Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am completely struck with the problem: Let $f$ be a function of three variables having continuously partial derivatives. For each direction vector $h=(h_1,h_2,h_3)$ such that $h_1^1+h_2^2+h_3^3=1$, Let $D_hf(x,y,z)$ be the directional derivative of $f$ along $h$ at $(x,y,z)$. For a point $(x_0,y_0,z_0)$ is not zero, mmaximize $D_hf(x_0,y_0,z_0)$ as a function off $h$

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

The maximum of the directional derivative at a point is reached in the direction of the gradient of the function at that point. $$ D_hf(x_0,y_0,z_0)=h\cdot\nabla f(x_0,y_0,z_0)=\|\nabla f(x_0,y_0,z_0)\|\cos(\langle\nabla f(x_0,y_0,z_0),h\rangle). $$ The maximum is reached when $\cos(\langle\nabla f(x_0,y_0,z_0),h\rangle)=1$, that is, when $h$ is in the direction of $\nabla f(x_0,y_0,z_0)$, so that $$ h=\frac{\nabla f(x_0,y_0,z_0)}{\|\nabla f(x_0,y_0,z_0)\|}. $$

share|improve this answer
add comment

What you need is the following fact from Schwarz's inequality:

$$ |\langle x,y\rangle| \leq \|x\| \|y\| $$

with equality if and only if $x \parallel y$.

Hence since

$$ D_h f = \langle h, \nabla f\rangle \leq \|\nabla f\| \|h\| $$

we have that the maximizer only occurs when $h \parallel \nabla f $. (And obviously you want to choose the direction where the inner product is positive.)

In other words,

$$ h = \frac{ \nabla f}{\|\nabla f\|} $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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