# Drop a ball into a bowl in the shape of a paraboloid. How far from the bottom of the bowl will the ball come to rest?

This is a question I like to bring up in a multivariable calculus class from time to time:

A sphere of radius 4 is dropped into a bowl-shaped paraboloid given by $z = x^2+y^2$. How close will the sphere come to the vertex of the paraboloid?

The problem can be solved using standard optimization techniques discussed in multivariable calculus, such as Lagrange multipliers or finding the zeros of the derivative(s) of the appropriate objective function. It can also be easily reduced to a problem in 2D by exploiting symmetry or introducing polar coordinates, and even to a problem in one variable, but the obvious solution approaches still seem to require Lagrange multipliers or finding the zeros of the derivative(s) of the objective function.

I've wondered, though, if there is a way (particularly a clever one) to solve this problem that does not use calculus. Has anyone seen such a solution, or can anyone think of one?

(I'll refrain from posting how close the sphere gets to the bottom of the bowl for those who would enjoy working on the problem for its own sake.)

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As you say, this is really a 2-dimensional problem. We seek a point $(x,z)$ on the parabola $z=x^2$ such that the length of the normal from ($x,z)$ to the $z$-axis is $4$. The slope of the parabola is $2x$, so the slope of the normal is $-1/(2x)$. So the normal meets the $z$-axis at $(0,z+\frac{1}{2})$. The squared length of the line segment from $(x,z)$ to $(0,z+\frac{1}{2})$ is $x^2+\frac{1}{4}$, which is $16$ by hypothesis. Thus $x = \frac{3}{2}\sqrt 7$, and the height of the centre of the sphere is $z+\frac{1}{2} = x^2+\frac{1}{2} = \frac{65}{4}$.
Nice solution. This is the kind of thing I was looking for. (I get the height of the center of the sphere to be $\frac{65}{4}$, though.) – Mike Spivey Feb 4 '11 at 19:16
I was thinking about this question again. While I still like your answer, how would we show that the slope of a parabola is $2x$ without using calculus? (Clearly, it's obvious with calculus.) – Mike Spivey Feb 28 '11 at 19:25