Involute of a catenary Involute

the involute of catenary, tractrix.

the involute of catenary through vertex tractrix. in cartesian coordinates curve follows

x



=
t
−
tanh
⁡
(
t
)




y



=
sech
⁡
(
t
)
=


1

cosh
⁡
(
t
)









{\displaystyle {\begin{aligned}x&=t-\tanh(t)\\y&=\operatorname {sech} (t)={\frac {1}{\cosh(t)}}\end{aligned}}}

where t parameter , sech hyperbolic secant (1/cosh(t)).

derivative

with

r
(
s
)
=


(



sinh

−
1


⁡
(
s
)
,
cosh
⁡

(

sinh

−
1


⁡
(
s
)
)



)




{\displaystyle r(s)={\big (}\sinh ^{-1}(s),\cosh \left(\sinh ^{-1}(s)\right){\big )}}

we have

r
′

(
s
)
=



(
1
,
s
)


1
+

s

2







{\displaystyle r (s)={\frac {(1,s)}{\sqrt {1+s^{2}}}}}

and

r
(
t
)
−
t

r

′


(
t
)
=

(

sinh

−
1


⁡
(
t
)
−


t

1
+

t

2





,


1

1
+

t

2





)

.


{\displaystyle r(t)-tr^{\prime }(t)=\left(\sinh ^{-1}(t)-{\frac {t}{\sqrt {1+t^{2}}}},{\frac {1}{\sqrt {1+t^{2}}}}\right).}

substitute

t
=



1
−

y

2



y




{\displaystyle t={\frac {\sqrt {1-y^{2}}}{y}}}

to get

(

sech

−
1


⁡
(
y
)
−


1
−

y

2




,
y
)

.


{\displaystyle \left(\operatorname {sech} ^{-1}(y)-{\sqrt {1-y^{2}}},y\right).}