find the directional derivative of $f(x,y)=2sqrt{x}-y^{3}$ at the point $(1,3)$ in the direction toward the…

Explanation:

Step1: Find the direction vector

The vector from $(1,3)$ to $(3,1)$ is $\vec{v}=(3 - 1,1 - 3)=(2,-2)$. The unit - vector $\vec{u}$ in the direction of $\vec{v}$ is $\vec{u}=\frac{\vec{v}}{\vert\vec{v}\vert}$. Since $\vert\vec{v}\vert=\sqrt{2^{2}+(-2)^{2}}=\sqrt{4 + 4}=2\sqrt{2}$, then $\vec{u}=(\frac{2}{2\sqrt{2}},\frac{-2}{2\sqrt{2}})=(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$.

Step2: Find the partial derivatives

First, find the partial derivative with respect to $x$: $f_x(x,y)=\frac{2}{2\sqrt{x}}=\frac{1}{\sqrt{x}}$. Evaluated at $(1,3)$, $f_x(1,3) = 1$. Second, find the partial derivative with respect to $y$: $f_y(x,y)=-3y^{2}$. Evaluated at $(1,3)$, $f_y(1,3)=-3\times3^{2}=-27$.

Step3: Calculate the directional derivative

The directional derivative $D_{\vec{u}}f(x,y)=f_x(x,y)u_1 + f_y(x,y)u_2$. Substitute $x = 1,y = 3,u_1=\frac{1}{\sqrt{2}},u_2=-\frac{1}{\sqrt{2}},f_x(1,3)=1,f_y(1,3)=-27$ into the formula: $D_{\vec{u}}f(1,3)=1\times\frac{1}{\sqrt{2}}+(-27)\times(-\frac{1}{\sqrt{2}})=\frac{1 + 27}{\sqrt{2}}=\frac{28}{\sqrt{2}} = 14\sqrt{2}$.

Answer:

$14\sqrt{2}$