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### Section 2-7 : Directional Derivatives

7. Find the maximum rate of change of $$f\left( {x,y,z} \right) = {{\bf{e}}^{2x}}\cos \left( {y - 2z} \right)$$ at $$\left( {4, - 2,0} \right)$$ and the direction in which this maximum rate of change occurs.

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First, we’ll need the gradient and its value at $$\left( {4, - 2,0} \right)$$.

\begin{align*}\nabla f & = \left\langle {2{{\bf{e}}^{2x}}\cos \left( {y - 2z} \right), - {{\bf{e}}^{2x}}\sin \left( {y - 2z} \right),2{{\bf{e}}^{2x}}\sin \left( {y - 2z} \right)} \right\rangle \\ \nabla f\left( {4, - 2,0} \right) & = \left\langle {2{{\bf{e}}^8}\cos \left( { - 2} \right), - {{\bf{e}}^8}\sin \left( { - 2} \right),2{{\bf{e}}^8}\sin \left( { - 2} \right)} \right\rangle = \left\langle { - 2481.03,2710.58, - 5421.15} \right\rangle \end{align*} Show Step 2

Now, by the theorem in class we know that the direction in which the maximum rate of change at the point in question is simply the gradient at $$\left( {4, - 2,0} \right)$$, which we found in the previous step. So, the direction in which the maximum rate of change of the function occurs is,

$\nabla f\left( {4, - 2,0} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle { - 2481.03,2710.58, - 5421.15} \right\rangle }}$ Show Step 3

The maximum rate of change is simply the magnitude of the gradient in the previous step. So, the maximum rate of change of the function is,

$\left\| {\nabla f\left( {4, - 2,0} \right)} \right\| = \sqrt {{{\left( { - 2481.03} \right)}^2} + {{\left( {2710.58} \right)}^2} + {{\left( { - 5421.15} \right)}^2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6549.17}}$