# How gradient of a function be perpendicular to the surface?

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?

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, $$\vec u\perp \vec v=0\Leftrightarrow \vec u\cdot \vec v=0.$$

• Sorry for responding late. Im asking from the equation $\phi (x,y,z)$ =c\$ if you take partials both sides then i can tell the gradient is zero as well. Please elaborate where am i wrong ? – user187604 Oct 29 '18 at 14:43