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]