1 Curves
Definition 1: Length of a curve
Example 2: Length of the Helix
Solution. We compute \[\begin{align*} \dot{{\pmb{\gamma}}}(t) & = (-R\sin(t), R\cos(t) , H) \qquad \left\| \dot{{\pmb{\gamma}}}(t) \right\| = \sqrt{R^2 + H^2} \\ L({\pmb{\gamma}}) & = \int_0^{2\pi} \left\| \dot{{\pmb{\gamma}}}(u) \right\| \, du = 2 \pi \sqrt{R^2 + H^2} \end{align*}\]
Definition 3: Arc-Length of a curve
Example 4: Arc-length of Logarithmic Spiral
Solution. The arc-length starting from \(t_0\) is \[\begin{align*} \dot{{\pmb{\gamma}}}(t) & = e^{kt} ( k \cos(t) - \sin(t), k \sin(t) + \cos(t) ,0) \\ \left\| \dot{{\pmb{\gamma}}}(t) \right\|^2 & = (k^2 + 1) e^{2kt} \\ s(t) & = \int_{t_0}^t \left\| \dot{{\pmb{\gamma}}}(\tau) \right\| \, d \tau = \frac{\sqrt{k^2 + 1}}{k} ( e^{kt} - e^{k t_0} ) \,. \end{align*}\]
Definition 5: Unit-speed curve
Proposition 6
Proof
Definition 7: Reparametrization
Definition 8: Unit-speed reparametrization
Definition 9: Regular curve
Theorem 10: Existence of unit-speed reparametrization
Let \({\pmb{\gamma}}\) be a curve. They are equivalent:
- \({\pmb{\gamma}}\) is regular,
- \({\pmb{\gamma}}\) admits unit-speed reparametrization.
Theorem 11: Characterization of unit-speed reparametrizations
Let \({\pmb{\gamma}}\ \colon (a,b) \to \mathbb{R}^3\) be a regular curve. Let \(\widetilde{{\pmb{\gamma}}}\ \colon (\tilde{a},\tilde{b}) \to \mathbb{R}^3\) be a reparametrization of \({\pmb{\gamma}}\), that is, \[ {\pmb{\gamma}}(t) = \widetilde{{\pmb{\gamma}}}( \phi(t) ) , \quad \forall \, t \in (a,b) \] for some diffeomorphism \(\phi\ \colon (a,b) \to (\tilde{a},\tilde{b})\). We have
If \(\widetilde{{\pmb{\gamma}}}\) is unit-speed, there exists \(c \in \mathbb{R}\) such that \[ \phi(t) = \pm s(t) + c \,, \quad \forall \, t \in (a,b) \,. \tag{1.1}\]
If \(\phi\) is given by (1.1), then \(\widetilde{{\pmb{\gamma}}}\) is unit-speed.
Definition 12: Arc-length reparametrization
Example 13: Reparametrization by arc-length
Prove that \({\pmb{\gamma}}\) is regular, and reparametrize it by arc-length.
Solution. \({\pmb{\gamma}}\) is regular because \[ \dot{{\pmb{\gamma}}}(t) =(-5 \sin (t), 5 \cos (t), 12) \,, \qquad \|\dot{{\pmb{\gamma}}}(t)\| = 13 \neq 0 \] The arc-length of \({\pmb{\gamma}}\) starting from \(t_0 = 0\), and its inverse, are \[ s(t) = \int_0^t \left\| \dot{{\pmb{\gamma}}}(u) \right\| \, d u = 13 t \,, \qquad t(s) = \frac{s}{13} \,. \] The arc-length reparametrization of \({\pmb{\gamma}}\) is \[ \widetilde{{\pmb{\gamma}}}(s)= {\pmb{\gamma}}(t(s)) = \left(5 \cos \left(\frac{s}{13} \right), 5 \sin \left(\frac{s}{13} \right), \frac{12}{13} s\right) \,. \]
1.1 Curvature
Definition 14: Curvature of unit-speed curve
Example 15: Curvature of the Circle
Solution. First, check that \({\pmb{\gamma}}\) is unit-speed: \[ \dot{{\pmb{\gamma}}}(t) = \left( - \sin \left( \frac{t}{R} \right) , \cos \left( \frac{t}{R} \right), 0\right) \,, \qquad \left\| \dot{{\pmb{\gamma}}}(t) \right\| = 1 \, \] Now, compute second derivative and curvature \[\begin{align*} \ddot{{\pmb{\gamma}}}(t) & = \left( -\frac{1}{R} \cos \left( \frac{t}{R} \right) , - \frac{1}{R} \sin \left( \frac{t}{R} \right) ,0\right) \,, \\ \kappa(t) & = \left\| \ddot{{\pmb{\gamma}}}(t) \right\| = \frac{1}{R} \,. \end{align*}\]
Definition 16: Curvature of regular curve
Remark 17: Computing curvature of regular \({\pmb{\gamma}}\)
Compute the arc-length \(s(t)\) of \({\pmb{\gamma}}\) and its inverse \(t(s)\).
