Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

A cylindrical roller is exactly 12 inches long and its diameter is measured as \(6 \pm 0.005\) inches. Calculate its volume with an estimate for the absolute error and the relative error.

Step-by-Step Solution

Answer

The volume is \( 108\pi \), absolute error is \( 0.2827 \), and relative error is \( 0.000833 \).

Step 1: Understand the problem

We need to find the volume of a cylinder with a given length and an uncertain diameter, and estimate both the absolute and relative error of this volume calculation.

Step 2: Calculate the exact volume

The formula for the volume of a cylinder is given by \[ V = \pi r^2 h \]where \( r \) is the radius and \( h \) is the height (or length) of the cylinder. The radius \( r \) is half of the diameter, so\[ r = \frac{6}{2} = 3 \text{ inches} \]The volume then becomes:\[ V = \pi (3)^2 (12) = 108\pi \text{ cubic inches} \]

Step 3: Calculate the partial derivative with respect to diameter

We approximate the change in volume based on changes in diameter. Using the formula\[ \frac{dV}{dD} = \frac{\partial}{\partial D}(\pi (D/2)^2 h) \]Substitute \( D = 6 \) and \( h = 12 \):\[ \frac{dV}{dD} = \pi \times 2 \times \frac{D}{2} \times h \times \frac{1}{2} = \pi \times 3 \times 12 = 18\pi \]So the derivative\[ \frac{dV}{dD} = 18\pi \]

Step 4: Estimate the absolute error

The formula for absolute error is given by \[ \Delta V = \frac{dV}{dD} \times \Delta D \]Substitute the values:\[ \Delta D = 0.005 \]\[ \Delta V = 18\pi \times 0.005 = 0.09\pi \approx 0.2827 \text{ cubic inches} \]

Step 5: Estimate the relative error

The relative error is the ratio of the absolute error to the original volume:\[ \text{Relative Error} = \frac{\Delta V}{V} \]Substitute the values:\[ \text{Relative Error} = \frac{0.09\pi}{108\pi} \approx \frac{0.09}{108} \approx 0.000833 \]

Step 6: Conclusion

The calculated volume is \( 108\pi \) cubic inches, with an absolute error of approximately \( 0.2827 \) cubic inches and a relative error of approximately \( 0.000833 \).

Key Concepts

Cylinder Volume Calculation

In the world of geometry and calculus, calculating the volume of a cylinder is a fundamental exercise. To find the volume of a cylinder, we 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 the height (or length) of the cylinder. In our example, the diameter is given as 6 inches, and the length is 12 inches. The radius is half of the diameter, so:

• \( r = \frac{6}{2} = 3 \) inches

Using these values, the volume is calculated as:

• \( V = \pi (3)^2 (12) = 108\pi \) cubic inches

Understanding how to manipulate shapes into calculable formulas is a central part of geometry within calculus.

Error Estimation in Measurements

In many real-world scenarios, measurements are not exact, leading to potential errors in calculations. Here, the diameter of the cylinder has an uncertainty, given as \(6 \pm 0.005\) inches. It's necessary to determine the potential errors this uncertainty can introduce to our volume calculation.
Absolute error is calculated using:

• \( \Delta V = \frac{dV}{dD} \times \Delta D \)

In our example, with \( \Delta D = 0.005 \), the absolute error is:

• \( \Delta V = 18\pi \times 0.005 \approx 0.2827 \) cubic inches

Relative error compares the absolute error to the original volume:

• \( \text{Relative Error} = \frac{0.09\pi}{108\pi} \approx 0.000833 \)

Understanding these errors helps in evaluating the accuracy and reliability of our measurements.

Partial Derivatives

Partial derivatives are a critical concept in calculus when functions depend on multiple variables. They allow us to approximate changes in a function with respect to its inputs. In this exercise, the volume of the cylinder is influenced by changes in diameter. Using a partial derivative, we differentiate the volume \( V \) with respect to diameter \( D \):

• \( \frac{dV}{dD} = \frac{\partial}{\partial D}(\pi \left(\frac{D}{2}\right)^2 h) \)

which simplifies to \( 18\pi \) when substituting \( D = 6 \) and \( h = 12 \). This process allows us to estimate how small changes in \( D \) will affect \( V \). It's a useful tool for error estimation in measurements.

Geometry in Calculus

The interplay between geometry and calculus is prominent in problems involving shapes like cylinders. Calculus provides tools like derivatives and integrals that give us deeper insights and predictive capabilities. Here, the geometric properties of the cylinder (diameter and length) are seamlessly integrated into calculus concepts to solve for volume and error estimation.
Geometry helps visualize the problem while calculus allows us to manipulate those visual elements to find concrete numerical values. Using geometric shapes with calculus principles bridges abstract mathematical theories with practical real-world applications.

• Calculating over curves and surfaces.
• Estimating changes in measurements.
• Integrating calculus into everyday geometry problems.

By combining these disciplines, we form a profound understanding of not only shapes but how they behave under change or approximation.