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A way of studying curves

  1. Sep 4, 2004 #1
    Ok, bear in mind that I am DUMB when it comes to math... There, I said it...
    :cry:


    Ok, I was thinking about calculus and part of calculus is studying curves, identifying degree of curves, writing equations for curves, and stuff like that.

    Well it is simple to me to relate everything to a common ratio so that you could say that a curve is + or - of that ratio. Well what better ratio than a circle?

    So, using pi as a standard you could then relate all other curves to this constant of 3.14.

    If you were to put a circle onto paper, and then plot a given curve on a given point of the circle's perimeter (lets say, 0,5cm in a circle whose center is 0,0cm) you should be able to extrapolate that curve so that it returns to the same x coordinate of 0 and then measure the difference in the y coordinates after one revolution. Taking that difference of Y (lets say the difference was +2cm) you could then calculate the degree of curvature of the curve using the equation Curvature = y2 - y1/ 2pi(r) in units of cm away from the perimeter of the circle and distance covered in relation to the circle's perimeter. The ratio is discovered to be .0636cm outward from the center per every cm of the circle's perimeter.

    For instance, you could ask what position the curve is at with respect to the circle's perimeter after it has moved a quarter of the way around the circle. To do that, you would simply reduce the value of the perimeter to one quarter of its whole value and then work the equation out from there, (yielding .5 as an answer.)

    So hopefully ive explained this idea enough so that you can all understand it and tell me if im right or wrong of if there is a better way for whatever, thanks!
     
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  3. Sep 4, 2004 #2

    Hurkyl

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    I'm really not sure at all what you're trying to say.

    There's something called the "radius of curvature": the radius of the circle that is tangent to the curve at the specified point.
     
  4. Sep 4, 2004 #3
    Not ever curve is part of a circle. Get a large sheet of grid/ graph paper and plot some (x,y) pairs of [tex] y = 2^x [/tex], such as (-2,0.25), .(-1,0.5), .... (6,64). If you take a small enough peice of a curve you could represent it as part of a circle but it is much easier to represent the curve as part of a parabola [tex] ax^2+ bx + c [/tex] and the coefficients a,b,c are easy to determine.

    a is determined by first finding the second derivative and b could be found using the first derivative. For exact details search for infomation about taylor series.
     
  5. Sep 5, 2004 #4

    HallsofIvy

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    Why a parabola? A parabola is going to be no better an approximation than a circle. (Yes, Taylor series gives the coefficients quickly for a parabola while the coefficients for a circle might be a bit more complicated.)

    I agree with Hurkyl- it is probably "radius of convergence" (or its reciprocal, curvature) that Mattius is looking for.
     
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