Problem 37 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

A right circular cylinder is inscribed in a sphere of radius \(5 .\) (a) Express the volume of the cylinder as a function of its radius, \(r\). (b) Express the surface area of the cylinder as a function of its radius, \(r\)

Step-by-Step Solution

Answer

The solution to part (a) is \(V = \pi r^3 \sqrt{5},\) and for part (b) is \(A = 2\pi r^2 (1 + \sqrt{5}).\)

Step 1: Farmer Become Herder

In this exercise, the height (h) of the cylinder corresponds to the diameter (d) of the sphere that inscribes it. Because a diameter equals two radii, we have \(h = 2 \times 5 = 10.\) Now, we can express h in terms of r using the Pythagorean theorem in a right triangle formed by the radius of the sphere, the radius of the base of the cylinder and the height of the cylinder: \(h^2 = r^2 + (2r)^2\). Solving for h gives \(h = \sqrt{4r^2 + r^2} = \sqrt{5r^2}.\)

Step 2: Expressing Cylinder's Volume as a Function of r

The volume of a cylinder is given by \(V = \pi r^2 h.\) We can substitute h from the previous step into this formula, yielding \(V = \pi r^2 \sqrt{5r^2}.\) That simplifies to \(V = \pi r^2 \times r\sqrt{5} = \pi r \sqrt{5r^2}^3. \) Also, after simplifying, we get the volume as a function of r that is \(V = \pi r^3 \sqrt{5}.\)

Step 3: Expressing Cylinder's Surface Area as a Function of r

The surface area of a cylinder is given by \(A = 2\pi r^2 + 2\pi r h.\) As in the previous steps, substitute \(h = \sqrt{5r^2}\) into this formula, yielding \(A = 2\pi r^2 + 2\pi r \sqrt{5r^2}.\) The surface area, thus, as a function of r is \(A = 2\pi r^2 + 2\pi r^2 \sqrt{5} = 2\pi r^2 (1 + \sqrt{5}).\)

Key Concepts

Volume of a Cylinder

The volume of a cylinder is a key measurement indicating how much space is enclosed within the cylinder. To find this volume, you use the formula \[ V = \pi r^2 h, \]where \(V\) represents the volume, \(r\) is the radius of the cylinder's base, and \(h\) is its height. In scenarios where a cylinder is inscribed within a sphere, we derive the height using geometric relationships.Here, when a cylinder is inside a sphere of radius 5, the height of the cylinder also needs to be derived to express volume as a function of its radius. By restructuring the height using the Pythagorean theorem (explained later), we find that \[ h = \sqrt{5r^2}. \]Substituting this into the volume formula gives \[ V = \pi r^2 \sqrt{5r^2} = \pi r^3 \sqrt{5}. \]This relation shows how the volume varies directly with the cube of the radius, considering the transformation of the height in relation to the sphere.

Surface Area of a Cylinder

Explaining the surface area of a cylinder helps us grasp how much material is required to cover the outside of the cylinder completely. The surface area includes both the circular bases and the curved surface. The formula to calculate this is \[ A = 2\pi r^2 + 2\pi r h, \]where \(A\) is the surface area, \(r\) is the radius of the base, and \(h\) is the height.For a cylinder inscribed in a sphere, we already found the height through the Pythagorean theorem to be \(h = \sqrt{5r^2}\). By substituting \(h\) with \(\sqrt{5r^2}\) in the surface area formula, we get \[ A = 2\pi r^2 + 2\pi r \sqrt{5r^2}. \]Simplifying gives \[ A = 2\pi r^2 (1 + \sqrt{5}). \]This expression illustrates how the surface area scales with the radius, acknowledging the special geometric conditions of the inscribed cylinder.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry, relating the lengths of a right triangle's sides. It states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:\[ c^2 = a^2 + b^2, \]where \(c\) is the hypotenuse, and \(a\) and \(b\) are the other sides.In the context of a cylinder inscribed in a sphere, this theorem helps determine the cylinder's height. Imagine the right triangle where:

• One leg is the radius of the cylinder's base, \(r\).
• The other leg is half the height of the cylinder, \(h/2\).
• The hypotenuse is the radius of the sphere, \(5\).

Applying the theorem gives\[ 5^2 = r^2 + \left(\frac{h}{2}\right)^2. \]Solving this helps us express \(h\) as a function of \(r\), leading to\[ h = \sqrt{5r^2}. \]This application demonstrates how the Pythagorean theorem connects geometry with algebra, crucial for solving such problems.