Problem 32 - University Calculus: Early Transcendentals

Question

What is the rate of change of the volume of a ball \(\left(V=(4 / 3) \pi r^{3}\right)\) with respect to the radius when the radius is \(r=2 ?\)

Step-by-Step Solution

Answer

The rate of change is \(16\pi\) when the radius is 2.

Step 1: Understanding the Problem

We are asked to find the rate of change of the volume of a sphere with respect to its radius. This requires us to find the derivative of the volume formula with respect to the radius.

Step 2: Writing the Volume Formula

The volume of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3 \]

Step 3: Calculating the Derivative

To find the rate of change, we need to differentiate the volume with respect to the radius, \( r \). The derivative \( \frac{dV}{dr} \) is: \[ \frac{dV}{dr} = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) \] Using the power rule \( \frac{d}{dr}(r^n) = n \cdot r^{n-1} \), the derivative becomes: \[ \frac{dV}{dr} = \frac{4}{3} \pi \cdot 3r^2 = 4 \pi r^2 \]

Step 4: Substitute the Radius Value

Now that we have \( \frac{dV}{dr} = 4 \pi r^2 \), substitute \( r = 2 \) to find the specific rate of change: \[ \frac{dV}{dr} = 4 \pi (2)^2 = 4 \pi \cdot 4 = 16 \pi \]

Step 5: Interpreting the Result

When \( r = 2 \), the rate of change of the volume with respect to the radius is \( 16 \pi \). This means that at a radius of 2 units, the volume increases by \( 16 \pi \) cubic units for each unit increase in the radius.

Key Concepts

Volume of a Sphere

The volume of a sphere is a measure of the space it occupies in three dimensions. To calculate this volume, we use a formula that involves the radius of the sphere, a line segment from the center to any point on the surface. Using the formula

• \[ V = \frac{4}{3} \pi r^3 \]

we can find the volume by substituting for \(r\), the radius.

This formula shows a direct relationship between the radius and the volume. If the radius increases, the volume does too, but it happens quite rapidly due to the radius being cubed. This relationship highlights the nonsimple manner in which small changes in the radius can significantly affect the sphere's volume.

Differentiation

Differentiation is a fundamental concept in calculus that involves calculating how a function changes at any point. When differentiating a function, you are computing its derivative, which represents the function's rate of change. For the volume of a sphere, we're interested in how its volume changes as the radius changes. This requires us to find the derivative of the volume function with respect to the radius.

For example, if you have the volume of a sphere function

• \( V = \frac{4}{3} \pi r^3 \)

and you want to know how the volume changes as \( r \) changes, you'll differentiate this function. Differentiation will provide a mathematical expression that tells you exactly how much the volume changes for each small change in \( r \).

Derivative

The derivative is a mathematical tool used to measure how a function's output changes as its input changes. In our context, the derivative of the volume of a sphere function,

• \( \frac{dV}{dr} = 4 \pi r^2 \)

tells us the rate of change of volume with respect to the radius. This expression means that for every unit increase in radius, the volume increases by \(4\pi r^2\) cubic units.

Understanding how to calculate a derivative and applying it to the volume function involves using rules like the power rule, which helps simplify finding derivatives of polynomial terms. This rule states that for any \(r^n\), its derivative is \(n \cdot r^{n-1}\).

Using the power rule makes it straightforward to differentiate functions and find rates, such as in our sphere example where knowing the derivative helps understand how even a tiny change in radius could drastically change the sphere's volume.

Geometry

Geometry is vital in understanding the spatial relations and properties of objects like spheres. A sphere is a perfectly symmetrical geometric shape, where every point on its surface is the same distance from its center.

In problems like finding the rate of change of volume, knowing the geometric properties of a sphere is essential. By understanding how volume, surface area, and radius interact, we are better prepared to set up and solve calculus problems.

One crucial geometric aspect is realizing how drastically the volume can change with slight radius adjustments, which a linear appreciation of geometric shapes might overlook. In contrast to straightforward shapes like cubes, a sphere showcases how calculus and geometry overlap—revealing that geometry requires more than visual perception; it also needs mathematical understanding.