Question

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

Answer 1

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`