Using Properties of the Gradient

1. Find the direction for which the directional derivative of \(f(x,y)=x^{2}-xy+y^{2}\) at \((1,-2)\) is a maximum.
2. Find the maximum value of the directional derivative.

Solution (a) We begin by finding the gradient of \(f\) at \((1,-2)\). \[ \begin{array}{rcl@{\qquad}l} {\bf\nabla\! }f(x,y) &=&(2x-y)\mathbf{i}+(2y-x)\mathbf{j} & \color{#0066A7}{{\bf\nabla}\! f(x,y)=f_{x}(x,y) \mathbf{i}+f_{y}(x,y)\mathbf{j}} \\ {\bf\nabla\! }f(1,-2) &=&4\mathbf{i}-5\mathbf{j} & \color{#0066A7}{{x=1; y=-2}} \end{array} \]

The direction \(\mathbf{u}\) for which \(D_{\mathbf{u}}f(1,-2)\) is maximum occurs when \({\bf\nabla}\! f\) and the unit vector \(\mathbf{u}\) have the same direction. That is, the directional derivative is a maximum when the direction is \(\mathbf{u}=\dfrac{4\sqrt{41}}{41}\mathbf{i}-\dfrac{5\sqrt{41}}{ 41}\mathbf{j}\).

(b) The maximum value of the directional derivative at \((1,-2)\) equals the magnitude of the gradient, namely \[ \Vert {\bf\nabla}\! f(1,-2)\Vert =\sqrt{16+25}=\sqrt{41} \]