Problem 28 - Thomas Calculus

Question

Inflating a balloon The volume \(V=(4 / 3) \pi r^{3}\) of a spherical balloon changes with the radius. a. At what rate \(\left(\mathrm{ft}^{3} / \mathrm{ft}\right)\) does the volume change with respect to the radius when \(r=2 \mathrm{ft} ?\) b. By approximately how much does the volume increase when the radius changes from 2 to 2.2 \(\mathrm{ft} ?\)

Step-by-Step Solution

Answer

a) The rate of volume change at \( r=2 \) ft is \( 16\pi \) ft³/ft. b) The volume increases approximately by \( 3.2\pi \) ft³.

Step 1: Understand the formula

The volume \( V \) of a spherical balloon is a function of its radius \( r \) and is given by the formula \( V = \frac{4}{3} \pi r^3 \). This means the volume changes as the radius changes.

Step 2: Differentiate the volume formula

To find how the volume changes with respect to the radius, we need to differentiate the volume function \( V = \frac{4}{3} \pi r^3 \) with respect to \( r \). This gives us the rate of change of volume, \( \frac{dV}{dr} \).

Step 3: Calculate the derivative

Differentiate \( V = \frac{4}{3} \pi r^3 \) with respect to \( r \):\[ \frac{dV}{dr} = 4 \pi r^2 \]This expression gives the rate of change of the volume with respect to the radius at any point \( r \).

Step 4: Substitute the radius into the derivative

Substitute \( r = 2 \) ft into \( \frac{dV}{dr} = 4 \pi r^2 \):\[ \frac{dV}{dr} = 4 \pi (2)^2 = 16 \pi \]Thus, the rate of change of the volume with respect to the radius at \( r = 2 \) ft is \( 16\pi \) ft³/ft.

Step 5: Estimate the increase in volume

To estimate the increase in volume as the radius changes from 2 ft to 2.2 ft, use the derivative to approximate the change:\[ \Delta V \approx \frac{dV}{dr} \cdot \Delta r \]\[ \Delta V \approx 16\pi \cdot (2.2 - 2) = 16\pi \cdot 0.2 = 3.2\pi \] ft³.

Key Concepts

Understanding Rate of Change

In mathematics, the rate of change is a measure of how a quantity changes in relation to another quantity. In the context of the balloon exercise, the rate of change refers to how the volume of the balloon changes as the radius changes. It tells us how much more or less volume we can expect with a small increase or decrease in the radius.
This is like observing how fast or slow a car travels; here, the goal is to see how fast the volume grows as the balloon gets bigger. The precise value of this rate is found by differentiating the volume formula with respect to the radius, giving the derivative \( \frac{dV}{dr} \).
This derivative helps us find the exact rate at which volume increases for any given radius, such as when \( r = 2 \) ft.

The Basics of Spherical Volume

Volume in a spherical shape, like our balloon, depends on the radius. The volume formula is \( V = \frac{4}{3} \pi r^3 \). This equation explains the relationship between the radius of a sphere and the space it occupies.
When the radius increases, the volume increases, following this cubic relationship. The rasied power of 3 indicates that small changes in the radius result in much larger changes in volume.

• Simple formula: directly connects radius and volume
• Cube relationship: means bigger changes as radius grows
• Requires understanding the geometric properties of spheres

Understanding this relationship is crucial for calculating how the volume reacts as the balloon inflates.

Exploring Radius Change

When working with radius change, it refers to the alteration of the radius over a given range of measurements. For the balloon, we're focusing on how the radius changes from 2 feet to 2.2 feet.
This small increase seems trivial but can significantly impact the volume due to the cubic nature of the volume formula \( V = \frac{4}{3} \pi r^3 \). Even a small change in the radius can lead to a noticeable increase in volume.

• Initial radius: 2 feet
• New radius: 2.2 feet
• Impact: Causes a substantial rise in volume

Analyzing this change helps us use the derivative to approximate changes in volume more accurately.

Volume Differentiation and Its Impact

Volume differentiation is the process of finding out how a change in the radius impacts the overall volume. By differentiating the volume formula \( V = \frac{4}{3} \pi r^3 \), we find the expression \( \frac{dV}{dr} = 4 \pi r^2 \), which reveals how sensitive the volume is to changes in the radius.
This derivative becomes increasingly useful because it allows us to quickly estimate changes in volume without recalculating the entire volume for every small radius change. By substituting \( r = 2 \) ft into this derivative, we find that the rate of change of volume is \( 16\pi \) ft³/ft.

• Useful: Estimates volume changes effectively
• Calculation: Simplified by using differentiation
• Result: Gives rate of volume change at any point

In our specific case, this calculated rate helps us predict the volume increase as the radius expands.