Vaizdas:Kreivispav136.jpg 136 pav. Tegu lankas kreivės
M
0
M
{\displaystyle M_{0}M}
y
=
f
(
x
)
{\displaystyle y=f(x)\;}
(a, b) . Nustatysime kreivės lanko ilgį. Paimsime ant kreivės AB taškus
M
0
{\displaystyle M_{0}}
M
1
{\displaystyle M_{1}}
M
2
{\displaystyle M_{2}}
M
i
−
1
{\displaystyle M_{i-1}}
M
i
{\displaystyle M_{i}}
M
n
−
1
{\displaystyle M_{n-1}}
M
{\displaystyle M}
M
0
M
1
M
2
.
.
.
M
i
−
1
M
i
.
.
.
M
n
−
1
M
,
{\displaystyle M_{0}M_{1}M_{2}...M_{i-1}M_{i}...M_{n-1}M,}
M
0
M
{\displaystyle M_{0}M}
P
n
{\displaystyle P_{n}}
Lanko ilgiu
M
0
M
{\displaystyle M_{0}M}
s ), prie kurios artėja lanko ilgis, artėjant prie nulio lanko atkarpų ilgiams
M
i
−
1
M
i
{\displaystyle M_{i-1}M_{i}}
M
0
M
1
M
2
.
.
.
M
i
−
1
M
i
.
.
.
M
n
−
1
M
.
{\displaystyle M_{0}M_{1}M_{2}...M_{i-1}M_{i}...M_{n-1}M.}
Pažymėsime, kad šitas apibrėžimas lanko ilgio betkokios kreivės analoginis apibrėžimui apskritimo ilgio.
Yra įrodyta, kad jeigu atkarpoje [a , b ] funkcija
f
(
x
)
{\displaystyle f(x)\;}
f
′
(
x
)
{\displaystyle f'(x)\;}
y
=
f
(
x
)
,
{\displaystyle y=f(x),\;}
[
a
;
f
(
a
)
]
{\displaystyle [a;\;f(a)]}
[
b
;
f
(
b
)
]
,
{\displaystyle [b;\;f(b)],}
lim
M
0
M
→
0
i
l
g
.
M
0
M
˘
i
l
g
.
M
0
M
¯
=
1.
{\displaystyle \lim _{M_{0}M\to 0}{\frac {ilg.\;{\breve {M_{0}M}}}{ilg.\;{\overline {M_{0}M}}}}=1.}
Vaizdas:Kreivispav137.jpg 137 pav. Šita teorema lengvai gali būti įrodyta apskritimui (panagrinėkime lanką AB , kurio centrinis kampas lygus
2
α
{\displaystyle 2\alpha }
2
R
α
{\displaystyle 2R\alpha }
2
R
sin
α
{\displaystyle 2R\sin \alpha }
lim
α
→
0
A
B
˘
A
B
¯
=
lim
α
→
0
2
R
α
2
R
sin
α
=
1
{\displaystyle \lim _{\alpha \to 0}{\frac {\breve {AB}}{\overline {AB}}}=\lim _{\alpha \to 0}{\frac {2R\alpha }{2R\sin \alpha }}=1}
Panagrinėkime sekantį klausimą. Tegu mes turime ant plokšumos kreivę, kurios lygtis
y
=
f
(
x
)
.
{\displaystyle y=f(x).\;}
Vaizdas:Kreivispav138.jpg 138 pav. Tegu
M
0
(
x
0
;
y
0
)
{\displaystyle M_{0}(x_{0};y_{0})}
M
(
x
;
y
)
{\displaystyle M(x;y)}
s ilgį lanko
M
0
M
{\displaystyle M_{0}M}
Kintant abscisei x taško M ilgis s lanko kis, t. y. s yra funkcija x . Rasime išvestinę s nuo x .
Duosime x priaugimą
Δ
x
{\displaystyle \Delta x}
s gaus priaugimą
Δ
s
=
ilg.
M
M
1
˘
.
