Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.1 - The Tangent and Velocity Problems - 2.1 Exercises - Page 82: 8

Answer

(a) (i) $v_{ave} = 6.0~cm/s$ (ii) $v_{ave} = -4.712~cm/s$ (iii) $v_{ave} = -6.134~cm/s$ (iv) $v_{ave} = -6.268~cm/s$ (b) We could estimate that the instantaneous velocity at $t = 1$ is $-6.3~cm/s$

Work Step by Step

(a) We can find the displacement at $t=1$: $s = 2~sin~\pi t+ 3~cos~\pi t$ $s = 2~sin~\pi (1)+ 3~cos~\pi (1)$ $s = -3~cm$ (i) We can find the displacement at $t=2$: $s = 2~sin~\pi t+ 3~cos~\pi t$ $s = 2~sin~\pi (2)+ 3~cos~\pi (2)$ $s = 3~cm$ We can find the average velocity: $v_{ave} = \frac{3~cm-(-3~cm)}{2~s-1~s} = 6.0~cm/s$ (ii) We can find the displacement at $t=1.1$: $s = 2~sin~\pi t+ 3~cos~\pi t$ $s = 2~sin~\pi (1.1)+ 3~cos~\pi (1.1)$ $s = -3.471203537635~cm$ We can find the average velocity: $v_{ave} = \frac{-3.471203537635~cm-(-3~cm)}{1.1~s-1~s} = -4.712~cm/s$ (iii) We can find the displacement at $t=1.01$: $s = 2~sin~\pi t+ 3~cos~\pi t$ $s = 2~sin~\pi (1.01)+ 3~cos~\pi (1.01)$ $s = -3.06134~cm$ We can find the average velocity: $v_{ave} = \frac{-3.06134~cm-(-3~cm)}{1.01~s-1~s} = -6.134~cm/s$ (iv) We can find the displacement at $t=1.001$: $s = 2~sin~\pi t+ 3~cos~\pi t$ $s = 2~sin~\pi (1.001)+ 3~cos~\pi (1.001)$ $s = -3.006268~cm$ We can find the average velocity: $v_{ave} = \frac{-3.006268~cm-(-3~cm)}{1.001~s-1~s} = -6.268~cm/s$ (b) As the time interval gets smaller, the average velocity gets closer to the instantaneous velocity at $t = 1$. We could estimate that the instantaneous velocity at $t = 1$ is $-6.3~cm/s$
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