Question

the directional derivative of f(xyz) = xy^3z^2 at point p(2,-1.4), in the direction of vector a= <1,2,-1> is 192/root(6)

the vector is -16/root(6), 192/root(6), 16/root(6)

in what direction does f increase most rapidly from the point p? (the answer is a unit vector)

Answer 1

\[Du|_f = 192/\sqrt(6)\]
\[<-16/\sqrt(6), 192 / \sqrt(6), 16/ \sqrt(6)>\]

Answer 2

the most rapidly increas direction for a function is it's gradient direction. So find the grad f.

Answer 3

i'm pretty sure the gradient is the vector above, no?

Answer 4

\[grad f=(y ^{3}z ^{2},3xy ^{2}z ^{2},2xy ^{3}z)\]
at point p(2,-1,4) grad f =(-16,96,-16)
check just in case, maybe i made some calculation error....

Answer 5

is the gradient:
\[Deltaf= (y^3x^2)i+ (3y^2xz^3)j+(2zxy^3)k\]

Answer 6

should be z^2 in the j component

Answer 7

and z^2 instead of x^2 in the i one

Answer 8

it will increase most rapidly at the direction which is parallel to our gradient vector. now our gradient vector is just <df/dx,df/dy,df/dz>=<y^3z^2,3xy^2z^2,2xy^3z>, now at point (2,-1,4), well have:
=<-16,96,-16>
now we just need to turn this to a unit vector which is parallel to the gradient vector, so we just multiply by the reciprocal of the norm of this gradient. so the norm is just:
=sqrt(9728)=16sqrt(38)
so the answer will be:
=<-16/16sqrt(38),96/16sqrt(38),16/sqrt(38)>=<-1/sqrt(38),6/sqrt(38),1/sqrt(38)>
this should be the direction of our unit vector

Answer 9

you lost me around here:
now we just need to turn this to a unit vector which is parallel to the gradient vector, so we just multiply by the reciprocal of the norm of this gradient. so the norm is just:
=sqrt(9728)=16sqrt(38)
so the answer will be:
=<-16/16sqrt(38),96/16sqrt(38),16/sqrt(38)>=<-1/sqrt(38),6/sqrt(38),1/sqrt(38)>
this should be the direction of our unit vector

Answer 10

at point p(2,-1,4) grad f =(-16,96,-16)
this is not unit vector. To make it unit vector you have to devide it by it's length. The length of the vector v=(x,y,z) is:\[|v|=\sqrt{x ^{2}+y ^{2}+z ^{2}}\]

Answer 11

in your case the length is:\[length=\sqrt{(-16)^{2}+96^{2}+(-16)^{2}}\]

Answer 12

how is that different from the first vector i gave you?

Answer 13

i'm really confused

Answer 14

if its in the direction of a = <1,2,-1> then isn't the unit vector
\[1/\sqrt(6) <-16, 96, -16>\]

Answer 15

it is not in the direction of a

Answer 16

it is in the direction of grad f

Answer 17

wait, no that's wrong
i took the gradient <-16, 96, -16> and dotted it with the unit vector of a to get
\[1/\sqrt(6) <-16,192,-16>\]

Answer 18

so i no longer care about a?

Answer 19

this way you get the derivative in the direction of vector a, but you were asking about the direction where f increas is maximum. And this maximum increas direction is direction of the grad f

Answer 20

ohhhhhhhh so i have to get the unit vector of f ohhh okay i think it's starting to make sense haha

Answer 21

unit of grad f

Answer 22

so the unit vector of =
\[1/\sqrt(9728) <-16, 96, -16>\]
?

Answer 23

yes, you just can simplify this a bit

Answer 24

i'm not sure how to simplify it /blush

Answer 25

oh.. 16x16.. ohh.... hahha

Answer 26

9728=256*38=16*16*38

Answer 27

so it's
\[1/\sqrt(38) <-1, 6, -1>\]?

Answer 28

good job

Answer 29

yayyyy thank you :D :D :D