My goal is to get an analytical solution of x ( λ\lambda in the problem), when ϵ\epsilon (f[x] in the problem) is maximum (use inverse form).

I found that if I can use trigonometric identity to change Sin[x] and Cos[x] to Sin[2x] or Cos[2x], then I can achieve my goal. Here is the code that I wrote, but I found it is not very good, because for the third line I copy and paste from the evaluation cell.

So, can anyone think of an easier way to change ϵ\epsilon in terms of 2λ\lambda, or can provide another way to achieve the goal?

f[x_] = (R Ï‰^2 Sin[x] Cos[x])/(g – Ï‰^2 (Cos[x])^2 R);

f'[x] // FullSimplify;

Solve[2 R Ï‰^2 (R Ï‰^2 + (-2 g + R Ï‰^2) Cos[2 x]) == 0, Cos[2 x]][[1]] // FullSimplify;

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3

You don’t have to copy anything, try der = f'[x] // FullSimplify Solve[der == 0, Cos[2 x]][[1]] // FullSimplify

– Kuba

Apr 16 ’14 at 6:31

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1 Answer

1

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Another way is to use Maximize rather than solving for the zero of the derivative.

f[x_] = (r Ï‰^2 Sin[x] Cos[x])/(g – Ï‰^2 (Cos[x])^2 r);

as = {r > 0, Ï‰ > 0, g > 0, r Ï‰^2 < g};
You can see that
f[x] // TrigReduce
(* -((r Ï‰^2 Sin[2 x])/(-2 g + r Ï‰^2 + r Ï‰^2 Cos[2 x])) *)
therefore you can make a simple substitution and, assuming the angle is very small,
sol = Simplify[Maximize[(f[x] // TrigReduce) /. {Cos[2 x] -> y, Sin[2 x] -> Sqrt[1 – y^2]},

y],

Assumptions -> as]

(* {(r Ï‰^2)/(2 Sqrt[g (g – r Ï‰^2)]), {y -> (r Ï‰^2)/(2 g – r Ï‰^2)}} *)

Thanks a lot. Your way is exactly what I want to get.

– Lawerance

Apr 16 ’14 at 7:13