Problem 2 - The Calculus with Analytic Geometry

Question

Suppose a right-circular cylinder has a constant height of \(10.00\) in. If \(V\) in. \({ }^{3}\) is the volume of the right-circular cylinder, and \(r\) in. is the radius of its base, find the average rate of change of \(V\) with respect to \(r\) as \(r\) changes from (a) \(5.00\) to \(5.40\); (b) \(5.00\) to 5.10; (c) \(5.00\) to 5.01. (d) Find the instantaneous rate of change of \(V\) with respect to \(r\) when \(r\) is \(5.00\).

Step-by-Step Solution

Answer

The average rates of change for (a), (b), and (c) are \( 104\pi \), \( 101\pi \), and \( 100.1\pi \) respectively. The instantaneous rate of change when \( r = 5.00 \) is \( 100\pi \).

Step 1: Define the volume formula for the cylinder

The volume of a right-circular cylinder is given by the formula: \[ V = \pi r^2 h \] where \( V \) is the volume, \( r \) is the radius of the base, and \( h \) is the height. Given that height \( h = 10.00 \) inches.

Step 2: Substitute height into the volume formula

Substitute \( h = 10.00 \) into the volume formula:\[ V = \pi r^2 \times 10.00 \]So, the volume formula simplifies to:\[ V = 10\pi r^2 \]

Step 3: Calculate average rate of change for (a)

The average rate of change of \( V \) with respect to \( r \) from \( r = 5.00 \) to \( r = 5.40 \):\[ \ \Delta V = V(5.40) - V(5.00) \]\[ V(5.40) = 10\pi (5.40)^2 \] and \[ V(5.00) = 10\pi (5.00)^2 \]\[ \ \Delta r = 5.40 - 5.00 = 0.40 \]So the average rate of change is\[ \ \text{Average rate of change} = \ \frac{V(5.40) - V(5.00)}{5.40 - 5.00} \]

Step 4: Calculate specific values for (a)

Calculate volumes:\[ V(5.40) = 10\pi (5.40)^2 = 10\pi \times 29.16 = 291.6\pi \]\[ V(5.00) = 10\pi (5.00)^2 = 10\pi \times 25 = 250\pi \]So,\[ \ \Delta V = 291.6\pi - 250\pi = 41.6\pi \]\[ \ \text{Average rate of change} = \ \frac{41.6\pi}{0.40} = 104\pi \]

Step 5: Repeat for (b) and (c)

(b) From \( r = 5.00 \) to \( r = 5.10 \):\[ \ \Delta V = V(5.10) - V(5.00) = 10\pi (5.10)^2 - 10\pi (5.00)^2 \]\[ V(5.10) = 10\pi (5.10)^2 = 10\pi \times 26.01 = 260.1\pi \]\[ \ \Delta V = 260.1 \pi - 250\pi = 10.1 \pi \]\[ \text{Average rate of change} = \ \frac{10.1\pi}{0.10}= 101\pi \](c) From \( r = 5.00 \) to \( r = 5.01 \):\[ \ \Delta V = V(5.01) - V(5.00) = 10\pi (5.01)^2 - 10\pi (5.00)^2 \]\[ V(5.01) = 10\pi (5.01)^2 = 10\pi \times 25.1001 = 251.001\pi \]\[ \ \Delta V = 251.001 \pi - 250\pi = 1.001 \pi \]\[ \text{Average rate of change} = \ \frac{1.001\pi}{0.01}= 100.1\pi \]

Step 6: Calculate instantaneous rate of change for (d)

The instantaneous rate of change of \( V \) with respect to \( r \) is given by the derivative of \( V \):\[ \frac{dV}{dr} = \frac{d}{dr}(10\pi r^2) \]Using the power rule:\[ \frac{dV}{dr} = 10\pi \times 2r = 20\pi r \]When \( r = 5.00 \):\[ \frac{dV}{dr} = 20\pi \times 5.00 = 100\pi \]

Key Concepts

Volume of Cylinder

The volume of a cylinder is an important fundamental concept in geometry and calculus. For a right-circular cylinder, the volume (V) is given by the formula: \[ V = \pi r^2 h \] where \( V \) is the volume, \( r \) is the radius of the cylinder's base, and \( h \) is the height of the cylinder. In this exercise, we are dealing with a cylinder that has a constant height of \(10\) inches. We substitute this height into the volume formula to simplify it:\[ V = 10\pi r^2 \]This formula tells us that the volume of the cylinder changes as the square of the radius of its base changes. Knowing this relation helps understand how the volume responds to changes in the radius.

Average Rate of Change

The average rate of change of a function measures how a quantity changes on average between two points. It is calculated as the difference in the function values at those two points divided by the difference in the points themselves. For our cylinder's volume, we want to find out how the volume changes as the radius (\( r \)) changes from one value to another. Mathematically, if \( r \) changes from \( r_1 \) to \( r_2 \), the average rate of change of the volume (\( V \)) is given by:\[ \text{Average rate of change} = \frac{V(r_2) - V(r_1)}{r_2 - r_1} \]For the specific intervals given in the exercise, such as from \(r = 5.00\) to \(r = 5.40\), we calculate the volumes at these radii and then use the formula. This provides a deeper understanding of how much the volume changes on average with respect to changes in the radius.

Differentiation

Differentiation is a key concept in calculus that allows us to find the instantaneous rate of change of a function. It extends the idea of finding the average rate of change by making the interval between two points infinitesimally small. If we have a function V(r) representing the volume of our cylinder, we can find the instantaneous rate of change of V with respect to \( r \) by taking the derivative of V. For the volume formula \( V = 10\pi r^2 \), its derivative with respect to \( r \), denoted as \( \frac{dV}{dr} \), is found using differentiation rules. Specifically, using the power rule:\[ \frac{dV}{dr} = 10\pi \cdot 2r = 20\pi r \]This derivative tells us how the volume changes instantaneously with respect to changes in the radius. For example, when \(r = 5.00\), we substitute to find \( \frac{dV}{dr} = 20\pi \cdot 5 = 100\pi \).

Power Rule

The power rule is a fundamental differentiation tool in calculus used to find the derivative of functions of the form \( f(x) = x^n \). According to the power rule, the derivative of \( f(x) \) with respect to \( x \) is \( f'(x) = nx^{n-1} \).Consider our volume function for the cylinder: \( V = 10\pi r^2 \). To find \( \frac{dV}{dr} \), we can apply the power rule. Here,\(n = 2\) and the constant \( 10\pi \) is a coefficient that remains unaffected by differentiation:\[ \frac{dV}{dr} = 10\pi \cdot 2r^{2-1} = 20\pi r \]This application of the power rule simplifies the process of finding how the volume changes with respect to the radius. It's a crucial skill when handling polynomial functions and helps understand variations in physical quantities like volume.