91Ó°ÊÓ

The gradient of a function, represented as \( abla f(x, y) \), is the vector of its partial derivatives. For the function \( f(x, y) = \sin x \cos y \), the partial derivatives are:\( f_x = \frac{\partial}{\partial x}(\sin x \cos y) = \cos x \cos y \) \( f_y = \frac{\partial}{\partial y}(\sin x \cos y) = -\sin x \sin y \) Thus, \( abla f(x, y) = \langle \cos x \cos y, -\sin x \sin y \rangle \).

Evaluate the gradient at the point \( P = \left( \frac{\pi}{4}, \frac{\pi}{3} \right) \):\( abla f\left( \frac{\pi}{4}, \frac{\pi}{3} \right) = \langle \cos\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{3}\right), -\sin\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{3}\right) \rangle \)Using trigonometric values,\( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)Thus, the gradient at P is \( abla f\left( \frac{\pi}{4}, \frac{\pi}{3} \right) = \langle \frac{\sqrt{2}}{2} \cdot \frac{1}{2}, -\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} \rangle = \langle \frac{\sqrt{2}}{4}, -\frac{\sqrt{6}}{4} \rangle \).

The directional derivative of \( f \) in the direction of a vector \( \vec{v} \) is defined as:\[ D_{\vec{v}}f = abla f \cdot \frac{\vec{v}}{\|\vec{v}\|} \]First, normalize \( \vec{v} = \langle 1, 1 \rangle \):\[ \|\vec{v}\| = \sqrt{1^2 + 1^2} = \sqrt{2} \]\( \frac{\vec{v}}{\|\vec{v}\|} = \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \)Now, compute the dot product:\( D_{\vec{v}}f = \langle \frac{\sqrt{2}}{4}, -\frac{\sqrt{6}}{4} \rangle \cdot \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle = \frac{\sqrt{2}}{4} \cdot \frac{1}{\sqrt{2}} + (-\frac{\sqrt{6}}{4}) \cdot \frac{1}{\sqrt{2}} \)\[ D_{\vec{v}}f = \frac{1}{4} - \frac{\sqrt{3}}{4} = \frac{1 - \sqrt{3}}{4} \]

For the direction toward point \( Q = (0,0) \) from \( P = \left( \frac{\pi}{4}, \frac{\pi}{3} \right) \), the direction vector is \( \vec{u} = \left\langle 0 - \frac{\pi}{4}, 0 - \frac{\pi}{3} \right\rangle = \left\langle -\frac{\pi}{4}, -\frac{\pi}{3} \right\rangle \).Normalize \( \vec{u} \):\[ \|\vec{u}\| = \sqrt{\left(-\frac{\pi}{4}\right)^2 + \left(-\frac{\pi}{3}\right)^2} = \sqrt{\frac{\pi^2}{16} + \frac{\pi^2}{9}} = \sqrt{\frac{25\pi^2}{144}} = \frac{5\pi}{12} \]\( \frac{\vec{u}}{\|\vec{u}\|} = \left\langle -\frac{3}{5}, -\frac{4}{5} \right\rangle \) (from simplifying \( \frac{\left(-\frac{\pi}{4}\right)}{\frac{5\pi}{12}} \) and \( \frac{\left(-\frac{\pi}{3}\right)}{\frac{5\pi}{12}} \))Now compute the dot product with the gradient:\( D_{\vec{u}}f = \langle \frac{\sqrt{2}}{4}, -\frac{\sqrt{6}}{4} \rangle \cdot \langle -\frac{3}{5}, -\frac{4}{5} \rangle \)\[ D_{\vec{u}}f = \frac{\sqrt{2}}{4} \cdot -\frac{3}{5} + \left(-\frac{\sqrt{6}}{4}\right) \cdot -\frac{4}{5} = -\frac{3\sqrt{2}}{20} + \frac{4\sqrt{6}}{20} \]\[ D_{\vec{u}}f = \frac{4\sqrt{6} - 3\sqrt{2}}{20} \]