Section 13.7 : Directional Derivatives
5. Determine \({D_{\vec u}}f\left( {3, - 1,0} \right)\) for \(f\left( {x,y,z} \right) = 4x - {y^2}{{\bf{e}}^{3x\,z}}\) in the direction of \(\vec v = \left\langle { - 1,4,2} \right\rangle \).
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Start SolutionOkay, we know we need the gradient so let’s get that first.
\[\nabla f = \left\langle {4 - 3z{y^2}{{\bf{e}}^{3x\,z}}, - 2y{{\bf{e}}^{3x\,z}}, - 3x{y^2}{{\bf{e}}^{3x\,z}}} \right\rangle \]Because we also know that we’ll eventually need this evaluated at the point we may as well get that as well.
\[\nabla f\left( {3, - 1,0} \right) = \left\langle {4,2, - 9} \right\rangle \] Show Step 2Also recall that we need to make sure that the direction vector is a unit vector. It is (hopefully) pretty clear that this vector is not a unit vector so let’s convert it to a unit vector.
\[\left\| {\vec v} \right\| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 2 \right)}^2}} = \sqrt {21} \hspace{0.25in}\vec u = \frac{{\vec v}}{{\left\| {\vec v} \right\|}} = \frac{1}{{\sqrt {21} }}\left\langle { - 1,4,2} \right\rangle = \left\langle { - \frac{1}{{\sqrt {21} }},\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}} \right\rangle \] Show Step 3The directional derivative is then,
\[{D_{\vec u}}f\left( {3, - 1,0} \right) = \left\langle {4,2, - 9} \right\rangle \centerdot \left\langle { - \frac{1}{{\sqrt {21} }},\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{14}}{{\sqrt {21} }}}}\]