Compute the arc-length reparametrization \[ \widetilde{{\pmb{\gamma}}}(s) = {\pmb{\gamma}}(t(s)) \,. \]
Compute the curvature of \(\widetilde{{\pmb{\gamma}}}\) \[ \widetilde{\kappa}(s) = \left\| \ddot{\widetilde{{\pmb{\gamma}}}}(s) \right\|\,. \]
The curvature of \({\pmb{\gamma}}\) is \[ \kappa (t)= \widetilde{\kappa}(s(t)) \,. \]
Definition 18: Hyperbolic functions
Theorem 19: Properties of Hyperbolic Functions
Example 20: Curvature of the Catenary
Question. Consider the Catenary curve \[ {\pmb{\gamma}}(t) = ( t, \cosh(t) ) \,, \quad t \in \mathbb{R}\,. \]
- Prove that \({\pmb{\gamma}}\) is regular.
- Compute the arc-length reparametrization of \({\pmb{\gamma}}\).
- Compute the curvature of \(\widetilde{{\pmb{\gamma}}}\).
- Compute the curvature of \({\pmb{\gamma}}\).
Solution.
\({\pmb{\gamma}}\) is regular because \[\begin{align*} \dot{{\pmb{\gamma}}}(t) & = (1 , \sinh(t)) \\ \left\| \dot{{\pmb{\gamma}}} \right\| & = \sqrt{1 + {\sinh}^2 (t)} = \cosh (t) \geq 1 \end{align*}\]
The arc-length of \({\pmb{\gamma}}\) starting at \(t_0 = 0\) is \[ s(t) = \int_0^t \left\| \dot{{\pmb{\gamma}}}(u) \right\| \, du = \int_0^t \cosh (u) \, du = \sinh (t) \] where we used that \(\sinh(0) = 0\). Moreover, \[\begin{align*} s = \sinh(t) \quad & \iff \quad s = \frac{e^t - e^{-t}}{2} \\ \quad & \iff \quad e^{2t} - 2s e^{t} - 1 = 0 \end{align*}\] Substitute \(y = e^t\) to obtain \[\begin{align*} e^{2t} - 2s e^{t} - 1 = 0 \quad & \iff \quad y^{2} - 2s y - 1 = 0 \\ \quad & \iff \quad y_{\pm} = s \pm \sqrt{1+s^2} \,. \end{align*}\] Notice that \[ y_{+} = s + \sqrt{1 + s^2} \geq s + \sqrt{s^2} = s + |s| \geq 0 \] by definition of absolute value. Therefore, \[\begin{align*} e^t = y_+ = s + \sqrt{1 + s^2 } \,\, \implies \,\, t(s) = \log \left( s + \sqrt{1 + s^2} \right) \end{align*}\] The arc-length reparametrization of \({\pmb{\gamma}}\) is \[ \widetilde{{\pmb{\gamma}}}(s) = {\pmb{\gamma}}( t(s) ) = \left( \log \left( s + \sqrt{ 1+ s^2} \right) , \sqrt{1 + s^2} \right) \]
Compute the curvature of \(\widetilde{{\pmb{\gamma}}}\) \[\begin{align*} \dot{\widetilde{{\pmb{\gamma}}}}(s) & = \left( \frac{1}{ \sqrt{1 + s^2} } , \frac{s}{ \sqrt{1 + s^2}} \right) \\ \ddot{\widetilde{{\pmb{\gamma}}}}(s) & = \left( - \frac{s}{ (1 + s^2)^{3/2} } , \frac{1}{(1 + s^2)^{3/2} } \right) \\ \widetilde{\kappa}(s) & = \left\| \ddot{\widetilde{{\pmb{\gamma}}}}(s) \right\|= \frac{1}{1+s^2} \end{align*}\]
Recalling that \(s(t) = \sinh(t)\), the curvature of \({\pmb{\gamma}}\) is \[ \kappa (t) = \widetilde{\kappa}(s(t)) = \frac{1}{1+ \sinh^2(t)} = \frac{1}{\cosh^2(t)} \,. \]
Definition 21: Vector product
Theorem 22: Geometric Properties of vector product
Let \(\mathbf{u},\mathbf{v}\in \mathbb{R}^3\) be linearly independent. Then
- \(\mathbf{u}\times \mathbf{v}\) is orthogonal to the plane spanned by \(\mathbf{u},\mathbf{v}\)
- \(\| \mathbf{u}\times \mathbf{v}\|\) is the area of the parallelogram with sides \(\mathbf{u},\mathbf{v}\)
- The triple \((\mathbf{u},\mathbf{v},\mathbf{u}\times \mathbf{v})\) is a positive basis of \(\mathbb{R}^3\)
Theorem 23
Theorem 24
Theorem 25: Curvature formula
Example 26: Curvature of the Helix
Question. Consider the Helix of radius \(R>0\) and rise \(H\), \[ {\pmb{\gamma}}(t) = ( R\cos(t) , R\sin(t) , Ht) \,. \]
- Prove that \({\pmb{\gamma}}\) is regular.
