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: 50

Answer

The average rate of change of V with respect to t: (i) Over time interval [4,11]: $-6.229RNAco/mL.da$ (ii) Over time interval [8,11]: $-2.867RNAco/mL.da$ (iii) Over time interval [11,15]: $-1.05RNAco/mL.da$ (iv) Over time interval [11,22]: $-0.527RNAco/mL.da$ $V'(11)\approx-1.829RNAco/mL.da$

Work Step by Step

(a) The average rate of change of V with respect to t over a time interval $[a,b]$ is calculated as follows: $$\frac{V(b)-V(a)}{b-a}$$ The unit of $V(a)$ and $V(b)$ is RNAco/mL, the unit of $a$ and $b$ is da (days), so the unit of the rate of change is $RNAco/mL.da$ Therefore, the average rate of change of V with respect to t is (i) Over time interval [4,11] $$\frac{V(11)-V(4)}{11-4}=\frac{9.4-53}{11-4}\approx-6.229RNAco/mL.da$$ (ii) Over time interval [8,11] $$\frac{V(11)-V(8)}{11-8}=\frac{9.4-18}{11-8}\approx-2.867RNAco/mL.da$$ (iii) Over time interval [11,15] $$\frac{V(15)-V(11)}{15-11}=\frac{5.2-9.4}{15-11}=-1.05RNAco/mL.da$$ (iv) Over time interval [11,22] $$\frac{V(22)-V(11)}{22-11}=\frac{3.6-9.4}{22-11}\approx-0.527RNAco/mL.da$$ b) To estimate $V'(11)$, we would calculate the rate of change over time interval $[8.15]$: $$V'(11)=\frac{V(15)-V(8)}{15-8}=\frac{5.2-18}{15-8}\approx-1.829RNAco/mL.da$$
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