目录
- T1
- T2
- T3
T1
(1)设 E 3 E^3 E3中曲线 C C C的所有切线过一个定点，证明 C C C是直线.
(2) 证明：所有主法线过定点的曲线是圆.
证明：(1) 设 P 0 P_0 P0是弧长参数曲线 C : r = r ( s ) C:\mathbf{r}=\mathbf{r}(s) C:r=r(s)的切线所过的定点，其位置向量
为 p 0 . \mathbf{p}_0. p0.
方法一：
由假设，
r ( s ) − p 0 = λ ( s ) t ( s ) , \mathbf r(s)-\mathbf p_0=\lambda(s)\mathbf t(s), r(s)−p0=λ(s)t(s),
其中 λ ( s ) = ⟨ r ( s ) − p 0 , t ( s ) ⟩ \lambda(s)=\langle\mathbf{r}(s)-\mathbf{p}_{0},\mathbf{t}(s)\rangle λ(s)=⟨r(s)−p0,t(s)⟩是一个光滑函数.对上式两边求导，有
t ( s ) = λ ( s ) ′ t ( s ) + λ ( s ) t ˙ ( s ) = λ ( s ) ′ t ( s ) + λ ( s ) κ ( s ) n ( s ) . \mathbf t(s)=\lambda(s)'\mathbf t(s)+\lambda(s)\dot{\mathbf t}(s)=\lambda(s)'\mathbf t(s)+\lambda(s)\kappa(s)\mathbf n(s). t(s)=λ(s)′t(s)+λ(s)t˙(s)=λ(s)′t(s)+λ(s)κ(s)n(s).
由于 t ( s ) , n ( s ) \mathbf{t}(s),\mathbf{n}(s) t(s),n(s)处处线性无关，有
λ ( s ) ′ = 1 , λ ( s ) κ ( s ) = 0. \lambda(s)'=1,\quad\lambda(s)\kappa(s)=0. λ(s)′=1,λ(s)κ(s)=0.
从 而 , λ ( s ) \lambda ( s) λ(s) 不 恒 为 0. 因 此 , 由 κ ( s ) \kappa ( s) κ(s) 的 连 续 性 , 知 κ ( s ) ≡ 0. \kappa ( s) \equiv 0. κ(s)≡0. 故 , C , C ,C 是 直 线
方 法 二 :
由 假 设 , r ( s ) − p 0 / / t ( s ) ⇔ ( r ( s ) − p 0 ) ∧ t ( s ) = 0. \mathbf{r} ( s) - \mathbf{p} _{0}/ / \mathbf{t} ( s) \Leftrightarrow ( \mathbf{r} ( s) - \mathbf{p} _{0}) \wedge \mathbf{t} ( s) = 0. r(s)−p0//t(s)⇔(r(s)−p0)∧t(s)=0. 若 p 0 \mathbf{p} _{0} p0 不 在 C C C上 , 则 由 习 题 一 第 2 题 ( 2 ) , C ( 2) , C (2),C 是 一 条 过 P 0 P_{0} P0 的 直 线 , 矛 盾 . 因 此 , P 0 , P_{0} ,P0在 C C C 上 . 设 r ( s 0 ) = p 0 . \mathbf{r} ( s_{0}) = \mathbf{p} _{0}. r(s0)=p0. 由 习 题 一 第 2 题 ( 2 ) , r ( s ) ( s > s 0 ) ( 2) , \mathbf{r} ( s) ( s> s_{0}) (2),r(s)(s>s0) 和 r ( s ) ( s < s 0 ) \mathbf{r} ( s) ( s< s_{0}) r(s)(s<s0) 是 两 条 射 线 . 而 r ( s ) \mathbf{r} ( s) r(s) 在 s = s 0 s= s_{0} s=s0 处
可微，即：只有一条切线，故 C C C 必然是一条直线.
E 3 E^3 E3中所有切线过一个定点的曲线是直线
(2)设弧长参数曲线 C : r = r ( s ) C:\mathbf{r}=\mathbf{r}(s) C:r=r(s)的主法线过定点 P 0 P_0 P0 (其位置向量为 p 0 ) . \mathbf{p}_0). p0).则
有(注意：假设了主法线存在，故曲率恒不为 0)
r ( s ) − p 0 = λ ( s ) n ( s ) , \mathbf r(s)-\mathbf p_0=\lambda(s)\mathbf n(s), r(s)−p0=λ(s)n(s),
其中 λ ( s ) = ⟨ r ( s ) − p 0 , n ( s ) ⟩ \lambda(s)=\langle\mathbf{r}(s)-\mathbf{p}_0,\mathbf{n}(s)\rangle λ(s)=⟨r(s)−p0,n(s)⟩是一个光滑函数.对上式两边求导并应用 Frenet 公
式，有
t ( s ) = λ ( s ) ′ n ( s ) − λ ( s ) κ ( s ) t ( s ) + λ ( s ) τ ( s ) b ( s ) . \mathbf{t}(s)=\lambda(s)'\mathbf{n}(s)-\lambda(s)\kappa(s)\mathbf{t}(s)+\lambda(s)\tau(s)\mathbf{b}(s). t(s)=λ(s)′n(s)−λ(s)κ(s)t(s)+λ(s)τ(s)b(s).
