Problem 14 - Calculus - AP Edition

Question

A spherical balloon is inflated so that its volume is increasing at the rate of \(3 \mathrm{ft}^{3} / \mathrm{min} .\) How fast is the diameter of the balloon increasing when the radius is \(1 \mathrm{ft} ?\)

Step-by-Step Solution

Answer

The diameter of the balloon is increasing at \( \frac{3}{2\pi} \text{ ft/min} \) when the radius is 1 ft.

Step 1: Understand the Given Information

We are given the rate at which the volume of the balloon is increasing: \( \frac{dV}{dt} = 3 \text{ ft}^3/\text{min} \). We need to find the rate at which the diameter of the balloon is increasing when the radius is 1 ft.

Step 2: Establish the Relationship Between Volume and Radius

Use the formula for the volume of a sphere: \[ V = \frac{4}{3}\pi r^3 \]where \( V \) is the volume and \( r \) is the radius of the sphere.

Step 3: Differentiate to Find the Rate of Change of the Radius

Differentiate the volume formula with respect to time, \( t \):\[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \]Here, we will solve for \( \frac{dr}{dt} \), which is the rate of change of the radius with respect to time.

Step 4: Substitute Known Values and Solve for \(\frac{dr}{dt}\)

Substitute the given values \( \frac{dV}{dt} = 3 \text{ ft}^3/\text{min} \) and \( r = 1 \text{ ft} \) into the differentiated equation:\[ 3 = 4\pi (1)^2 \frac{dr}{dt} \]Now solve for \( \frac{dr}{dt} \):\[ 3 = 4\pi \frac{dr}{dt} \]\[ \frac{dr}{dt} = \frac{3}{4\pi} \text{ ft/min} \]

Step 5: Relate Radius Change to Diameter Change

The diameter \( D \) of the sphere is twice the radius, \( D = 2r \). Thus, the rate of change of the diameter, \( \frac{dD}{dt} \), is twice the rate of change of the radius:\[ \frac{dD}{dt} = 2 \times \frac{dr}{dt} = 2 \times \frac{3}{4\pi} = \frac{3}{2\pi} \text{ ft/min} \]

Key Concepts

Understanding a Spherical Balloon

A spherical balloon is a perfectly round object that can expand or contract. When inflated, its volume grows. In math, this expansion is modeled as a change in volume as the balloon gets larger. For a spherical balloon, the mathematical term for its size is the volume, described by the formula: \[ V = \frac{4}{3}\pi r^3 \] where \( V \) is the volume and \( r \) is the radius, which is the distance from the center of the balloon to its surface. Understanding this volume formula is essential, as it helps to determine how much space the balloon occupies. As a balloon inflates, the radius increases, affecting its volume. This relationship is key when dealing with problems involving rates of change and is why understanding the formula is the first step in tackling such questions.

Volume Differentiation of a Sphere

Volume differentiation involves calculating how the volume of an object changes over time. For a balloon, this means differentiating the volume formula with respect to time, \( t \). The aim is to understand how quickly the radius, and subsequently, the volume grows.To do this, we apply calculus by taking the derivative of the equation: \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \] In this formula, \( \frac{dV}{dt} \) is the rate at which the volume changes over time, while \( \frac{dr}{dt} \) is the rate of change of the radius. This differentiation step lets us link the speed of the radius' growth with the change in volume. Understanding volume differentiation helps us bridge the gap between static measurements and dynamic processes, making it crucial for solving related rates problems.

The Rate of Change Explained

The rate of change is a measure of how much a quantity changes compared to time. In our balloon example, we know how fast the volume is increasing, \( \frac{dV}{dt} = 3 \text{ ft}^3/\text{min} \), and we're asked to find how fast the diameter is growing. The diameter's rate of change relies on determining \( \frac{dr}{dt} \), the rate of change of the radius. By substituting known values into the derivative equation: \[ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \] we solve for \( \frac{dr}{dt} \), which is then used to calculate the change in diameter: \[ \frac{dD}{dt} = 2 \times \frac{dr}{dt} \] In this case, it's crucial to understand that changes in the radius and diameter are directly linked through their rate of change. Clear grasp of this concept allows for the successful solution of related rates problems, efficiently predicting how sizes and speeds will evolve.