曲线的弧长与曲率
目录
- T1
- T2
- T3
- T4
- T5
- T6
T1
求 下 列 曲 线 的 弧 长 与 曲 率 :
(1) y = a x 2 ; {\:y}=ax^{2}; y=ax2;
解：
参数表达式
r
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\mathbf{r}(t)=(t,at^2)(t\in\mathbb{R}).
r(t)=(t,at2)(t∈R). 直接计算，有
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\mathbf{r}'(t)=(1,2at),\quad\mathbf{r}''(t)=(0,2a),\quad|\mathbf{r}'(t)|=\sqrt{1+4a^2t^2}.
r′(t)=(1,2at),r′′(t)=(0,2a),∣r′(t)∣=1+4a2t2.因此，弧长(作为
t
t
t 的函数)为(注意
a
≠
0.
)
a\neq0.)
a=0.)
s
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∫
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1
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∣
a
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log
∣
2
∣
a
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4
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∣
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s=\int_0^t|\mathbf{r}^{\prime}(u)|du=\int_0^t\sqrt{1+4a^2u^2}du=\frac12t\sqrt{1+4a^2t^2}+\frac1{4|a|}\log|2|a|t+\sqrt{1+4a^2t^2}|.
s=∫0t∣r′(u)∣du=∫0t1+4a2u2du=21t1+4a2t2+4∣a∣1log∣2∣a∣t+1+4a2t2∣.
积分运算过程：
令 tan θ = 2 ∣ a ∣ u \tan\theta=2|a|u tanθ=2∣a∣u.则 1 + 4 a 2 u 2 = sec θ \sqrt1+4a^2u^2=\sec\theta 1+4a2u2=secθ.从而 , ∫ 1 + 4 a 2 u 2 d u = 1 2 ∣ a ∣ ∫ sec 3 θ d θ . ,\int\sqrt{1+4a^2u^2}du=\frac1{2|a|}\int\sec^3\theta d\theta. ,∫1+4a2u2du=2∣a∣1∫sec3θdθ.
I : = ∫ sec 3 θ d θ = ∫ ( sec θ tan 2 θ + sec θ ) d θ = ∫ tan θ d ( sec θ ) + sec θ d θ = tan θ sec θ − ∫ sec 3 θ d θ + ∫ sec θ d θ = tan θ sec θ − I + log ∣ sec θ + tan θ ∣ = 1 2 ( tan θ sec θ + log ∣ sec θ + tan θ ∣ ) + C = 1 2 ( 2 ∣ a ∣ u 1 + 4 a 2 u 2 + log ∣ 2 ∣ a ∣ u + 1 + 4 a 2 u 2 ∣ ) + C . \begin{aligned} I:=\int\sec^3\theta d\theta &=\int(\sec\theta\tan^2\theta+\sec\theta)d\theta\\ &=\int\tan\theta d(\sec\theta)+\sec\theta d\theta \\ &=\tan\theta\sec\theta-\int\sec^3\theta d\theta+\int\sec\theta d\theta \\ &=\tan\theta\sec\theta-I+\log|\sec\theta+\tan\theta|\\ &=\frac12(\tan\theta\sec\theta+\log|\sec\theta+\tan\theta|)+C \\ &=\frac12(2|a|u\sqrt{1+4a^2u^2}+\log|2|a|u+\sqrt{1+4a^2u^2}|)+C.\\\end{aligned} I:=∫sec3θdθ=∫(secθtan2θ+secθ)dθ=∫tanθd(secθ)+secθdθ=tanθsecθ−∫sec3θdθ+∫secθdθ=tanθsecθ−I+log∣secθ+tanθ∣=21(tanθsecθ+log∣secθ+tanθ∣)+C=21(2∣a∣u1+4a2u2+log∣2∣a∣u+1+4a2u2∣)+C.
