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?
