Re: Finding intersection of two curves/ Chord that cuts a circle in ratio 1:3
- To: mathgroup at smc.vnet.net
- Subject: [mg41071] Re: Finding intersection of two curves/ Chord that cuts a circle in ratio 1:3
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Thu, 1 May 2003 04:57:57 -0400 (EDT)
- Organization: Universitaet Leipzig
- References: <b8o2ad$p38$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
Plot[Evaluate[Integrate[Sqrt[1 - x^2], {x, 0, S}] - Pi/8], {S, 0, 1}]
and
FindRoot[Evaluate[Integrate[Sqrt[1 - x^2], {x, 0, S}] == Pi/8], {S,
4/10}]
Regards
Jens
Sujai wrote:
>
> I feel like I should know this, but am stuck:
>
> Am trying to find the point along the radius in a circle where, if I
> draw a chord perpendicular to the radius, I get a segment that is 1/4th
> of the total area of the circle.
>
> For a unit circle (am only working in one quadrant for simplicity), this
> would be the point S along the radius, where:
>
> Integrate [Sqrt(1 - x^2), {x, 0, S}] == Pi/8
>
> I used the following code to visualize what the solution would be
> (approximately 0.4), but am getting stuck at the analytical answer.
>
> \!\(Plot[{Integrate[\@\((1 - x^2)\), {x, 0, s}], Pi/8}, {s, 0, 1}]\)
>
> thanks
>
> - sujai
>
> --
> [remove duplicate letters in eedduu for my email address]