Answer
Vector along the gradient: $\mathbf{V}=12 \mathbf{i}-\mathbf{j}$
Unit vector: $\mathbf{v}=\frac{1}{\sqrt{145}}(12 \mathbf{i}-\mathbf{j})$
Rate of change: $\frac{\sqrt{145}}{4}=\|\nabla f(2,4)\|$
Work Step by Step
$f$ increases most rapidly in the positive direction of $\nabla f(x, y)$
\[
\begin{array}{c}
3=f_{x}(x, y) \\
3=f_{x}(2,4) \\
-\frac{1}{y}=f_{y}(x, y) \\
-\frac{1}{4}=f_{y}(2,4) \\
\nabla f(2,4)=\frac{1}{4}(12 \mathbf{i}-\mathbf{j})
\end{array}
\]
Vector along the gradient: $\mathbf{V}=12 \mathbf{i}-\mathbf{j}$
Unit vector: $\mathbf{v}=\frac{1}{\sqrt{145}}(12 \mathbf{i}-\mathbf{j})$
Rate of change: $\frac{\sqrt{145}}{4}=\|\nabla f(2,4)\|$