则有 ∫ 1 + 4 a 2 u 2 d u = 1 2 u 1 + 4 a 2 u 2 + 1 4 ∣ a ∣ log ∣ 2 ∣ a ∣ u + 1 + 4 a 2 u 2 ∣ + C . \\\int\sqrt{1+4a^2u^2}du=\frac12u\sqrt{1+4a^2u^2}+\frac1{4|a|}\log|2|a|u+\sqrt{1+4a^2u^2}|+C. ∫1+4a2u2du=21u1+4a2u2+4∣a∣1log∣2∣a∣u+1+4a2u2∣+C.
从而得到曲率
κ
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2
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a
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\kappa(t)=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^{\frac32}}=\frac{2a}{(1+4a^2t^2)^{\frac32}}.
κ(t)=(x′(t)2+y′(t)2)23x′(t)y′′(t)−x′′(t)y′(t)=(1+4a2t2)232a.
T2
求 下 列 曲 线 的 弧 长 与 曲 率 :
x
2
a
2
+
y
2
b
2
=
1
;
\frac {x^2}{a^2}+ \frac {y^2}{b^2}= 1;
a2x2+b2y2=1;
解：
不妨设
a
>
0
a>0
a>0且
b
>
0.
b>0.
b>0.则椭圆曲线(去掉点
(
a
,
0
)
)
(a,0))
(a,0))的参数表达式为
r ( t ) = ( a cos t , b sin t ) ( 0 < t < 2 π ) . \mathbf{r}(t)=(a\cos t,b\sin t)(0<t<2\pi). r(t)=(acost,bsint)(0<t<2π).
直接计算，有
r ′ ( t ) = ( − a sin t , b cos t ) , r ′ ′ ( t ) = ( − a cos t , − b sin t ) . \mathbf{r}'(t)=(-a\sin t,b\cos t),\quad\mathbf{r}''(t)=(-a\cos t,-b\sin t). r′(t)=(−asint,bcost),r′′(t)=(−acost,−bsint).
弧长为
s = ∫ 0 t a 2 sin 2 u + b 2 cos 2 u d u = { a t , a = b ; 第一类椭圆积分 , a ≠ b . s=\int_0^t\sqrt{a^2\sin^2u+b^2\cos^2u}du=\left\{\begin{array}{cc}at,&a=b;\\\text{第一类椭圆积分},&a\neq b.\end{array}\right. s=∫0ta2sin2u+b2cos2udu={at,第一类椭圆积分,a=b;a=b.
曲率
κ ( t ) = x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3 2 = a b ( a 2 sin 2 t + b 2 cos 2 t ) 3 2 . \kappa(t)=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^{\frac32}}=\frac{ab}{(a^2\sin^2t+b^2\cos^2t)^{\frac32}}. κ(t)=(x′(t)2+y′(t)2)23x′(t)y′′(t)−x′′(t)y′(t)=(a2sin2t+b2cos2t)23ab.
T3
求 下 列 曲 线 的 弧 长 与 曲 率 :
r
(
t
)
=
(
a
cosh
t
,
b
sinh
t
)
(
t
∈
R
)
;
\mathbf{r} ( t) = ( a\cosh t, b\sinh t) ( t\in \mathbb{R} ) ;
r(t)=(acosht,bsinht)(t∈R);
解：
直接计算
r
′
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t
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=
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a
sinh
t
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b
cosh
t
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r
′
′
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t
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=
(
a
cosh
t
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b
sinh
t
)
\mathbf{r}^{\prime}(t)=(a\sinh t,b\cosh t),\quad\mathbf{r}^{\prime\prime}(t)=(a\cosh t,b\sinh t)
r′(t)=(asinht,bcosht),r′′(t)=(acosht,bsinht)
弧长
s
=
∫
0
t
a
2
sinh
2
u
+
b
2
cosh
2
u
d
u
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s=\int_0^t\sqrt{a^2\sinh^2u+b^2\cosh^2u}du.
s=∫0ta2sinh2u+b2cosh2udu.
