Chapter 5.6, Problem 12E

To determine

To show: the circumferences the droplet’s radius increases at a constant rate.

Explanation of Solution

Given information :

A droplet of mist is a perfect sphere and through condensation, the droplet picks up moisture at a rate proportional to its surface area.

All variables are differentiable functions of t .

Proof:

We have to show the circumferences the droplet’s radius increases at a constant rate.

Surface area of a sphere is $4\pi {\text{R}}^2$ .

Let,

  $\frac{d\text{V}}{dt}=4\pi {\text{R}}^2\cdot \text{K}$

Where k is at a constant rate.

Therefore,

Volume of a sphere,

  $\text{V=}\frac{4}{3}\pi r^3$

Differentiate with respect to t .

  $\begin{array}{l}\frac{d\text{V}}{dt}=\frac{4}{3}\pi \cdot 3r^2\frac{dr}{dt}\\ \frac{d\text{V}}{dt}=4\pi r^2\frac{dr}{dt}\end{array}$

Putting the value of $\frac{d\text{V}}{dt}=4\pi {\text{R}}^2\cdot \text{K}$ in above equation,

  $4\pi {\text{R}}^2\cdot \text{K}=4\pi r^2\frac{dr}{dt}$

Therefore,

  $\frac{dr}{dt}=\text{K}$

Hence, the radius is changing with respect to time at a constant rate.