Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.2 Derivatives and Integrals of Vector Functions - 13.2 Exercises - Page 900: 33


$\cos ^{-1}\dfrac{1}{\sqrt 6}\approx 66^\circ$

Work Step by Step

As we have from the given curves $r_1(t)=\lt t,t^2,t^3 \gt $ and $r_2(t)= \lt \sin t,\sin 2t,t \gt $ Tangent vector $r'(t)$ can be found by taking the derivative of $r_1(t)$ and $r_2(t)$ with respect to $t$. This yields,$r_1'(t)=\lt 1,2t,3t^2 \gt $ ,$r_2'(t)= \lt \cos t,2 \cos t,1 \gt $ $r_1'(0)=\lt 1,0,0 \gt $ and $r_2'(t)= \lt 1,2,1 \gt $ Consider $\theta$ be the angle between the two curves. Thus, $\cos \theta=\dfrac{\lt 1,0,0 \gt \cdot\lt 1,2,1 \gt}{\sqrt {1+0+0}\sqrt {1+4+1}}$ or, $ \theta=\cos ^{-1}[\dfrac{\lt 1,0,0 \gt \cdot\lt 1,2,1 \gt}{\sqrt {1+0+0}\sqrt {1+4+1}}]$ or, $=\cos ^{-1}\dfrac{1}{\sqrt 6}\approx 66^\circ$
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