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 October 27th, 2017, 08:10 PM #1 Newbie   Joined: Oct 2017 From: Springfield, MA Posts: 1 Thanks: 0 Equation for a _________ curve perpendicular to a line at a given point on the line Hi, I'm not sure whether calculus is the right thing for this, but I believe it is. I only have a loose memory of it from high school - 10 years ago, but I am nearly certain this is possible. Practically, my project starts with the idea of a triangle, and the idea is to bow out (or pinch in) the edge of the triangle at approximately the halfway point of its two edges, but the "pinch" needs to be a smooth curve, and the endpoints of either side of "triangle" can't move. This point could be represented by the line $\displaystyle y=4/3x$ and the two endpoints of the curve are (0,4) and (3,0) So we'd start with the bowed-in shape, be at the triangle at some point, then start to bow out. I'd take this shape and use the area between two chosen points on the y-axis (using integration) to figure out the area of the resulting subdivision. The point of the intersecting line and the "bow" is to add more area near the bottom of this shape when the user begins, and as the user of the project uses it, start to make the area more even by increasing the placement of that third point, ultimately creating a more consistent area when we subdivide the shape's height. The defining factor to determine that third point I'm calling a "confidence factor". It may be easier to think of this like a bow string would when pulled (or pushed), except that my angle of force is a 45ish degree angle (it's actually tan-1(4/3) up quadrant 1 starting at (0,0), and we want a rounded bowstring. Here's what I know: I'm looking for come up with a way to represent a curve that crosses (3,0) and (0,4) as well as a third point, determined by some intersecting point on the line represented by y=4/3x. That intersecting point - the confidence factor) - should be able to be described as a single number, with the Cartesian coordinates able to be derived using basic trig as follows. Assume 0 <= r <= 5 and r is the the "confidence factor" $\displaystyle x = r * \cos( \tan-1(4/3) ) \\ y = r * \sin( \tan-1(4/3) )$ I've tried quadratic, logarithmic, and exponential regressions, but I can't seem to find a way to make it work in all cases. I imagine I'll need to do some domain segmentation here, but that's where I'm stuck. I want it to be a smooth curve - much like that bow string would be if you pulled it back on a pulley (rather than the pinched triangle an actual bow string creates when using an arrow). Initially, I thought this would be quadratic, possibly with a transform across each axis (essentially rotating it 45 degrees) but I couldn't find a way to get the curve to hit all three points without transforming it back to the exact same curve each time. I feel like I'm missing a fundamental understanding of multidimensional analysis, trig, or calc here. Any ideas? Last edited by skipjack; October 27th, 2017 at 08:37 PM. Tags curve, equation, line, perpendicular, point Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mamort Applied Math 0 October 15th, 2014 05:33 AM bilano99 Algebra 2 September 1st, 2012 07:34 AM bilano99 Algebra 2 August 29th, 2012 06:48 AM bilano99 Algebra 3 August 27th, 2012 08:09 AM math12345 Algebra 3 April 3rd, 2011 10:54 AM

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