- Compute the curvature of \({\pmb{\gamma}}\).
Solution.
\({\pmb{\gamma}}\) is regular because \[\begin{align*} \dot{{\pmb{\gamma}}}(t) & = ( -R\sin(t) , R\cos(t) , H) \\ \left\| \dot{{\pmb{\gamma}}}(t) \right\| & = \sqrt{R^2 + H^2} \geq R > 0 \end{align*}\]
Compute the curvature using the formula: \[\begin{align*} \ddot{{\pmb{\gamma}}}(t) & = ( -R\cos(t) , -R\sin(t) , 0) \\ \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}& = \left( RH\sin(t), -RH\cos(t), R^2 \right) \\ \left\| \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}} \right\| & = R\sqrt{R^2 + H^2 } \\ \kappa (t) & = \frac{ \left\| \dot{{\pmb{\gamma}}}(t) \times \ddot{{\pmb{\gamma}}}(t) \right\| }{ \left\| \dot{{\pmb{\gamma}}}(t) \right\|^3 } = \frac{ R }{ R^2 + H^2 } \end{align*}\]
Example 27: Calculation of curvature
Question. Define the curve \[ {\pmb{\gamma}}(t)=\left(\frac{8}{5} \cos (t), 1-2 \sin (t), \frac{6}{5} \cos (t)\right) \,. \]
- Prove that \({\pmb{\gamma}}\) is regular.
- Compute the curvature of \({\pmb{\gamma}}\).
Solution.
\({\pmb{\gamma}}\) is regular because \[ \dot{{\pmb{\gamma}}}=\left(-\frac{8}{5} \sin (t),-2 \cos (t),-\frac{6}{5} \sin (t)\right) \,, \qquad \|\dot{{\pmb{\gamma}}}\| =2 \neq 0 \,. \]
Compute the curvature using the formula: \[ \begin{aligned} & \ddot{{\pmb{\gamma}}}=\left(-\frac{8}{5} \cos (t), 2 \sin (t),-\frac{6}{5} \cos (t)\right) &\,\,\,\,& \|\dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}\|=4 \\ & \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}=\left(-\frac{12}{5}, 0, \frac{16}{5}\right) &\,\,\,\,& \kappa = \frac{1}{2} \,. \end{aligned} \]
Example 28: Different curves, same curvature
Solution. Curvatures of \({\pmb{\gamma}}\) and \({\pmb{\eta}}\) were already computed: \[ \kappa^{\pmb{\gamma}}= \frac{1}{2}\,, \quad \kappa^{\pmb{\eta}}= \frac{S}{S^2 + H^2} \,. \] Imposing that \(\kappa^{\pmb{\gamma}}= \kappa^{\pmb{\eta}}\), we get \[ \frac12 = \frac{S}{S^2 + H^2} \quad \implies \quad H^2 = 2S - S^2 \,. \] Choosing \(S=1\) and \(H=1\) yields \(\kappa^{\pmb{\gamma}}= \kappa^{\pmb{\eta}}\).
1.2 Frenet frame and torsion
Definition 29: Frenet frame of unit-speed curve
Let \({\pmb{\gamma}}\colon (a,b) \to \mathbb{R}^3\) be unit-speed, with \(\kappa \neq 0\).
