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Chapter 3: Problem 30

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

Short Answer

Expert verified
The drop's radius increases at a constant rate because \( \frac{dr}{dt} = k \), a constant.

Step by step solution

01

Understanding the Surface Area of a Sphere

The surface area of a sphere is given by the formula \( S = 4\pi r^2 \), where \( r \) is the radius of the sphere.
02

Identifying the Rate of Increase

We know that the rate at which the drop picks up moisture is proportional to its surface area. Let's express this as \( \frac{dV}{dt} = k \cdot S \), where \( V \) is the volume, \( k \) is the proportionality constant, and \( S \) is the surface area of the sphere.
03

Express Volume in Terms of Radius

The volume of the sphere is given by \( V = \frac{4}{3} \pi r^3 \). Now, differentiate this volume with respect to time to find the rate of change of the volume: \( \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \).
04

Relate Rate of Volume Change to Surface Area

Since \( \frac{dV}{dt} = k \cdot S \), we substitute \( S = 4\pi r^2 \) into the expression: \( \frac{dV}{dt} = k \cdot 4\pi r^2 \).
05

Equate Two Expressions for dV/dt

We have two expressions for \( \frac{dV}{dt} \): \( \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \) and \( \frac{dV}{dt} = k \cdot 4\pi r^2 \). Set them equal: \( 4\pi r^2 \cdot \frac{dr}{dt} = k \cdot 4\pi r^2 \).
06

Solve for dr/dt

Cancel out \( 4\pi r^2 \) from both sides of the equation to isolate \( \frac{dr}{dt} \): \( \frac{dr}{dt} = k \). This shows that \( \frac{dr}{dt} \) is constant.

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Key Concepts

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

Surface Area
The surface area of a sphere is an important concept when dealing with three-dimensional objects like a drop of mist. For a sphere with radius \( r \), the surface area \( S \) can be calculated using the formula:
  • \( S = 4\pi r^2 \).
This formula tells us that the surface area of a sphere is proportional to the square of its radius.
This means, as the radius increases, the surface area increases quadratically. It's crucial to understand the relationship between surface area and volume because the rate at which moisture is absorbed by a mist drop is linked to its surface area. So, in this context, the "rate of change" of the volume due to condensation ties directly into understanding how surface area impacts this growth.
Volume of a Sphere
The volume of a sphere describes how much space it occupies, and is calculated by the formula:
  • \( V = \frac{4}{3} \pi r^3 \).
This volume formula shows us that the space occupied by a sphere is proportional to the cube of its radius.
Thus, any small increase in the radius leads to a much larger increase in volume.
In the exercise, the rate of moisture accumulation depends on this volume, which in turn is changing due to changes in the surface area.
The volume increase due to condensation is directly affecting the sphere鈥檚 radius, further explaining how these geometric properties are interconnected. By understanding this volume formula, predicting how the fall of condensation impacts the sphere's dimensions becomes clearer.
Differentiation
Differentiation is a mathematical concept that helps us find the rate at which one quantity changes with respect to another. In this exercise, we use differentiation to relate changes in volume to changes in radius over time.
For differentiation within context, consider the volume of a sphere:
  • \( V = \frac{4}{3} \pi r^3 \).
By differentiating this equation with respect to time \( t \), we obtain:
  • \( \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \).
This expression links the rate of change of volume \( \frac{dV}{dt} \) to the rate of change of the radius \( \frac{dr}{dt} \), showing us how alterations in the sphere's shape facilitate a constant expansion of the radius.
Differentiation simplifies complex relationships, allowing mathematicians and scientists to derive important conclusions about growth rates and other dynamic systems.

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