865

Solution (a) Since the partial derivatives of $f$, namely, $$f_{x}(x,y)=2xy\qquad \hbox{and}\qquad f_{y}(x,y)=x^{2}+2y$$

are continuous, the function $f$ is differentiable. So, we can use formula (1) with the unit vector $\mathbf{u}=\cos \dfrac{\pi }{4}\mathbf{i}+\sin \dfrac{\pi }{4}\mathbf{j} =\dfrac{\sqrt{2}}{2}\mathbf{i}+\dfrac{\sqrt{2}}{2 }\mathbf{j}$. Then $$\begin{eqnarray*} D_{\mathbf{u}}f(x,y)& =&f_{x}(x,y)\cos \theta +f_{y}(x,y)\sin \theta \\ &=& 2xy\cos \dfrac{\pi }{4}+(x^{2}+2y)\sin \dfrac{\pi }{4} \qquad \color{#0066A7}{f_x(x,y)=2xy;} \\ &&\hspace{11.50pc}\color{#0066A7}{f_y (x,y)=x^{2}+2y;\theta =\dfrac{\pi }{4}} \\ &=& \sqrt{2}xy+\dfrac{\sqrt{2}}{2}(x^{2}+2y) \end{eqnarray*}$$

(b) $D_{\mathbf{u}}f(1,2)=2\sqrt{2}+\dfrac{\sqrt{2}}{2}(1+4)=\dfrac{9\sqrt{2}}{2}$.

(c) When we are at the point $( 1,2)$ and moving in the direction $\mathbf{u}=\dfrac{\sqrt{2}}{2}\mathbf{i}+\dfrac{\sqrt{2}}{2} \mathbf{j}$, the function is changing at a rate of approximately 6.364 units per unit length.