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Chapter 1: Problem 45

Suppose that the mass of a spherical mothball decreases with time, due to evaporation, at a rate that is proportional to its surface area. Assuming that it always retains the shape of a sphere, it can be shown that the radius \(r\) of the sphere decreases linearly with the time \(t\). (a) If, at a certain instant, the radius is \(0.80 \mathrm{mm}\) and 4 days later it is \(0.75 \mathrm{mm},\) find an equation for \(r\) (in millimeters) in terms of the elapsed time \(t\) (in days). (b) How long will it take for the mothball to completely evaporate?

Short Answer

Expert verified
(a) \( r(t) = 0.80 - 0.0125t \); (b) 64 days.

Step by step solution

01

Understand the Proportional Relationship

The problem states that the rate of decrease of the mothball's mass is proportional to its surface area. Given that the mass of a sphere is also a function of its radius, this implies that the radius decreases in a linear manner with respect to time.
02

Establish the Linear Equation

We can write the linear relationship of the radius with time as \( r(t) = r_0 - kt \), where \( r_0 \) is the initial radius (0.80 mm), and \( k \) is a constant that determines the rate of decrease of the radius with time.
03

Calculate the Rate of Decrease

We know that after 4 days, the radius is 0.75 mm. Using this information, we substitute into the equation: \( 0.75 = 0.80 - 4k \). Simplifying, we find \( k = \frac{0.80 - 0.75}{4} = 0.0125 \).
04

Write the Equation of r in Terms of t

Substitute \( k \) back into the linear equation: \( r(t) = 0.80 - 0.0125t \). This now represents the radius of the mothball as a function of time.
05

Determine Evaporation Time

For complete evaporation, the radius \( r \) must be zero. Set \( r(t) = 0 \) and solve for \( t \): \( 0 = 0.80 - 0.0125t \). Solving gives \( 0.0125t = 0.80 \) or \( t = \frac{0.80}{0.0125} = 64 \). So, it will take 64 days for the mothball to completely evaporate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations play a crucial role in modeling change. In the context of the mothball evaporation problem, a differential equation could describe how the radius of a sphere decreases over time. Here, the equation utilizes the relationship between the change in mass (or radius) and time. Since the problem specifies that the rate of mass change is proportional to the surface area, it implies a differential relationship. This equation is of a simple form because it manifests as a linear decrease over time. This simplification is key to making predictions using straightforward mathematical methods.
Linear Models
Linear models are mathematical descriptions where the relationship between two variables forms a straight line. They are simple, yet powerful, as they allow predictions and insights into how one quantity changes with another. In the mothball scenario, we observe a linear relationship between the radius and time because the mothball's radius decreases steadily over days. The linear equation formulated in the problem is given by \( r(t) = r_0 - kt \). Here:
  • \( r(t) \): radius at time \( t \)
  • \( r_0 \): initial radius
  • \( k \): constant rate of decrease
By establishing this linear framework, the future state (or absence) of the mothball is easily predicted.
Evaporation Rate
The evaporation rate is central to understanding how quickly the mothball loses mass. In this case, the rate is linked directly to the surface area of the sphere. As the evaporation progresses, the surface area changes, influencing how fast the radius shrinks. When modeling such situations, it's essential to think about what factors affect the rate鈥攊n this instance, surface area. The given problem simplifies reality by maintaining the sphere shape constant, allowing us to use current surface area to predict changes effectively in the radius over time.
Proportional Relationships
Proportional relationships are foundational in many mathematical models. They describe scenarios where one quantity changes at a constant rate relative to another. For the mothball problem, the rate at which mass diminishes is proportional to the sphere's surface area. That is, as the surface area declines, so does the rate of mass loss. This relationship is crucial because it tightly couples the physical properties of an object (surface area) with its temporal properties (time), leading to a predictable pattern described by a simple linear equation. Recognizing such patterns aids in forming accurate models for real-world scenarios.

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