Calculte growth is some direction question

[khdani]

i know that the maximal growth is the gradiend of a function.

the formula is:

<br /> <br /> \triangledown f=f_x&#039;\vec{i}+f_y&#039;\vec{j}+f_z&#039;\vec{k}<br /> <br />



so when i am given a function

f(x,y,2x^2+y^2)=3x-5y

i am was told that it growth on point M(1,2,6) in direction
\hat{a}=(\frac{1}{3},\frac{2}{3},\frac{2}{3}) is 1

what is the gradient of f (maximal growth).



i tried to get it like this:

first of all i need a function which looks like this f(x,y,z)=...

in order to find the gradient

i don't know how to do the gradient of a function which i was given.



if i substiute the point

f(1,2,6)=3-10=7



i know i should write

<br /> <br /> (grad f(1,2,6)\dot \hat{a}=1<br /> <br />



?

 

[tiny-tim]

Hi khdani!

(have a grad: ∇ and a curly d: ∂ and try using the X² tag just above the Reply box )

first of all i need a function which looks like this f(x,y,z)=...

That's right …

or better still, f(x,y,z) = g(x,y) where g = 3x - 5y.

Now z = 2x² + y²,

so, using the chain rule, df/dx = … ?, df/dy = … ?, df/dz = … ?

 

[khdani]

"so, using the chain rule, df/dx = … ?, df/dy = … ?, df/dz = … "

what?

you say i would substitute z with2x^2+y^2
so i get
f(x,y,z)=3x-5y
so f&#039;_x=3
f&#039;_y=-5
f&#039;_z=0