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: 30

Answer

$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right){|_{t = - 2}} = 197$

Work Step by Step

We have $f\left( {x,y,z} \right) = xz - {y^2}$ and ${\bf{r}}\left( t \right) = \left( {t,{t^3},1 - t} \right)$. So, $\nabla f = \left( {z, - 2y,x} \right)$, ${\ \ \ \ }$ $\nabla {f_{{\bf{r}}\left( t \right)}} = \left( {1 - t, - 2{t^3},t} \right)$ ${\bf{r}}'\left( t \right) = \left( {1,3{t^2}, - 1} \right)$ At $t=-2$, we get $\nabla {f_{{\bf{r}}\left( { - 2} \right)}} = \left( {3,16, - 2} \right)$ and ${\bf{r}}'\left( { - 2} \right) = \left( {1,12, - 1} \right)$. Using the Chain Rule for Paths (Theorem 2 of Section 15.5): $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right){|_{t = - 2}} = \nabla {f_{{\bf{r}}\left( { - 2} \right)}}\cdot{\bf{r}}'\left( { - 2} \right)$ $ = \left( {3,16, - 2} \right)\cdot\left( {1,12, - 1} \right)$ $ = 197$
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