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Unit vector in direction of max increase of f(x,y,z)

by musicmar
Tags: direction, increase, unit, vector
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musicmar
#1
Oct11-10, 09:18 PM
P: 100
1. The problem statement, all variables and given/known data
Find the unit vector e at P=(0,0,1) pointing in the direction along which f(x,y,z)=xz+e-x2+y increases most rapidly.

3. The attempt at a solution
In order to find the direction where f increases most rapidly, I found the second derivative of f.
I don't know how to put the curly d's in here, but

<(d2f/dx2,d2f/dy2,d2f/dz2>=<4e-x2+y,e-x2+y,0>

The second derivative should be zero where f increases the most rapidly, but I'm not sure what do do with the point or how to set the second derivative equal to zero from this point.
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Kevin_Axion
#2
Oct11-10, 09:25 PM
P: 920
Here just click on this and copy this code:

[tex]\frac{\partial^2f}{\partial x^2},\frac{\partial^2f}{\partial y^2},\frac{\partial^2f}{\partial z^2}=4e^{-x^2+y},e^{-x^2+y},0[/tex] or you can just write [tex]\nabla^2 f[/tex]
musicmar
#3
Oct11-10, 09:33 PM
P: 100
But that doesn't help me answer the question.

Kevin_Axion
#4
Oct11-10, 09:36 PM
P: 920
Unit vector in direction of max increase of f(x,y,z)

I know I'm only in the 11th grade and I know very little multi-variable calculus. I was just making the question more presentable so people who have taken this course will have a better reception and hence will answer your question.
musicmar
#5
Oct11-10, 09:41 PM
P: 100
Well, thanks for showing me how to enter partial derivatives, anyway.


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