# 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?

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Do you mean that the vector normal to the surface at a point $p$, extended as a line outward from $p$, intersects the $z$-axis? I think this would be true almost everywhere, but not necessarily everywhere (for example, I don't think it will work on a circle $\{(x,y,f(\sqrt{x^2+y^2})):x^2+y^2=r^2\}$ when $r$ is a local extrema of $f$, because the normal vector will be parallel to the $z$-axis). Unless you have monotonicity conditions of $f$. –  anon May 21 '12 at 18:32
task : to prove that all normal vectors intersect the axis of rotation. –  sleepy May 22 '12 at 11:10

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.

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o thanks for detailed answer! –  sleepy May 22 '12 at 14:17
$Q = (z_x, z_y, -1)$ =) –  sleepy May 23 '12 at 9:10

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

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