I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 13.7 : Directional Derivatives
6. Find the maximum rate of change of \(f\left( {x,y} \right) = \sqrt {{x^2} + {y^4}} \) at \(\left( { - 2,3} \right)\) 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 \(\left( { - 2,3} \right)\).
\[\nabla f = \left\langle {\frac{x}{{\sqrt {{x^2} + {y^4}} }},\frac{{2{y^3}}}{{\sqrt {{x^2} + {y^4}} }}} \right\rangle \hspace{0.5in}\nabla f\left( { - 2,3} \right) = \left\langle {\frac{{ - 2}}{{\sqrt {85} }},\frac{{54}}{{\sqrt {85} }}} \right\rangle \] 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 \(\left( { - 2,3} \right)\), which we found in the previous step. So, the direction in which the maximum rate of change of the function occurs is,
\[\nabla f\left( { - 2,3} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\left\langle {\frac{{ - 2}}{{\sqrt {85} }},\frac{{54}}{{\sqrt {85} }}} \right\rangle }}\] 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,
\[\left\| {\nabla f\left( { - 2,3} \right)} \right\| = \sqrt {\frac{4}{{85}} + \frac{{2916}}{{85}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\sqrt {\frac{{584}}{{17}}} = 5.8611}}\]