由于 t ( s ) , n ( s ) , b ( s ) \mathbf{t}(s),\mathbf{n}(s),\mathbf{b}(s) t(s),n(s),b(s)处处线性无关，有
1 − λ ( s ) κ ( s ) = 0 , λ ( s ) ′ = 0 , λ ( s ) τ ( s ) = 0. \begin{matrix}1-\lambda(s)\kappa(s)=0,&\lambda(s)'=0,&\lambda(s)\tau(s)=0.\end{matrix} 1−λ(s)κ(s)=0,λ(s)′=0,λ(s)τ(s)=0.
由 λ ( s ) ′ = 0 \lambda(s)^\prime=0 λ(s)′=0,知 λ ( s ) = λ \lambda(s)=\lambda λ(s)=λ是常数.而由 1 − λ ( s ) κ ( s ) = 0 1-\lambda(s)\kappa(s)=0 1−λ(s)κ(s)=0,有 λ ≠ 0 \lambda\neq0 λ=0且 κ ( s ) = 1 λ \kappa(s)=\frac1\lambda κ(s)=λ1 是常数.而 λ ( s ) τ ( s ) = 0 \lambda(s)\tau(s)=0 λ(s)τ(s)=0,故 τ ( s ) ≡ 0. \tau(s)\equiv0. τ(s)≡0.因此，由定理 3.1 (p.22), C C C是平面曲线.而由例2.2(p.17),知 C C C 是圆.
主法线过定点的曲线是圆
T2
设 T ( X ) = X T + P \mathcal{T}(X)=X\mathbf{T}+P T(X)=XT+P是 E 3 E^3 E3的一个合同变换，det T = − 1. r ( t ) \mathbf{T}=-1.\mathbf{r}(t) T=−1.r(t)是 E 3 E^3 E3的则曲线.求曲线 r ~ ( t ) = T ∘ r ( t ) \widetilde{\mathbf{r}}(t)=\mathcal{T}\circ\mathbf{r}(t) r (t)=T∘r(t)与曲线 r ( t ) \mathbf{r}(t) r(t)的弧长参数、曲率、挠率间的关系.
证明：设 s ~ \widetilde{s} s 是 r ~ ~ ( t ) \widetilde{\tilde{\mathbf{r}}}(t) r~ (t)的弧长参数.由于 T T T 保 持 距 离 并 且 r ~ ′ ( t ) = r ′ ( t ) T \widetilde{\mathbf{r} } ^\prime ( t) = \mathbf{r} ^\prime ( t) \mathbf{T} r ′(t)=r′(t)T,有
d s ~ d t = ∣ r ~ ′ ( t ) ∣ = ∣ r ′ ( t ) ∣ = d s d t . \frac{d\widetilde{s}}{dt}=|\widetilde{\mathbf{r}}'(t)|=|\mathbf{r}'(t)|=\frac{ds}{dt}. dtds =∣r ′(t)∣=∣r′(t)∣=dtds.
故 s ~ − s = \widetilde{s}-s= s −s=常数，即： r ( t ) \mathbf{r}(t) r(t)与 r ~ ( t ) \widetilde{\mathbf{r}}(t) r (t)的弧长参数相同.因此，下面将 s s s看作 r ~ ( t ) \widetilde{\mathbf{r}}(t) r (t)的弧
长参数. 弧长参数曲线 r ~ ( s ) _\text{弧长参数曲线 }\widetilde{\mathbf{r}}(s) 弧长参数曲线 r (s)的 Frenet 标架为
t ~ ( s ) = d r ~ ( s ) d s = d r ( s ) d s T = t T , \widetilde{\mathbf{t}}(s)=\frac{d\widetilde{\mathbf{r}}(s)}{ds}=\frac{d\mathbf{r}(s)}{ds}\mathbf{T}=\mathbf{tT}, t (s)=dsdr (s)=dsdr(s)T=tT,
d t ~ ( s ) d s = d t d s T , \frac{d\widetilde{\mathbf{t}}(s)}{ds}=\frac{d\mathbf{t}}{ds}\mathbf{T}, dsdt (s)=dsdtT,
因此， n ~ ( s ) = n ( s ) T \widetilde{\mathbf{n} } ( s) = \mathbf{n} ( s) \mathbf{T} n (s)=n(s)T, κ ~ ( s ) = κ ( s ) . \widetilde{\kappa } ( s) = \kappa ( s) . κ (s)=κ(s).由于 det T = − 1 \mathbf{T}=-1 T=−1,有
b ~ ( s ) = t ~ ( s ) ∧ n ~ ( s ) = ( t ( s ) T ) ∧ ( n ( s ) T ) = det T ( t ( s ) ∧ n ( s ) ) T = − b ( s ) T . \widetilde{\mathbf{b}}(s)=\widetilde{\mathbf{t}}(s)\wedge\widetilde{\mathbf{n}}(s)=(\mathbf{t}(s)\mathbf{T})\wedge(\mathbf{n}(s)\mathbf{T})=\det\mathbf{T}(\mathbf{t}(s)\wedge\mathbf{n}(s))\mathbf{T}=-\mathbf{b}(s)\mathbf{T}. b (s)=t (s)∧n (s)=(t(s)T)∧(n(s)T)=detT(t(s)∧n(s))T=−b(s)T.
