posted by Erica on .
The velocity v of a freefalling skydiver is well modeled by the differential equation
where m is the mass of the skydiver, g is the gravitational constant, and k is the drag coefficient determined by the position of the driver during the dive.
(a)Find the general solution of the differential equation.
(b) Calculate the limit of v(t) as t -> INF.
I just started with dv/dt=g-(k/m)v^2 I don't really know where to go from there. Tried integrating dv/(g-(k/m)v^2) but I get stucked.
The answer for (a) is supposed to be v(t) = sqrt(m/(kv)*(Ce^(2*sqrt(kg/m)*t)-1)/(Ce^(2*sqrt(kg/m)*t)+1)
and for (b)sqrt (mg/k)
you have dv/dt = F(v)
use separation of variables to get
v'/F(v) = 1
dv/F(v) = dx
Now, F(v) = g-(k/m)v^2 = k/m (a^2-v^2)
where a^2 = gm/k
That integrate into 1/a arctanh(v/a) = t
which is just a big exponential expression like what you show.
arctanh(z) = 1/2 ln((1+z)/(1-z))
The details get messy. If you go to
you can enter the equation (sorry, no letter constants, but you can use y'=1-2*y**2 to see the method). It will show you t as a function of v, with logs as above. Convert to v as a function of t to encounter the exponentials you want.