Question

In Section 1.3 we saw that the autonomous differential equation m dv dt = mg − kv, where k is a positive constant and g is the acceleration due to gravity, is a model for the velocity v of a body of mass m that is falling under the influence of gravity. Because the term −kv represents air resistance, the velocity of a body falling from a great height does not increase without bound as time t increases. Use a phase portrait of the differential equation to find the limiting, or terminal, velocity of the body.lim v(t -> infinity)= ?????

Answer 1

Answer:

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

Step-by-step explanation:

Firstly its necessary to put the autonomous DE in its normal form to get

$\frac{dv}{dt} = g - \frac{kv}{m}$

To get the critical point we solve

g - $\frac{kv}{m} = 0$

$\frac{kv}{m} = g--> v = \frac{mg}{k}$

The critical point is $v = \frac{mg}{k}$

$\frac{dv}{dt} > 0 for v < \frac{mg}{k}$

and

$\frac{dv}{dt} < 0 for v > \frac{mg}{k}$

from the phase portrait we observe that the critical point

v = $\frac{mg}{k}$  is stable

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