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: 48

Answer

(a) The average rate of growth (i) from 2006 to 2008: $2120lo/ye$ (ii) from 2008 to 2010: $89lo/ye$ b) The instantaneous rate of growth is $346.5lo/ye$ c) The instantaneous rate of growth is $346.5lo/ye$

Work Step by Step

(a) The average rate of growth (i) from 2006 to 2008: $$\frac{N(2008)-N(2006)}{2008-2006}=\frac{16680-12440}{2008-2006}=2120(lo/ye)$$ (ii) from 2008 to 2010: $$\frac{N(2010)-N(2008)}{2010-2008}=\frac{16858-16680}{2010-2008}=89(lo/ye)$$ The unit of $N$ is $lo$ (locations), and the unit of the year is $ye$ (year). Therefore, the unit of the average rate of growth is $lo/ye$. We can conclude that there is a steep drop in the growth of locations from the period 2006-2008 to the period 2008-2010. (b) The average rate of growth (i) from 2010 to 2012: $$\frac{N(2012)-N(2010)}{2012-2010}=\frac{18066-16858}{2012-2010}=604(lo/ye)$$ (ii) from 2008 to 2010: $89(lo/ye)$ The instantaneous rate of growth in 2010 would be the average of 2 average rates of growth from 2008 to 2010 and from 2010 to 2012: $$\frac{89+604}{2}=346.5(lo/ye)$$ The unit is also $lo/ye$. (c) To measure the slope of the tangent at 2010, we would calculate the average rate of growth from 2008 to 2012: $$m=\frac{N(2012)-N(2008)}{2012-2008}=\frac{18066-16680}{4}=346.5$$ The slope of the above tangent is also the instantaneous rate of growth in 2010, which is $346.5lo/ye$
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