Aptarimas:Matematika/Kreivis

teisingas ir neteisingas sprendimo budas

• Nustatyti kreivį cikloidės

$\phi (t)=x=a(t-\sin t),\quad \psi (t)=y=a(1-\cos t)$

jos laisvai pasirenkame taške (x; y).

Sprendimas.

$\phi '={\frac {dx}{dt}}=a(1-\cos t),\quad \phi ''={\frac {d^{2}x}{dt^{2}}}=a\sin t,\quad \psi '={\frac {dy}{dt}}=a\sin t,\psi ''={\frac {d^{2}y}{dt^{2}}}=a\cos t.$

Įstatydami gautas išraiškas į formulę (3), randame:

$({\text{neteisingas}})\;\;K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|{\frac {a\cos t}{a\sin t}}\right|}{\left[1+\left({\frac {a\sin t}{a(1-\cos t)}}\right)^{2}\right]^{3 \over 2}}}={\frac {\frac {\cos t}{\sin t}}{\left[1+\left({\frac {\sin t}{(1-\cos t)}}\right)^{2}\right]^{3 \over 2}}}={\frac {\frac {\cos t}{\sin t}}{\left[{\frac {1}{(1-\cos t)^{2}}}((1-\cos t)^{2}+\sin ^{2}t)\right]^{3 \over 2}}}=$

$={\frac {\frac {\cos t}{\sin t}}{\left[{\frac {1}{(1-\cos t)^{2}}}(1-2\cos t+\cos ^{2}t+\sin ^{2}t)\right]^{3 \over 2}}}={\frac {\frac {\cos t}{\sin t}}{\left[{\frac {1}{(1-\cos t)^{2}}}(2-2\cos t)\right]^{3 \over 2}}}={\frac {\frac {\cos t}{\sin t}}{\left[{\frac {2}{1-\cos t}}\right]^{3 \over 2}}}=$

$={\frac {\cos t}{\sin t}}\left({\frac {\sqrt {1-\cos t}}{\sqrt {2}}}\right)^{3}={\frac {\cos t}{\sin t}}\sin ^{3}{\frac {t}{2}}.$

$({\text{teisingas}})\;\;K={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{\left[(\phi ')^{2}+(\psi ')^{2}\right]^{3 \over 2}}}={\frac {|a(1-\cos t)a\cos t-a\sin t\cdot a\sin t|}{\left[a^{2}(1-\cos t)^{2}+a^{2}\sin ^{2}t\right]^{3 \over 2}}}={\frac {|a^{2}\cos t-a^{2}\cos ^{2}t-a^{2}\sin ^{2}t|}{\left[a^{2}-2a^{2}\cos t+a^{2}\cos ^{2}t+a^{2}\sin ^{2}t\right]^{3 \over 2}}}=$

$={\frac {|a^{2}\cos t-a^{2}|}{\left[2a^{2}-2a^{2}\cos t\right]^{3 \over 2}}}={\frac {a^{2}|\cos t-1|}{2^{3 \over 2}a^{3}\left[1-\cos t\right]^{3 \over 2}}}={\frac {1}{2^{3 \over 2}a\left[1-\cos t\right]^{1 \over 2}}}={\frac {1}{2^{3 \over 2}a\cdot {\sqrt {2}}\sin {\frac {t}{2}}}}={\frac {1}{8a|\sin {\frac {t}{2}}|}}.$

vadovelio formule teisinga, o isvesta neteisinga nors vadovelis ir siulo isvesti taip

Apskaičiavimas kreivio linijos, užrašytos lygtimi polinėse koordinatėse

Tegu kreivė užrašyta lygtimi pavidalo

$\rho =f(\theta ).\quad (1)$

Užrašysime formules perėjimo iš polinių koordinačių į dekartines:

$x=\rho \cos \theta ,\quad y=\rho \sin \theta .\quad (2)$

Jeigu į šitas formules įstatyti vietoje $\rho$ jo išraišką per $\theta ,$ t. y. $f(\theta ),\;$ tai gausime:

$x=f(\theta )\cos \theta ,\quad y=f(\theta )\sin \theta .\quad (3)$

Paskutines lygtis galima nagrinėti kaip parametrines lygtis kreivės (1), be kita ko parametras yra $\theta .$

Tada

${\frac {dx}{d\theta }}={\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta ,\quad {\frac {dy}{d\theta }}={\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta ,$

${\frac {d^{2}x}{d\theta ^{2}}}=({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta -{\frac {d\rho }{d\theta }}\sin \theta )-({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )={\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta ,$

${\frac {d^{2}y}{d\theta ^{2}}}=({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta +{\frac {d\rho }{d\theta }}\cos \theta )+({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )={\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta +2{\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta .$

Įstatant paskutinę išraišką į formulę (1) praeito skyriaus, gausime formulę apskaičiavimui kreivio kreivės polinėse koordinatėse:

$K={\frac {|\psi ''\phi '-\psi '\phi ''|}{[\phi '^{2}+\psi '^{2}]^{3 \over 2}}}={\frac {\left|{\frac {d^{2}y}{d\theta ^{2}}}{\frac {dx}{d\theta }}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}\right|}{\left[\left({\frac {dx}{d\theta }}\right)^{2}+\left({\frac {dy}{d\theta }}\right)^{2}\right]^{3 \over 2}}}=$

$={\frac {|({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta +2{\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )-({\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )|}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )^{2}]^{3 \over 2}}}=$

$={\frac {({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta {\frac {d\rho }{d\theta }}\cos \theta +2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta -2{\frac {d\rho }{d\theta }}\cos \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta )}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )^{2}]^{3 \over 2}}}-$

$-{\frac {({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta {\frac {d\rho }{d\theta }}\sin \theta -2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta {\frac {d\rho }{d\theta }}\sin \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta +2{\frac {d\rho }{d\theta }}\sin \theta \rho \cos \theta +\rho \cos \theta \rho \cos \theta )}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )^{2}]^{3 \over 2}}}=$

$={\frac {{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta {\frac {d\rho }{d\theta }}\cos \theta +2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta -2{\frac {d\rho }{d\theta }}\cos \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )^{2}]^{3 \over 2}}}+$

$+{\frac {-{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta {\frac {d\rho }{d\theta }}\sin \theta +2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta {\frac {d\rho }{d\theta }}\sin \theta +{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta \rho \cos \theta -\rho \cos \theta \rho \cos \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )^{2}]^{3 \over 2}}}=$

$={\frac {\rho ^{2}\sin ^{2}\theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )^{2}]^{3 \over 2}}}+$

$+{\frac {-\rho ^{2}\cos ^{2}\theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )^{2}]^{3 \over 2}}}={\frac {|\rho ^{2}\sin ^{2}\theta -\rho ^{2}\cos ^{2}\theta |}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )^{2}]^{3 \over 2}}}=$

$={\frac {|\rho ^{2}\sin ^{2}\theta -\rho ^{2}\cos ^{2}\theta |}{[(({\frac {d\rho }{d\theta }})^{2}\cos ^{2}\theta -2{\frac {d\rho }{d\theta }}\cos \theta (-\rho \sin \theta )+\rho ^{2}\sin ^{2}\theta )+(({\frac {d\rho }{d\theta }})^{2}\sin ^{2}\theta +2{\frac {d\rho }{d\theta }}\sin \theta \rho \cos \theta +\rho ^{2}\cos ^{2}\theta )]^{3 \over 2}}}=$

$={\frac {|\rho ^{2}\sin ^{2}\theta -\rho ^{2}\cos ^{2}\theta |}{[4{\frac {d\rho }{d\theta }}\cos(\theta )\rho \sin(\theta )]^{3 \over 2}}}={\frac {|-\rho ^{2}(\cos ^{2}\theta -\sin ^{2}\theta )|}{\rho {\sqrt {\rho }}[4{\frac {d\rho }{d\theta }}\cos(\theta )\sin(\theta )]^{3 \over 2}}}={\frac {|-\rho \cos(2\theta )|}{{\sqrt {\rho }}[2{\frac {d\rho }{d\theta }}\sin(2\theta )]^{3 \over 2}}}={\frac {|-{\sqrt {\rho }}\cos(2\theta )|}{[2{\frac {d\rho }{d\theta }}\sin(2\theta )]^{3 \over 2}}}=$

$={\frac {|-{\sqrt {\rho }}\cos(2\theta )|}{2{\frac {d\rho }{d\theta }}\sin(2\theta ){\sqrt {2{\frac {d\rho }{d\theta }}\sin(2\theta )}}}}.$

Originali (vadovėlio) formulė yra:

$K={\frac {|\rho ^{2}+2\rho '^{2}-\rho \rho ''|}{(\rho ^{2}+\rho '^{2})^{3/2}}}.\quad (4)$

Pavyzdžiai

Vaizdas:Kreivispav144.jpg

144 pav.

