Kreivio skaičiavimas be evoliutės ir evolventės neturi prasmės. Kreivis susijęs su jų spinduliu. Rusiškai kreivis vadinasi кривизна .
Vaizdas:Kreivispav139.jpg 139 pav. Vaizdas:Kreivispav140.jpg 140 pav. Vienas iš elementų, charakterizuojančių formą kreivės, yra laipsnis jos išlinkimo, išsilenkimo.
Tegu mes turime kreivę, kuri nekerta pati savęs ir turi tam tikrą liestinę kiekviename taške. Pravesime liestines prie kreivės kokiuose nors dviejuose jos taškuose A ir B ir pažymėsime per
α
{\displaystyle \alpha }
A į tašką B (139 pav.). Šitas kampas vadinasi kreivumo kampu (углом смежности ) lanko AB . Pas du lankus, turinčius vienodą ilgį, daugiau išlinkęs tas lankas, pas kurią kreivumo kampas didesnis (139 ir 140 pav.).
Iš kitos pusės, nagrinėjant lankus skirtingo ilgio, mes negalime įvertinti laipsnį jų išlinkimo tiktai atitinkančiu kreivumo kampu. Iš čia seka, kad pilna charakteristika išlinkimo kreivės bus santykis kreivumo kampo su ilgiu atitinkančios tiesės.
Apibrėžimas 1. Vidutiniu kreiviu
K
v
i
d
{\displaystyle K_{vid}}
A
B
˘
{\displaystyle {\breve {AB}}}
α
{\displaystyle \alpha }
K
v
i
d
=
α
A
B
˘
.
{\displaystyle K_{vid}={\frac {\alpha }{\breve {AB}}}.}
Vaizdas:Kreivispav141.jpg 141 pav. Vienai ir tai pačiai kreivei vidutinis kreivis jos skirtingų dalių (lankų) gali būti skirtingas; pavyzdžiui, kreivei parodytai paveiksle 141, vidutinis kreivis lanko
A
B
˘
{\displaystyle {\breve {AB}}}
A
1
B
1
˘
,
{\displaystyle {\breve {A_{1}B_{1}}},}
A , įvesime apibrėžimą kreivio kreivės duotame taške.
Apibrėžimas 2. Kreiviu
K
A
{\displaystyle K_{A}}
A vadinasi riba vidutinio kreivio lanko
A
B
˘
,
{\displaystyle {\breve {AB}},}
B artėja (mes tariam, kad dydis ribos nepriklauso nuo to, iš kokios pusės nuo taško A mes imame kintamą tašką B ant kreivės) prie taško A ):
K
A
=
lim
B
→
A
K
v
i
d
=
lim
A
B
→
0
=
α
A
B
˘
.
{\displaystyle K_{A}=\lim _{B\to A}K_{vid}=\lim _{AB\to 0}={\frac {\alpha }{\breve {AB}}}.}
Pastaba . Pažymėsime, kad betkokiai kreivei kreivis skirtingose jos taškuose, bus skirtingas. Tai mes pamatysime žemiau.Vaizdas:Kreivispav142.jpg 142 pav. Apskritimui spindulio r : 1) nustatyti vidutinį kreivį lanko AB , atitinkantį centriam kamui
α
{\displaystyle \alpha }
A . Sprendimas . 1) Akivaizdu, kad kreivumo kampas lanko
A
B
˘
{\displaystyle {\breve {AB}}}
α
{\displaystyle \alpha }
α
r
{\displaystyle \alpha r}
K
v
i
d
=
α
α
r
,
{\displaystyle K_{vid}={\frac {\alpha }{\alpha r}},}
arba
K
v
i
d
=
1
r
.
{\displaystyle K_{vid}={\frac {1}{r}}.}
2) Kreivis taške A lygus
K
=
lim
α
→
0
α
α
r
=
1
r
.
{\displaystyle K=\lim _{\alpha \to 0}{\frac {\alpha }{\alpha r}}={\frac {1}{r}}.}
Tokiu budu, vidutinis kreivis lanko apskritimo spindulio r nepriklauso nuo ilgio ir padeties lanko, visiems lankams jis lygus
1
r
.
{\displaystyle {\frac {1}{r}}.}
1
r
.
{\displaystyle {\frac {1}{r}}.}
Įvesime formulę apskaičiavimui kreivio duotos linijos betkokiame jos taške
M
(
x
;
y
)
{\displaystyle M(x;y)}
y
=
f
(
x
)
(
1
)
{\displaystyle y=f(x)\quad (1)}
ir kad funkcija
f
(
x
)
{\displaystyle f(x)\;}
Vaizdas:Kreivispav143.jpg 143 pav. Pravesime liestines kreivės taškuose M ir
M
1
{\displaystyle M_{1}}
x ir
x
+
Δ
x
{\displaystyle x+\Delta x}
ϕ
{\displaystyle \phi }
ϕ
+
Δ
ϕ
{\displaystyle \phi +\Delta \phi }
Ilgį kreivės
M
0
M
˘
,
{\displaystyle {\breve {M_{0}M}},}
M
0
{\displaystyle M_{0}}
s ; tada
Δ
s
=
M
0
M
1
˘
−
M
0
M
˘
,
{\displaystyle \Delta s={\breve {M_{0}M_{1}}}-{\breve {M_{0}M}},}
|
Δ
s
|
=
M
M
1
˘
.
{\displaystyle |\Delta s|={\breve {MM_{1}}}.}
Kaip betarpiškai matosi iš pav. 143, gretimumo kampas, atitinkantis lankui
M
M
1
˘
{\displaystyle {\breve {MM_{1}}}}
|
Δ
ϕ
|
=
Δ
ϕ
,
{\displaystyle |\Delta \phi |=\Delta \phi ,}
Δ
ϕ
>
0
{\displaystyle \Delta \phi >0}
ϕ
{\displaystyle \phi }
ϕ
+
Δ
ϕ
,
{\displaystyle \phi +\Delta \phi ,}
|
Δ
ϕ
|
.
{\displaystyle |\Delta \phi |.}
Pagal apibrėžimą vidutinio kreivio kreivės srityje
M
M
1
{\displaystyle MM_{1}}
K
v
i
d
=
|
Δ
ϕ
|
|
Δ
s
|
=
|
Δ
ϕ
Δ
s
|
.
{\displaystyle K_{vid}={\frac {|\Delta \phi |}{|\Delta s|}}=\left|{\frac {\Delta \phi }{\Delta s}}\right|.}
Kad gauti kreivį taške M , reikia rasti ribą gautos išraiškos su sąlyga, kad ilgis lanko
M
M
1
˘
{\displaystyle {\breve {MM_{1}}}}
K
=
lim
Δ
s
→
0
=
|
Δ
ϕ
Δ
s
|
.
{\displaystyle K=\lim _{\Delta s\to 0}=\left|{\frac {\Delta \phi }{\Delta s}}\right|.}
Kadangi dydžiai
ϕ
{\displaystyle \phi }
s abu priklauso nuo x (yra funkcijos nuo x ), tai, dėl to,
ϕ
{\displaystyle \phi }
s . Mes galime laikyti, kad šita funkcija užduota parametriškai su x parametro pagalba. Tada
lim
Δ
s
→
0
=
Δ
ϕ
Δ
s
=
d
ϕ
d
s
{\displaystyle \lim _{\Delta s\to 0}={\frac {\Delta \phi }{\Delta s}}={\frac {d\phi }{ds}}}
ir, todėl,
K
=
|
d
ϕ
d
s
|
(
2
)
.
{\displaystyle K=\left|{\frac {d\phi }{ds}}\right|\quad (2).}
Skaičiavimui
d
ϕ
d
s
{\displaystyle {\frac {d\phi }{ds}}}
d
ϕ
d
s
=
d
ϕ
d
x
d
s
d
x
.
{\displaystyle {\frac {d\phi }{ds}}={\frac {\frac {d\phi }{dx}}{\frac {ds}{dx}}}.}
Kad išreikšti išvestinę
d
ϕ
d
x
{\displaystyle {\frac {d\phi }{dx}}}
y
=
f
(
x
)
,
{\displaystyle y=f(x),\;}
tan
ϕ
=
d
y
d
x
{\displaystyle \tan \phi ={\frac {dy}{dx}}}
ϕ
=
arctan
d
y
d
x
.
{\displaystyle \phi =\arctan {\frac {dy}{dx}}.}
Diferencijuodami pagal x paskutinę lygybę, turėsime:
d
ϕ
d
x
=
d
(
arctan
d
y
d
x
)
d
x
=
d
2
y
d
x
2
1
+
(
d
y
d
x
)
2
.
{\displaystyle {\frac {d\phi }{dx}}={\frac {d(\arctan {\frac {dy}{dx}})}{dx}}={\frac {\frac {d^{2}y}{dx^{2}}}{1+\left({\frac {dy}{dx}}\right)^{2}}}.}
(Arba
ϕ
′
(
x
)
=
(
arctan
(
y
′
)
)
′
=
y
″
1
+
(
y
′
)
2
.
{\displaystyle \phi '(x)=(\arctan(y'))'={\frac {y''}{1+(y')^{2}}}.}
Kas liečia išvestinę
d
s
d
x
,
{\displaystyle {\frac {ds}{dx}},}
radome
d
s
d
x
=
1
+
(
d
y
d
x
)
2
.
{\displaystyle {\frac {ds}{dx}}={\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}.}
Todėl
d
ϕ
d
s
=
d
ϕ
d
x
d
s
d
x
=
d
2
y
d
x
2
1
+
(
d
y
d
x
)
2
1
+
(
d
y
d
x
)
2
=
d
2
y
d
x
2
[
1
+
(
d
y
d
x
)
2
]
3
2
,
{\displaystyle {\frac {d\phi }{ds}}={\frac {\frac {d\phi }{dx}}{\frac {ds}{dx}}}={\frac {\frac {\frac {d^{2}y}{dx^{2}}}{1+\left({\frac {dy}{dx}}\right)^{2}}}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}}={\frac {\frac {d^{2}y}{dx^{2}}}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}},}
arba, kadangi
K
=
|
d
ϕ
d
s
|
,
{\displaystyle K=\left|{\frac {d\phi }{ds}}\right|,}
K
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
.
(
3
)
{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}.\quad (3)}
Iš to seka, kad betkokiame taške kreivės, kur egzistuoja ir netrūki antra išvestinė
d
2
y
d
x
2
,
{\displaystyle {\frac {d^{2}y}{dx^{2}}},}
Nustatyti kreivį parabolės
y
2
=
2
p
x
{\displaystyle y^{2}=2px}
a) jos laisvai pasirenktame taške M (x; y);
b) taške
M
1
(
0
;
0
)
{\displaystyle M_{1}(0;0)}
c) taške
M
2
(
p
2
;
p
)
.
{\displaystyle M_{2}({\frac {p}{2}};p).}
Sprendimas . Randame pirmą ir antrą išvestines funkcijos
y
=
2
p
x
{\displaystyle y={\sqrt {2px}}}
d
y
d
x
=
(
2
p
x
)
′
=
(
2
p
x
)
′
2
2
p
x
=
p
2
p
x
;
d
2
y
d
x
2
=
(
p
2
p
x
)
′
=
−
p
(
2
p
x
)
′
2
(
2
p
x
)
3
2
=
−
p
2
(
2
p
x
)
3
2
.
{\displaystyle {\frac {dy}{dx}}=({\sqrt {2px}})'={\frac {(2px)'}{2{\sqrt {2px}}}}={\frac {p}{\sqrt {2px}}};\quad {\frac {d^{2}y}{dx^{2}}}=\left({\frac {p}{\sqrt {2px}}}\right)'=-{\frac {p(2px)'}{2(2px)^{3 \over 2}}}=-{\frac {p^{2}}{(2px)^{3 \over 2}}}.}
Įstatydami gautas išraiškas į formulę (3), gausime:
a)
K
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
−
p
2
(
2
p
x
)
3
2
|
[
1
+
(
p
2
p
x
)
2
]
3
2
=
p
2
(
2
p
x
)
3
2
[
1
+
p
2
2
p
x
]
3
2
=
p
2
(
2
p
x
+
p
2
)
3
2
;
{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|-{\frac {p^{2}}{(2px)^{3 \over 2}}}\right|}{\left[1+\left({\frac {p}{\sqrt {2px}}}\right)^{2}\right]^{3 \over 2}}}={\frac {p^{2}}{(2px)^{3 \over 2}\left[1+{\frac {p^{2}}{2px}}\right]^{3 \over 2}}}={\frac {p^{2}}{(2px+p^{2})^{3 \over 2}}};}
b)
K
x
=
0
,
y
=
0
=
p
2
(
2
p
⋅
0
+
p
2
)
3
2
=
p
2
p
3
=
1
p
;
{\displaystyle K_{x=0,\;y=0}={\frac {p^{2}}{(2p\cdot 0+p^{2})^{3 \over 2}}}={\frac {p^{2}}{p^{3}}}={\frac {1}{p}};}
c)
K
x
=
p
2
,
y
=
p
=
p
2
(
2
p
⋅
p
2
+
p
2
)
3
2
=
p
2
(
2
p
2
)
3
2
=
p
2
2
3
p
3
=
1
2
2
p
.
{\displaystyle K_{x={\frac {p}{2}},\;y=p}={\frac {p^{2}}{(2p\cdot {\frac {p}{2}}+p^{2})^{3 \over 2}}}={\frac {p^{2}}{(2p^{2})^{3 \over 2}}}={\frac {p^{2}}{{\sqrt {2^{3}}}p^{3}}}={\frac {1}{2{\sqrt {2}}p}}.}
Nustatyti kreivį tiesės
y
=
a
x
+
b
{\displaystyle y=ax+b}
(x; y) . Sprendimas.
y
′
=
(
a
x
+
b
)
′
=
a
,
y
″
=
a
′
=
0.
{\displaystyle y'=(ax+b)'=a,\quad y''=a'=0.}
Pasinaudojant formule (3), gauname:
K
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
y
″
|
[
1
+
(
y
′
)
2
]
3
2
=
|
0
|
[
1
+
a
2
]
3
2
=
0.
{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|y''|}{[1+(y')^{2}]^{3 \over 2}}}={\frac {|0|}{[1+a^{2}]^{3 \over 2}}}=0.}
Tokiu budu, tiesė yra "linija nulinio kreivio". Šita gi rezultatą lengvai galima gauti betarpiškai iš kreivio apibrėžimo.
Apskaičiuokime kreivį bet kuriame grandininės linijos
y
=
a
cosh
x
a
{\displaystyle y=a\cosh {\frac {x}{a}}}
cosh
x
=
e
x
+
e
−
x
2
=
e
2
x
+
1
2
e
x
{\displaystyle \cosh x={\frac {e^{x}+e^{-x}}{2}}={\frac {e^{2x}+1}{2e^{x}}}}
(
cosh
x
)
′
=
sinh
x
=
e
x
−
e
−
x
2
=
e
2
x
−
1
2
e
x
{\displaystyle (\cosh x)'=\sinh x={\frac {e^{x}-e^{-x}}{2}}={\frac {e^{2x}-1}{2e^{x}}}}
(
sinh
x
)
′
=
cosh
x
{\displaystyle (\sinh x)'=\cosh x}
Kadangi
1
+
[
f
′
(
x
)
]
2
=
1
+
[
(
a
cosh
x
a
)
′
]
2
=
1
+
[
(
x
a
)
′
a
sinh
x
a
]
2
=
1
+
[
1
a
⋅
a
sinh
x
a
]
2
=
1
+
sinh
2
x
a
=
cosh
2
x
a
=
y
2
a
2
,
{\displaystyle 1+[f'(x)]^{2}=1+[(a\cosh {\frac {x}{a}})']^{2}=1+[({\frac {x}{a}})'a\sinh {\frac {x}{a}}]^{2}=1+[{\frac {1}{a}}\cdot a\sinh {\frac {x}{a}}]^{2}=1+\sinh ^{2}{\frac {x}{a}}=\cosh ^{2}{\frac {x}{a}}={\frac {y^{2}}{a^{2}}},}
f
′
(
x
)
=
(
a
cosh
x
a
)
′
=
(
x
a
)
′
a
sinh
x
a
=
1
a
⋅
a
sinh
x
a
=
sinh
x
a
,
{\displaystyle f'(x)=(a\cosh {\frac {x}{a}})'=({\frac {x}{a}})'a\sinh {\frac {x}{a}}={\frac {1}{a}}\cdot a\sinh {\frac {x}{a}}=\sinh {\frac {x}{a}},}
f
″
(
x
)
=
(
sinh
x
a
)
′
=
1
a
cosh
x
a
=
y
a
2
,
{\displaystyle f''(x)=(\sinh {\frac {x}{a}})'={\frac {1}{a}}\cosh {\frac {x}{a}}={\frac {y}{a^{2}}},}
tai kreivis yra lygus
k
=
f
″
(
x
)
[
1
+
(
f
′
(
x
)
)
2
]
3
/
2
=
1
a
cosh
x
a
[
1
+
(
sinh
x
a
)
2
]
3
/
2
=
1
a
cosh
x
a
[
cosh
2
x
a
]
3
/
2
=
y
a
2
(
y
2
a
2
)
3
/
2
=
y
a
2
y
3
a
3
=
a
3
y
a
2
y
3
=
a
y
2
=
a
a
2
cosh
2
x
a
=
1
a
cosh
2
x
a
.
{\displaystyle k={\frac {f''(x)}{[1+(f'(x))^{2}]^{3/2}}}={\frac {{\frac {1}{a}}\cosh {\frac {x}{a}}}{[1+(\sinh {\frac {x}{a}})^{2}]^{3/2}}}={\frac {{\frac {1}{a}}\cosh {\frac {x}{a}}}{[\cosh ^{2}{\frac {x}{a}}]^{3/2}}}={\frac {\frac {y}{a^{2}}}{({\frac {y^{2}}{a^{2}}})^{3/2}}}={\frac {\frac {y}{a^{2}}}{\frac {y^{3}}{a^{3}}}}={\frac {a^{3}y}{a^{2}y^{3}}}={\frac {a}{y^{2}}}={\frac {a}{a^{2}\cosh ^{2}{\frac {x}{a}}}}={\frac {1}{a\cosh ^{2}{\frac {x}{a}}}}.}
Nustatyti kreivį parabolės
y
=
x
2
{\displaystyle y=x^{2}}
x
0
=
0
{\displaystyle x_{0}=0}
x
1
=
5.
{\displaystyle x_{1}=5.}
C
0
(
x
2
;
y
2
)
{\displaystyle C_{0}(x_{2};\;y_{2})}
C
1
(
x
3
;
y
3
)
.
{\displaystyle C_{1}(x_{3};\;y_{3}).}
C
0
{\displaystyle C_{0}}
R
0
{\displaystyle R_{0}}
C
1
{\displaystyle C_{1}}
R
2
{\displaystyle R_{2}}
R
0
{\displaystyle R_{0}}
M
0
(
x
0
;
y
0
)
{\displaystyle M_{0}(x_{0};y_{0})}
C
0
(
x
2
;
y
2
)
{\displaystyle C_{0}(x_{2};y_{2})}
R
1
{\displaystyle R_{1}}
M
1
(
x
1
;
y
1
)
{\displaystyle M_{1}(x_{1};y_{1})}
C
1
(
x
3
;
y
3
)
{\displaystyle C_{1}(x_{3};y_{3})}
Sprendimas .
y
′
=
(
x
2
)
′
=
2
x
,
y
″
=
(
2
x
)
′
=
2.
{\displaystyle y'=(x^{2})'=2x,\quad y''=(2x)'=2.}
K
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
2
|
[
1
+
(
2
x
)
2
]
3
2
=
2
(
1
+
4
x
2
)
3
2
.
{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}.}
Kreivis taške
M
0
{\displaystyle M_{0}}
K
M
0
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
2
|
[
1
+
(
2
x
)
2
]
3
2
=
2
(
1
+
4
x
2
)
3
2
=
2
(
1
+
4
⋅
0
2
)
3
2
=
2.
{\displaystyle K_{M_{0}}={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}={\frac {2}{(1+4\cdot 0^{2})^{3 \over 2}}}=2.}
R
0
=
1
K
M
0
=
1
2
=
0.5.
{\displaystyle R_{0}={\frac {1}{K_{M_{0}}}}={\frac {1}{2}}=0.5.}
Kreivis taške
M
1
{\displaystyle M_{1}}
K
M
1
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
2
|
[
1
+
(
2
x
)
2
]
3
2
=
2
(
1
+
4
x
2
)
3
2
=
{\displaystyle K_{M_{1}}={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}=}
=
2
(
1
+
4
⋅
5
2
)
3
2
=
2
101
3
2
=
2
1030301
=
2
1015.037438
=
0.00197037.
{\displaystyle ={\frac {2}{(1+4\cdot 5^{2})^{3 \over 2}}}={\frac {2}{101^{3 \over 2}}}={\frac {2}{\sqrt {1030301}}}={\frac {2}{1015.037438}}=0.00197037.}
R
1
=
1
K
M
1
=
1
2
1030301
=
1030301
2
=
507.5187189.
{\displaystyle R_{1}={\frac {1}{K_{M_{1}}}}={\frac {1}{\frac {2}{\sqrt {1030301}}}}={\frac {\sqrt {1030301}}{2}}=507.5187189.}
Parabolės evoliutės lanko ilgis iš taško
C
0
{\displaystyle C_{0}}
C
1
{\displaystyle C_{1}}
L
=
R
1
−
R
0
=
1030301
2
−
1
2
=
507.5187189
−
0.5
=
507.0187189.
{\displaystyle L=R_{1}-R_{0}={\frac {\sqrt {1030301}}{2}}-{\frac {1}{2}}=507.5187189-0.5=507.0187189.}
Nustatyti kreivį parabolės
y
=
x
2
{\displaystyle y=x^{2}}
x
0
=
0
{\displaystyle x_{0}=0}
x
1
=
5.
{\displaystyle x_{1}=5.}
C
0
(
x
2
;
y
2
)
{\displaystyle C_{0}(x_{2};\;y_{2})}
C
1
(
x
3
;
y
3
)
{\displaystyle C_{1}(x_{3};\;y_{3})}
L
=
∫
a
b
1
+
(
y
′
)
2
d
x
.
