keisti
Vaizdas:Kreivispav145.jpg 145 pav. Apibrėžimas. Dydis R , priešingas kreiviui K linijos duotame taške M , vadinasi kreivio spinduliu šitos linijos nagrinėjamame taške:
R
=
1
K
,
(
1
)
{\displaystyle R={\frac {1}{K}},\quad (1)}
arba
R
=
[
1
+
(
d
y
d
x
)
2
]
3
2
|
d
2
y
d
x
2
|
.
(
2
)
{\displaystyle R={\frac {\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}{\left|{\frac {d^{2}y}{dx^{2}}}\right|}}.\quad (2)}
Nubrėžkime iš taško M normalę kreivės (145 pav.), nukreiptą į kreivės įlenkimo pusę, ir atidėsime ant šitos normalės atkarpą MC , lygią spinduliui R kreivio kreivės taške M . Taškas C vadinasi kreivio centru duotos kreivės taške M , apskritimas spindulio R su centru taške C (pereinantis per tašką M ) vadinasi apskritimu kreivio duotos kreivės taške M .
Iš apibrėžimo apskritimo kreivio seka, kad duotame taške kreivis kreivės ir kreivis apskritimo lygūs tarpusavyje.
Įvesime formules, nustatančias centro koordinates kreivio.
Tegu kreivė užrašyta lygtimi
y
=
f
(
x
)
.
(
3
)
{\displaystyle y=f(x).\quad (3)}
Vaizdas:Kreivispav146.jpg 146 pav. Užfiksuosime ant kreivės tašką M (x; y) ir nustatysime koordinates
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
lygtį normalės kreivės taške M :
Y
−
y
=
−
1
y
′
(
X
−
x
)
.
(
4
)
{\displaystyle Y-y=-{\frac {1}{y'}}(X-x).\quad (4)}
(Čia X ir Y - dabartinės koordinatės normalės taško.)
(Paaiškinimui, paimkime, kreivės tašką
M
(
x
M
;
y
M
)
,
{\displaystyle M(x_{M};\;y_{M}),}
y
−
y
M
=
−
1
f
′
(
x
M
)
(
x
−
x
M
)
,
{\displaystyle y-y_{M}=-{\frac {1}{f'(x_{M})}}(x-x_{M}),}
arba, kas visiškai tas pats (
y
′
(
x
M
)
=
f
′
(
x
M
)
=
d
y
d
x
|
x
M
{\displaystyle y'(x_{M})=f'(x_{M})={\frac {{\text{d}}y}{{\text{d}}x}}|_{x_{M}}}
y
M
−
y
=
−
1
f
′
(
x
M
)
(
x
M
−
x
)
;
{\displaystyle y_{M}-y=-{\frac {1}{f'(x_{M})}}(x_{M}-x);}
toliau mums reikia tik tų kreivės normalės taškų, kurie yra kreivio centrai, tai yra
C
(
α
;
β
)
{\displaystyle C(\alpha ;\;\beta )}
y
=
β
{\displaystyle y=\beta }
x
=
α
{\displaystyle x=\alpha }
β
−
y
M
=
−
1
f
′
(
x
M
)
(
α
−
x
M
)
;
{\displaystyle \beta -y_{M}=-{\frac {1}{f'(x_{M})}}(\alpha -x_{M});}
ta sąlyga yra
(
α
−
x
M
)
2
+
(
β
−
y
M
)
2
=
R
2
;
{\displaystyle (\alpha -x_{M})^{2}+(\beta -y_{M})^{2}=R^{2};}
toliau bus pažymėta, kad
x
M
=
x
{\displaystyle x_{M}=x}
y
M
=
y
{\displaystyle y_{M}=y}
f
′
(
x
M
)
=
y
′
{\displaystyle \;f'(x_{M})=y'}
Kadangi taškas
C
(
α
;
β
)
{\displaystyle C(\alpha ;\;\beta )}
β
−
y
=
−
1
y
′
(
α
−
x
)
.
(
5
)
{\displaystyle \beta -y=-{\frac {1}{y'}}(\alpha -x).\quad (5)}
Toliau, taškas
C
(
α
;
β
)
{\displaystyle C(\alpha ;\;\beta )}
M (x; y) atstumu, lygiu kreivio spinduliui R :
(
α
−
x
)
2
+
(
β
−
y
)
2
=
R
2
.
(
6
)
{\displaystyle (\alpha -x)^{2}+(\beta -y)^{2}=R^{2}.\quad (6)}
Sprendžiant kartu lygtis (5) ir (6), nustatysime
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
β
−
y
=
−
1
y
′
(
α
−
x
)
,
{\displaystyle \beta -y=-{\frac {1}{y'}}(\alpha -x),}
(
α
−
x
)
2
+
(
β
−
y
)
2
=
(
α
−
x
)
2
+
(
−
1
y
′
(
α
−
x
)
)
2
=
R
2
,
{\displaystyle (\alpha -x)^{2}+(\beta -y)^{2}=(\alpha -x)^{2}+\left(-{\frac {1}{y'}}(\alpha -x)\right)^{2}=R^{2},}
(
α
−
x
)
2
+
1
y
′
2
(
α
−
x
)
2
=
R
2
,
{\displaystyle (\alpha -x)^{2}+{\frac {1}{y'^{2}}}(\alpha -x)^{2}=R^{2},}
(
y
′
2
+
1
)
(
α
−
x
)
2
y
′
2
=
R
2
,
{\displaystyle {\frac {(y'^{2}+1)(\alpha -x)^{2}}{y'^{2}}}=R^{2},}
(
α
−
x
)
2
=
(
y
′
)
2
1
+
(
y
′
)
2
R
2
;
{\displaystyle (\alpha -x)^{2}={\frac {(y')^{2}}{1+(y')^{2}}}R^{2};}
iš čia
α
−
x
=
±
(
y
′
)
2
1
+
(
y
′
)
2
R
2
,
{\displaystyle \alpha -x=\pm {\sqrt {{\frac {(y')^{2}}{1+(y')^{2}}}R^{2}}},}
α
=
x
±
y
′
1
+
y
′
2
R
;
{\displaystyle \alpha =x\pm {\frac {y'}{\sqrt {1+y'^{2}}}}R;}
(
β
−
y
)
2
=
R
2
−
(
α
−
x
)
2
=
R
2
−
(
y
′
)
2
1
+
(
y
′
)
2
R
2
,
{\displaystyle (\beta -y)^{2}=R^{2}-(\alpha -x)^{2}=R^{2}-{\frac {(y')^{2}}{1+(y')^{2}}}R^{2},}
(
β
−
y
)
2
=
1
+
(
y
′
)
2
−
(
y
′
)
2
1
+
(
y
′
)
2
R
2
,
{\displaystyle (\beta -y)^{2}={\frac {1+(y')^{2}-(y')^{2}}{1+(y')^{2}}}R^{2},}
(
β
−
y
)
2
=
1
1
+
(
y
′
)
2
R
2
,
{\displaystyle (\beta -y)^{2}={\frac {1}{1+(y')^{2}}}R^{2},}
(
0
<
β
−
y
=
1
1
+
(
y
′
)
2
R
,
{\displaystyle 0<\beta -y={\frac {1}{\sqrt {1+(y')^{2}}}}R,}
β
=
y
∓
1
1
+
y
′
2
R
,
{\displaystyle \beta =y\mp {\frac {1}{\sqrt {1+y'^{2}}}}R,}
o kadangi
R
=
(
1
+
y
′
2
)
3
/
2
|
y
″
|
,
{\displaystyle R={\frac {(1+y'^{2})^{3/2}}{|y''|}},}
α
=
x
±
y
′
1
+
y
′
2
R
=
x
±
y
′
1
+
y
′
2
⋅
(
1
+
y
′
2
)
3
|
y
″
|
=
x
±
y
′
(
1
+
y
′
2
)
|
y
″
|
,
{\displaystyle \alpha =x\pm {\frac {y'}{\sqrt {1+y'^{2}}}}R=x\pm {\frac {y'}{\sqrt {1+y'^{2}}}}\cdot {\frac {\sqrt {(1+y'^{2})^{3}}}{|y''|}}=x\pm {\frac {y'(1+y'^{2})}{|y''|}},}
β
=
y
∓
1
1
+
y
′
2
R
=
y
∓
1
1
+
y
′
2
⋅
(
1
+
y
′
2
)
3
|
y
″
|
=
y
∓
1
+
y
′
2
|
y
″
|
.
{\displaystyle \beta =y\mp {\frac {1}{\sqrt {1+y'^{2}}}}R=y\mp {\frac {1}{\sqrt {1+y'^{2}}}}\cdot {\frac {\sqrt {(1+y'^{2})^{3}}}{|y''|}}=y\mp {\frac {1+y'^{2}}{|y''|}}.}
Kad išspręsti klausimą apie tai, viršutinius ar apatinius ženklus reikia imti paskutiniuose formulėse, reikia panagrinėti atvėjį
y
″
>
0
{\displaystyle y''>0}
y
″
<
0
{\displaystyle y''<0}
y
″
>
0
{\displaystyle y''>0}
β
>
y
{\displaystyle \beta >y}
|
y
″
|
=
y
″
{\displaystyle |y''|=y''}
α
=
x
−
y
′
(
1
+
y
′
2
)
y
″
,
(
7
)
{\displaystyle \alpha =x-{\frac {y'(1+y'^{2})}{y''}},\quad (7)}
β
=
y
+
1
+
y
′
2
y
″
.