• Nustatyti kreivį Archimedo spiralės $\rho =a\theta \;\;(a>0)$ laisvai pasirenkame taške (144 pav.).

Sprendimas.

${\frac {d\rho }{d\theta }}=a;\quad {\frac {d^{2}\rho }{d\theta ^{2}}}=0.$

Iš to seka,

$K={\frac {|\rho ^{2}+2\rho '^{2}-\rho \rho ''|}{(\rho ^{2}+\rho '^{2})^{3/2}}}={\frac {|a^{2}\theta ^{2}+2a^{2}-a\theta \cdot 0|}{(a^{2}\theta ^{2}+a^{2})^{3/2}}}={\frac {|a^{2}\theta ^{2}+2a^{2}|}{a^{3}(\theta ^{2}+1)^{3/2}}}={\frac {\theta ^{2}+2}{a(\theta ^{2}+1)^{3/2}}}.$

Pastebėsime, kad su didelėmis reikšmėmis $\theta$ turi vietą apytikslės lygybės: ${\frac {\theta ^{2}+2}{\theta ^{2}}}\approx 1,\;\;{\frac {\theta ^{2}+1}{\theta ^{2}}}\approx 1;$ todėl, pakeičiant praeitoje formulėje $\theta ^{2}+2$ į $\theta ^{2}$ ir $\theta ^{2}+1$ į $\theta ^{2},$ gauname apytikslę formulę (didelėms reikšmėms $\theta$):

$K\approx {\frac {1}{a}}{\frac {\theta ^{2}}{(\theta ^{2})^{3/2}}}={\frac {1}{a\theta }}.$

Tokiu budu, su didelėmis reikšmėmis $\theta$ Archimedo spiralė turi apytiksliai tą patį kreivį, kaip ir apskritimas spindulio $a\theta$.

Pavyzdžiui, jei $a=1,$ $\theta =1$, tuomet:

$K={\frac {|\rho ^{2}+2\rho '^{2}-\rho \rho ''|}{(\rho ^{2}+\rho '^{2})^{3/2}}}={\frac {\theta ^{2}+2}{a(\theta ^{2}+1)^{3/2}}}={\frac {1^{2}+2}{1\cdot (1^{2}+1)^{3/2}}}={\frac {3}{2^{3/2}}}={\frac {3}{\sqrt {8}}}={\frac {3}{2.828427125}}=1.060660172;$

$K={\frac {|-{\sqrt {\rho }}\cos(2\theta )|}{[2{\frac {d\rho }{d\theta }}\sin(2\theta )]^{3 \over 2}}}={\frac {|-{\sqrt {a\theta }}\cos(2\theta )|}{[2a\sin(2\theta )]^{3 \over 2}}}={\frac {|-{\sqrt {1\cdot 1}}\cos(2\cdot 1)|}{[2\cdot 1\sin(2\cdot 1)]^{3 \over 2}}}=$

$={\frac {|-\cos(2)|}{[2\sin(2)]^{3 \over 2}}}={\frac {|-(-0.416146836)|}{[2\cdot 0.909297426]^{3 \over 2}}}={\frac {0.416146836}{[1.818594854]^{3 \over 2}}}={\frac {0.416146836}{2.452471316}}=0.16968469.$

Bandymas 2

Apskaičiavimas kreivio linijos, užrašytos lygtimi polinėse koordinatėse

Tegu kreivė užrašyta lygtimi pavidalo

$\rho =f(\theta ).\quad (1)$

Užrašysime formules perėjimo iš polinių koordinačių į dekartines:

$x=\rho \cos \theta ,\quad y=\rho \sin \theta .\quad (2)$

Jeigu į šitas formules įstatyti vietoje $\rho$ jo išraišką per $\theta ,$ t. y. $f(\theta ),\;$ tai gausime:

$x=f(\theta )\cos \theta ,\quad y=f(\theta )\sin \theta .\quad (3)$

Paskutines lygtis galima nagrinėti kaip parametrines lygtis kreivės (1), be kita ko parametras yra $\theta .$

Tada

${\frac {dy}{dx}}={\frac {\frac {dy}{d\theta }}{\frac {dx}{d\theta }}},$

${\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {\frac {dy}{d\theta }}{\frac {dx}{d\theta }}}\right)={\frac {d}{d\theta }}\left({\frac {\frac {dy}{d\theta }}{\frac {dx}{d\theta }}}\right){\frac {d\theta }{dx}}={\frac {{\frac {dx}{d\theta }}{\frac {d}{d\theta }}({\frac {dy}{d\theta }})-{\frac {dy}{d\theta }}{\frac {d}{d\theta }}({\frac {dx}{d\theta }})}{({\frac {dx}{d\theta }})^{2}}}{\frac {d\theta }{dx}}={\frac {{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}{({\frac {dx}{d\theta }})^{2}}}{\frac {d\theta }{dx}}={\frac {{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}{({\frac {dx}{d\theta }})^{2}}}{\frac {1}{\frac {dx}{d\theta }}}={\frac {{\frac {dx}{d\theta }}{\frac {d^{2}y}{d\theta ^{2}}}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}}{({\frac {dx}{d\theta }})^{3}}}.$

${\frac {dx}{d\theta }}={\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta ,\quad {\frac {dy}{d\theta }}={\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta ,$

${\frac {d^{2}x}{d\theta ^{2}}}=({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta -{\frac {d\rho }{d\theta }}\sin \theta )-({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )={\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta ,$

${\frac {d^{2}y}{d\theta ^{2}}}=({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta +{\frac {d\rho }{d\theta }}\cos \theta )+({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )={\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta +2{\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta .$

Įstatant paskutinę išraišką į formulę (1) praeito skyriaus, gausime formulę apskaičiavimui kreivio kreivės polinėse koordinatėse:

$K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|{\frac {\psi ''\phi '-\psi '\phi ''}{(\phi ')^{3}}}\right|}{\left[1+\left({\frac {\psi '}{\phi '}}\right)^{2}\right]^{3 \over 2}}}={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{(\phi ')^{3}\left[{\frac {1}{(\phi ')^{2}}}((\phi ')^{2}+(\psi ')^{2})\right]^{3 \over 2}}}={\frac {\left|\psi ''\phi '-\psi '\phi ''\right|}{(\phi ')^{3}\cdot ({\frac {1}{(\phi ')^{2}}})^{3/2}\left[(\phi ')^{2}+(\psi ')^{2}\right]^{3 \over 2}}}=$

$={\frac {|\psi ''\phi '-\psi '\phi ''|}{[\phi '^{2}+\psi '^{2}]^{3 \over 2}}}={\frac {\left|{\frac {d^{2}y}{d\theta ^{2}}}{\frac {dx}{d\theta }}-{\frac {dy}{d\theta }}{\frac {d^{2}x}{d\theta ^{2}}}\right|}{\left[\left({\frac {dx}{d\theta }}\right)^{2}+\left({\frac {dy}{d\theta }}\right)^{2}\right]^{3 \over 2}}}=$

$={\frac {|({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta +2{\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )-({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta )|}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=$

$={\frac {({\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta {\frac {d\rho }{d\theta }}\cos \theta +2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta -2{\frac {d\rho }{d\theta }}\cos \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta )}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}-$

$-{\frac {({\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta {\frac {d\rho }{d\theta }}\sin \theta -2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta -\rho \cos \theta {\frac {d\rho }{d\theta }}\sin \theta +{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta -2{\frac {d\rho }{d\theta }}\sin \theta \rho \cos \theta -\rho \cos \theta \rho \cos \theta )}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=$

$={\frac {{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta {\frac {d\rho }{d\theta }}\cos \theta +2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta -2{\frac {d\rho }{d\theta }}\cos \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}+$

$+{\frac {-{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta {\frac {d\rho }{d\theta }}\sin \theta +2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta {\frac {d\rho }{d\theta }}\sin \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta +2{\frac {d\rho }{d\theta }}\sin \theta \rho \cos \theta +\rho \cos \theta \rho \cos \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=$

$={\frac {2{\frac {d\rho }{d\theta }}\cos \theta {\frac {d\rho }{d\theta }}\cos \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\sin \theta \rho \sin \theta +\rho \sin \theta \rho \sin \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}+$

$+{\frac {2{\frac {d\rho }{d\theta }}\sin \theta {\frac {d\rho }{d\theta }}\sin \theta -{\frac {d^{2}\rho }{d\theta ^{2}}}\cos \theta \rho \cos \theta +\rho \cos \theta \rho \cos \theta }{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=$

