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July 2

Practically drawing tangent to an inflection point (calculus)

I want to practically (by pencil and paper) draw a tangent to an inflection point. Please tell me about the method of drawing a tangent to an inflection point, because drawing a tangent to the other points is easier and I know it, but no books contain the method to draw the tangent on the inflection point. reference: https://wiki.riteme.site/wiki/Second_derivative (The point, on which the tangent crosses the curve graph of the function, is called an inflection point.) Ravijoshi99 (talk) 10:52, 2 July 2014 (UTC)[reply]

It's inflection point, and I don't really see the difficulty. Could you explain? — Arthur Rubin (talk) 14:44, 2 July 2014 (UTC)[reply]

I don't know what context you are using "inflation point" in. If you mean inflection point then I don't understand the problem. If you have the equation (or a numerical approximation) then the points at which the second derivative are zero contain the set of inflection points, but also contain undulation points (see Inflection_point#A_necessary_but_not_sufficient_condition). The second derivative is the first derivative of the first derivative, which implies that if you can take (general) derivitives, then you should be able to calculate the points of inflection, and the slope of the curve at those points. A slope and a point define a line. So, anyone should be able to construct a line (segment) with the appropriate slope and point of intersection. Choosing the end-points is a matter of accuracy, the larger the segment, the more accurate, generally, will be the phyical line...but there may be better choices depending on the graph paper used (lattice points which are closer to the line than the x and y axes intercepts). If you wish a purely physical method, then it depends on the line. Three common methods are: 1) to draw a line colinear with the curve at the point of inflection (parallel and on top of) or 2) to draw a series line segements from points on the curve from where the grid lines intersect the curve to the point of inflection, and then take their limit or average or 3) to draw line segments from two points on the curve equally far from the point, and take their limit as they both approach the point (as the distance diminishes to zero).173.189.75.163 (talk) 16:26, 2 July 2014 (UTC)[reply]

Oh gee, it sounds like they want method 3. If the curvature changes rapidly, they would want to place the two points a very short distance on either side of the inflection point. If it changes more gradually, then a wider placement might provide better accuracy. Of course, this assumes they already know exactly where the inflection point is, which may not be the case with a very gradual change in curvature. The curve could also have a "flat" (linear segment) in it, in which case there is no single inflection point. StuRat (talk) 16:40, 2 July 2014 (UTC)[reply]