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, \( D_uf(a)=-\frac{1}{6\sqrt{241}} \)