# Calculte growth is some direction question

1. Aug 12, 2009

### khdani

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

the formula is:

$$\triangledown f=f_x'\vec{i}+f_y'\vec{j}+f_z'\vec{k}$$

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 dont 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

$$(grad f(1,2,6)\dot \hat{a}=1$$

????

2. Aug 12, 2009

### tiny-tim

Hi khdani!

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

3. Aug 12, 2009

### khdani

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

what???

you say i would substitute z with$$2x^2+y^2$$
so i get
f(x,y,z)=3x-5y
so $$f'_x=3$$
$$f'_y=-5$$
$$f'_z=0$$