{\displaystyle \Delta s={\text{ilg.}}\;{\breve {MM_{1}}}.}
M
M
1
¯
{\displaystyle {\overline {MM_{1}}}}
lim
Δ
x
→
0
Δ
s
Δ
x
,
{\displaystyle \lim _{\Delta x\to 0}{\frac {\Delta s}{\Delta x}},}
Δ
M
M
1
Q
{\displaystyle \Delta MM_{1}Q}
(
M
M
1
¯
)
2
=
(
Δ
x
)
2
+
(
Δ
y
)
2
.
{\displaystyle ({\overline {MM_{1}}})^{2}=(\Delta x)^{2}+(\Delta y)^{2}.}
Padauginsime ir padalinsime kairę dalį iš
Δ
s
2
{\displaystyle \Delta s^{2}}
(
M
M
1
¯
Δ
s
)
2
Δ
s
2
=
(
Δ
x
)
2
+
(
Δ
y
)
2
.
{\displaystyle \left({\frac {\overline {MM_{1}}}{\Delta s}}\right)^{2}\Delta s^{2}=(\Delta x)^{2}+(\Delta y)^{2}.}
Padalinsime visus narius lygybės iš
Δ
x
2
{\displaystyle \Delta x^{2}}
(
M
M
1
¯
Δ
s
)
2
(
Δ
s
Δ
x
)
2
=
1
+
(
Δ
y
Δ
x
)
2
.
{\displaystyle \left({\frac {\overline {MM_{1}}}{\Delta s}}\right)^{2}\left({\frac {\Delta s}{\Delta x}}\right)^{2}=1+\left({\frac {\Delta y}{\Delta x}}\right)^{2}.}
Rasime ribą kairės ir dešinės dalies kai
Δ
x
→
0.
{\displaystyle \Delta x\to 0.}
lim
M
M
1
¯
→
0
M
M
1
¯
Δ
s
=
1
{\displaystyle \lim _{{\overline {MM_{1}}}\to 0}{\frac {\overline {MM_{1}}}{\Delta s}}=1}
lim
Δ
x
→
0
Δ
y
Δ
x
=
d
y
d
x
,
{\displaystyle \lim _{\Delta x\to 0}{\frac {\Delta y}{\Delta x}}={\frac {dy}{dx}},}
(
d
s
d
x
)
2
=
1
+
(
d
y
d
x
)
2
{\displaystyle \left({\frac {ds}{dx}}\right)^{2}=1+\left({\frac {dy}{dx}}\right)^{2}}
arba
d
s
d
x
=
1
+
(
d
y
d
x
)
2
.
(
1
)
{\displaystyle {\frac {ds}{dx}}={\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}.\quad (1)}
Lanko diferencialui gausime sekančią išraišką:
d
s
=
1
+
(
d
y
d
x
)
2
d
x
(
2
)
{\displaystyle ds={\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}dx\quad (2)}
arba
d
s
=
d
x
2
+
d
y
2
.
(
2
′
)
{\displaystyle ds={\sqrt {dx^{2}+dy^{2}}}.\quad (2')}
Griežtai kalbant, formulė (2') teisinga tik tam atvejui, kada
d
x
>
0.
{\displaystyle dx>0.}
d
x
<
0
{\displaystyle dx<0}
d
s
=
−
d
x
2
+
d
y
2
.
{\displaystyle ds=-{\sqrt {dx^{2}+dy^{2}}}.}
|
d
s
|
=
d
x
2
+
d
y
2
.
{\displaystyle |ds|={\sqrt {dx^{2}+dy^{2}}}.}
Mes gavome diferencialo išraišką lanko ilgio tam atvejui, kada kreivė apibūdinta lygtimi
y
=
f
(
x
)
.
{\displaystyle y=f(x).\;}
Jeigu kreivė apibūdinta parametriškai:
x
=
ϕ
(
t
)
,
y
=
ψ
(
t
)
,
{\displaystyle x=\phi (t),\quad y=\psi (t),}
tai
d
x
=
ϕ
′
(
t
)
d
t
,
d
y
=
ψ
′
(
t
)
d
t
,
{\displaystyle {\text{d}}x=\phi '(t){\text{d}}t,\quad {\text{d}}y=\psi '(t){\text{d}}t,}
ir išraiška (2') įgauna pavidalą
d
s
=
[
ϕ
′
(
t
)
]
2
+
[
ψ
′
(
t
)
]
2
d
t
.
