Given f(x,y,z)=x^(2)y^(3)z^(6), in what direction is f(x,y,z) increasing the most rapidly at the point P(1,-1,1)? What is its rate of increase in that direction

2. Originally Posted by bobby77
Given f(x,y,z)=x^(2)y^(3)z^(6), in what direction is f(x,y,z) increasing the most rapidly at the point P(1,-1,1)? What is its rate of increase in that direction
$
\nabla f=[2\,x\,y^3\,z^6,\ 3\,x^2\,y^2\,z^6,\ 6\,x^2\,y^3\,z^5]
$

so at the point in question:

$
\nabla f=[-2,\ 3,\ -6]
$
,

so the unit vector in the direction that $f$ is increasing most rapidly is:

$\frac{\nabla f}{|\nabla f|}=[-2/7,\ 3/7,\ -6/7]$,

and the rate of increase in this direction is $|\nabla f|=7$

RonL

3. Originally Posted by bobby77
Given f(x,y,z)=x^(2)y^(3)z^(6), in what direction is f(x,y,z) increasing the most rapidly at the point P(1,-1,1)? What is its rate of increase in that direction
That is the gradient at the point.

$\nabla f = 2xy^3z^6\bold{i}+3x^2y^2z^6\bold{j}+6x^2y^3z^5\bol d{k}$
Substitute, in the values of $x,y,z$ to get the optimal vector it increases by.

$2(1)(-1)^3(1)^6\bold{i}+3(1)^2(-1)^2(1)^6\bold{j}+6(1)^2(-1)^3(1)^5\bold{k}=-2\bold{i}+3\bold{j}-6\bold{k}$
The norm of this vector is $\sqrt{2^2+3^2+6^2}=7$.
Thus, the unit vector is,
$\bold{u}=-\frac{2}{7}\bold{i}+\frac{3}{7}\bold{j}-\frac{6}{7}\bold{k}$
This is the direction of maximal increase.