Question

Suppose the model in Problem 40 is modified so that air resistance is proportional to v²,that is,$\mathit{m}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{m}\mathit{g}\mathbf{-}\mathit{k}{\mathit{v}}^{\mathbf{2}}$See Problem 17 in Exercises 1.3. Use a phase portrait to find the terminal velocity of the body. Explain your reasoning.

Short answer

The terminal velocity of the given differential equation is.$\underset{t\to \infty }{\lim }v=\sqrt{\frac{mg}{k}}$

Step-by-step solution

Find the value of v

The given differential equation, can be written as

$\begin{array}{c}\frac{dv}{dt}=\frac{mg-k{v}^2}{m}\\ \frac{dv}{dt}=g-\frac{k{v}^2}{m}\end{array}$

To calculate the phase portrait, first we need to calculate the value of v such that

$\begin{array}{c}\frac{dv}{dt}=0\\ g-\frac{k{v}^2}{m}=0\\ \frac{k{v}^2}{m}=g\\ v^2=\frac{mg}{k}\\ v=\pm \sqrt{\frac{mg}{k}}\end{array}$

Since a free-falling object with velocity pointing downward does not make sense for the velocity to be negative. So the range for V is $[0,\infty )$and the only critical point is the positive square root,$v=\sqrt{\frac{mg}{k}}$

The phase portrait

The phase portrait for the given differential equation $\frac{dv}{dt}=g-\frac{k{v}^2}{m}$, where $k=1,m=1$and is$g=9.8$ shown below,