Problem 20 - Calculus Volume 1

Question

For the following exercises, draw and label diagrams to help solve the related-rates problems. The radius of a sphere increases at a rate of 1 m/sec. Find the rate at which the volume increases when the radius is 20 m.

Step-by-Step Solution

Answer

The volume of the sphere increases at approximately 5026.55 cubic meters per second when the radius is 20 m.

Step 1: Understand the problem

We need to find the rate at which the volume of a sphere increases (\(\frac{dV}{dt}\)) given that the radius (\(r\)) increases at 1 m/sec, and we need the rate at a particular radius (20 m).

Step 2: Write down known formulas

The formula for the volume of a sphere is \(V = \frac{4}{3}\pi r^3\). We are given \(\frac{dr}{dt} = 1\) m/sec and need to find \(\frac{dV}{dt}\) when \(r = 20\) m.

Step 3: Differentiate the volume formula with respect to time

Differentiate both sides of the volume formula with respect to time \(t\): \(\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = 4\pi r^2 \frac{dr}{dt}\). This uses the chain rule.

Step 4: Substitute known values into the differentiated formula

Substitute \(r = 20\)and \(\frac{dr}{dt} = 1\) into the differentiated equation: \(\frac{dV}{dt} = 4\pi (20)^2 (1) = 1600 \pi\).

Step 5: Final calculation

Calculate the final numeric answer for \(\frac{dV}{dt}\): \(\frac{dV}{dt} = 1600 \pi \approx 5026.55\) cubic meters per second.

Key Concepts

Sphere Volume

The volume of a sphere is a three-dimensional measure of how much space it occupies. Spheres are perfectly round, and their volume is calculated via a precise geometric formula. This formula is expressed as:

• \( V = \frac{4}{3}\pi r^3 \)

Here, \( V \) stands for volume, \( r \) is the radius of the sphere, and \( \pi \) is a mathematical constant approximately equal to 3.14159. When calculating the volume of a sphere, it is crucial to remember that all dimensions must be in consistent units, typically meters in practical applications.

The formula shows that the volume is directly related to the cube of the radius (\( r^3 \) ), which means the volume grows rapidly as the radius increases. This is especially important in related rates problems, where the radius changes over time.

Differentiation

Differentiation is a fundamental concept in calculus that deals with computing derivatives. It is the method used to find the rate at which a quantity changes. For the volume of a sphere, we're interested in how quickly the volume changes as the radius changes.

• The derivative with respect to time, \( \frac{dV}{dt} \), tells us how the volume changes over time.
• When differentiating the volume \( V = \frac{4}{3}\pi r^3 \), we use rules such as the power rule and the chain rule to find the derivative of the volume concerning time.

Differentiation is essential because it allows us to make predictions and understand the dynamics of the volume as the sphere grows.

Chain Rule for Derivatives

The chain rule is a critical tool used in differentiation when dealing with composite functions. In our problem, the volume \( V = \frac{4}{3}\pi r^3 \) is dependent on the radius \( r \), which itself can change with time.

When using the chain rule, we differentiate step by step. First, differentiate \( r^3 \) to get \( 3r^2 \), then multiply by the derivative of \( r \) concerning time, \( \frac{dr}{dt} \). Therefore, applying the chain rule to differentiate \( V \) with respect to time gives:

• \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \)

This equation helps us link the rate of change of volume to the rate of change of the radius, explaining how these two rates are related.

Rate of Change

The concept of rate of change refers to how a quantity changes over time. For a sphere, this could mean how its volume increases as the radius expands. In the problem, we know \( \frac{dr}{dt} = 1 \) m/sec, meaning the radius is increasing steadily.

We aim to find \( \frac{dV}{dt} \), the rate at which the volume increases when the radius is 20 meters. Using the relationship:

• \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \)

Substituting \( r = 20 \) m and \( \frac{dr}{dt} = 1 \) m/sec shows us the rate at which the volume changes at that moment. Thus, the calculation becomes:

• \( \frac{dV}{dt} = 4\pi (20)^2 \times 1 = 1600 \pi \)

Finally, converting \( 1600 \pi \) cubic meters per second gives approximately 5026.55 cubic meters per second. This tells us how fast the sphere's volume grows at precisely that radius size.