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: 32

Answer

$\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right){|_{t = \pi /3}} = 1$

Work Step by Step

We have $f\left( {x,y} \right) = {\tan ^{ - 1}}\frac{y}{x}$ and ${\bf{r}}\left( t \right) = \left( {\cos t,\sin t} \right)$. Using the Derivative of Inverse Trigonometric Functions (Section 7.8 on page 371): ${f_x} = \frac{1}{{{{\left( {y/x} \right)}^2} + 1}}\left( { - \frac{y}{{{x^2}}}} \right) = - \frac{y}{{{x^2} + {y^2}}}$ ${f_y} = \frac{1}{{{{\left( {y/x} \right)}^2} + 1}}\left( { \frac{1}{x}} \right) = \frac{x}{{{x^2} + {y^2}}}$ So, $\nabla f = \left( { - \frac{y}{{{x^2} + {y^2}}},\frac{x}{{{x^2} + {y^2}}}} \right)$ $\nabla {f_{{\bf{r}}\left( t \right)}} = \left( { - \sin t,\cos t} \right)$ ${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t} \right)$ Using the Chain Rule for Paths (Theorem 2 of Section 15.5): $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = \nabla {f_{{\bf{r}}\left( t \right)}}\cdot{\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t} \right)\cdot\left( { - \sin t,\cos t} \right) = 1$ Since $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right) = 1$ for all $t$, we have $\frac{d}{{dt}}f\left( {{\bf{r}}\left( t \right)} \right){|_{t = \pi /3}} = 1$.
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