曲率
κ ( t ) = x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3 2 = − a b ( a 2 sinh 2 t + b 2 cosh 2 t ) 3 2 . \kappa(t)=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^{\frac{3}{2}}}=-\frac{ab}{(a^2\sinh^2t+b^2\cosh^2t)^{\frac{3}{2}}}. κ(t)=(x′(t)2+y′(t)2)23x′(t)y′′(t)−x′′(t)y′(t)=−(a2sinh2t+b2cosh2t)23ab.
T4
求 下 列 曲 线 的 弧 长 与 曲 率 :
r
(
t
)
=
(
t
,
a
cosh
t
a
)
\mathbf{r} ( t) = ( t, a\cosh \frac ta)
r(t)=(t,acoshat)
(
a
>
0
)
(
t
∈
R
)
.
( a> 0) ( t\in \mathbb{R} ) .
(a>0)(t∈R).
解：
直接计算
r
′
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t
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=
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1
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sinh
t
a
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r
′
′
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=
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0
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a
cosh
t
a
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\mathbf{r}'(t)=(1,\sinh\frac ta),\quad\mathbf{r}''(t)=(0,\frac1a\cosh\frac ta).
r′(t)=(1,sinhat),r′′(t)=(0,a1coshat).
弧长
s
=
∫
0
t
1
+
sinh
2
u
a
d
u
=
∫
0
t
cosh
u
a
d
u
=
a
sinh
t
a
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s=\int_0^t\sqrt{1+\sinh^2\frac ua}du=\int_0^t\cosh\frac uadu=a\sinh\frac ta.
s=∫0t1+sinh2audu=∫0tcoshaudu=asinhat.
曲率
κ ( t ) = x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3 2 = cosh t a a ( 1 + sinh 2 t a ) 3 2 . \kappa(t)=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^{\frac32}}=\frac{\cosh\frac ta}{a(1+\sinh^2\frac ta)^{\frac32}}. κ(t)=(x′(t)2+y′(t)2)23x′(t)y′′(t)−x′′(t)y′(t)=a(1+sinh2at)23coshat.
T5
设曲线
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\mathbf{r}(t)=(x(t),y(t))
r(t)=(x(t),y(t)),证明它的曲率是
κ
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\kappa(t)=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^{\frac32}}.
κ(t)=(x′(t)2+y′(t)2)23x′(t)y′′(t)−x′′(t)y′(t).
证明：首先，
r ′ ( t ) = ( x ′ ( t ) , y ′ ( t ) ) , r ′ ′ ( t ) = ( x ′ ′ ( t ) , y ′ ′ ( t ) ) . \mathbf{r}'(t)=(x'(t),y'(t)),\quad\mathbf{r}''(t)=(x''(t),y''(t)). r′(t)=(x′(t),y′(t)),r′′(t)=(x′′(t),y′′(t)).
设 s s s是曲线 r ( t ) (t) (t)的弧长参数.则 s = s ( t ) s=s(t) s=s(t)与 t = t ( s ) t=t(s) t=t(s)互为反函数.
d s d t = ∣ r ′ ( t ) ∣ = x ′ ( t ) 2 + y ′ ( t ) 2 , d t d s = 1 ∣ r ′ ( t ) ∣ = 1 x ′ ( t ) 2 + y ′ ( t ) 2 , \dfrac{ds}{dt}=|\mathbf{r}'(t)|=\sqrt{x'(t)^2+y'(t)^2},\:\dfrac{dt}{ds}=\dfrac{1}{|\mathbf{r}'(t)|}=\dfrac{1}{\sqrt{x'(t)^2+y'(t)^2}}, dtds=∣r′(t)∣=x′(t)2+y′(t)2,dsdt=∣r′(t)∣1=x′(t)2+y′(t)21,
d 2 t d s 2 = d d t ( d t d s ) d t d s = − x ′ ( t ) x ′ ′ ( t ) + y ′ ( t ) y ′ ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 2 . \frac{d^2t}{ds^2}=\frac d{dt}(\frac{dt}{ds})\frac{dt}{ds}=-\frac{x'(t)x''(t)+y'(t)y''(t)}{(x'(t)^2+y'(t)^2)^2}. ds2d2t=dtd(dsdt)dsdt=−(x′(t)2+y′(t)2)2x′(t)x′′(t)+y′(t)y′′(t).