The tangent vector to \({\pmb{\gamma}}\) at \({\pmb{\gamma}}(t)\) is \[ \mathbf{t}(t)= \dot{{\pmb{\gamma}}}(t) \,. \]
The principal normal vector to \({\pmb{\gamma}}\) at \({\pmb{\gamma}}(t)\) is \[ \mathbf{n}(t) = \frac{\ddot{{\pmb{\gamma}}}(t)}{\kappa (t)} \,. \]
The binormal vector to \({\pmb{\gamma}}\) at \({\pmb{\gamma}}(t)\) is \[ \mathbf{b}(t) = \dot{{\pmb{\gamma}}}(t) \times \mathbf{n}(t) \,. \]
The Frenet frame of \({\pmb{\gamma}}\) at \({\pmb{\gamma}}(t)\) is the triple \[ \{ \mathbf{t}(t), \mathbf{n}(t), \mathbf{b}(t)\} \,. \]
Theorem 30: Frenet frame is orthonormal basis
Definition 31: Torsion of unit-speed curve with \(\kappa \neq 0\)
Definition 32: Torsion of regular curve with \(\kappa \neq 0\)
Example 33: Curvature and torsion of Helix with Frenet frame
Question. Consider the Helix of radius \(R>0\) and rise \(H\) \[ {\pmb{\gamma}}(t) = ( R\cos(t), R\sin(t),t H)\,, \quad \, t \in \mathbb{R}\,. \]
- Compute the arc-length reparametrization \(\widetilde{{\pmb{\gamma}}}\) of \({\pmb{\gamma}}\).
- Compute Frenet frame, curvature and torsion of \(\widetilde{{\pmb{\gamma}}}\).
- Compute curvature and torsion \({\pmb{\gamma}}\).
Solution.
The arc-length of \({\pmb{\gamma}}\) starting at \(t_0 = 0\), and its inverse, are \[\begin{align*} \dot{{\pmb{\gamma}}}(t) & = ( -R\sin(t), R\cos(t), H ) \\ \left\| \dot{{\pmb{\gamma}}} \right\| & = \rho, \qquad \rho := \sqrt{R^2 + H^2} \\ s(t) & = \int_0^t \left\| \dot{{\pmb{\gamma}}}(u) \right\| \, du = \rho t \,, \qquad t(s) = \frac{s}{\rho} \,. \end{align*}\] The arc-length reparametrization \(\widetilde{{\pmb{\gamma}}}\) of \({\pmb{\gamma}}\) is \[ \widetilde{{\pmb{\gamma}}}(s) = {\pmb{\gamma}}( t(s)) = \left( R \cos \left( \frac{s}{\rho} \right) ,R \sin \left( \frac{s}{\rho} \right) , \frac{H s }{\rho} \right) \,. \]
Compute the tangent vector to \(\widetilde{{\pmb{\gamma}}}\) and its derivative \[\begin{align*} \widetilde{\mathbf{t}}(s) & = \dot{\widetilde{{\pmb{\gamma}}}}= \frac{1}{\rho} \left( - R \sin \left( \frac{s}{\rho} \right) , R \cos \left( \frac{s}{\rho} \right) , H \right) \\ \dot{\widetilde{\mathbf{t}}}(s) & = \frac{R}{\rho^2} \left( -\cos \left( \frac{s}{\rho} \right) , -\sin \left( \frac{s}{\rho} \right) , 0 \right) \end{align*}\] The curvature of \(\widetilde{{\pmb{\gamma}}}\) is \[\begin{align*} \widetilde{\kappa}(s) & = \| \ddot{\widetilde{{\pmb{\gamma}}}}(s) \| = \| \dot{\widetilde{\mathbf{t}}}(s) \| = \frac{R}{R^2 + H^2}\,. \end{align*}\] The principal normal vector and binormal are \[\begin{align*} \widetilde{\mathbf{n}}(s) & = \frac{\widetilde{\mathbf{t}}}{\widetilde{\kappa}} = \left( -\cos \left( \frac{s}{\rho} \right) , -\sin \left( \frac{s}{\rho} \right) , 0 \right) \\ \widetilde{\mathbf{b}}(s) & = \widetilde{\mathbf{t}}\times \widetilde{\mathbf{n}} = \frac{1}{\rho} \left( H \sin \left( \frac{s}{\rho} \right) , - H \cos \left( \frac{s}{\rho} \right) , R \right) \,. \end{align*}\] We are