Chapter 13 - Parametric And Polar Curves; Conic Sections - 13.6 Directional Derivatives And Gradients - Exercises Set 13.6 - Page 968: 8

$${D_{\bf{u}}}f\left( {\frac{1}{2},\frac{1}{3},\pi } \right) = \frac{1}{{12}} - \frac{1}{{12}}\pi $$

$$\eqalign{ & f\left( {x,y,z} \right) = \sin xyz;\,\,\,\,\,\,P\left( {\frac{1}{2},\frac{1}{3},\pi } \right);\,\,\,\,\,\,\,{\bf{u}} = \frac{1}{{\sqrt 3 }}{\bf{i}} - \frac{1}{{\sqrt 3 }}{\bf{j}} + \frac{1}{{\sqrt 3 }}{\bf{k}} \cr & {\text{Calculate the partial derivatives }}{f_x}\left( {x,y,z} \right){\text{,}}\,{f_y}\left( {x,y,z} \right)\,{\text{ and}}{\text{,}}\,\,\,{f_z}\left( {x,y,z} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left( {\sin xyz} \right) \cr & {f_x}\left( {x,y,z} \right) = \cos xyz\left( {yz} \right) \cr & {f_x}\left( {x,y,z} \right) = yz\cos xyz \cr & \cr & {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left( {\sin xyz} \right) \cr & {f_y}\left( {x,y,z} \right) = \cos xyz\left( {xz} \right) \cr & {f_y}\left( {x,y,z} \right) = xz\cos xyz \cr & \cr & and \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left( {\sin xyz} \right) \cr & {f_z}\left( {x,y,z} \right) = \cos xyz\left( {xy} \right) \cr & {f_y}\left( {x,y,z} \right) = xy\cos xyz \cr & \cr & {\text{Calculate the gradient of }}f\left( {x,y,z} \right) \cr & \nabla f\left( {x,y,z} \right) = {f_x}\left( {x,y,z} \right){\bf{i}} + {f_y}\left( {x,y,z} \right){\bf{j}} + {f_y}\left( {x,y,z} \right){\bf{k}} \cr & \nabla f\left( {x,y,z} \right) = yz\cos xyz{\bf{i}} + xz\cos xyz{\bf{j}} + xy\cos xyz{\bf{k}} \cr & \nabla f\left( {x,y,z} \right) = \left( {yz{\bf{i}} + xz{\bf{j}} + xy{\bf{k}}} \right)\cos xyz \cr & {\text{evaluate the gradient at the given point }}P\left( {\frac{1}{2},\frac{1}{3},\pi } \right) \cr & \nabla f\left( {\frac{1}{2},\frac{1}{3},\pi } \right) = \left( {\frac{1}{3}\pi {\bf{i}} + \frac{1}{2}\pi {\bf{j}} + \frac{1}{6}{\bf{k}}} \right)\cos \left( {\frac{\pi }{6}} \right) \cr & \nabla f\left( {\frac{1}{2},\frac{1}{3},\pi } \right) = \frac{{\sqrt 3 }}{2}\left( {\frac{1}{3}\pi {\bf{i}} + \frac{1}{2}\pi {\bf{j}} + \frac{1}{6}{\bf{k}}} \right) \cr & \nabla f\left( {\frac{1}{2},\frac{1}{3},\pi } \right) = \left( {\frac{{\sqrt 3 }}{6}\pi {\bf{i}} + \frac{{\sqrt 3 }}{4}\pi {\bf{j}} + \frac{{\sqrt 3 }}{{12}}{\bf{k}}} \right) \cr & \cr & {\text{Calculate the directional derivative }}{D_{\bf{u}}}f\left( {x,y,z} \right){\text{ at }} \cr & P\left( {\frac{1}{2},\frac{1}{3},\pi } \right){\text{ in the direction of }}{\bf{u}} \cr & {D_{\bf{u}}}f\left( {x,y,z} \right) = \nabla f\left( {x,y,z} \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {\frac{1}{2},\frac{1}{3},\pi } \right) = \nabla f\left( {\frac{1}{2},\frac{1}{3},\pi } \right) \cdot {\bf{u}} \cr & {D_{\bf{u}}}f\left( {\frac{1}{2},\frac{1}{3},\pi } \right) = \left( {\frac{{\sqrt 3 }}{6}\pi {\bf{i}} + \frac{{\sqrt 3 }}{4}\pi {\bf{j}} + \frac{{\sqrt 3 }}{{12}}{\bf{k}}} \right) \cdot \left( {\frac{1}{{\sqrt 3 }}{\bf{i}} - \frac{1}{{\sqrt 3 }}{\bf{j}} + \frac{1}{{\sqrt 3 }}{\bf{k}}} \right) \cr & {\text{solving the dot product}} \cr & {D_{\bf{u}}}f\left( {\frac{1}{2},\frac{1}{3},\pi } \right) = \left( {\frac{{\sqrt 3 }}{6}\pi } \right)\left( {\frac{1}{{\sqrt 3 }}} \right) + \left( {\frac{{\sqrt 3 }}{4}\pi } \right)\left( { - \frac{1}{{\sqrt 3 }}} \right) + \left( {\frac{{\sqrt 3 }}{{12}}} \right)\left( {\frac{1}{{\sqrt 3 }}} \right) \cr & {D_{\bf{u}}}f\left( {\frac{1}{2},\frac{1}{3},\pi } \right) = \frac{1}{6}\pi - \frac{1}{4}\pi + \frac{1}{{12}} \cr & {D_{\bf{u}}}f\left( {\frac{1}{2},\frac{1}{3},\pi } \right) = \frac{1}{{12}} - \frac{1}{{12}}\pi \cr} $$