Re: Finding intersection of two curves/ Chord that cuts a circle in ratio 1:3
- To: mathgroup at smc.vnet.net
- Subject: [mg41086] Re: [mg41055] Finding intersection of two curves/ Chord that cuts a circle in ratio 1:3
- From: Bobby Treat <drmajorbob+MathGroup3528 at mailblocks.com>
- Date: Thu, 1 May 2003 05:00:41 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Here's one method:
area = Simplify@Integrate [2Sqrt[1 - x^2], {x, s, 1}]
FullSimplify[area == Pi/4]
FindRoot[area == Pi/4, {s, 0.3}]
and here's another, using theta = half the angle subtended by the chord:
theta - Sin[theta]Cos[theta];
area = % // TrigReduce
FindRoot[area == Pi/4, {theta, Pi/4.}]
Cos[theta] /. %
Bobby
-----Original Message-----
From: Sujai <sujai at uiuc.eedduu>
To: mathgroup at smc.vnet.net
Subject: [mg41086] [mg41055] Finding intersection of two curves/ Chord that cuts
a circle in ratio 1:3
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
--
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