$={\frac {|2({\frac {d\rho }{d\theta }})^{2}\cos ^{2}\theta +2({\frac {d\rho }{d\theta }})^{2}\sin ^{2}\theta -\rho {\frac {d^{2}\rho }{d\theta ^{2}}}\sin ^{2}\theta -\rho {\frac {d^{2}\rho }{d\theta ^{2}}}\cos ^{2}\theta +\rho ^{2}\sin ^{2}\theta +\rho ^{2}\cos ^{2}\theta |}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=$

$={\frac {|2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}+\rho ^{2}|}{[({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )^{2}+({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta )^{2}]^{3 \over 2}}}=$

$={\frac {|\rho ^{2}+2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}|}{[(({\frac {d\rho }{d\theta }})^{2}\cos ^{2}\theta -2{\frac {d\rho }{d\theta }}\cos(\theta )\rho \sin(\theta )+\rho ^{2}\sin ^{2}\theta )+(({\frac {d\rho }{d\theta }})^{2}\sin ^{2}\theta +2{\frac {d\rho }{d\theta }}\sin(\theta )\rho \cos(\theta )+\rho ^{2}\cos ^{2}\theta )]^{3 \over 2}}}=$

$={\frac {|\rho ^{2}+2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}|}{[({\frac {d\rho }{d\theta }})^{2}\cos ^{2}\theta +({\frac {d\rho }{d\theta }})^{2}\sin ^{2}\theta +\rho ^{2}\sin ^{2}\theta +\rho ^{2}\cos ^{2}\theta )]^{3 \over 2}}}={\frac {|\rho ^{2}+2({\frac {d\rho }{d\theta }})^{2}-\rho {\frac {d^{2}\rho }{d\theta ^{2}}}|}{[({\frac {d\rho }{d\theta }})^{2}+\rho ^{2}]^{3 \over 2}}}={\frac {|\rho ^{2}+2\rho '^{2}-\rho \rho ''|}{(\rho ^{2}+\rho '^{2})^{3/2}}}.$

Originali (vadovėlio) formulė yra:

$K={\frac {|\rho ^{2}+2\rho '^{2}-\rho \rho ''|}{(\rho ^{2}+\rho '^{2})^{3/2}}}.\quad (4)$

Tai hiperbolės lanko ilgis

${\frac {dy}{dx}}=({\sqrt {2x}})'={\frac {1}{2}}\cdot {\frac {(2x)'}{\sqrt {2x}}}={\frac {1}{\sqrt {2x}}};$

$L=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx=\int _{1}^{7}{\sqrt {1+\left({\frac {1}{\sqrt {2x}}}\right)^{2}}}dx=\int _{1}^{7}{\sqrt {1+{\frac {1}{2x}}}}dx=$

$=\left[x{\sqrt {{\frac {1}{2x}}+1}}+{\frac {1}{4}}\ln \left(2{\sqrt {2}}x{\sqrt {{\frac {1}{x}}+2}}+4x+2\right)\right]|_{1}^{7}=$

$=\left[x{\sqrt {{\frac {1}{2x}}+1}}+{\frac {1}{4}}\ln \left(4x{\sqrt {{\frac {1}{2x}}+1}}+4x+2\right)\right]|_{1}^{7}=$

$=\left[7{\sqrt {{\frac {1}{2\cdot 7}}+1}}+{\frac {1}{4}}\ln \left(4\cdot 7\cdot {\sqrt {{\frac {1}{2\cdot 7}}+1}}+4\cdot 7+2\right)\right]-\left[{\sqrt {{\frac {1}{2}}+1}}+{\frac {1}{4}}\ln \left(4{\sqrt {{\frac {1}{2}}+1}}+4+2\right)\right]=$

$=\left[7{\sqrt {{\frac {1}{14}}+1}}+{\frac {1}{4}}\ln \left(28\cdot {\sqrt {{\frac {1}{14}}+1}}+28+2\right)\right]-\left[{\sqrt {1.5}}+{\frac {1}{4}}\ln(4{\sqrt {1.5}}+6)\right]=$

$=\left[7{\sqrt {\frac {15}{14}}}+{\frac {1}{4}}\ln \left(28\cdot {\sqrt {\frac {15}{14}}}+30\right)\right]-\left[1.224744871+{\frac {1}{4}}\ln(4.898979486+6)\right]=$

$=7{\sqrt {1.071428571}}+{\frac {1}{4}}\ln \left(28\cdot {\sqrt {1.071428571}}+30\right)-\left[1.224744871+{\frac {2.38866916}{4}}\right]=$

$=7\cdot 1.035098339+{\frac {1}{4}}\ln(28\cdot 1.035098339+30)-[1.224744871+0.597167289]=$

$=7.245688373+{\frac {\ln(28.98275349+30)}{4}}-1.821912161=$

$=7.245688373+{\frac {4.077245087}{4}}-1.821912161=$

$=7.245688373+1.019311272-1.821912161=6.443087484.$

Pasinaudojome internetiniu integratoriumi http://integrals.wolfram.com/index.jsp?expr=Sqrt%5B1%2B+1%2F%282x%29%5D+&random=false.

Neteisingas sprendimas ieškant $C_{1}$

• Nustatyti kreivį parabolės $y=x^{2}$ taškuose $x_{0}=0$ ir $x_{1}=5.$ Rasti prabolės evoliutės lanko ilgį iš taško $C_{0}(x_{2};\;y_{2})$ iki taško $C_{1}(x_{3};\;y_{3})$ naudojantis kreivės lanko ilgio skaičiavimo formule

$L=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx.$

Taškas $C_{0}$ yra spindulio $R_{0}$ centras, o taškas $C_{1}$ yra spindulio $R_{2}$ centras. Spindulys $R_{0}$ yra atkarpa iš taško $M_{0}(x_{0};y_{0})$ iki taško $C_{0}(x_{2};y_{2})$. Spindulys $R_{1}$ yra atkarpa iš taško $M_{1}(x_{1};y_{1})$ iki taško $C_{1}(x_{3};y_{3})$.

Sprendimas. $y'=(x^{2})'=2x,\quad y''=(2x)'=2.$

$K={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}.$

Kreivis taške $M_{0}$ yra lygus:

$K_{M_{0}}={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}={\frac {2}{(1+4\cdot 0^{2})^{3 \over 2}}}=2.$

$R_{0}={\frac {1}{K_{M_{0}}}}={\frac {1}{2}}=0.5.$

Kreivis taške $M_{1}$ yra lygus:

$K_{M_{1}}={\frac {\left|{\frac {d^{2}y}{dx^{2}}}\right|}{\left[1+\left({\frac {dy}{dx}}\right)^{2}\right]^{3 \over 2}}}={\frac {|2|}{[1+(2x)^{2}]^{3 \over 2}}}={\frac {2}{(1+4x^{2})^{3 \over 2}}}=$

$={\frac {2}{(1+4\cdot 5^{2})^{3 \over 2}}}={\frac {2}{101^{3 \over 2}}}={\frac {2}{\sqrt {1030301}}}={\frac {2}{1015.037438}}=0.00197037.$

$R_{1}={\frac {1}{K_{M_{1}}}}={\frac {1}{\frac {2}{\sqrt {1030301}}}}={\frac {\sqrt {1030301}}{2}}=507.5187189.$

Dabar užrašysime parabolės normalės lygtį taške $M_{1}(5;25)$:

$y-y_{M_{1}}=-{\frac {1}{y'}}(x-x_{M_{1}}),$

$y-25=-{\frac {1}{2x}}(x-5),$

$y=-{\frac {1}{2}}+{\frac {5}{2x}}+25=2.5x+24.5.$

Toliau rasime spindulio $R_{1}={\frac {\sqrt {1030301}}{2}}$ centro $C_{1}(x_{3};\;y_{3})$ koordinates. Žinome, kad

$(x_{3}-x_{1})^{2}+(y_{3}-y_{1})^{2}=R_{1}^{2},$

${\sqrt {(x_{3}-5)^{2}+(y_{3}-25)^{2}}}=R_{1},$

$(x_{3}-5)^{2}+(y_{3}-25)^{2}=\left({\frac {\sqrt {1030301}}{2}}\right)^{2},$

$(x_{3}-5)^{2}+(y_{3}-25)^{2}={\frac {1030301}{4}}=257575.25.$

Išsprendę lygčių sistemą rasime $x_{3}$ taško $C_{1}$ koordinatę:

${\begin{cases}y-25=-{\frac {1}{2x}}(x-5),&\\(x-5)^{2}+(y-25)^{2}={\frac {1030301}{4}};&\end{cases}}$

keitimo budu gauname:

