Problem 35 - Calculus

Question

A closed cylindrical can of fixed volume \(V\) has radius \(r\) (a) Find the surface area, \(S,\) as a function of \(r\) (b) What happens to the value of \(S\) as \(r \rightarrow \infty ?\) (c) Sketch a graph of \(S\) against \(r,\) if \(V=10 \mathrm{cm}^{3}\)

Step-by-Step Solution

Answer

(a) \( S(r) = \frac{2V}{r} + 2\pi r^2 \); (b) As \( r \rightarrow \infty \), \( S(r) \rightarrow 2\pi r^2 \), so it increases indefinitely. (c) At \( V=10 \), sketch shows an initial decrease and then a sharp increase in surface area.

Step 1: Understand the Problem

We have a cylindrical can with a fixed volume and a variable radius. We need to express the surface area of the cylinder as a function of the radius and analyze its behavior as the radius changes.

Step 2: Use the Volume Formula

The volume of the cylinder is given by the formula \( V = \pi r^2 h \). Since the volume \( V \) is fixed, we can express the height \( h \) in terms of the radius \( r \): \( h = \frac{V}{\pi r^2} \).

Step 3: Derive the Surface Area Formula

The surface area of a closed cylinder includes the lateral surface area and two circular ends. The formula for surface area is \( S = 2\pi r h + 2\pi r^2 \). Substitute \( h = \frac{V}{\pi r^2} \) into the surface area formula to get:\[ S = 2\pi r \left(\frac{V}{\pi r^2}\right) + 2\pi r^2 = \frac{2V}{r} + 2\pi r^2 \]

Step 4: Simplify Surface Area as a Function of r

Now we have the surface area as a function of the radius: \( S(r) = \frac{2V}{r} + 2\pi r^2 \). This function will help us find the behavior of the surface area as \( r \) changes.

Step 5: Analyze S(r) as r Approaches Infinity

As \( r \rightarrow \infty \), the term \( \frac{2V}{r} \rightarrow 0 \) because the volume term becomes negligible. The term \( 2\pi r^2 \) will dominate, so \( S(r) \sim 2\pi r^2 \), which means the surface area increases indefinitely as the radius increases.

Step 6: Sketch the Graph for V=10 cm³

Substitute \( V = 10 \) cm³ into the surface area function to sketch the graph: \( S(r) = \frac{20}{r} + 2\pi r^2 \). At small values of \( r \), the graph is dominated by the \( \frac{20}{r} \) term, which decreases as \( r \) increases, while the \( 2\pi r^2 \) term increases and dominates as \( r \) becomes large. The graph should reflect this behavior, showing a decrease at first and then increasing sharply.

Key Concepts

Volume of Cylinder

The volume of a cylinder is an important concept in geometry, and it is calculated using a specific formula. For a cylinder, the volume formula is given by: \[ V = \pi r^2 h \]where:

• \( V \) is the volume of the cylinder,
• \( r \) is the radius of the circular base,
• \( h \) is the height of the cylinder.

This formula expresses how the size of the base and the height contribute to the total capacity of the cylinder.
To maintain a certain volume, adjustments made to either the radius or height will require compensations to the other dimension. For instance, with a fixed volume and an increasing radius, the height must decrease to keep the volume consistent, as shown with the derived formula for height: \( h = \frac{V}{\pi r^2} \).
It's vital to grasp this relationship to understand the impact of changing dimensions on a cylinder's volume.

Function Analysis

Function analysis involves examining how a mathematical function behaves with changing variables. Here, the surface area, \( S \), of a cylinder is expressed as a function of its radius, \( r \): \[ S(r) = \frac{2V}{r} + 2\pi r^2 \]To understand its behavior, consider the function's terms:

• The term \( \frac{2V}{r} \) represents the influence of the cylinder's fixed volume on the surface area. As the radius \( r \) becomes large, this term approaches zero because division by a larger radius diminishes this part's contribution.
• The term \( 2\pi r^2 \) models the contribution of the curved surface and the two ends of the cylinder. As the radius increases, this square term grows significantly.

Function analysis helps us to predict that as \( r \) increases to infinity, the surface area \( S \) becomes dominated by the \( 2\pi r^2 \) term. Thus, the surface area increases without bound. Understanding these behaviors is crucial when working with such functions in calculus and geometry.

Graph Sketching

Graph sketching showcases the visual behavior of a mathematical function. When we sketch the graph of the function \( S(r) = \frac{2V}{r} + 2\pi r^2 \), we observe how the surface area changes as the radius varies.
To make a meaningful graph, it's helpful to note the behavior at key points:

• At very small values of \( r \), the \( \frac{2V}{r} \) term is significant, causing the surface area to start high.
• As \( r \) increases, this term rapidly diminishes, and the \( 2\pi r^2 \) term takes over, causing the graph to dip initially, then sharply rise.

When \( V = 10 \text{ cm}^3 \), these behaviors reflect in the curve. It signifies a steep drop initially followed by a dramatic upward trajectory as \( r \) continues to grow.
Sketching this graph allows us to better visualize the transition of domination between these two components of the function, providing a powerful tool for analysis and prediction in mathematical modeling.