Problem 45 - University Calculus: Early Transcendentals

Question

Write a differential formula that estimates the given change in volume or surface area. The change in the volume \(V=(4 / 3) \pi r^{3}\) of a sphere when the radius changes from \(r_{0}\) to \(r_{0}+d r\)

Step-by-Step Solution

Answer

The change in volume is estimated by \( dV = 4 \pi r_0^2 \cdot dr \).

Step 1: Identify Variables and Formula

We are given the volume formula for a sphere, which is \( V = \frac{4}{3} \pi r^3 \). The task is to find a differential formula to estimate the change in volume when the radius changes from \( r_0 \) to \( r_0 + dr \).

Step 2: Differentiate the Volume Formula

To estimate the change in volume, we need the differential of the volume function. Find the derivative of \( V \) with respect to \( r \), which is \( \frac{dV}{dr} = 4 \pi r^2 \).

Step 3: Apply the Differential Formula

The differential change in volume, \( dV \), is given by \( dV = \frac{dV}{dr} \cdot dr \). Substitute the derivative we found: \( dV = 4 \pi r^2 \cdot dr \).

Step 4: Substitute the Initial Radius

To find the change in volume at an initial radius \( r_0 \), substitute \( r_0 \) into the differential formula: \( dV = 4 \pi r_0^2 \cdot dr \). This formula estimates the change in volume for a small change \( dr \) in radius.

Key Concepts

Derivative

Understanding derivatives is crucial in differential calculus. The derivative of a function represents the rate at which the function's value changes as its input changes. Think of it like this: if you have a graph of a function, the derivative at a point is the slope of the tangent line to the graph at that point.

In the context of our problem, we are working with the volume of a sphere, given by the formula \( V = \frac{4}{3} \pi r^3 \). To find how the volume changes as the radius changes, we need to take the derivative of this volume function with respect to the radius \( r \).

• The derivative of \( V \) with respect to \( r \) is \( \frac{dV}{dr} = 4 \pi r^2 \).
• This derivative tells us how fast the volume \( V \) increases with a tiny change in the radius \( r \).

The derivative makes it easy to estimate small changes in the volume when you know small changes in the radius.

Volume of a Sphere

The formula for the volume of a sphere is intrinsically related to its radius. The formula is given by \( V = \frac{4}{3} \pi r^3 \). Let's break it down a bit:

• \( V \) stands for volume, which is the amount of space inside the sphere.
• \( \pi \) is a constant approximately equal to 3.14159, crucial in calculations involving circles and spheres.
• \( r \) is the radius, or the distance from the center of the sphere to any point on its surface.

The volume of a sphere increases dramatically with even a small increase in the radius because the radius is cubed in the formula. This means that the change in volume is not linear; instead, it accelerates as the radius grows. Understanding how the volume grows with the radius helps us see why the derivative is necessary to capture these changes accurately.

Differential Formula

The differential formula estimates the change in a function's output as its input changes. In our problem, we are trying to estimate how the volume of a sphere changes as its radius changes slightly. We already found the derivative to be \( \frac{dV}{dr} = 4 \pi r^2 \).

With the derivative, the differential formula can be used:

• Start with \( dV = \frac{dV}{dr} \cdot dr \), where \( dV \) represents the estimated change in volume and \( dr \) is the change in the radius.
• Substitute \( \frac{dV}{dr} \) with \( 4 \pi r^2 \) to make the formula \( dV = 4 \pi r^2 \cdot dr \).
• To estimate the change at a specific radius, replace \( r \) with the initial radius, \( r_0 \), when using this formula.

This approach provides a powerful way to estimate how slight changes in the radius affect the volume, enabling better predictions and understanding in practical applications.