Question

Terminal Velocity In Section 1.3 we saw that the autonomousdifferential equation$\mathit{m}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{m}\mathit{g}\mathbf{-}\mathit{k}\mathit{v}$,where k is a positive constant and g is the acceleration due togravity, is a model for the velocity v of a body of mass m that isfalling under the influence of gravity. Because the termrepresents air resistance, the velocity of a body falling from a greatheight does not increase without bound as time t increases. Use aphase portrait of the differential equation to find the limiting, orterminal, velocity of the body. Explain your reasoning.

Short answer

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

Step-by-step solution

Find the value of v

The given differential equation, can be written as

$\begin{array}{c}\frac{dv}{dt}=\frac{mg-kv}{m}\\ \frac{dv}{dt}=g-\frac{kv}{m}\end{array}$

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

$\begin{array}{c}\frac{dv}{dt}=0\\ g-\frac{kv}{m}=0\\ \frac{kv}{m}=g\\ v=\frac{mg}{k}\end{array}$

The phase portrait

The phase portrait for the given differential equation is shown below,

From the phase portrait, we see that$\frac{mg}{k}$ is an asymptotically stable critical point. Thus,

$\underset{t\to \infty }{\lim }v=\frac{mg}{k}$

Hence, the terminal velocity of the given differential equation is.$\underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{v}\mathbf{=}\frac{\mathbf{m}\mathbf{g}}{\mathbf{k}}$