Problem 8 - Calculus: Early Transcendentals

Question

Use differentials to estimate the amount of paint needed to apply a coat of paint \(0.02 \mathrm{~cm}\) thick to a sphere with diameter 40 meters. (Recall that the volume of a sphere of radius \(r\) is \(V=(4 / 3) \pi r^{3}\). Notice that you are given that \(d r=0.02 .)\)

Step-by-Step Solution

Answer

Estimated paint required is \(0.32\pi\) cubic meters.

Step 1: Identify Variables

First, identify the given information and variables. We have a sphere with a diameter of 40 meters. Therefore, the radius \( r \) is \( 20 \) meters since radius is half of the diameter. We also know \( dr = 0.02 \) cm, which needs to be converted to meters for consistency, so \( dr = 0.0002 \) meters.

Step 2: Recall Volume Formula

Recall the formula for the volume of a sphere: \[ V = \frac{4}{3} \pi r^3 \]. This formula gives us the volume in terms of the radius.

Step 3: Differentiate the Volume Formula

Find the differential of the volume \( V \) with respect to \( r \), which is denoted as \( dV \). The differential \( dV \) represents a small change in the volume, which corresponds to the thin layer of paint.:\[ dV = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) \cdot dr = 4 \pi r^2 \, dr \].

Step 4: Substitute Known Values

Substitute the known values for \( r \) and \( dr \) into the differential equations: \[ dV = 4 \pi (20)^2 \times 0.0002 \].

Step 5: Calculate the Differential Volume

Compute the value for \( dV \):\[ dV = 4 \pi \times 400 \times 0.0002 = 4 \pi \times 0.08 = 0.32 \pi \]. This value is the estimated volume of paint needed in cubic meters.

Key Concepts

Volume of a Sphere

Understanding the volume of a sphere is crucial, especially when dealing with problems like estimating the amount of paint needed for coating a sphere. The sphere's volume can be determined using the formula: \[ V = \frac{4}{3} \pi r^3 \]where \( V \) stands for volume, \( r \) is the radius, and \( \pi \) is a constant (approximately 3.14159). This formula essentially calculates the space that the sphere occupies.
In practical applications, knowing the volume helps in understanding how much material will cover the outer surface or how much space it would occupy. For instance, when a sphere has a radius of 20 meters, inserting this value into the formula gives the volume that the sphere covers.

Differential Calculus

Differential calculus is a branch of calculus focused on how functions change. In simpler terms, it helps us understand the relationship between changing variables. The key concepts in differential calculus involve derivatives and differentiating functions.
In the context of this paint problem, the derivative of the sphere's volume with respect to its radius helps us estimate the volume of a thin layer of paint. By differentiating the volume formula:\[ \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \], we calculate the rate of change of volume concerning its radius. Multiplying by \( dr \), the change in radius, gives us the differential \( dV \), a small change in volume.

• Helps calculate small changes in physical quantities.
• Improves estimation of material usage or measurement variations.

This makes differential calculus an essential tool in various applications, especially for precision measurements.

Applications of Calculus

Calculus, including its differential branch, has wide-ranging applications in various fields. It provides the tools to find rates of change and make precise estimations in numerous scenarios.
For example, in engineering, calculus helps engineers design systems and predict behavior under different conditions. In our exercise, calculus helps estimate the volume of paint needed to cover a sphere. By calculating the differential \( dV \), one can accurately determine the paint volume needed for a sphere's thin coating.

• Facilitates optimal resource allocation.
• Helps in designing and predicting changes in physical systems.
• Used extensively in scientific research and data analysis.

Thus, the ability of calculus to aid in making precise and accurate calculations is vital, contributing significantly not only in mathematics but in practical, real-world applications.