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Chapter 4: Problem 12

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

Short Answer

Expert verified
Under the given conditions, the radius of the droplet increases at a constant rate.

Step by step solution

01

Establish relations

We start by understanding the relationship between the volume \(V\) of the droplet and its radius \(r\). The volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\). The surface area \(A\) is \(A = 4\pi r^2\). It's mentioned in the problem that the rate of moisture accumulation is proportional to the surface area, which gives us \(\frac{dV}{dt} = k A = k 4\pi r^2\). Here, \(k\) is the proportionality constant.
02

Differentiate the Volume formula

Differentiating the volume of a sphere with respect to time \(t\), we get \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\).
03

Equate and Simplify

Equate the two values of \(\frac{dV}{dt}\) from Step 1 and Step 2. This results in \(4\pi r^2 \frac{dr}{dt}= k 4\pi r^2\). Dividing both sides by \(4\pi r^2\) gives us \(\frac{dr}{dt} = k\). This implies that the rate of increase of radius is a constant, irrespective of its current size.

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Key Concepts

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

Rate of Change
In differential calculus, the rate of change is a key concept that helps us understand how one quantity changes in relation to another. It is often expressed as a derivative. In the context of our raindrop problem, we are interested in how the droplet's radius changes over time.

The rate of change of the volume with respect to time is given by \( \frac{dV}{dt} \). This represents how quickly the droplet is gaining or losing volume. The droplet's radius change is described by \( \frac{dr}{dt} \), which indicates how the radius of the droplet increases or decreases as time progresses.

In problems involving rate of change, we often relate these rates to the physical properties of the object. This allows us to build a mathematical model that shows how the droplet's characteristics evolve over time. By differentiating the volume equation, we can directly link the rate of volume change to the rate of radius change, allowing us to understand the overall growth dynamics of the sphere.
Volume and Surface Area of Sphere
A sphere's volume and surface area are essential concepts, particularly when dealing with problems involving growth or accumulation. The formula for the volume \( V \) of a sphere is \( V = \frac{4}{3} \pi r^3 \). This tells us how much space a spherical object occupies.

On the other hand, the surface area \( A \) is given by \( A = 4 \pi r^2 \), indicating how much surface the sphere has to interact with its environment. Understanding these formulas is crucial, especially when a problem, like our raindrop scenario, describes a situation where changes in volume are directly tied to changes in surface area.

In this exercise, the droplet is absorbing moisture at a rate proportional to its surface area, meaning the more surface area it has, the faster it can accumulate moisture. These relationships help us understand the role of geometry in dynamic changes and growth processes.
Proportional Relationships
Proportional relationships describe how two variables change together in a consistent ratio. If one variable is changing, the other changes systematically at a related scale.

In the raindrop problem, the rate at which moisture is picked up by the droplet is directly proportional to its surface area. This gives us the mathematical expression \( \frac{dV}{dt} = k A \), where \( k \) is a constant of proportionality. This formula tells us that the volume change over time scales with the surface area.

By isolating the effects of different factors (like surface area and radius), proportional relationships offer clarity on how one quantity impacts another. This concept allows us to deduce that the radius of our raindrop increases uniformly over time, as indicated by the constant rate \( \frac{dr}{dt} = k \). Understanding how these relationships form and function is crucial in predicting and modeling real-world phenomena.

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