Answer
(a) $-\nabla f(\bf {x})$
(b) $\lt -12,92 \gt$, -$4\sqrt{538}$
Work Step by Step
(a) Formula to calculate the directional derivative: $D_uf=\nabla f(x,y) \cdot
u$
or, $D_uf=║\nabla f(x,y)║║u║\cos \theta$
Minimum value of $\cos \theta=-1$ when $\theta= 180^\circ$ = $\pi$
Thus, $D_uf=-|\nabla f(x)|$ or, $-\nabla f(\bf {x})$
(b) Formula to calculate the maximum rate of change of $f$: $D_uf=║\nabla f(x,y)$║
$\nabla f(x,y)=\lt 4x^3y-2xy^3,x^4-3x^2y^2 \gt$
$\nabla f(2,-3)=\lt 12,-92 \gt$
$-\nabla f(2,-3)=\lt -12,92 \gt$
$║-\nabla f(2,-3)║=-\sqrt{(-12)^2+(92)^2}=-4\sqrt538$
So, the direction of the fastest decrease at (2,-3) is $\lt -12,92 \gt$ and the rate of change in this direction is -$4\sqrt{538}$.
