Answer
(a) $-\nabla f(\bf {x})$
(b) $\lt -12,92 \gt$
Work Step by Step
(a) Our aim is to determine the direction derivative of $f(x,y)$, we have : $D_uf=|\nabla f(x,y)|$
Formula to calculate the directional derivative: $D_uf=\nabla f(x,y) \cdot
u$
It can also be written as: $D_uf=|\nabla f(x,y)||u| \cos \theta$
Since, the minimum value of $\cos \theta$ is $-1$ when $\theta =\pi or 180^\circ$
This implies $D_uf=-|\nabla f(x)|$ or, $-\nabla f(\bf {x})$
(b) Our aim is to determine the maximum rate of change of $f(x,y)$.In order to find this, we have : $D_uf=|\nabla f(x,y)|$
Given: $f(x,y)=x^4y-x^2y^3$
$\nabla f(x,y)=\lt 4x^3y-2xy^3,x^4-3x^2y^2 \gt$
From the given data, we have $f(x,y)=f(2,-3)$
$\nabla f(2,-3)=\lt 12,-92 \gt$
$|\nabla f(2,-3)|=\sqrt{(12)^2+(-92)^2}=\sqrt {\dfrac{9}{25}}=\dfrac{3}{5}$
From part (a), we have $D_uf=-|\nabla f(x)|=-\lt 12,-92 \gt =\lt -12,92 \gt$
Therefore, the direction of $f(x,y)$ is:$\lt -12,92 \gt$
