Problem 31 - University Calculus: Early Transcendentals

Question

A spherical balloon is inflated with helium at the rate of \(100 \pi \mathrm{ft}^{3} / \mathrm{min}\). How fast is the balloon's radius increasing at the instant the radius is \(5 \mathrm{ft} ?\) How fast is the surface area increasing?

Step-by-Step Solution

Answer

The radius is increasing at 1 ft/min and the surface area at 40π ft²/min.

Step 1: Understand the Problem

We are given a spherical balloon with volume increasing at a rate of \(100 \pi \ \mathrm{ft}^3 / \mathrm{min}\), and we need to find how fast the radius and surface area are increasing when the radius is \(5 \mathrm{ft}\). This is a related rates problem in calculus.

Step 2: Write Down Known Formulas

The formulas we are using are: 1. Volume of a sphere, \( V = \frac{4}{3} \pi r^3 \) 2. Surface area of a sphere, \( A = 4 \pi r^2 \). We also have \( \frac{dV}{dt} = 100\pi \).

Step 3: Differentiate the Volume Formula

Differentiate both sides of the volume formula \( V = \frac{4}{3} \pi r^3 \) with respect to time \( t \). This gives us: \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \], where \( \frac{dV}{dt} = 100\pi \).

Step 4: Solve for \( \frac{dr}{dt} \)

Substitute the known values and solve for \( \frac{dr}{dt} \). When \( r = 5 \mathrm{ft} \), \[ 100\pi = 4\pi (5)^2 \frac{dr}{dt} \] \[ 100\pi = 100\pi \frac{dr}{dt} \] \[ \frac{dr}{dt} = 1 \mathrm{ft/min} \].

Step 5: Differentiate the Surface Area Formula

Differentiate both sides of the surface area formula \( A = 4 \pi r^2 \) with respect to time \( t \). This gives us: \[ \frac{dA}{dt} = 8 \pi r \frac{dr}{dt} \].

Step 6: Solve for \( \frac{dA}{dt} \)

Use the value of \( \frac{dr}{dt} \) found earlier to find \( \frac{dA}{dt} \). When \( r = 5 \mathrm{ft} \) and \( \frac{dr}{dt} = 1 \mathrm{ft/min} \): \[ \frac{dA}{dt} = 8 \pi (5)(1) = 40 \pi \mathrm{ft}^2/\mathrm{min} \].

Key Concepts

Spherical Balloon

A spherical balloon is a perfect example of a three-dimensional geometric object that is both practical and mathematically interesting. As opposed to other shapes, a spherical balloon means we're dealing with the same distance from the center in all directions, which is what defines a sphere.

• The sphere is very symmetrical. This makes calculations involving volume and surface area more straightforward.
• In real-world applications, such as a balloon being inflated, the uniformity of the sphere simplifies how we describe changes over time like volume and surface area.

In this exercise, the focus is on understanding how certain measurements, like the radius and surface area, change as the balloon inflates faster. This is where calculus enters the picture to help us describe these changing quantities accurately.

Volume of a Sphere

The volume of a sphere gives us an idea of the space the sphere occupies. To calculate the volume, we use the formula: \[ V = \frac{4}{3} \pi r^3 \]where \( r \) is the radius of the sphere. Here, every inch the radius increases, the volume grows by a factor of the radius cubed.

• This cubic relationship is key to understanding why small changes in radius can produce significant changes in volume.
• In the problem, the rate of increase of the sphere's volume helps us figure out how quick the radius is growing.

By knowing the volume change rate as \( 100\pi \ \text{ft}^3/\text{min} \), we can apply calculus, specifically differentiation with respect to time, to find how fast the radius changes at a specific point.

Surface Area of a Sphere

The surface area of a sphere describes how much surface is available on the outside. The formula used to calculate it is: \[ A = 4 \pi r^2 \]This relation with the radius means the surface area grows with the square of the radius.

• The quadratic relationship shows how the surface area changes at a different rate than the volume as the radius increases.
• Knowing the increase rate of the surface area helps us understand how quickly a surface can be covered.

In the task, we differentiated the surface area formula with time to find out how its change rate relates to the balloon's radius change rate. This measure indicates the rapidity with which the outer boundary of our spherical balloon expands.

Differentiation

Differentiation is a fundamental tool in calculus used to determine how a function changes as its input changes. It involves finding the derivative, which provides a rate of change.

• In the context of this problem, differentiation tells us how the radius and surface area of the balloon change with time.
• When differentiating the volume formula with respect to time, we use the chain rule to relate changes in radius to changes in volume.

By differentiating the volume and surface area formulas with respect to time, we derived expressions for \( \frac{dr}{dt} \) and \( \frac{dA}{dt} \), respectively. This derivation step is essential to solve related rates problems, which involve multiple variables changing over time.