Problem 50 - Algebra and Trigonometry with Analytic Geometry

Question

A spherical pill has diameter 1 centimeter. A second pill in the shape of a right circular cylinder is to be manufactured with the same volume and twice the surface area of the spherical pill. (a) If \(r\) is the radius and \(h\) is the height of the cylindrical pill, show that \(6 r^{2} h=1\) and \(r^{2}+r h=1\). Conclude that \(6 r^{3}-6 r+1=0\). (b) The positive solutions of \(6 r^{3}-6 r+1=0\) are approximately \(0.172\) and \(0.903\). Find the corresponding heights, and interpret these results.

Step-by-Step Solution

Answer

The cubic equation relates to the geometry of the cylinder. Two scenarios exist: small radius with large height and large radius with small height.

Step 1: Calculate Volume of Spherical Pill

First, we calculate the volume of a spherical pill using the formula for the volume of a sphere: \( V = \frac{4}{3}\pi r^3 \). The radius of the sphere is half of the diameter, so \( r = \frac{1}{2} \) cm. Substituting this into the formula, we get the volume as \( V = \frac{4}{3} \pi \left( \frac{1}{2} \right)^3 = \frac{1}{6} \pi \).

Step 2: Calculate Surface Area of Spherical Pill

Next, we find the surface area of the spherical pill. The formula for the surface area of a sphere is \( A = 4\pi r^2 \). With \( r = \frac{1}{2} \), the surface area becomes \( A = 4\pi \left( \frac{1}{2} \right)^2 = \pi \).

Step 3: Volume and Surface Area Constraints for Cylindrical Pill

The cylindrical pill must have the same volume as the spherical pill (\( \frac{1}{6} \pi \)) and twice the surface area of the spherical pill (\( 2\pi \)). The volume for a cylinder is \( V = \pi r^2 h \), thus \( \pi r^2 h = \frac{1}{6} \pi \). This simplifies to \( r^2 h = \frac{1}{6} \) or equivalently \( 6r^2 h = 1 \).

Step 4: Surface Area Equation

The surface area of a cylinder is \( A = 2\pi r(r + h) \). Given this surface area must be twice that of the sphere, or \( 2\pi \), we have the equation \( 2\pi r(r + h) = 2\pi \). Simplifying, \( r(r + h) = 1 \), which expands to \( r^2 + rh = 1 \).

Step 5: Solve to Find Cubic Equation

We now have two equations: (1) \( 6 r^2 h = 1 \) and (2) \( r^2 + rh = 1 \). By eliminating \( h \), substitute \( h = \frac{1 - r^2}{r} \) from the second equation into the first to get \( 6 r^2 \left(\frac{1 - r^2}{r}\right) = 1 \). Simplify to find a single cubic equation: \( 6 r^3 - 6 r + 1 = 0 \).

Step 6: Height Calculation for Given Solutions

The cubic equation \( 6r^3 - 6r + 1 = 0 \) has approximate solutions \( r_1 = 0.172 \) and \( r_2 = 0.903 \). For each solution, find \( h \) using \( h = \frac{1 - r^2}{r} \). For \( r_1 = 0.172 \), \( h = \frac{1 - (0.172)^2}{0.172} \approx 33.67 \). For \( r_2 = 0.903 \), \( h = \frac{1 - (0.903)^2}{0.903} \approx 0.218 \).

Step 7: Interpretation of Solutions

The values of \( r \) and \( h \) represent two possible geometric configurations for the cylindrical pill. A small radius results in a tall cylinder, while a larger radius gives a short cylinder. Both satisfy the volume and surface area conditions.

Key Concepts

Volume of a Sphere

When it comes to calculating the volume of a sphere, the formula to remember is \[ V = \frac{4}{3} \pi r^3 \]This represents how much three-dimensional space a sphere occupies. The variable \( r \) here denotes the radius of the sphere. You can think of the radius as the distance from the center of the sphere out to its surface.
For a sphere with a fixed diameter, the radius is half the diameter.

• The larger the radius, the greater the volume.
• Volume grows rapidly as the radius increases, given the cube in the formula.

In the original problem, the spherical pill has a diameter of 1 cm, which makes the radius 0.5 cm. Substituting this into the formula yields a volume of \( \frac{1}{6} \pi \) cubic centimeters. This computed value is crucial as it sets the volume that the cylindrical pill must match.

Surface Area of a Sphere

The surface area of a sphere gives you an idea of how much external area the sphere covers. The formula used is:\[ A = 4\pi r^2 \]The variable \( r \) still stands for the radius.
Understanding surface area becomes essential when considering anything that involves a coating, paint, or any application spread over the sphere.

• Just like with volume, increasing the radius significantly increases the surface area.
• The surface area depends on the square of the radius, indicating a fast growth in area with increased size.

For our spherical pill with radius 0.5 cm, the surface area calculates to \( \pi \) square centimeters. In the problem, this surface area is used to define the constraints for designing the cylindrical pill.

Volume of a Cylinder

Calculating the volume of a cylinder involves a different set of parameters. You employ the formula:\[ V = \pi r^2 h \]Here, \( r \) represents the radius of the cylinder's base, and \( h \) is the height of the cylinder.
This formula shows how the volume is dependent on both the circular base area \( \pi r^2 \) and the height.

• If either radius or height increases, the volume will increase, highlighting the relationship between these dimensions.
• The base area squared means any small change in radius affects the volume significantly.

Given the volume must match that of the spherical pill, the formula simplifies to \( 6 r^2 h = 1 \) in the original exercise, setting a direct relationship between \( r \), \( h \), and the constrained volume.

Surface Area of a Cylinder

For a cylinder, the surface area is determined by both its circular ends and its curved surface. The formula is:\[ A = 2\pi r (r + h) \]This translates to the area on the top and bottom, given by \( 2\pi r^2 \), and the side area, \( 2\pi r h \).
The surface area represents all the exposed surfaces of the cylinder.

• Since it includes both height and radius, changes in these dimensions will impact the total area.
• In scenarios like the stated exercise, the surface area constraint helps derive specific geometry of the forms.

In our specific problem, the cylinder's surface area must be twice that of the sphere, leading to an equation and eventual solution that guides the dimensions \( r \) and \( h \) such that they satisfy this constraint.