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: 47

Answer

a) The average rate of change of C with respect to t: (i) Over [1.0,2.0]: $-0.15mg/mL.hours$ (ii) Over [1.5,2.0]: $-0.12mg/mL.hours$ (i) Over [2.5,2.0]: $-0.12mg/mL.hours$ (i) Over [3.0,2.0]: $-0.11mg/mL.hours$ (b) The instantaneous rate of change at $t=2$ is $-0.12$. The unit is mg/mL.hours.

Work Step by Step

(a) The average rate of change of C with respect to t over a time interval $[a,b]$ is calculated as follows: $$\frac{C(b)-C(a)}{b-a}$$ The unit of $C(a)$ and $C(b)$ is mg/mL, the unit of $a$ and $b$ is hours, so the unit of the rate of change is $mg/mL.hours$ Therefore, the average rate of change of C with respect to t is (i) Over time interval [1.0,2.0] $$\frac{C(2.0)-C(1.0)}{2.0-1.0}=\frac{0.18-0.33}{2.0-1.0}=\frac{-0.15}{1.0}=-0.15(mg/mL.hours)$$ (ii) Over time interval [1.5,2.0] $$\frac{C(2.0)-C(1.5)}{2.0-1.5}=\frac{0.18-0.24}{2.0-1.5}=\frac{-0.06}{0.5}=-0.12(mg/mL.hours)$$ (iii) Over time interval [2.0,2.5] $$\frac{C(2.5)-C(2.0)}{2.5-2.0}=\frac{0.12-0.18}{2.5-2.0}=\frac{-0.06}{0.5}=-0.12(mg/mL.hours)$$ (iv) Over time interval [2.0,3.0] $$\frac{C(3.0)-C(2.0)}{3.0-2.0}=\frac{0.07-0.18}{3.0-2.0}=\frac{-0.11}{1.0}=-0.11(mg/mL.hours)$$ b) The instantaneous rate of change at $t=2$ is actually the derivative of $C(t)$ at $t=2$, or in other words, $C'(2)$. From the values found in part a), we would look for the closest time interval from $2.0$, which are [1.5,2.0] and [2.0,2.5]. We can see that both of them, the average rate of change is $-0.12$. Therefore, we would estimate that the instantaneuous rate of change at $t=2$, or $C'(2)=-0.12mg/mL.hours$
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