Recent content by MednataMiza

Well, I haven't estimated it also, just did it as an example. Do a little research :)

The answer to this question is the first result from a google search query. Please, do your homework. Reference Needs["PlotLegends`"] ShowLegend[ ContourPlot[(3 (2 + m t (4 + m t)) - 4 (1 + m t) Cos[\[Pi] x] - 2 Cos[2 \[Pi] x] - 4 (1 + m t) Cos[\[Pi] y] - 2 Cos[2 \[Pi] y] +...

http://www.compileonline.com/compile_fortran_online.php

I think this post should not be here ... And, yes, you can do that by reading the documentation first: http://www.mathworks.com/help/symbolic/mupad_ref/ode-solve.html

What do you see when you look at this plot ? Or you can simply take the table of values of the Tan function and analyse them.

You can use something fancy like this: Plot[Im@Exp[-I x], {x, -5, 5}, PlotStyle -> Thick, ColorFunction -> "StarryNightColors"]

You are welcome :)

You should check the initial settings in r, because I see errors there. For example a missing 't' in the first part :) Check again the problem and repost what is actually given.

Well, using the simple Sum[k[t],{t,0,4Pi}] I, again, receive 26.8967.

Well, I am probably using wrong/incorrect formula. Sticking to your notebook we get 26.8967 as an answer. Can you verify that ?

Something like this should do the trick: Sum[Limit[b, t -> \[Omega]], {\[Omega], 0, 4 Pi}] where 'b' is Where did you get this formula ?

If I am not mistaken \kappa = \left | \frac{f''(x)}{(1 + f'(x)^{2})^{\frac{3}{2}}} \right | is the formula one would use to calculate the curvature. Following this we would end up right here: Well, in theory you can apply //Simplify, but Mathematica has done this already, so in my opinion...

If you have defined a function using the "standard" form, id est, f[x_] := you do not want the x_ on the right hand side of the definition. Now to your real problem: You forgot to specify that the derivatives are also functions of t :) r[t_] := {4 + Cos[20 t]*Cos[t], 4 + Cos[20 t]*Sin[t], 0.4...

I will help you with a couple o'plots: First: Plot of all flow lines / you got that right / Second: Plot of all flow lines + plot of the equation which runs through the point (0,3,2)

Yes, it is :)