Re: Hypocycloid Area

• To: mathgroup at smc.vnet.net
• Subject: [mg27329] Re: Hypocycloid Area
• From: "Allan Hayes" <hay at haystack.demon.co.uk>
• Date: Wed, 21 Feb 2001 03:17:02 -0500 (EST)
• References: <96t8mk\$pq1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```The path traced out by a point on a circle of radius r rolling inside a
circle of radius R is given by
f[t_,r_,R_] = (R-r){Cos[t], Sin[t]} + r {Cos[(1-r/R)t],Sin[(1-R/r)t]}

So an n-cusped hypocycloid is given by

f[t_,n_, R_] = R((1-1/n){Cos[t], Sin[t]} +  {Cos[(1-n)t],Sin[(1-n)t]}/n)

Check this graphically:

ParametricPlot[Evaluate[f[t, 5, 3]], {t, 0, 2Pi},
AspectRatio -> Automatic,
Epilog -> Circle[{0, 0}, 3],
PlotRange -> All];

The square of the distance from the origin to a point on the hypocycloid is

(R^2*(2 - 2*n + n^2 + 2*(-1 + n)*Cos[n*t]))/n^2

The area of the hypocycloid is

{Element[n,Integers] &&n>0}]

(R^2*(n*(2 - 2*n + n^2)*Pi + (-1 + n)*Sin[2*n*Pi]))/n^3

More simplification is needed to get rid of the Sin.

Simplify[%,Element[n, Integers]]

((2 - 2*n + n^2)*Pi*R^2)/n^2

And after

Expand[%]

Pi*R^2 + (2*Pi*R^2)/n^2 - (2*Pi*R^2)/n

we see that as n->Infinity the area, as expected, -> the area of a circle
--
Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

"Lizveth Robles" <lrobles at unimail.uninorte.edu.co> wrote in message
news:96t8mk\$pq1 at smc.vnet.net...
> Hello,
>
> I have seen an answer you posted about the area of the epicycloid, using
polar
> coordinates. I'm trying to do the same thing with the hypocycloid with 5
cuspids
> but I don't get anywhere. What am I doing wrong?  Am I missing something?
>
>
> Lizveth
>
>
> _________________________________
> Visita nuestro sitio Web en: http://uninorte.edu.co
>

```

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