(
7
)
{\displaystyle \beta =y+{\frac {1+y'^{2}}{y''}}.\quad (7)}
Analogišku budu galima parodyti, kad formulės (7) bus teisingos ir atveju
y
″
<
0
{\displaystyle y''<0}
Jeigu kreivė užrašyta parametrinėmis lygtimis
x
=
ϕ
(
t
)
,
y
=
ψ
(
t
)
,
{\displaystyle x=\phi (t),\quad y=\psi (t),}
tai centro koordinates lengva gauti iš formulių (7), įstačius į jas vietoje
y
′
{\displaystyle y'}
y
″
{\displaystyle y''}
y
′
=
y
t
′
x
t
′
;
y
″
=
y
t
″
x
t
′
−
y
t
′
x
t
″
x
t
′
3
.
{\displaystyle y'={\frac {y_{t}'}{x_{t}'}};\quad y''={\frac {y_{t}''x_{t}'-y_{t}'x_{t}''}{x_{t}'^{3}}}.}
Tada
α
=
x
−
y
t
′
x
t
′
(
1
+
(
y
t
′
x
t
′
)
2
)
y
t
″
x
t
′
−
y
t
′
x
t
″
x
t
′
3
=
x
−
y
t
′
x
t
′
(
1
+
(
y
t
′
x
t
′
)
2
)
⋅
x
t
′
3
y
t
″
x
t
′
−
y
t
′
x
t
″
=
x
−
y
t
′
x
t
′
⋅
(
x
t
′
)
2
+
(
y
t
′
)
2
(
x
t
′
)
2
⋅
x
t
′
3
y
t
″
x
t
′
−
y
t
′
x
t
″
=
x
−
y
′
(
x
′
2
+
y
′
2
)
y
″
x
′
−
y
′
x
″
,
(
7
′
)
{\displaystyle \alpha =x-{\frac {{\frac {y_{t}'}{x_{t}'}}(1+({\frac {y_{t}'}{x_{t}'}})^{2})}{\frac {y_{t}''x_{t}'-y_{t}'x_{t}''}{x_{t}'^{3}}}}=x-{\frac {y_{t}'}{x_{t}'}}(1+({\frac {y_{t}'}{x_{t}'}})^{2})\cdot {\frac {x_{t}'^{3}}{y_{t}''x_{t}'-y_{t}'x_{t}''}}=x-{\frac {y_{t}'}{x_{t}'}}\cdot {\frac {(x_{t}')^{2}+(y_{t}')^{2}}{(x_{t}')^{2}}}\cdot {\frac {x_{t}'^{3}}{y_{t}''x_{t}'-y_{t}'x_{t}''}}=x-{\frac {y'(x'^{2}+y'^{2})}{y''x'-y'x''}},\quad (7')}
β
=
y
+
1
+
(
y
t
′
x
t
′
)
2
y
t
″
x
t
′
−
y
t
′
x
t
″
x
t
′
3
=
y
+
(
1
+
(
y
t
′
x
t
′
)
2
)
⋅
x
t
′
3
y
t
″
x
t
′
−
y
t
′
x
t
″
=
y
+
(
x
t
′
)
2
+
(
y
t
′
)
2
(
x
t
′
)
2
⋅
(
x
t
′
)
3
y
t
″
x
t
′
−
y
t
′
x
t
″
=
y
+
x
′
(
x
′
2
+
y
′
2
)
y
″
x
′
−
y
′
x
″
.
(
7
′
)
{\displaystyle \beta =y+{\frac {1+({\frac {y_{t}'}{x_{t}'}})^{2}}{\frac {y_{t}''x_{t}'-y_{t}'x_{t}''}{x_{t}'^{3}}}}=y+(1+({\frac {y_{t}'}{x_{t}'}})^{2})\cdot {\frac {x_{t}'^{3}}{y_{t}''x_{t}'-y_{t}'x_{t}''}}=y+{\frac {(x_{t}')^{2}+(y_{t}')^{2}}{(x_{t}')^{2}}}\cdot {\frac {(x_{t}')^{3}}{y_{t}''x_{t}'-y_{t}'x_{t}''}}=y+{\frac {x'(x'^{2}+y'^{2})}{y''x'-y'x''}}.\quad (7')}
Jeigu taške
M
1
(
x
;
y
)
{\displaystyle M_{1}(x;y)}
C
1
(
α
;
β
)
{\displaystyle C_{1}(\alpha ;\beta )}
evoliute atžvilgiu pirmos.
Tokiu budu, geometrinė vieta koordinačių centrų duotos linijos vadinasi jos evoliute . Atžvilgiu savo evoliutės duota linija vadinasi evolvente arba involiute . Jeigu duota kreivė
y
=
f
(
x
)
,
{\displaystyle y=f(x),\;}
x . Eliminavę iš šitų lygčių parametrą x (jeigu tai įmanoma), gausime betarpišką priklausomybę tarp dabartinių koordinačių evoliutės
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
x
=
ϕ
(
t
)
,
{\displaystyle x=\phi (t),}
y
=
ψ
(
t
)
,
{\displaystyle y=\psi (t),}
x
{\displaystyle x}
y
{\displaystyle y}
x
′
{\displaystyle x'}
y
′
{\displaystyle y'}
x
″
{\displaystyle x''}
y
″
{\displaystyle y''}
t ).
Polinėse koordinatėse kreivio centro koordinatės
(
x
c
,
y
c
)
,
{\displaystyle (x_{c},\;y_{c}),}
φ
,
{\displaystyle \varphi ,}
ρ
=
f
(
φ
)
,
{\displaystyle \rho =f(\varphi ),}
x
c
=
ρ
cos
φ
−
(
ρ
2
+
ρ
′
2
)
(
ρ
cos
φ
+
ρ
′
sin
φ
)
ρ
2
+
2
ρ
′
2
−
ρ
ρ
″
,
{\displaystyle x_{c}=\rho \cos \varphi -{\frac {(\rho ^{2}+\rho '^{2})(\rho \cos \varphi +\rho '\sin \varphi )}{\rho ^{2}+2\rho '^{2}-\rho \rho ''}},}
y
c
=
ρ
sin
φ
−
(
ρ
2
+
ρ
′
2
)
(
ρ
sin
φ
−
ρ
′
cos
φ
)
ρ
2
+
2
ρ
′
2
−
ρ
ρ
″
.
{\displaystyle y_{c}=\rho \sin \varphi -{\frac {(\rho ^{2}+\rho '^{2})(\rho \sin \varphi -\rho '\cos \varphi )}{\rho ^{2}+2\rho '^{2}-\rho \rho ''}}.}
Šios kreivio centro koordinatės
(
x
c
,
y
c
)
{\displaystyle (x_{c},\;y_{c})}
M
(
x
;
y
)
,
{\displaystyle M(x;y),}
ρ
{\displaystyle \rho }
ϕ
{\displaystyle \phi }
Šitos formulės gali būti užrašytos taip:
x
c
=
x
−
R
d
y
d
s
,
{\displaystyle x_{c}=x-R{\frac {dy}{ds}},}
y
c
=
y
+
R
d
x
d
s
;
{\displaystyle y_{c}=y+R{\frac {dx}{ds}};}
čia R apskaičiuojamas pagal (1) formulę (R=1/K). Pavyzdžiui, polinėse koordinatėse
d
x
d
φ
=
ρ
′
cos
φ
−
ρ
sin
φ
,
d
y
d
φ
=
ρ
′
sin
φ
+
ρ
cos
φ
;
{\displaystyle {\frac {dx}{d\varphi }}=\rho '\cos \varphi -\rho \sin \varphi ,\;\;{\frac {dy}{d\varphi }}=\rho '\sin \varphi +\rho \cos \varphi ;}
d
s
=
ρ
′
2
+
ρ
2
d
φ
;
{\displaystyle ds={\sqrt {\rho '^{2}+\rho ^{2}}}d\varphi ;}
K
=
ρ
2
+
2
ρ
′
2
−
ρ
ρ
″
(
ρ
2
+
ρ
′
2
)
3
/
2
.
{\displaystyle K={\frac {\rho ^{2}+2\rho '^{2}-\rho \rho ''}{(\rho ^{2}+\rho '^{2})^{3/2}}}.}
Vaizdas:Kreivispav147.jpg 147 pav. Nustatyti krevio centro koordinates parabolės
y
2
=
2
p
x
{\displaystyle y^{2}=2px}
a) laisvai pasirenktame taške M (x; y); b) taške
M
0
(
0
;
0
)
{\displaystyle M_{0}(0;0)}
M
1
(
p
2
;
p
)
.