left to compute the torsion of \(\widetilde{{\pmb{\gamma}}}\): \[\begin{align*} \dot{\widetilde{\mathbf{b}}}(s) & = \frac{H}{\rho^2} \left( \cos \left( \frac{s}{\rho} \right) , \sin \left( \frac{s}{\rho} \right) , 0 \right) \\ \dot{\widetilde{\mathbf{b}}}(s) & \cdot \widetilde{\mathbf{n}}(s) = - \frac{H}{\rho^2} \\ \widetilde{\tau}(s) & = - \dot{\widetilde{\mathbf{b}}}(s) \cdot \widetilde{\mathbf{n}}(s) = \frac{H}{\rho^2} = \frac{H}{R^2 + H^2} \end{align*}\]
The curvature and torsion of \({\pmb{\gamma}}\) are \[\begin{align*} \kappa(t) & = \widetilde{\kappa}(s(t)) = \frac{R}{R^2 + H^2} \\ \tau(t) & = \widetilde{\tau}(s(t)) = \frac{H}{R^2 + H^2} \end{align*}\]
Theorem 34: Torsion formula
Example 35: Torsion of the Helix with formula
Question. Consider the Helix of radius \(R>0\) and rise \(H>0\) \[ {\pmb{\gamma}}(t) = ( R\cos(t) , R\sin(t) , Ht) \,, \quad t \in \mathbb{R}\,. \]
- Prove that \({\pmb{\gamma}}\) is regular with non-vanishing curvature.
- Compute the torsion of \({\pmb{\gamma}}\).
Solution.
\({\pmb{\gamma}}\) is regular with non-vanishing curvature, since \[ \left\| \dot{{\pmb{\gamma}}}(t) \right\| = \sqrt{R^2 + H^2} \geq R > 0 \,, \qquad \kappa = \frac{R}{R^2 + H^2} > 0 \,. \]
We compute the torsion using the formula: \[\begin{align*} \dot{{\pmb{\gamma}}}(t) & = ( -R\sin(t) , R\cos(t) , H) \\ \ddot{{\pmb{\gamma}}}(t) & = ( -R\cos(t) , -R\sin(t) , 0) \\ \dddot{{\pmb{\gamma}}}(t) & = ( R\sin(t) , -R\cos(t) , 0) \\ \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}& = \left( RH\sin(t), -RH\cos(t), R^2 \right) \\ \left\| \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}} \right\| & = R\sqrt{R^2 + H^2 } \\ (\dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}) \cdot \dddot{{\pmb{\gamma}}}& = R^2 H \\ \tau (t) & = \frac{ ( \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}) \cdot \dddot{{\pmb{\gamma}}}}{ \left\| \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}} \right\|^2 } = \frac{ H }{ R^2 + H^2 } \end{align*}\]
Example 36: Calculation of torsion
Solution. Resuming calculations from Example 27, \[ \begin{aligned} \dddot{{\pmb{\gamma}}}& =\left(\frac{8}{5} \sin (t), 2 \cos (t), \frac{6}{5} \sin (t)\right) \\ ( \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}) \cdot \dddot{{\pmb{\gamma}}}& = \frac{96}{25} \sin (t)-\frac{96}{25} \sin (t) = 0 \\ \tau(t) & = \frac{ ( \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}) \cdot \dddot{{\pmb{\gamma}}}}{ \left\| \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}} \right\|^2 } = 0 \end{aligned} \]
Theorem 37: General Frenet frame formulas
Example 38: Twisted cubic
Question. Let \({\pmb{\gamma}}\colon \mathbb{R}\to \mathbb{R}^3\) be the twisted cubic \[ {\pmb{\gamma}}(t) = (t,t^2,t^3 ) \,. \]
- Is \({\pmb{\gamma}}\) regular/unit-speed? Justify your answer.
- Compute the curvature and torsion of \({\pmb{\gamma}}\).
- Compute the Frenet frame of \({\pmb{\gamma}}\).
Solution.
