Answer:
√3/2
Explanation:
The directional derivative at the given point is gotten using the formula;
∇f(x,y)•u where u is the unit vector in that direction.
∇f(x,y) = f/x i + f/y j
Given the function f(x, y) = y cos(xy),
f/x = -y²sin(xy) and
f/y = -xysin(xy)+cos(xy)
∇f(x,y) = -y²sin(xy) i + (cos(xy)-xysin(xy)) j
∇f(x,y) at (0,1) will give;
∇f(0,1) = -0sin0 i + cos0j
∇f(0,1) = 0i+j
The unit vector in the direction of angle θ is given as u = cosθ i + sinθ j
u = cos(π/3)i+ sin(π/3)j
u = 1/2 i + √3/2 j
Taking the dot product of both vectors;
∇f(x,y)•u = (0i+j)•(1/2 i + √3/2 j)
Note that i.i = j.j = 1 and i.j = 0
∇f(x,y)•u = 0 + √3/2
∇f(x,y)•u = √3/2
The directional derivative of [tex]f[/tex] at the given point in the direction indicated is [tex]\frac{\sqrt{3}}{2}[/tex].
The directional derivative is represented by the following formula:
[tex]\nabla_{\vec v} f = \nabla f(x_{o},y_{o}) \cdot \vec v[/tex] (1)
Where:
The gradient of [tex]f[/tex] is calculated below:
[tex]\nabla f (x_{o},y_{o}) = \left[\begin{array}{cc}\frac{\partial f}{\partial x} (x_{o}, y_{o}) \\\frac{\partial f}{\partial y} (x_{o}, y_{o})\end{array}\right][/tex] (2)
Where [tex]\frac{\partial f}{\partial x}[/tex] and [tex]\frac{\partial f}{\partial y}[/tex] are the partial derivatives with respect to [tex]x[/tex] and [tex]y[/tex], respectively.
If we know that [tex](x_{o}, y_{o}) = (0, 1)[/tex], then the gradient is:
[tex]\nabla f(x_{o}, y_{o}) = \left[\begin{array}{cc}-y^{2}\cdot \sin xy\\\cos xy -x\cdot y\cdot \sin xy\end{array}\right][/tex]
[tex]\nabla f (x_{o}, y_{o}) = \left[\begin{array}{cc}-1^{2}\cdot \sin 0\\\cos 0-0\cdot 1\cdot \sin 0\end{array}\right][/tex]
[tex]\nabla f (x_{o}, y_{o}) = \left[\begin{array}{cc}0\\1\end{array}\right][/tex]
If we know that [tex]\vec v = \cos \frac{\pi}{3}\,\hat{i} + \sin \frac{\pi}{3} \,\hat{j}[/tex], then the directional derivative is:
[tex]\Delta_{\vec v} f = \left[\begin{array}{cc}0\\1\end{array}\right]\cdot \left[\begin{array}{cc}\cos \frac{\pi}{3} \\\sin \frac{\pi}{3} \end{array}\right][/tex]
[tex]\nabla_{\vec v} f = (0)\cdot \cos \frac{\pi}{3} + (1)\cdot \sin \frac{\pi}{3}[/tex]
[tex]\nabla_{\vec v} f = \frac{\sqrt{3}}{2}[/tex]
The directional derivative of [tex]f[/tex] at the given point in the direction indicated is [tex]\frac{\sqrt{3}}{2}[/tex]. [tex]\blacksquare[/tex]
To learn more on directional derivatives, we kindly invite to check this verified question: https://brainly.com/question/9964491
