Update: H/t David K for pointing out that my assumption that I can force $a^2+b^2=r^2$ is wrong. This led to analyzing a cubic equation which is now moot, but I think the bulk of the question remains sensible.

I think I'm making some mistakes in the below, but I'm not sure where. Even if there are mistakes, I'm thinking there might not be any practical point to it, but I'm curious.

I think it's possible to find the slope of a tangent line without using limits in some cases. Further it's possible to generalize the procedure to many common curves. I am having trouble with a proof of this and I'm not sure where I am going wrong.

Proof we can find the slope of a tangent line to a parabola without taking a limit:

1) First we consider a circle. If we know the center of a circle and we know a point on that circle, we know the equation of a line connecting the center to that point. The slope of the line perpendicular to this line is the slope of the tangent line to the circle at that point. So we have already proven the thesis in the case of a circle.

2) A circle is tangent to some other curve if it only intersects that curve at a single point.

3) A circle can be constructed to be tangent to certain points on certain curves. Tangent circles share tangent lines with the curves to which they are tangent. So the procedure above can find tangents to those curves by constructing the correct tangent circle.

Parabola case: Consider $y=x^2$.

We want a circle such that $(x-a)^2+(x^2-b)^2=r^2$

So: $$x^4+(1-2b)x^2-2ax+(a^2+b^2-r^2)=0$$

We either have 2 complex roots or no real solutions. Assuming we have any real solutions, having a unique one requires that we have a repeated root.

So to find the tangent to the curve at $x_0$ we want to divide the quartic by $(x-x_0)^2$. We'd want to have the remainder terms be zero and the discriminant of the resulting quotient to be negative.

  • 1
    $\begingroup$ If you select $r,$ $a,$ and $b$ so that $a^2+b^2=r^2,$ then $(x-a)^2+(x^2-b)^2=r^2$ is the equation of a circle that passes through $(x,y)=(0,0),$ and the only possible point of tangency between this circle and the parabola $y=x^2$ is at $x=0.$ $\endgroup$ – David K Apr 22 at 18:52
  • $\begingroup$ Good eye. thanks! Will try to work that in there. $\endgroup$ – TurlocTheRed Apr 22 at 18:53
  • $\begingroup$ How is this generalizable for any curve?. Because often the procedure of finding an osculating circle of a curve require limit's. $\endgroup$ – yousef magableh Apr 22 at 19:11
  • $\begingroup$ This process would only be applicable to those curves for which you can find a tangent circle, so many curves, but probably not all. The real distinguishing feature is you don't need any terms to get arbitrarily small. I'm thinking that any $r$ smaller than the radius of the osculating circle should due. I think there are algebraic means of finding the minimum radius of osculating circles for some curves without using limits, if so I think that might lead to a more restrictive criterion. $\endgroup$ – TurlocTheRed Apr 22 at 19:24

This is very creative, but I think it's overly complicated. You could do something very similar by using lines instead of circles.

Consider the point $P=(a,a^2)$ on the parabola $y=x^2$. Which lines through $P$ only intersect the parabola once? The line is determined by its slope, so it has equation $y=m(x-a)+a^2$. This intersects $y=x^2$ where \begin{gather*} x^2 = m(x-a)+a^2 \\ \iff x^2 -mx + (am-a^2) = 0 \end{gather*} The last equation is quadratic in $x$ and has discriminant $$ \Delta = m^2 - 4(1)(am-a^2) = m^2-4am + 4a^2 = (m-2a)^2 $$ So there is a single solution if and only if $m=2a$. Thus the line is tangent to the parabola when $m=2a$.

For a general polynomial curve $y=f(x)$, you'd look for slopes $m$ such that the equation $f(x) = m(x-a) + f(a)$ (an equation in $x$) has multiple roots. You can probably also extend it to algebraic curves in the plane, too. However, I'm not sure it provides any more insight than ordinary calculus.

  • $\begingroup$ Actually, the line parallel to the axis of a parabola also intersects it in one (finite) point only. In this case it's implicitly discarded as no lines of the form $x=a$ were considered. $\endgroup$ – Berci Apr 22 at 19:08
  • $\begingroup$ Excellent solution. I think I might end up accepting this one. I'm also not sure if it offers anything over regular methods. I was asked why the derivative is the slope of the tangent line. The derivative is usually defined in terms of limits. The tangent line is offered as the limit case of a sequence of secants. I wanted to take a purely geometric approach without using limits to find the tangent line and then show it is equal to the limit definition of the derivative. $\endgroup$ – TurlocTheRed Apr 22 at 19:32
  • $\begingroup$ Thought occurs to me this approach to taking a derivative might have some benefits numerically. I need to look into if it bypasses the problems of machine epsilon. $\endgroup$ – TurlocTheRed Apr 23 at 20:18

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