UkusakazaL

2021-09-09

At what points is the direction of fastest change of the function $f\left(x,y\right)={x}^{2}+{y}^{2}-4x-8y$ is $i+j$.

d2saint0

The required course is established by  $\mathrm{\nabla }f\left(x,y\right)$
$\mathrm{\nabla }f\left(x,y\right)=<\frac{\partial f}{\partial x},\frac{\partial f}{\partial }y\right)>$
$\frac{\partial f}{\partial x}=\frac{\partial }{\partial x}\left({x}^{2}+{y}^{2}-4x-8y\right)$
$=2x-4$
$\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left({x}^{2}+{y}^{2}-4x-8y\right)$
$=2y-8$
Thus,
$\mathrm{\nabla }f\left(x,y\right)=<\frac{\partial f}{\partial x},\frac{\partial f}{\partial }y\right)>$
$=<2x-4,2y-8>$
gradf(x,y) has to be parallel to i+j
The direction vector of i+j is $<1,1>$
Thus,
$<2x-4,2y-8>$ has to be parallel to <1, 1>
Since in $<1,1>$ the coordinates are equal, 2x-4=2y-8
$2x-4=2y-8$
$2\left(x-2\right)=2\left(y-4\right)$
$x-2=y-4$
$y=x+2$

Do you have a similar question?