{\displaystyle M_{1}\left({\frac {p}{2}};\;p\right).}
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
p
x
)
′
=
1
2
⋅
(
2
p
x
)
′
2
p
x
=
p
2
p
x
;
{\displaystyle {\frac {dy}{dx}}=({\sqrt {2px}})'={\frac {1}{2}}\cdot {\frac {(2px)'}{\sqrt {2px}}}={\frac {p}{\sqrt {2px}}};}
d
2
y
d
x
2
=
(
p
2
p
x
)
′
=
−
1
2
⋅
p
(
2
p
x
)
′
(
2
p
x
)
3
2
=
−
p
2
(
2
p
x
)
3
2
;
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=\left({\frac {p}{\sqrt {2px}}}\right)'=-{\frac {1}{2}}\cdot {\frac {p(2px)'}{(2px)^{3 \over 2}}}=-{\frac {p^{2}}{(2px)^{3 \over 2}}};}
a)
α
=
x
−
y
′
(
1
+
y
′
2
)
y
″
=
x
−
p
2
p
x
(
1
+
p
2
2
p
x
)
−
p
2
(
2
p
x
)
3
2
=
x
−
p
2
p
x
+
p
3
(
2
p
x
)
3
/
2
−
p
2
(
2
p
x
)
3
2
=
x
+
(
2
p
x
)
3
2
p
2
p
2
p
x
+
p
=
{\displaystyle \alpha =x-{\frac {y'(1+y'^{2})}{y''}}=x-{\frac {{\frac {p}{\sqrt {2px}}}(1+{\frac {p^{2}}{2px}})}{-{\frac {p^{2}}{(2px)^{3 \over 2}}}}}=x-{\frac {{\frac {p}{\sqrt {2px}}}+{\frac {p^{3}}{(2px)^{3/2}}}}{-{\frac {p^{2}}{(2px)^{3 \over 2}}}}}=x+{\frac {(2px)^{3 \over 2}}{p^{2}}}{\frac {p}{\sqrt {2px}}}+p=}
=
x
+
2
p
x
p
+
p
=
3
x
+
p
,
{\displaystyle =x+{\frac {2px}{p}}+p=3x+p,}
β
=
y
+
1
+
y
′
2
y
″
=
2
p
x
+
1
+
p
2
2
p
x
−
p
2
(
2
p
x
)
3
2
=
2
p
x
+
(
−
(
2
p
x
)
3
2
p
2
)
+
p
2
2
p
x
(
−
(
2
p
x
)
3
2
p
2
)
=
{\displaystyle \beta =y+{\frac {1+y'^{2}}{y''}}={\sqrt {2px}}+{\frac {1+{\frac {p^{2}}{2px}}}{-{\frac {p^{2}}{(2px)^{3 \over 2}}}}}={\sqrt {2px}}+\left(-{\frac {(2px)^{3 \over 2}}{p^{2}}}\right)+{\frac {p^{2}}{2px}}\left(-{\frac {(2px)^{3 \over 2}}{p^{2}}}\right)=}
=
2
p
x
−
(
2
p
x
)
3
2
p
2
−
2
p
x
=
−
(
2
p
x
)
3
2
p
2
=
−
(
2
x
)
3
2
p
;
{\displaystyle ={\sqrt {2px}}-{\frac {(2px)^{3 \over 2}}{p^{2}}}-{\sqrt {2px}}=-{\frac {(2px)^{3 \over 2}}{p^{2}}}=-{\frac {(2x)^{3 \over 2}}{\sqrt {p}}};}
b) kai
x
=
0
{\displaystyle x=0}
α
=
3
⋅
0
+
p
=
p
,
β
=
−
(
2
⋅
0
)
3
2
p
=
0
;
{\displaystyle \alpha =3\cdot 0+p=p,\;\;\beta =-{\frac {(2\cdot 0)^{3 \over 2}}{\sqrt {p}}}=0;}
c) kai
x
=
p
2
{\displaystyle x={\frac {p}{2}}}
α
=
3
⋅
p
2
+
p
=
5
p
2
,
β
=
−
(
2
⋅
p
2
)
3
2
p
=
−
p
.
{\displaystyle \alpha =3\cdot {\frac {p}{2}}+p={\frac {5p}{2}},\;\;\beta =-{\frac {(2\cdot {\frac {p}{2}})^{3 \over 2}}{\sqrt {p}}}=-p.}
Vaizdas:Kreivispav148.jpg 148 pav. Rasti lygtį evoliutės parabolės
y
2
=
2
p
x
.
{\displaystyle y^{2}=2px.}
Sprendimas . Pagrindu pirmo pavyzdžio turime bet kokiam taškui (x; y) parabolės:
d
y
d
x
=
(
2
p
x
)
′
=
p
2
p
x
;
{\displaystyle {\frac {dy}{dx}}=({\sqrt {2px}})'={\frac {p}{\sqrt {2px}}};}
d
2
y
d
x
2
=
(
p
2
p
x
)
′
=
−
p
2
(
2
p
x
)
3
2
;
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=\left({\frac {p}{\sqrt {2px}}}\right)'=-{\frac {p^{2}}{(2px)^{3 \over 2}}};}
α
=
x
−
y
′
(
1
+
y
′
2
)
y
″
=
x
−
p
2
p
x
(
1
+
p
2
2
p
x
)
−
p
2
(
2
p
x
)
3
2
=
x
−
p
2
p
x
+
p
3
(
2
p
x
)
3
/
2
−
p
2
(
2
p
x
)
3
2
=
x
+
(
2
p
x
)
3
2
p
2
p
2
p
x
+
p
=
{\displaystyle \alpha =x-{\frac {y'(1+y'^{2})}{y''}}=x-{\frac {{\frac {p}{\sqrt {2px}}}(1+{\frac {p^{2}}{2px}})}{-{\frac {p^{2}}{(2px)^{3 \over 2}}}}}=x-{\frac {{\frac {p}{\sqrt {2px}}}+{\frac {p^{3}}{(2px)^{3/2}}}}{-{\frac {p^{2}}{(2px)^{3 \over 2}}}}}=x+{\frac {(2px)^{3 \over 2}}{p^{2}}}{\frac {p}{\sqrt {2px}}}+p=}
=
x
+
2
p
x
p
+
p
=
3
x
+
p
,
{\displaystyle =x+{\frac {2px}{p}}+p=3x+p,}
β
=
y
+
1
+
y
′
2
y
″
=
2
p
x
+
1
+
p
2
2
p
x
−
p
2
(
2
p
x
)
3
2
=
2
p
x
+
(
−
(
2
p
x
)
3
2
p
2
)
+
p
2
2
p
x
(
−
(
2
p
x
)
3
2
p
2
)
=
{\displaystyle \beta =y+{\frac {1+y'^{2}}{y''}}={\sqrt {2px}}+{\frac {1+{\frac {p^{2}}{2px}}}{-{\frac {p^{2}}{(2px)^{3 \over 2}}}}}={\sqrt {2px}}+\left(-{\frac {(2px)^{3 \over 2}}{p^{2}}}\right)+{\frac {p^{2}}{2px}}\left(-{\frac {(2px)^{3 \over 2}}{p^{2}}}\right)=}
=
2
p
x
−
(
2
p
x
)
3
2
p
2
−
2
p
x
=
−
(
2
p
x
)
3
2
p
2
=
−
(
2
x
)
3
2
p
.
{\displaystyle ={\sqrt {2px}}-{\frac {(2px)^{3 \over 2}}{p^{2}}}-{\sqrt {2px}}=-{\frac {(2px)^{3 \over 2}}{p^{2}}}=-{\frac {(2x)^{3 \over 2}}{\sqrt {p}}}.}
Eliminuojant iš šitų lygčių parametrą x , gausime:
α
=
3
x
+
p
,
{\displaystyle \alpha =3x+p,}
α
−
p
=
3
x
,
{\displaystyle \alpha -p=3x,}
x
=
α
−
p
3
;
{\displaystyle x={\frac {\alpha -p}{3}};}
β
=
−
(
2
x
)
3
2
p
,
{\displaystyle \beta =-{\frac {(2x)^{3 \over 2}}{\sqrt {p}}},}
β
=
−
(
2
⋅
α
−
p
3
)
3
2
p
,
{\displaystyle \beta =-{\frac {(2\cdot {\frac {\alpha -p}{3}})^{3 \over 2}}{\sqrt {p}}},}
β
=
−
8
27
(
α
−
p
)
3
2
p
,
{\displaystyle \beta =-{\sqrt {\frac {8}{27}}}{\frac {(\alpha -p)^{3 \over 2}}{\sqrt {p}}},}
β
2
=
8
27
p
(
α
−
p
)
3
.
{\displaystyle \beta ^{2}={\frac {8}{27p}}(\alpha -p)^{3}.}
Tai - lygtis pusiaukubinės parabolės (148 pav.).
Rasti lygtį evoliutės elipsės, užrašytos parametrinėmis lygtimis:
x
=
a
cos
t
,
y
=
b
sin
t
.
{\displaystyle x=a\cos t,\quad y=b\sin t.}
Sprendimas . Apskaičiuojame išvestines nuo x ir y per t :
x
′
=
−
a
sin
t
,
y
′
=
b
cos
t
;
{\displaystyle x'=-a\sin t,\quad y'=b\cos t;}
x
″
=
−
a
cos
t
,
y
″
=
−
b
sin
t
.