\({\pmb{\gamma}}\) is regular, but not-unit speed, because \[\begin{align*} & \dot {\pmb{\gamma}}(t) = (1 , 2t , 3t^2) \\ & \left\| \dot {\pmb{\gamma}}(t) \right\| = \sqrt{1 + 4t^2 + 9t^4} \geq 1 \qquad \left\| \dot{\pmb{\gamma}}(1) \right\| = \sqrt{14} \neq 1 \end{align*}\]
Compute the following quantities \[\begin{align*} & \ddot{{\pmb{\gamma}}}= (0,2,6t) &\,& \left\| \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}} \right\| = 2 \sqrt{ 1 + 9t^2 + 9t^4 } \\ & \dddot {\pmb{\gamma}}= (0,0,6) &\,& (\dot {\pmb{\gamma}}\times \ddot {\pmb{\gamma}}) \cdot \dddot {\pmb{\gamma}}= 12 \\ & \dot {\pmb{\gamma}}\times \ddot {\pmb{\gamma}}= (6t^2, -6t, 2 ) \end{align*}\] Compute curvature and torsion using the formulas: \[\begin{align*} \kappa(t) & = \frac{ \left\| \dot {\pmb{\gamma}}\times \ddot {\pmb{\gamma}} \right\| }{\left\| \dot {\pmb{\gamma}} \right\|^3} = \frac{ 2 \sqrt{ 1 + 9t^2 + 9t^4 } }{ (1 + 4t^2 + 9t^4)^{3/2} } \\ \tau(t) & = \frac{ (\dot {\pmb{\gamma}}\times \ddot {\pmb{\gamma}}) \cdot \dddot {\pmb{\gamma}}}{ \left\| \dot {\pmb{\gamma}}\times \ddot {\pmb{\gamma}} \right\|^2 } = \frac{3}{1 + 9t^2 + 9t^4} \,. \end{align*}\]
By the Frenet frame formulas and the above calculations, \[\begin{align*} \mathbf{t}& = \frac{\dot{{\pmb{\gamma}}}}{\left\| \dot{{\pmb{\gamma}}} \right\|} = \frac{1}{\sqrt{1 + 4t^2 + 9t^4}} \ (1 , 2t , 3t^2) \\ \mathbf{b}& = \frac{ \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}}{ \left\| \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}} \right\| } = \frac{1}{\sqrt{ 1 + 9t^2 + 9t^4}} (3t^2, -3t, 1 ) \\ \mathbf{n}& = \mathbf{b}\times \mathbf{t} = \frac{(−9t^3 − 2t,1 − 9t^4,6t^3 + 3t)}{\sqrt{ 1 + 9t^2 + 9t^4} \, \sqrt{1 + 4t^2 + 9t^4}} \end{align*}\]
1.3 Frenet-Serret equations
Theorem 39: Frenet-Serret equations
Definition 40: Rigid motion
Theorem 41: Fundamental Theorem of Space Curves
Let \(\kappa, \tau \ \colon (a,b) \to \mathbb{R}\) be smooth, with \(\kappa>0\). Then:
There exists a unit-speed curve \({\pmb{\gamma}}\ \colon (a,b) \to \mathbb{R}^3\) with curvature \(\kappa(t)\) and torsion \(\tau(t)\).
Suppose that \(\widetilde{{\pmb{\gamma}}}\ \colon (a,b) \to \mathbb{R}^3\) is a unit-speed curve whose curvature \(\widetilde{\kappa}\) and torsion \(\widetilde{\tau}\) satisfy \[ \widetilde{\kappa}(t) = \kappa(t) \,, \quad \widetilde{\tau}(t) = \tau(t) \,, \quad \forall \, t \in (a,b) \,. \] There exists a rigid motion \(M \colon \mathbb{R}^3 \to \mathbb{R}^3\) such that \[ \widetilde{{\pmb{\gamma}}}(t) = M ({\pmb{\gamma}}(t)) \,, \qquad \forall \, t \in (a,b) \,. \]
Example 42: Application of FTSC
Question. Consider the curve \[ {\pmb{\gamma}}( t ) = \left( \sqrt{3}\, t - \sin ( t ) , \sqrt{3} \sin ( t ) + t , 2 \cos ( t ) \right)\,. \]
Calculate the curvature and torsion of \({\pmb{\gamma}}\).
The helix of radius \(R\) and rise \(H\) is parametrized by \[ {\pmb{\eta}}(t) = (R \cos(t), R\sin(t), Ht) \,. \] Recall that \({\pmb{\eta}}\) has curvature and torsion \[ \kappa^{{\pmb{\eta}}} = \frac{R}{R^2 + H^2} \,, \qquad \tau^{{\pmb{\eta}}} = \frac{H}{R^2 + H^2} \,. \] Prove that there exist a rigid motion \(M \colon \mathbb{R}^3 \to \mathbb{R}^3\) such that \[ {\pmb{\gamma}}(t) = M({\pmb{\eta}}(t)) \,, \quad \forall \, t \in \mathbb{R}\,. \tag{1.2}\]
Solution.
