Find the directional derivative of $f(x,y)=x\sin y$ at the point $\left( 2, \dfrac{\pi }{3}\right)$ in the direction of $\mathbf{a}=3\mathbf{i}+4 \mathbf{j}$. Interpret the result as a slope.

Solution To find a directional derivative of a function $f$ in the direction of $\mathbf{a}$, where $\mathbf{a}$ is a nonzero vector, we must first find the unit vector in the direction of $\mathbf{a}$. The unit vector $\mathbf{u}$ in the direction of $\mathbf{a}$ is $$\begin{equation*} \mathbf{u}=\frac{\mathbf{a}}{\Vert \mathbf{a}\Vert }=\dfrac{3\mathbf{i}+4 \mathbf{j}}{\sqrt{9+16}}=\frac{3}{5}\mathbf{i}+\frac{4}{5}\mathbf{j} \end{equation*}$$

The partial derivatives of $f$ $$f_{x}(x,y)=\sin y\qquad \hbox{and}\qquad f_{y}(x,y)=x\cos y$$

are continuous throughout the plane, so $f$ is differentiable. Using the unit vector $\mathbf{u}=\dfrac{3}{5}\mathbf{i}+\dfrac{4}{5}\mathbf{j=}\cos \theta \mathbf{\mathbf{i}+}\sin \theta \mathbf{{j}}$, we find

Figure 4 $f( x,y) =x\sin y$

At the point $\left( 2,\dfrac{\pi }{3}\right)$, $$\begin{equation*} D_{\mathbf{u}}f\!\left( 2,\frac{\pi }{3}\right) =\dfrac{3}{5}\sin \dfrac{\pi }{3}+\dfrac{4}{5}\!\left( 2\cos \dfrac{\pi }{3}\right) =\dfrac{3}{5}\!\left( \frac{\sqrt{3}}{2}\right) +\dfrac{8}{5}\!\left(\dfrac{1}{2}\right)=\frac{3\sqrt{3}+8}{10} \end{equation*}$$

The slope of the tangent line to the curve $C$ that is the intersection of the surface $f( x,y) =x\sin y$ and the plane perpendicular to the $xy$-plane that contains the line through the point $\left( 2,\dfrac{\pi }{3}, 0\right)$ in the direction $\mathbf{a}$ is approximately 1.32. See Figure 4.