# Directional derivative

• Mar 14th 2009, 10:16 AM
Chris0724
Directional derivative
Given

F( x, y ,z) = 2x + 3y^2 + yz,

find Directional derivative of F at ( 2 , -1, 5) along the direction given by line x = y = ( z / 2).

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Directional derivative Formula = ∇f · u

i grad the F and achieve ( 2, 6y + z , y).

then i sub ( 2 , -1, 5) which is the point into it <--- is this correct ?

what is the u then ?

many thanks (Talking)
• Mar 14th 2009, 10:40 AM
HallsofIvy
Quote:

Originally Posted by Chris0724
Given

F( x, y ,z) = 2x + 3y^2 + yz,

find Directional derivative of F at ( 2 , -1, 5) along the direction given by line x = y = ( z / 2).

-----------------------------------------------------------------------
Directional derivative Formula = ∇f · u

i grad the F and achieve ( 2, 6y + z , y).

then i sub ( 2 , -1, 5) which is the point into it <--- is this correct ?

what is the u then ?

many thanks (Talking)

Yes, that is correct. u is the unit vector pointing in the given direction.

However, there are two answers to this problem: just a line does not define a "direction". There are two unit vectors pointing along the "along" this line in opposite directions. The derivative in one direction is the negative of the derivative in the opposite direction.
• Mar 15th 2009, 09:35 AM
Chris0724
Quote:

Originally Posted by HallsofIvy
However, there are two answers to this problem: just a line does not define a "direction". There are two unit vectors pointing along the "along" this line in opposite directions. The derivative in one direction is the negative of the derivative in the opposite direction.

hi,

i still have problem finding u. i don't know wat is the value of x , y , z...
• Mar 15th 2009, 09:37 AM
Jhevon
Quote:

Originally Posted by Chris0724
hi,

i still have problem finding u. i don't know wat is the value of x , y , z...

find it from the line. you were given the symmetric form of the equation of a line. can you find the direction vector for the line? the unit vector you want is in that direction
• Mar 15th 2009, 09:42 AM
Chris0724
Quote:

Originally Posted by Jhevon
find it from the line. you were given the symmetric form of the equation of a line. can you find the direction vector for the line? the unit vector you want is in that direction

oh i got it.... is it ( 1, 1, 0.5) ? Is that call the cartesian coordinate as well?
• Mar 15th 2009, 09:48 AM
Jhevon
Quote:

Originally Posted by Chris0724
oh i got it.... is it ( 1, 1, 0.5) ? Is that call the cartesian coordinate as well?

no, that's not it.

also, note that whatever vector you find, its negative will also work. this is what HallsofIvy meant by two answers.
• Mar 15th 2009, 09:55 AM
Chris0724
Quote:

Originally Posted by Jhevon
no, that's not it.

also, note that whatever vector you find, its negative will also work. this is what HallsofIvy meant by two answers.

so i will achieve ( 2, 6y + z , y) . [ ( 1, 1, 0.5) / |( 1, 1, 0.5)| ] ?
• Mar 15th 2009, 09:57 AM
Jhevon
Quote:

Originally Posted by Chris0724
so i will achieve ( 2, 6y + z , y) . [ ( 1, 1, 0.5) / |( 1, 1, 0.5)| ] ?

no, your unit vector is wrong, i said that

find the correct one then plug it into the formula you mentioned.
• Mar 15th 2009, 01:04 PM
HallsofIvy
No, Jhevon, his unit vector is right. It is (1, 1, 0.5) divided by its length and that is what he meant by "(1, 1, 0.5)/|(1, 1, 0.5)
|.
Chris0724, you can go ahead and put in the point (2, -1, 5) for x, y, and z.
• Mar 15th 2009, 01:19 PM
Jhevon
Quote:

Originally Posted by HallsofIvy
No, Jhevon, his unit vector is right. It is (1, 1, 0.5) divided by its length and that is what he meant by "(1, 1, 0.5)/|(1, 1, 0.5)

Chris0724, you can go ahead and put in the point (2, -1, 5) for x, y, and z.

am i forgetting my calc 3? i would think that the direction vector for the line x = y = z/2 is <1,1,2> and so it should be <1,1,2>/|<1,1,2>|. maybe i'm drunk again (Drunk)
• Mar 15th 2009, 02:42 PM
HallsofIvy
Quote:

Originally Posted by Jhevon
am i forgetting my calc 3? i would think that the direction vector for the line x = y = z/2 is <1,1,2> and so it should be <1,1,2>/|<1,1,2>|. maybe i'm drunk again (Drunk)

You are right and I am wrong. How foolish of me. Given x= y= z/2 and setting each to t, x= t, y= t, z= 2t are parametric equations and <1, 1, 2> is the vector.