Problem 230 - Calculus Volume 1

Question

A spherical balloon is being inflated at a constant rate. If the volume of the balloon changes from \(36 \pi\) in. \(^{3}\) to \(288 \pi\) in. \(^{3}\) between time \(t=30\) and \(t=60\) seconds, find the net change in the radius of the balloon during that time.

Step-by-Step Solution

Answer

The net change in the radius is 3 inches.

Step 1: Understand the Volume Formula for a Sphere

The volume of a sphere is given by the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere. This formula will help us relate the change in volume to the change in radius.

Step 2: Find the Initial Radius

Given the initial volume \( V_1 = 36\pi \) cubic inches at \( t = 30 \) seconds, we need to find the initial radius \( r_1 \). Plug \( V_1 \) into the volume formula: \[ \frac{4}{3} \pi r_1^3 = 36\pi \].Divide both sides by \( \pi \) and solve for \( r_1^3 \):\[ \frac{4}{3} r_1^3 = 36 \].Multiply by 3/4 to isolate \( r_1^3 \):\[ r_1^3 = \frac{3}{4} \times 36 = 27 \].Take the cube root to find \( r_1 \):\( r_1 = 3 \) inches.

Step 3: Find the Final Radius

With the final volume \( V_2 = 288\pi \) cubic inches at \( t = 60 \) seconds, find the final radius \( r_2 \). Using the volume formula:\[ \frac{4}{3} \pi r_2^3 = 288\pi \].Divide both sides by \( \pi \) and solve for \( r_2^3 \):\[ \frac{4}{3} r_2^3 = 288 \].Multiply by 3/4 to isolate \( r_2^3 \):\[ r_2^3 = \frac{3}{4} \times 288 = 216 \].Take the cube root to find \( r_2 \):\( r_2 = 6 \) inches.

Step 4: Calculate the Net Change in Radius

To find the net change in the radius of the balloon, subtract the initial radius from the final radius:\[ \Delta r = r_2 - r_1 = 6 - 3 \].Thus, \( \Delta r = 3 \) inches.

Key Concepts

Volume of a Sphere

To understand the problem, we first need to grasp the concept of the volume of a sphere. The formula for the volume of a sphere is given by \[ V = \frac{4}{3} \pi r^3 \]where \( V \) stands for volume and \( r \) is the radius of the sphere.

• This formula tells us that the volume of a sphere is directly tied to the cube of its radius.
• The \( \pi \) symbol is a constant, approximately equal to 3.14159, which is essential in calculations involving circles.

This formula helps us determine how a change in the radius affects the overall volume. As the radius increases or decreases, the volume will do the same, giving insight into the expansion or contraction of the sphere over time.

Radius Calculation

To find the radius from a given volume, we can rearrange the formula for the volume of a sphere. If we know the volume, we can solve for the radius.

• Re-arrange the formula to solve for the radius: \[ r^3 = \frac{3}{4\pi} V \]


• From here, we can isolate \( r \) and then take the cube root to find its value.

For example, given the initial volume of the balloon as \( 36\pi \) cubic inches, substitute this value into the formula and solve for the radius:

• \( r_1^3 = \frac{3}{4\pi} \times 36\pi = 27 \)


• Taking the cube root, \( r_1 = 3 \) inches.

This calculation gives us the initial radius at a specific point in time.

Cube Root Calculation

The cube root is an essential calculation when working with volumes and radii, especially in this context.

• It involves finding a number that, when multiplied by itself three times, gives the original number.

For instance, to find the radius from the volume via cube root, follow the steps:

• Convert the volume equation for the sphere to \( r^3 \)


• Calculate \( r^3 \) after isolating it on one side of the equation.


• Take the cube root to arrive at \( r \).

In calculating the final radius, given \( r_2^3 = 216 \), taking the cube root gives us \( r_2 = 6 \). This step shows the importance of reversing operations, finding the measurement of one dimension based on a three-dimensional quality.

Constant Rate of Change

The term 'constant rate of change' implies that the attribute being measured increases or decreases steadily over time. In this exercise, the spherical balloon's volume increases consistently from \( 36 \pi \) to \( 288 \pi \) cubic inches between \( t = 30 \) and \( t = 60 \) seconds.

• This change is important for calculating net change in the radius accurately.


• Understanding this concept aids in knowing how volume and time correspond.

By tracking the radius's progression over time, we see that:

• The radius changes from 3 inches to 6 inches.
• This change represents a difference, or net change, in radius of 3 inches.

These calculations demonstrate how mathematics can be used to model real-world physical changes, providing a clear picture of growth or shrinkage over time.