Question

Compute the directional derivative aong, $u$ , at the indicated points .

(a). $f(x,y)=x\sqrt{y-3}$  $,u=(-1,6)$  $a=(2,12)$

(b). $f(x,y,z)=\frac{1}{x+2y-3z}$ , $u=(12,-9,-4)$ $a=(1,1,-1)$

Answer 1

solution

Note : The directional derivative along u is :$$

$D_uf(x,y)=F_x(x,y)cos(\theta)+F_y(x,y)sin(\theta)$

$D_uf(a)=F_x(a).v_1+F_y(x,y).v_2$

where $v(x,y)=\frac{u}{||u||}=v_1i+v_2j$

(a). $f(x,y)=x\sqrt{y-3}$  $,u=(-1,6)$  $a=(2,12)$

Let's find the unit vector of u : $v=\frac{u}{||u|}=-\frac{1}{\sqrt{37}}+\frac{6}{\sqrt{37}}$

Then find $F_x(x,y)=\sqrt{y-3}=>F_x(a)=3$

                $F_y(x,y)=\frac{x}{2\sqrt{y-3}}=>F_y(a)=\frac{1}{3}$

Then, the directional derivative is $D_uf(a)=-\frac{3}{\sqrt{37}}+\frac{2}{\sqrt{37}}=-\frac{1}{\sqrt{37}}$

(b)=$f(x,y,z)=\frac{1}{x+2y-3z}$  $u=(12,-9,-4)$   $a=(1,1,-1)$

Let's find the unit vector of $v=\frac{u}{||u|}=\frac{1}{\sqrt{241}}i+\frac{1}{\sqrt{241}} j-\frac{1}{\sqrt{241}}$

Let's find the partial derivative of f at ponit $a=(1,1,-1)$

$F_x(x,y,z)=-\frac{1}{(x+2y-3z)^2}=>F_x(a)=-\frac{1}{36}$

$F_y(x,y,z)=-\frac{2}{(x+2y-3x)^2}=>F_y(a)=-\frac{2}{36}$

$F_z(x,y,z)=\frac{3}{(x+2y-3z)^2}=>F_z(a)=\frac{3}{36}$

Then, the derictional derivative is : $D_uf(a)=-\frac{1}{6\sqrt{241}}$

Answe

(a). Therefore, the directional derivative is: $D_uf(a)=-\frac{3}{\sqrt{37}}+\frac{2}{\sqrt{37}}=-\frac{1}{\sqrt{37}}$

(b).Therefore, $D_uf(a)=-\frac{1}{6\sqrt{241}}$