[net.math] Calculus problem

mse@iham1.UUCP (Scott Erickson) (03/31/84)

Help!, Anyone that could provide insight to this problem, please respond
	via mail.  Disregard after April 3.  The answer is given, I need
	to know how to work the problem.  Thanks in advance.


	A body starting from rest falls under the attraction of gravity but
	encounters resistance proportional to the square of its velocity.
	Show that if the body could continue to fall indefinitely under these
	same conditions, its velocity would approach a limiting value, and
	find the distance it would fall in time t.

The answers are:

       ________
a) lim v = / mg / k

                              ________
   b)     s = (m/k) ln[cosh(t/ gk / m )]


Where:

	v = velocity
	s = distance
	m = mass
	g = acceleration due to gravity
	t = time
	k = proportionality constant

gwyn@brl-vgr.ARPA (Doug Gwyn ) (03/31/84)

Your problem boils down to two assumed initial conditions:

	s(t=0) = 0

	v(t=0) = 0	where v = ds / dt

and one force equation:

	(F = )  m a  =  m g - k v^2	where a = dv / dt = d^2s / dt^2

(sorry about the superscripts)

In terms of s(t) this is a second-order non-linear ordinary differential
equation with two equations of constraint (boundary conditions):

	s(0) = 0

	s'(0) = 0	where ' denotes time-derivative

	m s''(t) = m g - k (s'(t))^2

What you do at this point depends on why you are asking the question.
I myself would just look up the solution and save time.  Since you
have the solution for s(t), all you technically need to do is just
demonstrate (by substitution) that it satisfies the three constraints.
If this is for a course in ODE, you should use some of the tricks
taught in the course to methodically synthesize the answer.

Note that IF there is a limiting velocity, its value is trivial to
derive, since then a == 0 and the force equation is trivial to solve
for v.