For the function $f(x,y)=\sqrt{x^{2}+y^{2}}$, graph the level curve containing the point $(3,4)$ and graph the gradient ${\bf\nabla\! }f(x,y)$ at this point.

Solution See Figure 8(a). The graph of the equation $z=\sqrt{x^{2}+y^{2}}$ is the upper half of a circular cone whose traces are circles. So, the level curves are concentric circles centered at $( 0,0)$. Because $f(3,4)=\sqrt{9+16}=5$, the level curve through $(3,4)$ is the circle $x^{2}+y^{2}=25$. Since $$\begin{equation*} {\bf\nabla}\! f(x,y)=\frac{x}{\sqrt{x^{2}+y^{2}}}\mathbf{i}+\frac{y}{\sqrt{ x^{2}+y^{2}}}\mathbf{j} \end{equation*}$$

the gradient at $(3,4)$ is $$\begin{equation*} {\bf\nabla}\! f(3,4)=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j} \end{equation*}$$

This vector is orthogonal to the level curve $x^{2}+y^{2}=25$, as shown in Figure 8(b).

Figure 8