Question

Find the directional derivatives of the following functions at the given points P in the direction of the vectors v.
f(x,y,z) =√xyz , P (3,2,6) , v =<-1,-2,2>

Answer 1

find the derivatives of the function x, y and z seperately, by keeping the other 2 variables constant in each case. Think of a smart way to keep the other 2 variables constant!

Answer 2

ok, i tried to work out. is the ans -1?

Answer 3

what is you function?\[f(x,y,z)=\sqrt{xyz}\]or\[f(x,y,z)=\sqrt xyz\]?

Answer 4

the first one

Answer 5

\[D_{\vec v}=\nabla f(3,2,6)\cdot\frac{\vec v}{\|\vec v\|}\]I don't think the answer is one but I haven't done it yet...

Answer 6

is there an advanced equation editor available to super users that has cool features like gradient?

Answer 7

negative one

Answer 8

@mathstina @TuringTest is right

Answer 9

@mathstina first fing the derivative of the function with respect to x, by substituting the y and x values

Answer 10

ok

Answer 11

oh yeah, I do get -1 :)

Answer 12

ok thanks a lot!

Answer 13

@TuringTest @mathstina no its not -1, atleast as per my calculations

Answer 14

I will show my work and you can spot a mistake if you see one...

Answer 15

ok

Answer 16

\[\nabla f=\frac12\langle\sqrt{\frac{yz}x},\sqrt{\frac{xz}y},\sqrt{\frac{xy}z}\rangle\implies\nabla f(3,2,6)=\frac12\langle2,3,1\rangle\]\[\frac{\vec v}{\|\vec v\|}=\frac13\langle-1,-2,2\rangle\]\[\nabla f(3,2,6)\cdot\frac{\vec v}{\|\vec v\|}=\frac16(-2-6+2)=-1\]

Answer 17

Great!

Answer 18

@TuringTest @mathstina im sorry, your right, i made a mistake by putting all -ves as +ves and vice versa. Bravo you two

Answer 19

cheers!

Answer 20

Its ok. u too got the same ans

Answer 21

i got it as 1 previously, now i have -1 :)