91Ó°ÊÓ

Question: Find the directional derivatives of the function f(x, y) = 8 - x^2/2 - y^2 at the point (2, 0) and in the directions of the unit vectors \(\mathbf{u} = (1/\sqrt{2}, 1/\sqrt{2})\), \(\mathbf{v} = (-1/\sqrt{2}, 1/\sqrt{2})\), and \(\mathbf{w} = (0, 1)\), and explain what they represent. To compute the directional derivatives at the point (2, 0), we plug in the values for x and y into the gradient we found in Step 1: $$ \nabla f(2, 0) = (-2, 0) $$ Now, we can find the directional derivatives in each of the given directions: 1. In the direction of \(\mathbf{u}\): $$ D_{\mathbf{u}}f(2, 0) = \nabla f(2, 0) \cdot \mathbf{u} = (-2, 0) \cdot \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = -\frac{2}{\sqrt{2}} = -\sqrt{2} $$ 2. In the direction of \(\mathbf{v}\): $$ D_{\mathbf{v}}f(2, 0) = \nabla f(2, 0) \cdot \mathbf{v} = (-2, 0) \cdot \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = \frac{2}{\sqrt{2}} = \sqrt{2} $$ 3. In the direction of \(\mathbf{w}\): $$ D_{\mathbf{w}}f(2, 0) = \nabla f(2, 0) \cdot \mathbf{w} = (-2, 0) \cdot (0, 1) = 0 $$ Interpretation: The directional derivatives give us the rate of change of the function f(x, y) in the direction of the given unit vectors at the point (2, 0). For the unit vector \(\mathbf{u}\), the function is decreasing at a rate of \(-\sqrt{2}\) at this point. For the unit vector \(\mathbf{v}\), the function is increasing at a rate of \(\sqrt{2}\) at this point. And for the unit vector \(\mathbf{w}\), the function is not changing at all in that direction.