Finding the maximal $r$ so that the circle sits on top of the graph and touches it at the extremum point

Question

I came across this problem. I guess it is not necessary to be solved fully by hand (since I don't know if that's possible really). I mean tools like Wolfram Alpha can be used. Also, I don't know if it can be solved exactly in radicals, or if we should look for a numeric solution. Anyway, here's the problem.

We are looking for the maximum radius $r$ so that this circle there sits on top of the graph of $f(x) = x^3 - x$ and touches the graph at its extremum point (that's the only point they have in common).

I did some computations by hand. I took the center of the circle which has coordinates $(1/\sqrt{3}, -2/(3\sqrt{3}) + r)$ and I wrote the equation of the "lower semi-circle" and I came up with a function $g(x)$ whose graph is that lower semi-circle. That function is defined for

$x \in [1/\sqrt{3}-r,1/\sqrt{3}+r]$

I then put a condition on it to be above $f(x)$ and so I think it boils down to finding the max $r > 0$ such that this inequality

$r-2/(3\sqrt{3}) - \sqrt{r^2 - (x-1/\sqrt{3})^2} \ge x^3 - x$

is satisfied for every $x \in [1/\sqrt{3}-r,1/\sqrt{3}+r]$.

The function in the LHS here is $g(x)$.

But I have no idea how to continue further i.e. how to find this max value for $r$.

In geogebra I can see that 0.28 is OK for r but 0.30 is already too much. So the answer is some number in between these two.

I don't know if my approach is OK. Maybe I should think in the opposite way, "drop" a larger circle which sits on top of $f(x)$ (its center won't lie on the line $x=1/\sqrt{3}$), touches the graph at 2 points, and then put some conditions on it to see when these two points will merge into one point (when we're decreasing the circle's radius).

So I don't know if my 1st approach will lead me anywhere, maybe there's some smarter trick to be used here.

Also, if anyone could share a general approach to solving such problems (again, using WA might be OK, numerical solution might also be OK) that would be great.

Answer 1

Let

$$f(x)=x^3-x$$

As indicated in the comments, your issue deals with the osculating circle in a certain point which is the minimum .

In order to manage this circle, you have to know that its center belongs to a curve called the evolute of the initial curve $(C)$, which is obtained as the envelope of the normals to the curve as shown on the graphics below :

The generic equation of the normal at point (a,f(a)) is :

$$y-(a^3-a)=-\frac{1}{3a^3-1}(x-a) \iff (x-a)+(3a^3-1)y-(3a^5-4a^3+a)=0 \tag{1}$$

The technique now to get the envelope is to "play" between (1) and its derivative with respect to parameter $a$, $$-1+6ay-(15a^4-12a^2+1)=0\tag{2}$$

The verb "to play" has to be interpreted as considering (1)+(2) as a sytem of 2 equations in two unknowns yielding a couple of solutions :

$$x=\varphi(a), \ \ y=\psi(a)$$

which are plainly parametric equations of the involute (blue curve).

I just had a look at your exchanges with Intelligenti Pauca. So my question is now : what do you need to know more ?

Answer 2

Let's continue along the path outlined in the question. The equation of the circle $c(r)$ tangent to the graph of $f(x)$ at $x=1/\sqrt3$ and with radius $r$ is: $$ \left(y+\frac{2}{3 \sqrt{3}}-r\right)^2+\left(x-\frac{1}{\sqrt{3}}\right)^2=r^2. $$ We want to choose $r$ so that it has no other intersection with $y=x^3-x$, i.e. the system of the two equations has no other solution. Eliminating $y$ we get an equation of sixth degree for $x$: $$ -2 r x^3+2 r x-\frac{4 r}{3 \sqrt{3}}+x^6-2 x^4+\frac{4 x^3}{3 \sqrt{3}}+2 x^2-\frac{10 x}{3 \sqrt{3}}+\frac{13}{27}=0. $$ We may factor out $(x-1/\sqrt3)^2$ obtaining a simpler equation: $$ -2 r x-\frac{4 r}{\sqrt{3}}+x^4+\frac{2 x^3}{\sqrt{3}}-x^2-\frac{4 x}{3 \sqrt{3}}+\frac{13}{9}=0. $$ This can be solved for $r$: $$ r(x)=\frac{9 x^4+6 \sqrt{3} x^3-9 x^2-4 \sqrt{3} x+13}{6 \left(3 x+2 \sqrt{3}\right)}. $$ The graph of $r(x)$ in a neighbourhood of $x=1/\sqrt3$ shows a minimum: if $r$ has a value lower than that minimum, then the equation has no solution. The limiting case is the value of $r$ giving exactly the minimum: this can be found by solving $r'(x)=0$, which has $x\approx0.6592689252$ as the only positive solution. Substituting this into $r(x)$ we get the limiting value of the radius: $$ r\approx0.281725069920. $$