由于平面曲线的 Frenet 标架和曲率与(同向的容许)参数选择无关，故
t ( t ) : = t ( s ( t ) ) = d r ( t ) d t d t d s = ( x ′ ( t ) x ′ ( t ) 2 + y ′ ( t ) 2 , y ′ ( t ) x ′ ( t ) 2 + y ′ ( t ) 2 ) . \mathbf{t}(t):=\mathbf{t}(s(t))=\frac{d\mathbf{r}(t)}{dt}\frac{dt}{ds}=(\frac{x'(t)}{\sqrt{x'(t)^2+y'(t)^2}},\frac{y'(t)}{\sqrt{x'(t)^2+y'(t)^2}}). t(t):=t(s(t))=dtdr(t)dsdt=(x′(t)2+y′(t)2x′(t),x′(t)2+y′(t)2y′(t)).
从而，
t ˙ ( s ( t ) ) = r ′ ′ ( t ) ( d t d s ) 2 + r ′ ( t ) d 2 t d s 2 = ( − y ′ ( t ) ( x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 2 , x ′ ( t ) ( x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 2 ) n ( t ) : = n ( s ( t ) ) = ( − y ′ ( t ) x ′ ( t ) 2 + y ′ ( t ) 2 , x ′ ( t ) x ′ ( t ) 2 + y ′ ( t ) 2 ) . \dot{\mathbf{t}}(s(t))=\mathbf{r}''(t)(\frac{dt}{ds})^2+\mathbf{r}'(t)\frac{d^2t}{ds^2}=(-\frac{y'(t)(x'(t)y''(t)-x''(t)y'(t))}{(x'(t)^2+y'(t)^2)^2},\frac{x'(t)(x'(t)y''(t)-x''(t)y'(t))}{(x'(t)^2+y'(t)^2)^2})\\\mathbf{n}(t):=\mathbf{n}(s(t))=(-\frac{y'(t)}{\sqrt{x'(t)^2+y'(t)^2}},\frac{x'(t)}{\sqrt{x'(t)^2+y'(t)^2}}). t˙(s(t))=r′′(t)(dsdt)2+r′(t)ds2d2t=(−(x′(t)2+y′(t)2)2y′(t)(x′(t)y′′(t)−x′′(t)y′(t)),(x′(t)2+y′(t)2)2x′(t)(x′(t)y′′(t)−x′′(t)y′(t)))n(t):=n(s(t))=(−x′(t)2+y′(t)2y′(t),x′(t)2+y′(t)2x′(t)).
所以，
κ ( t ) = κ ( s ( t ) ) = ⟨ t ˙ ( s ( t ) ) , n ( s ( t ) ) ⟩ = x ′ ( t ) y ′ ′ ( t ) − x ′ ′ ( t ) y ′ ( t ) ( x ′ ( t ) 2 + y ′ ( t ) 2 ) 3 2 . \kappa(t)=\kappa(s(t))=\langle\dot{\mathbf{t}}(s(t)),\mathbf{n}(s(t))\rangle=\frac{x'(t)y''(t)-x''(t)y'(t)}{(x'(t)^2+y'(t)^2)^{\frac32}}. κ(t)=κ(s(t))=⟨t˙(s(t)),n(s(t))⟩=(x′(t)2+y′(t)2)23x′(t)y′′(t)−x′′(t)y′(t).
t
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\mathbf{t}(t)=\frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}=(\frac{x'(t)}{\sqrt{x'(t)^2+y'(t)^2}},\frac{y'(t)}{\sqrt{x'(t)^2+y'(t)^2}}),
t(t)=∣r′(t)∣r′(t)=(x′(t)2+y′(t)2x′(t),x′(t)2+y′(t)2y′(t)),
n
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=
=
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−
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x
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+
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\mathbf{n}(t)==\frac{\mathbf{r}''(t)}{|\mathbf{r}''(t)|}=(-\frac{y'(t)}{\sqrt{x'(t)^2+y'(t)^2}},\frac{x'(t)}{\sqrt{x'(t)^2+y'(t)^2}}).
n(t)==∣r′′(t)∣r′′(t)=(−x′(t)2+y′(t)2y′(t),x′(t)2+y′(t)2x′(t)).
