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Section 2-7 : 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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First, 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 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 $$\left( { - 2,3} \right)$$, 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 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,

$\left\| {\nabla f\left( { - 2,3} \right)} \right\| = \sqrt {\frac{4}{{85}} + \frac{{2916}}{{85}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\sqrt {\frac{{584}}{{17}}} = 5.8611}}$