Respuesta :
Answer:
following are the solution to the given points:
Step-by-step explanation:
In point a:
[tex]\vec{v} = -\vec{1 i} +\vec{1j}\\\\|\vec{v}| = \sqrt{-1^2+1^2}[/tex]
[tex]=\sqrt{1+1}\\\\=\sqrt{2}[/tex]
calculating unit vector:
[tex]\frac{\vec{v}}{|\vec{v}|} = \frac{-1i+1j}{\sqrt{2}}[/tex]
the point Q is at a distance h from P(6,6) Here, h=0.1
[tex]a=-6+O.1 \times \frac{-1}{\sqrt{2}}\\\\= 5.92928 \\\\b= 6+O.1 \times \frac{-1}{\sqrt{2}} \\\\= 6.07071[/tex]
the value of Q= (5.92928 ,6.07071 )
In point b:
Calculating the directional derivative of [tex]f (x, y) = \sqrt{x+3y}[/tex] at P in the direction of [tex]\vec{v}[/tex]
[tex]f_{PQ} (P) =\fracx{f(Q)-f(P)}{h}\\\\[/tex]
[tex]=\frac{f(5.92928 ,6.07071)-f(6,6)}{0.1}\\\\=\frac{\sqrt{(5.92928+ 3 \times 6.07071)}-\sqrt{(6+ 3\times 6)}}{0.1}\\\\= \frac{0.197651557}{0.1}\\\\= 1.97651557[/tex]
[tex]\vec{v} = 1.97651557[/tex]
In point C:
Computing the directional derivative using the partial derivatives of f.
[tex]f_x(x,y)= \frac{1}{2 \sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{2 \sqrt{22}}\\\\f_x(x,y)= \frac{1}{\sqrt{x+3y}}\\\\ f_x (6,6)= \frac{1}{\sqrt{22}}\\\\f_{(PQ)}(P)= (f_x \vec{i} + f_y \vec{j}) \cdot \frac{\vec{v}}{|\vec{v}|}\\\\= (\frac{1}{2 \sqrt{22}}\vec{i} + \frac{1}{\sqrt{22}} \vec{j}) \cdot \frac{-1}{\sqrt{2}}\vec{i} + \frac{1}{\sqrt{2}} \vec{j}[/tex]
