Thread: Differential Equation direction fields

The equation for a stone falling under gravity with air resistance,
dv/dt = g - kv^2, v > 0,
where k is a constant drag coefficient. Pay particular attention to the direction along the line v = sqrt{g/k} = vt. What is the significance of vt?
Sketch solution graphs which start at t = 0,
(i) for an initial velocity > vt and
(ii) for an initial velocity < vt.
I know about putting in values to draw the direction fields, but this question has thrown many people in our class and our lecturer said it was easy, so any help would be greatly appreciated.

is terminal velocity ... note that when .

for , the speed will increase at a decreasing rate, with as .

for , the speed will decrease at a decreasing rate, with as .

hope that helps with your sketch.

Originally Posted by njr008

I know about putting in values to draw the direction fields, but this question has thrown many people in our class and our lecturer said it was easy, so any help would be greatly appreciated.

Hey, if I shoot a bullet straight up, how fast is it going when it hits the ground? Not too right? Same way if I start up there. If I shoot it down faster than terminal velocity, it will slow down, if I shoot it slower, it will speed up. So the plots look like the one I drew below: asymptotic to terminal velocity right? Here's the Mathematica code if you want to experiment with it:

Code:

line = Graphics[{Dashed, 
     Line[{{0, N[Sqrt[16/0.05]]}, 
       {5, N[Sqrt[16/0.05]]}}]}]; 
slow = NDSolve[{Derivative[1][v][t] == 
      g - k*v[t]^2 /. {g -> 16, k -> 0.05}, 
    v[0] == 1}, v, {t, 0, 5}]
slowplot = Plot[v[t] /. Flatten[slow], 
   {t, 0, 5}]
fast = NDSolve[{Derivative[1][v][t] == 
      g - k*v[t]^2 /. {g -> 16, k -> 0.05}, 
    v[0] == 40}, v, {t, 0, 5}]
fastplot = Plot[v[t] /. Flatten[fast], 
   {t, 0, 5}, PlotRange -> All]
Show[{slowplot, fastplot, line}, 
  PlotRange -> {{0, 5}, {0, 50}}]