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

May 29, 2016

The local linearization in point ${p}_{0} = \left(2 , 9 , 6\right)$ is given by
$- 3 \left(x - 2\right) + \frac{9 - y}{3} + z - 6 = 0$

#### Explanation:

The tangent plane to the surface $S \left(x . y , z\right) = z - x \sqrt{y} = 0$ in the point ${p}_{0} = \left(2 , 9 , 2 \sqrt{9}\right) = \left(2 , 9 , 6\right)$ is obtained as follows.
The first step is the normal vector to $S \left(2 , 9 , 6\right)$. The vector ${\vec{v}}_{0}$ is obtained with the calculation of $\nabla S \left(x , y , z\right) = \left(\frac{\partial S \left(x , y , z\right)}{\partial x} , \frac{\partial S \left(x , y , z\right)}{\partial y} , \frac{\partial S \left(x , y , z\right)}{\partial z}\right)$
in ${p}_{0}$ giving
${\vec{v}}_{0} = {\left(- \sqrt{y} , - \left(\frac{x}{2 \sqrt{y}}\right) , 1\right)}_{0} = \left(- 3 , - \left(\frac{1}{3}\right) , 1\right)$
Now, the tangent plane or the so called local linearization is given by
$\left(p - {p}_{0}\right) . {\vec{v}}_{0} = 0 \to - 3 \left(x - 2\right) + \frac{9 - y}{3} + z - 6 = 0$