Matematika/Išvestinė polinėje koordinačių sistemoje
keisti
Pav. 90. Tegu turime lygtį tiesės polinėse koordinatėse:
ρ
=
f
(
θ
)
.
(
1
)
{\displaystyle \rho =f(\theta ).\quad (1)}
Parašysime formules pereimo iš polinių koordinačių į stačiakampes dekartines:
x
=
ρ
cos
θ
,
y
=
ρ
sin
θ
.
{\displaystyle x=\rho \cos \theta ,\quad y=\rho \sin \theta .}
Įstatę čia vietoje
ρ
{\displaystyle \rho }
θ
{\displaystyle \theta }
x
=
f
(
θ
)
cos
θ
,
y
=
f
(
θ
)
sin
θ
.
(
2
)
{\displaystyle x=f(\theta )\cos \theta ,\quad y=f(\theta )\sin \theta .\quad (2)}
Lygtis (2) yra parametrinė lygtis duotos kreivės, be kita ko parametras yra polinis kampas
θ
{\displaystyle \theta }
Jeigu per
ϕ
{\displaystyle \phi }
M
(
ρ
,
θ
)
{\displaystyle M(\rho ,\;\theta )}
tan
ϕ
=
d
y
d
x
=
d
y
d
θ
d
x
d
θ
,
{\displaystyle \tan \phi ={\frac {{\text{d}}y}{{\text{d}}x}}={\frac {\frac {{\text{d}}y}{{\text{d}}\theta }}{\frac {{\text{d}}x}{{\text{d}}\theta }}},}
tan
ϕ
=
d
(
ρ
sin
θ
)
d
θ
d
(
ρ
cos
θ
)
d
θ
=
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
.
(
3
)
{\displaystyle \tan \phi ={\frac {\frac {{\text{d}}(\rho \sin \theta )}{{\text{d}}\theta }}{\frac {{\text{d}}(\rho \cos \theta )}{{\text{d}}\theta }}}={\frac {{\frac {{\text{d}}\rho }{{\text{d}}\theta }}\sin \theta +\rho \cos \theta }{{\frac {{\text{d}}\rho }{{\text{d}}\theta }}\cos \theta -\rho \sin \theta }}.\quad (3)}
Pažymėsime per
μ
{\displaystyle \mu }
μ
=
180
∘
−
θ
−
(
180
∘
−
ϕ
)
=
ϕ
−
θ
,
{\displaystyle \mu =180^{\circ }-\theta -(180^{\circ }-\phi )=\phi -\theta ,}
tan
μ
=
tan
(
ϕ
−
θ
)
=
tan
ϕ
−
tan
θ
1
+
tan
ϕ
tan
θ
.
{\displaystyle \tan \mu =\tan(\phi -\theta )={\frac {\tan \phi -\tan \theta }{1+\tan \phi \tan \theta }}.}
Įstatę čia vietoje
tan
ϕ
{\displaystyle \tan \phi }
tan
μ
=
tan
(
ϕ
−
θ
)
=
tan
ϕ
−
tan
θ
1
+
tan
ϕ
tan
θ
=
ρ
′
sin
θ
+
ρ
cos
θ
ρ
′
cos
θ
−
ρ
sin
θ
−
sin
θ
cos
θ
1
+
ρ
′
sin
θ
+
ρ
cos
θ
ρ
′
cos
θ
−
ρ
sin
θ
⋅
sin
θ
cos
θ
=
(
ρ
′
sin
θ
+
ρ
cos
θ
)
cos
θ
−
(
ρ
′
cos
θ
−
ρ
sin
θ
)
sin
θ
(
ρ
′
cos
θ
−
ρ
sin
θ
)
cos
θ
(
ρ
′
cos
θ
−
ρ
sin
θ
)
cos
θ
+
(
ρ
′
sin
θ
+
ρ
cos
θ
)
sin
θ
(
ρ
′
cos
θ
−
ρ
sin
θ
)
cos
θ
=