T6
设曲线 C C C在极坐标 ( r , θ ) (r,\theta) (r,θ)下的表示为 r = f ( θ ) r=f(\theta) r=f(θ),证明 C C C的曲率是
κ ( θ ) = f 2 ( θ ) + 2 ( d f d θ ) 2 − f ( θ ) d 2 f d θ 2 ( f 2 ( θ ) + ( d f d θ ) 2 ) 3 2 \kappa(\theta)=\frac{f^2(\theta)+2(\frac{df}{d\theta})^2-f(\theta)\frac{d^2f}{d\theta^2}}{(f^2(\theta)+(\frac{df}{d\theta})^2)^{\frac32}} κ(θ)=(f2(θ)+(dθdf)2)23f2(θ)+2(dθdf)2−f(θ)dθ2d2f
证明：
曲线
C
C
C有参数表示式
r
(
θ
)
=
(
f
(
θ
)
cos
θ
,
f
(
θ
)
sin
θ
)
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\mathbf{r}(\theta)=(f(\theta)\cos\theta,f(\theta)\sin\theta).
r(θ)=(f(θ)cosθ,f(θ)sinθ).直接计算，有
r
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=
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cos
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f
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sin
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f
′
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sin
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+
f
(
θ
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cos
θ
\mathbf{r}'(\theta)=(f'(\theta)\cos\theta-f(\theta)\sin\theta,f'(\theta)\sin\theta+f(\theta)\cos\theta
r′(θ)=(f′(θ)cosθ−f(θ)sinθ,f′(θ)sinθ+f(θ)cosθ
r ′ ′ ( θ ) = ( f ′ ′ ( θ ) cos θ − 2 f ′ ( θ ) sin θ − f ( θ ) cos θ , f ′ ′ ( θ ) sin θ + 2 f ′ ( θ ) cos θ − f ( θ ) sin θ ) . \mathbf{r}^{\prime\prime}(\theta)=(f^{\prime\prime}(\theta)\cos\theta-2f^{\prime}(\theta)\sin\theta-f(\theta)\cos\theta,f^{\prime\prime}(\theta)\sin\theta+2f^{\prime}(\theta)\cos\theta-f(\theta)\sin\theta). r′′(θ)=(f′′(θ)cosθ−2f′(θ)sinθ−f(θ)cosθ,f′′(θ)sinθ+2f′(θ)cosθ−f(θ)sinθ).
从而
κ ( θ ) = x ′ ( θ ) y ′ ′ ( θ ) − x ′ ′ ( θ ) y ′ ( θ ) ( x ′ ( θ ) 2 + y ′ ( θ ) 2 ) 3 2 = f 2 ( θ ) + 2 f ′ ( θ ) 2 − f ( θ ) f ′ ′ ( θ ) ( f 2 ( θ ) + ( f ′ ( θ ) ) 2 ) 3 2 . \kappa(\theta)=\frac{x'(\theta)y''(\theta)-x''(\theta)y'(\theta)}{(x'(\theta)^2+y'(\theta)^2)^{\frac32}}=\frac{f^2(\theta)+2f'(\theta)^2-f(\theta)f''(\theta)}{(f^2(\theta)+(f'(\theta))^2)^{\frac32}}. κ(θ)=(x′(θ)2+y′(θ)2)23x′(θ)y′′(θ)−x′′(θ)y′(θ)=(f2(θ)+(f′(θ))2)23f2(θ)+2f′(θ)2−f(θ)f′′(θ).