When I was in my first year of Prepa classes it was not at the program but we have to see it on an example and our maths teacher did it with hypocycloïd and epicycloïd too for fun, well it was very interesting.
Well I find on the net this "exercize" :
A goat is tied to a stake planted against a circular dovecote of 1m radius which is in the middle of a meadow. The rope has a length which allows the goat just reached the point diametrically opposite to the attachment, the rope then being stretched against the semi-circumference. Calculate the area of grassy land accessible to the goat.
I know how to calculate an integral for this kind of curves, but I have not idea or way which work to find polar (and parametric too but I think its not the best here) equations.
Can you help me, or (and it could be very appriciate) tell me a method which works nice for the most of cases ?
Thank you in advance
PS : No idea for the best tags...
May be it is the Involute in Geometry.Involute
In terms of ψ ( angle between radius vector and involute arc), the polar
coordinates parametric equations of Involute are r=r b secψ,θ=(tanψ−ψ)
If $ x=r/r_b $ the ODE for radius x in terms of θ is:
$$ x ′′ =−1/(x^ 2 −1)^{ 3/2 }$$
Its simple differential equation is
$$ r \cos\psi = r_b $$ where $\psi$ is angle radius vector makes to arc of involute.
$$ r^2- r_b^2 = 2 \,r_b s $$
where $r$ is radius in polar coordinates $r_b $ is radius of base circle, and $ s$ is arc length curve traced by taut rope
where $ s$ is unwound rope length = tangent length = $ r^2 = s^2 + (\theta r_b) ^2 $
This leads to
$$ \theta = s/r_b - tan ^{-1} (s/r_b)$$ or
$$ \theta = \tan \psi - \psi $$
$$ \theta = \sqrt{ (r/r_b)^2 -1 } - tan ^{-1}\sqrt { (r/r_b)^2 -1 }$$
Obtaining $r$ as direct function of $ \theta $ not possible.