Directional Vectors question

u = (0,2,-1)
g(x,y,z) = (x+2y+3z)^(3/2)
Dug(1,1,2)=?

so I take the partial derivative

g_x(x,y,z) =3(x+2y+3z)^(1/2)/2 = 9/2
g_y(x,y,z) =3(x+2y+3z)^(1/2) = 9
g_z(x,y,z) =9(x+2y+3z)^(1/2)/2 = 27/2

why and how can I tell to use:
U/||U||

I get U = (0,2/(2)^(1/2), -1/(5)^(1/2)))

Then I just sub into the point

and come up with (9/2, 9, 27/2)

then I just go

(9/2)(0) + 18/(2)^(1/2) - 27/(5)^(1/2) = 9(2)^(1/2) - (27/5^(1/2))

specific question,

how can I know when to use and why do I have to:
U/||U||
and finally is this answer correct?

Question

Directional Vectors question

u = (0,2,-1)
g(x,y,z) = (x+2y+3z)^(3/2)
Dug(1,1,2)=?

so I take the partial derivative

g_x(x,y,z) =3(x+2y+3z)^(1/2)/2 => 9/2
g_y(x,y,z) =3(x+2y+3z)^(1/2)     => 9 
g_z(x,y,z) =9(x+2y+3z)^(1/2)/2 => 27/2

why and how can I tell to use: 
U/||U||

I get U = (0,2/(2)^(1/2), -1/(5)^(1/2)))

Then I just sub into the point

and come up with (9/2, 9, 27/2)

then I just go

(9/2)(0) + 18/(2)^(1/2) - 27/(5)^(1/2)  = 9(2)^(1/2) - (27/5^(1/2))

specific question,

how can I know when to use and why do I have to:
 U/||U|| 
and finally is this answer correct?

experimentx · Answer

could you make it short? what is the question??

australopithecus · Answer

how can I know when to use and why do I have to: U/||U|| and finally is this answer correct?

experimentx · Answer

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australopithecus · Answer

?

experimentx · Answer

unit vector along the maximum rate of change?

australopithecus · Answer

apparently u isn't in the correct form so it has to be changed using the formula

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