Normals of an implicit surface

Question

I have a function $f\colon\mathbb{R}^3\to\mathbb{R}$. For two numbers $a$ and $b$, I define the function $g$ by $$ g(\rho,\theta,\phi)=f\bigl(\rho\cos(a\theta)\sin(b\phi), \rho\sin(a\theta)\sin(b\phi), \rho\cos(b\phi)\bigr). $$ With the help of a computer, I derive the implicit surface $g(\rho,\theta,\phi)=0$. I consider that the points of this surface are spherical coordinates, and I convert them to Cartesian coordinates: $(\rho,\theta,\phi) \mapsto (\rho\cos\theta\sin\phi, \rho\sin\theta\sin\phi, \rho\cos\phi)$. In this way I get a new surface. My problem is to get the normals of this surface. How to get the normals? I have the gradient of $f$.

Answer 1

The normal to a surface $S\subset \Bbb R^3$ defined by $S=F^{-1}(\{0\})$ for a smooth function $F:\Bbb R^3\to\Bbb R$ with $0$ as a regular value is just $\nabla F$. This means that at any point $p\in S$, the normal is $\nabla_p F$ (you can scale it if you want it to be unitary), and the tangent space $T_pS$ is just $\nabla_p F^\perp$.

Now you just have to compute $\nabla g(p)$ in terms of $\nabla f(p)$, $a$ and $b$. Note that $g$ is of the form $g=f\circ h$ with $h$ defined by $$h(\rho,\theta,\phi)=\bigl(\rho\cos(a\theta)\sin(b\phi), \rho\sin(a\theta)\sin(b\phi), \rho\cos(b\phi)\bigr).$$ Using the chain rule you have for all $u\in\Bbb R^3$ $$d_p g(u)=d_{h(p)}f\circ d_p h(u)= \langle \nabla f(h(p)), d_ph(u)\rangle.$$ Compute $d_ph(u)$, develop the scalar product and write it in the form $$a_1u_1+a_2u_2+a_3u_3$$ and you will have $\nabla g(p)=(a_1,a_2,a_3)$.