目录
- T1
T1
给定弧长参数曲线 r ( s ) \mathbf{r}(s) r(s),它的曲率和挠率分别是 κ = κ ( s ) , τ = τ ( s ) ; r ( s ) \kappa=\kappa(s),\tau=\tau(s);\mathbf{r}(s) κ=κ(s),τ=τ(s);r(s) 的单位切向量 t ( s ) (s) (s)可看作单位球面 S 2 S^2 S2上的一条曲线，称为曲线 r ( s ) (s) (s)的切线像，证明：曲线 r ~ ( s ) : = t ( s ) \widetilde{\mathbf{r}}(s):=\mathbf{t}(s) r (s):=t(s)的曲率、挠率分别是
κ ~ ( s ) = 1 + ( τ κ ) 2 , \widetilde{\kappa}(s)=\sqrt{1+(\frac{\tau}{\kappa})^2}, κ (s)=1+(κτ)2,
τ ~ ( s ) = d d s ( τ κ ) κ ( 1 + ( τ κ ) 2 ) . \widetilde{\tau}(s)=\frac{\frac{d}{ds}(\frac{\tau}{\kappa})}{\kappa(1+(\frac{\tau}{\kappa})^{2})}. τ (s)=κ(1+(κτ)2)dsd(κτ).
证明：
r ~ ′ ′ ( s ) = κ ˙ n + κ n ˙ = − κ 2 t + κ ˙ n + κ τ b , \widetilde{\mathbf{r}}''(s)=\dot{\kappa}\mathbf{n}+\kappa\dot{\mathbf{n}}=-\kappa^2\mathbf{t}+\dot{\kappa}\mathbf{n}+\kappa\tau\mathbf{b}, r ′′(s)=κ˙n+κn˙=−κ2t+κ˙n+κτb,
r ~ ′ ′ ′ ( s ) = − 3 κ κ ˙ t + ( κ ¨ − κ 3 − κ τ 2 ) n + ( 2 κ ˙ τ + κ τ ˙ ) b . \widetilde{\mathbf{r}}'''(s)=-3\kappa\dot{\kappa}\mathbf{t}+(\ddot{\kappa}-\kappa^3-\kappa\tau^2)\mathbf{n}+(2\dot{\kappa}\tau+\kappa\dot{\tau})\mathbf{b}. r ′′′(s)=−3κκ˙t+(κ¨−κ3−κτ2)n+(2κ˙τ+κτ˙)b.
从而，
r ~ ′ ( s ) ∧ r ~ ′ ′ ( s ) = κ 2 τ t + κ 3 b , ∣ r ~ ′ ( s ) ∧ r ~ ′ ′ ( s ) ∣ = κ 2 κ 2 + τ 2 , \widetilde{\mathbf{r}}'(s)\wedge\widetilde{\mathbf{r}}''(s)=\kappa^2\tau\mathbf{t}+\kappa^3\mathbf{b},\quad|\widetilde{\mathbf{r}}'(s)\wedge\widetilde{\mathbf{r}}''(s)|=\kappa^2\sqrt{\kappa^2+\tau^2}, r ′(s)∧r ′′(s)=κ2τt+κ3b,∣r ′(s)∧r ′′(s)∣=κ2κ2+τ2,
有
κ ~ ( s ) = ∣ r ~ ′ ( s ) ∧ r ~ ′ ′ ( s ) ∣ ∣ r ~ ′ ( s ) ∣ 3 = κ 2 κ 2 + τ 2 κ 3 = 1 + ( τ κ ) 2 . \widetilde{\kappa}(s)=\frac{|\widetilde{\mathbf{r}}'(s)\wedge\widetilde{\mathbf{r}}''(s)|}{|\widetilde{\mathbf{r}}'(s)|^3}=\frac{\kappa^2\sqrt{\kappa^2+\tau^2}}{\kappa^3}=\sqrt{1+(\frac{\tau}{\kappa})^2}. κ (s)=∣r ′(s)∣3∣r ′(s)∧r ′′(s)∣=κ3κ2κ2+τ2=1+(κτ)2.
由于
( r ~ ′ ( s ) , r ~ ′ ′ ( s ) , r ~ ′ ′ ′ ( s ) ) = κ 3 ( 2 κ ˙ τ + κ τ ˙ ) − 3 κ 3 κ ˙ τ = κ 3 ( κ ˙ τ − κ τ ˙ ) , (\widetilde{\mathbf r}'(s),\widetilde{\mathbf r}''(s),\widetilde{\mathbf r}'''(s))=\kappa^3(2\dot{\kappa}\tau+\kappa\dot{\tau})-3\kappa^3\dot{\kappa}\tau=\kappa^3(\dot{\kappa}\tau-\kappa\dot{\tau}), (r ′(s),r ′′(s),r ′′′(s))=κ3(2κ˙τ+κτ˙)−3κ3κ˙τ=κ3(κ˙τ−κτ˙),
故有
τ ~ ( s ) = ( r ~ ′ ( s ) , r ~ ′ ′ ( s ) , r ~ ′ ′ ′ ( s ) ) ∣ r ~ ′ ( s ) ∧ r ~ ′ ′ ( s ) ∣ 2 = κ 3 ( κ ˙ τ − κ τ ˙ ) κ 4 ( κ 2 + τ 2 ) = κ ˙ τ − κ τ ˙ κ 2 κ + τ 2 κ = d d s ( τ κ ) κ ( 1 + ( τ κ ) 2 ) . \widetilde{\tau}(s)=\frac{(\widetilde{\mathbf{r}}'(s),\widetilde{\mathbf{r}}''(s),\widetilde{\mathbf{r}}'''(s))}{|\widetilde{\mathbf{r}}'(s)\wedge\widetilde{\mathbf{r}}''(s)|^2}=\frac{\kappa^3(\dot{\kappa}\tau-\kappa\dot{\tau})}{\kappa^4(\kappa^2+\tau^2)}=\frac{\frac{\dot{\kappa}\tau-\kappa\dot{\tau}}{\kappa^2}}{\kappa+\frac{\tau^2}{\kappa}}=\frac{\frac{d}{ds}(\frac{\tau}{\kappa})}{\kappa(1+(\frac{\tau}{\kappa})^2)}. τ (s)=∣r ′(s)∧r ′′(s)∣2(r ′(s),r ′′(s),r ′′′(s))=κ4(κ2+τ2)κ3(κ˙τ−κτ˙)=κ+κτ2κ2κ˙τ−κτ˙=κ(1+(κτ)2)dsd(κτ).