$(x-5)^{2}+\left(-{\frac {1}{2x}}(x-5)\right)^{2}={\frac {1030301}{4}},$

$(x-5)^{2}+{\frac {1}{4x^{2}}}(x-5)^{2}={\frac {1030301}{4}},$

$x^{2}-10x+25+{\frac {x^{2}-10x+25}{4x^{2}}}={\frac {1030301}{4}},$

$x^{2}-10x+25+{\frac {1}{4}}-{\frac {10}{4x}}+{\frac {25}{4x^{2}}}={\frac {1030301}{4}},$

$x^{4}-10x^{3}+25x^{2}+{\frac {x^{2}}{4}}-{\frac {10x}{4}}+{\frac {25}{4}}={\frac {1030301x^{2}}{4}},$

$x^{4}-10x^{3}+25x^{2}+{\frac {x^{2}}{4}}-{\frac {1030301x^{2}}{4}}-{\frac {10x}{4}}+{\frac {25}{4}}=0,$

$x^{4}-10x^{3}+25.25x^{2}-257575.25x^{2}-2.5x+6.25=0,$

$x^{4}-10x^{3}-257550x^{2}-2.5x+6.25=0.$

Tai yra ketvirto laipsnio lygtis, kurios sprendinius rasime pasinaudodami internetu (http://www.1728.org/quartic.htm):

$x=512.5184773519854;\;0.00492131702699794;\;-502.51846764418235;\;-0.004931024830113984.$

Vadinsi spindulio $R_{1}$ centro $C_{1}(x_{3};y_{3})$ abscisės $x_{3}$ koordinatė turi buti viena iš keturių pateiktų. Kadangi vien spindulys $R_{1}=507.5187189$, tai centro abscisės koordinatė $x_{3}$ yra arba $x_{3}=512.5184773519854$ arba $x_{3}=-502.51846764418235$. Bet kadangi spindulys $R_{1}$ yra parabolės liestinės normalė ir spindulio $R_{1}$ galas (centras $C_{1}$) priklauso parabolės evoliutei, tai

$x_{3}=-502.51846764418235.$

Žinodami $x_{3}$, įstatę į parabolės evoliutės lygtį, randame (sekančiame pavyzdyje pateiktas parabolės evoliutės lygties radimas):

$y_{3}={\frac {3x_{3}^{2 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}={\frac {3\cdot (-502.5184676)^{2 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}={\frac {3\cdot 252524.8103^{1 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}=$

$={\frac {3\cdot 63.20741333}{16^{1 \over 3}}}+{\frac {1}{2}}={\frac {189.62224}{2.5198421}}+{\frac {1}{2}}=75.25163581+0.5=75.75163581.$

Arba per Pitagoro teoremą randame:

$y_{p}={\sqrt {R_{1}^{2}-(x_{3}-x_{1})^{2}}}={\sqrt {507.5187189^{2}-(-502.5184676-5)^{2}}}={\sqrt {257575.25-257574.995}}=$

$={\sqrt {0.255045}}=0.505019801;$

$y_{3}=y_{1}+y_{p}=25+0.505019801=25.5050198.$

Per Pitagoro teoremą gautas atsakymas panašus į tikrąjį, nes parabolės $y=x^{2}$ taške $M_{1}(5;25)$ spindulys $R_{1}$ yra statmenas parabolės liestinei taške $M_{1}(5;25)$ ir beveik lygiagretus Ox ašiai. O parabolės liestinė taške $M_{1}(5;25)$ yra beveik lygiagreti Oy ašiai. Taigi, radome spindulio $R_{1}$ centrą $C_{1}(-502.5184676;\;25.5050198),$ kuris nedaug skiriasi nuo teisingo $C_{1}(-500;\;75.5).$

Toliau rasime spindulio $R_{0}={\frac {1}{2}}$ centro $C_{0}(x_{2};\;y_{2})$ koordinates. Žinome, kad

$(x_{2}-x_{0})^{2}+(y_{2}-y_{0})^{2}=R_{0}^{2},$

${\sqrt {(x_{2}-0)^{2}+(y_{2}-0)^{2}}}=R_{0},$

$(x_{2}-0)^{2}+(y_{2}-0)^{2}=\left({\frac {1}{2}}\right)^{2},$

$(x_{3}-0)^{2}+(y_{3}-0)^{2}={\frac {1}{4}}.$

Išsprendę lygčių sistemą rasime $x_{2}$ (taško $C_{0}$ koordinate):

${\begin{cases}y-0=-{\frac {1}{2x}}(x-0),&\\(x-0)^{2}+(y-0)^{2}={\frac {1}{4}};&\end{cases}}$

keitimo budu gauname:

$(x-0)^{2}+\left(-{\frac {1}{2x}}(x-0)\right)^{2}={\frac {1}{4}},$

$x^{2}+{\frac {1}{4x^{2}}}(x-0)^{2}={\frac {1}{4}},$

$x^{2}+{\frac {x^{2}}{4x^{2}}}={\frac {1}{4}},$

$x^{2}+{\frac {1}{4}}={\frac {1}{4}},$

$x^{2}={\frac {1}{4}}-{\frac {1}{4}},$

$x=0.$

Taigi, $x_{2}=0.$ Žinodami $x_{2}$, įstatę į parabolės evoliutės lygtį, randame (sekančiame pavyzdyje pateiktas parabolės evoliutės lygties radimas):

$y_{2}={\frac {3x_{2}^{2 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}={\frac {3\cdot 0^{2 \over 3}}{16^{1 \over 3}}}+{\frac {1}{2}}={\frac {1}{2}}=0.5.$

Arba per Pitagoro teoremą randame:

$y_{d}={\sqrt {R_{0}^{2}-(x_{2}-x_{0})^{2}}}={\sqrt {0.5^{2}-(0-0)^{2}}}={\sqrt {0.25-0}}={\sqrt {0.25}}=0.5;$

$y_{3}=y_{0}+y_{d}=0+0.5=0.5.$

Radome spindulio $R_{0}$ centrą $C_{0}(0;\;{\frac {1}{2}}).$

Bet dėl kažkokių priežasčių sprendžiant ketvirtojo laipsnio lygtį nebuvo gautos visiškai teisingos $C_{1}$ koordinatės, o tik panašios.

blogas įrodymas

Kreivės užrašytos parametriškai

$x=\phi (t),\quad y=\psi (t),\quad z=\omega (t),$

kreivio apskaičiavimo formulė yra:

$K^{2}={\frac {1}{R^{2}}}={\frac {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}{[(\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2}]^{3}}};$

$K={\frac {1}{R}}={\frac {\sqrt {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}}{((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3 \over 2}}};$

$R={\frac {((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3 \over 2}}{\sqrt {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}}}.$

Galima naudotis ir šita formule:

$K^{2}={\frac {1}{R^{2}}}={\frac {((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})((\phi '')^{2}+(\psi '')^{2}+(\omega '')^{2})-(\phi '\phi ''+\psi '\psi ''+\omega '\omega '')^{2}}{((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3}}}.$

Įrodymas

Kreivis yra kampo $\alpha$ tarp erdvinės linijos liestinių skirtinguose taškuose M ir N santykis su tos kreivės lanko ilgiu ds tarp tų taškų. Kamui $\alpha$ artėjant prie nulio, artėja ir krevės lanko ilgis prie nulio.

Kreivė užrašyta parametriškai

$x=\phi (t),\quad y=\psi (t),\quad z=\omega (t),$

taške $M_{1}(x_{1};y_{1};z_{1})$ turi liestinės vektorių:

${\vec {a}}=\{\phi '(x_{1});\psi '(y_{1});\omega '(z_{1})\}.$

Taškas $M_{2}(x_{2};y_{2};z_{2})$ turi liestinės vektorių:

${\vec {b}}=\{\phi '(x_{2});\psi '(y_{2});\omega '(z_{2})\}.$

Kampas tarp liestinių taškuose $M_{1}$ ir $M_{2}$ yra lygus kamui tarp vektorių ${\vec {a}}$ ir ${\vec {b}}$:

$\alpha =\arcsin {\frac {\|\mathbf {a} \times \mathbf {b} \|}{\|\mathbf {a} \|\|\mathbf {b} \|}}=\arcsin {\frac {\sqrt {(\psi '(x_{1})\omega '(x_{2})-\omega '(x_{1})\psi '(x_{2}))^{2}+(\phi '(x_{1})\omega '(x_{2})-\omega '(x_{1})\phi '(x_{2}))^{2}+(\phi '(x_{1})\psi '(x_{2})-\psi '(x_{1})\phi '(x_{2}))^{2}}}{{\sqrt {(\phi '(x_{1}))^{2}+(\psi '(y_{1}))^{2}+(\omega '(z_{1}))^{2}}}{\sqrt {\phi '(x_{2}))^{2}+(\psi '(y_{2}))^{2}+(\omega '(z_{2}))^{2}}}}}.$

