University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 42


$r(t)=sec t -t-1$

Work Step by Step

Here, the derivative is given as: $r'(t)=sec t \tan t-1$, thus, the general form of the function is: $r(t)=sec t -t+C$ The graph of the given function passes through $(0,0)$ Therefore, $f(0)=0 \implies 0=sec(0)-0+C$ Thus, $C=-1$ Hence, $r(t)=sec t -t-1$
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