Answer
Vector along the gradient: $(4 \mathbf{i}-3 \mathbf{j})\frac{1}{5}=\mathbf{V}$
Unit vector:$(4 \mathbf{i}-3 \mathbf{j})\frac{1}{5}=\mathbf{v}$
Rate of change: $1=\|\nabla f(4,-3)\|$
Work Step by Step
$f$ increases most rapidly in the positive direction of $\nabla f(x, y)$
\[
\begin{array}{l}
\frac{x}{\sqrt{y^{2}+x^{2}}}=f_{x}(x, y) \\
\frac{4}{5}=f_{x}(4,-3) \\
f_{y}(x, y)=\frac{y}{\sqrt{x^{2}+y^{2}}} \\
f_{y}(4,-3)=-\frac{3}{5}
\end{array}
\]
\[
\frac{1}{5}(4 \mathbf{i}-3 \mathbf{j})=\nabla f(4,-3)
\]
Vector along the gradient: $(4 \mathbf{i}-3 \mathbf{j})\frac{1}{5}=\mathbf{V}$
Unit vector:$(4 \mathbf{i}-3 \mathbf{j})\frac{1}{5}=\mathbf{v}$
Rate of change: $1=\|\nabla f(4,-3)\|$
