Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises - Page 151: 57

Answer

(a) $S'(T)$ is the rate at which the oxygen solubility changes at the given temperature T. The units are $\frac{(mg/L)}{^{\circ}F}$ (b) $S'(16) \approx -0.2~\frac{(mg/L)}{^{\circ}F}$ When the temperature is $16^{\circ}F$, the oxygen solubility is changing at a rate of $-0.2~\frac{(mg/L)}{^{\circ}F}$

Work Step by Step

(a) $S'(T)$ is the rate at which the oxygen solubility changes at the given temperature T. The units are $\frac{(mg/L)}{^{\circ}F}$ (b) On the graph, we can see that the slope at $T = 16^{\circ}F$ is approximately equal to the average rate of change over the interval $[0, 40^{\circ}F]$ We can estimate $S'(16)$: $S'(16) \approx \frac{15~mg/L-7~mg/L}{0^{\circ}F - 40^{\circ}F} = -0.2~\frac{(mg/L)}{^{\circ}F}$ When the temperature is $16^{\circ}F$, the oxygen solubility is changing at a rate of $-0.2~\frac{(mg/L)}{^{\circ}F}$
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.