Directional Derivatives — Question 3

Question 3 – Solution

Strategy. Use $D_{u}f=\|\nabla f\|\cos\alpha$, where $\alpha$ is the angle from the gradient to the unit direction.

Step 1: Extremes Since $\|\nabla f\|=\sqrt{29}$, $$\boxed{D_{\max}=\sqrt{29}\text{ in direction }\frac 1{\sqrt{29}}\left\langle-2,5\right\rangle},$$ $$\boxed{D_{\min}=-\sqrt{29}\text{ in direction }\frac 1{\sqrt{29}}\left\langle 2,-5\right\rangle}.$$

Step 2: Zero directions Unit vectors perpendicular to $\left\langle-2,5\right\rangle$ are $$\boxed{u=\pm\frac 1{\sqrt{29}}\left\langle 5,2\right\rangle}.$$

Step 3: Prescribed angle For $u=\left\langle\sqrt 3/2,1/2\right\rangle$, $$D_{u}f=-2\frac{\sqrt 3}{2}+5\frac 12 =\boxed{\frac 52-\sqrt 3}.$$