{\displaystyle \tan \mu =\tan(\phi -\theta )={\frac {\tan \phi -\tan \theta }{1+\tan \phi \tan \theta }}={\frac {{\frac {\rho '\sin \theta +\rho \cos \theta }{\rho '\cos \theta -\rho \sin \theta }}-{\frac {\sin \theta }{\cos \theta }}}{1+{\frac {\rho '\sin \theta +\rho \cos \theta }{\rho '\cos \theta -\rho \sin \theta }}\cdot {\frac {\sin \theta }{\cos \theta }}}}={\frac {\frac {(\rho '\sin \theta +\rho \cos \theta )\cos \theta -(\rho '\cos \theta -\rho \sin \theta )\sin \theta }{(\rho '\cos \theta -\rho \sin \theta )\cos \theta }}{\frac {(\rho '\cos \theta -\rho \sin \theta )\cos \theta +(\rho '\sin \theta +\rho \cos \theta )\sin \theta }{(\rho '\cos \theta -\rho \sin \theta )\cos \theta }}}=}
=
(
ρ
′
sin
θ
+
ρ
cos
θ
)
cos
θ
−
(
ρ
′
cos
θ
−
ρ
sin
θ
)
sin
θ
(
ρ
′
cos
θ
−
ρ
sin
θ
)
cos
θ
+
(
ρ
′
sin
θ
+
ρ
cos
θ
)
sin
θ
=
ρ
′
sin
θ
cos
θ
+
ρ
cos
2
θ
−
ρ
′
cos
θ
sin
θ
+
ρ
sin
2
θ
ρ
′
cos
2
θ
−
ρ
sin
θ
cos
θ
+
ρ
′
sin
2
θ
+
ρ
cos
θ
sin
θ
=
{\displaystyle ={\frac {(\rho '\sin \theta +\rho \cos \theta )\cos \theta -(\rho '\cos \theta -\rho \sin \theta )\sin \theta }{(\rho '\cos \theta -\rho \sin \theta )\cos \theta +(\rho '\sin \theta +\rho \cos \theta )\sin \theta }}={\frac {\rho '\sin \theta \cos \theta +\rho \cos ^{2}\theta -\rho '\cos \theta \sin \theta +\rho \sin ^{2}\theta }{\rho '\cos ^{2}\theta -\rho \sin \theta \cos \theta +\rho '\sin ^{2}\theta +\rho \cos \theta \sin \theta }}=}
=
ρ
cos
2
θ
+
ρ
sin
2
θ
ρ
′
cos
2
θ
+
ρ
′
sin
2
θ
=
ρ
ρ
′
,
{\displaystyle ={\frac {\rho \cos ^{2}\theta +\rho \sin ^{2}\theta }{\rho '\cos ^{2}\theta +\rho '\sin ^{2}\theta }}={\frac {\rho }{\rho '}},}
arba
ρ
′
tan
μ
=
ρ
,
{\displaystyle \rho '\tan \mu =\rho ,}
ρ
θ
′
=
ρ
tan
μ
=
ρ
cot
μ
.
(
4
)
{\displaystyle \rho '_{\theta }={\frac {\rho }{\tan \mu }}=\rho \cot \mu .\quad (4)}
Tokiu budu, išvestinė spindulio-vektoriaus poliniu kampu lygi ilgiui spindulio-vektoriaus, padaugintam iš kotangento kampo tarp spindulio-vektoriaus ir liestinės kreivės duotame taške.
Parodyti, kad liestinė logoritminės spiralės
ρ
=
e
a
θ
{\displaystyle \rho =e^{a\theta }}
μ
{\displaystyle \mu }
Sprendimas . Iš spiralės lygties randame:
ρ
′
=
a
e
a
θ
.
{\displaystyle \rho '=ae^{a\theta }.}
cot
μ
=
ρ
′
ρ
=
a
e
a
θ
e
a
θ
=
a
,
{\displaystyle \cot \mu ={\frac {\rho '}{\rho }}={\frac {ae^{a\theta }}{e^{a\theta }}}=a,}
t. y.
μ
=
arccot
a
=
const
.