{\displaystyle L=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx.}
Taškas
C
0
{\displaystyle C_{0}}
R
0
{\displaystyle R_{0}}
C
1
{\displaystyle C_{1}}
R
2
{\displaystyle R_{2}}
R
0
{\displaystyle R_{0}}
M
0
(
x
0
;
y
0
)
{\displaystyle M_{0}(x_{0};y_{0})}
C
0
(
x
2
;
y
2
)
{\displaystyle C_{0}(x_{2};y_{2})}
R
1
{\displaystyle R_{1}}
M
1
(
x
1
;
y
1
)
{\displaystyle M_{1}(x_{1};y_{1})}
C
1
(
x
3
;
y
3
)
{\displaystyle C_{1}(x_{3};y_{3})}
Sprendimas .
y
′
=
(
x
2
)
′
=
2
x
,
y
″
=
(
2
x
)
′
=
2.
{\displaystyle y'=(x^{2})'=2x,\quad y''=(2x)'=2.}
K
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
2
|
[
1
+
(
2
x
)
2
]
3
2
=
2
(
1
+
4
x
2
)
3
2
.
{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}.}
Kreivis taške
M
0
{\displaystyle M_{0}}
K
M
0
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
2
|
[
1
+
(
2
x
)
2
]
3
2
=
2
(
1
+
4
x
2
)
3
2
=
2
(
1
+
4
⋅
0
2
)
3
2
=
2.
{\displaystyle K_{M_{0}}={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}={\frac {2}{(1+4\cdot 0^{2})^{3 \over 2}}}=2.}
R
0
=
1
K
M
0
=
1
2
=
0.5.
{\displaystyle R_{0}={\frac {1}{K_{M_{0}}}}={\frac {1}{2}}=0.5.}
Kreivis taške
M
1
{\displaystyle M_{1}}
K
M
1
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
2
|
[
1
+
(
2
x
)
2
]
3
2
=
2
(
1
+
4
x
2
)
3
2
=
{\displaystyle K_{M_{1}}={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}=}
=
2
(
1
+
4
⋅
5
2
)
3
2
=
2
101
3
2
=
2
1030301
=
2
1015.037438
=
0.00197037.
{\displaystyle ={\frac {2}{(1+4\cdot 5^{2})^{3 \over 2}}}={\frac {2}{101^{3 \over 2}}}={\frac {2}{\sqrt {1030301}}}={\frac {2}{1015.037438}}=0.00197037.}
R
1
=
1
K
M
1
=
1
2
1030301
=
1030301
2
=
507.5187189.
{\displaystyle R_{1}={\frac {1}{K_{M_{1}}}}={\frac {1}{\frac {2}{\sqrt {1030301}}}}={\frac {\sqrt {1030301}}{2}}=507.5187189.}
Dabar užrašysime parabolės normalės lygtį taške
M
1
(
5
;
25
)
{\displaystyle M_{1}(5;25)}
y
−
y
M
1
=
−
1
y
′
(
x
M
1
)
(
x
−
x
M
1
)
,
{\displaystyle y-y_{M_{1}}=-{\frac {1}{y'(x_{M_{1}})}}(x-x_{M_{1}}),}
y
−
y
M
1
=
−
1
2
x
M
1
(
x
−
x
M
1
)
,
{\displaystyle y-y_{M_{1}}=-{\frac {1}{2x_{M_{1}}}}(x-x_{M_{1}}),}
y
−
25
=
−
1
2
⋅
5
(
x
−
5
)
,
{\displaystyle y-25=-{\frac {1}{2\cdot 5}}(x-5),}
y
−
25
=
−
1
10
(
x
−
5
)
,
{\displaystyle y-25=-{\frac {1}{10}}(x-5),}
y
=
−
x
10
+
1
2
+
25
=
−
0.1
x
+
25.5.
{\displaystyle y=-{\frac {x}{10}}+{\frac {1}{2}}+25=-0.1x+25.5.}
Toliau rasime spindulio
R
1
=
1030301
2
{\displaystyle R_{1}={\frac {\sqrt {1030301}}{2}}}
C
1
(
x
3
;
y
3
)
{\displaystyle C_{1}(x_{3};\;y_{3})}
(
x
3
−
x
1
)
2
+
(
y
3
−
y
1
)
2
=
R
1
2
,
{\displaystyle (x_{3}-x_{1})^{2}+(y_{3}-y_{1})^{2}=R_{1}^{2},}
(
x
3
−
5
)
2
+
(
y
3
−
25
)
2
=
R
1
,
{\displaystyle {\sqrt {(x_{3}-5)^{2}+(y_{3}-25)^{2}}}=R_{1},}
(
x
3
−
5
)
2
+
(
y
3
−
25
)
2
=
(
1030301
2
)
2
,
{\displaystyle (x_{3}-5)^{2}+(y_{3}-25)^{2}=\left({\frac {\sqrt {1030301}}{2}}\right)^{2},}
(
x
3
−
5
)
2
+
(
y
3
−
25
)
2
=
1030301
4
=
257575.25.
{\displaystyle (x_{3}-5)^{2}+(y_{3}-25)^{2}={\frac {1030301}{4}}=257575.25.}
Išsprendę lygčių sistemą rasime
x
3
{\displaystyle x_{3}}
C
1
{\displaystyle C_{1}}
{
y
−
25
=
−
1
10
(
x
−
5
)
,
(
x
−
5
)
2
+
(
y
−
25
)
2
=
1030301
4
;
{\displaystyle {\begin{cases}y-25=-{\frac {1}{10}}(x-5),&\\(x-5)^{2}+(y-25)^{2}={\frac {1030301}{4}};&\end{cases}}}
keitimo budu gauname:
(
x
−
5
)
2
+
(
−
1
10
(
x
−
5
)
)
2
=
1030301
4
,
{\displaystyle (x-5)^{2}+\left(-{\frac {1}{10}}(x-5)\right)^{2}={\frac {1030301}{4}},}
(
x
−
5
)
2
+
1
100
(
x
−
5
)
2
=
1030301
4
,
{\displaystyle (x-5)^{2}+{\frac {1}{100}}(x-5)^{2}={\frac {1030301}{4}},}
x
2
−
10
x
+
25
+
x
2
−
10
x
+
25
100
=
1030301
4
,
{\displaystyle x^{2}-10x+25+{\frac {x^{2}-10x+25}{100}}={\frac {1030301}{4}},}
x
2
−
10
x
+
25
+
0.01
x
2
−
0.1
x
+
0.25
=
257575.25
,
{\displaystyle x^{2}-10x+25+0.01x^{2}-0.1x+0.25=257575.25,}
x
2
−
10
x
+
25
+
0.01
x
2
−
0.1
x
+
0.25
−
257575.25
=
0
,
{\displaystyle x^{2}-10x+25+0.01x^{2}-0.1x+0.25-257575.25=0,}
x
2
+
0.01
x
2
−
10
x
−
0.1
x
−
257575
+
25
=
0
,
{\displaystyle x^{2}+0.01x^{2}-10x-0.1x-257575+25=0,}
1.01
x
2
−
10.1
x
−
257550
=
0
,
{\displaystyle 1.01x^{2}-10.1x-257550=0,}
101
x
2
−
1010
x
−
25755000
=
0
,
{\displaystyle 101x^{2}-1010x-25755000=0,}
Tai yra kvadratinė lygtis, kurios sprendiniai yra:
x
=
−
b
+
b
2
−
4
a
c
2
a
=
−
(
−
1010
)
+
(
−
1010
)
2
−
4
⋅
101
⋅
(
−
25755000
)
2
⋅
101
=
{\displaystyle x={\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}={\frac {-(-1010)+{\sqrt {(-1010)^{2}-4\cdot 101\cdot (-25755000)}}}{2\cdot 101}}=}
=
1010
+
1020100
+
10405020000
202
=
1010
+
10406040100
202
=
1010
+
102010
202
=
103020
202
=
510
;
{\displaystyle ={\frac {1010+{\sqrt {1020100+10405020000}}}{202}}={\frac {1010+{\sqrt {10406040100}}}{202}}={\frac {1010+102010}{202}}={\frac {103020}{202}}=510;}
x
=
−
b
−
b
2
−
4
a
c
2
a
=
−
(
−
1010
)
−
(
−
1010
)
2
−
4
⋅
101
⋅
(
−
25755000
)
2
⋅
101
=
{\displaystyle x={\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}={\frac {-(-1010)-{\sqrt {(-1010)^{2}-4\cdot 101\cdot (-25755000)}}}{2\cdot 101}}=}
=
1010
−
1020100
+
10405020000
202
=
1010
−
10406040100
202
=
1010
−
102010
202
=
−
101000
202
=
−
500.
{\displaystyle ={\frac {1010-{\sqrt {1020100+10405020000}}}{202}}={\frac {1010-{\sqrt {10406040100}}}{202}}={\frac {1010-102010}{202}}={\frac {-101000}{202}}=-500.}
Kadangi spindulys
R
1
{\displaystyle R_{1}}
R
1
{\displaystyle R_{1}}
C
1
{\displaystyle C_{1}}
x
3
=
−
500.
{\displaystyle x_{3}=-500.}
Žinodami
x
3
{\displaystyle x_{3}}
y
3
=
3
x
3
2
3
16
1
3
+
1
2
=
3
⋅
(
−
500
)
2
3
16
1
3
+
1
2
=
3
⋅
250000
1
3
16
1
3
+
1
2
=
{\displaystyle y_{3}={\frac {3x_{3}^{2 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}={\frac {3\cdot (-500)^{2 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}={\frac {3\cdot 250000^{1 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}=}
=
3
⋅
15625
1
3
+
0.5
=
3
⋅
25
+
0.5
=
75
+
0.5
=
75.5.
{\displaystyle =3\cdot 15625^{1 \over 3}+0.5=3\cdot 25+0.5=75+0.5=75.5.}
Arba per Pitagoro teoremą randame:
y
p
=
R
1
2
−
(
x
3
−
x
1
)
2
=
507.5187189
2
−
(
−
500
−
5
)
2
=
257575.25
−
255025
=
2550.25
=
50.5
;
{\displaystyle y_{p}={\sqrt {R_{1}^{2}-(x_{3}-x_{1})^{2}}}={\sqrt {507.5187189^{2}-(-500-5)^{2}}}={\sqrt {257575.25-255025}}={\sqrt {2550.25}}=50.5;}
y
3
=
y
1
+
y
p
=
25
+
50.5
=
75.5.
{\displaystyle y_{3}=y_{1}+y_{p}=25+50.5=75.5.}
Taigi, radome spindulio
R
1
{\displaystyle R_{1}}
C
1
(
−
500
;
75.5
)
.
{\displaystyle C_{1}(-500;\;75.5).}
Toliau rasime spindulio
R
0
=
1
2
{\displaystyle R_{0}={\frac {1}{2}}}
C
0
(
x
2
;
y
2
)
{\displaystyle C_{0}(x_{2};\;y_{2})}
(
x
2
−
x
0
)
2
+
(
y
2
−
y
0
)
2
=
R
0
2
,
{\displaystyle (x_{2}-x_{0})^{2}+(y_{2}-y_{0})^{2}=R_{0}^{2},}
(
x
2
−
0
)
2
+
(
y
2
−
0
)
2
=
R
0
,
{\displaystyle {\sqrt {(x_{2}-0)^{2}+(y_{2}-0)^{2}}}=R_{0},}
(
x
2
−
0
)
2
+
(
y
2
−
0
)
2
=
(
1
2
)
2
,
{\displaystyle (x_{2}-0)^{2}+(y_{2}-0)^{2}=\left({\frac {1}{2}}\right)^{2},}
(
x
3
−
0
)
2
+
(
y
3
−
0
)
2
=
1
4
.
{\displaystyle (x_{3}-0)^{2}+(y_{3}-0)^{2}={\frac {1}{4}}.}
Išsprendę lygčių sistemą rasime
x
2
{\displaystyle x_{2}}
C
0
{\displaystyle C_{0}}
{
y
−
0
=
−
1
2
x
(
x
−
0
)
,
(
x
−
0
)
2
+
(
y
−
0
)
2
=
1
4
;
{\displaystyle {\begin{cases}y-0=-{\frac {1}{2x}}(x-0),&\\(x-0)^{2}+(y-0)^{2}={\frac {1}{4}};&\end{cases}}}
keitimo budu gauname:
(
x
−
0
)
2
+
(
−
1
2
x
(
x
−
0
)
)
2
=
1
4
,
{\displaystyle (x-0)^{2}+\left(-{\frac {1}{2x}}(x-0)\right)^{2}={\frac {1}{4}},}
x
2
+
1
4
x
2
(
x
−
0
)
2
=
1
4
,
{\displaystyle x^{2}+{\frac {1}{4x^{2}}}(x-0)^{2}={\frac {1}{4}},}
x
2
+
x
2
4
x
2
=
1
4
,
{\displaystyle x^{2}+{\frac {x^{2}}{4x^{2}}}={\frac {1}{4}},}
x
2
+
1
4
=
1
4
,
{\displaystyle x^{2}+{\frac {1}{4}}={\frac {1}{4}},}
x
2
=
1
4
−
1
4
,
{\displaystyle x^{2}={\frac {1}{4}}-{\frac {1}{4}},}
x
=
0.
{\displaystyle x=0.}
Taigi,
x
2
=
0.
{\displaystyle x_{2}=0.}
x
2
{\displaystyle x_{2}}
y
2
=
3
x
2
2
3
16
1
3
+
1
2
=
3
⋅
0
2
3
16
1
3
+
1
2
=
1
2
=
0.5.
{\displaystyle y_{2}={\frac {3x_{2}^{2 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}={\frac {3\cdot 0^{2 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}={\frac {1}{2}}=0.5.}
Arba per Pitagoro teoremą randame:
y
d
=
R
0
2
−
(
x
2
−
x
0
)
2
=
0.5
2
−
(
0
−
0
)
2
=
0.25
−
0
=
0.25
=
0.5
;
{\displaystyle y_{d}={\sqrt {R_{0}^{2}-(x_{2}-x_{0})^{2}}}={\sqrt {0.5^{2}-(0-0)^{2}}}={\sqrt {0.25-0}}={\sqrt {0.25}}=0.5;}
y
3
=
y
0
+
y
d
=
0
+
0.5
=
0.5.
{\displaystyle y_{3}=y_{0}+y_{d}=0+0.5=0.5.}
Radome spindulio
R
0
{\displaystyle R_{0}}
C
0
(
0
;
1
2
)
.
{\displaystyle C_{0}(0;\;{\frac {1}{2}}).}
Rasti lygtį evoliutės parabolės
y
=
x
2
.
{\displaystyle y=x^{2}.}
Rasti lanko ilgį L evoliutės šios parabolės naudojantis kreivės lanko ilgio skaičiavimo formule
L
=
∫
a
b
1
+
(
y
′
)
2
d
x
.
{\displaystyle L=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx.}
Rasti evoliutės lanko ilgį iš taško
M
0
(
0
;
1
2
)
{\displaystyle M_{0}(0;\;{\frac {1}{2}})}
M
1
(
−
500
;
75.5
)
.
{\displaystyle M_{1}(-500;\;75.5).}
Sprendimas . Turime bet kokiam taškui (x; y) parabolės kreivio centro koordinates
C
(
α
;
β
)
{\displaystyle C(\alpha ;\beta )}
d
y
d
x
=
(
x
2
)
′
=
2
x
;
{\displaystyle {\frac {dy}{dx}}=(x^{2})'=2x;}
d
2
y
d
x
2
=
(
2
x
)
′
=
2
;
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=(2x)'=2;}
α
=
x
−
y
′
(
1
+
y
′
2
)
y
″
=
x
−
2
x
(
1
+
(
2
x
)
2
)
2
=
x
−
x
(
1
+
4
x
2
)
=
x
−
x
−
4
x
3
=
−
4
x
3
.
{\displaystyle \alpha =x-{\frac {y'(1+y'^{2})}{y''}}=x-{\frac {2x(1+(2x)^{2})}{2}}=x-x(1+4x^{2})=x-x-4x^{3}=-4x^{3}.}
β
=
y
+
1
+
y
′
2
y
″
=
x
2
+
1
+
(
2
x
)
2
2
=
x
2
+
1
+
4
x
2
2
=
x
2
+
1
2
+
2
x
2
=
3
x
2
+
1
2
.
{\displaystyle \beta =y+{\frac {1+y'^{2}}{y''}}=x^{2}+{\frac {1+(2x)^{2}}{2}}=x^{2}+{\frac {1+4x^{2}}{2}}=x^{2}+{\frac {1}{2}}+2x^{2}=3x^{2}+{\frac {1}{2}}.}
(Pavyzdžiui, taške M (5; 25), turime
α
=
−
4
⋅
5
3
=
−
4
⋅
125
=
−
500
{\displaystyle \alpha =-4\cdot 5^{3}=-4\cdot 125=-500}
β
=
3
⋅
5
2
+
0.5
=
75
+
0.5
=
75.5
{\displaystyle \beta =3\cdot 5^{2}+0.5=75+0.5=75.5}
C (-500; 75.5)).
Eliminuojant iš šitų lygčių parametrą x , gausime:
α
=
−
4
x
3
,
{\displaystyle \alpha =-4x^{3},}
−
α
4
=
x
3
,
{\displaystyle -{\frac {\alpha }{4}}=x^{3},}
x
=
−
(
α
4
)
1
3
;
{\displaystyle x=-\left({\frac {\alpha }{4}}\right)^{1 \over 3};}
β
=
3
x
2
+
1
2
,
{\displaystyle \beta =3x^{2}+{\frac {1}{2}},}
β
=
3
(
−
(
α
4
)
1
3
)
2
+
1
2
=
3
(
α
4
)
2
3
+
1
2
,
{\displaystyle \beta =3\left(-\left({\frac {\alpha }{4}}\right)^{1 \over 3}\right)^{2}+{\frac {1}{2}}=3\left({\frac {\alpha }{4}}\right)^{2 \over 3}+{\frac {1}{2}},}
β
=
3
α
2
3
16
1
3
+
1
2
.
{\displaystyle \beta ={\frac {3\alpha ^{2 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}.}
Tai yra lygtis evoliutės. Čia
β
{\displaystyle \beta }
Oy ašies) reikšmė, o
α
{\displaystyle \alpha }
Ox ašies) reikšmė.
Rasime evoliutės lanko ilgį iš taško
M
0
(
0
;
1
2
)
{\displaystyle M_{0}(0;\;{\frac {1}{2}})}
M
1
(
−
500
;
75.5
)
:
{\displaystyle M_{1}(-500;\;75.5):}
d
β
d
α
=
(
3
α
2
3
16
1
3
+
1
2
)
′
=
2
3
⋅
3
α
−
1
3
16
1
3
=
2
16
1
3
α
1
3
;
{\displaystyle {\frac {d\beta }{d\alpha }}=\left({\frac {3\alpha ^{2 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}\right)'={\frac {2}{3}}\cdot {\frac {3\alpha ^{-{\frac {1}{3}}}}{16^{\frac {1}{3}}}}={\frac {2}{16^{1 \over 3}\alpha ^{1 \over 3}}};}
L
=
∫
a
b
1
+
(
d
β
d
α
)
2
d
α
=
∫
−
500
0
1
+
(
2
16
1
3
α
1
3
)
2
d
α
=
∫
−
500
0
1
+
4
256
1
3
α
2
3
d
α
=
∫
−
500
0
2
16
1
3
256
1
3
4
+
1
α
2
3
d
α
=
{\displaystyle L=\int _{a}^{b}{\sqrt {1+\left({\frac {d\beta }{d\alpha }}\right)^{2}}}d\alpha =\int _{-500}^{0}{\sqrt {1+\left({\frac {2}{16^{\frac {1}{3}}\alpha ^{1 \over 3}}}\right)^{2}}}d\alpha =\int _{-500}^{0}{\sqrt {1+{\frac {4}{256^{1 \over 3}\alpha ^{2 \over 3}}}}}d\alpha =\int _{-500}^{0}{\frac {2}{16^{1 \over 3}}}{\sqrt {{\frac {256^{1 \over 3}}{4}}+{\frac {1}{\alpha ^{2 \over 3}}}}}d\alpha =}
=
2
16
1
3
⋅
2
x
+
(
2
x
)
1
3
2
⋅
1
x
2
3
+
2
2
3
|
−
500
0
=
1
16
1
3
(
2
x
+
(
2
x
)
1
3
)
1
x
2
3
+
2
2
3
|
−
500
0
=
{\displaystyle ={\frac {2}{16^{1 \over 3}}}\cdot {\frac {2x+(2x)^{1 \over 3}}{2}}\cdot {\sqrt {{\frac {1}{x^{2 \over 3}}}+2^{2 \over 3}}}|_{-500}^{0}={\frac {1}{16^{1 \over 3}}}(2x+(2x)^{1 \over 3}){\sqrt {{\frac {1}{x^{2 \over 3}}}+2^{2 \over 3}}}|_{-500}^{0}=}
=
1
16
1
3
(
2
⋅
0
+
(
2
⋅
0
)
1
3
)
1
0
2
3
+
2
2
3
−
1
16
1
3
(
2
⋅
(
−
500
)
+
(
2
⋅
(
−
500
)
)
1
3
)
1
(
−
500
)
2
3
+
2
2
3
=
{\displaystyle ={\frac {1}{16^{1 \over 3}}}(2\cdot 0+(2\cdot 0)^{1 \over 3}){\sqrt {{\frac {1}{0^{2 \over 3}}}+2^{2 \over 3}}}-{\frac {1}{16^{1 \over 3}}}(2\cdot (-500)+(2\cdot (-500))^{1 \over 3}){\sqrt {{\frac {1}{(-500)^{2 \over 3}}}+2^{2 \over 3}}}=}
=
−
1
16
1
3
(
−
1000
+
(
−
1000
)
1
3
)
1
250000
1
3
+
4
1
3
=
−
1
16
1
3
(
−
1000
−
10
)
1
250000
1
3
+
4
1
3
=
{\displaystyle =-{\frac {1}{16^{1 \over 3}}}(-1000+(-1000)^{1 \over 3}){\sqrt {{\frac {1}{250000^{1 \over 3}}}+4^{1 \over 3}}}=-{\frac {1}{16^{1 \over 3}}}(-1000-10){\sqrt {{\frac {1}{250000^{1 \over 3}}}+4^{1 \over 3}}}=}
=
1
16
1
3
⋅
1010
1
250000
1
3
+
4
1
3
=
{\displaystyle ={\frac {1}{16^{1 \over 3}}}\cdot 1010{\sqrt {{\frac {1}{250000^{1 \over 3}}}+4^{1 \over 3}}}=}
=
0.396850263
⋅
1010
1
62.99605249
+
1.587401052
=
{\displaystyle =0.396850263\cdot 1010{\sqrt {{\frac {1}{62.99605249}}+1.587401052}}=}
=
0.396850263
⋅
1010
0.01587401
+
1.587401052
=
2.5198421
⋅
1010
1.603275062
=
{\displaystyle =0.396850263\cdot 1010{\sqrt {0.01587401+1.587401052}}=2.5198421\cdot 1010{\sqrt {1.603275062}}=}
=
0.396850263
⋅
1010
⋅
1.266204984
=
507.5187187.