τ ~ ( s ) = − ⟨ b ~ ˙ ( s ) , n ~ ( s ) ⟩ = − ⟨ − b ˙ ( s ) T , n ( s ) T ⟩ = ⟨ b ˙ ( s ) , n ( s ) ⟩ = − τ ( s ) . \widetilde{\tau}(s)=-\langle\dot{\widetilde{\mathbf{b}}}(s),\widetilde{\mathbf{n}}(s)\rangle=-\langle-\dot{\mathbf{b}}(s)\mathbf{T},\mathbf{n}(s)\mathbf{T}\rangle=\langle\dot{\mathbf{b}}(s),\mathbf{n}(s)\rangle=-\tau(s). τ (s)=−⟨b ˙(s),n (s)⟩=−⟨−b˙(s)T,n(s)T⟩=⟨b˙(s),n(s)⟩=−τ(s).
换回参数 t t t,有 κ ~ ( t ) = κ ( t ) , τ ~ ( t ) = − τ ( t ) . \widetilde\kappa(t)=\kappa(t),\widetilde\tau(t)=-\tau(t). κ (t)=κ(t),τ (t)=−τ(t).
T3
(1)设 r ( t ) (t) (t)是平面曲线，曲率为 κ ( t ) \kappa(t) κ(t),求曲线 r ~ ( t ) = r ( − t ) \widetilde{\mathbf{r}}(t)=\mathbf{r}(-t) r (t)=r(−t)的曲率；
( 2 ) 当 r ( t ) (2)当\mathbf{r}(\mathbf{t}) (2)当r(t) 是 E 3 E^3 E3 的曲线时，求曲线 r ~ ( t ) = r ( − t ) \widetilde{\mathbf{r}}(t)=\mathbf{r}(-t) r (t)=r(−t) 的曲率和挠率.
解：(1)设 r ( t ) = ( x ( t ) , y ( t ) ) \mathbf r(t)=(x(t),y(t)) r(t)=(x(t),y(t)),则 r ~ ( t ) = ( x ~ ( t ) , y ~ ( t ) ) = ( − x ( − t ) , − y ( − t ) ) \widetilde{\mathbf{r}}(t)=(\widetilde{x}(t),\widetilde{y}(t))=(-x(-t),-y(-t)) r (t)=(x (t),y (t))=(−x(−t),−y(−t))从而，
r ~ ′ ( t ) = ( x ~ ′ ( t ) , y ~ ′ ( t ) ) = ( − x ′ ( − t ) , − y ′ ( − t ) ) , \widetilde{\mathbf{r}}^{\prime}(t)=(\widetilde{x}^{\prime}(t),\widetilde{y}^{\prime}(t))=(-x^{\prime}(-t),-y^{\prime}(-t)), r ′(t)=(x ′(t),y ′(t))=(−x′(−t),−y′(−t)),
r ~ ′ ′ ( t ) = ( x ~ ′ ′ ( t ) , y ~ ′ ′ ( t ) ) = ( x ′ ′ ( − t ) , y ′ ′ ( − t ) ) . \widetilde{\mathbf{r}}''(t)=(\widetilde{x}''(t),\widetilde{y}''(t))=(x''(-t),y''(-t)). r ′′(t)=(x ′′(t),y ′′(t))=(x′′(−t),y′′(−t)).
