Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - Chapter Review Exercises - Page 835: 33

Answer

The directional derivative at $P = \left( {3, - 1} \right)$ in the direction of ${\bf{v}} = 2{\bf{i}} + {\bf{j}}$ is $ - \frac{{54}}{{\sqrt 5 }}$.

Work Step by Step

We are given $f\left( {x,y} \right) = {x^3}{y^4}$ and ${\bf{v}} = 2{\bf{i}} + {\bf{j}}$. First, we normalize the direction vector: ${\bf{u}} = \frac{{\bf{v}}}{{||{\bf{v}}||}} = \frac{{\left( {2,1} \right)}}{{\sqrt {\left( {2,1} \right)\cdot\left( {2,1} \right)} }} = \left( {\frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right)$ Next, we compute the gradient at $P = \left( {3, - 1} \right)$: $\nabla f = \left( {3{x^2}{y^4},4{x^3}{y^3}} \right)$, ${\ \ \ }$ $\nabla {f_P} = \left( {27, - 108} \right)$ Using Theorem 3 of Section 15.5, we obtain the directional derivative in the direction of ${\bf{v}}$ at $P = \left( {3, - 1} \right)$: ${D_{\bf{u}}}f\left( P \right) = \nabla {f_P}\cdot{\bf{u}}$ ${D_{\bf{u}}}f\left( P \right) = \left( {27, - 108} \right)\cdot\left( {\frac{2}{{\sqrt 5 }},\frac{1}{{\sqrt 5 }}} \right) = - \frac{{54}}{{\sqrt 5 }}$ So, the directional derivative at $P = \left( {3, - 1} \right)$ in the direction of ${\bf{v}} = 2{\bf{i}} + {\bf{j}}$ is $ - \frac{{54}}{{\sqrt 5 }}$.
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