Compute curvature and torsion with the formulas \[ \begin{aligned} & \dot {\pmb{\gamma}}(t) = \left( \sqrt{3} - \cos ( t ), \sqrt{3}\cos ( t ) + 1, -2 \sin ( t ) \right) \\ & \ddot {\pmb{\gamma}}(t) = \left( \sin ( t ),-\sqrt{3}\sin ( t ), -2 \cos ( t ) \right) \\ & \dddot {\pmb{\gamma}}(t) = \left( \cos ( t ),-\sqrt{3}\cos ( t ), 2 \sin ( t ) \right) \\ & \dot {\pmb{\gamma}}(t) \times \ddot {\pmb{\gamma}}(t) = {\small \left( -2 \left( \sqrt{3} + \cos ( t ) \right), 2 \left( \sqrt{3} \cos ( t ) - 1 \right), -4 \sin ( t ) \right) }\\ & \left\| \dot {\pmb{\gamma}}(t) \times \ddot {\pmb{\gamma}}(t) \right\|^2 = 32 \\ & \left\| \dot {\pmb{\gamma}}(t) \right\|^2 = 8 \\ & \left( \dot {\pmb{\gamma}}(t) \times \ddot {\pmb{\gamma}}(t) \right) \cdot \dddot {\pmb{\gamma}}( t ) = -8 \\ & \kappa ( t ) = \frac{ \left\| \dot{\pmb{\gamma}}\times \ddot {\pmb{\gamma}} \right\| }{ \left\| \dot{\pmb{\gamma}} \right\|^3 } = \dfrac{\sqrt{32}}{8^{\frac{3}{2}}} = \dfrac{1}{4} \\ & \tau ( t ) = \frac{ \left(\dot{\pmb{\gamma}}\times \ddot{\pmb{\gamma}}\right) \cdot \dddot {\pmb{\gamma}}}{ \left\| \dot{\pmb{\gamma}}\times \ddot{\pmb{\gamma}} \right\|^2 } = \dfrac{-8}{32} = -\dfrac{1}{4} \, . \end{aligned} \]
Equating \(\kappa = \kappa^{{\pmb{\eta}}}\) and \(\tau = \tau^{{\pmb{\eta}}}\), we obtain \[ \frac{R}{R^2 + H^2} = \frac{1}{4} \,, \qquad \frac{H}{R^2 + H^2} = - \frac{1}{4} \] Rearranging both equalities we get \[ R^2 + H^2 = 4 R \,, \qquad R^2 + H^2 = -4 H \,, \] from which we find the relation \(R = - H\). Substituting into \(R^2 + H^2 = -4 H\), we get \[ H = - 2 \,, \quad R = - H = 2 \,. \] For these values of \(R\) and \(H\) we have \(\kappa = \kappa^{{\pmb{\eta}}}\) and \(\tau = \tau^{{\pmb{\eta}}}\). By the FTSC, there exists a rigid motion \(M \colon \mathbb{R}^3 \to \mathbb{R}^3\) satisfying (1.2).
Theorem 43: Curves contained in a plane - Part I
Let \({\pmb{\gamma}}\colon (a,b) \to \mathbb{R}^3\) be regular with \(\kappa \neq 0\). They are equivalent:
The torsion of \({\pmb{\gamma}}\) satisfies \[ \tau(t) = 0 \,, \quad \forall \, t \in (a,b) \,. \]
\({\pmb{\gamma}}\) is contained in a plane: There exists a vector \(\mathbf{P} \in \mathbb{R}^3\) and a scalar \(d \in \mathbb{R}\) such that \[ {\pmb{\gamma}}(t)\cdot \mathbf{P} = d \,, \quad \forall \, t \in (a,b) \,. \]
Theorem 44: Curves contained in a plane - Part II
Example 45: A planar curve
Question. Consider the curve \[ {\pmb{\gamma}}(t) = ( t,2t,t^4) \,, \quad t > 0 \,. \]
Prove that \({\pmb{\gamma}}\) is regular.
Compute the curvature and torsion of \({\pmb{\gamma}}\).
Prove that \({\pmb{\gamma}}\) is contained in a plane. Compute the equation of such plane.
Solution.
\({\pmb{\gamma}}\) is regualar because \(\dot{{\pmb{\gamma}}}(t) = (1,2,4t^3) \neq {\pmb{0}}\).
