Finding the Gradient of a Function of Two Variables

1. Find the gradient of \(f(x,y)=x^{3}y\) at the point \((2,1)\).
2. Use the gradient to find the directional derivative of \(f\) at \((2,1)\) in the direction from \((2,1)\) to \((3,5)\).

Solution (a) The gradient of \(f\) at \((x,y)\) is \[ {\bf\nabla}\! f(x,y)=f_{x}(x,y)\mathbf{i}+f_{y}(x,y)\mathbf{j}=3x^{2}y \mathbf{i}+x^{3}\mathbf{j} \]

The gradient of \(f\) at \((2,1)\) is \[ {\bf\nabla}\! f(2,1)=12\mathbf{i}+8\mathbf{j} \]

(b) The unit vector \(\mathbf{u}\) from \((2,1)\) to \((3,5)\) is \[ \begin{equation*} \mathbf{u=}\dfrac{( 3-2) \mathbf{i} +( 5-1) \mathbf{j} }{\sqrt{( 3-2) ^{2}+( 5-1) ^{2}}}=\frac{\mathbf{i}+4 \mathbf{j}}{\sqrt{17}} \end{equation*} \]

We use formula (3) for the directional derivative to find \(D_{\mathbf{u} }(2,1).\) \[ \begin{equation*} D_{\mathbf{u}}(2,1)={\bf\nabla\! }f(2,1)\,{\bf\cdot}\, \mathbf{u}=(12\mathbf{i}+8 \mathbf{j})\,{\bf\cdot}\, \frac{\mathbf{i}+4\mathbf{j}}{\sqrt{17}}=\frac{44}{\sqrt{17} } \end{equation*} \]

The directional derivative of \(f\) at \((2,1)\) in the direction of \(\mathbf{u}\) is \(\dfrac{44}{\sqrt{17}}\)