Thread: 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

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

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.

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...

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

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?

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.

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)| ] ?

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.

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.

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

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

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.