Question

The radius of a sphere is measured to be 6 inches, with a possible error of \(0.02\) inch. Use differentials to approximate the possible error and the relative error in computing the volume of the sphere.

Short answer

The possible error in volume is found to be approximately \(0.45\) cubic inches and the relative error is approximately \(0.0075\), or \(0.75\% \).

Step-by-step solution

Differentiating the volume function

Start by differentiating the volume function with respect to \(r\). The derivative is \(V' = 4\pi r^2\).

Calculate the differential

Replace \(dr\) with the given possible error in radius measurement, \(0.02\) inch. The differential \(dV = V' dr = 4\pi r^2 dr\). Now substitute \(r = 6\) inches into \(dV\) to get the approximate error in the volume.

Calculate the Relative error

The relative error is given by the formula \( \frac{dV}{V}\). Substitute the values of \(dV\) and \(V\) with the values calculated in the previous steps to find the relative error.

Key concepts

Volume of a Sphere

Understanding the volume of a sphere is essential when dealing with three-dimensional objects. A sphere, like a basketball or a bubble, can be defined as a perfectly round object in the space, where every point on its surface is equidistant from its center. The volume of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
Here, \(V\) represents the volume, \(\pi\) is a mathematical constant (approximately 3.14159), and \(r\) is the radius of the sphere, notated as the distance from the center to any point on the surface.

• The exponent "3" signifies that the formula accounts for all three dimensions of the sphere.
• The constant \(\frac{4}{3}\) is specific to the geometry of spheres to ensure the formula yields the correct volume measurement.

Knowing this formula helps in calculating how much space the sphere occupies, which is crucial in fields like engineering, physics, and even daily tasks like calculating storage space.

Error Approximation

When measurements are involved, particularly in science and engineering, it's common to have minor errors due to various factors. Error approximation helps us understand how these small changes or errors in measurement can affect our calculations.
Using differentials, a mathematical concept in calculus, is a handy way to approximate the error. For a sphere’s volume, if the radius measurement has a possible error (\(dr\)), the differential \(dV\) represents how the volume will 'change' potentially due to this error. Here’s how it works:

• Start with the derivative of the volume formula related to the radius \(r\): \(V' = 4\pi r^2\).
• The differential or the approximate change in volume is \(dV = V' dr = 4\pi r^2 dr\).

This calculation allows us to predict how much the volume might increase or decrease in response to a small change in the radius. A practical use example would be assessing manufacturing tolerances where exact measurements are challenging.

Relative Error

Relative error gives us an insight into how significant a measurement error is, relative to the actual value of measurement. It’s a vital concept because it’s not just about how large the error is, but how it compares to the actual size of what you’re measuring.
The formula for relative error in the context of a sphere's volume is:
\[ \text{Relative Error} = \frac{dV}{V} \]

• \(dV\) is our differential, showing the approximate error in the volume.
• \(V\) is the regular volume calculated using the original measurement.

By dividing the approximate error \(dV\) by the actual volume \(V\), you get a ratio that helps you understand the error in a broader context. A small relative error suggests that the error is minor compared to the actual measurement, which often means our measurement is quite reliable. It's the perfect tool for reporting precision in scientific data or verifying the accuracy of technical designs.