Δ
θ
{\displaystyle \Delta \theta }
[1]
(
d
s
d
θ
)
2
=
ρ
2
+
(
d
ρ
d
θ
)
2
{\displaystyle \left({\frac {ds}{d\theta }}\right)^{2}=\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}}
(30 )
∴
d
s
d
θ
=
ρ
2
+
(
d
ρ
d
θ
)
2
{\displaystyle {\frac {ds}{d\theta }}={\sqrt {\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}}}}
In the notation of differentials this becomes
(31 )
d
s
=
[
ρ
2
+
(
d
ρ
d
θ
)
2
]
1
2
d
θ
{\displaystyle \ ds=\left[\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}\right]^{\frac {1}{2}}d\theta }
These relations between
ρ
{\displaystyle \rho }
ds, dp , and
d
θ
{\displaystyle d\theta }
ds and whose sides are
d
ρ
{\displaystyle d\rho }
ρ
d
θ
{\displaystyle \rho d\theta }
d
s
=
(
ρ
d
θ
)
2
+
(
d
ρ
)
2
,
{\displaystyle ds={\sqrt {(\rho d\theta )^{2}+(d\rho )^{2}}},}
and dividing by
d
θ
{\displaystyle d\theta }
(30 ) .
Denoting by
ψ
{\displaystyle \psi }
d
p
{\displaystyle dp}
d
s
{\displaystyle ds}
tan
ψ
=
ρ
d
θ
d
ρ
,
{\displaystyle \tan \psi =\rho {\frac {d\theta }{d\rho }},}
which is the same as (A .
Illustrative Example 1.
x
2
+
y
2
=
r
2
{\displaystyle x^{2}+y^{2}=r^{2}}
Solution. Differentiating,
d
y
d
x
=
−
x
y
{\displaystyle {\tfrac {dy}{dx}}=-{\tfrac {x}{y}}}
To find ds in terms of x we substitute in (27 ) , giving
d
s
=
[
1
+
x
2
y
2
]
1
2
d
x
=
[
y
2
+
x
2
y
2
]
1
2
d
x
=
[
r
2
y
2
]
1
2
d
x
=
r
d
y
r
2
−
x
2
.
{\displaystyle ds=\left[1+{\frac {x^{2}}{y^{2}}}\right]^{\frac {1}{2}}dx=\left[{\frac {y^{2}+x^{2}}{y^{2}}}\right]^{\frac {1}{2}}dx=\left[{\frac {r^{2}}{y^{2}}}\right]^{\frac {1}{2}}dx={\frac {rdy}{\sqrt {r^{2}-x^{2}}}}.}
To find ds in terms of y we substitute in (28 ), giving
d
s
=
[
1
+
y
2
x
2
]
1
2
d
y
=
[
x
2
+
y
2
x
2
]
1
2
d
y
=
[
r
2
x
2
]
1
2
=
r
d
y
r
2
−
y
2
.
{\displaystyle ds=\left[1+{\frac {y^{2}}{x^{2}}}\right]^{\frac {1}{2}}dy=\left[{\frac {x^{2}+y^{2}}{x^{2}}}\right]^{\frac {1}{2}}dy=\left[{\frac {r^{2}}{x^{2}}}\right]^{\frac {1}{2}}={\frac {rdy}{\sqrt {r^{2}-y^{2}}}}.}
Illustrative Example 2.
ρ
=
a
(
l
−
cos
θ
)
{\displaystyle \rho =a(l-\cos \theta )}
θ
{\displaystyle \theta }
Solution. Differentiating,
d
ρ
d
θ
=
a
sin
θ
{\displaystyle {\tfrac {d\rho }{d\theta }}=a\sin \theta }
Substituting in (31 ) , gives
d
s
=
[
a
2
(
1
−
cos
θ
)
2
+
a
2
sin
2
θ
]
1
2
d
θ
=
a
[
2
−
2
cos
θ
]
1
2
d
θ
=
a
[
4
sin
2
θ
2
]
1
2
d
θ
=
2
a
sin
θ
2
d
θ
.