{\displaystyle x''=-a\cos t,\quad y''=-b\sin t.}
Įstatant išraiškas išvestinių į formules (7'), gausime:
α
=
x
−
y
′
(
x
′
2
+
y
′
2
)
y
″
x
′
−
y
′
x
″
=
a
cos
t
−
b
cos
(
t
)
(
(
−
a
sin
t
)
2
+
(
b
cos
t
)
2
)
−
b
sin
(
t
)
(
−
a
sin
t
)
−
b
cos
(
t
)
(
−
a
cos
t
)
=
a
cos
t
−
b
cos
(
t
)
(
a
2
sin
2
t
+
b
2
cos
2
t
)
a
b
sin
2
t
+
a
b
cos
2
t
=
{\displaystyle \alpha =x-{\frac {y'(x'^{2}+y'^{2})}{y''x'-y'x''}}=a\cos t-{\frac {b\cos(t)((-a\sin t)^{2}+(b\cos t)^{2})}{-b\sin(t)(-a\sin t)-b\cos(t)(-a\cos t)}}=a\cos t-{\frac {b\cos(t)(a^{2}\sin ^{2}t+b^{2}\cos ^{2}t)}{ab\sin ^{2}t+ab\cos ^{2}t}}=}
=
a
cos
t
−
b
cos
(
t
)
(
a
2
sin
2
t
+
b
2
cos
2
t
)
a
b
=
a
cos
t
−
cos
(
t
)
(
a
2
sin
2
t
+
b
2
cos
2
t
)
a
=
a
cos
t
−
cos
(
t
)
(
a
sin
2
t
+
b
2
a
cos
2
t
)
=
{\displaystyle =a\cos t-{\frac {b\cos(t)(a^{2}\sin ^{2}t+b^{2}\cos ^{2}t)}{ab}}=a\cos t-{\frac {\cos(t)(a^{2}\sin ^{2}t+b^{2}\cos ^{2}t)}{a}}=a\cos t-\cos(t)(a\sin ^{2}t+{\frac {b^{2}}{a}}\cos ^{2}t)=}
=
a
cos
t
−
cos
(
t
)
a
sin
2
t
−
b
2
a
cos
3
t
=
a
cos
t
−
cos
(
t
)
a
sin
2
t
−
b
2
a
cos
3
t
=
a
cos
(
t
)
(
1
−
sin
2
t
)
−
b
2
a
cos
3
t
=
{\displaystyle =a\cos t-\cos(t)a\sin ^{2}t-{\frac {b^{2}}{a}}\cos ^{3}t=a\cos t-\cos(t)a\sin ^{2}t-{\frac {b^{2}}{a}}\cos ^{3}t=a\cos(t)(1-\sin ^{2}t)-{\frac {b^{2}}{a}}\cos ^{3}t=}
=
a
cos
3
t
−
b
2
a
cos
3
t
=
(
a
−
b
2
a
)
cos
3
t
.
{\displaystyle =a\cos ^{3}t-{\frac {b^{2}}{a}}\cos ^{3}t=\left(a-{\frac {b^{2}}{a}}\right)\cos ^{3}t.}
β
=
y
+
x
′
(
x
′
2
+
y
′
2
)
y
″
x
′
−
y
′
x
″
=
b
sin
t
+
−
a
sin
(
t
)
(
(
−
a
sin
t
)
2
+
(
b
cos
t
)
2
)
−
b
sin
(
t
)
(
−
a
sin
t
)
−
b
cos
(
t
)
(
−
a
cos
t
)
=
b
sin
t
−
a
sin
(
t
)
(
a
2
sin
2
t
+
b
2
cos
2
t
)
a
b
sin
2
t
+
a
b
cos
2
t
=
{\displaystyle \beta =y+{\frac {x'(x'^{2}+y'^{2})}{y''x'-y'x''}}=b\sin t+{\frac {-a\sin(t)((-a\sin t)^{2}+(b\cos t)^{2})}{-b\sin(t)(-a\sin t)-b\cos(t)(-a\cos t)}}=b\sin t-{\frac {a\sin(t)(a^{2}\sin ^{2}t+b^{2}\cos ^{2}t)}{ab\sin ^{2}t+ab\cos ^{2}t}}=}
=
b
sin
t
−
a
sin
(
t
)
(
a
2
sin
2
t
+
b
2
cos
2
t
)
a
b
=
b
sin
t
−
sin
(
t
)
(
a
2
sin
2
t
+
b
2
cos
2
t
)
b
=
{\displaystyle =b\sin t-{\frac {a\sin(t)(a^{2}\sin ^{2}t+b^{2}\cos ^{2}t)}{ab}}=b\sin t-{\frac {\sin(t)(a^{2}\sin ^{2}t+b^{2}\cos ^{2}t)}{b}}=}
=
b
sin
t
−
a
2
b
sin
3
t
−
b
cos
2
t
sin
t
=
(
1
−
cos
2
t
)
b
sin
t
−
a
2
b
sin
3
t
=
b
sin
3
t
−
a
2
b
sin
3
t
=
(
b
−
a
2
b
)
sin
3
t
.
{\displaystyle =b\sin t-{\frac {a^{2}}{b}}\sin ^{3}t-b\cos ^{2}t\sin t=(1-\cos ^{2}t)b\sin t-{\frac {a^{2}}{b}}\sin ^{3}t=b\sin ^{3}t-{\frac {a^{2}}{b}}\sin ^{3}t=\left(b-{\frac {a^{2}}{b}}\right)\sin ^{3}t.}
Vaizdas:Kreivispav149.jpg 149 pav. Eliminavę parametrą t , gauname lygtį evoliutės elipsės pavidalu
α
=
(
a
−
b
2
a
)
cos
3
t
,
{\displaystyle \alpha =\left(a-{\frac {b^{2}}{a}}\right)\cos ^{3}t,}
α
a
−
b
2
a
=
cos
3
t
,
{\displaystyle {\frac {\alpha }{a-{\frac {b^{2}}{a}}}}=\cos ^{3}t,}
(
α
a
−
b
2
a
)
1
3
=
cos
t
,
{\displaystyle \left({\frac {\alpha }{a-{\frac {b^{2}}{a}}}}\right)^{1 \over 3}=\cos t,}
(
α
a
−
b
2
a
)
2
3
=
cos
2
t
,
{\displaystyle \left({\frac {\alpha }{a-{\frac {b^{2}}{a}}}}\right)^{2 \over 3}=\cos ^{2}t,}
α
2
3
(
a
2
−
b
2
a
)
2
3
=
cos
2
t
;
{\displaystyle {\frac {\alpha ^{2 \over 3}}{({\frac {a^{2}-b^{2}}{a}})^{2 \over 3}}}=\cos ^{2}t;}
β
=
(
b
−
a
2
b
)
sin
3
t
,
{\displaystyle \beta =\left(b-{\frac {a^{2}}{b}}\right)\sin ^{3}t,}
β
b
−
a
2
b
=
sin
3
t
,
{\displaystyle {\frac {\beta }{b-{\frac {a^{2}}{b}}}}=\sin ^{3}t,}
(
β
b
−
a
2
b
)
1
3
=
sin
t
,
{\displaystyle \left({\frac {\beta }{b-{\frac {a^{2}}{b}}}}\right)^{1 \over 3}=\sin t,}
(
β
b
−
a
2
b
)
2
3
=
sin
2
t
,
{\displaystyle \left({\frac {\beta }{b-{\frac {a^{2}}{b}}}}\right)^{2 \over 3}=\sin ^{2}t,}
β
2
3
(
b
2
−
a
2
b
)
2
3
=
sin
2
t
;
{\displaystyle {\frac {\beta ^{2 \over 3}}{({\frac {b^{2}-a^{2}}{b}})^{2 \over 3}}}=\sin ^{2}t;}
cos
2
t
+
sin
2
t
=
α
2
3
(
a
2
−
b
2
a
)
2
3
+
β
2
3
(
b
2
−
a
2
b
)
2
3
,
{\displaystyle \cos ^{2}t+\sin ^{2}t={\frac {\alpha ^{2 \over 3}}{({\frac {a^{2}-b^{2}}{a}})^{2 \over 3}}}+{\frac {\beta ^{2 \over 3}}{({\frac {b^{2}-a^{2}}{b}})^{2 \over 3}}},}
1
=
α
2
3
(
a
2
−
b
2
a
)
2
3
+
β
2
3
(
b
2
−
a
2
b
)
2
3
,
{\displaystyle 1={\frac {\alpha ^{2 \over 3}}{({\frac {a^{2}-b^{2}}{a}})^{2 \over 3}}}+{\frac {\beta ^{2 \over 3}}{({\frac {b^{2}-a^{2}}{b}})^{2 \over 3}}},}
1
=
a
2
3
α
2
3
(
a
2
−
b
2
)
2
3
+
(
−
b
)
2
3
β
2
3
(
a
2
−
b
2
)
2
3
,
{\displaystyle 1={\frac {a^{2 \over 3}\alpha ^{2 \over 3}}{(a^{2}-b^{2})^{2 \over 3}}}+{\frac {(-b)^{2 \over 3}\beta ^{2 \over 3}}{(a^{2}-b^{2})^{2 \over 3}}},}
(
a
2
−
b
2
)
2
3
=
a
2
3
α
2
3
+
(
−
b
)
2
3
β
2
3
,
{\displaystyle (a^{2}-b^{2})^{2 \over 3}=a^{2 \over 3}\alpha ^{2 \over 3}+(-b)^{2 \over 3}\beta ^{2 \over 3},}
(
a
2
−
b
2
)
2
3
=
a
2
3
α
2
3
+
b
2
3
β
2
3
,
{\displaystyle (a^{2}-b^{2})^{2 \over 3}=a^{2 \over 3}\alpha ^{2 \over 3}+b^{2 \over 3}\beta ^{2 \over 3},}
(
α
b
)
2
3
+
(
β
a
)
2
3
=
(
a
2
−
b
2
a
b
)
2
3
;
{\displaystyle \left({\frac {\alpha }{b}}\right)^{2 \over 3}+\left({\frac {\beta }{a}}\right)^{2 \over 3}=\left({\frac {a^{2}-b^{2}}{ab}}\right)^{2 \over 3};}
arba
(
β
a
)
2
3
=
(
a
2
−
b
2
a
b
)
2
3
−
(
α
b
)
2
3
,
{\displaystyle \left({\frac {\beta }{a}}\right)^{2 \over 3}=\left({\frac {a^{2}-b^{2}}{ab}}\right)^{2 \over 3}-\left({\frac {\alpha }{b}}\right)^{2 \over 3},}
β
a
=
[
(
a
2
−
b
2
a
b
)
2
3
−
(
α
b
)
2
3
]
3
2
,
{\displaystyle {\frac {\beta }{a}}=\left[\left({\frac {a^{2}-b^{2}}{ab}}\right)^{2 \over 3}-\left({\frac {\alpha }{b}}\right)^{2 \over 3}\right]^{3 \over 2},}
β
=
a
[
(
a
2
−
b
2
a
b
)
2
3
−
(
α
b
)
2
3
]
3
2
.
