\gamma'' - (\gamma'\cdot \gamma'')\gamma'\| = \|\gamma'\times (\gamma''\times \gamma')\| = \|\gamma'\|\|\gamma''\times \gamma'\|\sin\theta\]

Note that \(\gamma'\) and \(\gamma''\times \gamma'\) are orthogonal by definition of cross product, so that \(\sin\theta = 1\)

\[\kappa = \frac{\|\gamma'\|\|\gamma''\times \gamma'\|}{\|\gamma'\|^4} = \frac{\|\gamma''\times \gamma'\|}{\|\gamma'\|^3}\]

Example: Compute curvature

(1) \(\gamma(t) = (\frac{(1+t)^{3/2} }{3},\frac{(1-t)^{3/2} }{3}, \frac{t}{\sqrt{2} })\)

\[\begin{align*} \gamma'(t) &= (\frac{1}{2}\sqrt{1+t}, -\frac{1}{2}\sqrt{1-t}, \frac{1}{\sqrt{2} })\\ \|\gamma'\| &= \sqrt{\frac{1+t}{4} + \frac{1-t}{4} + \frac{1}{2} } = 1\\ \gamma''(t) &= (\frac{1}{4}(1+t)^{-1/2}, \frac{1}{4}(1-t)^{-1/2}, 0)\\ \kappa(t) = \|\gamma''(t)\| &= \sqrt{(16(1+t))^{-1} + (16(1-t))^{-1} } \\ &= \sqrt{\frac{1-t + 1+t}{16(1+t)(1-t)} } \\ &= (8(1-t^2))^{-1/2} \end{align*}\]

(2) \(\gamma(t) = (\frac{4\cos t}{5}, 1-\sin t, \frac{-3\cos t}{5})\)

\[\begin{align*} \gamma'(t) &= (-\frac{4}{5}\sin t, -\cos t, \frac{3}{5}\sin t)\\ \|\gamma'(t)\| &= (\frac{16}{25}\sin^2 t + cos^2 t + \frac{9}{25}\sin^2 t)^{1/2} = 1\\ \gamma''(t) &= (-\frac{4}{5}\cos t, \sin t, \frac{3}{5}\cos t)\\ \kappa(t) = \|\gamma''(t)\| &= (\frac{16}{25}\cos^2 t + \sin^2 t + \frac{9}{25}\cos^2 t)^{1/2} = 1 \end{align*}\]

(3) \(\gamma(t) = (t, \cosh t)\)

\[\begin{align*} \gamma'(t) &= (1, \sinh t)\\ \|\gamma'(t)\| &= \sqrt{1 + \sinh^2} \neq 1\\ \gamma''(t) &= (0, \cosh t)\\ \kappa(t) &= \frac{\|\gamma''(t)\times \gamma'(t)\|}{\|\gamma'(t)\|^3}\\ &= (1 + \sinh^2)^{-3/2}\begin{vmatrix} i&j&k\\ 1&\sinh t&0\\ 0&\cosh t&0 \end{vmatrix} &= (1+\sinh^2)^{-3/2} \cosh t \end{align*}\]

(4) \(\gamma(t) = (\cos^3 t, \sin^3 t)\)

\[\begin{align*} \gamma'(t) &= (-3\cos^2 t \sin t, 3\sin^2 t \cos t)\\ \|\gamma'(t)\| &= (9\cos^4 t sin^2 t + 9 \sin^4 t \cos^2 t)^{1/2} = |3\sin t \cos t|\\ \gamma''(t) &= (6\cos t\sin^2 t - 3cos^3 t, 6\sin t\cos^2 t - 3\sin^3 t))\\ \|\gamma''(t) \times \gamma'(t)\| &= (-3\cos^2 t \sin t)(6\sin t\cos^2 t - 3\sin^3 t) - (3\sin^2 t \cos t)(6\cos t\sin^2 t - 3cos^3 t)\\ &= |-18\sin^2 t\cos^4 t + 9\sin^4 t \cos^2 t - 18\sin^4 t \cos^2 t + 9\sin^2 t \cos^4 t|\\ &= 9\sin^2 t\cos^2 t = |3\sin t \cos t|^2\\ \kappa(t) &= \frac{ |3\sin t \cos t|^2}{ |3\sin t \cos t|^3} = |3\sin t \cos t|^{-1} \end{align*}\]

Source code

Claim 1

For some regular curve \(\gamma\), and its curvature o\(\kappa\), if \(\forall t,\kappa(t) > 0\), then \(\kappa\) is smooth.

proof. Let \(\gamma\) be regular, wlog assume \(\gamma\) is unit-length so that \(\kappa = \|\gamma''\|\).
Note that \(\gamma\) is smooth, hence all of its components are smooth, add is smooth, and square root is smooth on \((0, \infty)\). Therefore, \(\kappa = \|\gamma''\|\) is smooth on \((0,\infty)\).