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Chapter 2 problem 66 aka. dropping a marble into an egg cup

January 28, 2010

Imagine an egg cup and a marble you decide to drop into it. What’s the biggest marble you can drop into the cup and have it touch only at the very bottom? This is a classic problem in analytic geometry and calculus.

In the picture,
the egg cup looks like a parabola, let’s say its cross section is the graph of y = x^2. The green marble touches the bottom, while the red one touches only on the sides of the cup.

Solution after the jump…

When the circle touches the parabola it necessarily touches at a point where the tangent line to the circle coincides with the tangent line to the parabola. The slope of this line at the point (a, a^2) on the parabola is 2a and its equation can be written as y = 2a(x-a)+a^2. On the other hand, if we follow the normal line at this point towards the Y-axis we get a radius for the circle (the length of which tells us how large the marble / circle is). The normal line has slope \frac{-1}{2a} and so its equation is y = \frac{-1}{2a}x + (a^2+\frac{1}{2}). The height on the Y-axis where this normal line intersects is then a^2 + \frac{1}{2}.

So what do we do from here? The radius of the circle is the length of normal segment from (a, a^2) to (0, a^2 + \frac{1}{2}) which is
\sqrt{ a^2 + (\frac{1}{2})^2} = \sqrt{ a^2 + \frac{1}{4}}. Now, if we think about shrinking the circle from the point where it touches the egg cup at some point (a,a^2) where a > 0 to the point where it touches at a=0 we’ll have the largest circle which does so.

This is a limit problem, and its rigorous solution is where calculus comes in. We need to understand \lim_{a \to 0} \sqrt{ a^2 + \frac{1}{4}}. By the limit laws (or continuity of the square root function) we can conclude that \lim_{a \to 0} \sqrt{ a^2 + \frac{1}{4}} = \sqrt{ 0^2+\frac{1}{4} } = \frac{1}{2}.

Thus the largest marble has a radius of one-half.

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From → math

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