How do you find the linearization at (-2,1) of #f(x,y) = x^2y^3 - 4sin(x+2y) #?

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Answer 1 · 2016-06-17T12:21:19

#8 (2 + x) - 4 ( y-1) +z-4=0#

Let us consider the surface

#f(x,y,z) = z-x^2y^3 + 4sin(x+2y)=0 #

#f(x,y,z)# is analytical, and at point #p_0={-2,1,4}#

the surface normal vector is given by

#grad f(x,y,z) = {f_x,f_y,f_z} = vec n#

or

#vec n={-2 x y^3 + 4 Cos(x + 2 y), -3 x^2 y^2 + 8 Cos(x + 2 y), 1}#

at point #p_0# we have

#vec n_0 = {8,-4,1}#

The tangent plane at point #p_0# is given by

#<< vec n_0,p-p_0 >> = 0# with #p = {x,y,z}#

So

#<< vec n_0,p-p_0 >> =8 (2 + x) - 4 ( y-1) +z-4 = 0#