Erdvinės linijos kreivį iš taško $M_{1}$ iki taško $M_{2}$ galima apskaičiuoti taip:

$K={\frac {\alpha }{\int _{M_{1}}^{M_{2}}((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})\mathbf {d} t}}={\frac {\arcsin {\frac {\|\mathbf {a} \times \mathbf {b} \|}{\|\mathbf {a} \|\|\mathbf {b} \|}}}{\int _{M_{1}}^{M_{2}}((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})\mathbf {d} t}}=$

$={\frac {\arcsin {\frac {\sqrt {(\psi '(x_{1})\omega '(x_{2})-\omega '(x_{1})\psi '(x_{2}))^{2}+(\phi '(x_{1})\omega '(x_{2})-\omega '(x_{1})\phi '(x_{2}))^{2}+(\phi '(x_{1})\psi '(x_{2})-\psi '(x_{1})\phi '(x_{2}))^{2}}}{{\sqrt {(\phi '(x_{1}))^{2}+(\psi '(y_{1}))^{2}+(\omega '(z_{1}))^{2}}}{\sqrt {\phi '(x_{2}))^{2}+(\psi '(y_{2}))^{2}+(\omega '(z_{2}))^{2}}}}}}{\int _{M_{1}}^{M_{2}}((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})\mathbf {d} t}}.$

Jei vektoriai liestinių normalizuoti, tuomet kampą tarp jų galima užrašyti taip:

$\alpha =\arcsin \|\mathbf {a} ^{\circ }\times \mathbf {b} ^{\circ }\|,$

čia $\mathbf {a} ^{\circ }={\frac {\vec {a}}{\sqrt {(\phi '(x_{1}))^{2}+(\psi '(y_{1}))^{2}+(\omega '(z_{1}))^{2}}}};\;\;\mathbf {b} ^{\circ }={\frac {\vec {b}}{\sqrt {(\phi '(x_{2}))^{2}+(\psi '(y_{2}))^{2}+(\omega '(z_{2}))^{2}}}}.$

Kai kampas $\alpha$ artėja į nulį, $\alpha \to 0,$ tada $M_{2}\to M_{1}$ ir kreivės lanko ilgis s tarp taškų $M_{1}$ ir $M_{2}$ artėja į nulį, $s\to 0$, tada turime:

$\lim _{M_{2}\to M_{1}}\int _{M_{1}}^{M_{2}}((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})\mathbf {d} t={\sqrt {(\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2}}}=\mathbf {d} s.$

Kai kampas $\alpha$ artėja į nulį iš formulės $\lim _{x\to 0}={\frac {\sin x}{x}}=1$ turime $\alpha =\sin \alpha$. Iš vektorių žinome, kad

$\sin \alpha ={\frac {\|\mathbf {a} \times \mathbf {b} \|}{\|\mathbf {a} \|\|\mathbf {b} \|}},$

todėl kreivis lygus:

$K={\frac {\lim _{\alpha \to 0}\alpha }{\lim _{M_{2}\to M_{1}}\int _{M_{1}}^{M_{2}}((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})\mathbf {d} t}}={\frac {\sin \alpha }{\mathbf {d} s}}={\frac {\frac {\|\mathbf {a} \times \mathbf {b} \|}{\|\mathbf {a} \|\|\mathbf {b} \|}}{\sqrt {(\phi '(x_{1}))^{2}+(\psi '(y_{1}))^{2}+(\omega '(z_{1}))^{2}}}}=$

$={\frac {\frac {\sqrt {(\psi '(x_{1})\omega '(x_{2})-\omega '(x_{1})\psi '(x_{2}))^{2}+(\phi '(x_{1})\omega '(x_{2})-\omega '(x_{1})\phi '(x_{2}))^{2}+(\phi '(x_{1})\psi '(x_{2})-\psi '(x_{1})\phi '(x_{2}))^{2}}}{{\sqrt {(\phi '(x_{1}))^{2}+(\psi '(y_{1}))^{2}+(\omega '(z_{1}))^{2}}}{\sqrt {(\phi '(x_{1}))^{2}+(\psi '(y_{1}))^{2}+(\omega '(z_{1}))^{2}}}}}{\sqrt {(\phi '(x_{1}))^{2}+(\psi '(y_{1}))^{2}+(\omega '(z_{1}))^{2}}}}=$

$={\frac {\sqrt {(\psi '(x_{1})\omega '(x_{2})-\omega '(x_{1})\psi '(x_{2}))^{2}+(\phi '(x_{1})\omega '(x_{2})-\omega '(x_{1})\phi '(x_{2}))^{2}+(\phi '(x_{1})\psi '(x_{2})-\psi '(x_{1})\phi '(x_{2}))^{2}}}{(\phi '(x_{1}))^{2}+(\psi '(y_{1}))^{2}+(\omega '(z_{1}))^{2})^{3 \over 2}}}.$

Skaityklyje negalime įstatyti vektoriaus ${\vec {a}}=\{\phi '(x_{1});\psi '(y_{1});\omega '(z_{1})\}$ koordinačių, nes kreivis tampa lygus nuliui.

Žinome, kad dvimatėje erdvėje ant plokštumos xOy kokios nors kreivės liestinės kampas su Ox ašimi yra $\lim _{\Delta x\to 0}{\frac {\Delta y}{\Delta x}}={\frac {dy}{dx}}=\tan \alpha .$ Jeigu vektoriaus $\{\phi '(x_{2});\psi '(y_{2});\omega '(z_{2})\}$ kiekvieną iš parametrinių funkcijų padalinsime iš $\Delta t$, kai $\Delta t$ yra parametro t atstumas iki kol bus pasiektas taškas $M_{1}(x_{1};y_{1};z_{1}),$ tai gausime vektorių $\lim _{\Delta t\to 0;\;t\to pozicija\;M_{1}}\{{\frac {\Delta \phi '(x_{2})}{\Delta t}};{\frac {\Delta \psi '(y_{2})}{\Delta t}};{\frac {\Delta \omega '(z_{2})}{\Delta t}}\}=\{\phi ''(x_{1});\psi ''(y_{1});\omega ''(z_{1})\}.$

Kai liestinės dvimatėje erdvėje kampas $\alpha$ yra surištas su $\tan \alpha$, tai ${\frac {dy}{dx}}$ savybės negali būti panaudotos vektoriui apibūdinti. Mūsų atveju, mes jau turime, kad $\sin \alpha =\alpha$ tarp vektorių, kai mažėja kampas $\alpha$. Todėl suradę prie ko artėja riba $\lim _{\Delta t\to 0}\{{\frac {\Delta \phi '(x_{2})}{\Delta t}};{\frac {\Delta \psi '(y_{2})}{\Delta t}};{\frac {\Delta \omega '(z_{2})}{\Delta t}}\},$ artėjant kamui $\alpha$ prie nulio, turime kampo išvestinę, kuri savo reikšme analogiška funkcijos kitimo greičio išvestinei tam tikrame taške. Kampo išvestinė reiškia, kiek linija nukrypsta nuo savo pradinės trajektorijos tam tikrame taške. Tiesės kampo išvestinė yra lygi nuliui, nes linija nenukrypsta nei kiek. Vektoriaus ${\vec {b}}$ ilgis, artėjant taškui $M_{2}$ prie taško $M_{1}$, tampa lygus vektoriaus ${\vec {a}}$ ilgiui, todėl $\lim _{M_{2}\to M_{1}}{\sqrt {(\phi '(x_{2}))^{2}+(\psi '(y_{2}))^{2}+(\omega '(z_{2}))^{2}}}={\sqrt {(\phi '(x_{1}))^{2}+(\psi '(y_{1}))^{2}+(\omega '(z_{1}))^{2}}}.$

Iš vektorių formulės $\cos \alpha ={\frac {\mathbf {a} \cdot \mathbf {b} }{\|\mathbf {a} \|\cdot \|\mathbf {b} \|}}$ turime:

$\sin ^{2}\alpha =1-\cos ^{2}\alpha ,$

arba Lagrandžo tapatumą:

$\|\mathbf {a} \times \mathbf {b} \|^{2}=\|\mathbf {a} \|^{2}\|\mathbf {b} \|^{2}-(\mathbf {a} \cdot \mathbf {b} )^{2},$

${\frac {\|\mathbf {a} \times \mathbf {b} \|}{\|\mathbf {a} \|\|\mathbf {b} \|}}={\sqrt {1-{\frac {(\mathbf {a} \cdot \mathbf {b} )^{2}}{\|\mathbf {a} \|^{2}\|\mathbf {b} \|^{2}}}}}.$

Siuloma normalė nėra statmena liestinei

Apskaičiavimas kreivio linijos, užrašytos parametriškai erdvėje

Kreivės užrašytos parametriškai

$x=\phi (t),\quad y=\psi (t),\quad z=\omega (t),$

kreivio apskaičiavimo formulė yra:

$K^{2}={\frac {1}{R^{2}}}={\frac {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}{[(\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2}]^{3}}};$

$K={\frac {1}{R}}={\frac {\sqrt {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}}{((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3 \over 2}}};$

$R={\frac {((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3 \over 2}}{\sqrt {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}}}.$

Galima naudotis ir šita formule:

$K^{2}={\frac {1}{R^{2}}}={\frac {((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})((\phi '')^{2}+(\psi '')^{2}+(\omega '')^{2})-(\phi '\phi ''+\psi '\psi ''+\omega '\omega '')^{2}}{((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3}}}.$

Kreivės liestinės vektorius taške $M(x_{M};y_{M};z_{M})$ yra ${\vec {a}}=\{\phi '(x_{M});\psi '(y_{M});\omega '(z_{M})\}.$ Šis liestinės vektorius yra lygiagretus kreivės liestinei taške $M(x_{M};y_{M};z_{M}).$

Erdvinės kreivės liestinės lygtis taške $M(x_{M};y_{M};z_{M})$ yra:

${\frac {x-x_{M}}{\phi '(x_{M})}}={\frac {y-y_{M}}{\psi '(y_{M})}}={\frac {z-z_{M}}{\omega '(z_{M})}};$

arba

${\begin{cases}x-x_{M}=t\phi '(x_{M}),&\\y-y_{M}=t\psi '(y_{M}),&\\z-z_{M}=t\omega '(z_{M}).&\end{cases}}$

Kreivės normalės vektorius (kreivio spindulio vektorius) taške $M(x_{M};y_{M};z_{M})$ yra ${\vec {b}}=\{{\frac {1}{\phi '(x_{M})}};{\frac {1}{\psi '(y_{M})}};{\frac {1}{\omega '(z_{M})}}\}$ tik toms parametrinėms funkcijoms kurių rodikliai p yra riboje $(1;\infty ),$ tai yra $t^{p};\;1<p,$ pavyzdžiui, $t^{2}$, $t^{3}$, $t^{3 \over 2}$, $t^{5}.$ Pavyzdžiui, funkcija užrašyta parametriškai $\phi (t)=t^{2},\;\psi (t)=t^{3},\;\omega (t)=t^{4},$ turi normalės vektorių ${\vec {b}}=\{{\frac {1}{2x_{M}}};{\frac {1}{3x_{M}^{2}}};{\frac {1}{4x_{M}^{3}}}\}.$

Kreivės normalės vektorius ${\vec {b}}=\{{\frac {1}{\pm \phi '(x_{M})}};{\frac {1}{\pm \psi '(y_{M})}};{\frac {1}{\pm \omega '(z_{M})}}\}$ taške $M(x_{M};y_{M};z_{M})$ yra visoms funkcijoms užrašytoms parametriškai. Minuso ženklas prie išvestinės vardiklyje pasirenkamas tuo atveju, jeigu rodiklis parametro yra $-1\leq p<1$. Pavyzdžiui, funkcija užrašyta parametriškai $\phi (t)=t^{1 \over 3},\;\psi (t)=t^{1 \over 2},\;\omega (t)=t^{3},$ turi taške $M(x_{M};y_{M};z_{M})$ tik vieną normalės vektorių statmeną kreivės liestinei taške $M(x_{M};y_{M};z_{M}),$ kuris yra:

${\vec {b}}=\{{\frac {1}{-\phi '(x_{M})}};{\frac {1}{-\psi '(y_{M})}};{\frac {1}{\omega '(z_{M})}}\};$

${\vec {b}}=\{{\frac {1}{-{\frac {1}{3{\sqrt {t^{2}}}}}}};{\frac {1}{-{\frac {1}{2{\sqrt {t}}}}}};{\frac {1}{3t^{2}}}\},$

${\vec {b}}=\{-3{\sqrt {t^{2}}};-2{\sqrt {t}};{\frac {1}{3t^{2}}}\},$

${\vec {b}}=\{-3{\sqrt {x_{M}^{2}}};-2{\sqrt {y_{M}}};{\frac {1}{3z_{M}^{2}}}\}.$

Jeigu funkcijos parametras yra $t^{p}$ ir $p=1$, tada pliuso ar minuso ženklą pasirinkti prie parametrinės funkcijos išvestinės priklauso nuo kitų funkcijų. Pavyzdžiui, parametrinėms funkcijoms $\phi (t)=t$, $\psi (t)=t^{2}$, $\omega (t)=t^{3}$ gausime kreivės užrašytos šitomis parametrinėmis funkcijomis normalės vektorių:

${\vec {b}}=\{{\frac {1}{-\phi '(x_{M})}};{\frac {1}{\psi '(y_{M})}};{\frac {1}{\omega '(z_{M})}}\};$

${\vec {b}}=\{{\frac {1}{-1}};{\frac {1}{2t}};{\frac {1}{3t^{2}}}\}.$

Kitas pavyzdis, kreivės užrašytos parametriškai $\phi (t)=t$, $\psi (t)=t^{1 \over 2}$, $\omega (t)=t^{3}$, normalės vektorius taške $M(x_{M};y_{M};z_{M})$ yra:

${\vec {b}}=\{{\frac {1}{\phi '(x_{M})}};{\frac {1}{-\psi '(y_{M})}};{\frac {1}{\omega '(z_{M})}}\};$

${\vec {b}}=\{{\frac {1}{1}};{\frac {1}{-{\frac {1}{2{\sqrt {t}}}}}};{\frac {1}{3t^{2}}}\};$

${\vec {b}}=\{1;-2{\sqrt {y_{M}}};{\frac {1}{3z_{M}^{2}}}\}.$

Todėl galime užrašyti kreivės normalės lygtį:

${\begin{cases}x-x_{M}={\frac {t}{\pm \phi '(x_{M})}},&\\y-y_{M}={\frac {t}{\pm \psi '(y_{M})}},&\\z-z_{M}={\frac {t}{\pm \omega '(z_{M})}};&\end{cases}}$

arba

${\frac {x-x_{M}}{\frac {1}{\pm \phi '(x_{M})}}}={\frac {y-y_{M}}{\frac {1}{\pm \psi '(y_{M})}}}={\frac {z-z_{M}}{\frac {1}{\pm \omega '(z_{M})}}},$

$\pm (x-x_{M})\phi '(x_{M})=\pm (y-y_{M})\psi '(y_{M})=\pm (z-z_{M})\omega '(z_{M}).$

Turime lygčių sistemą:

${\begin{cases}\pm (x-x_{M})\phi '(x_{M})=\pm (y-y_{M})\psi '(y_{M})=\pm (z-z_{M})\omega '(z_{M}),&\\(x-x_{M})^{2}+(y-y_{M})^{2}+(z-z_{M})^{2}={\frac {1}{K^{2}}};&\end{cases}}$

Gauname, kad

$y-y_{M}={\frac {\pm \phi '(x_{M})(x-x_{M})}{\pm \psi '(y_{M})}};$

$z-z_{M}={\frac {\pm \phi '(x_{M})(x-x_{M})}{\pm \omega '(z_{M})}}.$

Įstatę į antrą sistemos lygtį gauname:

$(x-x_{M})^{2}+\left({\frac {\pm \phi '(x_{M})(x-x_{M})}{\pm \psi '(y_{M})}}\right)^{2}+\left({\frac {\pm \phi '(x_{M})(x-x_{M})}{\pm \omega '(z_{M})}}\right)^{2}={\frac {1}{K^{2}}},$

$(x-x_{M})^{2}+{\frac {(\phi '(x_{M}))^{2}(x-x_{M})^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}(x-x_{M})^{2}}{(\omega '(z_{M}))^{2}}}={\frac {1}{K^{2}}},$

$(x-x_{M})^{2}\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)=R^{2}.$

Išsprendus kvadratinę lygtį

$(x^{2}-2xx_{M}+x_{M}^{2})\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)-R^{2}=0,$

$x^{2}\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)-2xx_{M}\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)+x_{M}^{2}\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)-R^{2}=0,$

surandama kreivio centro $C(x_{C};y_{C};z_{C})$ koordinatė $x_{C}.$ Analogiškai surandama ir $y_{C}$ kordinatė išsprendus lygtį:

