Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 4: Applications of Derivatives - Practice Exercises - Page 245: 82

Answer

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

Work Step by Step

Step 1. Given $\frac{d^3r}{dt^3}=-cos(t)$, we have $\frac{d^2r}{dt^2}=-sin(t)+C$, where $C$ is a constant. Step 2. Using the initial condition of $r''(0)=0$, we have $-sin(0)+C=0$ and $C=0$ Step 3. Continue step 1 to get $\frac{dr}{dt}=cos(t)+D$, where $D$ is a constant. Step 4. Using the initial condition of $r'(0)=0$, we have $cos(0)+D=0$ and $D=-1$ Step 5. Continue step 3 to get $r(t)=sin(t)-t+E$, where $E$ is a constant. Step 6. Using the initial condition of $r(0)=-1$, we have $sin(0)-(0)+E=-1$ and $E=-1$ Step 7. We conclude that $r(t)=sin(t)-t-1$
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