Directional Derivative Derivation

Question

I don't understand the part underlined in the derivation of the directional derivative. Why is the $\lim_{Q \to P}$ interchangeable with $\lim_{N \to P}$? I understand that the surfaces are getting closer together but doesn't whether or not N is getting closer to P depend on the direction in which one surface is moving closer to the other?

For instance N could just be hovering above P but as Q moves closer to P N moves more 'North west' than 'south west' (referring to how the graph appears on the page) and hence moves further away from P?

Answer 1

Because PN is defined as the norm of $P$. So as Q approaches P, N must approach Q. But since Q approaches P (as was just stated), then N must also approach P.

Symbolically, $Q\to P$ implies $N\to Q$, therefore $N\to P$.

The key point is to note that $\theta$ is not fixed.

Answer 2

The vectors $\hat{\mathbf{n}}$ and $\hat{\mathbf{s}}$ are fixed and constant.

The point $N$ is defined as being in direction $\hat{\mathbf{n}}$ from point $P$ and point $Q$ is in direction $\hat{\mathbf{s}}$ from point $P$. So you are restricted in the way $N$ and $Q$ can approach $P$. This answers your question exactly since $Q$ approach $P$ now explicitly requires $N$ to approach $P$ along this specific direction.

A better way to think about it is this: $N=P+\hat{\mathbf{n}}h$ and $Q=P+\hat{\mathbf{s}}h$ where $h$ is directly proportional to the distance between $N$ and $Q$. Now take the limit as $h$ goes to zero.

Answer 3

It seems that we have a function $$\phi:\quad{\mathbb R}^3\to{\mathbb R},\qquad (x,y,z)\mapsto \phi(x,y,z)$$ modeling air pressure. Given a point $P\in{\mathbb R}^3$ the equation $$\phi(x,y,z)=\phi(P)$$ defines the "isobaric surface" through $P$. When $\nabla \phi(P)\ne{\bf 0}$ this is a bona fide smooth surface whose unit normal at $P$ is given by $${\bf n}(P):={\nabla \phi(P)\over|\nabla \phi(P)|}\ ,$$ whereby ${\bf n}$ points into the direction of increasing pressure. It is customary to denote the directional derivative of $\phi$ in direction ${\bf n}$ by ${\partial\phi\over\partial n}$; whence $${\partial\phi\over\partial n}=\nabla\phi(P)\cdot{\bf n}=|\nabla\phi(P)|\ .$$

In addition we are given a certain unit vector ${\bf s}$ attached at $P$, and we want to know the directional derivative of $\phi$ at $P$ in direction ${\bf s}$. This directional derivative can be denoted by ${\partial\phi\over\partial s}$, and is given by $${\partial \phi\over\partial s}(P)=\nabla \phi(P)\cdot{\bf s}=|\nabla \phi(P)|\ {\bf n}\cdot{\bf s}={\partial\phi\over\partial n}\cos\theta\ .$$ Note that the above formulas have assumed that ${\bf n}$ and ${\bf s}$ are unit vectors.