Question

Find equations of the tangent plane and normal line to the surface z = arctan(y/x) at the point P(1,1,pi/4)
Please show work and thank you

Answer 1

First lets find the gradiant of,
$$\phi=\tan z -\frac yx$$
$$\nabla\phi=\left \langle \frac y{x^2},-\frac 1x,\sec ^2 z \right\rangle$$

since this is perpendicular to the surface-
$$z=\tan^{-1} \frac yx$$

the perpendicular vector to the surface at p=(1,1,pi/4)
$$\vec n=\nabla \phi_{(1,1,\frac \pi 4)}=\langle1,-1,2\rangle$$
equation of a plane with a perpendicular vector n, and a point p on the surface is,
$$\vec n \cdot(\vec r-\vec p)=0$$
where $$\vec r=\langle x,y,z \rangle$$

equation of the normal line can be taken from,
$$t(\vec r-\vec p)=\vec n$$