$y^{2}\left(1+{\frac {(\psi '(y_{M}))^{2}}{(\phi '(x_{M}))^{2}}}+{\frac {(\psi '(y_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)-2yy_{M}\left(1+{\frac {(\psi '(y_{M}))^{2}}{(\phi '(x_{M}))^{2}}}+{\frac {(\psi '(y_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)+y_{M}^{2}\left(1+{\frac {(\psi '(y_{M}))^{2}}{(\phi '(x_{M}))^{2}}}+{\frac {(\psi '(y_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)-R^{2}=0.$

Tokiu pačiu principu surandama ir $z_{C}$ koordinatė, išsprendžiant lygtį:

$z^{2}\left(1+{\frac {(\omega '(z_{M}))^{2}}{(\phi '(x_{M}))^{2}}}+{\frac {(\omega '(z_{M}))^{2}}{(\psi '(y_{M}))^{2}}}\right)-2zz_{M}\left(1+{\frac {(\omega '(z_{M}))^{2}}{(\phi '(x_{M}))^{2}}}+{\frac {(\omega '(z_{M}))^{2}}{(\psi '(y_{M}))^{2}}}\right)+z_{M}^{2}\left(1+{\frac {(\omega '(z_{M}))^{2}}{(\phi '(x_{M}))^{2}}}+{\frac {(\omega '(z_{M}))^{2}}{(\psi '(y_{M}))^{2}}}\right)-R^{2}=0.$

Norint surasti parametrines erdvinės kreivės evoliutės lygtis grįžtame prie sistemos:

${\begin{cases}\pm (x-x_{M})\phi '(x_{M})=\pm (y-y_{M})\psi '(y_{M})=\pm (z-z_{M})\omega '(z_{M}),&\\(x-x_{M})^{2}+(y-y_{M})^{2}+(z-z_{M})^{2}={\frac {1}{K^{2}}}.&\end{cases}}$

Iš kurios turime:

$(x-x_{M})^{2}+\left({\frac {\pm \phi '(x_{M})(x-x_{M})}{\pm \psi '(y_{M})}}\right)^{2}+\left({\frac {\pm \phi '(x_{M})(x-x_{M})}{\pm \omega '(z_{M})}}\right)^{2}={\frac {1}{K^{2}}},$

$(x-x_{M})^{2}+{\frac {(\phi '(x_{M}))^{2}(x-x_{M})^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}(x-x_{M})^{2}}{(\omega '(z_{M}))^{2}}}={\frac {1}{K^{2}}},$

$(x-x_{M})^{2}\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)=R^{2};$

$(x-x_{M})^{2}={\frac {R^{2}}{\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)}},$

$x-x_{M}=\pm {\sqrt {\frac {R^{2}}{\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)}}},$

$x=x_{M}\pm {\frac {R}{\sqrt {\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)}}},$

$x_{C}=x_{M}\pm {\frac {1}{K{\sqrt {\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)}}}}=$

$=x_{M}\pm {\frac {((\phi '(x_{M}))^{2}+(\psi '(y_{M}))^{2}+(\omega '(z_{M}))^{2})^{3 \over 2}}{{\sqrt {(\psi '(y_{M})\omega ''(z_{M})-\omega '(z_{M})\psi ''(y_{M}))^{2}+(\phi '(x_{M})\omega ''(z_{M})-\omega '(z_{M})\phi ''(x_{M}))^{2}+(\phi '(x_{M})\psi ''(y_{M})-\psi '(y_{M})\phi ''(x_{M}))^{2}}}{\sqrt {\left(1+{\frac {(\phi '(x_{M}))^{2}}{(\psi '(y_{M}))^{2}}}+{\frac {(\phi '(x_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)}}}};$

$\alpha (t)=x\pm {\frac {1}{K{\sqrt {\left(1+{\frac {(\phi ')^{2}}{(\psi ')^{2}}}+{\frac {(\phi ')^{2}}{(\omega ')^{2}}}\right)}}}}=x\pm {\frac {\frac {((\phi ')^{2}+(\psi ')^{2}+(\omega ')^{2})^{3 \over 2}}{\sqrt {(\psi '\omega ''-\omega '\psi '')^{2}+(\phi '\omega ''-\omega '\phi '')^{2}+(\phi '\psi ''-\psi '\phi '')^{2}}}}{\sqrt {\left(1+{\frac {(\phi ')^{2}}{(\psi ')^{2}}}+{\frac {(\phi ')^{2}}{(\omega ')^{2}}}\right)}}}.$

Ar pliuso ar minuso ženklą pasirinkti, reikia vadovautis tuom, kad erdvinės kreivės normalės vektorius ${\vec {b}}=\{{\frac {1}{\pm \phi '(x_{M})}};{\frac {1}{\pm \psi '(y_{M})}};{\frac {1}{\pm \omega '(z_{M})}}\}$ būtų tos pačios krypties arba bent jau lygiagretus vektoriui ${\vec {c}}=\{x_{C}-x_{M};y_{C}-y_{M};z_{C}-z_{M}\}.$ Jei lygiagretumo sąlyga išpildyta, bet krypties sąlyga neišpildyta, tuomet turėtume gauti išverstos evoliutės lygtį, kuri bus aplink kreivę, o ne kreivės viduje. Kad gauti evoliutės lygtį kuo panašesnę į plokščios (dvimatės) evoliutės lygtį, ir lygiagretumo sąlyga, ir krypties sąlyga turi būti išpildyta (kai abu vektoriai yra tos pačios krypties tuomet $|R_{2}-R_{1}|=|s_{2}-s_{1}|$; čia $|s_{2}-s_{1}|$ yra lanko ilgis evoliutės iš vieno centro taško iki kito centro taško).

Analogiškai turime ir parametrinę y evoliutės išraišką:

$y_{C}=y_{M}\pm {\frac {1}{K{\sqrt {\left(1+{\frac {(\psi '(y_{M}))^{2}}{(\phi '(x_{M}))^{2}}}+{\frac {(\psi '(y_{M}))^{2}}{(\omega '(z_{M}))^{2}}}\right)}}}};$

$\beta (t)=y\pm {\frac {1}{K{\sqrt {\left(1+{\frac {(\psi ')^{2}}{(\phi ')^{2}}}+{\frac {(\psi ')^{2}}{(\omega ')^{2}}}\right)}}}}.$

Taip pat surandama ir parametrinė z evoliutės lygties koordinatė:

$z_{C}=z_{M}\pm {\frac {1}{K{\sqrt {\left(1+{\frac {(\omega '(z_{M}))^{2}}{(\phi '(x_{M}))^{2}}}+{\frac {(\omega '(z_{M}))^{2}}{(\psi '(y_{M}))^{2}}}\right)}}}};$

$\gamma (t)=z\pm {\frac {1}{K{\sqrt {\left(1+{\frac {(\omega ')^{2}}{(\phi ')^{2}}}+{\frac {(\omega ')^{2}}{(\psi ')^{2}}}\right)}}}}.$

Taškas $C(x_{C};y_{C};z_{C})$ yra kreivės kreivio centro taškas, kuris su tašku $M(x_{M};y_{M};z_{M})$ sudaro spindulį R.

Labai keistas sutapimas

padalinę vektoriaus narius ${\vec {MC}}$ iš vektoriaus ${\vec {b}}$ narių turėtume gauti tas pačias x, y, z reikšmes gautame vektoriuje, taigi

${\frac {-2823.65612-5}{-{\frac {2}{3}}}}=2818.65612\cdot {\frac {3}{2}}=4227.98418,$

${\frac {166.432806-25}{\frac {1}{30}}}=141.432806\cdot 30=4242.98418,$

${\frac {143.3175405-125}{\frac {1}{225}}}=18.3175405\cdot 225=4121.446613$

kažkodėl gavome tik apytiksiai vienodas reikšmes (su kitomis $x_{C}$, $y_{C}$, $z_{C}$ reikšmėmis jos dar mažiau bus panašios, ypač su kita $z_{C}$ reikšme) ir keistą sutapimą, kad liekanos vienodos (.98418).

nesiprastina $\alpha (t)$

c) Erdvinės evoliutės parametrinės lygtys yra (bent jau, kai t kinta nuo 0 iki $\infty$):

$\alpha (t)=\phi (t)-{\frac {1}{K{\sqrt {1+{\frac {(\phi '(t))^{2}}{4(\psi '(t))^{2}}}+{\frac {(\phi '(t))^{2}}{4(\omega '(t))^{2}}}}}}}=t-{\frac {1}{K{\sqrt {1+{\frac {1^{2}}{4(2t)^{2}}}+{\frac {1^{2}}{4(3t^{2})^{2}}}}}}}=$

$=t-{\frac {1}{K{\sqrt {1+{\frac {1}{4\cdot 4t^{2}}}+{\frac {1}{4\cdot 9t^{4}}}}}}}=t-{\frac {1}{{\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}{\sqrt {1+{\frac {1}{16t^{2}}}+{\frac {1}{36t^{4}}}}}}}=$

