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Chapter 1: Problem 19

A large snowball is shaped into the form of a sphere. Starting at some time, which we can designate as \(t=0\), the snowball begins to melt. Assume for the sake of discussion that the snowball melts in such a manner that its shape remains spherical. Discuss the quantities that change with time as the snowball melts. Discuss an interpretation of "melting" as a rate. If possible, construct a mathematical model that describes the state of the snowball at any time \(t>0\).

Short Answer

Expert verified
The snowball's radius and volume decrease with time; it's modeled by \( \frac{dr}{dt} = -k \).

Step by step solution

01

Identify the Changing Quantities

As the snowball melts, its volume, surface area, and radius will change with time. The mass of the snowball will also decrease. Since it remains spherical, the relationships between these quantities will be maintained as the snowball shrinks.
02

Define the Relationship Between Quantities

The volume of a sphere is given by the formula \( V = \frac{4}{3}\pi r^3 \), where \( r \) is the radius. The surface area is \( A = 4\pi r^2 \). These formulas relate the radius to the volume and surface area, which helps in understanding how these change as the snowball melts.
03

Understand Melting as a Rate

Melting can be interpreted as a rate of volume decrease. If \( \frac{dV}{dt} \) represents the rate of decrease in volume, this rate is negative because the volume is shrinking over time. This rate may depend on external factors like temperature.
04

Construct a Mathematical Model

To model the melting process, we assume the rate of volume change \( \frac{dV}{dt} \) is a function of time or possibly a function of radius. A simple model could be \( \frac{dr}{dt} = -k \), where \( k \) is a positive constant representing the melting rate, leading to a linear decrease in radius over time.
05

Derive the Model Equations

Given \( V = \frac{4}{3} \pi r^3 \), the rate of change of volume with respect to time can be derived using the chain rule: \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 4\pi r^2 (-k) \). Solving this differential equation gives the radius as a function of time.

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Key Concepts

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

Volume of a Sphere
The volume of a sphere is a fundamental concept in geometry, and it applies to various real-life scenarios, like calculating the space occupied by a spherical snowball. The formula for the volume is \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere.
  • This formula indicates that the volume is directly related to the cube of the radius.
    The larger the radius, the larger the volume.
  • As the snowball melts, its radius decreases. Consequently, so does its volume, since they are interconnected.
  • Understanding these relationships helps in knowing how various physical properties change as conditions like temperature evolve.
In the context of a melting snowball, analyzing the change in volume over time is crucial. By identifying the volume as a function of the radius, we can better comprehend how the snowball's size diminishes with time.
Overall, the formula for the volume of a sphere provides insight into how spatial dimensions transform during melting.
Rate of Change
The rate of change is a key concept when studying melting processes. It refers to how a quantity varies with time. In the case of our snowball, we are interested in how its volume decreases at the melting rate.
  • The rate of change of volume, denoted as \( \frac{dV}{dt} \), helps quantify how quickly the snowball loses its mass.
  • This rate is negative, indicating that the volume constantly declines as time passes.
  • Factors like ambient temperature can influence this rate, making understanding it crucial for accurate predictions.
By focusing on the rate of change, you can see the melting process as a flow of volume reduction. This perspective turns a physical observation into a mathematical insight, facilitating predictions about future states of the snowball.
Mathematical Modeling
Mathematical modeling transforms real-world phenomena into mathematical expressions. For a melting snowball, modeling allows us to predict changes in size over time.
  • In our example, the model starts with the assumption that \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \), where \( \frac{dr}{dt} = -k \) and \( k \) is a constant representing the melting rate.
  • This model proposes that the snowball鈥檚 radius decreases linearly with time, depicted as a function of time.
  • Using this model, we can calculate the snowball鈥檚 radius at any given moment \( t > 0 \).
Mathematical models like this one are essential in scenarios where we aim to predict future behavior based on current trends. By substituting the equations into differential equations, we derive solutions that describe the snowball鈥檚 evolution over time.
Through such models, theoretical predictions meet practical applicability.

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Most popular questions from this chapter

Suppose \(y=\phi(x)\) is a solution of the differential equation \(d y / d x=\) \(y(a-b y)\), where \(a\) and \(b\) are positive constants. (a) By inspection find two constant solutions of the equation. (b) Using only the differential equation, find intervals on the \(y\)-axis on which a nonconstant solution \(y=\phi(x)\) is increasing: on which \(y=\phi(x)\) is decreasing. (c) Using only the differential equation, explain why \(y=a / 2 b\) is the \(y\)-coordinate of a point of inflection of the graph of a nonconstant solution \(y=\phi(x)\). (d) On the same coordinate axes, sketch the graphs of the two constant solutions found in part (b) and a graph of the nonconstant solution \(y=\phi(x)\) whose shape is suggested in parts (b) and (c).

After a mass \(m\) is attached to a spring, it stretches the spring \(s\) units and then hangs at rest in the equilibrium position, as shown in Figure \(1.19(b)\). After the spring/mass system has been set in motion, let \(x(t)\) denote the directed distance of the mass beyond the equilibrium position. Assume that the downward direction is positive and that the motion takes place in a vertical straight line through the center of gravity of the mass. Assume, too, that the only forces acting on the system are the weight \(m g\) of the mass and the restoring force of the stretched spring, which, by Hooke's law, is proportional to its total elongation. Determine a differential equation for the displacement \(x(t)\) for any time \(t\).

Determine by inspection at least one solution for the given differcntial equation. $$ y^{\prime}=2 x $$

Determine a region of the \(x y\)-plane for which the given differential equation would have a unique solution through a point \(\left(x_{0}, y_{0}\right)\) in the region. $$ x \frac{d y}{d x}=y $$

\(y^{\prime \prime}+25 y=0 ; \quad y=c_{1} \cos 5 x\)
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