Constructing a Cone and its Normal Vectors in Spherical Coordinates I am attempting to construct a right circular cone of maximum radius $R$ and angle $\theta$ in spherical coordinates, then find the normal vector of the surface of this cone at all points. Here's what I have:
$$\text{cone}(r, \theta, \phi) = 
\begin{cases} 
x & = & r\cos{\theta}\cos{\phi} \\ 
y & = & r\cos{\theta}\sin{\phi} & \\
z & = & r\sin{\theta}
\end{cases}
$$
$$
\text{Such That:}
\begin{cases}
0 \leq \phi \leq 2\pi \\
\theta = \text{constant} \\
0 \leq r \leq R
\end{cases}$$
For the normal vector, we know that the equation of a cone in cartesian coordinates is $~~x^2 + y^2 - z^2 = 0$. To find the normal vector to this surface, we take the gradient of the equation and convert it to spherical coordinates:
$$\nabla(x^2 + y^2 - z^2) = ~ <2x, 2y, -2z> ~ = ~2\cdot \text{cone}(r,-\theta,\phi)$$
Is this correct?
Although it may be correct, there is some part of my brain that doesn't fully grasp what I'm doing here. I think I lack a way of thinking of the construction of the normal vector geometrically. Can anyone give me some insight?
 A: As far as I can tell, your calculus is correct. The way to think about it (at least, I do), is to do it one dimension at a time. think about the normal vector to a particular grid-line on the surface of the cone, and then slide it around.
Also, remember that a cone is constructed physically from a pie (with my slice cut out ;)) rolled up. Thus, the vector that originates from the origin and goes out to the surface of the cone must be perpendicular at that point - and thus the normal. By substitution, the equation of the cone is:
$$x^2+y^2=z^2$$
$$r^2 \cos^2 \theta  \cos^2 \phi + r^2 \cos^2 \theta \sin^2 \phi = r^2 \sin^2 \theta$$
Cancel out the $r^2$ term.
$$ \cos^2 \theta \cos^2 \phi + \cos^2 \theta \sin^2 \phi  = \sin^2 \theta$$
Taking advantage of the trig identity: $\cos^2 \phi + \sin^2 \phi = 1$
$$ \cos^2 \theta = \sin^2 \theta$$
This is only true at 45 degree angles: $ \pi/4, 3\pi/4, ... $
A: Regarding your derivation: your answer does seem to be correct.
Regarding the intuition behind the gradient vector being the normal to a surface, it relies on two things:
First, that you can construct the equation for any plane, given it's normal vector,
$\vec{n}=p\vec{i}+q\vec{j}+r\vec{k}$
and the distance vector for arbitrary point $(a,b,c)$, $\vec{d}=(x-a)\vec{i}+(y-b)\vec{j}+(z-c)\vec{k}$, by taking the dot product between them, and observing that they are perpendicular: $\vec{n}\cdot\vec{d}=0$. This becomes the familiar plane equation $p(x-a)+q(y-b)+r(z-c)=0$.
The other thing we need, is that to find the tangent-plane to a surface, we find it's local linearization. 
At the point $(a,b,c)$, it's $f_x(a,b,c)(x-a)+f_y(a,b,c)(y-b)+f_z(a,b,c)(z-c)=0$.
Using the above two facts, we get that $\vec{n}=f_x(a,b,c)\vec{i}+f_y(a,b,c)\vec{j}+f_z(a,b,c)\vec{k}$, which is the gradient of $f$ at $(a,b,c)$, $\nabla f(a,b,c)$. Therefore, the normal vector to a surface is it's gradient vector. 
The above basically shows that since the gradient vector is normal to the tangent plane at an arbitrary point $(a,b,c)$ on the surface, it's also normal to the surface at that same point. That has to do with the fact, that if you "zoom-in" on the surface at $(a,b,c)$, it will be similar to a plane (in particular it's tangent plane), so the normal vectors are the same.
