91Ó°ÊÓ

First, we need to find the gradient of the function \(f(x, y) = x^2 - y^2\). The gradient is given by the partial derivatives of the function, that is, \(\nabla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). So, let's compute the partial derivatives: $$\frac{\partial f}{\partial x} = 2x$$ $$\frac{\partial f}{\partial y} = -2y$$ Now we can form the gradient vector as follows: $$\nabla f(x, y) = \left\langle 2x, -2y \right\rangle$$

We are given the direction vector as \(\left\langle\frac{3}{5},-\frac{4}{5}\right\rangle\). It's easy to check that this is a unit vector: $$\left\|\left\langle\frac{3}{5},-\frac{4}{5}\right\rangle\right\| = \sqrt{\left(\frac{3}{5}\right)^2 + \left(-\frac{4}{5}\right)^2} = \sqrt{\frac{9}{25} + \frac{16}{25}} = \sqrt{1} = 1$$ Since its magnitude is 1, it is indeed a unit vector.

Our goal is to compute the directional derivative, which is obtained by taking the dot product of the gradient of the function at point P and the unit direction vector: $$D_u f(P) = \nabla f(P) \cdot u$$ First, we'll find the gradient of the function at point P(-1, -3): $$\nabla f(-1, -3) = \left\langle 2(-1), -2(-3) \right\rangle = \left<-2, 6\right>$$ Now, let's compute the dot product: $$D_u f(P) = \left<-2, 6\right> \cdot \left\langle\frac{3}{5},-\frac{4}{5}\right\rangle = -2\left(\frac{3}{5}\right) + 6\left(-\frac{4}{5}\right) = -\frac{6}{5} - \frac{24}{5} = -\frac{30}{5} = -6$$ Therefore, the directional derivative of the function \(f(x, y) = x^2 - y^2\) at point P(-1, -3) in the direction of the vector \(\left\langle \frac{3}{5}, -\frac{4}{5} \right\rangle\) is -6.