{\displaystyle \beta =a\left[\left({\frac {a^{2}-b^{2}}{ab}}\right)^{2 \over 3}-\left({\frac {\alpha }{b}}\right)^{2 \over 3}\right]^{3 \over 2}.}
Čia
α
{\displaystyle \alpha }
β
{\displaystyle \beta }
α
{\displaystyle \alpha }
x kordinatė,
β
{\displaystyle \beta }
y koordinatė;
β
{\displaystyle \beta }
α
{\displaystyle \alpha }
Vaizdas:Kreivispav150.jpg 150 pav. Rasti parametrines lygtis evoliutės cikloidės
x
=
a
(
t
−
sin
t
)
,
{\displaystyle x=a(t-\sin t),}
y
=
a
(
1
−
cos
t
)
.
{\displaystyle y=a(1-\cos t).}
Sprendimas .
x
′
=
a
(
1
−
cos
t
)
;
y
′
=
a
sin
t
;
{\displaystyle x'=a(1-\cos t);\quad y'=a\sin t;}
x
″
=
a
sin
t
;
y
″
=
a
cos
t
.
{\displaystyle x''=a\sin t;\quad y''=a\cos t.}
Įstačius gautas išraiškas į formulę (7'), randame:
α
=
x
−
y
′
(
x
′
2
+
y
′
2
)
y
″
x
′
−
y
′
x
″
=
a
(
t
−
sin
t
)
−
a
sin
(
t
)
(
a
2
(
1
−
cos
t
)
2
+
(
a
sin
t
)
2
)
a
cos
(
t
)
a
(
1
−
cos
t
)
−
a
sin
(
t
)
a
sin
(
t
)
=
{\displaystyle \alpha =x-{\frac {y'(x'^{2}+y'^{2})}{y''x'-y'x''}}=a(t-\sin t)-{\frac {a\sin(t)(a^{2}(1-\cos t)^{2}+(a\sin t)^{2})}{a\cos(t)a(1-\cos t)-a\sin(t)a\sin(t)}}=}
=
a
(
t
−
sin
t
)
−
a
sin
(
t
)
(
a
2
(
1
−
2
cos
t
+
cos
2
t
)
+
a
2
sin
2
t
)
a
2
cos
t
−
a
2
cos
2
t
−
a
2
sin
2
t
=
a
(
t
−
sin
t
)
−
a
sin
(
t
)
(
a
2
−
2
a
2
cos
t
+
a
2
)
a
2
cos
t
−
a
2
=
{\displaystyle =a(t-\sin t)-{\frac {a\sin(t)(a^{2}(1-2\cos t+\cos ^{2}t)+a^{2}\sin ^{2}t)}{a^{2}\cos t-a^{2}\cos ^{2}t-a^{2}\sin ^{2}t}}=a(t-\sin t)-{\frac {a\sin(t)(a^{2}-2a^{2}\cos t+a^{2})}{a^{2}\cos t-a^{2}}}=}
=
a
(
t
−
sin
t
)
+
a
sin
(
t
)
(
2
a
2
−
2
a
2
cos
t
)
a
2
(
1
−
cos
t
)
=
a
(
t
−
sin
t
)
+
2
a
2
sin
(
t
)
(
1
−
cos
t
)
a
(
1
−
cos
t
)
=
a
(
t
−
sin
t
)
+
2
a
sin
t
=
{\displaystyle =a(t-\sin t)+{\frac {a\sin(t)(2a^{2}-2a^{2}\cos t)}{a^{2}(1-\cos t)}}=a(t-\sin t)+{\frac {2a^{2}\sin(t)(1-\cos t)}{a(1-\cos t)}}=a(t-\sin t)+2a\sin t=}
=
a
t
−
a
sin
t
+
2
a
sin
t
=
a
(
t
+
sin
t
)
.
{\displaystyle =at-a\sin t+2a\sin t=a(t+\sin t).}
β
=
y
+
x
′
(
x
′
2
+
y
′
2
)
y
″
x
′
−
y
′
x
″
=
a
(
1
−
cos
t
)
+
a
(
1
−
cos
t
)
(
a
2
(
1
−
cos
t
)
2
+
(
a
sin
t
)
2
)
a
cos
(
t
)
a
(
1
−
cos
t
)
−
a
sin
(
t
)
a
sin
(
t
)
=
{\displaystyle \beta =y+{\frac {x'(x'^{2}+y'^{2})}{y''x'-y'x''}}=a(1-\cos t)+{\frac {a(1-\cos t)(a^{2}(1-\cos t)^{2}+(a\sin t)^{2})}{a\cos(t)a(1-\cos t)-a\sin(t)a\sin(t)}}=}
=
a
(
1
−
cos
t
)
+
a
(
1
−
cos
t
)
(
a
2
(
1
−
2
cos
t
+
cos
2
t
)
+
a
2
sin
2
t
)
a
2
cos
t
−
a
2
cos
2
t
−
a
2
sin
2
t
=
a
(
1
−
cos
t
)
+
a
(
1
−
cos
t
)
(
a
2
−
2
a
2
cos
t
+
a
2
)
a
2
cos
t
−
a
2
=
{\displaystyle =a(1-\cos t)+{\frac {a(1-\cos t)(a^{2}(1-2\cos t+\cos ^{2}t)+a^{2}\sin ^{2}t)}{a^{2}\cos t-a^{2}\cos ^{2}t-a^{2}\sin ^{2}t}}=a(1-\cos t)+{\frac {a(1-\cos t)(a^{2}-2a^{2}\cos t+a^{2})}{a^{2}\cos t-a^{2}}}=}
=
a
(
1
−
cos
t
)
−
a
(
1
−
cos
t
)
(
2
a
2
−
2
a
2
cos
t
)
a
2
(
1
−
cos
t
)
=
a
(
1
−
cos
t
)
−
2
a
2
(
1
−
cos
t
)
(
1
−
cos
t
)
a
(
1
−
cos
t
)
=
{\displaystyle =a(1-\cos t)-{\frac {a(1-\cos t)(2a^{2}-2a^{2}\cos t)}{a^{2}(1-\cos t)}}=a(1-\cos t)-{\frac {2a^{2}(1-\cos t)(1-\cos t)}{a(1-\cos t)}}=}
=
a
(
1
−
cos
t
)
−
2
a
(
1
−
cos
t
)
=
−
a
(
1
−
cos
t
)
.
{\displaystyle =a(1-\cos t)-2a(1-\cos t)=-a(1-\cos t).}
Padarysime perdarymus kintamųjų, nustatę
α
=
ξ
−
π
a
,
{\displaystyle \alpha =\xi -\pi a,}
β
=
η
−
2
a
,
{\displaystyle \beta =\eta -2a,}
t
=
τ
−
π
;
{\displaystyle t=\tau -\pi ;}
tada lygtys evoliutės atrodys taip
ξ
=
a
(
t
+
sin
t
)
+
π
a
=
a
(
τ
−
π
+
sin
(
τ
−
π
)
)
+
π
a
=
a
(
τ
−
π
−
sin
τ
)
+
π
a
=
a
(
τ
−
sin
τ
)
,
{\displaystyle \xi =a(t+\sin t)+\pi a=a(\tau -\pi +\sin(\tau -\pi ))+\pi a=a(\tau -\pi -\sin \tau )+\pi a=a(\tau -\sin \tau ),}
η
=
−
a
(
1
−
cos
t
)
+
2
a
=
−
a
(
1
−
cos
(
τ
−
π
)
)
+
2
a
=
−
a
(
1
+
cos
τ
)
+
2
a
=
a
(
1
−
cos
τ
)
;
{\displaystyle \eta =-a(1-\cos t)+2a=-a(1-\cos(\tau -\pi ))+2a=-a(1+\cos \tau )+2a=a(1-\cos \tau );}
jos nustato koordinatėse
ξ
{\displaystyle \xi }
η
{\displaystyle \eta }
a . Tokiu budu, evoliute cikloidės yra tokia pati cikloidė, bet perstumta per ašį Ox dydžiu -
π
a
{\displaystyle \pi a}
Oy dydžiu -
2
a
{\displaystyle 2a}
Teorema 1. Duotosios kreivės normalė yra jos evoliutės liestinė. Įrodymas. Kampinis koeficientas liestinės evoliutės, nustatytos parametrinėmis lygtimis (7') praeito skyriaus, lygus
d
β
d
α
=
d
β
d
x
d
α
d
x
.
