Section 13.7 : 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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Start SolutionFirst, 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 2Now, 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 3The 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,
\[\left\| {\nabla f\left( {4, - 2,0} \right)} \right\| = \sqrt {{{\left( { - 2481.03} \right)}^2} + {{\left( {2710.58} \right)}^2} + {{\left( { - 5421.15} \right)}^2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{6549.17}}\]