obliczyc gradient w jednym wybranym punkcie i dorysować go do wykresu
stworzyć widget pozwalajacy na wybranie punktu w którym rysuje sie wybrany gradient $$ \nabla f = \left[ \frac{\partial f}{\partial x},\frac{\partial f}{\partial y} \right] $$



In [3]:

    
%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
nx = 44
ny = 44
x1,y1 = 5,5

X,Y = np.meshgrid(np.linspace(0,x1,nx),np.linspace(0,y1,ny))
X.shape









    Out[3]:





(44, 44)



In [21]:

    
def gradient_f_x(x,y):
    return( 2*np.cos(2*x)*np.cos(y) )
def gradient_f_y(x,y):
    return( -np.sin(2*x)*np.sin(y) )



In [22]:

    
import sympy
from sympy.abc import x,y



In [23]:

    
F_sympy = sympy.sin(2*x)*sympy.cos(y)
F_sympy.diff(y)









    Out[23]:





-sin(2*x)*sin(y)



In [32]:

    
gradient_f_x_z_sympy = sympy.lambdify((x,y),F_sympy.diff(x),"numpy")
gradient_f_y_z_sympy = sympy.lambdify((x,y),F_sympy.diff(y),"numpy")



In [31]:

    
gradient_f_x(1,2),gradient_f_x_z_sympy(1,2)









    Out[31]:





(0.34635637913638812, 0.34635637913638812)



In [43]:

    
gradient_f_x_z_sympy(1,2),gradient_f_y_z_sympy(1,2)









    Out[43]:





(0.34635637913638812, -0.82682181043180603)



In [44]:

    
f = lambda X_,Y_:np.sin(2*X_)*np.cos(Y_)
Z = f(X,Y)
plt.axes().set_aspect('equal')
plt.arrow(1,2,0.34635637, -0.826821810,head_width=0.4)
plt.contourf(X,Y,Z)









    Out[44]:





<matplotlib.contour.QuadContourSet at 0x7f2b058b07b8>



In [41]:

    
from ipywidgets import widgets
from ipywidgets import interact



In [42]:

    
@interact(x0=widgets.FloatSlider(min=0,max=5),\
          y0=widgets.FloatSlider(min=0,max=5) ) 
def myplot(x0,y0):
    plt.plot( x0,y0,'ro')
    plt.contourf(X,Y,Z)
    plt.show()



In [39]:

    
myplot(1,2)



In [ ]: