# How do you find the linear approximation L to f at the designated point P. compare the error in approximating f by L at the specified point Q with the distance between P and Q given f(x,y) = 1/sqrt(x^2+y^2), P(4,3) and Q(3.92, 3.01)?

May 31, 2016

$d = 0.0000142382$

#### Explanation:

The tangent plane $\Pi$ to a Surface $S$ in ${p}_{0} \in S$ is obtained as

$\Pi \to \left(p - {p}_{0}\right) . {\vec{n}}_{0} = 0$

where

$p = \left(x , y , z\right)$ and ${\vec{n}}_{0}$ is the normal vector to $S$ at ${p}_{0}$.

The normal vector to $S$ is computed as

$\vec{n} = \nabla S = \left(\frac{\partial S}{\partial x} , \frac{\partial S}{\partial y} , \frac{\partial S}{\partial z}\right)$.

Giving

$S \left(x , y , z\right) = z - \frac{1}{\sqrt{{x}^{2} + {y}^{2}}} = 0$ then
$\vec{n} = \left(\frac{x}{{x}^{2} + {y}^{2}} ^ \left(\frac{3}{2}\right) , \frac{y}{{x}^{2} + {y}^{2}} ^ \left(\frac{3}{2}\right) , 1\right)$
At ${p}_{0} = \left(4 , 3 , \frac{1}{\sqrt{{4}^{2} + {3}^{2}}}\right)$ gives
${\vec{n}}_{0} = \left(\frac{4}{125} , \frac{3}{125} , 1\right)$

$\Pi \to \frac{4}{125} \left(x - 4\right) + \frac{3}{125} \left(y - 3\right) + z - \frac{1}{5} = 0$
Given a point ${q}_{S} = \left(3.92 , 3.01 , 0.202334\right) \in S$
and a point ${q}_{\Pi} = \left(3.92 , 3.01 , 0.20232\right) \in \Pi$
Their distance is $\left\lVert {p}_{S} - {p}_{\Pi} \right\rVert = 0.0000142382$