{\displaystyle {\frac {d\beta }{d\alpha }}={\frac {\frac {d\beta }{dx}}{\frac {d\alpha }{dx}}}.}
Pastebėsime, kad [pagrindu tų pačių lygčių (7')]
d
α
d
x
=
(
x
−
y
′
(
x
′
2
+
y
′
2
)
y
″
x
′
−
y
′
x
″
)
′
=
1
−
(
y
′
x
′
2
+
y
′
y
′
2
y
″
x
′
−
y
′
x
″
)
′
=
1
−
(
y
′
⋅
1
2
+
y
′
y
′
2
y
″
−
y
′
⋅
0
)
′
=
1
−
(
y
′
+
y
′
3
y
″
)
′
=
{\displaystyle {\frac {d\alpha }{dx}}=\left(x-{\frac {y'(x'^{2}+y'^{2})}{y''x'-y'x''}}\right)'=1-\left({\frac {y'x'^{2}+y'y'^{2}}{y''x'-y'x''}}\right)'=1-\left({\frac {y'\cdot 1^{2}+y'y'^{2}}{y''-y'\cdot 0}}\right)'=1-\left({\frac {y'+y'^{3}}{y''}}\right)'=}
=
1
−
(
y
′
+
y
′
3
)
′
y
″
−
(
y
′
+
y
′
3
)
y
‴
y
″
2
=
1
−
(
y
″
+
(
y
′
3
)
′
)
y
″
−
y
′
y
‴
−
y
′
3
y
‴
y
″
2
=
{\displaystyle =1-{\frac {(y'+y'^{3})'y''-(y'+y'^{3})y'''}{y''^{2}}}=1-{\frac {(y''+(y'^{3})')y''-y'y'''-y'^{3}y'''}{y''^{2}}}=}
=
1
−
(
y
″
+
3
y
′
2
y
″
)
y
″
−
y
′
y
‴
−
y
′
3
y
‴
y
″
2
=
1
−
y
″
2
+
3
y
″
2
y
′
2
−
y
′
y
‴
−
y
′
3
y
‴
y
″
2
=
{\displaystyle =1-{\frac {(y''+3y'^{2}y'')y''-y'y'''-y'^{3}y'''}{y''^{2}}}=1-{\frac {y''^{2}+3y''^{2}y'^{2}-y'y'''-y'^{3}y'''}{y''^{2}}}=}
=
y
″
2
−
(
y
″
2
+
3
y
″
2
y
′
2
−
y
′
y
‴
−
y
′
3
y
‴
)
y
″
2
=
−
3
y
″
2
y
′
2
+
y
′
y
‴
+
y
′
3
y
‴
y
″
2
=
{\displaystyle ={\frac {y''^{2}-(y''^{2}+3y''^{2}y'^{2}-y'y'''-y'^{3}y''')}{y''^{2}}}={\frac {-3y''^{2}y'^{2}+y'y'''+y'^{3}y'''}{y''^{2}}}=}
=
−
3
y
″
2
y
′
2
−
y
′
y
‴
−
y
′
3
y
‴
y
″
2
=
−
y
′
3
y
″
2
y
′
−
y
‴
−
y
′
2
y
‴
y
″
2
,
(
1
)
{\displaystyle =-{\frac {3y''^{2}y'^{2}-y'y'''-y'^{3}y'''}{y''^{2}}}=-y'{\frac {3y''^{2}y'-y'''-y'^{2}y'''}{y''^{2}}},\quad (1)}
d
β
d
x
=
(
y
+
x
′
(
x
′
2
+
y
′
2
)
y
″
x
′
−
y
′
x
″
)
′
=
y
′
+
(
x
′
3
+
x
′
y
′
2
y
″
x
′
−
y
′
x
″
)
′
=
y
′
+
(
1
3
+
1
⋅
y
′
2
y
″
⋅
1
−
y
′
⋅
0
)
′
=
y
′
+
(
1
+
y
′
2
y
″
)
′
=
{\displaystyle {\frac {d\beta }{dx}}=\left(y+{\frac {x'(x'^{2}+y'^{2})}{y''x'-y'x''}}\right)'=y'+\left({\frac {x'^{3}+x'y'^{2}}{y''x'-y'x''}}\right)'=y'+\left({\frac {1^{3}+1\cdot y'^{2}}{y''\cdot 1-y'\cdot 0}}\right)'=y'+\left({\frac {1+y'^{2}}{y''}}\right)'=}
=
y
′
+
(
1
+
y
′
2
)
′
y
″
−
(
1
+
y
′
2
)
y
‴
y
″
2
=
y
′
+
(
0
+
2
y
′
y
″
)
y
″
−
y
‴
−
y
′
2
y
‴
y
″
2
=
{\displaystyle =y'+{\frac {(1+y'^{2})'y''-(1+y'^{2})y'''}{y''^{2}}}=y'+{\frac {(0+2y'y'')y''-y'''-y'^{2}y'''}{y''^{2}}}=}
=
y
′
+
2
y
′
y
″
2
−
y
‴
−
y
′
2
y
‴
y
″
2
=
y
′
y
″
2
+
2
y
′
y
″
2
−
y
‴
−
y
′
2
y
‴
y
″
2
=
{\displaystyle =y'+{\frac {2y'y''^{2}-y'''-y'^{2}y'''}{y''^{2}}}={\frac {y'y''^{2}+2y'y''^{2}-y'''-y'^{2}y'''}{y''^{2}}}=}
=
3
y
″
2
y
′
−
y
‴
−
y
′
2
y
‴
y
″
2
,
(
2
)
{\displaystyle ={\frac {3y''^{2}y'-y'''-y'^{2}y'''}{y''^{2}}},\quad (2)}
gauname santykį
d
β
d
α
=
−
1
y
′
.
{\displaystyle {\frac {d\beta }{d\alpha }}=-{\frac {1}{y'}}.}
Bet
y
′
{\displaystyle y'}
Teorema 2. Jeigu kai kurioje srityje
M
1
M
2
{\displaystyle M_{1}M_{2}}
kreivės spindulys kreivio kinta monotoniškai (t. y. arba tik didėja, arba tik mažėja), tai evoliutės lanko ilgio prieaugis šitoje srityje kreivės lygus (absoliučiu dydžiu) atitinkančiam prieaugiui kreivio spindulio duotosios kreivės. Įrodymas . Pagrindu formulės [
d
s
=
d
x
2
+
d
y
2
{\displaystyle ds={\sqrt {dx^{2}+dy^{2}}}}
d
s
=
d
α
2
+
d
β
2
,
{\displaystyle ds=d\alpha ^{2}+d\beta ^{2},}
kur ds - diferencialas evoliutės lanko ilgio; iš čia
(
d
s
d
x
)
2
=
(
d
α
d
x
)
2
+
(
d
β
d
x
)
2
.
{\displaystyle \left({\frac {ds}{dx}}\right)^{2}=\left({\frac {d\alpha }{dx}}\right)^{2}+\left({\frac {d\beta }{dx}}\right)^{2}.}
Įstatant čia išraiškas (1) ir (2), gausime:
(
d
s
d
x
)
2
=
(
d
α
d
x
)
2
+
(
d
β
d
x
)
2
=
(
−
y
′
3
y
″
2
y
′
−
y
‴
−
y
′
2
y
‴
y
″
2
)
2
+
(
3
y
″
2
y
′
−
y
‴
−
y
′
2
y
‴
y
″
2
)
2
=
{\displaystyle \left({\frac {ds}{dx}}\right)^{2}=\left({\frac {d\alpha }{dx}}\right)^{2}+\left({\frac {d\beta }{dx}}\right)^{2}=\left(-y'{\frac {3y''^{2}y'-y'''-y'^{2}y'''}{y''^{2}}}\right)^{2}+\left({\frac {3y''^{2}y'-y'''-y'^{2}y'''}{y''^{2}}}\right)^{2}=}
=
y
′
2
(
3
y
″
2
y
′
−
y
‴
−
y
′
2
y
‴
y
″
2
)
2
+
(
3
y
″
2
y
′
−
y
‴
−
y
′
2
y
‴
y
″
2
)
2
=
{\displaystyle =y'^{2}\left({\frac {3y''^{2}y'-y'''-y'^{2}y'''}{y''^{2}}}\right)^{2}+\left({\frac {3y''^{2}y'-y'''-y'^{2}y'''}{y''^{2}}}\right)^{2}=}
=
(
y
′
2
+
1
)
(
3
y
″
2
y
′
−
y
‴
−
y
′
2
y
‴
y
″
2
)
2
.
(
3
)
{\displaystyle =(y'^{2}+1)\left({\frac {3y''^{2}y'-y'''-y'^{2}y'''}{y''^{2}}}\right)^{2}.\quad (3)}
Rasime, toliau,
(
d
R
d
x
)
2
.
{\displaystyle \left({\frac {dR}{dx}}\right)^{2}.}
R
=
(
1
+
y
′
2
)
3
2
y
″
,
{\displaystyle R={\frac {(1+y'^{2})^{3 \over 2}}{y''}},}
R
2
=
(
1
+
y
′
2
)
3
y
″
2
.
