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Directional derivatives

by brendan
Tags: derivatives, directional
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brendan
#1
Apr3-09, 04:51 AM
P: 65
1. The problem statement, all variables and given/known data

Find a unit vector in the direction in which f(x,y,z) = (x-3y+4z)1/2 increases most rapidly at P(0,-3,0), and find the rate of increase of f in that direction

2. Relevant equations



3. The attempt at a solution

I've calculated the unit vector to be <0,-1,0>

and the gradient to be <1/6,-1/2,2/3>

To find the directional derivative we find the dot product of <0,-1,0> . <1/6,-1/2,2/3> = 1/2
Do I than multiply that by cos(0)?

regards
Brendan
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CompuChip
#2
Apr3-09, 05:11 AM
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How did you calculate that unit vector?
And what does it matter whether you multiply by cos(0)?
Limecat
#3
Apr3-09, 05:12 AM
P: 15
cos(0) is just 1...

HallsofIvy
#4
Apr3-09, 06:58 AM
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Directional derivatives

Quote Quote by brendan View Post
1. The problem statement, all variables and given/known data

Find a unit vector in the direction in which f(x,y,z) = (x-3y+4z)1/2 increases most rapidly at P(0,-3,0), and find the rate of increase of f in that direction

2. Relevant equations



3. The attempt at a solution

I've calculated the unit vector to be <0,-1,0>
No. That is the unit vector in the direction from (0,0,0) to (0,-3,0). You are asked for the unit vector in the direction in which f(x,y,z) increases most rapidly. A function increases most rapidly in the direction of its gradient vector.

and the gradient to be <1/6,-1/2,2/3>
That is correct. What is a unit vector in that direction?

To find the directional derivative we find the dot product of <0,-1,0> . <1/6,-1/2,2/3> = 1/2
Do I than multiply that by cos(0)?

regards
Brendan[/QUOTE]
Again, <0, -1, 0> is the wrong vector. The rate of increase in the direction of the gradient vector is just the dot product of the unit vector in that direction with the gradient vector and is just the length of the gradient vector.
brendan
#5
Apr3-09, 06:23 PM
P: 65
So If I use the gradient <1/6 ,-1/2 ,2/3> and find its unit vector which is

<sqrt(26)/936, -sqrt(26)/312, sqrt(26)/234 >

than find the dot product of them both <1/6 ,-1/2 ,2/3> . <sqrt(26)/936, -sqrt(26)/312, sqrt(26)/234 >


which is sqrt(26)/6 the magnitude of the gradient vector?


regards
Brendan
HallsofIvy
#6
Apr3-09, 07:04 PM
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Quote Quote by brendan View Post
So If I use the gradient <1/6 ,-1/2 ,2/3> and find its unit vector which is

<sqrt(26)/936, -sqrt(26)/312, sqrt(26)/234 >
No, that's not at all right. Did you divide by the length or multiply? 1/36+ 1/4+ 4/9= 1/36+ 9/36+ 15/36= 26/36. The length of the gradient is [itex]\sqrt{26}/6[/itex]. Dividing by that will put the square root in the denominator.

than find the dot product of them both <1/6 ,-1/2 ,2/3> . <sqrt(26)/936, -sqrt(26)/312, sqrt(26)/234 >

which is sqrt(26)/6 the magnitude of the gradient vector?
Of course, that is wrong now, but in any case you did not need to do that product. If [itex]\vec{v}[/itex] is a vector of length [itex]||\vec{v}||[/itex], then the unit vector in that direction, [itex]\vec{u}[/itex], is [itex]\vec{v}/||/vec{v}||[/itex] and the inner product of the two vectors is [itex]\vec{v}\cdot\vec{v}/||\vec{v}}||= ||\vec{v}||^2/||\vec{v}||= ||\vec{v}||[/itex]. The derivative in the direction of the gradient is, as I said before, the length of the gradient.

regards
Brendan
brendan
#7
Apr3-09, 09:05 PM
P: 65
Sorry,
the unit vector would be < 1/6 , -1/2 , 2/3 > / < sqrt(26)/6 , sqrt(26)/6 , sqrt(26)/6 >
regards
Brendan


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