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If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
If you are looking for some problems with solutions you can find some by clicking on the "Practice Problems" link above.
Section 13.7 : Directional Derivatives
For problems 1 – 4 determine the gradient of the given function.
- \(f\left( {x,y} \right) = {y^3}{x^5} + \ln \left( {xy} \right)\)
- \(f\left( {x,y} \right) = {{\bf{e}}^{\frac{x}{y}}} + {y^4}\sin \left( {xy} \right)\)
- \(\displaystyle f\left( {x,y,z} \right) = 4z - \frac{{{y^4}}}{{2{z^3}}} + \sqrt {{x^3}} \left( {z - 1} \right)\)
- \(\displaystyle f\left( {x,y,z} \right) = \cos \left( {\frac{{xy}}{z}} \right) + {z^3}{y^2}x\)
For problems 5 – 8 determine \({D_{\vec u}}f\) for the given function in the indicated direction.
- \(f\left( {x,y} \right) = \ln \left( {2xy} \right) - \sin \left( {{x^2} + {y^2}} \right)\) in the direction of \(\vec v = \left\langle {7, - 3} \right\rangle \)
- \(f\left( {x,y} \right) = 4{x^2}{y^3} - \sqrt {2x + 5y} \) in the direction of \(\vec v = \left\langle { - 1,4} \right\rangle \)
- \(\displaystyle f\left( {x,y,z} \right) = 8x{y^2} - \frac{{5{z^2}}}{x} + {y^4}\) in the direction of \(\vec v = \left\langle { - 4,1,2} \right\rangle \)
- \(\displaystyle f\left( {x,y,z} \right) = \frac{{3x}}{{{y^2} - {z^3}}} + 5{x^2} - 8y\) in the direction of \(\vec v = \left\langle {0,3, - 2} \right\rangle \)
- Determine \({D_{\vec u}}f\left( { - 1,4,6} \right)\) for \(f\left( {x,y,z} \right) = {{\bf{e}}^{x\,y{\,^2}}} + 4z{y^3}\) direction of \(\vec v = \left\langle {2, - 3,6} \right\rangle \).
- Determine \({D_{\vec u}}f\left( {8,1,2} \right)\) for \(\displaystyle f\left( {x,y,z} \right) = \ln \left( {\frac{x}{z}} \right) + \ln \left( {\frac{z}{y}} \right) + {y^2}x\) direction of \(\vec v = \left\langle {1,5,2} \right\rangle \).
For problems 11 – 13 find the maximum rate of change of the function at the indicated point and the direction in which this maximum rate of change occurs.
- \(f\left( {x,y} \right) = {{\bf{e}}^{4x\,y}}\) at \(\left( {6, - 2} \right)\)
- \(f\left( {x,y,z} \right) = {x^2}{y^4} - 3{z^2}x\) at \(\left( {1, - 6,3} \right)\)
- \(\displaystyle f\left( {x,y,z} \right) = \ln \left( {\frac{{2x + 3y}}{z}} \right)\) at \(\left( {2,7,4} \right)\)
- Given \(\displaystyle \vec u = \left\langle { - \frac{3}{5}, - \frac{4}{5}} \right\rangle \), \(\displaystyle \vec v = \left\langle {\frac{4}{{\sqrt {20} }},\frac{2}{{\sqrt {20} }}} \right\rangle \), \(\displaystyle \vec w = \left\langle { - \frac{3}{{\sqrt {11} }}, - \frac{2}{{\sqrt {11} }}} \right\rangle \), \(\displaystyle {D_{\vec u}}\left( { - 1,4} \right) = \frac{{14}}{5}\) and \(\displaystyle {D_{\vec v}}\left( { - 1,4} \right) = - \frac{{22}}{{\sqrt {20} }}\) determine the value of \({D_{\vec w}}\left( { - 1,4} \right)\).
- Given \(\displaystyle \vec u = \left\langle {\frac{1}{{\sqrt {15} }},\frac{4}{{\sqrt {15} }}} \right\rangle \), \(\displaystyle \vec v = \left\langle { - \frac{3}{{\sqrt {34} }}, - \frac{5}{{\sqrt {34} }}} \right\rangle \), \(\displaystyle \vec w = \left\langle { - \frac{1}{{\sqrt 2 }},\frac{1}{{\sqrt 2 }}} \right\rangle \), \(\displaystyle {D_{\vec u}}\left( {0,1} \right) = \frac{{18}}{{\sqrt {15} }}\) and \(\displaystyle {D_{\vec v}}\left( {0,1} \right) = - \frac{{40}}{{\sqrt {34} }}\) determine the value of \({D_{\vec w}}\left( {0,1} \right)\).