{\displaystyle ds=[a^{2}(1-\cos \theta )^{2}+a^{2}\sin ^{2}\theta ]^{\frac {1}{2}}d\theta =a[2-2\cos \theta ]^{\frac {1}{2}}d\theta =a\left[4\sin ^{2}{\frac {\theta }{2}}\right]^{\frac {1}{2}}d\theta =2a\sin {\frac {\theta }{2}}d\theta .}
↑
lim
Δ
θ
→
0
chord
P
Q
Δ
θ
=
lim
Δ
θ
→
0
Δ
s
Δ
θ
=
d
s
d
θ
{\displaystyle \lim _{\Delta \theta \to 0}{\frac {{\mbox{chord}}PQ}{\Delta \theta }}=\lim _{\Delta \theta \to 0}{\frac {\Delta s}{\Delta \theta }}={\frac {ds}{d\theta }}}
By (G ), §90
lim
Δ
→
0
sin
Δ
θ
Δ
θ
=
1
{\displaystyle \lim _{\Delta \to 0}{\frac {\sin \Delta \theta }{\Delta \theta }}=1}
By §22
lim
Δ
θ
→
0
1
−
cos
Δ
θ
Δ
θ
=
lim
Δ
θ
→
0
2
sin
2
Δ
θ
2
Δ
θ
=
lim
Δ
θ
→
0
sin
Δ
θ
2
⋅
sin
Δ
θ
2
Δ
θ
2
=
0
⋅
1
=
0
{\displaystyle \lim _{\Delta \theta \to 0}{\frac {1-\cos \Delta \theta }{\Delta \theta }}=\lim _{\Delta \theta \to 0}{\frac {2\sin ^{2}{\frac {\Delta \theta }{2}}}{\Delta \theta }}=\lim _{\Delta \theta \to 0}\sin {\frac {\Delta \theta }{2}}\cdot {\frac {\sin {\frac {\Delta \theta }{2}}}{\frac {\Delta \theta }{2}}}=0\cdot 1=0}
By 39 , §1§22 .
![{\displaystyle \ ds=\left[\rho ^{2}+\left({\frac {d\rho }{d\theta }}\right)^{2}\right]^{\frac {1}{2}}d\theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfdf4fce3c6dacb64d515471aa2157151154c4e4)
![{\displaystyle ds=\left[1+{\frac {x^{2}}{y^{2}}}\right]^{\frac {1}{2}}dx=\left[{\frac {y^{2}+x^{2}}{y^{2}}}\right]^{\frac {1}{2}}dx=\left[{\frac {r^{2}}{y^{2}}}\right]^{\frac {1}{2}}dx={\frac {rdy}{\sqrt {r^{2}-x^{2}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67ca57cdf0c2c4bf4e75ad54330b3e0ff7e2d0e1)
![{\displaystyle ds=\left[1+{\frac {y^{2}}{x^{2}}}\right]^{\frac {1}{2}}dy=\left[{\frac {x^{2}+y^{2}}{x^{2}}}\right]^{\frac {1}{2}}dy=\left[{\frac {r^{2}}{x^{2}}}\right]^{\frac {1}{2}}={\frac {rdy}{\sqrt {r^{2}-y^{2}}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5742a669df0b774d4fd60565a59e44d8a3d5b196)
![{\displaystyle ds=[a^{2}(1-\cos \theta )^{2}+a^{2}\sin ^{2}\theta ]^{\frac {1}{2}}d\theta =a[2-2\cos \theta ]^{\frac {1}{2}}d\theta =a\left[4\sin ^{2}{\frac {\theta }{2}}\right]^{\frac {1}{2}}d\theta =2a\sin {\frac {\theta }{2}}d\theta .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8dd4cb412e1e0e7fffb6e3bfcb1e1a7bf55444a)