{\displaystyle ds={\sqrt {[\phi '(t)]^{2}+[\psi '(t)]^{2}}}dt.}
Rasti hiperbolės
y
=
2
x
{\displaystyle y={\sqrt {2x}}}
L , kai x kinta nuo 1 iki 7. Sprendimas .
d
y
d
x
=
(
2
x
)
′
=
1
2
⋅
(
2
x
)
′
2
x
=
1
2
x
;
{\displaystyle {\frac {dy}{dx}}=({\sqrt {2x}})'={\frac {1}{2}}\cdot {\frac {(2x)'}{\sqrt {2x}}}={\frac {1}{\sqrt {2x}}};}
L
=
∫
a
b
1
+
(
y
′
)
2
d
x
=
∫
1
7
1
+
(
1
2
x
)
2
d
x
=
∫
1
7
1
+
1
2
x
d
x
=
{\displaystyle 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=}
=
[
x
1
2
x
+
1
+
1
4
ln
(
2
2
x
1
x
+
2
+
4
x
+
2
)
]
|
1
7
=
{\displaystyle =\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}=}
=
[
x
1
2
x
+
1
+
1
4
ln
(
4
x
1
2
x
+
1
+
4
x
+
2
)
]
|
1
7
=
{\displaystyle =\left[x{\sqrt {{\frac {1}{2x}}+1}}+{\frac {1}{4}}\ln \left(4x{\sqrt {{\frac {1}{2x}}+1}}+4x+2\right)\right]|_{1}^{7}=}
=
[
7
1
2
⋅
7
+
1
+
1
4
ln
(
4
⋅
7
⋅
1
2
⋅
7
+
1
+
4
⋅
7
+
2
)
]
−
[
1
2
+
1
+
1
4
ln
(
4
1
2
+
1
+
4
+
2
)
]
=
{\displaystyle =\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]=}
=
[
7
1
14
+
1
+
1
4
ln
(
28
⋅
1
14
+
1
+
28
+
2
)
]
−
[
1.5
+
1
4
ln
(
4
1.5
+
6
)
]
=
{\displaystyle =\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]=}
=
[
7
15
14
+
1
4
ln
(
28
⋅
15
14
+
30
)
]
−
[
1.224744871
+
1
4
ln
(
4.898979486
+
6
)
]
=
{\displaystyle =\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
1.071428571
+
1
4
ln
(
28
⋅
1.071428571
+
30
)
−
[
1.224744871
+
2.38866916
4
]
=
{\displaystyle =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
⋅
1.035098339
+
1
4
ln
(
28
⋅
1.035098339
+
30
)
−
[
1.224744871
+
0.597167289
]
=
{\displaystyle =7\cdot 1.035098339+{\frac {1}{4}}\ln(28\cdot 1.035098339+30)-[1.224744871+0.597167289]=}
=
7.245688373
+
ln
(
28.98275349
+
30
)
4
−
1.821912161
=
{\displaystyle =7.245688373+{\frac {\ln(28.98275349+30)}{4}}-1.821912161=}
=
7.245688373
+
4.077245087
4
−
1.821912161
=
{\displaystyle =7.245688373+{\frac {4.077245087}{4}}-1.821912161=}
=
7.245688373
+
1.019311272
−
1.821912161
=
8.264999645
−
1.821912161
=
6.443087484.
{\displaystyle =7.245688373+1.019311272-1.821912161=8.264999645-1.821912161=6.443087484.}
Pasinaudojome internetiniu integratoriumi http://integrals.wolfram.com/index.jsp?expr=Sqrt%5B1%2B+1%2F%282x%29%5D+&random=false .
Patikriname ar tiesės ilgis iš taško
M
1
(
1
;
2
)
{\displaystyle M_{1}(1;{\sqrt {2}})}
M
2
(
7
;
14
)
{\displaystyle M_{2}(7;{\sqrt {14}})}
L
T
=
(
7
−
1
)
2
+
(
14
−
2
)
2
=
6
2
+
(
3.741657387
−
1.414213562
)
2
=
36
+
2.327443824
2
=
{\displaystyle L_{T}={\sqrt {(7-1)^{2}+({\sqrt {14}}-{\sqrt {2}})^{2}}}={\sqrt {6^{2}+(3.741657387-1.414213562)^{2}}}={\sqrt {36+2.327443824^{2}}}=}
=
36
+
5.416994756
=
41.41699476
=
6.435603682.
