Problem 84 - Calculus Volume 1

Question

For the following exercises, use differentials to estimate the maximum and relative error when computing the surface area or volume. A spherical golf ball is measured to have a radius of \(5 \mathrm{mm},\) with a possible measurement error of 0.1 \(\mathrm{mm} .\) What is the possible change in volume?

Step-by-Step Solution

Answer

The possible change in volume is approximately \( 31.42 \) cubic millimeters.

Step 1: Identify the Formula for Volume of a Sphere

The volume \( V \) of a sphere with radius \( r \) is given by the formula \( V = \frac{4}{3}\pi r^3 \). This formula calculates the volume based on the radius of the sphere.

Step 2: Calculate the Differential of Volume

The differential \( dV \) can be calculated using the derivative of the volume with respect to the radius. That is \( dV = \frac{dV}{dr} \cdot dr \). First, find the derivative: \( \frac{dV}{dr} = 4\pi r^2 \). Thus, \( dV = 4\pi r^2 \cdot dr \).

Step 3: Substitute Known Values into the Differential Equation

Substitute \( r = 5 \) mm and \( dr = 0.1 \) mm into the equation \( dV = 4\pi r^2 \cdot dr \). This becomes \( dV = 4\pi (5)^2 (0.1) \).

Step 4: Calculate the Differential for Change in Volume

Perform the calculations: \[ dV = 4\pi (25)(0.1) = 10\pi \]. Thus, the possible change in volume is approximately \( 31.42 \) cubic millimeters.

Key Concepts

Volume of a Sphere

To understand why a sphere's volume is calculated with the formula \( V = \frac{4}{3}\pi r^3 \), it's helpful to consider the geometry of a sphere. A sphere is a perfectly round object where every point on its surface is equidistant from its center.
The formula \( V = \frac{4}{3}\pi r^3 \) gives the space inside the sphere in terms of its radius \( r \). This cubic factor \( r^3 \) accounts for the three-dimensional space the sphere occupies. The constant \( \frac{4}{3} \pi \) ensures that the volume is a precise measurement based on mathematical derivations from integral calculus.
If you're ever asked to calculate the volume of a sphere, just remember you'll need the radius, and then apply it to this formula to find the total space inside.

Measurement Error

Measurement error can affect any physical measurement, including the size of a sphere such as a golf ball. It is crucial to account for this when making calculations, particularly for volume or surface area.
When given a radius, acknowledging a measurement error means recognizing that the measurement could be off by a small increment, such as \( \pm 0.1 \) mm in our exercise.

• This error suggests that the actual radius could be anywhere between 4.9 mm to 5.1 mm.
• As a result, the calculated volume may vary slightly.

Understanding and accounting for possible measurement error ensures more precise estimations and highlights the importance of tolerance in mathematical and engineering applications.

Calculus Problems

Calculus offers tools to handle variations and changes, especially useful in real-world problems like measuring a sphere's volume with possible errors. These problems often require the use of differentials.
Differentials help us estimate the impact of small changes in measurements. In this exercise, we use the differential \( dV \) to represent the change in the sphere's volume due to a minor deviation in its radius. This method makes calculus a powerful tool in approximating changes without computing an entirely new volume calculation.
When you approach calculus problems involving differentials, consider what small initial changes might mean for the overall calculation. This allows for effective problem-solving in mathematical and practical scenarios.

Derivative

In the context of calculus, a derivative represents an instantaneous rate of change or how a function changes at a specific point. For a sphere's volume, the derivative \( \frac{dV}{dr} \) shows how a slight increase in radius affects the total volume.
To obtain this, differentiate the volume formula \( V = \frac{4}{3} \pi r^3 \) with respect to \( r \), resulting in \( \frac{dV}{dr} = 4\pi r^2 \). This derivative indicates how rapidly the volume changes per unit of change in radius.

• The derivative forms the foundation for calculating the differential \( dV \), which estimates the change in volume given a small change in radius \( dr \).
• Recognizing this connection between a derivative and the practical change it represents is crucial for solving calculus-based problems.

By using derivatives, we can make better predictions about how systems change in response to small inputs, which is invaluable for calculus and beyond.