Directional derivatives


by brendan
Tags: derivatives, directional
brendan
brendan is offline
#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
Phys.Org News Partner Science news on Phys.org
Review: With Galaxy S5, Samsung proves less can be more
Making graphene in your kitchen
Study casts doubt on climate benefit of biofuels from corn residue
CompuChip
CompuChip is offline
#2
Apr3-09, 05:11 AM
Sci Advisor
HW Helper
P: 4,301
How did you calculate that unit vector?
And what does it matter whether you multiply by cos(0)?
Limecat
Limecat is offline
#3
Apr3-09, 05:12 AM
P: 15
cos(0) is just 1...

HallsofIvy
HallsofIvy is offline
#4
Apr3-09, 06:58 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890

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
brendan is offline
#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
HallsofIvy is offline
#6
Apr3-09, 07:04 PM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,890
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
brendan is offline
#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


Register to reply

Related Discussions
estimating partial derivatives/directional derivatives Calculus & Beyond Homework 1
Directional derivatives Calculus & Beyond Homework 9
Directional derivatives Calculus & Beyond Homework 4
Directional derivatives Introductory Physics Homework 3
Directional Derivatives General Math 2