{\displaystyle ={\sqrt {36+5.416994756}}={\sqrt {41.41699476}}=6.435603682.}
Rasti lanko ilgį iš taško
M
0
(
0
;
5
2
)
{\displaystyle M_{0}(0;\;{\frac {5}{2}})}
M
1
(
−
20
;
27.5
)
{\displaystyle M_{1}(-20;\;27.5)}
y
=
x
2
16
+
5
2
.
{\displaystyle y={\frac {x^{2}}{16}}+{\frac {5}{2}}.}
Sprendimas .Rasime parabolės lanko ilgį iš taško
M
0
(
0
;
5
2
)
{\displaystyle M_{0}(0;\;{\frac {5}{2}})}
M
1
(
−
20
;
27.5
)
:
{\displaystyle M_{1}(-20;\;27.5):}
d
y
d
x
=
(
x
2
16
+
5
2
)
′
=
2
x
16
=
x
8
;
{\displaystyle {\frac {dy}{dx}}=({\frac {x^{2}}{16}}+{\frac {5}{2}})'={\frac {2x}{16}}={\frac {x}{8}};}
L
=
∫
a
b
1
+
(
y
′
)
2
d
x
=
∫
−
20
0
1
+
(
x
8
)
2
d
x
=
∫
−
20
0
1
+
x
2
64
d
x
=
∫
−
20
0
1
8
64
+
x
2
d
x
=
{\displaystyle L=\int _{a}^{b}{\sqrt {1+(y')^{2}}}dx=\int _{-20}^{0}{\sqrt {1+\left({\frac {x}{8}}\right)^{2}}}dx=\int _{-20}^{0}{\sqrt {1+{\frac {x^{2}}{64}}}}dx=\int _{-20}^{0}{\frac {1}{8}}{\sqrt {64+x^{2}}}dx=}
=
1
8
(
x
2
64
+
x
2
+
64
2
ln
|
x
+
x
2
+
64
|
)
|
−
20
0
=
{\displaystyle ={\frac {1}{8}}\left({\frac {x}{2}}{\sqrt {64+x^{2}}}+{\frac {64}{2}}\ln \left|x+{\sqrt {x^{2}+64}}\right|\right)|_{-20}^{0}=}
=
1
8
[
0
2
64
+
0
2
+
64
2
ln
|
0
+
0
2
+
64
|
]
−
1
8
[
−
20
2
64
+
(
−
20
)
2
+
64
2
ln
|
−
20
+
(
−
20
)
2
+
64
|
]
=
{\displaystyle ={\frac {1}{8}}\left[{\frac {0}{2}}{\sqrt {64+0^{2}}}+{\frac {64}{2}}\ln \left|0+{\sqrt {0^{2}+64}}\right|\right]-{\frac {1}{8}}\left[{\frac {-20}{2}}{\sqrt {64+(-20)^{2}}}+{\frac {64}{2}}\ln \left|-20+{\sqrt {(-20)^{2}+64}}\right|\right]=}
=
1
8
[
32
ln
|
64
|
]
−
1
8
[
−
10
464
+
32
ln
|
−
20
+
464
|
]
=
{\displaystyle ={\frac {1}{8}}\left[32\ln \left|{\sqrt {64}}\right|\right]-{\frac {1}{8}}\left[-10{\sqrt {464}}+32\ln \left|-20+{\sqrt {464}}\right|\right]=}
=
32
8
ln
|
8
|
−
1
8
[
−
215.4065923
+
32
ln
|
−
20
+
21.54065923
|
]
=
{\displaystyle ={\frac {32}{8}}\ln \left|8\right|-{\frac {1}{8}}[-215.4065923+32\ln \left|-20+21.54065923\right|]=}
=
4
ln
|
8
|
−
1
8
[
−
215.4065923
+
32
ln
|
1.540659229
|
]
=
{\displaystyle =4\ln |8|-{\frac {1}{8}}[-215.4065923+32\ln |1.540659229|]=}
=
4
ln
|
8
|
−
1
8
[
−
215.4065923
+
32
⋅
0.432210395
]
=
{\displaystyle =4\ln |8|-{\frac {1}{8}}[-215.4065923+32\cdot 0.432210395]=}
=
4
⋅
2.079441542
−
1
8
[
−
215.4065923
+
13.83073265
]
=
{\displaystyle =4\cdot 2.079441542-{\frac {1}{8}}[-215.4065923+13.83073265]=}
=
8.317766167
+
201.5758597
8
=
8.317766167
+
25.19698246
=
33.51474862.
