Question

A snowball melts at a rate proportional to its surface area. (a) Show that its radius shrinks at a constant rate. (b) If it melts to \(\frac{8}{27}\) its original volume in one hour, how long will it take to melt completely?

Short answer

The radius shrinks at a constant rate of \(-k\). It takes 3 hours to melt completely.

Step-by-step solution

Introduction to the Problem

We are given that a snowball melts at a rate proportional to its surface area. We need to show that its radius shrinks at a constant rate and determine how long it will take to melt completely if it reduces to \( \frac{8}{27} \) of its original volume in one hour.

Express Surface Area and Volume of a Sphere

The surface area \( A \) of a sphere with radius \( r \) is given by \( A = 4\pi r^2 \). The volume \( V \) of a sphere is given by \( V = \frac{4}{3}\pi r^3 \). We will use these formulas to express the rate of change of volume in terms of the radius.

Set Up the Rate of Change of Volume

The snowball melts at a rate proportional to its surface area, so we have the rate of change of the volume \( \frac{dV}{dt} = -kA = -k(4\pi r^2) \), where \( k \) is some positive constant.

Express \( \frac{dV}{dt} \) in Terms of \( r \)

Using \( V = \frac{4}{3}\pi r^3 \), differentiate both sides with respect to \( t \): \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \). Equating this to the rate of change of volume: \( 4\pi r^2 \frac{dr}{dt} = -4\pi kr^2 \).

Solve for \( \frac{dr}{dt} \)

From the equation \( 4\pi r^2 \frac{dr}{dt} = -4\pi kr^2 \), divide both sides by \( 4\pi r^2 \): \( \frac{dr}{dt} = -k \). This shows that the radius shrinks at a constant rate.

Calculate Original Volume and Reduced Volume

Let the original volume be \( V_0 = \frac{4}{3}\pi r_0^3 \). The volume after one hour is \( V = \frac{8}{27}V_0 = \frac{8}{27} \times \frac{4}{3}\pi r_0^3 \).

Relate Volumes to Radii

The reduced volume after one hour is \( \frac{4}{3}\pi r^3 = \frac{8}{27} \times \frac{4}{3}\pi r_0^3 \). Simplify to get \( r^3 = \frac{8}{27} r_0^3 \). Solving, \( r = \left(\frac{8}{27}\right)^{1/3} r_0 = \frac{2}{3} r_0 \).

Determine Total Melting Time

From \( r = r_0 - kt \) and knowing that after one hour \( r_0 - k \times 1 = \frac{2}{3}r_0 \), solve for \( k \): \( k = \frac{1}{3}r_0 \). To find when the snowball melts completely (\( r = 0 \)): \( 0 = r_0 - \frac{1}{3}r_0 \times t \), leading to \( t = 3 \) hours.

Key concepts

Rate of Change

Understanding the concept of rate of change is crucial in solving problems involving changing quantities over time. In calculus, the rate of change of a quantity is often represented by a derivative. For instance, if the volume of a snowball is changing as it melts, we describe its rate of change using the derivative \( \frac{dV}{dt} \). This signifies how fast the volume \( V \) changes with respect to time \( t \).

In the snowball melting problem, the rate at which the snowball loses volume is directly tied to its surface area. This is termed as being proportional, meaning \( \frac{dV}{dt} = -kA \), where \( k \) is a positive constant, and \( A \) is the surface area. By understanding this relationship, we learn that as the snowball’s surface area decreases, the rate of volume loss decreases as well.

The beauty of the rate of change in such problems lies in how it leads to further insights, like understanding the constant rate of radius reduction in the snowball.

Surface Area

Surface area plays a significant role in determining the rate at which the snowball melts. The surface area of a sphere is given by the formula \( A = 4\pi r^2 \). This formula tells us that the surface area depends on the square of the radius \( r \), making it essential for understanding how quickly the snowball melts.

Since the melting rate of the snowball is proportional to its surface area, understanding \( A \) allows us to set up the equation \( \frac{dV}{dt} = -kA \), where \( k \) is a constant. This relationship indicates that larger snowballs with greater surface areas will initially melt faster than smaller ones. However, as the snowball shrinks, the surface area decreases, thereby slowing down the rate of volume loss over time.

Thus, surface area is not just an abstract concept but a practical factor influencing the dynamics of the melting process.

Sphere Volume

The volume of a sphere is another critical concept introduced in the snowball problem. The formula for the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \), where \( r \) is the radius. This cubic relationship means that changes in the radius have a significant impact on the sphere's volume.

In our problem, calculating the change in volume over time helps us understand the melt rate. Given the initial and final volumes, you can determine how the radius—and therefore the volume—will shift as the snowball shrinks. When the problem states the volume reduces to \( \frac{8}{27} \) of its original size in one hour, it shows us that the radius shrinks to \( \frac{2}{3} \) of its original size. This is because the cube root of \( \frac{8}{27} \) is \( \frac{2}{3} \).

Understanding sphere volume is crucial for predicting how quickly a sphere of snow (or anything similar) will transform, especially when dealing with physical phenomena like melting.

Constant Rate

A key insight from the snowball exercise is the concept of constant rate concerning radius reduction. The problem shows that the radius diminishes at a constant rate over time. This is evident from the equation \( \frac{dr}{dt} = -k \), extracted by equating the derivatives we discussed.

To understand this better, note that while the volume and surface area decrease as the snowball melts, the rate at which the radius itself decreases remains uniform. This constancy is what allows us to solve for the time it takes for the snowball to completely melt, as shown by the calculation of \( t \), the time needed when \( r = 0 \).

Processes like these, which have constant rates, provide assurance that changes don’t occur erratically and that you can reliably predict the outcomes based on initial data. Knowing how to handle constant rates is a fundamental skill in calculus, aiding the analysis of real-world scenarios like this one.