normal to the surface of revolution Please, help me with my homework...
Prove that the normal to the surface of revolution $z = f(\sqrt{x^2+y^2})$
 intersect the axis of rotation.
first, I need to find $z_x$, $z_y$
$z_x$ = $x*\frac{f'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}$
$z_y$ = $y*\frac{f'(\sqrt{x^2+y^2})}{\sqrt{x^2+y^2}}$
Next, I need to find the equitation of tangential surface
$z_x(x_0, y_0)*(x-x_0) + z_y(x_0, y_0)*(y-y_0) = 0 $
Next, the normal :
$\frac{x-x_0}{z_x(x_0,y_0)}=\frac{y-y_0}{z_y(x_0,y_0)}$
And what's the next step? How to prove? 
 A: I find this stuff easier using parametric equations. For the surface of revolution, we can
use the equations
$$ S(u,v) = (u\cos v, u\sin v, f(u))$$
Then $z = f(\sqrt{x^2 + y^2})$, as in your problem.
Partial derivatives are:
$$S_u = ( \cos v, \sin v, f'(u))$$
$$S_v = ( -u \sin v, u \cos v, 0)$$
So the surface normal vector is in the direction
$$ Q(u,v) = S_u \times S_v = (-u f'(u) \cos v, -u f'(u) \sin v, u)$$
and so the normal line at $(u_0, v_0)$ is 
$$ L(t) = S(u_0, v_0) + tQ(u_0, v_0)$$
This simplifies to 
$$ L(t) = (1-tf'(u_0))(u_0\cos v_0, u_0 \sin v_0, 0) + (0, 0, tu_0 + f(u_0))$$
When $t = 1/f'(u_0)$, you can see that this gives you a point on the $z$-axis
A: I thought I'd give another answer that's more aligned with the work you already did.
You calculated the partial derivatives $z_x$ and $z_y$. So, the surface normal is in the direction of the vector $Q = (z_x, z_y, z_z) = (z_x, z_y, 1)$. So, as in my first answer, the normal line at the point $(x_0, y_0, z_0)$ has equation
$L(t) = ((x_0, y_0, z_0) + tQ$. As in my first answer, you can find a value of $t$ that makes the first two components of $L(t)$ zero. The details: find a value of t that makes the first component of $L(t)$ zero, and then just confirm that this makes the second component zero, also.
You calculated the equation of the tangent plane, too, but I don't see any need for that, really.
