Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 12 - Vectors and the Geometry of Space - 12.3 Exercises - Page 830: 31

Answer

$\theta =0^\circ$ at $(0,0)$ and $\theta \approx 8.1^\circ$ at $(1,1)$

Work Step by Step

$f(x)=x^2$ and $g(x)=x^3$ $x^2=x^3$ $ \implies$ $x=0,1$ $f'(x)=2x$ and $g'(x)=3x^2$ $f'(0)=0$ and $g'(0)=0$ $f'(1)=2$ and $g'(1)=3$ Thus, $a= \lt 1,2\gt$ and $b =\lt 1,3 \gt$ $ \theta =0 ^\circ$ at $x=0$ $ \theta = cos^{-1}\dfrac{a \cdot b}{|a||b|}=cos^{-1}\dfrac{1(1)+2(3)}{\sqrt {5}\sqrt {10}}$ $ \theta = cos^{-1}\dfrac{a \cdot b}{|a||b|}=cos^{-1}\dfrac{7}{\sqrt {50}}\approx 8.13^\circ$ at $x=1$ Hence, $\theta =0^\circ$ at $(0,0)$ and $\theta \approx 8.1^\circ$ at $(1,1)$
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