{\displaystyle \mu =\operatorname {arccot} a={\text{const}}.}
Rasti kampą
μ
{\displaystyle \mu }
ρ
{\displaystyle \rho }
M
(
6
;
2
3
)
{\displaystyle M(6;\;2{\sqrt {3}})}
x
2
+
y
2
=
8
x
,
<=>
(
x
−
4
)
2
+
y
2
=
16.
{\displaystyle x^{2}+y^{2}=8x,\;<=>\;(x-4)^{2}+y^{2}=16.}
(Įstačius šio M taško koordinates į apskritimo lygti gaunama tapatybė; vadinasi, taškas M priklauso apskritimui
x
2
+
y
2
=
8
x
.
{\displaystyle x^{2}+y^{2}=8x.}
Apskritimo
x
2
+
y
2
=
8
x
{\displaystyle x^{2}+y^{2}=8x}
R
=
4
{\displaystyle R=4}
Dydžiojo apskritimo lygtis
x
2
+
y
2
=
8
x
.
{\displaystyle x^{2}+y^{2}=8x.}
Sprendimas . Randame
x
2
+
y
2
=
ρ
2
;
{\displaystyle x^{2}+y^{2}=\rho ^{2};}
8
x
=
8
ρ
cos
θ
.
{\displaystyle 8x=8\rho \cos \theta .}
Dabar apskritimo lygtis perrašyta į polines koordinates atrodo taip:
x
2
+
y
2
=
8
x
;
{\displaystyle x^{2}+y^{2}=8x;}
ρ
2
=
8
ρ
cos
θ
,
{\displaystyle \rho ^{2}=8\rho \cos \theta ,}
ρ
=
8
cos
θ
.
{\displaystyle \rho =8\cos \theta .}
Randame
ρ
θ
{\displaystyle \rho _{\theta }}
ρ
′
=
(
8
cos
θ
)
′
=
−
8
sin
θ
.
{\displaystyle \rho '=(8\cos \theta )'=-8\sin \theta .}
Randame liestinės lygtį:
y
−
y
1
=
f
′
(
x
1
)
(
x
−
x
1
)
;
{\displaystyle y-y_{1}=f'(x_{1})(x-x_{1});}
y
2
=
8
x
−
x
2
;
{\displaystyle y^{2}=8x-x^{2};}
y
=
8
x
−
x
2
;
{\displaystyle y={\sqrt {8x-x^{2}}};}
y
′
=
(
8
x
−
x
2
)
′
=
1
2
⋅
8
−
2
x
8
x
−
x
2
=
4
−
x
8
x
−
x
2
;
{\displaystyle y'=({\sqrt {8x-x^{2}}})'={\frac {1}{2}}\cdot {\frac {8-2x}{\sqrt {8x-x^{2}}}}={\frac {4-x}{\sqrt {8x-x^{2}}}};}
f
′
(
x
1
)
=
f
′
(
6
)
=
y
′
|
x
=
6
=
4
−
6
8
⋅
6
−
6
2
=
−
2
48
−
36
=
−
2
12
=
−
2
2
3
=
−
1
3
;
{\displaystyle f'(x_{1})=f'(6)=y'|_{x=6}={\frac {4-6}{\sqrt {8\cdot 6-6^{2}}}}={\frac {-2}{\sqrt {48-36}}}=-{\frac {2}{\sqrt {12}}}=-{\frac {2}{2{\sqrt {3}}}}=-{\frac {1}{\sqrt {3}}};}
y
−
2
3
=
−
1
3
(
x
−
6
)
;
{\displaystyle y-2{\sqrt {3}}=-{\frac {1}{\sqrt {3}}}(x-6);}
y
=
−
1
3
x
+
6
3
+
2
3
=
−
1
3
x
+
6
+
2
⋅
3
⋅
3
3
=
−
1
3
x
+
6
+
2
⋅
3
3
=
−
1
3
x
+
12
3
.
