Problem 7 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Consider the solid right cylinder with a xed height of 10 inches and a variable radius. Let \(V(r)\) be the volume of the cylinder as a function of \(r\), the radius, given in inches. Interpret \(d V / d r\) geometrically. Explain why your answer makes sense by looking at \(\Delta V\) geometrically.

Step-by-Step Solution

Answer

The derivative \(d V / d r\) represents the rate of change of the volume with respect to the radius, and geometrically, it is equivalent to the volume of a cylindrical shell with small thickness \(\Delta r\). This makes sense because as the radius changes by a small amount, the change in volume corresponds to the volume of this thin cylindrical shell.

Step 1: Establish the formula for volume

The volume of a cylinder is given by the formula \(V = \pi r^2h\), where \(r\) is the radius and \(h\) is the height. Here, \(h = 10\) is a constant, and the volume is being considered as a function of the radius \(r\), so we can write the formula as \(V(r) = \pi r^2 \cdot 10\).

Step 2: Calculate the derivative of V with respect to r

The derivative of a function gives the rate of change of that function. In this case, we want to find \(d V / d r\), which represents the rate of change of volume with respect to the radius. Using the power rule for differentiation, where the derivative of \(x^n\) is \(n x^{n-1}\), we find that \(d V / d r = 2 \pi r \cdot 10\). This tells us how much the volume changes for a small change in the radius.

Step 3: Interpret \(d V / d r\) geometrically

Geometric interpretation: The derivative \(d V / d r\), is equivalent to the volume of a cylindrical shell with small thickness \(\Delta r\). As the radius changes by a small amount, the increase in volume is equivalent to the volume of this thin cylindrical shell.

Step 4: Explain why the answer makes sense by looking at \(\Delta V\) geometrically

As the radius \(r\) increases by a small amount \(\Delta r\), the change in volume \(\Delta V\) corresponds to the volume of a thin cylindrical shell with radius \(r\) and thickness \(\Delta r\). So, \(d V / d r\), which is the rate of change of the volume with respect to the radius, makes sense because it is simply adding the volume of these thin cylindrical shells caused by the small increases in the radius.

Key Concepts

Cylinder Volume Calculation

Understanding the volume of a cylinder is fundamental in calculus differentiation and geometry. A cylinder consists of a circular base and a fixed height. To calculate its volume, one uses the formula: \[ V = \pi r^2 h \]where:

• \(r\) is the radius of the cylinder's base.
• \(h\) is the height of the cylinder.
• \(\pi\) is approximately 3.14159.

For our specific problem, the height \(h\) is constant at 10 inches. Therefore, the volume function concerning the radius \(r\) is written as:\[ V(r) = \pi r^2 \cdot 10 \]This implies that the volume changes dynamically with different values of \(r\). By understanding this formula, you can grasp how altering the radius impacts the overall volume of the cylinder.

Rate of Change Interpretation

The rate of change is a crucial concept in calculus, showing how one quantity changes in response to another. For this cylinder example, we are interested in how the volume \(V(r)\) changes when the radius \(r\) experiences a small alteration. We express this rate of change as \(\frac{dV}{dr}\).
When differentiating \(V(r) = 10 \pi r^2\) with respect to \(r\), we apply the power rule, giving us:\[ \frac{dV}{dr} = 20 \pi r \]This result tells us how sensitively the volume responds to a change in the radius.

• For instance, an increase in \(r\) leads to a proportional increase in volume.
• The factor \(20 \pi r\) provides the exact rate of volume increase for each unit change in \(r\).

Understanding \(\frac{dV}{dr}\) offers insights into optimizing designs where space or material efficiency is critical.

Geometric Interpretation of Derivatives

The geometric interpretation of derivatives provides a visual understanding of changes in shapes and sizes. In this exercise, \(\frac{dV}{dr}\) represents the derivative that outlines the change in volume for an incremental increase in radius.
Consider a thin cylindrical shell formed by increasing the radius slightly by \(\Delta r\). The volume of this shell is characterized by the formula for the derivative:\[ \Delta V \approx 2 \pi r \cdot 10 \cdot \Delta r \]This equation implies:

• The thickness of the shell is \(\Delta r\).
• The volume of the shell is proportional to its thickness and the current radius.

Thus, \(\frac{dV}{dr} = 20 \pi r\) shows that this derivative not only calculates a numeric rate but translates into an actual geometric layer added to the cylinder's surface. It makes sense, as each small increase in \(r\) extends the body of the cylinder by this distinct volume.