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

1 Answer
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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