Question

The radius of a sphere is measured to be about 10 inches. Determine a tolerance \(\delta\) in this measurement that will ensure an error of less than \(0.01\) square inch in the calculated value of the surface area of the sphere.

Short answer

A tolerance \( \delta \approx 3.98 \times 10^{-5} \) inches ensures less than 0.01 square inch surface area error.

Step-by-step solution

Recall the formula for the surface area of a sphere

The surface area \( A \) of a sphere with radius \( r \) is given by the formula \( A = 4\pi r^2 \). This formula will be used to determine the relationship between changes in radius and changes in surface area.

Use Differentiation to Find Change in Surface Area

To find how a small change in radius \( \delta r \) affects the surface area, we differentiate \( A = 4\pi r^2 \) with respect to \( r \). This gives us \( \frac{dA}{dr} = 8\pi r \). Hence, the approximate change in area \( \delta A \) is \( \delta A = 8\pi r \cdot \delta r \).

Substitute Known Values and Solve for \( \delta r \)

We know that \( |\delta A| < 0.01 \). Substituting \( r = 10 \) inches and the expression for \( \delta A \) gives \( |8\pi \cdot 10 \cdot \delta r| < 0.01 \). Simplifying, we find \( 80\pi |\delta r| < 0.01 \).

Solve for Tolerance \( \delta r \)

To find \( \delta r \), we solve the inequality: \( |\delta r| < \frac{0.01}{80\pi} \). This simplifies to \( \delta r < \frac{0.01}{251.327} \approx 3.98 \times 10^{-5} \).

Conclusion

Therefore, the tolerance in the measurement of the radius that will ensure an error of less than 0.01 square inch in the surface area is approximately \( 3.98 \times 10^{-5} \) inches.

Key concepts

Surface Area Calculation

Understanding how to calculate the surface area of a sphere is a foundational aspect of geometry. The surface area of a sphere with radius \( r \) is calculated using the formula \( A = 4\pi r^2 \). This formula essentially tells us that the surface area is proportional to the square of the radius. If we increase the radius, the surface area increases significantly. For example, doubling the radius will quadruple the surface area. This relationship highlights the importance of precise measurements when calculating surface areas for spheres. In practical applications, knowing the formula allows us to predict changes in surface areas when the radius is slightly altered, which is crucial when small tolerances are necessary.

Measurement Error Analysis

Anytime we measure objects in the real world, there can be small inaccuracies or "errors." In the context of calculating surface areas, even a tiny error in measuring the radius of a sphere can lead to a significant error in the calculated surface area. Measurement error analysis helps us to estimate how these small deviations in measurements affect the outcome. For instance:

• In our exercise, the task is to determine a radius measurement error that results in an error of less than \(0.01\) square inch in the surface area.
• We approach this by setting up an inequality using the derived formula for surface area changes: \( |8\pi r \cdot \delta r| < 0.01 \).

This analysis leads us to conclude the tolerance level for measuring the radius, ensuring precise surface area calculations.

Differential Approximation

Differential approximation is a mathematical concept used to estimate how small changes in inputs lead to changes in outputs. In simple terms, it helps us understand the ripple effect of tiny variations. In the given problem, we use differential approximation to find the approximate change in the surface area with a small change in the radius. After differentiating the surface area formula \( A = 4\pi r^2 \), we get \( \frac{dA}{dr} = 8\pi r \). This derivative tells us how the surface area changes with a small change in the radius. By multiplying the derivative by the small change \( \delta r \), we approximate the change in surface area \( \delta A = 8\pi r \cdot \delta r \). This approach is essential for cases where precision is key, allowing us to control and limit the error in our calculations, like ensuring the error in surface area stays under \(0.01\) square inches.