{\displaystyle =0.396850263\cdot 1010\cdot 1.266204984=507.5187187.}
Pasinaudojome internetiniu integratoriumi http://integrals.wolfram.com/index.jsp?expr=sqrt%5B%28256%5E%281%2F3%29%29%2F4%2B1%2F%28x%5E%282%2F3%29%29%5D&random=false .
Bet tikrasis lanko ilgis yra
L
=
507.0187187
{\displaystyle L=507.0187187}
x
=
−
0.001
{\displaystyle x=-0.001}
1
16
1
3
(
2
x
+
(
2
x
)
1
3
)
1
x
2
3
+
2
2
3
=
1
16
1
3
(
−
0.002
+
(
−
0.002
)
1
3
)
1
(
−
0.001
)
2
3
+
2
2
3
=
{\displaystyle {\frac {1}{16^{1 \over 3}}}(2x+(2x)^{1 \over 3}){\sqrt {{\frac {1}{x^{2 \over 3}}}+2^{2 \over 3}}}={\frac {1}{16^{1 \over 3}}}(-0.002+(-0.002)^{1 \over 3}){\sqrt {{\frac {1}{(-0.001)^{2 \over 3}}}+2^{2 \over 3}}}=}
=
1
16
1
3
(
−
0.002
−
0.125992105
)
1
(
−
0.001
)
2
3
+
2
2
3
=
1
2.5198421
⋅
(
−
0.127992105
)
⋅
1
0.01
+
2
2
3
=
{\displaystyle ={\frac {1}{16^{1 \over 3}}}(-0.002-0.125992105){\sqrt {{\frac {1}{(-0.001)^{2 \over 3}}}+2^{2 \over 3}}}={\frac {1}{2.5198421}}\cdot (-0.127992105)\cdot {\sqrt {{\frac {1}{0.01}}+2^{2 \over 3}}}=}
=
0.396850263
⋅
(
−
0.127992105
)
⋅
100
+
1.587401052
=
−
0.0507937
⋅
101.5874011
=
{\displaystyle =0.396850263\cdot (-0.127992105)\cdot {\sqrt {100+1.587401052}}=-0.0507937\cdot {\sqrt {101.5874011}}=}
=
−
0.0507937
⋅
10.07905755
=
−
0.51195263.
{\displaystyle =-0.0507937\cdot 10.07905755=-0.51195263.}
Nustatyti krevio centro taškų
C
1
(
x
1
;
y
1
)
{\displaystyle C_{1}(x_{1};y_{1})}
C
2
(
x
2
;
y
2
)
{\displaystyle C_{2}(x_{2};y_{2})}
y
=
2
x
{\displaystyle y={\sqrt {2x}}}
O
(
0
;
0
)
{\displaystyle O(0;0)}
M
(
2
;
2
)
{\displaystyle M(2;2)}
Rasti kreivį taškuose
O
(
0
;
0
)
{\displaystyle O(0;0)}
M
(
2
;
2
)
.
{\displaystyle M(2;2).}
Rasti spindulio
R
1
{\displaystyle R_{1}}
O
(
0
;
0
)
{\displaystyle O(0;\;0)}
C
1
(
x
1
;
y
1
)
.
{\displaystyle C_{1}(x_{1};\;y_{1}).}
Rasti spindulio
R
2
{\displaystyle R_{2}}
M
(
2
;
2
)
{\displaystyle M(2;\;2)}
C
2
(
x
2
;
y
2
)
.
{\displaystyle C_{2}(x_{2};\;y_{2}).}
Rasti hiperbolės evoliutės lanko ilgį L iš taško
C
1
(
x
1
;
y
1
)
{\displaystyle C_{1}(x_{1};\;y_{1})}
C
2
(
x
2
;
y
2
)
.
{\displaystyle C_{2}(x_{2};\;y_{2}).}
L lygus spindulių
R
2
{\displaystyle R_{2}}
R
1
{\displaystyle R_{1}}
Rasti hiperbolės evolutės lygtį.
Rasti hiperbolės evoliutės lanko ilgį L iš taško
C
1
(
x
1
;
y
1
)
{\displaystyle C_{1}(x_{1};\;y_{1})}
C
2
(
x
2
;
y
2
)
{\displaystyle C_{2}(x_{2};\;y_{2})}
L
=
∫
a
b
1
+
(
y
′
)
2
d
x
.
{\displaystyle L=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx.}
Sprendimas . Įstatant reikšmes
d
y
d
x
{\displaystyle {\frac {dy}{dx}}}
d
2
y
d
x
2
{\displaystyle {\frac {d^{2}y}{dx^{2}}}}
d
y
d
x
=
(
2
x
)
′
=
1
2
⋅
(
2
x
)
′
2
x
=
1
2
x
;
{\displaystyle {\frac {dy}{dx}}=({\sqrt {2x}})'={\frac {1}{2}}\cdot {\frac {(2x)'}{\sqrt {2x}}}={\frac {1}{\sqrt {2x}}};}
d
2
y
d
x
2
=
(
p
2
x
)
′
=
−
1
2
⋅
(
2
x
)
′
(
2
x
)
3
2
=
−
1
(
2
x
)
3
2
;
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=\left({\frac {p}{\sqrt {2x}}}\right)'=-{\frac {1}{2}}\cdot {\frac {(2x)'}{(2x)^{3 \over 2}}}=-{\frac {1}{(2x)^{3 \over 2}}};}
α
=
x
−
y
′
(
1
+
y
′
2
)
y
″
=
x
−
1
2
x
(
1
+
1
2
x
)
−
1
(
2
x
)
3
2
=
x
−
1
1
+
1
2
x
−
1
2
x
=
x
+
2
x
+
1
=
3
x
+
1
;
{\displaystyle \alpha =x-{\frac {y'(1+y'^{2})}{y''}}=x-{\frac {{\frac {1}{\sqrt {2x}}}(1+{\frac {1}{2x}})}{-{\frac {1}{(2x)^{3 \over 2}}}}}=x-{\frac {1}{1+{\frac {1}{2x}}}}{-{\frac {1}{2x}}}=x+2x+1=3x+1;}
β
=
y
+
1
+
y
′
2
y
″
=
2
x
+
1
+
1
2
x
−
1
(
2
x
)
3
2
=
−
2
x
(
2
x
)
3
2
+
1
+
1
2
x
−
1
(
2
x
)
3
2
=
{\displaystyle \beta =y+{\frac {1+y'^{2}}{y''}}={\sqrt {2x}}+{\frac {1+{\frac {1}{2x}}}{-{\frac {1}{(2x)^{3 \over 2}}}}}={\frac {-{\frac {\sqrt {2x}}{(2x)^{3 \over 2}}}+1+{\frac {1}{2x}}}{-{\frac {1}{(2x)^{3 \over 2}}}}}=}
=
−
1
2
x
+
1
+
1
2
x
−
1
(
2
x
)
3
2
=
1
−
1
(
2
x
)
3
2
=
−
(
2
x
)
3
2
.
{\displaystyle ={\frac {-{\frac {1}{2x}}+1+{\frac {1}{2x}}}{-{\frac {1}{(2x)^{3 \over 2}}}}}={\frac {1}{-{\frac {1}{(2x)^{3 \over 2}}}}}=-(2x)^{3 \over 2}.}
Taško
C
1
(
x
1
;
y
1
)
{\displaystyle C_{1}(x_{1};y_{1})}
x
1
=
α
=
3
x
+
1
=
3
x
O
+
1
=
3
⋅
0
+
1
=
1
;
{\displaystyle x_{1}=\alpha =3x+1=3x_{O}+1=3\cdot 0+1=1;}
y
1
=
β
=
−
(
2
x
)
3
2
=
−
(
2
x
O
)
3
2
=
−
(
2
⋅
0
)
3
2
=
0.
{\displaystyle y_{1}=\beta =-(2x)^{3 \over 2}=-(2x_{O})^{3 \over 2}=-(2\cdot 0)^{3 \over 2}=0.}
Vadinasi kreivio centras taške
O
(
0
;
0
)
{\displaystyle O(0;0)}
C
1
(
1
;
0
)
{\displaystyle C_{1}(1;0)}
Taško
C
2
(
x
2
;
y
2
)
{\displaystyle C_{2}(x_{2};y_{2})}
x
2
=
α
=
3
x
+
1
=
3
x
M
+
1
=
3
⋅
2
+
1
=
7
;
{\displaystyle x_{2}=\alpha =3x+1=3x_{M}+1=3\cdot 2+1=7;}
y
2
=
β
=
−
(
2
x
)
3
2
=
−
(
2
x
M
)
3
2
=
−
(
2
⋅
2
)
3
2
=
−
4
3
=
−
4
4
=
−
8.
{\displaystyle y_{2}=\beta =-(2x)^{3 \over 2}=-(2x_{M})^{3 \over 2}=-(2\cdot 2)^{3 \over 2}=-{\sqrt {4^{3}}}=-4{\sqrt {4}}=-8.}
Kreivio centras taške
M
(
2
;
2
)
{\displaystyle M(2;2)}
C
2
(
7
;
−
8
)
{\displaystyle C_{2}(7;-8)}
Hiperbolės
y
=
2
x
{\displaystyle y={\sqrt {2x}}}
K
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
−
1
(
2
x
)
3
2
|
[
1
+
(
1
2
x
)
2
]
3
2
=
1
(
2
x
)
3
2
[
1
+
1
2
x
]
3
2
=
1
(
2
x
+
1
)
3
2
.
{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|-{\frac {1}{(2x)^{3 \over 2}}}\right|}{\left[1+\left({\frac {1}{\sqrt {2x}}}\right)^{2}\right]^{3 \over 2}}}={\frac {1}{(2x)^{3 \over 2}\left[1+{\frac {1}{2x}}\right]^{3 \over 2}}}={\frac {1}{(2x+1)^{3 \over 2}}}.}
Hiperbolės
y
=
2
x
{\displaystyle y={\sqrt {2x}}}
R
=
1
K
=
1
1
(
2
x
+
1
)
3
2
=
(
2
x
+
1
)
3
2
.
{\displaystyle R={\frac {1}{K}}={\frac {1}{\frac {1}{(2x+1)^{3 \over 2}}}}=(2x+1)^{3 \over 2}.}
Hiperbolės
y
=
2
x
{\displaystyle y={\sqrt {2x}}}
R
1
{\displaystyle R_{1}}
O
(
0
;
0
)
{\displaystyle O(0;\;0)}
C
1
(
1
;
0
)
{\displaystyle C_{1}(1;\;0)}
R
1
=
(
2
x
O
+
1
)
3
2
=
(
2
⋅
0
+
1
)
3
2
=
1.
{\displaystyle R_{1}=(2x_{O}+1)^{3 \over 2}=(2\cdot 0+1)^{3 \over 2}=1.}
Hiperbolės
y
=
2
x
{\displaystyle y={\sqrt {2x}}}
R
2
{\displaystyle R_{2}}
M
(
2
;
2
)
{\displaystyle M(2;\;2)}
C
2
(
7
;
−
8
)
{\displaystyle C_{2}(7;\;-8)}
R
2
=
(
2
x
M
+
1
)
3
2
=
(
2
⋅
2
+
1
)
3
2
=
5
3
=
125
=
5
5
=
11.18033989.
{\displaystyle R_{2}=(2x_{M}+1)^{3 \over 2}=(2\cdot 2+1)^{3 \over 2}={\sqrt {5^{3}}}={\sqrt {125}}=5{\sqrt {5}}=11.18033989.}
Arba
R
2
=
(
x
2
−
x
M
)
2
+
(
y
2
−
y
M
)
2
=
(
7
−
2
)
2
+
(
−
8
−
2
)
2
=
5
2
+
(
−
10
)
2
=
125
=
11.18033989.
{\displaystyle R_{2}={\sqrt {(x_{2}-x_{M})^{2}+(y_{2}-y_{M})^{2}}}={\sqrt {(7-2)^{2}+(-8-2)^{2}}}={\sqrt {5^{2}+(-10)^{2}}}={\sqrt {125}}=11.18033989.}
Hiperbolės evoliutės lanko ilgis L iš taško
C
1
(
1
;
0
)
{\displaystyle C_{1}(1;\;0)}
C
2
(
7
;
−
8
)
{\displaystyle C_{2}(7;\;-8)}
L
=
R
2
−
R
1
=
125
−
1
=
11.18033989
−
1
=
10.18033989.
{\displaystyle L=R_{2}-R_{1}={\sqrt {125}}-1=11.18033989-1=10.18033989.}
Rasime hiperbolės
y
=
2
x
{\displaystyle y={\sqrt {2x}}}
α
=
3
x
+
1
,
{\displaystyle \alpha =3x+1,}
α
−
1
=
3
x
,
{\displaystyle \alpha -1=3x,}
x
=
α
−
1
3
;
{\displaystyle x={\frac {\alpha -1}{3}};}
β
=
−
(
2
x
)
3
2
,
{\displaystyle \beta =-(2x)^{3 \over 2},}
β
=
−
(
2
⋅
α
−
1
3
)
3
2
,
{\displaystyle \beta =-\left(2\cdot {\frac {\alpha -1}{3}}\right)^{3 \over 2},}
β
=
−
8
27
(
α
−
1
)
3
,
{\displaystyle \beta =-{\frac {\sqrt {8}}{\sqrt {27}}}{\sqrt {(\alpha -1)^{3}}},}
Tai yra lygtis evoliutės hiperbolės
y
=
2
x
.
{\displaystyle y={\sqrt {2x}}.}
Toliau rasime hiperbolės evoliutės lanko ilgį L iš taško
C
1
(
1
;
0
)
{\displaystyle C_{1}(1;\;0)}
C
2
(
7
;
−
8
)
{\displaystyle C_{2}(7;\;-8)}
L
=
∫
a
b
1
+
(
y
′
)
2
d
x
,
{\displaystyle L=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx,}
x kinta nuo 1 iki 7, taigi:
d
β
d
α
=
(
−
8
27
(
α
−
1
)
3
)
′
=
−
8
27
⋅
3
2
α
−
1
=
−
2
3
α
−
1
;
{\displaystyle {\frac {d\beta }{d\alpha }}=\left(-{\frac {\sqrt {8}}{\sqrt {27}}}{\sqrt {(\alpha -1)^{3}}}\right)'=-{\frac {\sqrt {8}}{\sqrt {27}}}\cdot {\frac {3}{2}}{\sqrt {\alpha -1}}=-{\frac {\sqrt {2}}{\sqrt {3}}}{\sqrt {\alpha -1}};}
L
=
∫
a
b
1
+
(
β
′
)
2
d
α
=
∫
1
7
1
+
(
−
2
3
α
−
1
)
2
d
α
=
∫
1
7
1
+
2
3
(
α
−
1
)
d
α
=
∫
1
7
2
3
3
2
+
α
−
1
d
α
=
{\displaystyle L=\int _{a}^{b}{\sqrt {1+(\beta ')^{2}}}d\alpha =\int _{1}^{7}{\sqrt {1+(-{\frac {\sqrt {2}}{\sqrt {3}}}{\sqrt {\alpha -1}})^{2}}}d\alpha =\int _{1}^{7}{\sqrt {1+{\frac {2}{3}}(\alpha -1)}}d\alpha =\int _{1}^{7}{\sqrt {\frac {2}{3}}}{\sqrt {{\frac {3}{2}}+\alpha -1}}\;d\alpha =}
=
∫
1
7
2
3
3
−
2
2
+
α
d
α
=
∫
1
7
2
3
1
2
+
α
d
(
1
2
+
α
)
=
2
3
(
1
2
+
α
)
3
2
3
2
|
1
7
=
2
3
2
3
(
1
2
+
α
)
3
2
|
1
7
=
{\displaystyle =\int _{1}^{7}{\sqrt {\frac {2}{3}}}{\sqrt {{\frac {3-2}{2}}+\alpha }}\;d\alpha =\int _{1}^{7}{\sqrt {\frac {2}{3}}}{\sqrt {{\frac {1}{2}}+\alpha }}\;{\mathsf {d}}\left({\frac {1}{2}}+\alpha \right)={\sqrt {\frac {2}{3}}}{\frac {({\frac {1}{2}}+\alpha )^{3 \over 2}}{\frac {3}{2}}}|_{1}^{7}={\frac {2}{3}}{\sqrt {\frac {2}{3}}}({\frac {1}{2}}+\alpha )^{3 \over 2}|_{1}^{7}=}
=
8
27
(
1
2
+
7
)
3
2
−
8
27
(
1
2
+
1
)
3
2
=
{\displaystyle ={\sqrt {\frac {8}{27}}}({\frac {1}{2}}+7)^{3 \over 2}-{\sqrt {\frac {8}{27}}}({\frac {1}{2}}+1)^{3 \over 2}=}
=
0.296296296
⋅
(
7.5
)
3
2
−
0.296296296
⋅
(
1.5
)
3
2
=
{\displaystyle ={\sqrt {0.296296296}}\cdot (7.5)^{3 \over 2}-{\sqrt {0.296296296}}\cdot (1.5)^{3 \over 2}=}
=
0.544331054
⋅
(
421.875
)
1
2
−
0.544331054
⋅
(
3.375
)
1
2
=
{\displaystyle =0.544331054\cdot (421.875)^{1 \over 2}-0.544331054\cdot (3.375)^{1 \over 2}=}
=
0.544331054
⋅
20.53959591
−
0.544331054
⋅
1.837117307
=
11.18033989
−
1
=
10.18033989.
{\displaystyle =0.544331054\cdot 20.53959591-0.544331054\cdot 1.837117307=11.18033989-1=10.18033989.}
keisti
Tegu kreivė užduota parametriškai:
x
=
ϕ
(
t
)
,
y
=
ψ
(
t
)
.
{\displaystyle x=\phi (t),\quad y=\psi (t).}
Tada
y
′
=
d
y
d
x
=
ψ
′
(
t
)
ϕ
′
(
t
)
,
{\displaystyle y'={\frac {dy}{dx}}={\frac {\psi '(t)}{\phi '(t)}},}
y
″
=
d
2
y
d
x
2
=
ψ
″
ϕ
′
−
ψ
′
ϕ
″
(
ϕ
′
)
3
.
{\displaystyle y''={\frac {d^{2}y}{dx^{2}}}={\frac {\psi ''\phi '-\psi '\phi ''}{(\phi ')^{3}}}.}
Arba
d
y
d
x
=
d
y
d
t
d
x
d
t
,
{\displaystyle {\frac {dy}{dx}}={\frac {\frac {dy}{dt}}{\frac {dx}{dt}}},}
d
2
y
d
x
2
=
d
d
x
(
d
y
d
t
d
x
d
t
)
=
d
d
t
(
d
y
d
t
d
x
d
t
)
d
t
d
x
=
d
x
d
t
d
d
t
(
d
y
d
t
)
−
d
y
d
t
d
d
t
(
d
x
d
t
)
(
d
x
d
t
)
2
d
t
d
x
=
d
x
d
t
d
2
y
d
t
2
−
d
y
d
t
d
2
x
d
t
2
(
d
x
d
t
)
2
d
t
d
x
=
d
x
d
t
d
2
y
d
t
2
−
d
y
d
t
d
2
x
d
t
2
(
d
x
d
t
)
2
1
d
x
d
t
=
d
x
d
t
d
2
y
d
t
2
−
d
y
d
t
d
2
x
d
t
2
(
d
x
d
t
)
3
.
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}\right)={\frac {d}{dt}}\left({\frac {\frac {dy}{dt}}{\frac {dx}{dt}}}\right){\frac {dt}{dx}}={\frac {{\frac {dx}{dt}}{\frac {d}{dt}}({\frac {dy}{dt}})-{\frac {dy}{dt}}{\frac {d}{dt}}({\frac {dx}{dt}})}{({\frac {dx}{dt}})^{2}}}{\frac {dt}{dx}}={\frac {{\frac {dx}{dt}}{\frac {d^{2}y}{dt^{2}}}-{\frac {dy}{dt}}{\frac {d^{2}x}{dt^{2}}}}{({\frac {dx}{dt}})^{2}}}{\frac {dt}{dx}}={\frac {{\frac {dx}{dt}}{\frac {d^{2}y}{dt^{2}}}-{\frac {dy}{dt}}{\frac {d^{2}x}{dt^{2}}}}{({\frac {dx}{dt}})^{2}}}{\frac {1}{\frac {dx}{dt}}}={\frac {{\frac {dx}{dt}}{\frac {d^{2}y}{dt^{2}}}-{\frac {dy}{dt}}{\frac {d^{2}x}{dt^{2}}}}{({\frac {dx}{dt}})^{3}}}.}
Įstatydami gautas išraiškas į formulę (3) praeito skyrio, gausime:
K
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
ψ
″
ϕ
′
−
ψ
′
ϕ
″
(
ϕ
′
)
3
|
[
1
+
(
ψ
′
ϕ
′
)
2
]
3
2
=
|
ψ
″
ϕ
′
−
ψ
′
ϕ
″
|
(
ϕ
′
)
3
[
1
(
ϕ
′
)
2
(
(
ϕ
′
)
2
+
(
ψ
′
)
2
)
]
3
2
=
{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|{\frac {\psi ''\phi '-\psi '\phi ''}{(\phi ')^{3}}}\right|}{\left[1+\left({\frac {\psi '}{\phi '}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{(\phi ')^{3}\left[{\frac {1}{(\phi ')^{2}}}((\phi ')^{2}+(\psi ')^{2})\right]^{3 \over 2}}}=}
=
|
ψ
″
ϕ
′
−
ψ
′
ϕ
″
|
(
ϕ
′
)
3
⋅
(
1
(
ϕ
′
)
2
)
3
/
2
[
(
ϕ
′
)
2
+
(
ψ
′
)
2
]
3
2
=
|
ψ
″
ϕ
′
−
ψ
′
ϕ
″
|
[
(
ϕ
′
)
2
+
(
ψ
′
)
2
]
3
2
.