{\displaystyle =8.317766167+{\frac {201.5758597}{8}}=8.317766167+25.19698246=33.51474862.}
Pasinaudojome integralų lentele
∫
x
2
±
a
2
d
x
=
x
2
a
2
±
x
2
±
a
2
2
ln
|
x
+
x
2
±
a
2
|
+
C
.
{\displaystyle \int {\sqrt {x^{2}\pm a^{2}}}\;{\mathsf {d}}x={\frac {x}{2}}{\sqrt {a^{2}\pm x^{2}}}\pm {\frac {a^{2}}{2}}\ln \left|x+{\sqrt {x^{2}\pm a^{2}}}\right|+C.}
keisti
x
=
ρ
cos
θ
,
y
=
ρ
sin
θ
.
{\displaystyle x=\rho \cos \theta ,\quad y=\rho \sin \theta .}
ρ
=
f
(
θ
)
.
{\displaystyle \rho =f(\theta ).}
d
y
d
x
=
d
y
d
θ
d
x
d
θ
.
{\displaystyle {\frac {dy}{dx}}={\frac {\frac {dy}{d\theta }}{\frac {dx}{d\theta }}}.}
d
x
d
θ
=
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
,
d
y
d
θ
=
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
.
{\displaystyle {\frac {dx}{d\theta }}={\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta ,\quad {\frac {dy}{d\theta }}={\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta .}
d
x
=
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
d
θ
.
{\displaystyle dx=({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )d\theta .}
d
s
=
1
+
(
d
y
d
x
)
2
d
x
=
1
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
d
θ
=
{\displaystyle ds={\sqrt {1+\left({\frac {dy}{dx}}\right)^{2}}}dx={\sqrt {1+\left({\frac {{\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta }{{\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta }}\right)^{2}}}({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta )d\theta =}
=
(
d
ρ
d
θ
cos
θ
−
ρ
sin
θ
)
2
+
(
d
ρ
d
θ
sin
θ
+
ρ
cos
θ
)
2
d
θ
=
{\displaystyle ={\sqrt {\left({\frac {d\rho }{d\theta }}\cos \theta -\rho \sin \theta \right)^{2}+\left({\frac {d\rho }{d\theta }}\sin \theta +\rho \cos \theta \right)^{2}}}d\theta =}
=
[
(
d
ρ
d
θ
)
2
cos
2
θ
−
2
ρ
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ρ
d
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(
d
ρ
d
θ
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sin
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2
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d
ρ
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d
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{\displaystyle ={\sqrt {\left[\left({\frac {d\rho }{d\theta }}\right)^{2}\cos ^{2}\theta -2\rho {\frac {d\rho }{d\theta }}\sin(\theta )\cos(\theta )+\rho ^{2}\sin ^{2}\theta \right]+\left[\left({\frac {d\rho }{d\theta }}\right)^{2}\sin ^{2}\theta +2\rho {\frac {d\rho }{d\theta }}\sin(\theta )\cos(\theta )+\rho ^{2}\cos ^{2}\theta \right]}}d\theta =}
=
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{\displaystyle ={\sqrt {\left({\frac {d\rho }{d\theta }}\right)^{2}\cos ^{2}\theta +\rho ^{2}\sin ^{2}\theta +\left({\frac {d\rho }{d\theta }}\right)^{2}\sin ^{2}\theta +\rho ^{2}\cos ^{2}\theta }}\;d\theta ={\sqrt {\left({\frac {d\rho }{d\theta }}\right)^{2}+\rho ^{2}}}\;d\theta ={\sqrt {\rho ^{2}+(\rho ')^{2}}}\;d\theta .}
![{\displaystyle [a;\;f(a)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8d0ac8260624336d7fca8995216f503bbba7e10)
![{\displaystyle [b;\;f(b)],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d76d386a15a7b83643cfaa152a9ca0c458a1289)
