Problem 35 - Calculus Early Transcendentals

Question

Approximate the change in the volume of a sphere when its radius changes from \(r=5 \mathrm{ft}\) to \(r=5.1 \mathrm{ft}\left(V(r)=\frac{4}{3} \pi r^{3}\right)\).

Step-by-Step Solution

Answer

#tag_title# Calculate the new volume #tag_content# To calculate the volume of the sphere when its radius is 5.1 ft, plug r = 5.1 into the formula: \(V(5.1) = \frac{4}{3}\pi(5.1)^3\). #tag_title# Find the change in volume #tag_content# To find the approximate change in volume, subtract the initial volume from the new volume: \(\Delta V \approx V(5.1) - V(5)\). Calculate the change in volume and provide the answer.

Step 1: Calculate the initial volume

To calculate the volume of the sphere when its radius is 5 ft, plug r = 5 into the formula: \(V(5) = \frac{4}{3}\pi(5)^3\).

Step 2: Continue the solution

#tag_title# Calculate the new volume #tag_content# To calculate the volume of the sphere when its radius is 5.1 ft, plug r = 5.1 into the formula:

Step 3: Continue the solution

\(V(5.1) = \frac{4}{3}\pi(5.1)^3\).

Step 4: Continue the solution

#tag_title# Find the change in volume #tag_content# To find the approximate change in volume, subtract the initial volume from the new volume:

Step 5: Continue the solution

\(\Delta V \approx V(5.1) - V(5)\).

Key Concepts

Volume of a Sphere

The volume of a sphere is determined using the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere. This formula results from integral calculus and represents the three-dimensional space occupied by the sphere. It's important to understand that as the radius increases, the volume grows rapidly because the radius is raised to the third power.

• The factor of \( \pi \) (approximately 3.14159) acknowledges the sphere's circular symmetry, similar to its role in calculations involving circles.
• The \( \frac{4}{3} \) fraction is derived from geometric principles involving the calculation of three-dimensional shapes.

By substituting \( r = 5 \) ft into this formula, you calculate the initial volume of the sphere. This understanding is crucial for exploring how changes in radius affect the volume.

Rate of Change

In calculus, the rate of change of a function represents how rapidly a variable quantity changes over time or through different conditions. When considering the volume of a sphere, the rate of change of the volume with respect to the radius is a vital insight.
To find out how much the volume changes as the radius changes, you determine the derivative of the volume formula \( V(r) = \frac{4}{3} \pi r^3 \) with respect to \( r \). This derivative \( V'(r) \) tells us the rate at which the volume is increasing per unit increase in radius. The derivative here is \( V'(r) = 4 \pi r^2 \), highlighting that the rate is proportional to the square of the radius at any point.
Understanding how quickly the volume increases as the radius expands is essential when considering practical scenarios like inflating a balloon or bubbles growing underwater.

Differentiation

Differentiation is a fundamental aspect of calculus, which involves finding the derivative of a function. A derivative provides the instantaneous rate of change of a function concerning one of its variables. Here, it's notably used to detect slight modifications in a sphere's volume resulting from changes in its radius.
In this problem, we differentiate the sphere's volume formula to find the rate of change of volume in terms of the radius. Specifically:

• The original volume formula is \( V = \frac{4}{3} \pi r^3 \).
• Differentiating with respect to \( r \) gives \( V'(r) = 4 \pi r^2 \).

Differentiation allows us to predict how much the volume will change when the radius increases from 5 ft to 5.1 ft. By applying the derivative, \( V'(5) = 4 \pi (5)^2 \), we ascertain that this value reflects the slope of the volume function at \( r = 5 \), guiding us to the incremental volume change for a small increase in radius.