{\displaystyle y=-{\frac {1}{\sqrt {3}}}x+{\frac {6}{\sqrt {3}}}+2{\sqrt {3}}=-{\frac {1}{\sqrt {3}}}x+{\frac {6+2\cdot {\sqrt {3}}\cdot {\sqrt {3}}}{\sqrt {3}}}=-{\frac {1}{\sqrt {3}}}x+{\frac {6+2\cdot 3}{\sqrt {3}}}=-{\frac {1}{\sqrt {3}}}x+{\frac {12}{\sqrt {3}}}.}
Randame kampą
ϕ
{\displaystyle \phi }
Ox ašies:
tan
ϕ
=
k
=
−
1
3
;
{\displaystyle \tan \phi =k=-{\frac {1}{\sqrt {3}}};}
ϕ
=
arctan
(
−
1
3
)
=
−
π
6
=
−
0.523598775
{\displaystyle \phi =\arctan(-{\frac {1}{\sqrt {3}}})=-{\frac {\pi }{6}}=-0.523598775}
ϕ
=
−
30
∘
.
{\displaystyle \phi =-30^{\circ }.}
ϕ
=
330
∘
{\displaystyle \phi =330^{\circ }}
ϕ
=
2
π
−
π
6
=
12
π
−
π
6
=
11
π
6
=
5.75958653158
{\displaystyle \phi =2\pi -{\frac {\pi }{6}}={\frac {12\pi -\pi }{6}}={\frac {11\pi }{6}}=5.75958653158}
Vektorius
M
O
→
=
{
6
−
0
;
2
3
−
0
}
=
{
6
;
2
3
}
=
{
l
;
m
}
{\displaystyle {\vec {MO}}=\{6-0;\;2{\sqrt {3}}-0\}=\{6;\;2{\sqrt {3}}\}=\{l;\;m\}}
ρ
{\displaystyle \rho }
M
(
6
;
2
3
)
.
{\displaystyle M(6;\;2{\sqrt {3}}).}
Kai yra žinomi du tiesės taškai
M
(
6
;
2
3
)
{\displaystyle M(6;\;2{\sqrt {3}})}
O (0; 0), tada tiesės lygtis yra
x
−
x
0
l
=
y
−
y
0
m
{\displaystyle {\frac {x-x_{0}}{l}}={\frac {y-y_{0}}{m}}}
x
−
x
0
x
1
−
x
0
=
y
−
y
0
y
1
−
y
0
{\displaystyle {\frac {x-x_{0}}{x_{1}-x_{0}}}={\frac {y-y_{0}}{y_{1}-y_{0}}}}
x
−
x
1
x
1
−
x
0
=
y
−
y
1
y
1
−
y
0
.
{\displaystyle {\frac {x-x_{1}}{x_{1}-x_{0}}}={\frac {y-y_{1}}{y_{1}-y_{0}}}.}
Taigi, randame spindulio-vektoriaus
ρ
{\displaystyle \rho }
k
ρ
{\displaystyle k_{\rho }}
M :
x
−
x
0
l
=
y
−
y
0
m
;
{\displaystyle {\frac {x-x_{0}}{l}}={\frac {y-y_{0}}{m}};}
x
−
0
6
=
y
−
0
2
3
;
{\displaystyle {\frac {x-0}{6}}={\frac {y-0}{2{\sqrt {3}}}};}
2
3
x
6
=
y
;
{\displaystyle {\frac {2{\sqrt {3}}x}{6}}=y;}
y
=
3
x
3
=
x
3
;
{\displaystyle y={\frac {{\sqrt {3}}x}{3}}={\frac {x}{\sqrt {3}}};}
k
ρ
=
1
3
=
tan
θ
.
{\displaystyle k_{\rho }={\frac {1}{\sqrt {3}}}=\tan \theta .}
Arba galima daug paprasčiau rasti tiesės jungiančios tašką O (0; 0) ir tašką
M
(
6
;
2
3
)
{\displaystyle M(6;\;2{\sqrt {3}})}
tan
θ
=
y
M
x
M
=
2
3
6
=
3
3
=
1
3
=
0.577350269.
