How do you find the linearization at #(5,16)# of #f(x,y) = x sqrt(y)#?

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Answer 1 · 2016-08-05T13:51:27

#z = 5/8 (88 - 8 x - y)#

Consider the surface

#S(x,y,z) = f(x,y)-z=0#.

The linearization of #f(x,y)# at the point #x_0,y_0# is equivalent to determining the tangent plane to #S# at the point #{x_0,y_0,f(x_0,y_0)}#

The tangent plane to #S# is

#Pi_t -> << p-p_0, vec n >> = 0# where

#p_0 ={x_0,y_0,f(x_0,y_0)} ={5,16,20}#
#p = {x,y,z}#
#vec n = grad S = {f_x,f_y,f_z} = {x_0,1/2x_0/sqrt(y_0),1} = {5,5/8,1}#

so

#Pi_t-> (x-5)5+(y-16)5/8+(z-20)=0#

The linearization in the desired point is

#z = 5/8 (88 - 8 x - y)#