(
1
)
{\displaystyle ={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{(\phi ')^{3}\cdot ({\frac {1}{(\phi ')^{2}}})^{3/2}\left[(\phi ')^{2}+(\psi ')^{2}\right]^{3 \over 2}}}={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{\left[(\phi ')^{2}+(\psi ')^{2}\right]^{3 \over 2}}}.\quad (1)}
Nustatyti kreivį cikloidės
ϕ
(
t
)
=
x
=
a
(
t
−
sin
t
)
,
ψ
(
t
)
=
y
=
a
(
1
−
cos
t
)
{\displaystyle \phi (t)=x=a(t-\sin t),\quad \psi (t)=y=a(1-\cos t)}
jos laisvai pasirenktame taške (x ; y ).
Sprendimas.
ϕ
′
=
d
x
d
t
=
a
(
1
−
cos
t
)
,
ϕ
″
=
d
2
x
d
t
2
=
a
sin
t
,
ψ
′
=
d
y
d
t
=
a
sin
t
,
ψ
″
=
d
2
y
d
t
2
=
a
cos
t
.
{\displaystyle \phi '={\frac {dx}{dt}}=a(1-\cos t),\quad \phi ''={\frac {d^{2}x}{dt^{2}}}=a\sin t,\quad \psi '={\frac {dy}{dt}}=a\sin t,\psi ''={\frac {d^{2}y}{dt^{2}}}=a\cos t.}
Įstatydami gautas išraiškas į formulę (3), randame:
K
=
|
ψ
″
ϕ
′
−
ψ
′
ϕ
″
|
[
(
ϕ
′
)
2
+
(
ψ
′
)
2
]
3
2
=
|
a
(
1
−
cos
t
)
a
cos
t
−
a
sin
t
⋅
a
sin
t
|
[
a
2
(
1
−
cos
t
)
2
+
a
2
sin
2
t
]
3
2
=
|
a
2
cos
t
−
a
2
cos
2
t
−
a
2
sin
2
t
|
[
a
2
−
2
a
2
cos
t
+
a
2
cos
2
t
+
a
2
sin
2
t
]
3
2
=
{\displaystyle K={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{\left[(\phi ')^{2}+(\psi ')^{2}\right]^{3 \over 2}}}={\frac {|a(1-\cos t)a\cos t-a\sin t\cdot a\sin t|}{\left[a^{2}(1-\cos t)^{2}+a^{2}\sin ^{2}t\right]^{3 \over 2}}}={\frac {|a^{2}\cos t-a^{2}\cos ^{2}t-a^{2}\sin ^{2}t|}{\left[a^{2}-2a^{2}\cos t+a^{2}\cos ^{2}t+a^{2}\sin ^{2}t\right]^{3 \over 2}}}=}
=
|
a
2
cos
t
−
a
2
|
[
2
a
2
−
2
a
2
cos
t
]
3
2
=
a
2
|
cos
t
−
1
|
2
3
2
a
3
[
1
−
cos
t
]
3
2
=
1
2
3
2
a
[
1
−
cos
t
]
1
2
=
1
2
3
a
⋅
2
sin
t
2
=
1
4
a
|
sin
t
2
|
.
{\displaystyle ={\frac {|a^{2}\cos t-a^{2}|}{\left[2a^{2}-2a^{2}\cos t\right]^{3 \over 2}}}={\frac {a^{2}|\cos t-1|}{2^{3 \over 2}a^{3}\left[1-\cos t\right]^{3 \over 2}}}={\frac {1}{2^{3 \over 2}a\left[1-\cos t\right]^{1 \over 2}}}={\frac {1}{{\sqrt {2^{3}}}a\cdot {\sqrt {2}}\sin {\frac {t}{2}}}}={\frac {1}{4a|\sin {\frac {t}{2}}|}}.}
keisti
Tegu kreivė užrašyta lygtimi pavidalo
ρ
=
f
(
θ
)
.
(
1
)
{\displaystyle \rho =f(\theta ).\quad (1)}
Užrašysime formules perėjimo iš polinių koordinačių į dekartines:
x
=
ρ
cos
θ
,
y
=
ρ
sin
θ
.
(
2
)
{\displaystyle x=\rho \cos \theta ,\quad y=\rho \sin \theta .\quad (2)}
Jeigu į šitas formules įstatyti vietoje
ρ
{\displaystyle \rho }
θ
,
{\displaystyle \theta ,}
f
(
θ
)
,
{\displaystyle f(\theta ),\;}
x
=
f
(
θ
)
cos
θ
,
y
=
f
(
θ
)
sin
θ
.
(
3
)
{\displaystyle x=f(\theta )\cos \theta ,\quad y=f(\theta )\sin \theta .\quad (3)}
Paskutines lygtis galima nagrinėti kaip parametrines lygtis kreivės (1), be kita ko parametras yra
θ
.
{\displaystyle \theta .}
Tada
d
y
d
x
=
d
y
d
θ
d
x
d
θ
,
{\displaystyle {\frac {dy}{dx}}={\frac {\frac {dy}{d\theta }}{\frac {dx}{d\theta }}},}
d
2
y
d
x
2
=
d
d
x
(
d
y
d
θ
d
x
d
θ
)
=
d
d
θ
(
d
y
d
θ
d
x
d
θ
)
d
θ
d
x
=
d
x
d
θ
d
d
θ
(
d
y
d
θ
)
−
d
y
d
θ
d
d
θ
(
d
x
d
θ
)
(
d
x
d
θ
)
2
d
θ
d
x
=
d
x
d
θ
d
2
y
d
θ
2
−
d
y
d
θ
d
2
x
d
θ
2
(
d
x
d
θ
)
2
d
θ
d
x
=
d
x
d
θ
d
2
y
d
θ
2
−
d
y
d
θ
d
2
x
d
θ
2
(
d
x
d
θ
)
2
1
d
x
d
θ
=
d
x
d
θ
d
2
y
d
θ
2
−
d
y
d
θ
d
2
x
d
θ
2
(
d
x
d
θ
)
3
.
{\displaystyle {\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {\frac {dy}{d\theta }}{\frac {dx}{d\theta }}}\right)={\frac {d}{d\theta }}\left({\frac {\frac {dy}{d\theta }}{\frac {dx}{d\theta }}}\right){\frac {d\theta }{dx}}={\frac {{\frac {dx}{d\theta }}{\frac {d}{d\theta }}({\frac {dy}{d\theta }})-{\frac {dy}{d\theta }}{\frac {d}{d\theta }}({\frac {dx}{d\theta }})}{({\frac {dx}{d\theta }})^{2}}}{\frac {d\theta }{dx}}={\frac {{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}{({\frac {dx}{d\theta }})^{2}}}{\frac {d\theta }{dx}}={\frac {{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}{({\frac {dx}{d\theta }})^{2}}}{\frac {1}{\frac {dx}{d\theta }}}={\frac {{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}{({\frac {dx}{d\theta }})^{3}}}.}
d
x
d
θ
=
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
,
d
y
d
θ
=
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
,
{\displaystyle {\frac {dx}{d\theta }}={\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta ,\quad {\frac {dy}{d\theta }}={\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta ,}
d
2
x
d
θ
2
=
(
d
2
ρ
d
θ
2
cos
θ
−
d
ρ
d
θ
sin
θ
)
−
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
=
d
2
ρ
d
θ
2
cos
θ
−
2
d
ρ
d
θ
sin
θ
−
ρ
cos
θ
,
{\displaystyle {\frac {d^{2}x}{d\theta ^{2}}}=({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta -{\frac {d\rho }{d\theta }}\sin \theta )-({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )={\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta ,}
d
2
y
d
θ
2
=
(
d
2
ρ
d
θ
2
sin
θ
+
d
ρ
d
θ
cos
θ
)
+
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
=
d
2
ρ
d
θ
2
sin
θ
+
2
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
.
{\displaystyle {\frac {d^{2}y}{d\theta ^{2}}}=({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta +{\frac {d\rho }{d\theta }}\cos \theta )+({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )={\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta +2{\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta .}
Įstatant paskutines išraiškas į formulę (1) praeito skyriaus, gausime formulę apskaičiavimui kreivio kreivės polinėse koordinatėse:
K
=
|
d
2
y
d
x
2
|
[
1
+
(
d
y
d
x
)
2
]
3
2
=
|
ψ
″
ϕ
′
−
ψ
′
ϕ
″
(
ϕ
′
)
3
|
[
1
+
(
ψ
′
ϕ
′
)
2
]
3
2
=
|
ψ
″
ϕ
′
−
ψ
′
ϕ
″
|
(
ϕ
′
)
3
[
1
(
ϕ
′
)
2
(
(
ϕ
′
)
2
+
(
ψ
′
)
2
)
]
3
2
=
|
ψ
″
ϕ
′
−
ψ
′
ϕ
″
|
(
ϕ
′
)
3
⋅
(
1
(
ϕ
′
)
2
)
3
/
2
[
(
ϕ
′
)
2
+
(
ψ
′
)
2
]
3
2
=
{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|{\frac {\psi ''\phi '-\psi '\phi ''}{(\phi ')^{3}}}\right|}{\left[1+\left({\frac {\psi '}{\phi '}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{(\phi ')^{3}\left[{\frac {1}{(\phi ')^{2}}}((\phi ')^{2}+(\psi ')^{2})\right]^{3 \over 2}}}={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{(\phi ')^{3}\cdot ({\frac {1}{(\phi ')^{2}}})^{3/2}\left[(\phi ')^{2}+(\psi ')^{2}\right]^{3 \over 2}}}=}
=
|
ψ
″
ϕ
′
−
ψ
′
ϕ
″
|
[
ϕ
′
2
+
ψ
′
2
]
3
2
=
|
d
2
y
d
θ
2
d
x
d
θ
−
d
y
d
θ
d
2
x
d
θ
2
|
[
(
d
x
d
θ
)
2
+
(
d
y
d
θ
)
2
]
3
2
=
{\displaystyle ={\frac {|\psi ''\phi '-\psi '\phi ''|}{[\phi '^{2}+\psi '^{2}]^{3 \over 2}}}={\frac {\left|{\frac {d^{2}y}{d\theta ^{2}}}{\frac {dx}{d\theta }}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}\right|}{\left[\left({\frac {dx}{d\theta }}\right)^{2}+\left({\frac {dy}{d\theta }}\right)^{2}\right]^{3 \over 2}}}=}
=
|
(
d
2
ρ
d
θ
2
sin
θ
+
2
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
−
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
(
d
2
ρ
d
θ
2
cos
θ
−
2
d
ρ
d
θ
sin
θ
−
ρ
cos
θ
)
|
[
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
2
]
3
2
=
{\displaystyle ={\frac {|({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta +2{\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )-({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )|}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}
=
(
d
2
ρ
d
θ
2
sin
θ
d
ρ
d
θ
cos
θ
+
2
d
ρ
d
θ
cos
θ
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
d
ρ
d
θ
cos
θ
−
d
2
ρ
d
θ
2
sin
θ
ρ
sin
θ
−
2
d
ρ
d
θ
cos
θ
ρ
sin
θ
+
ρ
sin
θ
ρ
sin
θ
)
[
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
2
]
3
2
−
{\displaystyle ={\frac {({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta {\frac {d\rho }{d\theta }}\cos \theta +2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta -2{\frac {d\rho }{d\theta }}\cos \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta )}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}-}
−
(
d
2
ρ
d
θ
2
cos
θ
d
ρ
d
θ
sin
θ
−
2
d
ρ
d
θ
sin
θ
d
ρ
d
θ
sin
θ
−
ρ
cos
θ
d
ρ
d
θ
sin
θ
+
d
2
ρ
d
θ
2
cos
θ
ρ
cos
θ
−
2
d
ρ
d
θ
sin
θ
ρ
cos
θ
−
ρ
cos
θ
ρ
cos
θ
)
[
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
2
]
3
2
=
{\displaystyle -{\frac {({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta {\frac {d\rho }{d\theta }}\sin \theta -2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta {\frac {d\rho }{d\theta }}\sin \theta +{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta \rho \cos \theta -\rho \cos \theta \rho \cos \theta )}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}
=
d
2
ρ
d
θ
2
sin
θ
d
ρ
d
θ
cos
θ
+
2
d
ρ
d
θ
cos
θ
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
d
ρ
d
θ
cos
θ
−
d
2
ρ
d
θ
2
sin
θ
ρ
sin
θ
−
2
d
ρ
d
θ
cos
θ
ρ
sin
θ
+
ρ
sin
θ
ρ
sin
θ
[
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
2
]
3
2
+
{\displaystyle ={\frac {{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta {\frac {d\rho }{d\theta }}\cos \theta +2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta -2{\frac {d\rho }{d\theta }}\cos \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}+}
+
−
d
2
ρ
d
θ
2
cos
θ
d
ρ
d
θ
sin
θ
+
2
d
ρ
d
θ
sin
θ
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
d
ρ
d
θ
sin
θ
−
d
2
ρ
d
θ
2
cos
θ
ρ
cos
θ
+
2
d
ρ
d
θ
sin
θ
ρ
cos
θ
+
ρ
cos
θ
ρ
cos
θ
[
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
2
]
3
2
=
{\displaystyle +{\frac {-{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta {\frac {d\rho }{d\theta }}\sin \theta +2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta {\frac {d\rho }{d\theta }}\sin \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta +2{\frac {d\rho }{d\theta }}\sin \theta \rho \cos \theta +\rho \cos \theta \rho \cos \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}
=
2
d
ρ
d
θ
cos
θ
d
ρ
d
θ
cos
θ
−
d
2
ρ
d
θ
2
sin
θ
ρ
sin
θ
+
ρ
sin
θ
ρ
sin
θ
[
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
2
]
3
2
+
{\displaystyle ={\frac {2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}+}
+
2
d
ρ
d
θ
sin
θ
d
ρ
d
θ
sin
θ
−
d
2
ρ
d
θ
2
cos
θ
ρ
cos
θ
+
ρ
cos
θ
ρ
cos
θ
[
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
2
]
3
2
=
{\displaystyle +{\frac {2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta +\rho \cos \theta \rho \cos \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}
=
|
2
(
d
ρ
d
θ
)
2
cos
2
θ
+
2
(
d
ρ
d
θ
)
2
sin
2
θ
−
ρ
d
2
ρ
d
θ
2
sin
2
θ
−
ρ
d
2
ρ
d
θ
2
cos
2
θ
+
ρ
2
sin
2
θ
+
ρ
2
cos
2
θ
|
[
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
2
]
3
2
=
{\displaystyle ={\frac {|2({\frac {d\rho }{d\theta }})^{2}\cos ^{2}\theta +2({\frac {d\rho }{d\theta }})^{2}\sin ^{2}\theta -\rho {\frac {d^{2}\rho }{d\theta ^{2}}}\sin ^{2}\theta -\rho {\frac {d^{2}\rho }{d\theta ^{2}}}\cos ^{2}\theta +\rho ^{2}\sin ^{2}\theta +\rho ^{2}\cos ^{2}\theta |}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}
=
|
2
(
d
ρ
d
θ
)
2
−
ρ
d
2
ρ
d
θ
2
+
ρ
2
|
[
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
2
]
3
2
=
{\displaystyle ={\frac {|2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}+\rho ^{2}|}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}
=
|
ρ
2
+
2
(
d
ρ
d
θ
)
2
−
ρ
d
2
ρ
d
θ
2
|
[
(
(
d
ρ
d
θ
)
2
cos
2
θ
−
2
d
ρ
d
θ
cos
(
θ
)
ρ
sin
(
θ
)
+
ρ
2
sin
2
θ
)
+
(
(
d
ρ
d
θ
)
2
sin
2
θ
+
2
d
ρ
d
θ
sin
(
θ
)
ρ
cos
(
θ
)
+
ρ
2
cos
2
θ
)
]
3
2
=
{\displaystyle ={\frac {|\rho ^{2}+2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}|}{[(({\frac {d\rho }{d\theta }})^{2}\cos ^{2}\theta -2{\frac {d\rho }{d\theta }}\cos(\theta )\rho \sin(\theta )+\rho ^{2}\sin ^{2}\theta )+(({\frac {d\rho }{d\theta }})^{2}\sin ^{2}\theta +2{\frac {d\rho }{d\theta }}\sin(\theta )\rho \cos(\theta )+\rho ^{2}\cos ^{2}\theta )]^{3 \over 2}}}=}
=
|
ρ
2
+
2
(
d
ρ
d
θ
)
2
−
ρ
d
2
ρ
d
θ
2
|
[
(
d
ρ
d
θ
)
2
cos
2
θ
+
(
d
ρ
d
θ
)
2
sin
2
θ
+
ρ
2
sin
2
θ
+
ρ
2
cos
2
θ
)
]
3
2
=
|
ρ
2
+
2
(
d
ρ
d
θ
)
2
−
ρ
d
2
ρ
d
θ
2
|
[
(
d
ρ
d
θ
)
2
+
ρ
2
]
3
2
=
|
ρ
2
+
2
(
ρ
′
)
2
−
ρ
ρ
″
|
(
ρ
2
+
(
ρ
′
)
2
)
3
/
2
.
(
4
)
{\displaystyle ={\frac {|\rho ^{2}+2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}|}{[({\frac {d\rho }{d\theta }})^{2}\cos ^{2}\theta +({\frac {d\rho }{d\theta }})^{2}\sin ^{2}\theta +\rho ^{2}\sin ^{2}\theta +\rho ^{2}\cos ^{2}\theta )]^{3 \over 2}}}={\frac {|\rho ^{2}+2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}|}{[({\frac {d\rho }{d\theta }})^{2}+\rho ^{2}]^{3 \over 2}}}={\frac {|\rho ^{2}+2(\rho ')^{2}-\rho \rho ''|}{(\rho ^{2}+(\rho ')^{2})^{3/2}}}.\quad (4)}
Vaizdas:Kreivispav144.jpg 144 pav. Nustatyti kreivį Archimedo spiralės
ρ
=
a
θ
(
a
>
0
)
{\displaystyle \rho =a\theta \;\;(a>0)}
Sprendimas .
d
ρ
d
θ
=
a
;
d
2
ρ
d
θ
2
=
0.
{\displaystyle {\frac {d\rho }{d\theta }}=a;\quad {\frac {d^{2}\rho }{d\theta ^{2}}}=0.}
Iš to seka,
K
=
|
ρ
2
+
2
ρ
′
2
−
ρ
ρ
″
|
(
ρ
2
+
ρ
′
2
)
3
/
2
=
|
a
2
θ
2
+
2
a
2
−
a
θ
⋅
0
|
(
a
2
θ
2
+
a
2
)
3
/
2
=
|
a
2
θ
2
+
2
a
2
|
a
3
(
θ
2
+
1
)
3
/
2
=
θ
2
+
2
a
(
θ
2
+
1
)
3
/
2
.
{\displaystyle K={\frac {|\rho ^{2}+2\rho '^{2}-\rho \rho ''|}{(\rho ^{2}+\rho '^{2})^{3/2}}}={\frac {|a^{2}\theta ^{2}+2a^{2}-a\theta \cdot 0|}{(a^{2}\theta ^{2}+a^{2})^{3/2}}}={\frac {|a^{2}\theta ^{2}+2a^{2}|}{a^{3}(\theta ^{2}+1)^{3/2}}}={\frac {\theta ^{2}+2}{a(\theta ^{2}+1)^{3/2}}}.}
Pastebėsime, kad su didelėmis reikšmėmis
θ
{\displaystyle \theta }
θ
2
+
2
θ
2
≈
1
,
θ
2
+
1
θ
2
≈
1
;
{\displaystyle {\frac {\theta ^{2}+2}{\theta ^{2}}}\approx 1,\;\;{\frac {\theta ^{2}+1}{\theta ^{2}}}\approx 1;}
θ
2
+
2
{\displaystyle \theta ^{2}+2}
θ
2
{\displaystyle \theta ^{2}}
θ
2
+
1
{\displaystyle \theta ^{2}+1}
θ
2
,
{\displaystyle \theta ^{2},}
θ
{\displaystyle \theta }
K
≈
1
a
θ
2
(
θ
2
)
3
/
2
=
1
a
θ
.
{\displaystyle K\approx {\frac {1}{a}}{\frac {\theta ^{2}}{(\theta ^{2})^{3/2}}}={\frac {1}{a\theta }}.}
Tokiu budu, su didelėmis reikšmėmis
θ
{\displaystyle \theta }
a
θ
{\displaystyle a\theta }
keisti
Kreivės užrašytos parametriškai
x
=
ϕ
(
t
)
,
y
=
ψ
(
t
)
,
z
=
ω
(
t
)
,
{\displaystyle x=\phi (t),\quad y=\psi (t),\quad z=\omega (t),}
kreivio apskaičiavimo formulė yra:
K
2
=
1
R
2
=
(
ψ
′
ω
″
−
ω
′
ψ
″
)
2
+
(
ϕ
′
ω
″
−
ω
′
ϕ
″
)
2
+
(
ϕ
′
ψ
″
−
ψ
′
ϕ
″
)
2
[
(
ϕ
′
)
2
+
(
ψ
′
)
2
+
(
ω
′
)
2
]
3
;
{\displaystyle K^{2}={\frac {1}{R^{2}}}={\frac {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}{[(\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2}]^{3}}};}
K
=
1
R
=
(
ψ
′
ω
″
−
ω
′
ψ
″
)
2
+
(
ϕ
′
ω
″
−
ω
′
ϕ
″
)
2
+
(
ϕ
′
ψ
″
−
ψ
′
ϕ
″
)
2
(
(
ϕ
′
)
2
+
(
ψ
′
)
2
+
(
ω
′
)
2
)
3
2
;
{\displaystyle K={\frac {1}{R}}={\frac {\sqrt {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}}{((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3 \over 2}}};}
R
=
(
(
ϕ
′
)
2
+
(
ψ
′
)
2
+
(
ω
′
)
2
)
3
2
(
ψ
′
ω
″
−
ω
′
ψ
″
)
2
+
(
ϕ
′
ω
″
−
ω
′
ϕ
″
)
2
+
(
ϕ
′
ψ
″
−
ψ
′
ϕ
″
)
2
.
{\displaystyle R={\frac {((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3 \over 2}}{\sqrt {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}}}.}
Galima naudotis ir šita formule:
K
2
=
1
R
2
=
(
(
ϕ
′
)
2
+
(
ψ
′
)
2
+
(
ω
′
)
2
)
(
(
ϕ
″
)
2
+
(
ψ
″
)
2
+
(
ω
″
)
2
)
−
(
ϕ
′
ϕ
″
+
ψ
′
ψ
″
+
ω
′
ω
″
)
2
(
(
ϕ
′
)
2
+
(
ψ
′
)
2
+
(
ω
′
)
2
)
3
.
{\displaystyle K^{2}={\frac {1}{R^{2}}}={\frac {((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})((\phi '')^{2}+(\psi '')^{2}+(\omega '')^{2})-(\phi '\phi ''+\psi '\psi ''+\omega '\omega '')^{2}}{((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3}}}.}
Kreivės liestinės vektorius taške
M
(
x
M
;
y
M
;
z
M
)
{\displaystyle M(x_{M};y_{M};z_{M})}
a
→
=
{
ϕ
′
(
t
)
;
ψ
′
(
t
)
;
ω
′
(
t
)
}
.
{\displaystyle {\vec {a}}=\{\phi '(t);\psi '(t);\omega '(t)\}.}
M
(
x
M
;
y
M
;
z
M
)
.
{\displaystyle M(x_{M};y_{M};z_{M}).}
Erdvinės kreivės liestinės lygtis taške
M
(
x
M
;
y
M
;
z
M
)
{\displaystyle M(x_{M};y_{M};z_{M})}
t reikšmė turi būti tokia, kad ją įstčius į funkcijas gautusi taško
M
(
x
M
;
y
M
;
z
M
)
{\displaystyle M(x_{M};y_{M};z_{M})}
x
−
x
M
ϕ
′
(
t
)
=
y
−
y
M
ψ
′
(
t
)
=
z
−
z
M
ω
′
(
t
)
;
{\displaystyle {\frac {x-x_{M}}{\phi '(t)}}={\frac {y-y_{M}}{\psi '(t)}}={\frac {z-z_{M}}{\omega '(t)}};}
arba
{
x
−
x
M
=
t
ϕ
′
(
t
)
,
y
−
y
M
=
t
ψ
′
(
t
)
,
z
−
z
M
=
t
ω
′
(
t
)
.
{\displaystyle {\begin{cases}x-x_{M}=t\phi '(t),&\\y-y_{M}=t\psi '(t),&\\z-z_{M}=t\omega '(t).&\end{cases}}}
1. Kreivės normalės vektorius (kreivio spindulio vektorius) taške
M
(
x
M
;
y
M
;
z
M
)
{\displaystyle M(x_{M};y_{M};z_{M})}
b
→
=
{
−
2
3
ϕ
′
(
t
)
;
1
3
ψ
′
(
t
)
;
1
3
ω
′
(
t
)
}
{\displaystyle {\vec {b}}=\{-{\frac {2}{3\phi '(t)}};{\frac {1}{3\psi '(t)}};{\frac {1}{3\omega '(t)}}\}}
p yra riboje
[
1
;
∞
)
,
{\displaystyle [1;\infty ),}
t
p
,
1
≤
p
,
{\displaystyle t^{p},\;1\leq p,}
t
2
{\displaystyle t^{2}}
t
3
{\displaystyle t^{3}}
t
3
2
{\displaystyle t^{3 \over 2}}
t
5
.
{\displaystyle t^{5}.}
−
2
3
{\displaystyle -{\frac {2}{3}}}
t prirašoma
1
3
,
{\displaystyle {\frac {1}{3}},}
1
≤
p
,
t
p
.
{\displaystyle 1\leq p,\;t^{p}.}
ϕ
(
t
)
=
t
2
,
ψ
(
t
)
=
t
3
,
ω
(
t
)
=
t
4
,
{\displaystyle \phi (t)=t^{2},\;\psi (t)=t^{3},\;\omega (t)=t^{4},}
b
→
=
{
−
2
3
⋅
1
2
x
M
;
1
3
⋅
1
3
x
M
2
;
1
3
⋅
1
4
x
M
3
}
.
{\displaystyle {\vec {b}}=\{-{2 \over 3}\cdot {\frac {1}{2x_{M}}};{1 \over 3}\cdot {\frac {1}{3x_{M}^{2}}};{1 \over 3}\cdot {\frac {1}{4x_{M}^{3}}}\}.}
2. Kreivės užrašytos parametriškai
x
=
ϕ
(
t
)
,
y
=
ψ
(
t
)
,
z
=
ω
(
t
)
,
{\displaystyle x=\phi (t),\;y=\psi (t),\;z=\omega (t),}
b
→
=
{
1
±
ϕ
′
(
t
)
;
1
±
ψ
′
(
t
)
;
1
±
ω
′
(
t
)
}
.
{\displaystyle {\vec {b}}=\{{\frac {1}{\pm \phi '(t)}};{\frac {1}{\pm \psi '(t)}};{\frac {1}{\pm \omega '(t)}}\}.}
p
<
1
,
t
p
.
{\displaystyle p<1,\;t^{p}.}
ϕ
(
t
)
=
t
1
3
,
ψ
(
t
)
=
t
1
2
,
ω
(
t
)
=
t
,
{\displaystyle \phi (t)=t^{1 \over 3},\;\psi (t)=t^{1 \over 2},\;\omega (t)=t,}
a
→
=
{
ϕ
′
(
t
)
;
ψ
′
(
t
)
;
ω
′
(
t
)
}
=
{
t
−
2
3
3
;
1
2
t
;
1
}
.
{\displaystyle {\vec {a}}=\{\phi '(t);\psi '(t);\omega '(t)\}=\{{\frac {t^{-{\frac {2}{3}}}}{3}};{\frac {1}{2{\sqrt {t}}}};1\}.}
b
→
=
{
−
1
3
ϕ
′
(
t
)
;
−
1
3
ψ
′
(
t
)
;
2
3
ω
′
(
t
)
}
=
{
−
1
3
⋅
3
t
2
3
;
−
1
3
⋅
2
t
;
2
3
}
.
{\displaystyle {\vec {b}}=\{-{\frac {1}{3\phi '(t)}};-{\frac {1}{3\psi '(t)}};{\frac {2}{3\omega '(t)}}\}=\{-{\frac {1}{3}}\cdot 3t^{2 \over 3};-{\frac {1}{3}}\cdot 2{\sqrt {t}};{\frac {2}{3}}\}.}
Skaičius
2
3
{\displaystyle {\frac {2}{3}}}
Sudetingesnis pavyzdis, kai kreivė užrašyta parametriškai
ϕ
(
t
)
=
t
,
ψ
(
t
)
=
t
1
3
,
ω
(
t
)
=
t
3
,
{\displaystyle \phi (t)=t,\;\psi (t)=t^{1 \over 3},\;\omega (t)=t^{3},}
t
=
5
{\displaystyle t=5}
arctan
ψ
(
5
)
ϕ
(
5
)
=
arctan
5
1
3
5
=
18.8804412
∘
,
{\displaystyle \arctan {\frac {\psi (5)}{\phi (5)}}=\arctan {\frac {5^{1 \over 3}}{5}}=18.8804412^{\circ },}
arctan
ω
(
5
)
ϕ
(
5
)
=
arctan
5
3
5
=
arctan
25
=
87.70938996
∘
{\displaystyle \arctan {\frac {\omega (5)}{\phi (5)}}=\arctan {\frac {5^{3}}{5}}=\arctan 25=87.70938996^{\circ }}
90
−
18.88
=
71.12
<
87.709
{\displaystyle 90-18.88=71.12<87.709}
b
→
=
{
1
3
ϕ
′
(
t
)
;
2
−
3
ψ
′
(
t
)
;
1
3
ω
′
(
t
)
}
=
{
1
3
;
−
2
t
2
3
;
1
9
t
2
}
.
{\displaystyle {\vec {b}}=\{{\frac {1}{3\phi '(t)}};{\frac {2}{-3\psi '(t)}};{\frac {1}{3\omega '(t)}}\}=\{{\frac {1}{3}};-2t^{2 \over 3};{\frac {1}{9t^{2}}}\}.}
Bet galėtų būti ir toks (arba šis pavyzdis iš vis neturi normalės vektoriaus):
b
→
=
{
1
−
3
ϕ
′
(
t
)
;
1
−
3
ψ
′
(
t
)
;
2
3
ω
′
(
t
)
}
=
{
−
1
3
;
−
t
2
3
;
2
9
t
2
}
.
{\displaystyle {\vec {b}}=\{{\frac {1}{-3\phi '(t)}};{\frac {1}{-3\psi '(t)}};{\frac {2}{3\omega '(t)}}\}=\{-{\frac {1}{3}};-t^{2 \over 3};{\frac {2}{9t^{2}}}\}.}
Pavyzdžiui, funkcija užrašyta parametriškai
ϕ
(
t
)
=
t
1
3
,
ψ
(
t
)
=
t
1
2
,
ω
(
t
)
=
t
3
,
{\displaystyle \phi (t)=t^{1 \over 3},\;\psi (t)=t^{1 \over 2},\;\omega (t)=t^{3},}
b
→
=
{
1
−
3
ϕ
′
(
t
)
;
1
−
3
ψ
′
(
t
)
;
2
3
ω
′
(
t
)
}
;
{\displaystyle {\vec {b}}=\{{\frac {1}{-3\phi '(t)}};{\frac {1}{-3\psi '(t)}};{\frac {2}{3\omega '(t)}}\};}
b
→
=
{
−
1
3
⋅
1
1
3
t
2
3
;
1
3
⋅
1
−
1
2
t
;
2
3
⋅
1
3
t
2
}
,
{\displaystyle {\vec {b}}=\{-{\frac {1}{3}}\cdot {\frac {1}{\frac {1}{3{\sqrt[{3}]{t^{2}}}}}};{\frac {1}{3}}\cdot {\frac {1}{-{\frac {1}{2{\sqrt {t}}}}}};{\frac {2}{3}}\cdot {\frac {1}{3t^{2}}}\},}
b
→
=
{
−
1
3
⋅
3
t
2
3
;
1
3
⋅
(
−
2
)
t
;
2
3
⋅
1
3
t
2
}
,
{\displaystyle {\vec {b}}=\{-{\frac {1}{3}}\cdot 3t^{2 \over 3};{\frac {1}{3}}\cdot (-2){\sqrt {t}};{\frac {2}{3}}\cdot {\frac {1}{3t^{2}}}\},}
b
→
=
{
−
t
2
3
;
−
2
t
3
;
2
9
t
2
}
.
{\displaystyle {\vec {b}}=\{-{\sqrt[{3}]{t^{2}}};-{\frac {2{\sqrt {t}}}{3}};{\frac {2}{9t^{2}}}\}.}
Ar prirašyti
2
3
{\displaystyle {\frac {2}{3}}}
1
3
{\displaystyle {\frac {1}{3}}}
2
3
{\displaystyle {\frac {2}{3}}}
2
3
{\displaystyle {\frac {2}{3}}}
Kitas pavyzdis, kreivės užrašytos parametriškai
ϕ
(
t
)
=
t
{\displaystyle \phi (t)=t}
ψ
(
t
)
=
t
1
2
{\displaystyle \psi (t)=t^{1 \over 2}}
ω
(
t
)
=
t
3
{\displaystyle \omega (t)=t^{3}}
b
→
=
{
1
3
ϕ
′
(
t
)
;
2
−
3
ψ
′
(
t
)
;
1
3
ω
′
(
t
)
}
;
{\displaystyle {\vec {b}}=\{{\frac {1}{3\phi '(t)}};{\frac {2}{-3\psi '(t)}};{\frac {1}{3\omega '(t)}}\};}
b
→
=
{
1
3
;
2
3
⋅
1
−
1
2
t
;
1
3
⋅
1
3
t
2
}
;
{\displaystyle {\vec {b}}=\{{\frac {1}{3}};{\frac {2}{3}}\cdot {\frac {1}{-{\frac {1}{2{\sqrt {t}}}}}};{\frac {1}{3}}\cdot {\frac {1}{3t^{2}}}\};}
b
→
=
{
1
3
;
−
4
t
3
;
1
9
t
2
}
.
{\displaystyle {\vec {b}}=\{{\frac {1}{3}};-{\frac {4{\sqrt {t}}}{3}};{\frac {1}{9t^{2}}}\}.}
Dar vienas pavyzdis, kreivės užrašytos parametriškai
ϕ
(
t
)
=
t
{\displaystyle \phi (t)=t}
ψ
(
t
)
=
t
−
2
{\displaystyle \psi (t)=t^{-2}}
ω
(
t
)
=
t
−
3
{\displaystyle \omega (t)=t^{-3}}
b
→
=
{
2
3
ϕ
′
(
t
)
;
1
−
3
ψ
′
(
t
)
;
1
−
3
ω
′
(
t
)
}
=
{
2
3
;
−
1
3
⋅
1
(
−
2
)
t
−
3
;
−
1
3
⋅
1
(
−
3
)
t
−
4
}
=
{
2
3
;
t
3
6
;
t
4
9
}
,
{\displaystyle {\vec {b}}=\{{\frac {2}{3\phi '(t)}};{\frac {1}{-3\psi '(t)}};{\frac {1}{-3\omega '(t)}}\}=\{{\frac {2}{3}};-{\frac {1}{3}}\cdot {\frac {1}{(-2)t^{-3}}};-{\frac {1}{3}}\cdot {\frac {1}{(-3)t^{-4}}}\}=\{{\frac {2}{3}};{\frac {t^{3}}{6}};{\frac {t^{4}}{9}}\},}
bet tai reiškia, kad visur pliusai, o du minusus turi liestinės vektorius
a
→
=
{
1
;
−
2
t
−
3
;
−
3
t
−
4
}
,
{\displaystyle {\vec {a}}=\{1;-2t^{-3};-3t^{-4}\},}
Todėl galime užrašyti kreivės normalės lygtį:
{
x
−
x
M
=
2
t
±
3
ϕ
′
(
t
)
,
y
−
y
M
=
t
±
3
ψ
′
(
t
)
,
z
−
z
M
=
t
±
3
ω
′
(
t
)
;
{\displaystyle {\begin{cases}x-x_{M}={\frac {2t}{\pm 3\phi '(t)}},&\\y-y_{M}={\frac {t}{\pm 3\psi '(t)}},&\\z-z_{M}={\frac {t}{\pm 3\omega '(t)}};&\end{cases}}}
arba
x
−
x
M
2
±
3
ϕ
′
(
t
)
=
y
−
y
M
1
±
3
ψ
′
(
t
)
=
z
−
z
M
1
±
3
ω
′
(
t
)
,
{\displaystyle {\frac {x-x_{M}}{\frac {2}{\pm 3\phi '(t)}}}={\frac {y-y_{M}}{\frac {1}{\pm 3\psi '(t)}}}={\frac {z-z_{M}}{\frac {1}{\pm 3\omega '(t)}}},}
±
3
(
x
−
x
M
)
ϕ
′
(
t
)
2
=
±
3
(
y
−
y
M
)
ψ
′
(
t
)
=
±
3
(
z
−
z
M
)
ω
′
(
t
)
.
{\displaystyle \pm {\frac {3(x-x_{M})\phi '(t)}{2}}=\pm 3(y-y_{M})\psi '(t)=\pm 3(z-z_{M})\omega '(t).}
Turime lygčių sistemą:
{
±
3
2
(
x
−
x
M
)
ϕ
′
(
t
)
=
±
3
(
y
−
y
M
)
ψ
′
(
t
)
=
±
3
(
z
−
z
M
)
ω
′
(
t
)
,
(
x
−
x
M
)
2
+
(
y
−
y
M
)
2
+
(
z
−
z
M
)
2
=
1
K
2
.
{\displaystyle {\begin{cases}\pm {\frac {3}{2}}(x-x_{M})\phi '(t)=\pm 3(y-y_{M})\psi '(t)=\pm 3(z-z_{M})\omega '(t),&\\(x-x_{M})^{2}+(y-y_{M})^{2}+(z-z_{M})^{2}={\frac {1}{K^{2}}}.&\end{cases}}}
Parametro t reikšmė turi būti tokia, kad ją įstačius į funkcijas
ϕ
(
t
)
,
ψ
(
t
)
,
ω
(
t
)
{\displaystyle \phi (t),\;\psi (t),\;\omega (t)}
M
(
x
M
;
y
M
;
z
M
)
{\displaystyle M(x_{M};y_{M};z_{M})}
Gauname, kad
y
−
y
M
=
±
ϕ
′
(
t
)
(
x
−
x
M
)
±
2
ψ
′
(
t
)
;
{\displaystyle y-y_{M}={\frac {\pm \phi '(t)(x-x_{M})}{\pm 2\psi '(t)}};}
z
−
z
M
=
±
ϕ
′
(
t
)
(
x
−
x
M
)
±
2
ω
′
(
t
)
.
{\displaystyle z-z_{M}={\frac {\pm \phi '(t)(x-x_{M})}{\pm 2\omega '(t)}}.}
Įstatę į antrą sistemos lygtį gauname:
(
x
−
x
M
)
2
+
(
±
ϕ
′
(
t
)
(
x
−
x
M
)
±
2
ψ
′
(
t
)
)
2
+
(
±
ϕ
′
(
t
)
(
x
−
x
M
)
±
2
ω
′
(
t
)
)
2
=
1
K
2
,
{\displaystyle (x-x_{M})^{2}+\left({\frac {\pm \phi '(t)(x-x_{M})}{\pm 2\psi '(t)}}\right)^{2}+\left({\frac {\pm \phi '(t)(x-x_{M})}{\pm 2\omega '(t)}}\right)^{2}={\frac {1}{K^{2}}},}
(
x
−
x
M
)
2
+
(
ϕ
′
(
t
)
)
2
(
x
−
x
M
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
(
x
−
x
M
)
2
4
(
ω
′
(
t
)
)
2
=
1
K
2
,
{\displaystyle (x-x_{M})^{2}+{\frac {(\phi '(t))^{2}(x-x_{M})^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}(x-x_{M})^{2}}{4(\omega '(t))^{2}}}={\frac {1}{K^{2}}},}
(
x
−
x
M
)
2
(
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
)
=
R
2
.
{\displaystyle (x-x_{M})^{2}\left(1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}\right)=R^{2}.}
Išsprendus kvadratinę lygtį
(
x
2
−
2
x
x
M
+
x
M
2
)
(
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
)
−
R
2
=
0
,
{\displaystyle (x^{2}-2xx_{M}+x_{M}^{2})\left(1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}\right)-R^{2}=0,}
x
2
(
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
)
−
2
x
x
M
(
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
)
+
x
M
2
(
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
)
−
R
2
=
0
,
{\displaystyle x^{2}\left(1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}\right)-2xx_{M}\left(1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}\right)+x_{M}^{2}\left(1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}\right)-R^{2}=0,}
surandama kreivio centro
C
(
x
C
;
y
C
;
z
C
)
{\displaystyle C(x_{C};y_{C};z_{C})}
x
C
.
{\displaystyle x_{C}.}
Analogiškai surandama ir
y
C
{\displaystyle y_{C}}
x
−
x
M
=
±
2
ψ
′
(
t
)
(
y
−
y
M
)
±
ϕ
′
(
t
)
,
{\displaystyle x-x_{M}={\frac {\pm 2\psi '(t)(y-y_{M})}{\pm \phi '(t)}},}
z
−
z
M
=
±
ψ
′
(
t
)
(
y
−
y
M
)
±
ω
′
(
t
)
;
{\displaystyle z-z_{M}={\frac {\pm \psi '(t)(y-y_{M})}{\pm \omega '(t)}};}
(
±
2
ψ
′
(
t
)
(
y
−
y
M
)
±
ϕ
′
(
t
)
)
2
+
(
y
−
y
M
)
2
+
(
±
ψ
′
(
t
)
(
y
−
y
M
)
±
ω
′
(
t
)
)
2
=
1
K
2
,
{\displaystyle \left({\frac {\pm 2\psi '(t)(y-y_{M})}{\pm \phi '(t)}}\right)^{2}+(y-y_{M})^{2}+\left({\frac {\pm \psi '(t)(y-y_{M})}{\pm \omega '(t)}}\right)^{2}={\frac {1}{K^{2}}},}
4
(
ψ
′
(
t
)
)
2
(
y
−
y
M
)
2
(
ϕ
′
(
t
)
)
2
+
(
y
−
y
M
)
2
+
(
ψ
′
(
t
)
)
2
(
y
−
y
M
)
2
(
ω
′
(
t
)
)
2
=
1
K
2
,
{\displaystyle {\frac {4(\psi '(t))^{2}(y-y_{M})^{2}}{(\phi '(t))^{2}}}+(y-y_{M})^{2}+{\frac {(\psi '(t))^{2}(y-y_{M})^{2}}{(\omega '(t))^{2}}}={\frac {1}{K^{2}}},}
(
y
−
y
M
)
2
(
4
(
ψ
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
1
+
(
ψ
′
(
t
)
)
2
(
ω
′
(
t
)
)
2
)
=
R
2
;
{\displaystyle (y-y_{M})^{2}\left({\frac {4(\psi '(t))^{2}}{(\phi '(t))^{2}}}+1+{\frac {(\psi '(t))^{2}}{(\omega '(t))^{2}}}\right)=R^{2};}
išsprendžiant lygtį:
y
2
(
4
(
ψ
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
1
+
(
ψ
′
(
t
)
)
2
(
ω
′
(
t
)
)
2
)
−
2
y
y
M
(
4
(
ψ
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
1
+
(
ψ
′
(
t
)
)
2
(
ω
′
(
t
)
)
2
)
+
y
M
2
(
4
(
ψ
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
1
+
(
ψ
′
(
t
)
)
2
(
ω
′
(
t
)
)
2
)
−
R
2
=
0.
