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Directional derivatives 
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#1
Apr309, 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) = (x3y+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 


#2
Apr309, 05:11 AM

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P: 4,300

How did you calculate that unit vector?
And what does it matter whether you multiply by cos(0)? 


#3
Apr309, 05:12 AM

P: 15

cos(0) is just 1...



#4
Apr309, 06:58 AM

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P: 39,683

Directional derivatives
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. 


#5
Apr309, 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 


#6
Apr309, 07:04 PM

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#7
Apr309, 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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