I was given a problem for my physics class, and perhaps I have taken it too far, but I wish to finish what I started. I have a function (i warn you that this does not look pretty)

y[t]:= E^(1/2*(Sqrt[-4*g + b^2/m^2] - b/m)*t) (-((h*m)/2) + (-b*h + v)/
    Sqrt[-4*g + b^2/m^2]) + 
 1/2*E^(1/2 (-Sqrt[-4*g + b^2/m^2] - b/m) t)
   h*(1 - (b*h - 2*m*v)/Sqrt[b^2*h^2*m^2 - 4 g*h^2*m^2])

* as a point of reference, the only variable in the equation is t, all others (b, m, h, v, and g) are constants *

Now this equation describes the motion of a falling "particle" in a uniform gravitational field. So far so good. Now you can find the "terminal velocity" of this "particle" by the following performing the following, ideally that is.

* the actual equation in my physics text book looks like the following



Mathematica didn't very much like that. So what I did that I thought would help was to name

y1[t]:= D[y[t],t]


y1[t] = 1/2 E^(1/2 (Sqrt[-4 g + b^2/m^2] - b/m) t) (Sqrt[-4 g + b^2/m^2] - b/m)*(-((h*m)/2) + (-b*h + v)/Sqrt[-4*g + b^2/m^2]) + 
 1/4 E^(1/2 (-Sqrt[-4 g + b^2/m^2] - b/m)*t)
  *h*(-Sqrt[-4 g + b^2/m^2] - b/m) (1 - (b*h - 2*m*v)/Sqrt[
    b^2*h^2*m^2 - 4*g*h^2*m^2])

Where y1[t] would be the equivalent of y'[t] in the previously stated formula. So now what I want to be able to do is the derivative of y1[t] in terms of y1[t] instead of the terms from the entire function.

To repeat, what I get for D[my1[t],t] or D[mD[y[t],t],t], which I name y2[t] is

y2[t] := m*(1/4*E^(
    1/2*(Sqrt[-4*g + b^2/m^2] - b/m)*t)*(Sqrt[-4*g + b^2/m^2] - b/
      m)^2 (-((h*m)/2) + (-b*h + v)/Sqrt[-4*g + b^2/m^2]) + 
   1/8*E^(1/2*(-Sqrt[-4*g + b^2/m^2] - b/m)*t)
     h*(-Sqrt[-4*g + b^2/m^2] - b/m)^2 (1 - (b*h - 2*m*v)/Sqrt[
      b^2*h^2*m^2 - 4*g*h^2*m^2]))

Effectively what I want is to see y2[t] expressed in terms of y1[t]. Not with all the b's, m's, h's, v's, and g's. I get that there will be a few of those constants with a few binary operations or whatever, but all I really need is to see y1[t] terms expressed in the function y2[t].

  • $\begingroup$ Consider differentiating the right side of your first expression, equate it to zero, and Solve it for t. With this result, you can obtain y and related quantities at that point, if you wish. $\endgroup$ – bbgodfrey Sep 7 '15 at 23:59
  • $\begingroup$ Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. $\endgroup$ – bbgodfrey Sep 7 '15 at 23:59

The way you have translated the physics problem into Mathematica is the source of the problem.

Solve[D[m*D[y[t], t], t] == 0, D[y[t], t]]

If your remove D[y[t] from the above expression then Solve will yield a solution.

solution = Solve[D[m*D[y[t], t], t] == 0, t]

{{t -> (I π)/Sqrt[(b^2 - 4 g m^2)/m^2] + (1/Sqrt[((
    b^2 - 4 g m^2)/(
    m^2))])(Log[1/4 b Sqrt[-4 g + b^2/m^2] + b^2/(4 m) - (g m)/2] - 
      Log[(b^2 h)/4 - (b^3 h)/(2 Sqrt[-4 g + b^2/m^2] m) + (b g h m)/
        Sqrt[-4 g + b^2/m^2] + 1/4 b h Sqrt[-4 g + b^2/m^2] m + 
        1/2 g h m^2 - (b v)/2 + (b^2 v)/(2 Sqrt[-4 g + b^2/m^2] m) - (
        g m v)/Sqrt[-4 g + b^2/m^2]])}}

Now you can insert this solution into the expression for D[y[t]]

D[y[t],t] /. solution[[1, 1]]

This is fairly long so I will not copy it in this answer.


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