The document contains a series of problems related to the geometry and calculus of a martini glass and a ball submerged in it. I will break down the key questions and provide detailed steps and explanations for solving them.

Problem Overview

The problem involves a right-circular cone (the martini glass) and a ball being submerged in it. We want to determine the radius $R$ of the ball that causes the maximum volume of liquid to overflow from the glass.

Question 1: Geometry of the Glass

1.1 Coordinates of Points A and B

• Point O is the bottom of the glass.
• Point A is at the rim, and Point B is at the height $h$ of the glass.

Using trigonometry:

• Coordinate of Point A: $(h \tan(\theta), h)$
• Coordinate of Point B: $(0, h)$

Distances:

• $|AB| = h \sec(\theta)$
• $|OB| = h$

1.2 Geometry of Cases (i) and (ii)

• Case (i): Ball fully submerged.
• Case (ii): Ball partially submerged, not resting on the rim.

For the coordinates:

• Point C (Center of Ball): $(h \tan(\theta) - R, h - R)$
• Point E (Top of Ball): $(h \tan(\theta) - R, h - R + R)$ = $(h \tan(\theta) - R, h)$
• Point F (Bottom of Ball): $(h \tan(\theta) - R, h - R - R)$ = $(h \tan(\theta) - R, h - 2R)$

Distance $|OD|$: Using right triangle properties, $|OD|$ can be calculated similarly.

1.3 Geometry of Case (iii)

In this case, the ball is resting on the rim:

• Point C (Center of Ball) remains the same.
• Point F (Bottom of Ball): Position changes based on ball resting on the rim.

Question 2: Displaced Volume of Liquid

2.1 Displaced Liquid, Case (i)

The volume of the liquid displaced when the ball is fully submerged: $V(i)(R) = \frac{4}{3} \pi R^3$

2.2 Volume of a Spherical Dome

The volume of a spherical dome (spherical cap) of height $H$: $V_{\text{dome}} = \frac{\pi H^2(3R - H)}{3}$

2.3 Displaced Liquid, Case (ii)

For $R_1 < R < R_2$: $V(ii)(R) = \frac{\pi R^3}{3}(2 + 3\sin(\theta))$

2.4 Displaced Liquid, Case (iii)

For $R \geq R_2$: $V(iii)(R) = \frac{\pi}{3} \left[ R\sqrt{R^2 - h^2} + \text{higher order terms} \right]$

Finding the Maximum Displaced Volume

To find the radius $R$ that maximizes the volume of liquid displaced:

1. Differentiate the volume function $V(R)$ for each case.
2. Set the derivative to zero to find critical points.
3. Analyze intervals to determine where the function is increasing or decreasing.

Conclusion

For each question, use the appropriate geometric and calculus principles to derive the necessary equations and conditions for the radius $R$ that maximizes the displaced liquid volume.

Next Steps

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Tip: Always sketch the geometric configuration to better understand the relationships between the different elements in the problem.