Calculus 8th Edition

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

Chapter 3 - Applications of Differentiation - 3.9 Antiderivatives - 3.9 Exercises - Page 283: 56

Answer

$$ a(t)=v^{\prime}(t)= 3 \cos t-2 \sin t, \quad s(0)=0 , \quad v(0)=4 $$ The position of the particle is $$ s(t)=-3 \cos t+ 2 \sin t +2t +6 $$

Work Step by Step

$$ a(t)=v^{\prime}(t)= 3 \cos t-2 \sin t, \quad s(0)=0 , \quad v(0)=4 $$ The general anti-derivative of $ v^{\prime}(t)= 3 \cos t-2 \sin t $ is $$ v(t)= 3 \sin t+ 2 \cos t +C $$ To determine C we use the fact that $v(0)=-2$: $$ v(0)= 3 \sin (0)+ 2 \cos (0) +C =4 $$ $ \Rightarrow $ $$ 2+C=4 \quad \Rightarrow \quad C=2, $$ so $$ v(t)= 3 \sin t+ 2 \cos t +2 $$ Since $s^{\prime }(t)=v(t)$, we antidifferentiate again and obtain $$ s(t)=-3 \cos t+ 2 \sin t +2t +D $$ To determine D we use the fact that $s(0)=0$: $$ s(0)=-3 \cos (0)+ 2 \sin (0) +2(0) +D =3 $$ $ \Rightarrow $ $$ -3+D=3 \quad \Rightarrow \quad D=6, $$ so the position of the particle is $$ s(t)=-3 \cos t+ 2 \sin t +2t +6 $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.