UkusakazaL

2021-09-09

At what points is the direction of fastest change of the function

d2saint0

Skilled2021-09-10Added 89 answers

The required course is established by $\mathrm{\nabla}f(x,y)$

$\mathrm{\nabla}f(x,y)=<\frac{\partial f}{\partial x},\frac{\partial f}{\partial}y)>$

$\frac{\partial f}{\partial x}=\frac{\partial}{\partial x}({x}^{2}+{y}^{2}-4x-8y)$

$=2x-4$

$\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}({x}^{2}+{y}^{2}-4x-8y)$

$=2y-8$

Thus,

$\mathrm{\nabla}f(x,y)=<\frac{\partial f}{\partial x},\frac{\partial f}{\partial}y)>$

$=<2x-4,2y-8>$

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

$2x-4=2y-8$

$2(x-2)=2(y-4)$

$x-2=y-4$

$y=x+2$