I recently re-posted a blog I wrote 12 years ago about a way to find the tangent to the curve of a conic without employing calculus that was one of my favorite math "tricks". In response I got a serious question from Brandon@_nilradical who asked, "anything "similar" for cubics, quartics, etc?". Being busy and wanting to reply I passed off a hasty, "Nothing quite so simple." and went back to mowing. Later I felt guilty about dismissing the question, so I thought I would fess up to what little I have found in playing with the idea "Beyond the Quadratics."
(If you are not familiar with the point substitution approach to finding tangents, take a moment to read the older post, it's pretty brief.)
My first excursion was to try a simple cubic and see if I could figure out how to apply the same polar idea. What could be easier than y=x3 so I picked the point (1,1) and set out exploring. I first had to decide how to replace the three x's in the right side, and decided that I would replace y= x3 with (y+1)/2 =(x)(1)(x+1)/2
Ok, we got a tangent, but it was a tangent quadratic, not a tangent line. So I decided to press on and apply the idea recursively into the new parabola. The parabola simplified to y=x2 +x - 1, so I began to substitute into that using the point (1,1) again. This leads to (y+1)/2 = (1)(x) + (x+1)/2-1
I didn't even bother to simply, just entered into the Desmos calculator and ....Eureka!!!
So how could we extract this and explain with a little calculus? Well, if we begin by saying we want to create a parabola at the point (1,1) with the same slope (3) as the cubic there, we would begin with y=x2 + bx +c and since the derivative of that, y'=2x+b must equal 3, setting 3=2(1)+b we see that b must be 1 also. Now we just plug (1,1) into the equation for y=x2 + x +c and we quickly find that indeed, the calculus will give us y=x2 +x - 1 as the parabola tangent to the point (1,1) with the same slope as y=x3
At this point I had no idea what this descending cascade of polar approaches to a tangent would do with something really complicated, but I barged ahead and created something minorly absurd.
So I started typing into the calculator creating as I go along and came up with x3y = y2 +x+x4
Desmos responds with :
Ok, this looks like fun. The first problem is finding a nice point with integers and a non-zero slope.... and (-1,0) jumps out because it should eliminate some congestion substituting y=0 in some places.
I begin by the same approach of using equal parts of variable and constant wherever possible, and write out (-1)(x)(x+1)/2 (y+0)/2= 0y +(x-1)/2 + (-1)(-1)(x)(x) and we get a cubic with two infinite discontinuities, but one of the branches slashes through the point we seek
At this point I'm convinced that our descending iteration of polars will proceed to a line tangent at the same point... and I found it interesting that even picking the equation out of my head, the point I chose also had a slope of 3 at that point.
I'm not sure I have any idea how useful these techniques might be for those forging father than my simple experiences in math can anticipate, but if you are one who knows more about this, share what you know.
I posted tongue-in-cheek at the end of the first post that the challenge of proving that it always worked was a homework assignment for calculus students, due by Wednesday. In response, Thomas Morgan replied: "I can verify that this will work in general. Multivariable calculus tells us that to compute the tangent vector at a point, one need only compute the partial derivatives at that point. It should be clear from the method description that the new curve intersects the old curve at the point under consideration. Two different curves that intersect at a point are tangent at that point if their tangent vector points in the same direction, so we can scale our tangent if necessary. Using partial derivatives, in addition to the sum rule for derivatives, reduces the effort to computing single-variable derivatives for two cases. In the first case, consider the tangent of c*x^(2n) at the point x=t. Computing the derivative directly yields 2n*c*x^(2n-1). Using your method, we first transform c*x^(2n) to c*x^n*t^n. Its derivative is n*c*x^(n-1)*t^n, which at the point x=t only differs from the first calculation by a factor of 2. We similarly compute the derivative of c*x^(2n+1) at the point x=t. Direct calculation yields (2n+1)*c*x^(2n). Using your method, we first transform c*x^(2n+1) into (c*x^(n+1)*t^n + c*x^n*t^(n+1))/2. Its derivative is ((n+1)*c*x^n*t^n + n*c*x^(n-1)*t^(n+1))/2, which at the point x=t only differs from the first calculation by a factor of 2. Thus we see that the tangent vector to the curve given by your transformation is the same vector as the tangent vector of the original curve, compressed by a factor of 2. This shows that the new curve is tangent to the original curve, and by induction, we are done."
Thank you for the response Thomas.
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