Section 13.7 : Directional Derivatives
4. Determine \({D_{\vec u}}f\) for \(f\left( {x,y,z} \right) = {x^2}{y^3} - 4xz\) in the direction of \(\vec v = \left\langle { - 1,2,0} \right\rangle \).
Show All Steps Hide All Steps
Start SolutionOkay, we know we need the gradient so let’s get that first.
\[\nabla f = \left\langle {2x{y^3} - 4z,3{x^2}{y^2}, - 4x} \right\rangle \] 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.
\[\left\| {\vec v} \right\| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 0 \right)}^2}} = \sqrt 5 \hspace{0.5in}\vec u = \frac{{\vec v}}{{\left\| {\vec v} \right\|}} = \frac{1}{{\sqrt 5 }}\left\langle { - 1,2,0} \right\rangle = \left\langle { - \frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }},0} \right\rangle \] Show Step 3The directional derivative is then,
\[{D_{\vec u}}f = \left\langle {2x{y^3} - 4z,3{x^2}{y^2}, - 4x} \right\rangle \centerdot \left\langle { - \frac{1}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }},0} \right\rangle = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{\sqrt 5 }}\left( {4z - 2x{y^3} + 6{x^2}{y^2}} \right)}}\]