{\displaystyle R^{2}={\frac {(1+y'^{2})^{3}}{y''^{2}}}.}
Diferencijuodami per x abi dalis šitos lygybės, gausime po to atitinkančius virsmus
d
(
R
2
)
d
x
=
2
R
d
R
d
x
=
(
(
1
+
y
′
2
)
3
y
″
2
)
′
=
(
(
1
+
y
′
2
)
3
)
′
y
″
2
−
(
1
+
y
′
2
)
3
(
y
″
2
)
′
(
y
″
2
)
2
=
{\displaystyle {\frac {d(R^{2})}{dx}}=2R{\frac {dR}{dx}}=({\frac {(1+y'^{2})^{3}}{y''^{2}}})'={\frac {((1+y'^{2})^{3})'y''^{2}-(1+y'^{2})^{3}(y''^{2})'}{(y''^{2})^{2}}}=}
=
3
(
1
+
y
′
2
)
2
(
1
−
y
′
2
)
′
y
″
2
−
(
1
+
y
′
2
)
3
(
2
y
″
y
‴
)
y
″
4
=
3
(
1
+
y
′
2
)
2
(
−
2
y
′
y
″
)
y
″
2
−
(
1
3
+
3
⋅
1
2
⋅
y
′
2
+
3
⋅
1
⋅
(
y
′
2
)
2
+
(
y
′
2
)
3
)
(
2
y
″
y
‴
)
y
″
4
=
{\displaystyle ={\frac {3(1+y'^{2})^{2}(1-y'^{2})'y''^{2}-(1+y'^{2})^{3}(2y''y''')}{y''^{4}}}={\frac {3(1+y'^{2})^{2}(-2y'y'')y''^{2}-(1^{3}+3\cdot 1^{2}\cdot y'^{2}+3\cdot 1\cdot (y'^{2})^{2}+(y'^{2})^{3})(2y''y''')}{y''^{4}}}=}
=
3
(
1
+
2
y
′
2
+
y
′
4
)
(
−
2
y
′
y
″
)
y
″
2
−
2
y
″
y
‴
(
1
+
3
y
′
2
+
3
y
′
4
+
y
′
6
)
y
″
4
=
−
2
y
′
y
″
y
″
2
(
3
+
6
y
′
2
+
3
y
′
4
)
−
2
y
″
y
‴
(
1
+
3
y
′
2
+
3
y
′
4
+
y
′
6
)
y
″
4
=
{\displaystyle ={\frac {3(1+2y'^{2}+y'^{4})(-2y'y'')y''^{2}-2y''y'''(1+3y'^{2}+3y'^{4}+y'^{6})}{y''^{4}}}={\frac {-2y'y''y''^{2}(3+6y'^{2}+3y'^{4})-2y''y'''(1+3y'^{2}+3y'^{4}+y'^{6})}{y''^{4}}}=}
=
−
2
y
″
⋅
y
′
y
″
2
(
3
+
6
y
′
2
+
3
y
′
4
)
+
y
‴
(
1
+
3
y
′
2
+
3
y
′
4
+
y
′
6
)
y
″
4
=
−
2
⋅
(
3
y
′
y
″
2
+
6
y
′
3
y
″
2
+
3
y
′
5
y
″
2
)
+
(
y
‴
+
3
y
′
2
y
‴
+
3
y
′
4
y
‴
+
y
′
6
y
‴
)
y
″
3
;
{\displaystyle =-2y''\cdot {\frac {y'y''^{2}(3+6y'^{2}+3y'^{4})+y'''(1+3y'^{2}+3y'^{4}+y'^{6})}{y''^{4}}}=-2\cdot {\frac {(3y'y''^{2}+6y'^{3}y''^{2}+3y'^{5}y''^{2})+(y'''+3y'^{2}y'''+3y'^{4}y'''+y'^{6}y''')}{y''^{3}}};}
arba
d
(
R
2
)
d
x
=
2
R
d
R
d
x
=
2
(
1
+
y
′
2
)
3
2
y
″
(
(
1
+
y
′
2
)
3
2
y
″
)
′
=
{\displaystyle {\frac {d(R^{2})}{dx}}=2R{\frac {dR}{dx}}=2{\frac {(1+y'^{2})^{3 \over 2}}{y''}}\left({\frac {(1+y'^{2})^{3 \over 2}}{y''}}\right)'=}
=
2
(
1
+
y
′
2
)
3
2
y
″
(
(
1
+
y
′
2
)
3
2
)
′
y
″
−
(
1
+
y
′
2
)
3
2
y
‴
y
″
2
=
2
(
1
+
y
′
2
)
3
2
y
″
3
2
(
1
+
y
′
2
)
1
2
(
1
+
y
′
2
)
′
y
″
−
(
1
+
y
′
2
)
3
2
y
‴
y
″
2
=
{\displaystyle =2{\frac {(1+y'^{2})^{3 \over 2}}{y''}}{\frac {((1+y'^{2})^{3 \over 2})'y''-(1+y'^{2})^{3 \over 2}y'''}{y''^{2}}}=2{\frac {(1+y'^{2})^{3 \over 2}}{y''}}{\frac {{3 \over 2}(1+y'^{2})^{1 \over 2}(1+y'^{2})'y''-(1+y'^{2})^{3 \over 2}y'''}{y''^{2}}}=}
=
2
(
1
+
y
′
2
)
3
2
y
″
3
2
(
1
+
y
′
2
)
1
2
(
2
y
′
y
″
)
y
″
−
(
1
+
y
′
2
)
3
2
y
‴
y
″
2
=
2
⋅
(
1
+
y
′
2
)
3
2
y
″
⋅
3
(
1
+
y
′
2
)
1
2
y
′
y
″
2
−
(
1
+
y
′
2
)
3
2
y
‴
y
″
2
=
{\displaystyle =2{\frac {(1+y'^{2})^{3 \over 2}}{y''}}{\frac {{3 \over 2}(1+y'^{2})^{1 \over 2}(2y'y'')y''-(1+y'^{2})^{3 \over 2}y'''}{y''^{2}}}=2\cdot {\frac {(1+y'^{2})^{3 \over 2}}{y''}}\cdot {\frac {3(1+y'^{2})^{1 \over 2}y'y''^{2}-(1+y'^{2})^{3 \over 2}y'''}{y''^{2}}}=}
=
2
⋅
3
(
1
+
y
′
2
)
2
y
′
y
″
2
−
(
1
+
y
′
2
)
3
y
‴
y
″
3
=
2
⋅
(
1
+
y
′
2
)
2
⋅
3
y
′
y
″
2
−
(
1
+
y
′
2
)
y
‴
y
″
3
=
{\displaystyle =2\cdot {\frac {3(1+y'^{2})^{2}y'y''^{2}-(1+y'^{2})^{3}y'''}{y''^{3}}}=2\cdot (1+y'^{2})^{2}\cdot {\frac {3y'y''^{2}-(1+y'^{2})y'''}{y''^{3}}}=}
=
2
⋅
(
1
+
y
′
2
)
2
⋅
3
y
′
y
″
2
−
y
‴
−
y
′
2
y
‴
y
″
3
.
{\displaystyle =2\cdot (1+y'^{2})^{2}\cdot {\frac {3y'y''^{2}-y'''-y'^{2}y'''}{y''^{3}}}.}
Dalindami abi lygybės dalis iš
2
R
=
2
(
1
+
y
′
2
)
3
2
y
″
,
{\displaystyle 2R={\frac {2(1+y'^{2})^{3 \over 2}}{y''}},}
d
R
d
x
=
(
1
+
y
′
2
)
1
2
3
y
′
y
″
2
−
y
‴
−
y
′
2
y
‴
y
″
2
.
{\displaystyle {\frac {dR}{dx}}=(1+y'^{2})^{1 \over 2}{\frac {3y'y''^{2}-y'''-y'^{2}y'''}{y''^{2}}}.}
Pakėlę kvadratu, gausime:
(
d
R
d
x
)
2
=
(
1
+
y
′
2
)
(
3
y
′
y
″
2
−
y
‴
−
y
′
2
y
‴
y
″
2
)
2
.
(
4
)
{\displaystyle \left({\frac {dR}{dx}}\right)^{2}=(1+y'^{2})\left({\frac {3y'y''^{2}-y'''-y'^{2}y'''}{y''^{2}}}\right)^{2}.\quad (4)}
Lygyndami lygybes (3) ir (4), randame:
(
d
R
d
x
)
2
=
(
d
s
d
x
)
2
,
{\displaystyle \left({\frac {dR}{dx}}\right)^{2}=\left({\frac {ds}{dx}}\right)^{2},}
iš kur
d
R
d
x
=
∓
d
s
d
x
.
{\displaystyle {\frac {dR}{dx}}=\mp {\frac {ds}{dx}}.}
Vaizdas:Kreivispav151deltas.jpg 151 pav. Pagal sąlyga
d
R
d
x
{\displaystyle {\frac {dR}{dx}}}
R tiktai didėja arba tiktai mažėja), todėl, ir
d
s
d
x
{\displaystyle {\frac {ds}{dx}}}
d
R
d
x
≤
0
,
d
s
d
x
≥
0
{\displaystyle {\frac {dR}{dx}}\leq 0,\;\;{\frac {ds}{dx}}\geq 0}
d
R
d
x
=
−
d
s
d
x
.
{\displaystyle {\frac {dR}{dx}}=-{\frac {ds}{dx}}.}
Tegu taškas
M
1
{\displaystyle M_{1}}
x
1
{\displaystyle x_{1}}
M
2
{\displaystyle M_{2}}
x
2
{\displaystyle x_{2}}
s (x ) ir R (x ) atkarpoje
[
x
1
;
x
2
]
{\displaystyle [x_{1};x_{2}]}
s
(
x
2
)
−
s
(
x
1
)
R
(
x
2
)
−
R
(
x
1
)
=
(
d
s
d
x
)
x
=
ξ
(
d
R
d
x
)
x
=
ξ
=
−
1
,
{\displaystyle {\frac {s(x_{2})-s(x_{1})}{R(x_{2})-R(x_{1})}}={\frac {\left({\frac {ds}{dx}}\right)_{x=\xi }}{\left({\frac {dR}{dx}}\right)_{x=\xi }}}=-1,}
kur
ξ
{\displaystyle \xi }
x
1
{\displaystyle x_{1}}
x
2
{\displaystyle x_{2}}
x
1
<
ξ
<
x
2
{\displaystyle x_{1}<\xi <x_{2}}
Įvesime reikšmes (151 pav.):
s
(
x
2
)
=
s
2
,
s
(
x
1
)
=
s
1
,
R
(
x
2
)
=
R
2
,
R
(
x
1
)
=
R
1
.
{\displaystyle s(x_{2})=s_{2},\quad s(x_{1})=s_{1},\quad R(x_{2})=R_{2},\quad R(x_{1})=R_{1}.}
Tada
s
2
−
s
1
R
2
−
R
1
=
−
1
,
{\displaystyle {\frac {s_{2}-s_{1}}{R_{2}-R_{1}}}=-1,}
s
2
−
s
1
=
−
(
R
2
−
R
1
)
.
{\displaystyle s_{2}-s_{1}=-(R_{2}-R_{1}).}
|
s
2
−
s
1
|
=
|
R
2
−
R
1
|
.
