1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '14 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 '14 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 '14 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 '14 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 '14 at 1:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.