Chapter 15 - Differentiation in Several Variables - 15.5 The Gradient and Directional Derivatives - Exercises - Page 801: 2

(a) $\nabla f = \left( {{f_x},{f_y}} \right) = \left( {y{{\rm{e}}^{xy}},x{{\rm{e}}^{xy}}} \right)$, $\nabla {f_{{\bf{r}}\left( t \right)}} = \left( {\left( {1 + t} \right){{\rm{e}}^{{t^3}\left( {1 + t} \right)}},{t^3}{{\rm{e}}^{{t^3}\left( {1 + t} \right)}}} \right)$ ${\bf{r}}'\left( t \right) = \left( {3{t^2},1} \right)$ (b) $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \left( {3{t^2} + 4{t^3}} \right){{\rm{e}}^{{t^3}\left( {1 + t} \right)}}$ (c) $f\left( t \right) = {{\rm{e}}^{{t^3}\left( {1 + t} \right)}}$ $f'\left( t \right) = {{\rm{e}}^{{t^3}\left( {1 + t} \right)}}\left( {3{t^2} + 4{t^3}} \right)$ The result agrees with part (b).

(a) We are given $f\left( {x,y} \right) = {{\rm{e}}^{xy}}$ and ${\bf{r}}\left( t \right) = \left( {{t^3},1 + t} \right)$. So, $\nabla f = \left( {{f_x},{f_y}} \right) = \left( {y{{\rm{e}}^{xy}},x{{\rm{e}}^{xy}}} \right)$, $\nabla {f_{{\bf{r}}\left( t \right)}} = \left( {\left( {1 + t} \right){{\rm{e}}^{{t^3}\left( {1 + t} \right)}},{t^3}{{\rm{e}}^{{t^3}\left( {1 + t} \right)}}} \right)$ ${\bf{r}}'\left( t \right) = \left( {3{t^2},1} \right)$ (b) Using the Chain Rule for Paths we have $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \nabla {f_{{\bf{r}}\left( t \right)}}\cdot{\bf{r}}'\left( t \right)$ $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \left( {\left( {1 + t} \right){{\rm{e}}^{{t^3}\left( {1 + t} \right)}},{t^3}{{\rm{e}}^{{t^3}\left( {1 + t} \right)}}} \right)\cdot\left( {3{t^2},1} \right)$ $ = 3{t^2}\left( {1 + t} \right){{\rm{e}}^{{t^3}\left( {1 + t} \right)}} + {t^3}{{\rm{e}}^{{t^3}\left( {1 + t} \right)}}$ $ = \left( {3{t^2} + 4{t^3}} \right){{\rm{e}}^{{t^3}\left( {1 + t} \right)}}$ (c) Write out the composite $f\left( {{\bf{r}}\left( t \right)} \right)$: $f\left( {x,y} \right) = {{\rm{e}}^{xy}}$ $f\left( t \right) = {{\rm{e}}^{{t^3}\left( {1 + t} \right)}}$ Taking the derivative with respect to $t$ gives $f'\left( t \right) = {{\rm{e}}^{{t^3}\left( {1 + t} \right)}}\left( {3{t^2} + 4{t^3}} \right)$ The result agrees with part (b).