{\displaystyle y^{2}\left({\frac {4(\psi '(t))^{2}}{(\phi '(t))^{2}}}+1+{\frac {(\psi '(t))^{2}}{(\omega '(t))^{2}}}\right)-2yy_{M}\left({\frac {4(\psi '(t))^{2}}{(\phi '(t))^{2}}}+1+{\frac {(\psi '(t))^{2}}{(\omega '(t))^{2}}}\right)+y_{M}^{2}\left({\frac {4(\psi '(t))^{2}}{(\phi '(t))^{2}}}+1+{\frac {(\psi '(t))^{2}}{(\omega '(t))^{2}}}\right)-R^{2}=0.}
Tokiu pačiu principu surandama ir
z
C
{\displaystyle z_{C}}
z
2
(
4
(
ω
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
(
ω
′
(
t
)
)
2
(
ψ
′
(
t
)
)
2
+
1
)
−
2
z
z
M
(
4
(
ω
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
(
ω
′
(
t
)
)
2
(
ψ
′
(
t
)
)
2
+
1
)
+
z
M
2
(
4
(
ω
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
(
ω
′
(
t
)
)
2
(
ψ
′
(
t
)
)
2
+
1
)
−
R
2
=
0.
{\displaystyle z^{2}\left({\frac {4(\omega '(t))^{2}}{(\phi '(t))^{2}}}+{\frac {(\omega '(t))^{2}}{(\psi '(t))^{2}}}+1\right)-2zz_{M}\left({\frac {4(\omega '(t))^{2}}{(\phi '(t))^{2}}}+{\frac {(\omega '(t))^{2}}{(\psi '(t))^{2}}}+1\right)+z_{M}^{2}\left({\frac {4(\omega '(t))^{2}}{(\phi '(t))^{2}}}+{\frac {(\omega '(t))^{2}}{(\psi '(t))^{2}}}+1\right)-R^{2}=0.}
Norint surasti parametrines erdvinės kreivės evoliutės lygtis grįžtame prie sistemos:
{
±
3
2
(
x
−
x
M
)
ϕ
′
(
t
)
=
±
3
(
y
−
y
M
)
ψ
′
(
t
)
=
±
3
(
z
−
z
M
)
ω
′
(
t
)
,
(
x
−
x
M
)
2
+
(
y
−
y
M
)
2
+
(
z
−
z
M
)
2
=
1
K
2
.
{\displaystyle {\begin{cases}\pm {\frac {3}{2}}(x-x_{M})\phi '(t)=\pm 3(y-y_{M})\psi '(t)=\pm 3(z-z_{M})\omega '(t),&\\(x-x_{M})^{2}+(y-y_{M})^{2}+(z-z_{M})^{2}={\frac {1}{K^{2}}}.&\end{cases}}}
Iš kurios turime:
(
x
−
x
M
)
2
+
(
±
ϕ
′
(
t
)
(
x
−
x
M
)
±
2
ψ
′
(
t
)
)
2
+
(
±
ϕ
′
(
t
)
(
x
−
x
M
)
±
2
ω
′
(
t
)
)
2
=
1
K
2
,
{\displaystyle (x-x_{M})^{2}+\left({\frac {\pm \phi '(t)(x-x_{M})}{\pm 2\psi '(t)}}\right)^{2}+\left({\frac {\pm \phi '(t)(x-x_{M})}{\pm 2\omega '(t)}}\right)^{2}={\frac {1}{K^{2}}},}
(
x
−
x
M
)
2
+
(
ϕ
′
(
t
)
)
2
(
x
−
x
M
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
(
x
−
x
M
)
2
4
(
ω
′
(
t
)
)
2
=
1
K
2
,
{\displaystyle (x-x_{M})^{2}+{\frac {(\phi '(t))^{2}(x-x_{M})^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}(x-x_{M})^{2}}{4(\omega '(t))^{2}}}={\frac {1}{K^{2}}},}
(
x
−
x
M
)
2
(
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
)
=
R
2
;
{\displaystyle (x-x_{M})^{2}\left(1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}\right)=R^{2};}
(
x
−
x
M
)
2
=
R
2
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
,
{\displaystyle (x-x_{M})^{2}={\frac {R^{2}}{1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}}},}
x
−
x
M
=
±
R
2
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
,
{\displaystyle x-x_{M}=\pm {\sqrt {\frac {R^{2}}{1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}}}},}
x
=
x
M
±
R
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
,
{\displaystyle x=x_{M}\pm {\frac {R}{\sqrt {1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}}}},}
x
C
=
x
M
±
1
K
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
=
{\displaystyle x_{C}=x_{M}\pm {\frac {1}{K{\sqrt {1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}}}}}=}
=
x
M
±
(
(
ϕ
′
(
t
)
)
2
+
(
ψ
′
(
t
)
)
2
+
(
ω
′
(
t
)
)
2
)
3
2
(
ψ
′
(
t
)
ω
″
(
t
)
−
ω
′
(
t
)
ψ
″
(
t
)
)
2
+
(
ϕ
′
(
t
)
ω
″
(
t
)
−
ω
′
(
t
)
ϕ
″
(
t
)
)
2
+
(
ϕ
′
(
t
)
ψ
″
(
t
)
−
ψ
′
(
t
)
ϕ
″
(
t
)
)
2
(
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
)
,
{\displaystyle =x_{M}\pm {\frac {((\phi '(t))^{2}+(\psi '(t))^{2}+(\omega '(t))^{2})^{3 \over 2}}{{\sqrt {(\psi '(t)\omega ''(t)-\omega '(t)\psi ''(t))^{2}+(\phi '(t)\omega ''(t)-\omega '(t)\phi ''(t))^{2}+(\phi '(t)\psi ''(t)-\psi '(t)\phi ''(t))^{2}}}{\sqrt {\left(1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}\right)}}}},}
kad gauti koordinatę
x
C
{\displaystyle x_{C}}
t reikšmė turi būti tokia, su kuria būtų gautas taškas
M
(
x
M
;
y
M
;
z
M
)
{\displaystyle M(x_{M};y_{M};z_{M})}
x
=
ϕ
(
t
)
,
y
=
ψ
(
t
)
,
z
=
ω
(
t
)
;
{\displaystyle x=\phi (t),\;y=\psi (t),\;z=\omega (t);}
α
(
t
)
=
x
±
1
K
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
=
ϕ
(
t
)
±
R
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
=
{\displaystyle \alpha (t)=x\pm {\frac {1}{K{\sqrt {1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}}}}}=\phi (t)\pm {\frac {R}{\sqrt {1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}}}}=}
=
ϕ
±
(
(
ϕ
′
)
2
+
(
ψ
′
)
2
+
(
ω
′
)
2
)
3
2
(
ψ
′
ω
″
−
ω
′
ψ
″
)
2
+
(
ϕ
′
ω
″
−
ω
′
ϕ
″
)
2
+
(
ϕ
′
ψ
″
−
ψ
′
ϕ
″
)
2
1
+
(
ϕ
′
)
2
4
(
ψ
′
)
2
+
(
ϕ
′
)
2
4
(
ω
′
)
2
.
{\displaystyle =\phi \pm {\frac {\frac {((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3 \over 2}}{\sqrt {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}}}{\sqrt {1+{\frac {(\phi ')^{2}}{4(\psi ')^{2}}}+{\frac {(\phi ')^{2}}{4(\omega ')^{2}}}}}}.}
Ar pliuso ar minuso ženklą pasirinkti, reikia vadovautis tuo, kad erdvinės kreivės normalės vektorius
b
→
=
{
2
±
3
ϕ
′
(
t
)
;
1
±
3
ψ
′
(
t
)
;
1
±
3
ω
′
(
t
)
}
{\displaystyle {\vec {b}}=\{{\frac {2}{\pm 3\phi '(t)}};{\frac {1}{\pm 3\psi '(t)}};{\frac {1}{\pm 3\omega '(t)}}\}}
c
→
=
{
x
C
−
x
M
;
y
C
−
y
M
;
z
C
−
z
M
}
.
{\displaystyle {\vec {c}}=\{x_{C}-x_{M};y_{C}-y_{M};z_{C}-z_{M}\}.}
|
R
2
−
R
1
|
=
|
s
2
−
s
1
|
{\displaystyle |R_{2}-R_{1}|=|s_{2}-s_{1}|}
|
s
2
−
s
1
|
{\displaystyle |s_{2}-s_{1}|}
Analogiškai turime ir parametrinę y evoliutės išraišką:
y
C
=
y
M
±
1
K
4
(
ψ
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
1
+
(
ψ
′
(
t
)
)
2
(
ω
′
(
t
)
)
2
;
{\displaystyle y_{C}=y_{M}\pm {\frac {1}{K{\sqrt {{\frac {4(\psi '(t))^{2}}{(\phi '(t))^{2}}}+1+{\frac {(\psi '(t))^{2}}{(\omega '(t))^{2}}}}}}};}
β
(
t
)
=
ψ
(
t
)
±
1
K
4
(
ψ
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
1
+
(
ψ
′
(
t
)
)
2
(
ω
′
(
t
)
)
2
.
{\displaystyle \beta (t)=\psi (t)\pm {\frac {1}{K{\sqrt {{\frac {4(\psi '(t))^{2}}{(\phi '(t))^{2}}}+1+{\frac {(\psi '(t))^{2}}{(\omega '(t))^{2}}}}}}}.}
Taip pat surandama ir parametrinė z evoliutės lygties koordinatė:
z
C
=
z
M
±
1
K
4
(
ω
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
(
ω
′
(
t
)
)
2
(
ψ
′
(
t
)
)
2
+
1
;
{\displaystyle z_{C}=z_{M}\pm {\frac {1}{K{\sqrt {{\frac {4(\omega '(t))^{2}}{(\phi '(t))^{2}}}+{\frac {(\omega '(t))^{2}}{(\psi '(t))^{2}}}+1}}}};}
γ
(
t
)
=
ω
(
t
)
±
1
K
4
(
ω
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
(
ω
′
(
t
)
)
2
(
ψ
′
(
t
)
)
2
+
1
.
{\displaystyle \gamma (t)=\omega (t)\pm {\frac {1}{K{\sqrt {{\frac {4(\omega '(t))^{2}}{(\phi '(t))^{2}}}+{\frac {(\omega '(t))^{2}}{(\psi '(t))^{2}}}+1}}}}.}
Taškas
C
(
x
C
;
y
C
;
z
C
)
{\displaystyle C(x_{C};y_{C};z_{C})}
M
(
x
M
;
y
M
;
z
M
)
{\displaystyle M(x_{M};y_{M};z_{M})}
R . Apskaičiuosime kreivį sraigtinės linijos
r
=
i
a
cos
t
+
j
a
sin
t
+
k
a
m
t
{\displaystyle \mathbf {r} =\mathbf {i} a\cos t+\mathbf {j} a\sin t+\mathbf {k} amt}
laisvai pasirinktame taške.
Sprendimas :
d
r
d
t
=
−
i
a
sin
t
+
j
a
cos
t
+
k
a
m
,
{\displaystyle {\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}=-\mathbf {i} a\sin t+\mathbf {j} a\cos t+\mathbf {k} am,}
d
2
r
d
t
2
=
−
i
a
cos
t
−
j
a
sin
t
,
{\displaystyle {\frac {{\text{d}}^{2}\mathbf {r} }{{\text{d}}t^{2}}}=-\mathbf {i} a\cos t-\mathbf {j} a\sin t,}
d
r
d
t
×
d
2
r
d
t
2
=
|
i
j
k
−
a
sin
t
a
cos
t
a
m
−
a
cos
t
−
a
sin
t
0
|
=
|
a
cos
t
a
m
−
a
sin
t
0
|
i
−
|
−
a
sin
t
a
m
−
a
cos
t
0
|
j
+
|
−
a
sin
t
a
cos
t
−
a
cos
t
−
a
sin
t
|
k
=
{\displaystyle {\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\times {\frac {{\text{d}}^{2}\mathbf {r} }{{\text{d}}t^{2}}}={\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\-a\sin t&a\cos t&am\\-a\cos t&-a\sin t&0\end{vmatrix}}={\begin{vmatrix}a\cos t&am\\-a\sin t&0\end{vmatrix}}\mathbf {i} -{\begin{vmatrix}-a\sin t&am\\-a\cos t&0\end{vmatrix}}\mathbf {j} +{\begin{vmatrix}-a\sin t&a\cos t\\-a\cos t&-a\sin t\end{vmatrix}}\mathbf {k} =}
=
i
(
a
cos
t
⋅
0
−
a
m
⋅
(
−
a
sin
t
)
)
−
j
(
−
a
sin
t
⋅
0
−
a
m
⋅
(
−
a
cos
t
)
)
+
k
(
−
a
sin
t
⋅
(
−
a
sin
t
)
−
a
cos
t
⋅
(
−
a
cos
t
)
)
=
{\displaystyle =\mathbf {i} (a\cos t\cdot 0-am\cdot (-a\sin t))-\mathbf {j} (-a\sin t\cdot 0-am\cdot (-a\cos t))+\mathbf {k} (-a\sin t\cdot (-a\sin t)-a\cos t\cdot (-a\cos t))=}
=
i
a
2
m
sin
t
−
j
a
2
m
cos
t
+
k
a
2
,
{\displaystyle =\mathbf {i} a^{2}m\sin t-\mathbf {j} a^{2}m\cos t+\mathbf {k} a^{2},}
(
d
r
d
t
×
d
2
r
d
t
2
)
2
=
(
a
2
m
sin
t
)
2
+
(
−
a
2
m
cos
t
)
2
+
(
a
2
)
2
=
a
4
m
2
sin
2
t
+
a
4
m
2
cos
2
t
+
a
4
=
a
4
m
2
+
a
4
=
a
4
(
m
2
+
1
)
,
{\displaystyle \left({\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\times {\frac {{\text{d}}^{2}\mathbf {r} }{{\text{d}}t^{2}}}\right)^{2}=(a^{2}m\sin t)^{2}+(-a^{2}m\cos t)^{2}+(a^{2})^{2}=a^{4}m^{2}\sin ^{2}t+a^{4}m^{2}\cos ^{2}t+a^{4}=a^{4}m^{2}+a^{4}=a^{4}(m^{2}+1),}
(
d
r
d
t
)
2
=
(
−
a
sin
t
)
2
+
(
a
cos
t
)
2
+
(
a
m
)
2
=
a
2
sin
2
t
+
a
2
cos
2
t
+
a
2
m
2
=
a
2
+
a
2
m
2
=
a
2
(
1
+
m
2
)
.
{\displaystyle \left({\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\right)^{2}=(-a\sin t)^{2}+(a\cos t)^{2}+(am)^{2}=a^{2}\sin ^{2}t+a^{2}\cos ^{2}t+a^{2}m^{2}=a^{2}+a^{2}m^{2}=a^{2}(1+m^{2}).}
Gauname,
K
2
=
1
R
2
=
[
d
r
d
t
×
d
2
r
d
t
2
]
2
{
(
d
r
d
t
)
2
}
3
=
[
d
r
d
t
×
d
2
r
d
t
2
]
2
(
d
r
d
t
⋅
d
r
d
t
)
3
=
a
4
(
m
2
+
1
)
(
a
2
(
1
+
m
2
)
)
3
=
a
4
(
m
2
+
1
)
a
6
(
1
+
m
2
)
3
=
1
a
2
(
1
+
m
2
)
2
,
{\displaystyle K^{2}={\frac {1}{R^{2}}}={\frac {\left[{\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\times {\frac {{\text{d}}^{2}\mathbf {r} }{{\text{d}}t^{2}}}\right]^{2}}{\{\left({\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\right)^{2}\}^{3}}}={\frac {\left[{\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\times {\frac {{\text{d}}^{2}\mathbf {r} }{{\text{d}}t^{2}}}\right]^{2}}{\left({\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\cdot {\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\right)^{3}}}={\frac {a^{4}(m^{2}+1)}{(a^{2}(1+m^{2}))^{3}}}={\frac {a^{4}(m^{2}+1)}{a^{6}(1+m^{2})^{3}}}={\frac {1}{a^{2}(1+m^{2})^{2}}},}
iš kur
R
=
a
(
1
+
m
2
)
=
const.
{\displaystyle R=a(1+m^{2})={\text{const.}}}
Tokiu budu, sraigtinė linija turi pastovų spindulį kreivio.
Duota kreivė užrašyta parametrinėmis lygtimis:
x
=
ϕ
(
t
)
=
t
,
y
=
ψ
(
t
)
=
t
2
,
z
=
ω
(
t
)
=
t
3
.
{\displaystyle x=\phi (t)=t,\quad y=\psi (t)=t^{2},\quad z=\omega (t)=t^{3}.}
Rasti:
a) šios kreivės kreivį taškuose
M
(
5
;
25
;
125
)
{\displaystyle M(5;25;125)}
N
(
2
;
4
;
8
)
{\displaystyle N(2;4;8)}
b) šios kreivės kreivio centro koordinates
C
(
x
C
;
y
C
;
z
C
)
{\displaystyle C(x_{C};y_{C};z_{C})}
M
(
5
;
25
;
125
)
{\displaystyle M(5;25;125)}
c) erdvinę šios kreivės evoliutės lygtį;d) evoliutės lanko ilgį iš taško
N
(
2
;
4
;
8
)
{\displaystyle N(2;4;8)}
M
(
5
;
25
;
125
)
{\displaystyle M(5;25;125)}
L
=
R
M
−
R
N
;
{\displaystyle L=R_{M}-R_{N};}
e) evoliutės lanko ilgį iš taško
N
(
2
;
4
;
8
)
{\displaystyle N(2;4;8)}
M
(
5
;
25
;
125
)
{\displaystyle M(5;25;125)}
L
=
∫
t
0
T
(
α
′
(
t
)
)
2
+
(
β
′
(
t
)
)
2
+
(
γ
′
(
t
)
)
2
d
t
.
{\displaystyle L=\int _{t_{0}}^{T}{\sqrt {(\alpha '(t))^{2}+(\beta '(t))^{2}+(\gamma '(t))^{2}}}\;\mathbf {d} t.}
Sprendimas. a) Randame pirmos ir antro eilės išvestines ir kreivį
ϕ
′
(
t
)
=
t
′
=
1
,
{\displaystyle \phi '(t)=t'=1,}
ψ
′
(
t
)
=
(
t
2
)
′
=
2
t
,
{\displaystyle \psi '(t)=(t^{2})'=2t,}
ω
′
(
t
)
=
(
t
3
)
′
=
3
t
2
;
{\displaystyle \omega '(t)=(t^{3})'=3t^{2};}
ϕ
″
(
t
)
=
1
′
=
0
,
{\displaystyle \phi ''(t)=1'=0,}
ψ
″
(
t
)
=
(
2
t
)
′
=
2
,
{\displaystyle \psi ''(t)=(2t)'=2,}
ω
″
(
t
)
=
(
3
t
2
)
′
=
6
t
;
{\displaystyle \omega ''(t)=(3t^{2})'=6t;}
K
=
1
R
=
(
ψ
′
ω
″
−
ω
′
ψ
″
)
2
+
(
ϕ
′
ω
″
−
ω
′
ϕ
″
)
2
+
(
ϕ
′
ψ
″
−
ψ
′
ϕ
″
)
2
(
(
ϕ
′
)
2
+
(
ψ
′
)
2
+
(
ω
′
)
2
)
3
2
=
{\displaystyle K={\frac {1}{R}}={\frac {\sqrt {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}}{((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3 \over 2}}}=}
=
(
2
t
⋅
6
t
−
3
t
2
⋅
2
)
2
+
(
1
⋅
6
t
−
3
t
2
⋅
0
)
2
+
(
1
⋅
2
−
2
t
⋅
0
)
2
(
1
2
+
(
2
t
)
2
+
(
3
t
2
)
2
)
3
2
=
{\displaystyle ={\frac {\sqrt {(2t\cdot 6t-3t^{2}\cdot 2)^{2}+(1\cdot 6t-3t^{2}\cdot 0)^{2}+(1\cdot 2-2t\cdot 0)^{2}}}{(1^{2}+(2t)^{2}+(3t^{2})^{2})^{3 \over 2}}}=}
=
(
12
t
2
−
6
t
2
)
2
+
(
6
t
)
2
+
2
2
(
1
+
4
t
2
+
9
t
4
)
3
2
=
36
t
4
+
36
t
2
+
4
(
1
+
4
t
2
+
9
t
4
)
3
2
;
{\displaystyle ={\frac {\sqrt {(12t^{2}-6t^{2})^{2}+(6t)^{2}+2^{2}}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}={\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}};}
randame kreivį taške
M
(
5
;
25
;
125
)
:
{\displaystyle M(5;25;125):}
K
M
=
36
t
4
+
36
t
2
+
4
(
1
+
4
t
2
+
9
t
4
)
3
2
=
36
⋅
5
4
+
36
⋅
5
2
+
4
(
1
+
4
⋅
5
2
+
9
⋅
5
4
)
3
2
=
36
⋅
625
+
36
⋅
25
+
4
(
1
+
4
⋅
25
+
9
⋅
625
)
3
2
=
{\displaystyle K_{M}={\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}={\frac {\sqrt {36\cdot 5^{4}+36\cdot 5^{2}+4}}{(1+4\cdot 5^{2}+9\cdot 5^{4})^{3 \over 2}}}={\frac {\sqrt {36\cdot 625+36\cdot 25+4}}{(1+4\cdot 25+9\cdot 625)^{3 \over 2}}}=}
=
22500
+
900
+
4
(
1
+
100
+
5625
)
3
2
=
23404
5726
3
2
=
{\displaystyle ={\frac {\sqrt {22500+900+4}}{(1+100+5625)^{3 \over 2}}}={\frac {\sqrt {23404}}{5726^{3 \over 2}}}=}
=
152.9836593
75.67033765
3
=
152.9836593
433288.3534
=
0.000353075
,
{\displaystyle ={\frac {152.9836593}{75.67033765^{3}}}={\frac {152.9836593}{433288.3534}}=0.000353075,}
R
M
=
1
K
M
=
1
0.000353075
=
2832.252513
;
{\displaystyle R_{M}={\frac {1}{K_{M}}}={\frac {1}{0.000353075}}=2832.252513;}
randame kreivį taške
N
(
2
;
4
;
8
)
{\displaystyle N(2;4;8)}
K
N
=
36
t
4
+
36
t
2
+
4
(
1
+
4
t
2
+
9
t
4
)
3
2
=
36
⋅
2
4
+
36
⋅
2
2
+
4
(
1
+
4
⋅
2
2
+
9
⋅
2
4
)
3
2
=
36
⋅
16
+
36
⋅
4
+
4
(
1
+
4
⋅
4
+
9
⋅
16
)
3
2
=
{\displaystyle K_{N}={\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}={\frac {\sqrt {36\cdot 2^{4}+36\cdot 2^{2}+4}}{(1+4\cdot 2^{2}+9\cdot 2^{4})^{3 \over 2}}}={\frac {\sqrt {36\cdot 16+36\cdot 4+4}}{(1+4\cdot 4+9\cdot 16)^{3 \over 2}}}=}
=
576
+
144
+
4
(
1
+
16
+
144
)
3
2
=
724
161
3
2
=
{\displaystyle ={\frac {\sqrt {576+144+4}}{(1+16+144)^{3 \over 2}}}={\frac {\sqrt {724}}{161^{3 \over 2}}}=}
=
26.90724809
12.68857754
3
=
26.90724809
2042.860984
=
0.013171355
,
{\displaystyle ={\frac {26.90724809}{12.68857754^{3}}}={\frac {26.90724809}{2042.860984}}=0.013171355,}
R
N
=
1
K
N
=
1
0.013171355
=
75.92233055.
