University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.4 - The Derivative as a Rate of Change - Exercises - Page 146: 22


$(C)$ represent position $s$, $(B)$ represent velocity $v$ and $(A)$ represent acceleration $a$.

Work Step by Step

To recognize the curve $(a)$ and its derivative curve $(a')$, what I would do first is to look for corresponding points: change-of-sign points in $(a')$ should correspond to change-of-direction points in $(a)$. The reason is that - As $a'\gt0$, the graph of $a$ moves upward. - As $a'\lt0$, the graph of $a$ moves downward. So a change in the sign of $a'$ will signify a change in the direction of $(a)$ at that exact same point. The blue curve $(A)$ and the yellow curve $(B)$ represent such a pair. At the point $(A)$ changes sign from negative to positive, $(B)$ changes from moving downward to upward. So $(A)$ is the derivative of $(B)$. But still that leaves the relationship between the red curve $(C)$ and the other two in question. A clue we can see in $(C)$ is that it is continuously moving downward. The derivative of $(C)$, as a result, must be negative all the time, which we can relate to the curve $(B)$. Another reason to believe that $(B)$ is the derivative $(C)$ is that for the first half, $(C)$ decreases more and more steeply, corresponding to the downward trend in $(B)$. For the second half, $(C)$ decreases less and less steeply, corresponding to the upward trend in $(B)$. Therefore, $(C)$ represent position $s$, $(B)$ represent velocity $v$ and $(A)$ represent acceleration $a$.
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