How gradient of a function be perpendicular to the surface?

Question

Suppose we have $\phi (x,y,z) = c$ here c is a constant and $\phi (x,y,z)$ is a surface . Now we take differential of both sides and say $d\phi = \nabla \phi . dr$ =0. Now i say how can i prove this that $\nabla \phi $ is perpendicular to dr or the unit vectors of the surface ? As $\nabla \phi$ is also zero (if we take partial of the both sides of the very first equation. Then gradient of $\phi $ is zero as well. How can we conclude that $\nabla \phi$ is perpendicular to unit vector of $\phi$ ? If gradient can be zero as well the dot product then also can be zero . Where am i wrong?

Answer 1

This follows from the definition of derivative as a linear transformation, together with the dot product:

assuming that $\phi$ is differentiable at $(x,y,z),$ we have

$D\phi(xy,z):\mathbb R^3\to \mathbb R:(r_1,r_2,r_3)\to$

$\frac{\partial \phi}{\partial x}(x,y,z)r_1+\frac{\partial \phi}{\partial x}(x,y,z)r_2+\frac{\partial \phi}{\partial x}(x,y,z)r_3=$

$\nabla(x,y,z)\cdot (r_1,r_2,r_3)$, using the defintion of the dot product.

Now the right hand side of your equation is $c$, and so its derivtative is zero.

Therefore, $\nabla(x,y,z)\cdot (r_1,r_2,r_3)=0$, which means that $\nabla(x,y,z)\perp (r_1,r_2,r_3)$, using the fact that in general, $\vec u\perp \vec v=0\Leftrightarrow \vec u\cdot \vec v=0.$