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Section 2-7 : Directional Derivatives

3. Determine \({D_{\vec u}}f\) for \(\displaystyle f\left( {x,y} \right) = \cos \left( {\frac{x}{y}} \right)\) in the direction of \(\vec v = \left\langle {3, - 4} \right\rangle \).

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Okay, we know we need the gradient so let’s get that first.

\[\nabla f = \left\langle { - \frac{1}{y}\sin \left( {\frac{x}{y}} \right),\frac{x}{{{y^2}}}sin\left( {\frac{x}{y}} \right)} \right\rangle \] 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.

\[\left\| {\vec v} \right\| = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 4} \right)}^2}} = \sqrt {25} = 5\hspace{0.5in}\vec u = \frac{{\vec v}}{{\left\| {\vec v} \right\|}} = \frac{1}{5}\left\langle {3, - 4} \right\rangle = \left\langle {\frac{3}{5}, - \frac{4}{5}} \right\rangle \] Show Step 3

The directional derivative is then,

\[\begin{align*}{D_{\vec u}}f & = \left\langle { - \frac{1}{y}\sin \left( {\frac{x}{y}} \right),\frac{x}{{{y^2}}}\sin\left( {\frac{x}{y}} \right)} \right\rangle \centerdot \left\langle {\frac{3}{5}, - \frac{4}{5}} \right\rangle \\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{3}{{5y}}\sin \left( {\frac{x}{y}} \right) - \frac{{4x}}{{5{y^2}}}\sin\left( {\frac{x}{y}} \right) = - \frac{1}{5}\left( {\frac{3}{y} + \frac{{4x}}{{{y^2}}}} \right)sin\left( {\frac{x}{y}} \right)}}\end{align*}\]