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?


1 Answer 1


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?

  • $\begingroup$ that will just be a unit vector(1,0)(that i used in calculation) at $\hat{i}$ direction, isn't it? why would I want to find that out anyway? $\endgroup$
    – Chris Aung
    Nov 12, 2014 at 1:46
  • $\begingroup$ No you don't know that the maximum value of the grad will always be pointing in the i direction. The vector (1,0) would yield the vector $(df(P)/dx,0)$ every time and you don't know that is the maximum for the derivative at that point. $\endgroup$
    – Rammus
    Nov 12, 2014 at 1:51
  • $\begingroup$ but the maximum directional derivative at point P, $D_\hat{i}(P)$ is always achieved by setting $\theta$ to $0$ or is it not? $\endgroup$
    – Chris Aung
    Nov 12, 2014 at 1:56
  • $\begingroup$ Try to find a unit vector $g$ in your dot product is $||g|| = 1$ and that satisfies $\cos(\theta)=0$. $\endgroup$
    – Rammus
    Nov 12, 2014 at 1:57
  • $\begingroup$ The $\cos(\theta)$ refers to the angle between the two vectors and does not relate to the axis, if that is confusing you. $\theta =0$ does not imply the vector is $(1,0)$. $\endgroup$
    – Rammus
    Nov 12, 2014 at 1:59

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