Balling or Bowling?

Since both the paraboloid and ball are completely rotationally symmetrical about the z-axis, we can simply consider this problem in two dimensions with a circle rolling on a parabola.

The parabola would have the formula $z=x^2$ (where z is the vertical axis and x is the horizontal axis). Consider a circle such that the lowest point on the circle is at (0,0). This circle would have the formula $x^2+(z-r)^2=r^2$ , where r is the radius. This is simply taking the formula for a circle and moving it up by r units.

Solving these two equations, we can find where the circle intersects the parabola:

$\begin{aligned}z=x^2&\qquad(1)\\x^2+(z-r)^2=r^2&\qquad(2)\\ \text{substituting (1) into (2):}\\ z+(z-r)^2&=r^2\\ z+z^2-2zr+r^2&=r^2\\ z^2-2rz+z&=0\\ z(z-2r+1)&=0\\ z=0,&\quad z=2r-1\end{aligned}$

The first solution of z=0 is trivial, as we already know the circle is resting on (0,0), but the second shows that it is possible for the circle to intersect the parabola somewhere other than the origin, which we do not want. The circle needs to be tangent to the parabola.

It is, however, possible for $z=2r-1$ to be negative, in which case this solution will not be a solution (because the parabola only exists for positive values of z), and the circle will in fact be only touching the parabola and not cutting through it. Therefore:

$2r-1\leq0\\ 2r\leq1\\ \boxed{r\leq0.5}$

We can use the formula for finding the curvature of a function $y(x)$

$\kappa =\frac { \left| y'' \right| }{ { \left( 1+{ \left( y' \right) }^{ 2 } \right) }^{ \frac { 3 }{ 2 } } }$

which works out for $y=x^2$ to be

$\dfrac{2}{(1+2x)^\frac{3}{2}}$

so that the radius of curvature, which is the inverse of the curvature, at $x=0$ is $\dfrac{1}{2}$

The surface of the ball satisfies the equation $x^2 + y^2 + (z_{\text{ball}}-r)^2 = r^2;$ we want this to coincide with the surface of the parabola to second order in $x,y$ , or to first order in $z$ : $x^2 + y^2 + (z-r)^2 = r^2 + O(z^2);$ substitute the equation of the parabola: $z + (z-r)^2 = r^2 + O(z^2);$ $z(1 - 2r) = O(z^2);$ $1 - 2r = O(z).$ This requires $r = \tfrac12$ .

Due to symmetry we can consider the problem in two dimensions. Assume for a moment the circle is too large to touch the bottom of the parabola. It will contact the parabola at two points. CP is a radius and hence normal to the parabola. The centre of the circle C is the y-intercept of the normal to the parabola, hence determining the equation of the normal will give us the location of the centre. If we consider the parabola $y=x^2$ in parametric form, we have parametric equations $x=\frac{p}{2},y=\frac{p^{2}}{4}$

Differentiating $y=x^2$ , we get $y'=2x$ .

Taking the negative reciprocal and substituting $x=\frac{p}{2}$ gives the gradient of the normal at the point P: $m=-\frac{1}{p}$

Putting this and the coordinates of P into the point-gradient formula for a line gives the parametric equation of the normal: $x+py=\frac{p^3}{4}+\frac{p}{2}$

Substituting $x=0$ and dividing throughout by $p$ gives the y-intercept (and the locus of C): $y=\frac{p^2}{4}+\frac{1}{2}$

Now as we make the circle smaller P and P' will come closer together. The circle will touch bottom when these two points coincide, that is, there is a single point of contact. This is the point (0, 0), and obviously our parameter $p$ will also be zero.

Substituting $p=0$ into the above equation will give the location of the centre at $y=\frac{1}{2}$ .

Hence the radius of the circle is $\frac{1}{2}$ .

As any physics student knows, the maximum radius of curvature of a parabola is twice the focal length. For a parabola of the form y = a(x-b)^2+c the focal length is 1/4a . Here a=1 so the focal length is 1/4 and the radius of curvature is 1/2. The sphere must be therefore of a maximum radius 1/2.

Begin by visualizing a basic 2-dimensional parabola of y = x^2 and recall that any x value greater than 1 will result in a y value greater than that input x value. This effectively means that it’s “height” will be greater than its “width” at those points beyond x = 1. For a circle to rest perfectly in the bottom of the parabola, it would need to have even height and width, and so the maximum value for diameter would be 1 which would make the radius 0.5.

For the parabola the equation is: z(parabola) = x^2.

For the circle (the sphere in 2D) with radius r and centre at a distance r above the origin the equation is:

x^2 + (z - r)^2 = r^2.