{\displaystyle R_{N}={\frac {1}{K_{N}}}={\frac {1}{0.013171355}}=75.92233055.}
b) Rasime kreivės kreivio centro koordinates priklausančias nuo taško
M
(
5
;
25
;
125
)
{\displaystyle M(5;25;125)}
M
(
5
;
25
;
125
)
{\displaystyle M(5;25;125)}
K
M
=
23404
5726
3
2
=
0.000353075
,
{\displaystyle K_{M}={\frac {\sqrt {23404}}{5726^{3 \over 2}}}=0.000353075,}
x
C
=
x
M
±
1
K
M
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
=
5
±
2832.252513
1
+
1
2
4
(
2
⋅
5
)
2
+
1
2
4
(
3
⋅
5
2
)
2
=
{\displaystyle x_{C}=x_{M}\pm {\frac {1}{K_{M}{\sqrt {1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}}}}}=5\pm {\frac {2832.252513}{\sqrt {1+{\frac {1^{2}}{4(2\cdot 5)^{2}}}+{\frac {1^{2}}{4(3\cdot 5^{2})^{2}}}}}}=}
=
5
±
2832.252513
1
+
1
4
⋅
10
2
+
1
4
⋅
75
2
=
5
±
2832.252513
1
+
1
4
⋅
100
+
1
4
⋅
5625
=
{\displaystyle =5\pm {\frac {2832.252513}{\sqrt {1+{\frac {1}{4\cdot 10^{2}}}+{\frac {1}{4\cdot 75^{2}}}}}}=5\pm {\frac {2832.252513}{\sqrt {1+{\frac {1}{4\cdot 100}}+{\frac {1}{4\cdot 5625}}}}}=}
=
5
±
2832.252513
1
+
1
400
+
1
22500
=
5
±
2832.252513
1
+
0.0025
+
0.000044444
=
{\displaystyle =5\pm {\frac {2832.252513}{\sqrt {1+{\frac {1}{400}}+{\frac {1}{22500}}}}}=5\pm {\frac {2832.252513}{\sqrt {1+0.0025+0.000044444}}}=}
=
5
±
2832.252513
1.002544444
=
5
±
2832.252513
1.001271414
=
{\displaystyle =5\pm {\frac {2832.252513}{\sqrt {1.002544444}}}=5\pm {\frac {2832.252513}{1.001271414}}=}
=
5
±
2828.65612
=
2833.65612
;
−
2823.65612
;
{\displaystyle =5\pm 2828.65612=2833.65612;\;-2823.65612;}
pastebime, kad
x
C
=
2833.65612
>
R
M
=
2832.252513
,
{\displaystyle x_{C}=2833.65612>R_{M}=2832.252513,}
x
C
=
−
2823.65612
{\displaystyle x_{C}=-2823.65612}
y
C
=
y
M
±
1
K
M
4
(
ψ
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
1
+
(
ψ
′
(
t
)
)
2
(
ω
′
(
t
)
)
2
=
25
±
2832.252513
4
(
2
⋅
5
)
2
1
2
+
1
+
(
2
⋅
5
)
2
(
3
⋅
5
2
)
2
=
{\displaystyle y_{C}=y_{M}\pm {\frac {1}{K_{M}{\sqrt {{\frac {4(\psi '(t))^{2}}{(\phi '(t))^{2}}}+1+{\frac {(\psi '(t))^{2}}{(\omega '(t))^{2}}}}}}}=25\pm {\frac {2832.252513}{\sqrt {{\frac {4(2\cdot 5)^{2}}{1^{2}}}+1+{\frac {(2\cdot 5)^{2}}{(3\cdot 5^{2})^{2}}}}}}=}
=
25
±
2832.252513
400
1
+
1
+
100
75
2
=
25
±
2832.252513
400
+
1
+
100
5625
=
{\displaystyle =25\pm {\frac {2832.252513}{\sqrt {{\frac {400}{1}}+1+{\frac {100}{75^{2}}}}}}=25\pm {\frac {2832.252513}{\sqrt {400+1+{\frac {100}{5625}}}}}=}
=
25
±
2832.252513
400
+
1
+
0.017777777
=
25
±
2832.252513
401.017777777
=
25
±
2832.252513
20.02542828
=
{\displaystyle =25\pm {\frac {2832.252513}{\sqrt {400+1+0.017777777}}}=25\pm {\frac {2832.252513}{\sqrt {401.017777777}}}=25\pm {\frac {2832.252513}{20.02542828}}=}
=
25
±
2832.252513
20.02542828
=
25
±
141.432806
=
166.432806
;
−
116.432806
;
{\displaystyle =25\pm {\frac {2832.252513}{20.02542828}}=25\pm 141.432806=166.432806;\;-116.432806;}
taške M (5; 25; 125) koordinatė
y
C
=
166.432806
{\displaystyle y_{C}=166.432806}
M kreivės normalė eina į [parabolės (kuri yra projekcija kreivės į xOy plokštumą)] vidų ir kyla į viršų;
z
C
=
z
M
±
1
K
M
4
(
ω
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
(
ω
′
(
t
)
)
2
(
ψ
′
(
t
)
)
2
+
1
=
125
±
2832.252513
4
(
3
⋅
5
2
)
2
1
2
+
(
3
⋅
5
2
)
2
(
2
⋅
5
)
2
+
1
=
{\displaystyle z_{C}=z_{M}\pm {\frac {1}{K_{M}{\sqrt {{\frac {4(\omega '(t))^{2}}{(\phi '(t))^{2}}}+{\frac {(\omega '(t))^{2}}{(\psi '(t))^{2}}}+1}}}}=125\pm {\frac {2832.252513}{\sqrt {{\frac {4(3\cdot 5^{2})^{2}}{1^{2}}}+{\frac {(3\cdot 5^{2})^{2}}{(2\cdot 5)^{2}}}+1}}}=}
=
125
±
2832.252513
4
⋅
75
2
1
+
75
2
100
+
1
=
125
±
2832.252513
4
⋅
5625
1
+
5625
100
+
1
=
{\displaystyle =125\pm {\frac {2832.252513}{\sqrt {{\frac {4\cdot 75^{2}}{1}}+{\frac {75^{2}}{100}}+1}}}=125\pm {\frac {2832.252513}{\sqrt {{\frac {4\cdot 5625}{1}}+{\frac {5625}{100}}+1}}}=}
=
125
±
2832.252513
22500
+
56.25
+
1
=
125
±
2832.252513
22557.25
=
125
±
2832.252513
150.1907121
=
{\displaystyle =125\pm {\frac {2832.252513}{\sqrt {22500+56.25+1}}}=125\pm {\frac {2832.252513}{\sqrt {22557.25}}}=125\pm {\frac {2832.252513}{150.1907121}}=}
=
125
±
18.85770747
=
143.8577075
;
106.1422925
;
{\displaystyle =125\pm 18.85770747=143.8577075;106.1422925;}
pagal pirmą taisyklę reikia, kad vektorius
M
C
→
{\displaystyle {\vec {MC}}}
b
→
=
{
−
2
3
ϕ
′
(
t
)
;
1
3
ψ
′
(
t
)
;
1
3
ω
′
(
t
)
}
=
{
−
2
3
⋅
1
;
1
3
⋅
(
2
⋅
5
)
;
1
3
(
3
⋅
5
2
)
}
=
{
−
2
3
;
1
30
;
1
225
}
;
{\displaystyle {\vec {b}}=\{-{\frac {2}{3\phi '(t)}};{\frac {1}{3\psi '(t)}};{\frac {1}{3\omega '(t)}}\}=\{-{\frac {2}{3\cdot 1}};{\frac {1}{3\cdot (2\cdot 5)}};{\frac {1}{3(3\cdot 5^{2})}}\}=\{-{\frac {2}{3}};{\frac {1}{30}};{\frac {1}{225}}\};}
padalinę vektoriaus narius
M
C
→
{\displaystyle {\vec {MC}}}
b
→
{\displaystyle {\vec {b}}}
x , y , z reikšmes gautame vektoriuje, taigi
−
2823.65612
−
5
−
2
3
=
2828.65612
⋅
3
2
=
4242.98418
,
{\displaystyle {\frac {-2823.65612-5}{-{\frac {2}{3}}}}=2828.65612\cdot {\frac {3}{2}}=4242.98418,}
166.432806
−
25
1
30
=
141.432806
⋅
30
=
4242.98418
,
{\displaystyle {\frac {166.432806-25}{\frac {1}{30}}}=141.432806\cdot 30=4242.98418,}
143.8577075
−
125
1
225
=
18.85770747
⋅
225
=
4242.98418
,
{\displaystyle {\frac {143.8577075-125}{\frac {1}{225}}}=18.85770747\cdot 225=4242.98418,}
106.1422925
−
125
1
225
=
−
18.85770747
⋅
225
=
−
4242.98418
{\displaystyle {\frac {106.1422925-125}{\frac {1}{225}}}=-18.85770747\cdot 225=-4242.98418}
randame, kad
z
C
=
143.8577075
{\displaystyle z_{C}=143.8577075}
c) Erdvinės evoliutės parametrinės lygtys yra (bent jau, kai t kinta nuo 0 iki
∞
{\displaystyle \infty }
α
(
t
)
=
ϕ
(
t
)
−
1
K
1
+
(
ϕ
′
(
t
)
)
2
4
(
ψ
′
(
t
)
)
2
+
(
ϕ
′
(
t
)
)
2
4
(
ω
′
(
t
)
)
2
=
t
−
1
K
1
+
1
2
4
(
2
t
)
2
+
1
2
4
(
3
t
2
)
2
=
{\displaystyle \alpha (t)=\phi (t)-{\frac {1}{K{\sqrt {1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}}}}}=t-{\frac {1}{K{\sqrt {1+{\frac {1^{2}}{4(2t)^{2}}}+{\frac {1^{2}}{4(3t^{2})^{2}}}}}}}=}
=
t
−
1
K
1
+
1
4
⋅
4
t
2
+
1
4
⋅
9
t
4
=
t
−
1
36
t
4
+
36
t
2
+
4
(
1
+
4
t
2
+
9
t
4
)
3
2
1
+
1
16
t
2
+
1
36
t
4
=
{\displaystyle =t-{\frac {1}{K{\sqrt {1+{\frac {1}{4\cdot 4t^{2}}}+{\frac {1}{4\cdot 9t^{4}}}}}}}=t-{\frac {1}{{\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}{\sqrt {1+{\frac {1}{16t^{2}}}+{\frac {1}{36t^{4}}}}}}}=}
=
t
−
1
36
t
4
+
36
t
2
+
4
(
1
+
4
t
2
+
9
t
4
)
3
2
36
t
4
+
4
t
2
+
1
36
t
4
=
t
−
6
t
2
(
1
+
4
t
2
+
9
t
4
)
3
2
36
t
4
+
36
t
2
+
4
36
t
4
+
4
t
2
+
1
=
{\displaystyle =t-{\frac {1}{{\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}{\sqrt {\frac {36t^{4}+4t^{2}+1}{36t^{4}}}}}}=t-{\frac {6t^{2}(1+4t^{2}+9t^{4})^{3 \over 2}}{{\sqrt {36t^{4}+36t^{2}+4}}{\sqrt {36t^{4}+4t^{2}+1}}}}=}
=
t
−
3
t
2
(
1
+
4
t
2
+
9
t
4
)
3
2
(
9
t
4
+
9
t
2
+
1
)
(
36
t
4
+
4
t
2
+
1
)
;
{\displaystyle =t-{\frac {3t^{2}(1+4t^{2}+9t^{4})^{3 \over 2}}{\sqrt {(9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1)}}};}
β
(
t
)
=
ψ
(
t
)
+
1
K
4
(
ψ
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
1
+
(
ψ
′
(
t
)
)
2
(
ω
′
(
t
)
)
2
=
t
2
+
1
K
4
(
2
t
)
2
1
2
+
1
+
(
2
t
)
2
(
3
t
2
)
2
=
{\displaystyle \beta (t)=\psi (t)+{\frac {1}{K{\sqrt {{\frac {4(\psi '(t))^{2}}{(\phi '(t))^{2}}}+1+{\frac {(\psi '(t))^{2}}{(\omega '(t))^{2}}}}}}}=t^{2}+{\frac {1}{K{\sqrt {{\frac {4(2t)^{2}}{1^{2}}}+1+{\frac {(2t)^{2}}{(3t^{2})^{2}}}}}}}=}
=
t
2
+
1
K
16
t
2
+
1
+
4
t
2
9
t
4
=
t
2
+
1
36
t
4
+
36
t
2
+
4
(
1
+
4
t
2
+
9
t
4
)
3
2
16
t
2
+
1
+
4
9
t
2
=
{\displaystyle =t^{2}+{\frac {1}{K{\sqrt {16t^{2}+1+{\frac {4t^{2}}{9t^{4}}}}}}}=t^{2}+{\frac {1}{{\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}{\sqrt {16t^{2}+1+{\frac {4}{9t^{2}}}}}}}=}
=
t
2
+
1
36
t
4
+
36
t
2
+
4
(
1
+
4
t
2
+
9
t
4
)
3
2
144
t
4
+
9
t
2
+
4
9
t
2
=
t
2
+
3
t
(
1
+
4
t
2
+
9
t
4
)
3
2
2
9
t
4
+
9
t
2
+
1
144
t
4
+
9
t
2
+
4
;
{\displaystyle =t^{2}+{\frac {1}{{\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}{\sqrt {\frac {144t^{4}+9t^{2}+4}{9t^{2}}}}}}=t^{2}+{\frac {3t(1+4t^{2}+9t^{4})^{3 \over 2}}{2{\sqrt {9t^{4}+9t^{2}+1}}{\sqrt {144t^{4}+9t^{2}+4}}}};}
γ
(
t
)
=
ω
(
t
)
+
1
K
4
(
ω
′
(
t
)
)
2
(
ϕ
′
(
t
)
)
2
+
(
ω
′
(
t
)
)
2
(
ψ
′
(
t
)
)
2
+
1
=
t
3
+
1
K
4
(
3
t
2
)
2
1
2
+
(
3
t
2
)
2
(
2
t
)
2
+
1
=
{\displaystyle \gamma (t)=\omega (t)+{\frac {1}{K{\sqrt {{\frac {4(\omega '(t))^{2}}{(\phi '(t))^{2}}}+{\frac {(\omega '(t))^{2}}{(\psi '(t))^{2}}}+1}}}}=t^{3}+{\frac {1}{K{\sqrt {{\frac {4(3t^{2})^{2}}{1^{2}}}+{\frac {(3t^{2})^{2}}{(2t)^{2}}}+1}}}}=}
=
t
3
+
1
K
36
t
4
+
9
t
4
4
t
2
+
1
=
t
3
+
1
36
t
4
+
36
t
2
+
4
(
1
+
4
t
2
+
9
t
4
)
3
2
144
t
4
+
9
t
2
+
4
4
=
{\displaystyle =t^{3}+{\frac {1}{K{\sqrt {36t^{4}+{\frac {9t^{4}}{4t^{2}}}+1}}}}=t^{3}+{\frac {1}{{\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}{\sqrt {\frac {144t^{4}+9t^{2}+4}{4}}}}}=}
=
t
3
+
1
9
t
4
+
9
t
2
+
1
(
1
+
4
t
2
+
9
t
4
)
3
2
144
t
4
+
9
t
2
+
4
=
t
3
+
(
1
+
4
t
2
+
9
t
4
)
3
2
9
t
4
+
9
t
2
+
1
144
t
4
+
9
t
2
+
4
.
{\displaystyle =t^{3}+{\frac {1}{{\frac {\sqrt {9t^{4}+9t^{2}+1}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}{\sqrt {144t^{4}+9t^{2}+4}}}}=t^{3}+{\frac {(1+4t^{2}+9t^{4})^{3 \over 2}}{{\sqrt {9t^{4}+9t^{2}+1}}{\sqrt {144t^{4}+9t^{2}+4}}}}.}
d) Turime, kad
K
=
1
R
=
36
t
4
+
36
t
2
+
4
(
1
+
4
t
2
+
9
t
4
)
3
2
;
{\displaystyle K={\frac {1}{R}}={\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}};}
randame kreivį taške
N
(
2
;
4
;
8
)
{\displaystyle N(2;4;8)}
t
=
2
{\displaystyle t=2}
K
N
=
1
R
N
=
36
⋅
2
4
+
36
⋅
2
2
+
4
(
1
+
4
⋅
2
2
+
9
⋅
2
4
)
3
2
=
{\displaystyle K_{N}={\frac {1}{R_{N}}}={\frac {\sqrt {36\cdot 2^{4}+36\cdot 2^{2}+4}}{(1+4\cdot 2^{2}+9\cdot 2^{4})^{3 \over 2}}}=}
=
36
⋅
16
+
36
⋅
4
+
4
(
1
+
4
⋅
4
+
9
⋅
16
)
3
2
=
576
+
144
+
4
(
1
+
16
+
144
)
3
2
=
{\displaystyle ={\frac {\sqrt {36\cdot 16+36\cdot 4+4}}{(1+4\cdot 4+9\cdot 16)^{3 \over 2}}}={\frac {\sqrt {576+144+4}}{(1+16+144)^{3 \over 2}}}=}
=
724
161
3
2
=
724
4173281
=
26.90724809
2042.860984
=
0.013171355
;
{\displaystyle ={\frac {\sqrt {724}}{161^{3 \over 2}}}={\frac {\sqrt {724}}{\sqrt {4173281}}}={\frac {26.90724809}{2042.860984}}=0.013171355;}
randame kreivio spindulio ilgį iš taško
N
(
2
;
4
;
8
)
{\displaystyle N(2;4;8)}
R
N
=
1
K
N
=
1
0.013171355
=
75.92233055
;
{\displaystyle R_{N}={\frac {1}{K_{N}}}={\frac {1}{0.013171355}}=75.92233055;}
toliau randame evoliutės lanko ilgį:
L
=
R
M
−
R
N
=
2832.252513
−
75.92233055
=
2756.330182.
