Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

The rate of change of volume \(V\) of a melting snowball is proportional to the surface area \(S\) of the ball; that is, \(d V / d t=-k S\), where \(k\) is a positive constant. If the radius of the ball at \(t=0\) is \(r=2\) and at \(t=10\) is \(r=0.5\), show that \(r=-\frac{3}{20} t+2 .\)

Step-by-Step Solution

Answer

The function is \(r=-\frac{3}{20} t+2\).

Step 1: Understand the problem

We are given a rate of change of volume of a snowball in the form \( \frac{dV}{dt} = -kS \). We know the radius of the sphere changes from 2 to 0.5 over 10 seconds, and we need to express this as a linear function of \( t \).

Step 2: Relate Volume and Surface Area to Radius

The volume of a sphere is given by \( V = \frac{4}{3} \pi r^3 \) and the surface area is \( S = 4 \pi r^2 \). Thus, \( \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \cdot \frac{dr}{dt} \).

Step 3: Set up the relationships

Since \( \frac{dV}{dt} = -kS \), substitute for \( \frac{dV}{dt} \) and \( S \) to get: \( 4 \pi r^2 \cdot \frac{dr}{dt} = -k \cdot 4 \pi r^2 \). Simplifying gives \( \frac{dr}{dt} = -k \).

Step 4: Determine the constant \(k\)

Integrate \( \frac{dr}{dt} = -k \) to get \( r(t) = -kt + C \). Using the initial condition \( r(0) = 2 \), set \( C = 2 \), so \( r(t) = -kt + 2 \).

Step 5: Use given radius change to find \(k\)

At \( t = 10 \), \( r = 0.5 \). Substitute these values into \( r(t) = -kt + 2 \) to get \( 0.5 = -10k + 2 \). Solving, \( 10k = 2 - 0.5 \) or \( 10k = 1.5 \). Thus, \( k = 0.15 \).

Step 6: Derive the final equation for \(r\)

Substitute \( k = 0.15 \) into \( r(t) = -kt + 2 \), resulting in \( r(t) = -0.15t + 2 \). Change \( 0.15 \) to fraction form \( \frac{3}{20} \) to match the required form: \( r(t) = -\frac{3}{20}t + 2 \).

Key Concepts

Rate of Change

In mathematics, the rate of change refers to how one quantity changes in relation to another. It's essentially a measure of how fast something is increasing or decreasing over time or through space. For our problem, we're examining the rate of change of the volume of a sphere–the melting snowball. This rate is given by the derivative \( \frac{dV}{dt} \), indicating how the volume \( V \) changes over time \( t \).

In the exercise, the rate of volume change is expressed as \( -kS \). The negative sign here shows that the volume is decreasing. The proportionality constant \( k \) ensures that the rate is adjusted with respect to the surface area \( S \) of the snowball. In nature, scenarios such as a melting ice cube or a draining tank often involve similar principles of changing rates.

This problem makes it clear that the rate of change can be understood through the lenses of both mathematics and practical, physical processes.

Surface Area

The surface area is the total area covered by the surface of a three-dimensional object. For a sphere, the formula is \( S = 4\pi r^2 \). This formula gives the surface area in terms of the sphere's radius \( r \). It's fascinating how such a simple formula can help us understand the surface behavior of spherical objects.

In the melting snowball problem, the surface area is pivotal. It's directly proportional to how fast the volume of the sphere changes, as shown in the equation \( \frac{dV}{dt} = -kS \). By identifying the surface area, we can understand how external conditions or characteristics might influence the melting rate of the snowball.

So, when analyzing physical incidents involving spheres, like the melting of a snowball, this surface area formula is an essential tool to predict or calculate behavior based on radius and other factors.

Volume of a Sphere

The volume of a sphere is calculated using the formula \( V = \frac{4}{3} \pi r^3 \). This equation allows us to understand how much space the sphere occupies. In the context of our exercise, knowing the volume is crucial since we're observing how it decreases over time as the snowball melts.

During the melting process, the rate of volume change is described through the derivative \( \frac{dV}{dt} \), and we relate this back to sphere volume using calculus. The volume decreases as the sphere's radius decreases, showing the nature of its interaction with external conditions.

Understanding the volume in relation to the sphere's radius offers insights into not just mathematical computations but also real-world physics, such as determining how long a snowball might take to melt under different conditions.

Integration

Integration is a core concept in calculus that involves finding a function when given its derivative. In simpler terms, it's the process of finding a whole from its parts, or an accumulation of change. In the context of this problem, we use integration to derive the function for the snowball's radius over time.

From the relationship \( \frac{dr}{dt} = -k \), integration allows us to solve for the radius \( r(t) \). The integration process here helps us transcribe a simple rate of change into a comprehensive equation: \( r(t) = -kt + C \), which is a linear equation representing radius as a function of time.

This integrated equation showcases how seemingly complex calculus operations can yield straightforward, interpretable results, showing the decreasing radius of our snowball as time progresses, with \( k \) as the rate constant. Integration ties mathematical change back to a tangible form and shows the power of calculus in modeling real-world behavior.