I am seeking the connection between two formulas that I saw to compute the directional derivative of function $f$ in the direction of a vector $\vec v$.
One of them is :
$$\nabla_{\vec v} f(\vec x_0) = \lim_{h\to 0} \frac{f(\vec x_0+h\vec v)−f(\vec x_0)}{h}$$
and the other one is
$$\nabla_{\vec v} f(x,y)=\nabla f\cdot \vec v$$
Also, will I have to use a unit vector to compute the directional derivative and the slope?
Thank you and cheers :)
Let $f: D\subseteq \Bbb R^n \to \Bbb R$ be differentiable. Then
$$\begin{align}\require{cancel}\nabla_{\vec v} f(\vec x_0) &= \lim_{h\to 0} \frac{f(\vec x_0 + h\vec v)-f(\vec x_0)}{h} \\ &= \lim_{h\to 0} \frac{\left(\color{red}{\cancel {\color{black}{f(\vec x_0)}}}+\nabla f(\vec x_0)\cdot (h\vec v) + o(h)\right) - \color{red}{\cancel {\color{black}{f(\vec x_0)}}}}{h} \\ &= \lim_{h\to 0} \frac{\color{red}{\cancel {\color{black}{h}}}\left[\nabla f(\vec x_0)\cdot (\vec v)\right]}{\color{red}{\cancel {\color{black}{h}}}} + \cancelto{0}{\lim_{h\to 0}\frac{o(h)}{h}} \\ &= \nabla f(\vec x_0) \cdot \vec v\end{align}$$
You don't have to use a unit vector to calculate the directional derivative, but the dd will only correspond to the geometric idea of slope if you use a unit vector $\vec v$.
Edit: I assume that you are familiar with Taylor's theorem. Recall that the first order Taylor expansion of a function $g: \Bbb R\to \Bbb R$ around $a$ is
$$g(a+h) = g(a) + g'(a)h + o(h)$$
Here $o(h)$ is a stand-in for the remainder function $g(a+h)-g(a)-g'(a)h$. This notation (called little oh notation) tells us that the remainder has the property $$\lim_{h\to 0}\frac{g(a+h)-g(a)-g'(a)h}{h} = 0.$$
For functions of a vector variable, there's a similar Taylor expansion:
$$f(\vec a + \vec h) = f(\vec a) + \nabla f(\vec a)\cdot \vec h + o(\|\vec h\|)$$
So what I'm doing above is replacing $f(\vec x_0 + h\vec v)$ with its first order Taylor expansion. Then two terms cancel, one tends to zero, and we're left with the identity you're looking for.
Another way to intuatively prove this, is take a scalar function F(x,y)
and a vector function R(t)= x(t) i + y(t)j
substituting x(t),y(t) into F yields F(x(t),y(t))
this obviously represents the values of F along the vector path
Differentiating F with respect to T using the chain rule
dF/dt= (dF/dx)(dx/dt) + (dF/dy)(dy/dt)
this represents the RATE OF CHANGE of F along the vector path R
Notice that the expression for dF/dt is identicle to a dot product between the vectors (df/dx, df/dy) and (dx/dt , dy/dt)
these vectors are identical to Grad F and R'(t)
so what does this mean? we have found that the vectors Grad F (dot) R'(t) represent the rate of change of F in the direction of R(t) well what is the direction of R(t)? clearly its R'(t) so this equation represents the rate of change of F at a point T in the direction of R'(t).
So all we have to do to get the directional derivative is to "artifically" construct a random function R(t) such that at a point T, which corresponds to a point (x,y), the direction of R(t) points in the direction of a random vector V that you want to find the derivative in the direction of. THIS IS IDENTICLE to assuming that at a point x,y the vector R'(t) is just a random vector V
thus you get Grad F (dot) V
corresponding to a rate of change of F in the direction of V
This is exactly HOW grad F is derived, as to maximise the equation grad F (dot) V.
then V must be in then direction of grad F ( as its a dot product) which means that the MAXIMUM rate of change would be a vector that points in the direction of grad F. substituting v = grad(f)/abs( grad f) into the directional derivative ewuation gets u that then MAXIMUM rate of change of F is abs(grad F) which is why the grad F has the property of its.magnitude being the maximum rate of change
I believe the following works. It's a more direct proof inspired by this paper.
Let $\vec{u}=(u_1,u_2)$ be a unit vector and $f:\mathbb{R}^2\rightarrow \mathbb{R}$ a function differentiable at $(a,b)$. The result is easy to prove when either $u_1$ or $u_2$ is zero (see the paper for more info), so I'll assume $u_1,u_2\neq 0$.
$$\frac{\partial f}{\partial \vec{u}}:=\lim_{h\rightarrow 0}\frac{f(a+hu_1,b+hu_2)-f(a,b)}{h}$$
$$=\lim_{h\rightarrow 0}\frac{f(a+hu_1,b+hu_2)-f(a,b+hu_2)}{hu_1}u_1+\lim_{h\rightarrow 0}\frac{f(a,b+hu_2)-f(a,b)}{hu_2}u_2$$
The right-most term is clearly equal to $\partial _yf(a,b)u_2$. Applying the Mean Value Theorem on the left-most term we have
$$=\lim_{h\rightarrow 0}\partial_xf(\alpha(h),b+hu_1)u_1+\partial _yf(a,b)u_2$$
where $\alpha(h)$ is between $a$ and $a+hu_1$, meaning $\alpha(h)\rightarrow a$ as $h\rightarrow 0$. Thus the above is equal to
$$\partial_xf(a,b)u_1+\partial _yf(a,b)u_2$$
as wanted.
The proof can be generalized to other dimensions by expanding $\partial f/\partial \vec{u}$ so that each term varies only in one of its dimensions. For example, for a function $f:\mathbb{R}^3\rightarrow \mathbb{R}$ we would begin the proof by expanding $\partial f/\partial \vec{u}$ as
$$\frac{f(a+hu_1,b+hu_2,c+hu_3)-f(a,b+hu_2,c+hu_3)}{hu_1}u_1 +\frac{f(a,b+hu_2,c+hu_3)-f(a,b,c+hu_3)}{hu_2}u_2 +\frac{f(a,b,c+hu_3)-f(a,b,c)}{hu_3}u_3$$
and we would need to use the Mean Value Theorem in all terms except the last.
