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How would you draw an accurate quadratic curve?

How would you draw (freehand, with compass and ruler) a pretty accurate quadratic curve, say \( y = 3x^2\), in the domain \([-20,20]\)?

Would you have to laboriously plot a lot of points corresponding to the domain that we are interested in?

Are there any tricks that we can use? For example, we know how to draw the circle \(x^2 + y^2 = 1 \) using a compass.

How about a cubic curve \( y = 4 x^3 \)? Would any of the same ideas apply?

Note by Calvin Lin
4 years, 3 months ago

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In calculus we had to freehand draw lots of curves up to quintic and what we always did was use basic derivatives to find the zeroes and where each curve is increasing/decreasing and concave up/concave down (this is relatively straightforward without using derivatives for quadratics if you don't know calculus) after that we would just plot those points where the "activity" of the curve changes (inflexion points and maximums and minimums) and the zeroes and the rest of the curve is surprisingly easy to freehand. With just a little bit of practice we all could draw curves that almost exactly matched what a graphing application would output to the projector

Brian Hogan - 4 years, 3 months ago

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We did that too. What fun.

Bob Krueger - 4 years, 3 months ago

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I was thinking of this problem when I saw this.

However, a better approach would be to use curvature.

In this case, the radius of curvature would be \(\frac {(36x^2 + 1)^{\frac {3} {2}}} {6}\) for any point \((x, 3x^2)\).

The circle of curvature would be centered at \((-36x^3 - x, 6x^2 + \frac {1} {6})\) (assuming I did all the math correctly).

Notice that in the formula for curvature does not have any cubic (or higher degree) roots. This is convenient for us because construction can generate all algebraic numbers of degree \(2\).

Picking rational values of \(x\), finding the curvature at \((x, f(x))\), and drawing the circle of curvature should be tedious but would produce an extremely accurate curve.

Joe Ill - 4 years, 3 months ago

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My thought would be to use the conic definition of a parabola, i.e. find the focus and directrix. Then one can pick arbitrary lengths, to which to set a compass, and then draw the circle of that radius from the focus and find the two points on that circle that same distance away from the directrix. This gets rid of the need for plugging in several values and plotting points that way.

It doesn't seem like this method would apply to a cubic, however.

Bob Krueger - 4 years, 3 months ago

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  1. know the intercepts
  2. know the vertical, horizontal, slant asymptote if there's any. Usually when there's a denominator
  3. determine maxima and minima by calculus
  4. inflection points by calculus
  5. finds more points :)))
oh and don't forget analytic geom ....it really helps if you know the general form of a circle, eclipse, parabola, hyperbola, etc. :P

Kalyph Dioquino - 4 years, 2 months ago

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