# Thread: calculte growth is some direction question..

1. ## calculte growth is some direction question..

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
$
$

????

2. $f(x,y,2x^2+y^2)=3x-5y$
i dont know how to do a erivative by x
by y
if it were
$f(x,y,z)=3x-5y$ then
f'x=3 f'y=-5 f'z=0
how the f(x,y,2x^2+y^2)=3x-5y
changes??

3. Originally Posted by transgalactic
$f(x,y,2x^2+y^2)=3x-5y$
i dont know how to do a erivative by x
by y
if it were
$f(x,y,z)=3x-5y$ then
f'x=3 f'y=-5 f'z=0
how the f(x,y,2x^2+y^2)=3x-5y
changes??
you may use the Jacobian Matrix
(fx,fy)=
(fx,fy,fz)*
(xx,xy
yx,yy
zx,zy)
=
(3,-5,0)
(1,0
0,1
4x,2y)
=(3,-5)
another easier method is that fx=fx*1+fz*zx=3
i know it may be very unclear, but i hope it will help.

4. yakobian i a method of tranforming into a new varibles
and its only for integral

5. Originally Posted by transgalactic
$f(x,y,2x^2+y^2)=3x-5y$
i dont know how to do a erivative by x
by y
if it were
$f(x,y,z)=3x-5y$ then
f'x=3 f'y=-5 f'z=0
how the f(x,y,2x^2+y^2)=3x-5y
changes??
??