The last two questions, differential equations ### The Rate of Evaporation of a Spherical RaindropIn this exercise, we explore the behavior of a spherical raindrop evaporating at a rate proportional to its surface area. The rate of change of the volume is given as \(-k\) times the surface area, where \(k\) is a positive constant of proportionality. We will derive the differential equations for the volume, radius, and surface area of the raindrop as functions of time.#### Problem StatementA spherical raindrop evaporates such that the rate of change of its volume (\(V\)) is proportional to its surface area (\(S\)) by a factor of \(-k\).\[\frac{dV}{dt} = -kS\]#### Given- \(dV/dt\) (Rate of change of volume)- \(dr/dt\) (Rate of change of radius)- \(dS/dt\) (Rate of change of surface area)#### Derivations1. **Volume of the drop:**The rate of change of the volume of the drop is given by:\[\frac{dV}{dt} = -K'\]2. **Radius of the drop:** The rate of change of the radius of the drop is given by:\[\frac{dr}{dt} = -k\]3. **Surface area of the drop:**The rate of change of the surface area of the drop is given by:\[\frac{dS}{dt} = -k\left(2\left(\frac{7}{3}\right)\left( \left(\frac{1}{3}\right)^{3} \right) \Pi \left(\left(\frac{2}{3}\right)V\left(\left(\frac{1}{3}\right)\right)\right)\right)\]This expression showcases how the surface area changes with respect to time, incorporating both the constant of proportionality \(k\) and the current volume \(V\).#### Notes on Equations- **Volume Rate Equation:** This is a simple linear relationship reflecting the proportionality between the surface area and the volume rate.- **Radius Rate Equation:** This indicates a steady change in the radius over time, given by the constant \(k\).- **Surface Area Equation:** This more complex derivative reflects the geometr

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A spherical raindrop evaporates at a rate proportional to its surface area with (positive) constant of proportionality k; i.e. the rate of change of the volume exactly equals -k times the surface area. Write differential equations for each of the quantities below as a function of time. For each case the right hand side should be a function of the dependent variable and the constant k. For example, the answer the first question should not depend on S or r. The volume of the drop: The radius of the drop: dv dt dr dt The surface area of the drop: = -K' E -k ds dt = Your answer includes 1 character that can't be graded. Delete your recent changes and use the pad tools to finish your answer. More information -K(2) (³) ₂ ( ( ³ ) ), (³), (³) II x

A spherical raindrop evaporates at a rate proportional to its surface area with (positive) constant of proportionality k; i.e. the rate of change of the volume exactly equals -k times the surface area. Write differential equations for each of the quantities below as a function of time. For each case the right hand side should be a function of the dependent variable and the constant k. For example, the answer the first question should not depend on S or r. The volume of the drop: The radius of the drop: dv dt dr dt The surface area of the drop: = -K' E -k ds dt = Your answer includes 1 character that can't be graded. Delete your recent changes and use the pad tools to finish your answer. More information -K(2) (³) ₂ ( ( ³ ) ), (³), (³) II x

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter2: Graphical And Tabular Analysis

Section2.4: Solving Nonlinear Equations

Problem 17E: Van der Waals Equation In Exercise 18 at the end of Section 2.3, we discussed the ideal gas law,...

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Question

The last two questions, differential equations

### The Rate of Evaporation of a Spherical Raindrop

In this exercise, we explore the behavior of a spherical raindrop evaporating at a rate proportional to its surface area. The rate of change of the volume is given as \(-k\) times the surface area, where \(k\) is a positive constant of proportionality. We will derive the differential equations for the volume, radius, and surface area of the raindrop as functions of time.

#### Problem Statement
A spherical raindrop evaporates such that the rate of change of its volume (\(V\)) is proportional to its surface area (\(S\)) by a factor of \(-k\).

\[
\frac{dV}{dt} = -kS
\]

#### Given
- \(dV/dt\) (Rate of change of volume)
- \(dr/dt\) (Rate of change of radius)
- \(dS/dt\) (Rate of change of surface area)

#### Derivations

1. **Volume of the drop:**

The rate of change of the volume of the drop is given by:

\[
\frac{dV}{dt} = -K'
\]

2. **Radius of the drop:**

The rate of change of the radius of the drop is given by:

\[
\frac{dr}{dt} = -k
\]

3. **Surface area of the drop:**

The rate of change of the surface area of the drop is given by:

\[
\frac{dS}{dt} = -k\left(2\left(\frac{7}{3}\right)\left( \left(\frac{1}{3}\right)^{3} \right) \Pi \left(\left(\frac{2}{3}\right)V\left(\left(\frac{1}{3}\right)\right)\right)\right)
\]

This expression showcases how the surface area changes with respect to time, incorporating both the constant of proportionality \(k\) and the current volume \(V\).

#### Notes on Equations

- **Volume Rate Equation:** This is a simple linear relationship reflecting the proportionality between the surface area and the volume rate.
- **Radius Rate Equation:** This indicates a steady change in the radius over time, given by the constant \(k\).
- **Surface Area Equation:** This more complex derivative reflects the geometr

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