![{\displaystyle ds={\sqrt {[\phi '(t)]^{2}+[\psi '(t)]^{2}}}dt.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c10fc0b691adfc0229e484d0ac1df98cb4707449)
![{\displaystyle =\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}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e82bb1f6bdbf9af2a5034c587b71be7ebc69705e)
![{\displaystyle =\left[x{\sqrt {{\frac {1}{2x}}+1}}+{\frac {1}{4}}\ln \left(4x{\sqrt {{\frac {1}{2x}}+1}}+4x+2\right)\right]|_{1}^{7}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e135fd5b60289669f6ec65503c62c0dc24c3c801)
![{\displaystyle =\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]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7100451cc858d89562bce5af02e4fc5f57ef1bbb)
![{\displaystyle =\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]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f73947eecf65ffa17740db01923778102e312f1a)
![{\displaystyle =\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]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2dc762f0355e6091bf2fb7fa5ce88234250de7f2)
![{\displaystyle =7{\sqrt {1.071428571}}+{\frac {1}{4}}\ln \left(28\cdot {\sqrt {1.071428571}}+30\right)-\left[1.224744871+{\frac {2.38866916}{4}}\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b7bfd6c0726b1c8d14985fe7a79376f4c34a9638)
![{\displaystyle =7\cdot 1.035098339+{\frac {1}{4}}\ln(28\cdot 1.035098339+30)-[1.224744871+0.597167289]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af32342135a2353563bb7e608bfaefb7d57423d7)
![{\displaystyle ={\frac {1}{8}}\left[{\frac {0}{2}}{\sqrt {64+0^{2}}}+{\frac {64}{2}}\ln \left|0+{\sqrt {0^{2}+64}}\right|\right]-{\frac {1}{8}}\left[{\frac {-20}{2}}{\sqrt {64+(-20)^{2}}}+{\frac {64}{2}}\ln \left|-20+{\sqrt {(-20)^{2}+64}}\right|\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1432121b8ed177c8dcde9d5b7a5d947b29b92795)
![{\displaystyle ={\frac {1}{8}}\left[32\ln \left|{\sqrt {64}}\right|\right]-{\frac {1}{8}}\left[-10{\sqrt {464}}+32\ln \left|-20+{\sqrt {464}}\right|\right]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08ef7e15d35e920b3327962f07b642d1f394dd54)
![{\displaystyle ={\frac {32}{8}}\ln \left|8\right|-{\frac {1}{8}}[-215.4065923+32\ln \left|-20+21.54065923\right|]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f584609c7bda480566b00675c1d74a4f31dde746)
![{\displaystyle =4\ln |8|-{\frac {1}{8}}[-215.4065923+32\ln |1.540659229|]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ccb2701d6db045b2ca9c897fa127861d95ac34b)
![{\displaystyle =4\ln |8|-{\frac {1}{8}}[-215.4065923+32\cdot 0.432210395]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67dd38d36d2c84c16b336c3c4c56480268624cff)
![{\displaystyle =4\cdot 2.079441542-{\frac {1}{8}}[-215.4065923+13.83073265]=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a0c9d929ebe5066fa4d052019f4efff9e39b0f2)
![{\displaystyle ={\sqrt {\left[\left({\frac {d\rho }{d\theta }}\right)^{2}\cos ^{2}\theta -2\rho {\frac {d\rho }{d\theta }}\sin(\theta )\cos(\theta )+\rho ^{2}\sin ^{2}\theta \right]+\left[\left({\frac {d\rho }{d\theta }}\right)^{2}\sin ^{2}\theta +2\rho {\frac {d\rho }{d\theta }}\sin(\theta )\cos(\theta )+\rho ^{2}\cos ^{2}\theta \right]}}d\theta =}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2dcee70b6b6df97279384696a6e2876a93486ff)
