# How to plot the Joukowski curve...?

Posted 8 years ago
5018 Views
|
7 Replies
|
3 Total Likes
|
 How to plot the Joukowski curve...? Thank you for your cooperation.
7 Replies
Sort By:
Posted 8 years ago
 Merci...
Posted 8 years ago
 You can get the area of the curve using differential forms. In[32]:= tw[t] = TensorWedge[reimz[t], dz[t]][[1, 2]] Out[32]= ((3 + 25 Cos[t]) (567 + 75 Cos[t] + 250 Sin[t]) (-15000 + 81289 Cos[t] + 55050 Cos[t]^2 + 5625 Cos[t]^3 - 6000 Sin[t] + 183500 Cos[t] Sin[t] + 37500 Cos[t]^2 Sin[t] + 62500 Cos[t] Sin[t]^2))/(16 (367 + 75 Cos[t] + 250 Sin[t])^3) + (5 (334 + 150 Cos[t] + 1335 Sin[t] + 375 Cos[t] Sin[t] + 1250 Sin[t]^2) (6000 Cos[t] + 50000 Cos[t]^2 + 206289 Sin[t] + 55050 Cos[t] Sin[t] + 5625 Cos[t]^2 Sin[t] + 233500 Sin[t]^2 + 37500 Cos[t] Sin[t]^2 + 62500 Sin[t]^3))/(16 (367 + 75 Cos[t] + 250 Sin[t])^3) In[33]:= 1/2 NIntegrate[tw[t], {t, 0, 2 \[Pi]}] Out[33]= 1.95895 
Posted 8 years ago
Posted 8 years ago
 In[1]:= a = 2; In[2]:= b[t_] = 3/10 + I + 5/2 Exp[I t]; In[3]:= z[t_] = 1/2 (b[t] + a^2/b[t]) Out[3]= 1/2 ((3/10 + I) + (5 E^(I t))/2 + 4/((3/10 + I) + (5 E^(I t))/2)) In[7]:= reimz[t_] = {Re[z[t]], Im[z[t]]} // ComplexExpand // FullSimplify Out[7]= {1/ 20 (3 + 25 Cos[t]) (1 + 200/(367 + 75 Cos[t] + 250 Sin[t])), -(1/ 2) + (5 Sin[t])/4 + (267 + 75 Cos[t])/( 367 + 75 Cos[t] + 250 Sin[t])} In[12]:= dz[t_] = D[reimz[t], t] // FullSimplify Out[12]= {-(( 250 (3 + 25 Cos[t]) (10 Cos[t] - 3 Sin[t]))/(367 + 75 Cos[t] + 250 Sin[t])^2) - 5/4 Sin[t] (1 + 200/(367 + 75 Cos[t] + 250 Sin[t])), 5/4 (Cos[t] - ( 600 (25 + 89 Cos[t] + 10 Sin[t]))/(367 + 75 Cos[t] + 250 Sin[t])^2)} In[16]:= NIntegrate[Sqrt[dz[t].dz[t]], {t, 0, 2 \[Pi]}] Out[16]= 9.40235 
Posted 8 years ago
 How to calculate the circumference of the curve?
Posted 8 years ago
Posted 8 years ago
 Define b as a function of t. Then define z as a function of t, using b[t]. Then do a ParametricPlot of the real and imaginary parts of z[t]