Explanation:
Given: [tex]f(x, y) = \sin(x + 2y)[/tex] P(-2, -3)
The unit vector in the direction of [tex]\theta = 3\pi/4[/tex] is given by
[tex]\vec{\textbf{u}} = \left(-\dfrac{\sqrt{2}}{2}\right)\hat{\textbf{i}} + \left(\dfrac{\sqrt{2}}{2}\right)\hat{\textbf{j}}[/tex]
The gradient of f(x, y) [tex]\nabla{f(x, y)}[/tex] is defined as
[tex]\nabla{f(x, y)} = \dfrac{\partial{f}}{\partial{x}}\hat{\textbf{i}} + \dfrac{\partial{f}}{\partial{y}}\hat{\textbf{j}}[/tex]
[tex]= \cos(x + 2y)\hat{\textbf{i}} + 2\cos(x+ 2y)\hat{\textbf{j}}[/tex]
so the gradient of f(x, y) at (-2,-3) is
[tex]\nabla{f(-2, -3)} = \cos(-8)\hat{\textbf{i}} + 2\cos(-8)\hat{\textbf{j}}[/tex]
Therefore, the directional derivative [tex]D_uf(x, y)[/tex] is
[tex]D_uf(x, y) = \nabla{f(x, y)}\cdot \vec{\textbf{u}}[/tex]
[tex]\:\:\:\:\:\:\:=\cos(-8)\left(-\frac{\sqrt{2}}{2}\right) + 2\cos(-8)\left(\frac{\sqrt{2}}{2}\right)[/tex]
[tex]\:\:\:\:\:\:\:\:=\cos(-8)\left(\frac{\sqrt{2}}{2}\right) = -0.103[/tex]
Note: the arguments of the cosines are in radian measure.