κ ~ ( t ) = x ~ ′ ( t ) y ~ ′ ′ ( t ) − x ~ ′ ′ ( t ) y ~ ′ ( t ) ( x ~ ′ ( t ) 2 + y ~ ′ ( t ) 2 ) 3 2 = − x ′ ( − t ) y ′ ′ ( − t ) − x ′ ′ ( − t ) y ′ ( − t ) ( x ′ ( − t ) 2 + y ′ ( − t ) 2 ) 3 2 = − κ ( − t ) . \widetilde{\kappa}(t)=\frac{\widetilde{x}'(t)\widetilde{y}''(t)-\widetilde{x}''(t)\widetilde{y}'(t)}{(\widetilde{x}'(t)^2+\widetilde{y}'(t)^2)^{\frac{3}{2}}}=-\frac{x'(-t)y''(-t)-x''(-t)y'(-t)}{(x'(-t)^2+y'(-t)^2)^{\frac{3}{2}}}=-\kappa(-t). κ (t)=(x ′(t)2+y ′(t)2)23x ′(t)y ′′(t)−x ′′(t)y ′(t)=−(x′(−t)2+y′(−t)2)23x′(−t)y′′(−t)−x′′(−t)y′(−t)=−κ(−t).
(2) 由 于 r ~ ′ ( t ) = − r ′ ( − t ) , r ~ ′ ′ ( t ) = r ′ ′ ( − t ) , r ~ ′ ′ ′ ( t ) = − r ′ ′ ′ ( − t ) \widetilde{\mathbf{r} } ^{\prime }( t) = - \mathbf{r} ^{\prime }( - t) , \widetilde{\mathbf{r} } ^{\prime \prime }( t) = \mathbf{r} ^{\prime \prime }( - t) , \widetilde{\mathbf{r} } ^{\prime \prime \prime }( t) = - \mathbf{r} ^{\prime \prime \prime }( - t) r ′(t)=−r′(−t),r ′′(t)=r′′(−t),r ′′′(t)=−r′′′(−t),则
κ ~ ( t ) = ∣ r ~ ′ ( t ) ∧ r ~ ′ ′ ( t ) ∣ ∣ r ~ ′ ( t ) ∣ 3 = ∣ − r ′ ( − t ) ∧ r ′ ′ ( − t ) ∣ ∣ − r ′ ( − t ) ∣ 3 = ∣ r ′ ( − t ) ∧ r ′ ′ ( − t ) ∣ ∣ r ′ ( − t ) ∣ 3 = κ ( − t ) , \widetilde{\kappa}(t)=\frac{|\widetilde{\mathbf{r}}'(t)\wedge\widetilde{\mathbf{r}}''(t)|}{|\widetilde{\mathbf{r}}'(t)|^3}=\frac{|-\mathbf{r}'(-t)\wedge\mathbf{r}''(-t)|}{|-\mathbf{r}'(-t)|^3}=\frac{|\mathbf{r}'(-t)\wedge\mathbf{r}''(-t)|}{|\mathbf{r}'(-t)|^3}=\kappa(-t), κ (t)=∣r ′(t)∣3∣r ′(t)∧r ′′(t)∣=∣−r′(−t)∣3∣−r′(−t)∧r′′(−t)∣=∣r′(−t)∣3∣r′(−t)∧r′′(−t)∣=κ(−t),
τ ~ ( t ) = ( r ~ ′ , r ~ ′ ′ , r ~ ′ ′ ′ ) ∣ r ~ ′ ( t ) ∧ r ~ ′ ′ ( t ) ∣ 2 = ( − r ′ ( − t ) , r ′ ′ ( − t ) , − r ′ ′ ′ ( − t ) ) ∣ − r ′ ( − t ) ∧ r ′ ′ ( − t ) ∣ 2 = ( r ′ ( − t ) , r ′ ′ ( − t ) , r ′ ′ ′ ( − t ) ) ∣ − r ′ ( − t ) ∧ r ′ ′ ( − t ) ∣ 2 = τ ( − t ) . \widetilde{\tau}(t)=\frac{(\widetilde{\mathbf{r}}^{\prime},\widetilde{\mathbf{r}}^{\prime\prime},\widetilde{\mathbf{r}}^{\prime\prime\prime})}{|\widetilde{\mathbf{r}}^{\prime}(t)\wedge\widetilde{\mathbf{r}}^{\prime\prime}(t)|^{2}} =\frac{(-\mathbf{r}^{\prime}(-t),\mathbf{r}^{\prime\prime}(-t),-\mathbf{r}^{\prime\prime\prime}(-t))}{|-\mathbf{r}^{\prime}(-t)\wedge\mathbf{r}^{\prime\prime}(-t)|^{2}}\\=\frac{(\mathbf{r}^{\prime}(-t),\mathbf{r}^{\prime\prime}(-t),\mathbf{r}^{\prime\prime\prime}(-t))}{|-\mathbf{r}^{\prime}(-t)\wedge\mathbf{r}^{\prime\prime}(-t)|^{2}}=\tau(-t). τ (t)=∣r ′(t)∧r ′′(t)∣2(r ′,r ′′,r ′′′)=∣−r′(−t)∧r′′(−t)∣2(−r′(−t),r′′(−t),−r′′′(−t))=∣−r′(−t)∧r′′(−t)∣2(r′(−t),r′′(−t),r′′′(−t))=τ(−t).