Making z the subject of the formula gives:

z(circle) = r + (r^2 - x^2)^1/2 or r - (r^2 - x^2)^1/2.

The solution with the - sign defines the lower half of the circle.

So consider only this solution.

When x = 0, z(circle) = r - (r^2 - 0^2)^1/2 = r - r = 0 (as expected)

We need to ensure that z(circle) > z(parabola) for all x > 0.

i.e. r - (r^2 - x^2)^1/2 > x^2

or r(1 - (1 - (x/r)^2)1/2 > x^2. A

Intuitively we can see that this will be so for all values of x if the inequality holds for small values of x close to 0.

Use the small value approximation for the square root for values of x close to 0 for the left hand side of inequality A:

As x -> 0, r(1 - (1 - (x/r)^2)1/2 -> r(1 - (1 - (1/2)(x/r)^2) = r(1 - 1 + (1/2)(x/r)^2) = x^2 / 2r

For small x, inequality A therefore becomes:

x^2 / 2r > x^2

Therefore: 2r < 1

Therefore: r < 0.5.

I compared the second derivatives of the circle and the parabola at the point (0,0).

Circle: second derivative at (0,0) is 1/r.

Parabola: second derivative at (0,0) is 2.

1/r >= 2

=> r <= 1/2

A simpler way to look at this problem would be as follows, We want to find a sphere with center on the +Z axis that intersects the given paraboloid at only the origin. Thus the equation of the sphere is $x^2 + y^2 + z^2 - 2zR = 0$ , where R is its radius. To find its intersection with the paraboloid, we substitute $z = x^2 + y^2$ .

After simplifying, we get $(x^2 + y^2)(x^2 + y^2 + 1-2R) =0$

One of the solutions of the above equation is the one that we already know viz. $(0,0,0)$ The other solution is given by $x^2 + y^2 = 2R-1$ .

Since, we want only one intersection, we don't want the above equation to have any solution. To do so, we need to observe that the LHS of the above equation is always positive in Real domain. Thus, if we somehow make the RHS to be negative, our work would be done.

Thus we want $2R-1 <0$

Subsequently, $R <1/2$

The problem is the same if we switch to 2D, with $y=x^2$ and $x^2+(y-r)^2=r^2$ . When do they have only one intersection?

Second derivatives must be equal.

The equation x² + y² = z can be thought of as equations of circles (x² + y² = r²) whose radius increases as you go up.

Let r = the radius of a circle at a certain z value.

Then z = r²

The rate at which the bowl curves as you go down is represented dz/dr = 2r

Let b = the radius of the ball radius.

Then b = r, and db/dr = 1

The radius of the ball is constant as it falls in the bowl.

For the ball to not get lodged in the upper parts of the bowl, then db/dr ≥ dz/dr (i.e. the bowl curves slower or as quickly as than the ball's descent)

1 ≥ 2r

r ≤ 0.5

Therefore the max value of r is 0.5

we can look at the derivative. We define Zo = -sqrt(r^2 - x^2) + r^2 for the equation of the circle, and Zu = x^2 for the parabolla. then we will have a break point for r>=0 (when the ball would stop fitting in the buttom of the parabolla), whenever the derivatives are equal: dZ0/dx = dZu /dx in x=0. Solving for r, we get r=1/2.

Since both equations are symmetrical we can consider two dimensions:

$\begin{aligned} z = x^2 \qquad(1) \\ z = r - \sqrt{r^2 - x^2} \qquad(2) \\ \end{aligned}$

The ball is not touches the paraboloid if (1) < (2):

$\begin{aligned} x^2 < r - \sqrt{r^2 - x^2} \\ (x^2 - r)^2 > r^2 - x^2 \\ x^2 (x^2 + (1 - 2r)) > 0 \\ x^2 + (1 - 2r) > 0 \\ x^2 > 2r - 1 \\ x > \sqrt{2r - 1} \\ \end{aligned}$

In order to this be always true except when (x = 0):

$\begin{aligned} 2r - 1 \leq 0 \\ \boxed{r \leq \frac{1}{2}} \end{aligned}$

This Sol is requires basic knowledge of calculus, so proceed only when u know the basics of differentiation.

There is a formula for finding radius of curvature which can be easily obtained with some use of coordinate geometry and limits. Here I state it without proof.

R = [1+(dy/dx)^2]^(3/2) / (d^2y/dx^2)

This problem can be made equivalent to a circle on a parabola, as the reason provided by the other Sol.

So the max. radius of the sphere which the parabola can accommodate will be equal to its radius of curvature at that point (here we take the point of contact to be origin).

By finding required derivatives, and we obtain, R=0.5