{\displaystyle \tan \theta ={\frac {y_{M}}{x_{M}}}={\frac {2{\sqrt {3}}}{6}}={\frac {\sqrt {3}}{3}}={\frac {1}{\sqrt {3}}}=0.577350269.}
Randame kampą
θ
{\displaystyle \theta }
ρ
{\displaystyle \rho }
M ir ašies Ox :
θ
=
arctan
k
ρ
=
arctan
1
3
=
π
6
=
0
,
523598775
{\displaystyle \theta =\arctan k_{\rho }=\arctan {\frac {1}{\sqrt {3}}}={\frac {\pi }{6}}=0,523598775}
Toliau randame kampą
μ
{\displaystyle \mu }
M su spinduliu-vektoriu
ρ
{\displaystyle \rho }
μ
=
ϕ
−
θ
=
−
π
6
−
π
6
=
−
π
3
=
2
π
−
π
3
=
6
π
−
π
3
=
5
π
3
=
5.235987756
{\displaystyle \mu =\phi -\theta =-{\frac {\pi }{6}}-{\frac {\pi }{6}}=-{\frac {\pi }{3}}=2\pi -{\frac {\pi }{3}}={\frac {6\pi -\pi }{3}}={\frac {5\pi }{3}}=5.235987756}
tan
μ
=
tan
(
ϕ
−
θ
)
=
tan
(
−
π
6
−
π
6
)
=
tan
−
π
3
=
tan
5
π
3
=
−
1.7320508.
{\displaystyle \tan \mu =\tan(\phi -\theta )=\tan(-{\frac {\pi }{6}}-{\frac {\pi }{6}})=\tan {\frac {-\pi }{3}}=\tan {\frac {5\pi }{3}}=-1.7320508.}
Patikriname tapatumus:
ρ
=
8
cos
θ
=
8
cos
π
6
=
8
⋅
3
2
=
6
,
92820323
;
{\displaystyle \rho =8\cos \theta =8\cos {\frac {\pi }{6}}=8\cdot {\frac {\sqrt {3}}{2}}=6,92820323;}
ρ
′
=
−
8
sin
θ
=
−
8
sin
π
6
=
−
8
⋅
1
2
=
−
4
;
{\displaystyle \rho '=-8\sin \theta =-8\sin {\frac {\pi }{6}}=-8\cdot {\frac {1}{2}}=-4;}
ρ
θ
′
=
ρ
tan
μ
=
6.92820323
−
1.7320508
=
−
4.00000001732.
{\displaystyle \rho '_{\theta }={\frac {\rho }{\tan \mu }}={\frac {6.92820323}{-1.7320508}}=-4.00000001732.}
Nors sprendžiant iš to, kad liestinės krypties koeficientas
k
ϕ
=
tan
ϕ
=
−
1
3
,
{\displaystyle k_{\phi }=\tan \phi =-{\frac {1}{\sqrt {3}}},}
OM krypties koeficientas lygus
k
θ
=
tan
θ
=
1
3
,
{\displaystyle k_{\theta }=\tan \theta ={\frac {1}{\sqrt {3}}},}
μ
{\displaystyle \mu }
M ir tiesės OM yra 90 laipsnių (grafiškai taip irgi atrodo įtikinamiau nei 60 laipsnių kampas
μ
{\displaystyle \mu }
Galima kitaip rasti kampą
μ
{\displaystyle \mu }
tan
μ
=
ρ
ρ
θ
′
=
8
cos
θ
−
8
sin
θ
=
−
cos
θ
sin
θ
=
−
cot
θ
=
−
1
tan
π
6
=
−
1
1
3
=
−
3
;
{\displaystyle \tan \mu ={\frac {\rho }{\rho '_{\theta }}}={\frac {8\cos \theta }{-8\sin \theta }}=-{\frac {\cos \theta }{\sin \theta }}=-\cot \theta =-{\frac {1}{\tan {\frac {\pi }{6}}}}=-{\frac {1}{\frac {1}{\sqrt {3}}}}=-{\sqrt {3}};}
μ
=
arctan
(
−
3
)
=
−
1.047197551
;
{\displaystyle \mu =\arctan(-{\sqrt {3}})=-1.047197551;}
μ
=
−
1.047197551
⋅
180
π
=
−
60
∘
.
{\displaystyle \mu =-1.047197551\cdot {\frac {180}{\pi }}=-60^{\circ }.}
