Firstly, I am aware that there are quite a few question regarding with "maximizing direction derivative" already being asked. But after scanning through, I am still not able to figure out my question thus posting it here.
Let's say we have a function of a surface, $f(x,y)$: then the gradient is ,$\bigtriangledown f = (\dfrac{\partial{f}}{\partial{x}},\dfrac{\partial{f}}{\partial{y}})$
From my understanding, the directional derivative at Point $P$ , $P(x,y)$ in the direction of $\hat{i}$ will be:
$D_{\hat{i}}(P) = \bigtriangledown{f}(P) \bullet \left(\begin{array}{cc} 1 \\ 0 \end{array}\right)$ (dot product of gradient at point P and vector pointing in $\hat{i}$ direction)
To find the direction of $\hat{i}$ that maximize $D_{\hat{i}}$, I use their dot product,
$D_{\hat{i}}(P) = \bigtriangledown{f}(P) \bullet \left(\begin{array}{cc} 1 \\ 0 \end{array}\right)= \|\bigtriangledown{f}\| \|\hat{i}\| cos(\theta)$
from the above equation, $D_{\hat{i}}(P)$ is larget when $\theta$ is $0$ ($cos(0) = 1$).
So would I be correct to say that when vector $\hat{i}$ is parallel , (ie,$cos(\theta)=1)$ it will give the maximum directional derivative? Then, it makes no sense to find out what is the direction of vector $\hat{i}$ that yield maximum $D_{\hat{i}}$ as everytime it will simply just be $0$ (Parallel)?
EDIT: my attempt to find $u$ which maximize $D_{u}(P)$
$\bigtriangledown{f} \bullet u = \|\bigtriangledown{f}\| \|u\| cos(\theta)$
Since we are talking about unit vector, $\|u\|$ will be $1$. Also $\theta=0$ for maximum directional derivative. Thus,
$\left(\begin{array}{cc} f_x \\ f_y \end{array}\right) \bullet \left(\begin{array}{cc} u_1 \\ u_2 \end{array}\right) = \sqrt{(f_x)^2+(f_y)^2}$
$f_x u_1 + f_y u_2 = \sqrt{(f_x)^2+(f_y)^2}$
but since I only have 1 equation and two unknowns $(u_1,u_2)$ how am I gonna get the components of $u$ $(u_1 and u_2)$ that gives maximum directional derivative?
Yes it will be the (unit) vector that is parallel to the derivative itself but you still have the task of finding that vector don't you?