$=t-{\frac {1}{{\frac {\sqrt {36t^{4}+36t^{2}+4}}{(1+4t^{2}+9t^{4})^{3 \over 2}}}{\sqrt {\frac {36t^{4}+4t^{2}+1}{36t^{4}}}}}}=t-{\frac {1}{{\frac {\sqrt {36t^{4}+36t^{2}+4}}{\sqrt {(1+4t^{2}+9t^{4})^{3}}}}{\sqrt {{\frac {27t^{4}}{36t^{4}}}+{\frac {9t^{4}+4t^{2}+1}{36t^{4}}}}}}}=$

$=t-{\frac {1}{{\frac {\sqrt {36t^{4}+36t^{2}+4}}{\sqrt {(1+4t^{2}+9t^{4})^{2}}}}{\sqrt {{\frac {3}{4(1+4t^{2}+9t^{4})}}+{\frac {1}{36t^{4}}}}}}}=t-{\frac {1}{{\frac {\sqrt {36t^{4}+36t^{2}+4}}{1+4t^{2}+9t^{4}}}{\sqrt {\frac {3\cdot 36t^{4}+4(1+4t^{2}+9t^{4})}{4(1+4t^{2}+9t^{4})\cdot 36t^{4}}}}}}=$

$=t-{\frac {12t^{2}(1+4t^{2}+9t^{4}){\sqrt {1+4t^{2}+9t^{4}}}}{{\sqrt {36t^{4}+36t^{2}+4}}{\sqrt {3\cdot 36t^{4}+4(1+4t^{2}+9t^{4})}}}}=t-{\frac {12t^{2}(1+4t^{2}+9t^{4})^{3 \over 2}}{{\sqrt {36t^{4}+36t^{2}+4}}{\sqrt {4+16t^{2}+144t^{4}}}}}=$

$=t-{\frac {3t^{2}(1+4t^{2}+9t^{4})^{3 \over 2}}{{\sqrt {9t^{4}+9t^{2}+1}}{\sqrt {1+4t^{2}+36t^{4}}}}};$

$\beta (t)=\psi (t)+{\frac {1}{K{\sqrt {{\frac {4(\psi '(t))^{2}}{(\phi '(t))^{2}}}+1+{\frac {(\psi '(t))^{2}}{(\omega '(t))^{2}}}}}}};$

$\gamma (t)=\omega (t)+{\frac {1}{K{\sqrt {{\frac {4(\omega '(t))^{2}}{(\phi '(t))^{2}}}+{\frac {(\omega '(t))^{2}}{(\psi '(t))^{2}}}+1}}}}.$

Vienoje vietoje vietoj t turi būti x, kad derivatorius duotų teisingai (kad $\beta '(t)$ gauti teisingai)

e) Randame išvestines (pasinaudodami derivatoriumi iš www.derivator.org):

funkcijai $\gamma '(t)$ įvedame į derivatorių ((1+4*x^2+9*x^4)^(3/2))/((9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^(1/2) ir gauname rezultatą

(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;

$\gamma '(t)=(t^{3}+{\frac {(1+4t^{2}+9t^{4})^{3 \over 2}}{{\sqrt {9t^{4}+9t^{2}+1}}{\sqrt {144t^{4}+9t^{2}+4}}}})'=$

$=3t^{2}+{\frac {{3 \over 2}{\sqrt {1+4t^{2}+9t^{4}}}(8t+36t^{3}){\sqrt {(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}-(1+4t^{2}+9t^{4})^{3 \over 2}\cdot {1 \over 2}((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{-{\frac {1}{2}}}((36t^{3}+18t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(576t^{3}+18t))}{({\sqrt {(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}})^{2}}}=$

$=3t^{2}+{\frac {{3 \over 2}{\sqrt {1+4t^{2}+9t^{4}}}(8t+36t^{3})(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)-(1+4t^{2}+9t^{4})^{3 \over 2}\cdot {1 \over 2}((36t^{3}+18t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(576t^{3}+18t))}{((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{\frac {1}{2}}(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}=$

$=3t^{2}+{\frac {3{\sqrt {1+4t^{2}+9t^{4}}}(4t+18t^{3})(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)-(1+4t^{2}+9t^{4})^{3 \over 2}((18t^{3}+9t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(288t^{3}+9t))}{((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{\frac {3}{2}}}};$

toliau randame $\beta '(t)$ įvedame į derivatorių (3*t*(1+4*x^2+9*x^4)^(3/2))/(4*(9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^(1/2) ir gaudami

((0+3*(0+t*1.5*(1+4*x^2+9*x^4)^0.5*(0+0+4*2*x+0+9*4*x^3)))*(4*(9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^0.5-3*t*(1+4*x^2+9*x^4)^1.5*0.5*(4*(9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^-0.5*(0+4*((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))))/(4*(9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^0.5^2;

$\beta '(t)=(t^{2}+{\frac {3t(1+4t^{2}+9t^{4})^{3 \over 2}}{2{\sqrt {9t^{4}+9t^{2}+1}}{\sqrt {144t^{4}+9t^{2}+4}}}})'=$

$=2t+{\frac {3({\frac {3t}{2}}{\sqrt {1+4t^{2}+9t^{4}}}(4\cdot 2t+9\cdot 4\cdot t^{3})){\sqrt {4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}-3t(1+4t^{2}+9t^{4})^{3 \over 2}\cdot {1 \over 2}(4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{-{\frac {1}{2}}}\cdot 4((9\cdot 4t^{3}+9\cdot 2t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(144\cdot 4t^{3}+9\cdot 2t))}{((4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{1 \over 2})^{2}}}=$

$=2t+{\frac {{\frac {9t}{2}}{\sqrt {1+4t^{2}+9t^{4}}}(4\cdot 2t+9\cdot 4\cdot t^{3}){\sqrt {4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}-3t(1+4t^{2}+9t^{4})^{3 \over 2}\cdot {1 \over 2}(4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{-{\frac {1}{2}}}\cdot 4((9\cdot 4t^{3}+9\cdot 2t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(144\cdot 4t^{3}+9\cdot 2t))}{4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}=$

$=2t+{\frac {{\frac {9t}{2}}{\sqrt {1+4t^{2}+9t^{4}}}(8t+36t^{3}){\sqrt {4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}-3t(1+4t^{2}+9t^{4})^{3 \over 2}(4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{-{\frac {1}{2}}}\cdot 2((36t^{3}+18t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(576t^{3}+18t))}{4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}=$

$=2t+{\frac {{\frac {9t}{2}}{\sqrt {1+4t^{2}+9t^{4}}}(8t+36t^{3}){\sqrt {4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}-6t(1+4t^{2}+9t^{4})^{3 \over 2}(4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{-{\frac {1}{2}}}((36t^{3}+18t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(576t^{3}+18t))}{4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)}}=$

$=2t+{\frac {{\frac {9t}{2}}{\sqrt {1+4t^{2}+9t^{4}}}(8t+36t^{3})\cdot 4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)-6t(1+4t^{2}+9t^{4})^{3 \over 2}((36t^{3}+18t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(576t^{3}+18t))}{(4(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{3 \over 2}}}=$

$=2t+{\frac {36t{\sqrt {1+4t^{2}+9t^{4}}}(4t+18t^{3})(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)-12t(1+4t^{2}+9t^{4})^{3 \over 2}((18t^{3}+9t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(288t^{3}+9t))}{8((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{3 \over 2}}}=$

$=2t+{\frac {9t{\sqrt {1+4t^{2}+9t^{4}}}(4t+18t^{3})(9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4)-3t(1+4t^{2}+9t^{4})^{3 \over 2}((18t^{3}+9t)(144t^{4}+9t^{2}+4)+(9t^{4}+9t^{2}+1)(288t^{3}+9t))}{2((9t^{4}+9t^{2}+1)(144t^{4}+9t^{2}+4))^{3 \over 2}}};$

kai ieškodami $\beta '(t)$ įvedame į derivatorių (1.5*(x^(2/3)+4*x^(8/3)+9*x^(14/3))^(3/2))/((9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^(1/2), tai gauname

((0+1.5*1.5*(x^0.666667+4*x^2.66667+9*x^4.66667)^0.5*(0.666667*x^-0.333333+0+4*2.66667*x^1.66667+0+9*4.66667*x^3.66667))*((9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^0.5-1.5*(x^0.666667+4*x^2.66667+9*x^4.66667)^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;

tuo tarpu [beieškant $\beta '(t)$] įvedant (1.5*x*(1+4*x^2+9*x^4)^1.5)/((9*x^4+9*x^2+1)*(144*x^4+9*x^2+4))^0.5 gauname

((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;