Problem 30 - University Calculus: Early Transcendentals

Question

Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop's radius increases at a constant rate.

Step-by-Step Solution

Answer

The radius increases at a constant rate.

Step 1: Understand the Relationship

First, recognize that the volume of the sphere, representing the mist drop, changes due to condensation. The relationship is such that the rate of change of volume is proportional to its surface area. The surface area of a sphere is given by the formula: \( A = 4\pi r^2 \). Since the problem states that condensation is proportional to the surface area, we can write this as \( \frac{dV}{dt} \propto 4\pi r^2 \).

Step 2: Express Volume in Terms of Radius

Express the volume \( V \) of the sphere in terms of the radius \( r \): \( V = \frac{4}{3}\pi r^3 \). We need this expression because the rate of change of the volume \( \frac{dV}{dt} \) is what we're analyzing in relation to the radius \( r \) and its change over time.

Step 3: Derive the Rate of Change of Volume

Differentiate the volume \( V = \frac{4}{3}\pi r^3 \) with respect to time \( t \) to find \( \frac{dV}{dt} \):\[\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}\]This result comes from applying the chain rule, where \( \frac{dr}{dt} \) is the rate of change of the radius with respect to time.

Step 4: Set Up Proportionality Condition

Given that \( \frac{dV}{dt} \) is proportional to \( 4\pi r^2 \), set \( \frac{dV}{dt} = k \cdot 4\pi r^2 \), where \( k \) is a constant of proportionality. Substitute this into the expression from Step 3:\[ 4\pi r^2 \cdot \frac{dr}{dt} = k \cdot 4\pi r^2 \]

Step 5: Solve for \( \frac{dr}{dt} \)

Since \( 4\pi r^2 \) appears on both sides of the equation, divide both sides by \( 4\pi r^2 \):\[\frac{dr}{dt} = k\]This shows that the rate of change of the radius, \( \frac{dr}{dt} \), is a constant \( k \), which is what we wanted to demonstrate.

Key Concepts

Proportionality

In mathematics, proportionality indicates a constant relationship between two quantities. If one quantity changes, the other changes at a consistent rate. This concept is often written as "one variable is proportional to another." In the case of our exercise with the mist drop, the rate at which moisture is added is directly proportional to the surface area of the sphere.
This means the formula expressing this relationship includes a constant factor, denoted as the proportionality constant. If we represent the rate at which volume changes as \( \frac{dV}{dt} \), and the surface area by \( 4\pi r^2 \), then

• We have \( \frac{dV}{dt} \propto 4\pi r^2 \)
• This can be rewritten as \( \frac{dV}{dt} = k \cdot 4\pi r^2 \), where \( k \) is the constant.

Understanding proportionality allows us to explore how changing one part of our system, such as the sphere's surface, directly impacts another, like the volume's rate of change.

Sphere Geometry

Sphere geometry is a branch of geometry that deals with properties and attributes of spheres, which are perfectly round 3D shapes. The key elements to understand here are the surface area and the volume of a sphere.
The surface area \( A \) of a sphere with radius \( r \) is calculated as:

• \( A = 4\pi r^2 \)

The volume \( V \) of a sphere is given by:

• \( V = \frac{4}{3}\pi r^3 \)

In problems involving spheres, knowing these formulas is crucial. For our exercise, the rate of change of the volume depends on the surface area, which itself depends on the radius. Understanding the relationship between these elements helps in solving problems that involve changes in sphere geometry.

Rate of Change

Rate of change refers to how quickly one variable changes in relation to another. It is a fundamental concept in calculus and is used to understand dynamic systems, such as our mist drop's growth.
In our exercise, the rate of change of volume \( \frac{dV}{dt} \) is essential in understanding how the drop evolves over time. The given relationship shows that this rate is proportional to the sphere's surface area. By differentiating the volume with respect to time, we arrive at a critical expression:

• \( \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \)

This equation is derived using the chain rule and helps to connect the change in the sphere's radius to changes in volume. Solving this equation highlights that the radius changes at a constant rate:\( \frac{dr}{dt} = k \).
This detail underlines the predictable nature of the system, where constants provide a steady rate of growth.