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Exaple Find the tangent plane and the normal line of the following surface goes through $(1,1,\frac{\pi}{4})$．

$\displaystyle z = \tan^{-1}\frac{y}{x} $

Answer Given the point $(x_0,y_0,f(x_0,y_0))$ and take a point $(x,y,z)$. Then we can form a vector $(x-x_0, y-y_0, z-f(x_0,y_0)$. Now $((x-x_0, y-y_0, z-f(x_0,y_0)\cdot (f_{x}(x_0,y_0), f_y(x_0,y_0),-1) = f_{x}(x-x_0) + f_y(x_0,y_0) - (z - f(x_0,y_0)) = 0$ Thus $(f_{x}(x_0,y_0), f_y(x_0,y_0),-1)$ is orthogonal to the tangent plane. Furthermore. this gives the equation of the tangent plane such as

$\displaystyle z - f(x_{0},y_{0}) = (f_{x}(x_{0},y_{0}), f_{y}(x_{0},y_{0})) \cdot (x-x_{0},y-y_{0}) $

or

$\displaystyle z = f(x_{0},y_{0}) + f_{x}(x_{0},y_{0})(x-x_{0}) + f_{y}(x_{0},y_{0})(y-y_{0})$

Apply this to the problem, we have

$\displaystyle f_{x}(x,y) = \frac{-\frac{y}{x^2}}{1+(\frac{y}{x})^2} = \frac{-y}{x^2 + y^2}$

$\displaystyle f_{y}(x,y) = \frac{\frac{1}{x}}{1+(\frac{y}{x})^2} = \frac{x}{x^2 + y^2}$

$f(1,1) = \tan^{-1}{1} = \frac{\pi}{4}$ Thus, the equation of the tangent plane is given by

$\displaystyle z = \frac{\pi}{4} - \frac{1}{2}(x-1) + \frac{1}{2}(y-1)$

Alternative solutoin Set $f(x,y,z) = \tan^{-1}{\frac{y}{x}} - z = 0$. Then

$\displaystyle \nabla f(x,y,z) = (\frac{\frac{-y}{x}}{1 + (\frac{y}{x})^2}, \fra... ...x}}{1 + (\frac{y}{x})^2}, -1) = (\frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}, -1)$

Thus,

$\displaystyle \nabla f(1,1,\frac{\pi}{4}) = (-\frac{1}{2}, \frac{1}{2}, -1)$

Note that $\nabla f$ is orthogonal to the surface $f(x,y,z) = 0$. Then any point on $(X,y,x)$ on the $\Gamma$, the vector $(x - 1, y - 1, z - \frac{\pi}{4})$ and $\nabla f(1,1,\frac{\pi}{4})$ are orthogonal．Thus, the equation of the tangent plane is

$\displaystyle (x - 1, y - 1, z - \frac{\pi}{4}) \cdot (-\frac{1}{2}, \frac{1}{2}, -1) = 0$

or

$\displaystyle -\frac{1}{2}(x-1) + \frac{1}{2}(y-1) - (z - \frac{\pi}{4}) = 0$

The normal line is parallel to $\nabla f$. Then by taking a point $(x,y,z)$ on the normal line, we have，

$\displaystyle (x - 1, y - 1, z - \frac{\pi}{4}) = t(-\frac{1}{2}, \frac{1}{2}, -1)$

or

$\displaystyle t = \frac{x-1}{-\frac{1}{2}} = \frac{y-1}{\frac{1}{2}} = \frac{z - \frac{\pi}{4}}{-1} $