{\displaystyle |s_{2}-s_{1}|=|R_{2}-R_{1}|.}
Visiškai taip pat įrodoma šita lygybė ir didėjant kreivio spinduliui.
Mes įrodėme teorėmas 1 ir 2 tam atvejui, kada kreivė užrašyta lygtimi pavidalu
y
=
f
(
x
)
.
{\displaystyle y=f(x).\;}
Jeigu kreivė užrašyta parametrinėmis lygtimis, tai šitos teoremos galioja, be to jų įrodymas vyksta visiškai analogiškai. Vaizdas:Kreivispav152.jpg 152 pav. Pastaba . Nurodysime sekantį paprastą mechaninį būdą sudarymo kreivės (evolventės) pagal jos evoliutę.Tegu lanksti liniuotė sulenkta pagal formą evoliutės
C
0
C
5
{\displaystyle C_{0}C_{5}}
C
0
{\displaystyle C_{0}}
M
5
M
0
{\displaystyle M_{5}M_{0}}
Pažymėsime, kad vienai evoliutei atitinka nesuskaičiuojama daugybė skirtingų evolvenčių (152 pav.).
Vaizdas:Kreivispav153.jpg 153 pav. Tegu turime apskritimą spindulio a (153 pav.). Paimsime tą iš evolvenčių šito apskritimo, kuri pereina per tašką
M
0
(
a
;
0
)
.
{\displaystyle M_{0}(a;\;0).}
Atsižvelgiant, kad
C
M
=
C
M
0
˘
=
a
t
,
{\displaystyle CM={\breve {CM_{0}}}=at,}
O
P
=
x
=
a
(
cos
t
+
t
sin
t
)
,
{\displaystyle OP=x=a(\cos t+t\sin t),}
P
M
=
y
=
a
(
sin
t
−
t
cos
t
)
.
{\displaystyle PM=y=a(\sin t-t\cos t).}
Pažymėsime, kad profilis danties dantuoto rato turi dažniausiai formą apskritimo evolventės.
Sprendimas .Duotasis apskritimas su spinduliu a yra evoliutė, ieškomos evolventės. Apskritimo parametrinės lygtys:
α
=
a
cos
t
,
{\displaystyle \alpha =a\cos t,}
β
=
a
sin
t
.
{\displaystyle \beta =a\sin t.}
Randame apskritimo liestinę taške
M
(
x
M
;
y
M
)
{\displaystyle M(x_{M};y_{M})}
k
=
β
α
=
a
sin
t
a
cos
t
=
tan
t
=
β
′
(
α
)
,
{\displaystyle k={\frac {\beta }{\alpha }}={\frac {a\sin t}{a\cos t}}=\tan t=\beta '(\alpha ),}
y
M
−
β
=
k
(
x
M
−
α
)
=
(
x
M
−
α
)
tan
t
.
{\displaystyle y_{M}-\beta =k(x_{M}-\alpha )=(x_{M}-\alpha )\tan t.\;}
β
=
y
M
−
(
x
M
−
α
)
tan
t
.
{\displaystyle \beta =y_{M}-(x_{M}-\alpha )\tan t.\;}
Toliau randame apskritimo normalės lygtį:
y
M
−
β
=
−
1
k
(
x
M
−
α
)
=
−
1
tan
t
(
x
M
−
α
)
,
{\displaystyle y_{M}-\beta =-{\frac {1}{k}}(x_{M}-\alpha )=-{\frac {1}{\tan t}}(x_{M}-\alpha ),}
β
=
−
y
M
+
1
tan
t
(
x
M
−
α
)
;
{\displaystyle \beta =-y_{M}+{\frac {1}{\tan t}}(x_{M}-\alpha );}
a
sin
t
=
−
y
M
+
cos
t
sin
t
(
x
M
−
a
cos
t
)
,
{\displaystyle a\sin t=-y_{M}+{\frac {\cos t}{\sin t}}(x_{M}-a\cos t),}
a
sin
t
=
−
y
M
+
cos
t
sin
t
x
M
−
a
cos
2
t
sin
t
,
{\displaystyle a\sin t=-y_{M}+{\frac {\cos t}{\sin t}}x_{M}-{\frac {a\cos ^{2}t}{\sin t}},}
y
M
=
−
a
sin
t
−
a
cos
2
t
sin
t
+
cos
t
sin
t
x
M
,
{\displaystyle y_{M}=-a\sin t-{\frac {a\cos ^{2}t}{\sin t}}+{\frac {\cos t}{\sin t}}x_{M},}
y
M
=
−
a
sin
2
t
+
a
cos
2
t
sin
t
+
cos
t
sin
t
x
M
,
{\displaystyle y_{M}=-{\frac {a\sin ^{2}t+a\cos ^{2}t}{\sin t}}+{\frac {\cos t}{\sin t}}x_{M},}
y
M
=
x
M
cos
t
−
a
sin
t
=
x
M
1
tan
t
−
a
sin
t
;
{\displaystyle y_{M}={\frac {x_{M}\cos t-a}{\sin t}}=x_{M}{\frac {1}{\tan t}}-{\frac {a}{\sin t}};}
y
M
sin
t
=
x
M
cos
t
−
a
,
{\displaystyle y_{M}\sin t=x_{M}\cos t-a,}
y
M
sin
t
+
a
=
x
M
cos
t
,
{\displaystyle y_{M}\sin t+a=x_{M}\cos t,}
x
M
=
y
M
sin
t
+
a
cos
t
.
{\displaystyle x_{M}={\frac {y_{M}\sin t+a}{\cos t}}.}
Tokiu budu gavome apskritimo evolventės lygtį, kuri yra analogiška parametrinėms lygtims
x
M
=
a
(
cos
t
+
t
sin
t
)
,
{\displaystyle x_{M}=a(\cos t+t\sin t),}
y
M
=
a
(
sin
t
−
t
cos
t
)
.
{\displaystyle y_{M}=a(\sin t-t\cos t).}
Pavyzdžiui, kai
t
=
1
{\displaystyle t=1}
a
=
1
{\displaystyle a=1}
x
M
=
a
(
cos
t
+
t
sin
t
)
=
1
⋅
(
cos
1
+
1
⋅
sin
1
)
=
cos
1
+
sin
1
=
0.540302305
+
0.841470984
=
1.381773291
,
{\displaystyle x_{M}=a(\cos t+t\sin t)=1\cdot (\cos 1+1\cdot \sin 1)=\cos 1+\sin 1=0.540302305+0.841470984=1.381773291,}
y
M
=
a
(
sin
t
−
t
cos
t
)
=
1
⋅
(
sin
1
−
1
⋅
cos
1
)
=
sin
1
−
cos
1
=
0.841470984
−
0.540302305
=
0.301168678
;
{\displaystyle y_{M}=a(\sin t-t\cos t)=1\cdot (\sin 1-1\cdot \cos 1)=\sin 1-\cos 1=0.841470984-0.540302305=0.301168678;}
y
M
=
x
M
cos
t
−
a
sin
t
=
1.381773291
⋅
cos
1
−
1
0.841470984
=
0.746575295
−
1
0.841470984
=
0.253424704
0.841470984
=
−
0.301168678
,
{\displaystyle y_{M}={\frac {x_{M}\cos t-a}{\sin t}}={\frac {1.381773291\cdot \cos 1-1}{0.841470984}}={\frac {0.746575295-1}{0.841470984}}={\frac {0.253424704}{0.841470984}}=-0.301168678,}
x
M
=
y
M
sin
t
+
a
cos
t
=
0.301168678
⋅
sin
1
+
1
cos
1
=
1.253424704
0.540302305
=
2.318038035.
{\displaystyle x_{M}={\frac {y_{M}\sin t+a}{\cos t}}={\frac {0.301168678\cdot \sin 1+1}{\cos 1}}={\frac {1.253424704}{0.540302305}}=2.318038035.}
Dėl tam tikrų priežasčių (galbūt x ir y laikymas kaip viena funkcija nuo kitos, o ne atvirkščiai):
x
M
=
y
M
sin
t
−
a
cos
t
=
0.301168678
⋅
sin
1
−
1
cos
1
=
0.253424704
−
1
0.540302305
=
0.253424704
−
1
0.540302305
=
−
0.746575296
0.540302305
=
−
1.381773292.
{\displaystyle x_{M}={\frac {y_{M}\sin t-a}{\cos t}}={\frac {0.301168678\cdot \sin 1-1}{\cos 1}}={\frac {0.253424704-1}{0.540302305}}={\frac {0.253424704-1}{0.540302305}}={\frac {-0.746575296}{0.540302305}}=-1.381773292.}
![{\displaystyle R={\frac {\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}{\left|{\frac {d^{2}y}{dx^{2}}}\right|}}.\quad (2)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cac7ed0c5f9985b41f4a690eb4c401019b83716)
![{\displaystyle {\frac {\beta }{a}}=\left[\left({\frac {a^{2}-b^{2}}{ab}}\right)^{2 \over 3}-\left({\frac {\alpha }{b}}\right)^{2 \over 3}\right]^{3 \over 2},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c10bcfd1b323b7dc822c24c75b00ba032f26676e)
![{\displaystyle \beta =a\left[\left({\frac {a^{2}-b^{2}}{ab}}\right)^{2 \over 3}-\left({\frac {\alpha }{b}}\right)^{2 \over 3}\right]^{3 \over 2}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd6422e043587864302953c27e6d96c227cf2ff0)
![{\displaystyle [x_{1};x_{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d111814e88fbf11c97a4fb2610e3706fc9a2d3b0)
