91Ó°ÊÓ

We need to find the gradient vector of the function, which is given as a vector with its components being the partial derivatives of the function with respect to \(x\) and \(y\). So, we have: \(\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\) Let's calculate the partial derivatives: \(\frac{\partial f}{\partial x} = \frac{\partial(3x^2 - y^2)}{\partial x} = 6x\) \(\frac{\partial f}{\partial y} = \frac{\partial(3x^2 - y^2)}{\partial y} = -2y\) So, the gradient vector is: \(\nabla f(x, y) = (6x, -2y)\)

We are given the angle \(\theta = \frac{2\pi}{6}\). A unit vector in this direction can be represented as: \(\bold{u} = (\cos{\theta}, \sin{\theta})\) Using our given angle, we find that: \(\bold{u} = (\cos{\frac{2\pi}{6}}, \sin{\frac{2\pi}{6}}) = (\cos{\frac{\pi}{3}}, \sin{\frac{\pi}{3}}) = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\)

Now that we have the gradient vector and the unit vector, we can compute the directional derivative of the function at point (-1, -4) in the direction given by the angle \(\frac{2\pi}{6}\). The directional derivative can be found by taking the dot product of the two vectors: \(D_{\bold{u}}f(x, y) = \nabla f(x, y) \cdot \bold{u}\) Using the gradient vector \(\nabla f(x, y) = (6x, -2y)\) at point (-1, -4), the gradient vector becomes: \(\nabla f(-1, -4) = (6(-1), -2(-4)) = (-6, 8)\) Now, compute the dot product: \(D_{\bold{u}}f(-1, -4) = (-6, 8) \cdot \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) = -6\left(\frac{1}{2}\right) + 8\left(\frac{\sqrt{3}}{2}\right) = -3 + 4\sqrt{3}\) So, the directional derivative at point (-1, -4) in the direction given by the angle \(\frac{2\pi}{6}\) is: \(D_{\bold{u}}f(-1, -4) = -3 + 4\sqrt{3}\)