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### Section 2-7 : Directional Derivatives

5. Determine $${D_{\vec u}}f\left( {3, - 1,0} \right)$$ for $$f\left( {x,y,z} \right) = 4x - {y^2}{{\bf{e}}^{3x\,z}}$$ direction of $$\vec v = \left\langle { - 1,4,2} \right\rangle$$.

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Okay, 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 2

Also 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 3

The directional derivative is then,

${D_{\vec u}}f\left( {3, - 1,0} \right) = \left\langle {4,2, - 9} \right\rangle \centerdot \left\langle { - \frac{1}{{\sqrt {21} }},\frac{4}{{\sqrt {21} }},\frac{2}{{\sqrt {21} }}} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{14}}{{\sqrt {21} }}}}$