How do you find the linearization at (2,9) of #f(x,y) = xsqrty#?

1 Answer
Cesareo R.
May 29, 2016

The local linearization in point #p_0=(2,9,6)# is given by
# -3 (x-2) + (9 - y)/3 + z - 6 = 0#

Explanation:

The tangent plane to the surface #S(x.y,z)=z-x sqrt(y) = 0# in the point #p_0=(2,9,2sqrt(9))=(2,9,6)# is obtained as follows.
The first step is the normal vector to #S(2,9,6)#. The vector #vec v_0 # is obtained with the calculation of #grad S(x,y,z) = ((partial S(x,y,z))/(partial x),(partial S(x,y,z))/(partial y),(partial S(x,y,z))/(partial z))#
in #p_0# giving
#vec v_0 = (-sqrt[y], -(x/(2 sqrt[y])), 1)_0 = (-3, -(1/3), 1)#
Now, the tangent plane or the so called local linearization is given by
#(p - p_0).vec v_0 = 0 -> -3 (x-2) + (9 - y)/3 + z - 6 = 0#

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