Discussions

Given the following definitions: The differential of x, denoted by dx, is defined by: dx = delta x The differential of y, denoted by dy, us defined by: dy = f'(x)dx For the curve f(x) = sin(x) between x = pi/6 and x = pi/4, find the exact values of: 1) The point P1 on the curve at x = pi/6 2) The point P2 on the curve at x = pi/4 3) f'(pi/6) 4) The equation of the tangent at P1: x = pi/6 5) The point P3 on the tangent when x = pi/4 6) dx 7) dy 8) delta y between P1 and P2 (note that dy > delta y) 9) the ratio, dy (from 7) divided by dx (from 6) and compare to (3) A picture is worth a thousand words. Cheers - Ian

We are all very familiar with the statement: dy = f'(x)dx - yes, that's "equal to". However, there seems to be a very deep rooted misconception about what this actually means and how that meaning is based on the actual definitions of these "differentials". And there is a good reason why "dy by dx" behaves like a fraction (i.e. diff(y) "divided by" diff(x)) when, in reality, "dy by dx" is not a fraction. In the "differential" definitions the variables dy and dx are NOT - yes you heard it right- NOT restricted to be infinitely small. dx can be as large as you like, and when you find the corresponding value of dy (also quite likely to be large) and then divide them ( dy "divided by" dx) - by the magic of similar triangles - you get the slope of the tangent. In fact, depending on f(x) and the value of dx, dy can actually be larger than "delta"y. Now having said that, the definitions of differentials, do not prevent dx and dy from being infinitely small, but the key is that they are not required to be so. Newton knew this and it was key to his wonderful discovery. As strange as this may sound to some, it is not fake news, but really is true. BTW Newton used the term "moment of x". "differential of x" is a Leibniz term. But they are very similar. "Moment" is a "differential" that is infinitely small. Cheers - Ian

Hi blackpenredpen For a long time I've been fascinated by the way you can tile the 80 80 20 isosceles with smaller isosceles/equilateral triangles, like this https://www.cut-the-knot.org/triangle/80-80-20/CID.shtml I also really like the geometric determination of sin 18 using recursive triangles, which you've epxlored on video. Have you looked at combining the two? I realised today that the 80 80 20 tiling very quickly and elegantly shows that y=2sin(10) has to satisfy y=1/(3-y^2) ==> y^3-3y^2+1. After adjusting for the factor of 2 this is the same cubic as you talk about on your sin 10 video. To see this, start with the isosceles tiling from cut-the-knot above, setting the base of the big 80 80 20 triangle to 1; this length then propagates through all the isosceles triangles in the diagram; and set y=2sin(10) as shown. Because ABC and BCD are similar we know their sides are in proportion so y/1 = 1/AB = 1/(AF+1). Now focussing on EFA we can add points G and H along FA so that FEH and AEG are congruent to each other and to BCD, so EG = EH =y; and because EGH is similar we know that GH = y^2, so AF = AG +FH - GH = 1+1-y^2 = 2-y^2. Substituting into y=1/(AF+1) gives y=1/(3-y^2) ==> y^3-3y+1=0, QED. This diagram also gives you the connection with cos40=sin50: the two sides of the big isosceles triangle ABC, namely AB and AC have to be equal. We know AB=3-y^2 as just shown and can see that AC=1+y+2sin50. Equating these gives sin50=1-y/2-(y^2)/2. Best regards Josh DAnziger