Compute the following quantities \[\begin{align*} & \left\| \dot{{\pmb{\gamma}}} \right\| = \sqrt{5 + 16 t^4} &\,& \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}= 12 \,(2t^2, -t^2, 0) \\ & \ddot{{\pmb{\gamma}}}= 12 \, (0,0,t^2) &\,& \left\| \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}} \right\| = 12 \sqrt{5} \, t^2 \\ & \dddot{{\pmb{\gamma}}}= 24 \, (0,0,t) &\,& (\dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}) \cdot \dddot{{\pmb{\gamma}}}= 0 \\ \end{align*}\] Compute curvature and torsion with the formulas \[\begin{align*} & \kappa(t) = \frac{\dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}}{\left\| \dot{{\pmb{\gamma}}} \right\|^3} = \frac{12 \sqrt{5} \, t^2}{\sqrt{ 5 + 16 t^4 }} \\ & \tau(t) = \frac{(\dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}) \cdot \dddot{{\pmb{\gamma}}}}{ \left\| \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}} \right\| } = 0 \,. \end{align*}\]
\({\pmb{\gamma}}\) lies in a plane because \(\tau = 0\). The binormal is \[ \mathbf{b}= \frac{\dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}}}{\left\| \dot{{\pmb{\gamma}}}\times \ddot{{\pmb{\gamma}}} \right\|} = \frac{1}{\sqrt{5}} \, (2,-1,0) \,. \] At \(t_0 = 0\) we have \({\pmb{\gamma}}\left( 0 \right) = {\pmb{0}}\). The equation of the plane containing \({\pmb{\gamma}}\) is then \(\mathbf{x}\cdot \mathbf{b}= 0\), which reads \[ \frac{2}{\sqrt5} x - \frac{1}{\sqrt5} y = 0 \quad \implies \quad 2x - y = 0 \,. \]
Theorem 46: Curves contained in a circle
Let \({\pmb{\gamma}}\colon (a,b) \to \mathbb{R}^3\) be unit-speed. They are equivalent:
\({\pmb{\gamma}}\) is contained in a circle of radius \(R>0\).
There exists \(R>0\) such that \[ \kappa (t) = \frac{1}{R} \,, \quad \tau(t) = 0\,, \quad \forall \, t \in (a,b)\,. \]
Example 47: A curve contained in a circle
Question. Consider the curve \[ {\pmb{\gamma}}(t) = \left( \frac45 \cos(t), 1 - \sin(t) , -\frac35 \cos(t) \right) \,. \]
- Prove that \({\pmb{\gamma}}\) is unit-speed.
- Compute Frenet frame, curvature and torsion of \({\pmb{\gamma}}\).
- Prove that \({\pmb{\gamma}}\) is part of a circle.
Solution.
\({\pmb{\gamma}}\) is unit-speed because \[\begin{align*} \dot{{\pmb{\gamma}}}(t) & = \left( -\frac45 \sin(t), - \cos(t) , \frac35 \sin(t) \right) \\ \left\| \dot{{\pmb{\gamma}}}(t) \right\|^2 & = \frac{16}{25} \sin^2(t) + \cos^2(t) + \frac{9}{25} \sin^2(t) = 1 \end{align*}\]
As \({\pmb{\gamma}}\) is unit-speed, the tangent vector is \(\mathbf{t}(t) = \dot{{\pmb{\gamma}}}(t)\). The curvature, normal, binormal and torsion are \[\begin{align*} \dot{\mathbf{t}} (t) & = \left( -\frac45 \cos(t), \sin(t) , \frac35 \cos(t) \right) \\ \kappa(t) & = \left\| \dot{\mathbf{t}}(t) \right\| = \frac{16}{25} \cos^2(t) + \sin^2(t) + \frac{9}{25} \cos^2(t) = 1 \\ \mathbf{n}(t)& = \frac{1}{\kappa(t)} \ddot{{\pmb{\gamma}}}(t) = \left( -\frac45 \cos(t), \sin(t) , \frac35 \cos(t) \right) \\ \mathbf{b}(t) & = \dot{{\pmb{\gamma}}}(t) \times \mathbf{n}(t) = \left( -\frac35, 0 ,-\frac45 \right) \\ \dot{\mathbf{b}} & = {\pmb{0}}\\ \tau & = - \dot{\mathbf{b}} \cdot \mathbf{n}= 0 \end{align*}\]
The curvature of \({\pmb{\gamma}}\) is constant and the torsion is zero. Therefore \({\pmb{\gamma}}\) is contained in a circle of radius \[ R = \frac{1}{\kappa} = 1 \,. \]