{\displaystyle L=R_{M}-R_{N}=2832.252513-75.92233055=2756.330182.}
e) Randame išvestines (pasinaudodami derivatoriumi iš www.derivator.org):funkcijai
γ
′
(
t
)
{\displaystyle \gamma '(t)}
(1.5*(1+4*x^2+9*x^4)^0.5*(0+0+4*2*x+0+9*4*x^3)*((9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^0.5-(1+4*x^2+9*x^4)^1.5*0.5*((9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^-0.5*((0+9*4*x^3+0+9*2*x+0)*(144*x^4+9*x^2+4)+(9*x^4+9*x^2+1)*(0+144*4*x^3+0+9*2*x+0)))/((9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^0.5^2;
γ
′
(
t
)
=
(
t
3
+
(
1
+
4
t
2
+
9
t
4
)
3
2
9
t
4
+
9
t
2
+
1
144
t
4
+
9
t
2
+
4
)
′
=
{\displaystyle \gamma '(t)=(t^{3}+{\frac {(1+4t^{2}+9t^{4})^{3 \over 2}}{{\sqrt {9t^{4}+9t^{2}+1}}{\sqrt {144t^{4}+9t^{2}+4}}}})'=}
=
3
t
2
+
3
2
1
+
4
t
2
+
9
t
4
(
8
t
+
36
t
3
)
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
−
(
1
+
4
t
2
+
9
t
4
)
3
2
⋅
1
2
(
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
)
−
1
2
(
(
36
t
3
+
18
t
)
(
144
t
4
+
9
t
2
+
4
)
+
(
9
t
4
+
9
t
2
+
1
)
(
576
t
3
+
18
t
)
)
(
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
)
2
=
{\displaystyle =3t^{2}+{\frac {{3 \over 2}{\sqrt {1+4t^{2}+9t^{4}}}(8t+36t^{3}){\sqrt {(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}-(1+4t^{2}+9t^{4})^{3 \over 2}\cdot {1 \over 2}((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{-{\frac {1}{2}}}((36t^{3}+18t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(576t^{3}+18t))}{({\sqrt {(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}})^{2}}}=}
=
3
t
2
+
3
2
1
+
4
t
2
+
9
t
4
(
8
t
+
36
t
3
)
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
−
(
1
+
4
t
2
+
9
t
4
)
3
2
⋅
1
2
(
(
36
t
3
+
18
t
)
(
144
t
4
+
9
t
2
+
4
)
+
(
9
t
4
+
9
t
2
+
1
)
(
576
t
3
+
18
t
)
)
(
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
)
1
2
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
=
{\displaystyle =3t^{2}+{\frac {{3 \over 2}{\sqrt {1+4t^{2}+9t^{4}}}(8t+36t^{3})(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)-(1+4t^{2}+9t^{4})^{3 \over 2}\cdot {1 \over 2}((36t^{3}+18t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(576t^{3}+18t))}{((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{\frac {1}{2}}(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}=}
=
3
t
2
+
3
1
+
4
t
2
+
9
t
4
(
4
t
+
18
t
3
)
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
−
(
1
+
4
t
2
+
9
t
4
)
3
2
(
(
18
t
3
+
9
t
)
(
144
t
4
+
9
t
2
+
4
)
+
(
9
t
4
+
9
t
2
+
1
)
(
288
t
3
+
9
t
)
)
(
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
)
3
2
;
{\displaystyle =3t^{2}+{\frac {3{\sqrt {1+4t^{2}+9t^{4}}}(4t+18t^{3})(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)-(1+4t^{2}+9t^{4})^{3 \over 2}((18t^{3}+9t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(288t^{3}+9t))}{((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{\frac {3}{2}}}};}
tuo tarpu [beieškant
β
′
(
t
)
{\displaystyle \beta '(t)}
((0+1.5*((1+4*x^2+9*x^4)^1.5+x*1.5*(1+4*x^2+9*x^4)^0.5*(0+0+4*2*x+0+9*4*x^3)))*((9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^0.5-1.5*x*(1+4*x^2+9*x^4)^1.5*0.5*((9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^-0.5*((0+9*4*x^3+0+9*2*x+0)*(144*x^4+9*x^2+4)+(9*x^4+9*x^2+1)*(0+144*4*x^3+0+9*2*x+0)))/((9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^0.5^2;
β
′
(
t
)
=
(
t
2
+
3
t
(
1
+
4
t
2
+
9
t
4
)
3
2
2
9
t
4
+
9
t
2
+
1
144
t
4
+
9
t
2
+
4
)
′
=
{\displaystyle \beta '(t)=(t^{2}+{\frac {3t(1+4t^{2}+9t^{4})^{3 \over 2}}{2{\sqrt {9t^{4}+9t^{2}+1}}{\sqrt {144t^{4}+9t^{2}+4}}}})'=}
2
t
+
1.5
(
(
1
+
4
t
2
+
9
t
4
)
1.5
+
1.5
t
(
1
+
4
t
2
+
9
t
4
)
0.5
(
4
⋅
2
t
+
9
⋅
4
t
3
)
)
(
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
)
0.5
−
1.5
t
(
1
+
4
t
2
+
9
t
4
)
1.5
⋅
0.5
(
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
)
−
0.5
(
(
9
⋅
4
t
3
+
9
⋅
2
t
)
(
144
t
4
+
9
t
2
+
4
)
+
(
9
t
4
+
9
t
2
+
1
)
(
144
⋅
4
t
3
+
9
⋅
2
t
)
)
(
(
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
)
0.5
)
2
=
{\displaystyle 2t+{\frac {1.5((1+4t^{2}+9t^{4})^{1.5}+1.5t(1+4t^{2}+9t^{4})^{0.5}(4\cdot 2t+9\cdot 4t^{3}))((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{0.5}-1.5t(1+4t^{2}+9t^{4})^{1.5}\cdot 0.5((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{-0.5}((9\cdot 4t^{3}+9\cdot 2t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(144\cdot 4t^{3}+9\cdot 2t))}{(((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{0.5})^{2}}}=}
2
t
+
1.5
(
(
1
+
4
t
2
+
9
t
4
)
1.5
+
1.5
t
(
1
+
4
t
2
+
9
t
4
)
0.5
(
8
t
+
36
t
3
)
)
(
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
)
0.5
−
0.75
t
(
1
+
4
t
2
+
9
t
4
)
1.5
(
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
)
−
0.5
(
(
36
t
3
+
18
t
)
(
144
t
4
+
9
t
2
+
4
)
+
(
9
t
4
+
9
t
2
+
1
)
(
576
t
3
+
18
t
)
)
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
=
{\displaystyle 2t+{\frac {1.5((1+4t^{2}+9t^{4})^{1.5}+1.5t(1+4t^{2}+9t^{4})^{0.5}(8t+36t^{3}))((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{0.5}-0.75t(1+4t^{2}+9t^{4})^{1.5}((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{-0.5}((36t^{3}+18t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(576t^{3}+18t))}{(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}=}
2
t
+
1.5
(
(
1
+
4
t
2
+
9
t
4
)
1.5
+
1.5
t
(
1
+
4
t
2
+
9
t
4
)
0.5
(
8
t
+
36
t
3
)
)
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
−
0.75
t
(
1
+
4
t
2
+
9
t
4
)
1.5
(
(
36
t
3
+
18
t
)
(
144
t
4
+
9
t
2
+
4
)
+
(
9
t
4
+
9
t
2
+
1
)
(
576
t
3
+
18
t
)
)
(
(
9
t
4
+
9
t
2
+
1
)
(
144
t
4
+
9
t
2
+
4
)
)
1.5
;
{\displaystyle 2t+{\frac {1.5((1+4t^{2}+9t^{4})^{1.5}+1.5t(1+4t^{2}+9t^{4})^{0.5}(8t+36t^{3}))(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)-0.75t(1+4t^{2}+9t^{4})^{1.5}((36t^{3}+18t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(576t^{3}+18t))}{((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{1.5}}};}
kad rasti
α
′
(
t
)
{\displaystyle \alpha '(t)}
((0+3*(2*x*(1+4*x^2+9*x^4)^1.5+x^2*1.5*(1+4*x^2+9*x^4)^0.5*(0+0+4*2*x+0+9*4*x^3)))*((9*x^4+9*x^2+1)*(36*x^4+4*x^2+1))^0.5-3*x^2*(1+4*x^2+9*x^4)^1.5*0.5*((9*x^4+9*x^2+1)*(36*x^4+4*x^2+1))^-0.5*((0+9*4*x^3+0+9*2*x+0)*(36*x^4+4*x^2+1)+(9*x^4+9*x^2+1)*(0+36*4*x^3+0+4*2*x+0)))/((9*x^4+9*x^2+1)*(36*x^4+4*x^2+1))^0.5^2;
α
′
(
t
)
=
(
t
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{\displaystyle \alpha '(t)=\left(t-{\frac {3t^{2}(1+4t^{2}+9t^{4})^{3 \over 2}}{\sqrt {(9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1)}}}\right)'=}
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{\displaystyle =1-{\frac {3(2t(1+4t^{2}+9t^{4})^{1.5}+1.5t^{2}(1+4t^{2}+9t^{4})^{0.5}(4\cdot 2t+9\cdot 4t^{3}))((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{0.5}-3t^{2}(1+4t^{2}+9t^{4})^{1.5}0.5((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{-0.5}((9\cdot 4t^{3}+9\cdot 2t)(36t^{4}+4t^{2}+1)+(9t^{4}+9t^{2}+1)(36\cdot 4t^{3}+4\cdot 2t))}{[((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{0.5}]^{2}}};}
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{\displaystyle =1-{\frac {(6t(1+4t^{2}+9t^{4})^{1.5}+4.5t^{2}(1+4t^{2}+9t^{4})^{0.5}(8t+36t^{3}))((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{0.5}-1.5t^{2}(1+4t^{2}+9t^{4})^{1.5}((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{-0.5}((36t^{3}+18t)(36t^{4}+4t^{2}+1)+(9t^{4}+9t^{2}+1)(144t^{3}+8t))}{(9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1)}}=}
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{\displaystyle =1-{\frac {(6t(1+4t^{2}+9t^{4})^{1.5}+9t^{2}(1+4t^{2}+9t^{4})^{0.5}(4t+18t^{3}))((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{0.5}-3t^{2}(1+4t^{2}+9t^{4})^{1.5}((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{-0.5}((18t^{3}+9t)(36t^{4}+4t^{2}+1)+(9t^{4}+9t^{2}+1)(72t^{3}+4t))}{(9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1)}}=}
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;
{\displaystyle =1-{\frac {(6t(1+4t^{2}+9t^{4})^{1.5}+9t^{2}(1+4t^{2}+9t^{4})^{0.5}(4t+18t^{3}))(9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1)-3t^{2}(1+4t^{2}+9t^{4})^{1.5}((18t^{3}+9t)(36t^{4}+4t^{2}+1)+(9t^{4}+9t^{2}+1)(72t^{3}+4t))}{((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{3 \over 2}}};}
![{\displaystyle {\frac {d\phi }{ds}}={\frac {\frac {d\phi }{dx}}{\frac {ds}{dx}}}={\frac {\frac {\frac {d^{2}y}{dx^{2}}}{1+\left({\frac {dy}{dx}}\right)^{2}}}{\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}}={\frac {\frac {d^{2}y}{dx^{2}}}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad9a7a45a8841cb93a2fe55083959a931a2ecddd)
![{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}.\quad (3)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7705f941866e3b9fbf20292db17a49ffbfb7f5df)
![{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|-{\frac {p^{2}}{(2px)^{3 \over 2}}}\right|}{\left[1+\left({\frac {p}{\sqrt {2px}}}\right)^{2}\right]^{3 \over 2}}}={\frac {p^{2}}{(2px)^{3 \over 2}\left[1+{\frac {p^{2}}{2px}}\right]^{3 \over 2}}}={\frac {p^{2}}{(2px+p^{2})^{3 \over 2}}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2f748c452e5a458a7fda98879ea6e63b34c7a2d)
![{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|y''|}{[1+(y')^{2}]^{3 \over 2}}}={\frac {|0|}{[1+a^{2}]^{3 \over 2}}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bb1f66922cc47d9b4e1ef6effa02e4c7f6618bf)
![{\displaystyle 1+[f'(x)]^{2}=1+[(a\cosh {\frac {x}{a}})']^{2}=1+[({\frac {x}{a}})'a\sinh {\frac {x}{a}}]^{2}=1+[{\frac {1}{a}}\cdot a\sinh {\frac {x}{a}}]^{2}=1+\sinh ^{2}{\frac {x}{a}}=\cosh ^{2}{\frac {x}{a}}={\frac {y^{2}}{a^{2}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4fe90da1957bb9af94aa0e9366b4c9eeecfa240)
![{\displaystyle k={\frac {f''(x)}{[1+(f'(x))^{2}]^{3/2}}}={\frac {{\frac {1}{a}}\cosh {\frac {x}{a}}}{[1+(\sinh {\frac {x}{a}})^{2}]^{3/2}}}={\frac {{\frac {1}{a}}\cosh {\frac {x}{a}}}{[\cosh ^{2}{\frac {x}{a}}]^{3/2}}}={\frac {\frac {y}{a^{2}}}{({\frac {y^{2}}{a^{2}}})^{3/2}}}={\frac {\frac {y}{a^{2}}}{\frac {y^{3}}{a^{3}}}}={\frac {a^{3}y}{a^{2}y^{3}}}={\frac {a}{y^{2}}}={\frac {a}{a^{2}\cosh ^{2}{\frac {x}{a}}}}={\frac {1}{a\cosh ^{2}{\frac {x}{a}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/056e0ceb46e74d93c345dc1e4e5105804cb841a0)
![{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a35996d8035d23c38c84ea5e57e289a1b06549ec)
![{\displaystyle K_{M_{0}}={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}={\frac {2}{(1+4\cdot 0^{2})^{3 \over 2}}}=2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f309fa5241b05f753b97f07cc375be774ade78d4)
![{\displaystyle K_{M_{1}}={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f605a27ade00c02ef5db7751849a7d6f542d2801)
![{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|-{\frac {1}{(2x)^{3 \over 2}}}\right|}{\left[1+\left({\frac {1}{\sqrt {2x}}}\right)^{2}\right]^{3 \over 2}}}={\frac {1}{(2x)^{3 \over 2}\left[1+{\frac {1}{2x}}\right]^{3 \over 2}}}={\frac {1}{(2x+1)^{3 \over 2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d65cbfbdc44d534e47343f4907ae3ccb7ea5a508)
![{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|{\frac {\psi ''\phi '-\psi '\phi ''}{(\phi ')^{3}}}\right|}{\left[1+\left({\frac {\psi '}{\phi '}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{(\phi ')^{3}\left[{\frac {1}{(\phi ')^{2}}}((\phi ')^{2}+(\psi ')^{2})\right]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d48854d99afc809c0583d6ef8b9ef4dd2adf0609)
![{\displaystyle ={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{(\phi ')^{3}\cdot ({\frac {1}{(\phi ')^{2}}})^{3/2}\left[(\phi ')^{2}+(\psi ')^{2}\right]^{3 \over 2}}}={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{\left[(\phi ')^{2}+(\psi ')^{2}\right]^{3 \over 2}}}.\quad (1)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d73204aa6441e47082a70b40b9e92d1458ba9ac)
![{\displaystyle K={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{\left[(\phi ')^{2}+(\psi ')^{2}\right]^{3 \over 2}}}={\frac {|a(1-\cos t)a\cos t-a\sin t\cdot a\sin t|}{\left[a^{2}(1-\cos t)^{2}+a^{2}\sin ^{2}t\right]^{3 \over 2}}}={\frac {|a^{2}\cos t-a^{2}\cos ^{2}t-a^{2}\sin ^{2}t|}{\left[a^{2}-2a^{2}\cos t+a^{2}\cos ^{2}t+a^{2}\sin ^{2}t\right]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c6002c32e429ec95fdd2dbe2a7eab1414281de1)
![{\displaystyle ={\frac {|a^{2}\cos t-a^{2}|}{\left[2a^{2}-2a^{2}\cos t\right]^{3 \over 2}}}={\frac {a^{2}|\cos t-1|}{2^{3 \over 2}a^{3}\left[1-\cos t\right]^{3 \over 2}}}={\frac {1}{2^{3 \over 2}a\left[1-\cos t\right]^{1 \over 2}}}={\frac {1}{{\sqrt {2^{3}}}a\cdot {\sqrt {2}}\sin {\frac {t}{2}}}}={\frac {1}{4a|\sin {\frac {t}{2}}|}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65ae84fd7980d1aa49f856833fb2b61e3e35174f)
![{\displaystyle K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|{\frac {\psi ''\phi '-\psi '\phi ''}{(\phi ')^{3}}}\right|}{\left[1+\left({\frac {\psi '}{\phi '}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{(\phi ')^{3}\left[{\frac {1}{(\phi ')^{2}}}((\phi ')^{2}+(\psi ')^{2})\right]^{3 \over 2}}}={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{(\phi ')^{3}\cdot ({\frac {1}{(\phi ')^{2}}})^{3/2}\left[(\phi ')^{2}+(\psi ')^{2}\right]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61ad11af725b026da1fe819575d7e25b0b1e1a64)
![{\displaystyle ={\frac {|\psi ''\phi '-\psi '\phi ''|}{[\phi '^{2}+\psi '^{2}]^{3 \over 2}}}={\frac {\left|{\frac {d^{2}y}{d\theta ^{2}}}{\frac {dx}{d\theta }}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}\right|}{\left[\left({\frac {dx}{d\theta }}\right)^{2}+\left({\frac {dy}{d\theta }}\right)^{2}\right]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/538ee15ccf0e2f7816d373dedf0bb1f28d8993c3)
![{\displaystyle ={\frac {|({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta +2{\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )-({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )|}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9dc54b517d9e9d4a4e29b976d36d22dccc9e9eac)
![{\displaystyle ={\frac {({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta {\frac {d\rho }{d\theta }}\cos \theta +2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta -2{\frac {d\rho }{d\theta }}\cos \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta )}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}-}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64ef08cee60100b7a5d6a8cffb7f3ce981e1a900)
![{\displaystyle -{\frac {({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta {\frac {d\rho }{d\theta }}\sin \theta -2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta {\frac {d\rho }{d\theta }}\sin \theta +{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta \rho \cos \theta -\rho \cos \theta \rho \cos \theta )}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/108877b38eb0049ce955246eca7fd396e934556c)
![{\displaystyle ={\frac {{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta {\frac {d\rho }{d\theta }}\cos \theta +2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta -2{\frac {d\rho }{d\theta }}\cos \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}+}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f58e88514a2e9e8d2ff9041c2bbe02bdddd4bc90)
![{\displaystyle +{\frac {-{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta {\frac {d\rho }{d\theta }}\sin \theta +2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta {\frac {d\rho }{d\theta }}\sin \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta +2{\frac {d\rho }{d\theta }}\sin \theta \rho \cos \theta +\rho \cos \theta \rho \cos \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/30b17ea202388d9972c4cc154e1df6454f8c166e)
![{\displaystyle ={\frac {2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}+}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50b7dc98c371f0bbce0e16f853daf73c2ecb0757)
![{\displaystyle +{\frac {2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta +\rho \cos \theta \rho \cos \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3400672a1cdbd6574f0ec0bd7aee522c6bec073)
![{\displaystyle ={\frac {|2({\frac {d\rho }{d\theta }})^{2}\cos ^{2}\theta +2({\frac {d\rho }{d\theta }})^{2}\sin ^{2}\theta -\rho {\frac {d^{2}\rho }{d\theta ^{2}}}\sin ^{2}\theta -\rho {\frac {d^{2}\rho }{d\theta ^{2}}}\cos ^{2}\theta +\rho ^{2}\sin ^{2}\theta +\rho ^{2}\cos ^{2}\theta |}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecde29bf0237ef0d531f70d15fa82be548f2faf7)
![{\displaystyle ={\frac {|2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}+\rho ^{2}|}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1df4db52c95347ad546aa0f6a82a221cf7b985c9)
![{\displaystyle ={\frac {|\rho ^{2}+2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}|}{[(({\frac {d\rho }{d\theta }})^{2}\cos ^{2}\theta -2{\frac {d\rho }{d\theta }}\cos(\theta )\rho \sin(\theta )+\rho ^{2}\sin ^{2}\theta )+(({\frac {d\rho }{d\theta }})^{2}\sin ^{2}\theta +2{\frac {d\rho }{d\theta }}\sin(\theta )\rho \cos(\theta )+\rho ^{2}\cos ^{2}\theta )]^{3 \over 2}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b828daeaa07bf522a19b179bb0cac19d84d3b040)
![{\displaystyle ={\frac {|\rho ^{2}+2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}|}{[({\frac {d\rho }{d\theta }})^{2}\cos ^{2}\theta +({\frac {d\rho }{d\theta }})^{2}\sin ^{2}\theta +\rho ^{2}\sin ^{2}\theta +\rho ^{2}\cos ^{2}\theta )]^{3 \over 2}}}={\frac {|\rho ^{2}+2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}|}{[({\frac {d\rho }{d\theta }})^{2}+\rho ^{2}]^{3 \over 2}}}={\frac {|\rho ^{2}+2(\rho ')^{2}-\rho \rho ''|}{(\rho ^{2}+(\rho ')^{2})^{3/2}}}.\quad (4)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2dd72f40bbf4011ebad5b6419a3937f1365af819)
![{\displaystyle K^{2}={\frac {1}{R^{2}}}={\frac {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}{[(\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2}]^{3}}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebd4f0e009ae394722d403671f2dabd0cc56c069)
![{\displaystyle {\vec {b}}=\{-{\frac {1}{3}}\cdot {\frac {1}{\frac {1}{3{\sqrt[{3}]{t^{2}}}}}};{\frac {1}{3}}\cdot {\frac {1}{-{\frac {1}{2{\sqrt {t}}}}}};{\frac {2}{3}}\cdot {\frac {1}{3t^{2}}}\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f3568b1ddbb81c4500e3768f0843665430cab207)
![{\displaystyle {\vec {b}}=\{-{\sqrt[{3}]{t^{2}}};-{\frac {2{\sqrt {t}}}{3}};{\frac {2}{9t^{2}}}\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99781ba260d56091d17eb79838fbd37d8d5f4743)
![{\displaystyle K^{2}={\frac {1}{R^{2}}}={\frac {\left[{\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\times {\frac {{\text{d}}^{2}\mathbf {r} }{{\text{d}}t^{2}}}\right]^{2}}{\{\left({\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\right)^{2}\}^{3}}}={\frac {\left[{\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\times {\frac {{\text{d}}^{2}\mathbf {r} }{{\text{d}}t^{2}}}\right]^{2}}{\left({\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\cdot {\frac {{\text{d}}\mathbf {r} }{{\text{d}}t}}\right)^{3}}}={\frac {a^{4}(m^{2}+1)}{(a^{2}(1+m^{2}))^{3}}}={\frac {a^{4}(m^{2}+1)}{a^{6}(1+m^{2})^{3}}}={\frac {1}{a^{2}(1+m^{2})^{2}}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e1c37e519b2abe41abd079f04de50a92974570d)
![{\displaystyle =1-{\frac {3(2t(1+4t^{2}+9t^{4})^{1.5}+1.5t^{2}(1+4t^{2}+9t^{4})^{0.5}(4\cdot 2t+9\cdot 4t^{3}))((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{0.5}-3t^{2}(1+4t^{2}+9t^{4})^{1.5}0.5((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{-0.5}((9\cdot 4t^{3}+9\cdot 2t)(36t^{4}+4t^{2}+1)+(9t^{4}+9t^{2}+1)(36\cdot 4t^{3}+4\cdot 2t))}{[((9t^{4}+9t^{2}+1)(36t^{4}+4t^{2}+1))^{0.5}]^{2}}};}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe42c989163f84debc390833758b9f8b146d89e1)
