RE: Finding intersection of two curves/ Chord that cuts a circle in ratio 1:3
- To: mathgroup at smc.vnet.net
- Subject: [mg41078] RE: [mg41055] Finding intersection of two curves/ Chord that cuts a circle in ratio 1:3
- From: "David Park" <djmp at earthlink.net>
- Date: Thu, 1 May 2003 04:59:40 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
Sujai,
CALCULATE first, PLOT second.
Here is the function for the area of a segment obtained from Integrate...
area[r_] = Integrate [Sqrt[1 - x^2], {x, 0, r}]
(1/2)*(r*Sqrt[1 - r^2] + ArcSin[r])
Next we calculate the radius value that gives an area of Pi/8 for one
quadrant. We use Mathematica's FindRoot.
rval = r /. FindRoot[area[r] == Pi/8, {r, 0.4}][[1]]
0.403973
In order to make a nice plot we will need the angle to the corner of the
segment.
angle = ArcSin[rval]
0.415856
Now, we can make the plot...
Needs["Graphics`FilledPlot`"]
Needs["Graphics`Colors`"]
FilledPlot[{rval, Sqrt[1 - x^2]}, {x, -Cos[angle], Cos[angle]},
Fills -> Salmon,
AspectRatio -> Automatic,
Axes -> None,
PlotRange -> {{-1, 1}, {-1, 1}}1.1,
Background -> Linen,
PlotLabel -> "Radius for 1/4 Area Segment",
ImageSize -> 400,
Epilog ->
{Circle[{0, 0}, 1, {Pi - angle, 2Pi + angle}],
Line[{{0, 0}, {0, rval}}],
Gray, Line[{{0, rval}, {0, 1}}],
Black,
Line[{{-1, 0}, {1, 0}}],
Text[1, {0.5, -0.1}],
Text[0.403973, {0.0, 0.2}, {-1, 0}]}];
David Park
djmp at earthlink.net
http://home.earthlink.net/~djmp/
From: Sujai [mailto:sujai at uiuc.eedduu]
To: mathgroup at smc.vnet.net
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]